Category: CBSE Class 11

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 14 Environmental Chemistry. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 14 Important Extra Questions Environmental Chemistry

Environmental Chemistry Important Extra Questions Very Short Answer Type

Question 1.
What is the percentage of CO₂ in pure dry air?
Answer:
About 0.03%.

Question 2.
Which gas was responsible for the Bhopal tragedy?
Answer:
Methyl isocyanate (MIC).

Question 3.
What is smog?
Answer:
A combination of smoke and fog.

Question 4.
Name three natural sources of air pollution.
Answer:

1. Volcanic eruptions
2. Forest fires
3. Pollen grains of flowers.

Question 5.
What is that compound formed when CO combines with blood?
Answer:

Question 6.
Name three gases that are major air pollutants.
Answer:

1. CO₂
2. NO₂
3. SO₂.

Question 7.
How does particulate help in cloud formation?
Answer:
The particulates work as nuclei for cloud formation.

Question 8.
Give one main reason for ozone depletion.
Answer:
The release of chlorofluorocarbons (CFCs) also known as freon.

Question 9.
Name two important sinks of CO.
Answer:
Oceans that dissolve it and plants use it for photosynthesis.

Question 10.
Which acids arc present in the added rain?
Answer:

1. H₂SO₄
2. HNO₃
3. HCl.

Question 11.
What is the nature of London smog?
Answer:
It is reducing in nature.

Question 12.
Name two herbicides used to kill weeds.
Answer:

1. Sodium chlorate (NaClO₃)
2. Sodium arsenite (Na₃AsO₃)

Question 13.
Name two gases that form acid rain.
Answer:

1. SO₂
2. NO₂.

Question 14.
What is the composition of photochemical smog?
Answer:
It is a mixture of irritation causing compounds NO₂, O₃, hydrocarbons, acrolein, formaldehyde, peroxyacetyl nitrate (PAN).

Question 15.
Which zone is called the ozonosphere?
Answer:
Stratospheric zone.

Question 16.
Name any four methods of waste management.
Answer:

1. Recycling
2. Burning and incineration
3. Sewage treatment
4. Dumping.

Question 17.
What is BOD?
Answer:
The amount of oxygen consumed by microorganisms in decomposing waste in a sample of sewage water is called BOD (Biochemical Oxygen Demand).

Question 18.
What is the size range of particulates?
Answer:
5 nm to 500,000 nm.

Question 19.
What type of aromatic compounds are present as particulates in the air?
Answer:
Poly cyclic aromatic hydrocarbons [PAH].

Question 20.
In which season and what time of day, there is London smog?
Answer:
In winter during morning hours.

Question 21.
Which disease is caused due to a hole in the ozone layer and why?
Answer:
Ultraviolet rays from the sun will reach the earth after passing through the ozone hole and cause skin cancer.

Question 22.
What are polar stratospherical clouds (PSC)?
Answer:
The special type of clouds present over Antarctica in winter are called polar stratospheric clouds.

Question 23.
What should be the tolerable limit of fluoride ions in drinking water? What happens if it is more than 10 ppm?
Answer:
One ppm or 1 mg dm^-3. The higher concentration is harmful to bones and teeth.

Question 24.
Which main compounds are causing damage to the ozone layer?
Answer:
NO and freons.

Question 25.
What is the composition of London smog?
Answer:
Fog of H₂SO₄ droplets deposited on the particulates.

Question 26.
Why photochemical smog is so called?
Answer:
Because it is formed as a result of the photochemical reaction between oxides of nitrogen and hydrocarbons.

Question 27.
In which season and what time of day, there is photochemical smog?
Answer:
In summer, in the afternoon.

Question 28.
What is the ‘nature of photochemical smog?
Answer:
It is oxidising in nature.

Question 29.
Name three methods used in green chemistry.
Answer:
Use of sunlight and microwaves, use of sound waves and use of enzymes.

Question 30.
Give three examples in which green chemistry has been applied.
Answer:

1. Dry cleaning of clothes
2. Bleaching of paper
3. Synthesising chemicals.

Question 31.
What are the main advantages of using green chemistry?
Answer:
It is a cost-effective approach that involves a reduction in material, energy consumption and waste generation.

Question 32.
What is the troposphere?
Answer:
The troposphere is the lowest region of the atmosphere (~ 10 km) in which man along with other organisms including plants exist.

Question 33.
What is the stratosphere?
Answer:
The stratosphere extends above the troposphere up to 50 km above; sea level.

Question 34.
Name three scientists who got Noble Prize in chemistry in 2005 for their work in reducing hazardous waste in creating new chemicals.
Answer:

1. Yves Chauvin
2. Robert H. Grubbs
3. Richard R. Schrock.

Question 35.
What is meant by metathesis?
Answer:
Metathesis is an example of how important basic science has been applied for the benefit of man, society and the environment.

Question 36.
In what regions of the atmosphere, the temperature increases with altitude and in which regions it decreases?
Answer:
Temperature increases with altitude in the stratosphere and thermosphere and it decreases with altitude in the troposphere and mesosphere.

Question 37.
What do you mean by ‘Inversion temperature’ in different regions of the atmosphere?
Answer:
When we go from one region of the atmosphere to the next adjoining region, the trend of temperature changes from increase to decrease or vice-versa is called inversion temperature.

Question 38.
What is the most important sink of CO pollutant?
Answer:
Soil micro-organisms.

Question 39.
What is the compound formed when CO combines with blood?
Answer:
Carboxyhaemoglobin (HbCO).

Question 40.
What is anoxia starvation in the body (due to CO poisoning)?
Answer:
Actual oxygen starvation in the body (due to CO poisoning).

Question 41.
How are the flue gases from industries freed from oxides of N and S?
Answer:
By scrubbing them with a cone. H₂SO₄ or with alkaline solutions of Ca(OH)₂ & Mg(OH)₂.

Question 42.
What is chlorosis?
Answer:
Slowing down the formation of chlorophyll in plants due to the presence of SO₂ as a pollutant is called chlorosis.

Question 43.
What is the cause of blue baby syndrome?
Answer:
Excess of nitrates in drinking water leads to methemoglobinemia which is commonly called ‘blue baby syndrome.

Question 44.
Give I.U.P.A.C. name of B.H.C.
Answer:
1, 2, 3, 4, 5, 6 Hexa- chloro cyclohexane.

Question 45.
What is meant by P.C.Bs.?
Answer:
P.C.Bs. are polychlorinated biphenyls. They are contaminants of water.

Question 46.
What is P.A.N.?
Answer:
Peroxy acetyl nitrate.

Question 47.
Name the oxidizing agent used in determining C.O.D.
Answer:
Potassium dichromate (K₂C₂O₇).

Question 48.
What are fungicides?
Answer:
The chemicals which check the growth of fungi are called fungicides.

Question 49.
When does rainwater become Acid rain?
Answer:
When the pH of rainwater becomes as low as 2 to 3.5, it forms acid rain.

Question 50.
What is meant by Taj Trapezium?
Answer:
Taj Trapezium is the area that includes the town of Agra, Firozabad, Mathura & Bharatpur. Under this plan, more than 2000 polluting industries lying inside the Trapezium would switch over to the use of natural gas or L.P.G. instead of coal or oil.

Environmental Chemistry Important Extra Questions Short Answer Type

Question 1.
Fish do not grow as well in warm water as in cold water. Why?
Answer:
The amount of dissolved oxygen in warm water is less than in cold water. For the fish to survive in water, the concentration of dissolved O₂ is 6 ppm which decreases in warm water.

Question 2.
Why does rainwater normally have a pH of about 5.6? When does it become acid rain?
Answer:
Normally rain has a pH of about 5.6 due to the dissolution of CO₂ of the atmosphere into it.
CO₂(g) + H₂0O(l) → H₂CO₃(al)
H2CO₃(aq) ⇌ 2H+ + CO₃^2-
When the pH of rain falls below 5.6, it becomes acid rain.

Question 3.
In which season the depletion of ozone in Antarctica takes place and when is it replenished?
Answer:
During spring seasons (i.e., in the months of September and October) depletion of ozone takes place and after spring (in the month of November) it is replenished.

Question 4.
What is the role of CO₂ in the greenhouse effect?
Answer:
The heat from the sun after being absorbed by the earth is remitted by the earth and absorbed by CO₂ and then radiated back to the earth, thereby warming it.

Question 5.
What are viable and non-viable particulates?
Answer:
Viable particulates are small-sized living organisms such as bacteria, fungi, moulds and algae etc. Non-viable particulates are formed by the disintegration of large-sized materials or condensation of small-sized particles or droplets e.g. mist, smoke, fume and dust.

Question 6.
Why there is ozone depletion mainly over Antarctica?
Answer:
This is because, in other parts of the stratosphere, chlorine-free radicals combine away but in Antarctica, the compounds formed are converted back to chlorine-free radicals which deplete the ozone layer.

Question 7.
Explain giving reasons. “The presence of CO reduces the amount of haemoglobin available in the blood for carrying oxygen to the body cells”.
Answer:
CO combines with haemoglobin of the red blood corpuscles (R.B.Cs) about 300 times more easily than oxygen to form carboxyhaemoglobin reversibly as follows:
Hb + CO ⇌ HbCO

Thus it is not able to combine with oxygen to form oxyhaemoglobin and transport of oxygen to different body cells cannot take place.

Question 8.
Why is acid rain considered a threat to the Taj Mahal?
Answer:
Taj Mahal is made of white marble (CaCO₃). The acid rain contains H₂SO₄ in a very dilute form which attacks marble, thereby pitting it, discolouring it and making it lustreless.
CaCO₃(s) + H₂SO₄ (aq) → CaSO₄(aq) + CO₂(g) + H₂O(l)

Question 9.
What are the harmful effects of carbon monoxide?
Answer:
1. Carbon mono-oxide when inhaled passed through the lungs into the blood where it reacts with the haemoglobin of the die R.B.C. to form a complex. Known as Carboxy haemoglobin the latter is not in a position to transport the inhaled oxygen to various parts of the body. This will cause suffocation & will ultimately lead to death.

2. A high concentration of carbon mono-oxide (100 ppm) or more will harmfully affect the plants causing leaf drop reduction in leaf size & pre-mature ageing etc.

Question 10.
What are the harmful effects of oxides of nitrogen?
Answer:

1. A high concentration of NO₂ in the atmosphere is harmful to plants resulting in leaf spotting, retardation of photosynthetic activity & also suppression the vegetation growth.
2. NO₂ results in respiratory problems in human beings & lead to bronchitis.
3. NO₂ has harmful effects on nylon, rayon & cotton yarns & also cause cracks in the rubber.
4. They also react with ozone present in the atmosphere & thus decrease the density of ozone.

Question 11.
What are the harmful effects of oxides of Sulphur?
Answer:

1. SO₂ is a major source of irritation for the eyes & respiratory tract & leads to serious diseases such as bronchitis & lung cancer even when it is about 2.5 ppm. in the atmosphere.
2. London smog & sulphurous acid smog which mainly consists of SO₂(0-40 ppm or above) is known as smog killer.
3. Plants are relatively more sensitive to the harmful effect of SO₂ as compared to animals. Constant exposure to air containing a high level of SO₂ kills leaf tissues, reduces plants productivity & also bleaches leaf pigments.
4. SO₂ has harmful effects on buildings & statues made up of marble & limestone.
5. Air polluted with oxides of sulphur also accelerates the corrosion of metals like copper, zinc, iron etc.

Question 12.
How do detergents cause water pollution?
Answer:

1. Detergents are non-biodegradable & they cause water pollution.
2. They inhibit the oxidation of organic substances present in wastewaters because they form a sort of Envelope around them.
3. They form stable foam in rivers that extend over several hundred meters of the river water.

Question 13.
Why is add rain considered a threat to the Taj Mahal?
Answer:
The air of the city Agra, where the Taj Mahal is located, contains a very high level of Sulphur & nitrogen oxides. The resulting acid rain reacts with the marble of the Taj Mahal. Causing pitting in this wonderful monument that has attracted people from around the world. As a result, the monument is being slowly eaten away & the marble is getting decoloured & lusterless.

Question 14.
Explain giving reasons “The presence of co-reduces the amount of haemoglobin available in the blood for carrying oxygen to the body cells?
Answer:
Carbon mono-oxide binds itself with the haemoglobin of the R.B.Cs. about 200 times more easily than oxygen to form carboxyhaemoglobin reversibly as follows:
Hb + CO ⇌ HbCO

The presence of CO, therefore, reduces the amount of haemoglobin available in the blood for the transport of oxygen to the body cells & therefore, with fewer O₂ levels, normal metabolism is impaired.

Question 15.
What do you understand by the greenhouse effect? What are the major greenhouse gases?
Answer:
The greenhouse effect is the phenomenon in which the earth’s atmosphere traps the heat from the sun, & prevents it from escaping into outer space. The warming of the earth or global warming due to re-emission of sun’s energy absorbed by the earth followed by its adsorption by CO₂ molecules & H₂O vapours present near the earth’s surface & then it’s radiation back to the earth is called “Greenhouse effect”. The major greenhouse gases are:

CO₂, CH₄, C.F.Cs & water vapours.

Question 16.
What do you understand by Mists, Smoke, Fumes & Dust?
Answer:
Mists: Mists are produced by particles of spray liquids & the condensation of vapours in the air. Examples are portions of herbicides & insecticides, that miss their targets & travel through the air to form mists.

Smoke: They are very small soot particles produced by burning & combustion of organic matter. Oil smoke, Tobacco smoke & Carbon smoke are typical examples.

Dust: It consists of fine particle produced during crushing, grinding & attribution of solid materials. Non-viable dust particulates in the atmosphere consist of ground limestone, sand tailing from floatation, pulverised coal, cement, fly ash & silica dust.

Question 17.
What is Pneumo coniosis? How does it occur?
Answer:
The smaller particulate pollutants are more likely to penetrate into the lungs. These fine particles are carcinogens. Inhalation of small particles irritates the lungs & exposure to such particles for long periods of time causes fibrosis of lung lining. This type of disease is termed “Pneumoconiosis”. This occurs due to exposure to such particles for a long time.

Question 18.
What measures should be taken to check pollution by sewage?
Answer:

1. Sewage must be churned by machines so that the large pieces may break into smaller ones & may get mixed thoroughly. The churned sewage is passed into a tank with a gentle slope. Heavier particles settle & the water flowing down is relatively pure.
2. Water must be sterilized with the help of chlorination. Chlorination is very essential, particularly in rainy session.
3. Treatment of water with alum, lime etc.

Question 19.
Discuss the water pollution caused by industrial wastes.
Answer:
The compounds of lead, mercury, cadmium, nickel, cobalt & zinc etc. Which are the products of chemical reactions; carried in the industrial units pollute water to a large extent & are responsible for many diseases. Mercury leads to Minamata disease, & lead poisoning leads to various types of deformities.

Question 20.
What remedial steps should be taken to save a person suffering from co-poisoning?
Answer:
Remedial treatment for CO poisoning:

1. Carry the patient into the fresh air immediately & do not allow him to walk.
2. Lose his clothes & take off his shoes.
3. Give artificial respiration if the patient is not able to breathe properly.
4. In the hospital the patient should be kept in a high-pressure chamber containing oxygen at 2-2.5 atm pressure under pressure, CO of carboxyhaemoglobin is replaced by O₂ & thus the transport of O₂ to different parts of the body starts.
   HbCO + O₂ ⇌ HbO₂ + CO

Question 21.
How can pollution due to nitrogen & Sulphur oxides be controlled?
Answer:
The following steps are taken to control NO₂ & SO₂ pollution:
(i) The catalytic converters should be used in the automobile exhausts which is the first stage of converting the oxides of nitrogen, to free N₂ or to a small amount of NH₃.
(ii) The fumes of gases coming from power plants or industrial, units & containing NO₂ & SO₂ are freed from these gases by scrubbing the fumes with H₂SO₄. The following reactions take place:

• I- step: NO₂ + SO₂ + H₂O → H₂SO₄ + NO
• II- step: NO + NO₂ → N₂O₃
• III- step: N₂O₃ + 2H₂SO₄ → 2NOHSO₄ + H₂O

the fumes of gases are thus freed from NO₂ & SO₂ & are released into the atmosphere. The reaction product NOH₂O₄ is decomposed to get H₂SO₄ which is then used again for scrubbing. As NO_x & SO_x are > acidic oxides, scrubbing can also be done with an alkaline solution such as Ca(OH)₂ & Mg(OH)₂.

Question 22.
What is groundwater pollution? How does it take place?
Answer:
Water below the surface of the earth is called groundwater. Most of the freshwater (> 90%) is present as groundwater. The remaining is present as groundwater. The remaining is present in lakes, ponds, rivers, streams etc. only 2% of water is present as soil moisture above the water table, which is needed for the growth of the plants.

Groundwater collected below the surface of the earth after passing through the pores of earthy materials which acts as a filter for it & is pure. It is for this reason that well water is spring water is used for drinking purpose in rural areas. However, due to the disposal of domestic wastes, industrial effluents, use of fertilizers & pesticides in agriculture, A number of harmful soluble substance dissolve into the rainwater & result in groundwater pollution especially where the water table is high.

Question 23.
How is the pollution of river water caused in India? What measures have been taken by the Government to check river pollution?
Answer:
The main reasons for pollution in rivers are as follows:

1. Industrial waste discharge: It includes those from paper, textiles, fertilizers, rayon, pesticides, detergents, drug industries & refineries.
2. Domestic sewage discharge: The government has enacted laws banning the discharge of industrial or domestic wastes into these rivers. It has started the following plans to clean up the water of these rivers.
   (a) Ganga Action Plan I & II
   (b) Yamuna Action Plan
   (c) Plan to clean Hoogly water.

Question 24.
What are the effects of oil pollution on Sea Water?
Answer:
Effects of oil pollution on Sea Water:

1. oil spills cause heavy damage to fishes oil coating makes them unable to respire & clogs their gill slits. Aromatic compounds present in them are poison for fishes.
2. Emulsified oil goes deep down into the sea damaging aquatic animals & plants.
3. Oil spills result in a reduction of dissolved oxygen.
4. The most affected by oil pollution are the sea birds. Natural, insulating oil & waxes which shield the birds from water are broken down by the spilt oil. As a result due to loss of insulation, they start shivering & are freezing to death, especially in winter.

Question 25.
Which is the permitted safety limit of fluoride & lead concentration with respect to international standards of drinking water?
Answer:
The fluoride concentration should be about 1 ppm in drinking water. This concentration is within agreed safety limits & has been shown to protect teeth against decay. A high concentration of fluoride is poisonous & are harmful to bones & teeth at levels over 10 ppm.

The lead concentration should be about 50 ppm in drinking water.

Question 26.
Define the term pesticides? What are three categories of pesticides?
Answer:
Pesticides are substances that are used to control the reproductive process of unwanted organisms.

Three main categories of pesticides are:

1. Insecticides: These are used to control insects & curb diseases (Malaria, Yellow fever) & protect crops e.g. D.D.T.
2. Herbicides: These are used to kill weeds eg. NaClO₃ & Na₂AsO₃
3. Fungicides: These are used to check the growth of fungi e.g. Methyl mercury.

Question 27.
Write four major pollutants of water, their source & effects.
Answer:

S.No. Major Pollutants of Water	Sources	Effects
1. Lead	Lead-acid batteries & leaded gasoline	Toxic to organisms causes phimosis
2. Acids	Mine drainages, Industrial wastes	Kills organisms, overgrowth of algae & aquatic weeds.
3. Detergents	Households & industries	Depletion of dissolved oxygen.
4. Pesticides & Insecticides	Agriculture & Mosquitoes repellants	Toxic to fishes, birds & mammals.

Question 28.
Discuss the mechanism of treatment of industrial wastes.
Answer:
The treatment of industrial waste depends upon the nature of the pollutant present. In order to ascertain it, the pH of a medium is first determined & the waste is then neutralized with the help of suitable acids or alkalies.

The chemical substances present in the industrial waste product, dissolved in water can be precipitated by suitable chemical reactions & removed later on from water. Quite recently, photo-catalysis & ion exchangers have been developed for the treatment of industrial wastes.

Question 29.
What measures are necessary to control soil pollution?
Answer:
In order to control soil pollution, the following measures are necessary:

1. Use of manures: Manures is semi-decayed organic matter which is added to the soil to maintain its fertility. These are mostly prepared from animal dung & another form of wastes. These are much better than the commonly used fertilizers.
2. Use of bio-fertilizers: These are organisms that are inoculated in order to bring about nutrient enrichment of the soil e.g. nitrogen-fixing bacteria & blue-green algae.
3. Proper sewerage System; A proper sewerage system must be employed & sewerage recycling plants must be installed in all town & cities.
4. Salvage & Recycling: Rag-picking removes a large no. of waste articles such s paper, polythene, cardboard rags, empty bottles & metallic, articles. These are subjected to recycling & this helps in checking soil pollution.

Question 30.
What are aerosols? Define the term green chemistry.
Answer:
Very small particles of solids & liquids dispersed in the air are called aerosols. Aerosol particles have a diameter of less than one micrometre (< 1μm). The particles of aerosols are so small that they remain suspended in the air.

Green Chemistry: The branch of chemistry which emphasizes the process & products that reduce or eliminate the use & generation of toxic/hazardous substances is called Green Chemistry.

Environmental Chemistry Important Extra Questions Long Answer Type

Question 1.
What is the difference between London (classical) smog and photochemical (Los Angeles) smog?
Answer:

Classical (London) smog	Photochemical (Los Angeles) Smog
1. This type of smog was first observed in London in 1952.	1. This type of smog was first observed in Los Angeles in 1950.
2. It is formed due to the presence of SO2 and humidity in the air which combines to form H2S04 fog which gets deposited on the particulates.	2. It is formed due to a photochemical reaction taking place when the air contains NO2 and hydrocarbons.
3. It involves smoke and fog.	3. It does not involve any smoke or fog. The word smog is a misnomer here.
4. It is formed in the months of winter particularly in the morning hours when the temperature is low.	4. It is formed during the months of summer during the afternoon when there is bright sunlight so that photochemical reaction can take place.
5. It causes a problem in the lungs.	5. It causes irritation in the eyes.
6. It is reducing in character.	6. It is oxidising in character.

Question 2.
(a) Mention some of the sources of soil pollution.
Answer:
Some of the sources of soil pollution are:

1. Industrial wastes.
2. Urban wastes
3. Agriculture pollutants include synthetic fertilizers, pesticides, insecticides like DDT, BHC, herbicides like NaClO₃, Na₃AsO₃, fungicides soil conditioners, farm wastes.
4. Radioactive pollutants.

(b) Mention some ways to control environmental pollution.
Answer:
Environmental pollutants like household waste and industrial waste can be controlled in the following manner.

1. Recycling: Used glass bottles, iron scrap, plastic waste, polythene bags, used newspapers and magazines can all be properly recycled.
2. Burning and incineration.
3. Sewage treatment
4. Digesting.
5. Dumping
6. Green chemistry can be used to reduce pollution.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 12 Important Extra Questions Organic Chemistry: Some Basic Principles and Techniques

Organic Chemistry: Some Basic Principles and Techniques Important Extra Questions Very Short Answer Type

Question 1.
What type, of hybridisation, is involved in
(i) planar and
Answer:
sp²

(ii) linear molecules?
Answer:
sp.

Question 2.
Arrange the following in increasing order of C – C bond strength:
C₂H₆, C₂H₄C₂H₂.
Answer:
C₂H₆ < C₂H₄ < C₂H₂.

Question 3.
Arrange the following in decreasing order of C — C bond length:
Answer:
C₂H₆ > C₂H₄ > C₂H₂.

Question 4.
What is the type of hybridisation of C atoms in benzene?
Answer:
It is an sp² type of hybridisation.

Question 5.
What are isomers?
Answer:
Compounds having the same molecular formula but different physical and chemical properties are called isomers.

Question 6.
Select electrophiles out of the following:
H⁺ Na⁺, Cl^–, C₂HSOH, AlCl₃, SO₃, CN^–, CH₃CH₂⁺,: CCl₂, R-X.
Answer:
H⁺, Na⁺, A1Cl3, SO3, CH3CH₂⁺,: CCl₂, R-X.

Question 7.
Select nucleophiles from the following.
BF₃ NH₃ OH^–, R-X, C₂H₅OH, H₃O⁺, NO₂, CN^–.
Answer:
NH₃, OH, C₂H₅OH, CN

Question 8.
Give the I.U.P.A.C. names of the following compounds

Answer:
2-Bromo-4 – methyl pent-3- one

Answer:
4-Methyl-2-nitro pent – 3 – one

Answer:
2 – Ethoxy – 4 – methoxypent – 3 – one

Answer:
2-Bromo-4-nitro pent-3-one

(v) (CH₃)₄C
Answer:
2, 2-Dimethylpropane

(vi) (CH₃)₂CHCOOH.
Answer:
2-Methyl propanoic acid.

Question 9.
What is a functional group?
Answer:
The atom or group of atoms present in a molecule that determines its chemical properties is called the functional group.

Question 10.
Arrange the following in increasing order of-I effect.
(i) -NO₂, -COOH, -F, -CN, – I.
Answer:
-I < -F < -COOH < -CN <NO₂.

Question 11.
Arrange the following in decreasing order of + I effect: CH₃-, D, (CH₃)₃C-, (CH₃)₂CH-, CH₃-CH2₂–
Answer:
(CH₃)₃C → (CH₃)₂CH → CH₃-CH₂ → CH₃ → D

Question 12.
Name the alkyl groups derived from isobutane.
Answer:
(CH₃)₂CH – CH₂– isobutyl and (CH₃)₃C – tertiary butyl.

Question 13.
What type of isomerism is shown by butane and isobutane.
Answer:
Chain or nuclear isomerism.

Question 14.
Write the tautomer of acetaldehyde and its I.U.P.A.C. name.
Answer:
CH₂ = CH-OH and its I.U.P.A.C. name is Eth-1-en-1-ol,

Question 15.
Give one example of functional isomerism.
Answer:
CH₃– CH₂– OH and CH₃ – O – CH₃.

Question 16.
Give one example of position isomerism.
Answer:
CH₃ – CH₂ – CH₂OH and CH₃CH(OH) CH₃ .

Question 17.
Draw the structure of the tautomer of phenol and write its I.U.P.A.C name.
Answer:

Question 18.
A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. What is the number of structural isomers possible?
Answer:
Four
1. CH₃CH₂CHCl₂
2. CH₃CHClCH₂Cl
3. CH₃CCl₂CH₃
4. ClCH₂ – CH₂ – CH₂Cl

Question 19.
Write the metamer of diethyl ether. What is its I.U.P.A.C. name?
Answer:

1. CH₃O CH₃ CH₂ CH₃ Its I.U.P.A.C. name is 1 -Methoxypropane.
2. 

Question 20.
Give the I.U.P.A.C. name of CH₂ = CH-CH (CH₃)₂
Answer:

Question 21.
How many cr and it bonds are present in each of the following molecules?
(a) H-C = CCH=CH-CH₃
Answer:
No. of σ_C-C = 4;
No. of σ_C-H = 6.

Total no. of σ bonds =10.
Total no. of π_C=C bonds = 2 + 1=3.

(b) CH₂=C=CH-CH₃.
Answer:
(b) No. of σ_C-C = 3 bonds = 3,
No. of σ_C-H = 6;

Total No. of , σ-bonds = 3 + 6 = 9
No. of π_C=C bonds = 2.

Question 22.
What is the shape of the following molecules
(a) H₃ C = O
Answer:
HCHO formaldehyde C: sp² hybridised; shape: Trigonal planar

(b) CH₃F
Answer:
CH₃F; C = sp³ hybridised shape: Tetrahedral

(c) HON.
Answer:
H-C ≡ N; C is sp hybridised, HCN is linear.

Question 23.
Write the T.U.P.A.C. name of

Answer:
Compound is

Question 24.
Give the condensed and bond-line structural formula for
(a) Penta-1, 4-dien
Answer:
CH2 = CH – CH2 – CH = CH2

(b) Hexa-1, 3, 5 triene.
Answer:
CH₂ = CH – CH = CH – CH = CH
Its bond line structure is

Question 25.
Write the I.U.P.A.C. names of HOOC-C ≡ C-COOH and

Answer:
But-2-yne – 1,4-dioic acid and 3-Methyl pentanenitrile.

Question 26.
How will you purify essential oils?
Answer:
Essential oils are volatile and are insoluble in water, Therefore, they are purified by steam distillation.

Question 27.
A liquid (1.0 g) has three components. Which technique will you employ to separate them?
Answer:
Column chromatography.

Question 28.
A reaction carried out using aniline as a reactant as well as a solvent. How will you remove unreacted aniline?
Answer:
By steam distillation.

Question 29.
How will you separate a mixture of O-nitrophenol and /Mutrophenol?
Answer:
Steam distillation O-Nitrophenol being volatile distils over along with water while p-nitrophenol being non-volatile remains in the flask.

Question 30.
How will you purify a liquid having non-volatile impurities?
Answer:
Simple distillation will give pure liquid while the non-volatile impurities in the flask as residue.

Question 31.
Suggest a method to purify
(i) Kerosene containing water
Answer:
By solvent extraction using a separating funnel.

(ii) a liquid that decomposes at its boiling point.
Answer:
Distillation under reduced pressure.

Question 32.
Suggest methods for the separation of the following mixtures:
(i) A mixture of liquid A,(B. Pt. =365 K) and liquid B (b.p.356 K)
Answer:
Fractional distillation B.Pts of two liquids differ only by just 9°.

(ii) A mixture of liquid C (b.p. 353 K) and liquid D (b.p. 413 K).
Answer:
Simple distillation, since B.Pts of two liquids, are wide apart.

Question 33.
Name two compounds that do not contain halogen but give a positive Beilstein test.
Answer:
Urea and thiourea give a positive Beilstein test due to the formation of volatile cupric cyanide.

Question 34.
Lassaigne’s test is not shown by diazonium salts. Why?
Answer:
Diazonium salts lose N2 on heating much before they have a chance to react with fused sodium metal.

Question 35.
What type of compounds are purified by sublimation?
Answer:
Substances whose vapour pressures become equal to atmospheric pressure much below their melting point.

Question 36.
How will you separate iodine from sodium chloride?
Answer:
By sublimation.

Question 37.
Name two methods that can be safely used to purify aniline.
Answer:
Vacuum distillation and steam distillation.

Question 38.
Define the term elution as applied to column chromatography.
Answer:
The process of extraction of different adsorbed compounds from the column by means of a suitable solvent is called elution.

Question 39.
Suggest a suitable technique of separating naphthalene from kerosene oil present in a mixture.
Answer:
Simple distillation.

Question 40.
What conclusion would you draw if during Lassaigne’s test a blood-red colouration is obtained?
Answer:
It shows the presence of N & S together in the compound.

Question 41.
What type of organic compounds cannot be Kjeldahlised?
Answer:
A compound containing an N atom in the ring or the presence of – NO₂, (nitro) and — N =N -(azo) in them.

Question 42.
Can we estimate oxygen in the organic compound?
Answer:
No. it is estimated indirectly by subtracting the percentage of all elements present in an organic compound from 100.

Question 43.
Why do we use copper spiral in Duma’s method?
Answer:
To reduce back any oxides of nitrogen formed during combustion to nitrogen.

Question 44.
Give two examples of adsorbents used in chromatography.
Answer:

1. Alumina gel (Al₂O₃) and
2. Silica gel SiO₂.

Question 45.
Is Lassaigne’s extract neutral, acidic or alkaline?
Answer:
It is alkaline.

Question 46.
The empirical formula of a compound is CH₂. It’s one mole has a mass of 42 g. What is its molecular formula?
Answer:
Molecular formula = n × empirical formula
= \(\frac{42}{14}\) × CH₂ = C₃H₆

Question 47.
In which C-C bond of CH₃CH₂CH₂Br, the inductive effect is expected to be the least?
Answer:
The magnitude of the inductive effect diminishes as the number of intervening bonds increases. The effect is least in the C₃-H bond.

Question 48.
Can you use K in place of Na for fusing an organic compound in Lassaigne’s test?
Answer:
No, because K is more reactive than Na.

Question 49.
Which solution is used to absorb CO₂ produced during combustion?
Answer:
KOH solution is used to absorb CO₂ gas.

Question 50.
What is the cause of geometrical isomerism in alkenes?
Answer:
Alkenes have a π-bond & the restricted rotation around the π-bond gives rise to geometrical isomerism.

Organic Chemistry: Some Basic Principles and Techniques Important Extra Questions Short Answer Type

Question 1.
Expand each of the following bond-line formulae to show all the atoms including carbon and hydrogen.

Answer:

Answer:

Answer:

Answer:

Question 2.
For each of the following compounds, write a more condensed and also their bond line formulae.

Answer:
Condensed formulae are
(CH3)2CH CH2 OH

Answer:
CH₃(CH₂)₅ CHBr CH₂ CHO

Answer:
HO(CH₂)₃ CH(CH₃) CH(CH₃)₂

Answer:
HOCH(CN)₂

Question 3.
What is the type of hybridisation of each carbon in the following compounds?
(a) CH₃Cl
Answer:

(b) (CH₃)₂CO
Answer:

(c) CH₃CN
Answer:

(d) HCONH₂
Answer:

(e) CH₃CH = CHCN.
Answer:

Question 4.
What is the shape of the following molecules:
(a) H₂C = O
Answer:
In H₂C = O; C is sp² hybridised, hence its shape is H trigonal planar

(b) CH₃F
Answer:
In CH₃ -F; C is sp³ hybridized
∴ it is tetrahedral

(c) H-C ≡ N?
Answer:
In H-C ≡ N; C is sp-hybridized, hence HCN is linear
H—C ≡ N.

Question 5.
Give the I.U.P. A.C. names of the following compounds:

Answer:
2-Ethylprop-2-en-l-ol

(ii) CH₃ – CH = CH COOH
Answer:
But-2-en-l-oic acid

(iii) (CH₃)₂C = CHCOCH₃
Answer:
4-Methylpent-3-en-2-one

Answer:
3-Chloropropanal

Answer:
3-Methylbutane-l-al

(vi) CH2 = CH – CN.
Answer:
Prop-2-en-1-nitrile.

Question 6.
Write the I.U.P.A.C. names of

Answer:
3-Ethyl-4-methylhept-5-en-2-one

Answer:
2-Ethyl-3-methylpent-2-en-1 -one.

Question 7.
Write 1.U.P.A.C. names of

Answer:
2-[2-methylclo but-l-enyl] ethanal

Answer:
2-(3-Oxobutyl) cyclohexane-1 one

Answer:
Methyl (2-oxo cyclopentane-1-carboxylate

Answer:
Cyclohex-2-en-l-ol

Answer:
2-Ethenyl-3-methyl-cyclohexa-l, 3-diene

Answer:
4-Formyl-2-oxo-cyclohexane-l carboxylic acid.

Question 8.
Draw the structures of
(i) Methyl t-butyl ether
Answer:

(ii) 2-Chloro-1, 1, 1-trifluoro ethane
Answer:
F3C – CH2Cl

(iii) 2-Methyl buta, 1, 3-diene
Answer:

(iv) Pent-2-en-l-ol
Answer:
CH₃ – CH₂ – CH = CH – CH₂OH

(v) Cyclo hex-2-en-l-ol
Answer:

(vi) l-Bromo-3-chloro cyclohex-1-ene
Answer:

Question 9.
Write I.U.P.A.C. names of

Answer:
(2-Isopropyl) benzene

Answer:
1-Phenylpropane-l-one

Answer:
1 -Phenylethan-1 -ol

Answer:
1, 3-Benzene dicarboxylic acid

Answer:
3-Phenylpropanal

Answer:
3-Boromo-4-hydroxybenzoic acid

Answer:
4-Hydroxy-3-methoxy benzaldehyde.

Question 10.
Write the structure of
(i) O-Ethyl anisole
Answer:

(ii) p— nitroaniline
Answer:

(iii) 4-Ethyl-I-fluoro-2-nitrobenzene.
Answer:

Question 11.
Which is more polar bond in the following pairs of molecules
(a) H₃C-H, H₃C-Br
Answer:
C —Br since Br is more electronegative than H.

(b) H₃C-NH₂, H₃C-OH
Answer:
C—O since O is more electronegative than N.

(c) H₃C-OH, H₃C-SH.
Answer:
C — O since O is more electronegative than S.

Question 12.
In which C-C bond of CH₃CH₂CH₂Br, the inductive effect is expected to be the least.
Answer:
The magnitude of the inductive effect decreases with distance from the active centre and hence this effect is least in C₂—C₃ bond.

Question 13.
Write the resonance structures of
(a) CH₃COO- and
Answer:

(b) CH₆H₅NH₂. Show the movement of electrons by curved arrows.
Answer:

Question 14.
Which of the following pairs of structures do not constitute resonance structures?

Answer:
They differ in the position of atoms and hence are not resonance structures.

Answer:
They are a pair of resonance structures as they differ in the position of electrons.

Answer:
They differ in the position of atoms and so are not resonance structures. They are tautomers.

(d) CH₃CH = CHCH₃ and CH₃CH₂CH=CH₂
Answer:
They are not resonance structures as they differ in the position of atoms.

Question 15.
Write the resonance structures of CH₂ = CH-CHO and arrange them in decreasing order of stability.
Answer:

The structure I is most stable as each C & O have octets completed and there is no charge on either of them.
II & III involve charge separation hence both are less stable than I. However II is more stable than III because in II more electronegative oxygen carries a negative charge.

Thus decreasing order of stability is I > II > III.

Question 16.
Using curved arrow notation show the formation, of reactive intermediates when the following covalent bonds undergo heterolytic fission.
(a) CH₃-SH₃
Answer:

(b) CH₃-CN
Answer:

(c) CH₃-CU
Answer:

Question 17.
Giving proper justification categorise the following molecules/ ions as nucleophiles or electrophiles:

Answer:

are all nucleophiles as each one of them has one or more lone pairs of electrons to donate.

are all electrophiles. All these species have a sextet of electrons around positive centres.

Question 18.
A mixture contains two components A and B. The solubilities of A and B in the water near its boiling point are 10 grams per 100 ml and 2g per 100 ml respectively. How will you separate A and B from this mixture?
Answer:
Fractional crystallisation. When the saturated hot solution of this mixture is allowed to cool, the less soluble component B crystallises out first leaving the more soluble component B in the mother liquor.

Question 19.
A mixture containing benzoic acid and nitrobenzene is given to you. Using an appropriate chemical reagent, how will you proceed to separate them?
Answer:
The mixture is shaken with a dilute solution of NaHCO₃ and extracted with-ether or chloroform when nitrobenzene goes into the organic layer; Distillation of this will yield nitrobenzene. The aqueous layer is acidified with dil. HCl and the solution are cooled.

Filteration gives benzoic acid.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + CO₂ + H₂O.
C₆H₅COONa + HCl (dil.) → C₆H₅COOH + NaCl.

Question 20.
The Rf value of A and B in a mixture determined by TLC in a solvent mixture are 0.65 and 0.42 respectively. If the mixture separated by column chromatography using the same solvent mixture as a mobile phase, which of the two components A or B will elute first? Explain.
Answer:
Since the Rf value of A is 0.65, therefore, it is less strongly adsorbed as compared to component B with an Rf value of 0.42. Therefore an extraction in column chromatography, A will elute first.

Question 21.
Without using column chromatography, how will you separate a mixture of camphor and benzoic acid?
Answer:
Sublimation cannot be used as both camphor and benzoic acid sublime on heating. Therefore, a chemical method using NaHCO₃ solution is used when benzoic acid dissolves leaving camphor behind. The filtrate is then cooled with dilute HCl to get benzoic acid.

Question 22.
0.12g of an organic compound containing phosphorus gave 0.22g of Mg2P207 by the usual analysis. Calculate the percentage of phosphorus in the compound.
Answer:
Mass the substance taken (w) = 0.12g
Wt. of Mg₂P₂O₇ formed (x) = 0.22 g
222 g of Mg₂P₂O₇contain phosphorus = 62 g.

% of phosphorus = \(\frac{62}{222} \times \frac{0.22}{0.12}\) × 100 = 51.20

Question 23.
Compare inductive & mesomeric effects.
Answer:

Inductive effect	Mesomeric effect
1. It operates in saturated gp. of compounds.	1. It occurs in unsaturated & especially in conjugated compounds.
2. It involves electrons in σ – bonds.	2. It involves electrons in π – bonds.
3. Electron pair is slightly displaced & there only partial charges are developed.	3. The electron pair is transferred completely with the result full positive & negative charges are created.
4. It is transmitted over only a quite short distance.	4. It is transmitted from one end to the other of quite large molecules provided conjugation (i.e. delocalised orbitals) is present through which it can proceed.

Question 24.
What is the difference between distillation, distillation under reduced pressure & steam distillation?
Answer:

Distillation	Distillation under reduced pressure	Steam distillation
This is used to separate volatile liquid from non-volatile liquid or solid separately.	This is used to purify liquids that decompose at or below their boiling points.	This is used for purifying substances that are steam volatile & immiscible with water.

Question 25.
How will you purify sugar which has impurities of sodium chloride?
Answer:
Sugar may be purified by the crystallization method. This can be purified by shaking the impure solid with hot ethanol at 345K. The sugar will dissolve whereas common salt remains insoluble. The hot solution is filtered, concentrated & allowed to cool when crystals of sugar will separate out. In this case, hot water has been used as a solvent. The purification of sugar would not have been possible since both sugar’& common salt are soluble in water.

Question 26.
Differentiate between Ionic & free radical reactions.
Answer:

Ionic reactions	Free radical reactions
1. These occur only rarely in the gas phase but mainly in a solution of polar solvents; the reaction is influenced by the polarity of the solvent.	1. These occur in gas phases or in non-polar solvents.

Question 27.
For each of the following compounds, write a more condensed formula & also their bond-line formula.

(b) HOCH2CH2CH2CHCH3CHCH3CH3

Answer:

Question 28.
Expand each of the following bond line formulae to show all the atoms including carbon & hydrogen.

Answer:

Answer:

Answer:

Question 29.
Explain why is (CH₃) C+ more stable than CH₃ C⁺ H₂ & C⁺ H₃ is the least stable cation?
Answer:
Hyperconjugation interaction in (CH₃)₃ C⁺ is greater than in C+⁺ H₃ C⁺ H₂ has 9 C -H bonds. In C H₃, the C -H bonds are in the nodal plane of the vacant 2p-orbital & hence cannot overlap with it.
Thus C⁺ H₃ is least stable.

Question 30.
The choice of the solvent is of great importance in crystallizing organic substances. What are the characteristics of a suitable solvent?
Answer:
A suitable solvent must have the following characteristics;

1. The impurities & pure compound must have a large difference in their solubilities.
2. The pure compound must have low solubility at room temperature but high solubility at its boiling point.
3. The impurity should either be insoluble at room temperature or must have high solubility so that crystallization may give a high yield.
4. The solvent should have an average boiling point.
5. The solvent should neither react with the compound nor with impurities.
6. The solvent should not be highly inflammable.

Organic Chemistry: Some Basic Principles and Techniques Important Extra Questions Long Answer Type

Question 1.
Explain the principle of steam distillation.
Answer:
Steam distillation: The process of steam distillation is employed in the purification of substance from non-volatile impurities provided the substance itself is volatile in steam and insoluble in water.

This method is based on the facts that

1. A liquid boils at a temperature when its vapour pressure becomes equal to the atmospheric pressure.
2. The vapour pressure of a mixture of two immiscible liquids is equal to the sum of the vapour pressures of the individual liquids.

In the actual process, steam is continuously passed through the impure organic liquid. Steam heats the liquid and it gets practically condensed to water. After some time mixture of the liquid and water begins to boil, because the vapour pressure of the mixture becomes equal to the atmospheric pressure.

Obviously, this happens at a temperature that is lower than the boiling point of the substance or that of water. Thus an organic compound boils below its boiling points and chances of decomposition avoided. For example, a mixture of aniline (b.p 453 K) with decomposition and water (b.p. 373 K) under normal atmospheric pressure boils at 371K. At this temperature the

Steam Distillation

water boils at 371 K. At this temperature, the vapour pressure of water is 717 mm and that of aniline is 43 mm and therefore the total pressure is equal, to 760 mm. Thus in steam distillation, the liquid gets distilled at a temperature lower than its boiling point and chances of decomposition avoided. The proportion of water and liquid in the mixture that distils over is given by the relation.

\(\frac{w_{1}}{w_{2}}=\frac{P_{1} \times 18}{P_{2} \times M}\)
where w₁ and w₂ stand for the masses of water and liquid that distils over. P₁ and P₂ are vapour pressure of water and of liquid at the distillation temperature and M is the molecular mass of the liquid.

Question 2.
Dehydrobromination of compounds (A) and (B) yield the same alkene (c) Alkene (c) Can regenerate (A) and (B) by the addition of HBr in the presence and absence of peroxide respectively. Hydrolysis of A and B give isomeric products (D) and (E) respectively. 1, 1-Diphenyl ethane is obtained on the reaction of (C) of benzene in the presence of H⁺ ions. Give structures of A to E with reactions.
Answer:
Alkene (C) on reaction with benzene in the presence of H+ ions gives 1, 1-Diphenyl ethane. Therefore C must be styrene as depicted below

Now dehydrobromination of A and B give the same alkene C, i.e.,
styrene.
∴ A and B must be isomeric alkyl bromide.

A and B can be obtained by the addition of HBr in the presence and absence of peroxide to styrene.

Hydrolysis of A and B give isomeric alcohols (D) & (E) as

Question 3.
What are reaction intermediates? How are they generated by bond fission?
Answer:
The species which are generated as a result of bond fission are called reaction intermediates. The important reaction intermediates are:
1. Free Radicals: A free radical may be defined as an atom or group of atoms having an impaired electron. These are obtained as a result of homolytic fission of covalent bonds.

These free radicals are neutral particles, extremely transient, (short-lived) and highly reactive. They get consumed as soon as they are formed. They pair up their electron with another electron from wherever it is available. They occur only as a reaction intermediate. Their presence is felt in reactions, but cannot be isolated in a free state. For example dissociation of Cl2 gas in the presence of Ultraviolet light produces free radicals.

The alkyl free radicals are obtained when free radical: Cl reacts with alkanes.

Free radical may be primary, secondary, tertiary depending upon whether, one, two or three carbon atom attached to the carbon atoms carrying the odd electron.

The stability is CH₃ < 1° < 2° < 3°.

2. Carbocation or carbonium ion: It is defined as a group of atoms that contain positively charged carbon having only six electrons. It is obtained by heterolytic fission of a covalent bond involving a carbon atom.

They are also classified as primary, secondary and tertiary depending upon whether one, two or three carbon atoms are attached to the carbon bearing the positive charge as:

Thus the order of stability if CH₃⁺ < 1° < 2° < 3°.

3. Carbanion: A carbanion may be defined as a species containing a carbon atom carrying a negative charge. These are generated by the atom in which the atom linked to carbon goes without the bonding electrons. As a result of this carbon acquires a negative charge. For example, the removal of hydrogen of methyl part of acetaldehyde molecule as H+ ion leaving both the electron on carbon.

They are also very reactive species. They are also classified as primary, secondary and tertiary depending upon whether one, two or three carbon atoms are attached to the carbon atom bearing negative
charge.

The order of stability is the reverse of free radicals and carbocations
CH₃ ^– > 1° > 2° > 3°.

(iv) Carbenes: The carbenes are reactive neutral species in which carbon atom has six electrons in the valency shell out of which two are shared. The simplest carbene is methylene (CH₂). It is formed when diazomethane is decomposed by the action of light.

It is very reactive. It reacts with alkenes by adding to the double bond forming cyclopropane.

 

Organic Chemistry: Some Basic Principles and Techniques Important Extra Questions Numerical Problems

Question 1.
0.395 g of an organic compound by various method for the estimation of sulphur gave 0.582g of BaS04. Calculate the percentage of Sulphur.
Answer:
Mass of BaSO₄ = 0.582 g
BaSO₄ = S
233 = 32

233g of BaSO₄ contain sulphur = 32g
0. 582 g of BaSO₄ contains sulphur

Question 2.
0.15g of an organic compound gave 0.12g of AgBr by carius method. Find the percentage of bromine in the compound.
Answer:
Mass of AgBr formed = 0.12g
188 g of AgBr contains bromine = 80g.

Therefore, 0.12g of AgBr will contain bromine
= \(\frac{80 \times 0.12}{188}\) = 0.051 g

Percentage of bromine = \(\frac{0.051}{0.15}\) × 100 = 34%

Question 3.
0.40 g of an organic compound gave 0.3g of AgBr by carius method. Find the percentage of bromine in the compound.
Answer:
Mass of compound = 0.40 g

Now, 188 g of AgBr contains bromine = 80g.

Question 4.
0.15 g of an organic compound gave 0.12g of silver bromide by the carius method. Find out the percentage of bromine in the compound.
Answer:
Here the mass of substance taken = 0.15g
Mass of AgBr formed = 0.12g
Now 1 mole of AgBr = 1 g. atom of Br
or
188 g = 80g of Br

188g of AgBr contains bromine = 80g.
Hence 0.12g of AgBr contain bromine
= \(\frac{80 \times 0.12}{188}\) = 0.051 g

Thus, 0.15g of compound contains 0.051g of Br
Percentage of Br = \(\frac{0.051}{0.15}\) × 100 = 34%

Question 5.
0.2595g of an organic compound when treated with carius method, gave 0.35g of BaSO₄. Calculate the percentage of Sulphur in the compound.
Answer:
Here the mass of substance taken = 0.35g
Mass of BaSO₄ ppt. formed = 0.35g

Now 1 mole of BaSO₄ = 1 g. atom of Sulphur
233 g of BaSO₄ = 32g of S
i. e. 233g of BaS04 contains 32g of Sulphur

Therefore 0.35g of BaSO₄ will contain Sulphur

Question 6.
0.12g of an organic compound containing Phosphorus gave 0.22g of Mg₂P₂O₇ by the usual analysis. Calculate the percentage of Phosphorus in the compound.
Answer:
Here the mass of organic compound taken = 0.12g
Mass of Mg₂P₂O₇ formed = 0.22g .
Now 1 mole of Mg₂P₂O₇ = 222 g of Mg₂P₂O₇
= 62g of Phosphorus

i. e. 222g of Mg₂P₂O₇ contains Phosphorus = 62g
Therefore 0.22g of Mg₂P₂O₇ contains Phosphorus
= \(\frac{62}{222}\) × 0.22

Hence Percentage of Phosphorus = \(\frac{62}{222} \times \frac{0.22}{0.12}\) × 100 = 51.2

Question 7.
In a Dumas nitrogen estimation, 0.303g of an organic compound gave 50 cm3 of nitrogen collected at 300k & 715 m.m. of pressure. Calculate the percentage of nitrogen in the compound, (vapour pressure of water at 300K = 15 m.m.).
Answer:
Vapour pres1 re of the gas = 715 – 15 = 700 mm.
V₁ = 50 cm3, V₂ =?, P₁ = 700 mm, P₂ = 760 mm,
T₁ = 300 K, T₂ = 273 K
Applying \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
or
V₂ = \(\frac{P_{1} V_{1} T_{2}}{P_{2} T_{2}}\)
= \(\frac{700 \times 50 \times 273}{760 \times 300}\)
= 41.9 cm³

22400 cm³ of nitrogen at S.T.P. weighs = 28g.
41.9 cm³ of nitrogen at S.T.P. weighs
= \(\frac{28}{22400}\) × 41.9 = 0.0524 g

Percentage of nitrogen = \(\frac{0.0524}{0.3}\) × 100 = 17.46

Question 8.
In an estimation of Sulphur by various method, 0.2175g of a compound gave 0.5825g barium sulphate. Calculate the percentage of sulphur in the compound.
Answer:
Mass of the compound = 0.2175 g
Mass of barium sulphate = 0.5825 g
Molecular mass of BaS04 = 137 + 32 + 64 = 233 g

233 g of BaSO₄ contains sulphur = 32g
0.5825g of BaSO₄ contains Sulphur
= \(\frac{32}{233}\) × 0.5825g

Percentage of Sulphur = \(\frac{32}{233} \times \frac{0.5825}{0.2175}\) × 100
= 36.78

Question 9.
0.515 g of an organic compound containing Phosphorus give 0.214g of magnesium Pyrophosphate in various method for the estimation of Phosphorus. Calculate the percentage of Phosphorus in the given organic compound.
Answer:
Mass of organic compound = 0.515 g
Mass of magnesium Pyrophosphate = 0.214 g
Molar mass of Mg₂P₂O₇ = 222 g

222 g of Mg₂P₂O₇ obtained from Phosphorus = 62g
0.214g of Mg₂P₂O₇ are obtained from Phosphorus
= \(\frac{62}{222}\) × 100 = 0.0597 g

Percentage of Phosphorus = \(\frac{0.0597}{0.515}\) × 100 = 11.6

Question 10.
In Duma’s method for estimation of Nitrogen 0.3g of an organic compound gave 50 mL of nitrogen collected at 300K temperature & 715 mm pressure. Calculate the percentage composition of nitrogen in the compound (Aqueous tension at 300K = 15 mm)
Answer:
Volume of nitrogen collected at 300K & 715 mm Pressure = 50 mL
Actual pressure = 715 – 15 = 700 mm

Volume of nitrogen at S.T.P = \(\frac{273 \times 700 \times 50}{300 \times 760}\) = 41.9 mL
22400 mL of nitrogen weighs = \(\frac{28 \times 41.9}{22400 g}\)

Percentage of nitrogen = \(\frac{28 \times 41.9 \times 100}{22400 \times 0.3}\) = 17.46

Question 11.
Ammonia produced when 0.75g of a substance was KJeldahlized, neutralized 30 cm³ of 0.25N H₂SO₄. Calculate the percentage of nitrogen in the compound.
Answer:
Mass of organic substance = 0.75g
Volume of H₂SO₄ used up = 30 cm³
Normality of sulphuric acid = 0.25 N 30 cm³

H₂SO₄ of normality 0.25 N = 30 mL of NH₃ solution of normality 0.25N
But 1000 cm³ of ammonia solution of normality 1 contains 14 g of nitrogen

Therefore, 30 cm³ of 0.25N ammonia solution will contain nitrogen
= \(\frac{14}{1000}\) × 30 × 0.25
Percentage of nitrogen = \(\frac{14}{1000}\) × \(\frac{30 \times 0.25}{0.75}\) × 100 = 14

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 11 The p-Block Elements. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 11 Important Extra Questions The p-Block Elements

The p-Block Elements Important Extra Questions Very Short Answer Type

Question 1.
do boron halides form additional compounds with amines?
Answer:
Boron halides are Lewis acids and hence accept a pair of electrons from amines to form additional compounds.

Question 2.
How does boron interact with NaOH?
Answer:
2B + 6NaOH → 2Na₃BO₃ + 3H₂.

Question 3.
What is the oxidation state of C in
(a) CO
Answer:
+ 2

(b) HCN
Answer:
+ 2

(c) H₂CO₃
Answer:
+ 4

(d) CaC₂.
Answer:
-1.

Question 4.
What is the state of hybridization of C in
(a) CO₃^2-
Answer:
sp²

(b) CCl₄
Answer:
sp³

(c) diamond
Answer:
sp³

(d) graphite?
Answer:
sp²

Question 5.
Give two examples of electron-deficient compounds.
Answer:
BF₃ and B₂H₆.

Question 6.
Arrange the following halides of boron in the increasing order of acidic character.
BF₃, BCl₃, BBr₃, BI₃.
Answer:
BF₃ < BCl₃ < BBr₃ < BI₃.

Question 7.
What is dry ice? Why is it so-called?
Answer:
Solid CO₂ is known as dry ice. It does not wet a piece of paper/cloth and sublimes without melting. Therefore, it is called dry ice.

Question 8.
Write balanced equations to show hydrolysis reactions of CO₃^2- and HCO₃^–.
Answer:
CO₃^2- + H₂O ⇌ OH^– + HCO₃^–
HCO₃^– + H₂O ⇌ OH^– + H₂CO₃.

Question 9.
Why boron does not form B³⁺ ion?
Answer:
Boron has a very high sum of the first three ionisation enthalpies. Hence it cannot lose three electrons to form a B³⁺ ion.

Question 10.
Which oxide of carbon is an anhydride of carbonic acid?
Answer:
CO₂, because H₂CO₃ acid decomposes to give H₂O and CO₂.

Question 11.
What happens when a borax solution is acidified? Write a balanced equation for the reaction.
Answer:
Boric acid is formed.
Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4H₃BO₃ (boric acid)

Question 12.
By means of a balanced equation show how B(OH)₃ behaves as an acid in water.
Answer:
B(OH)₃ + 2HOH → [B(OH)^– + H₃O⁺.

Question 13.
What happens when boric acid is heated?
Answer:

Question 14.
What are boranes?
Answer:
Stable covalent hydrides of boron like B₂H₆, B₄H₁₀, B₅H₉ on analogy with alkanes are called boranes.

Question 15.
Write a balanced equation for the preparation of boron by reduction of BBr₃.
Answer:
2BBr₃(g) + 3H₂(g) → 2B(s) + 6HBr(g)

Question 16.
What is carborundum? What is its common use?
Answer:
Silicon carbide (SiC). It is used as an abrasive.

Question 17.
What is Freon gas? To what use is it put?
Answer:
Freon gas is dichlorodifluoromethane CCl₂F₂. It is used as a coolant in refrigerators and in air-conditioners.

Question 18.
Give the name of the compound used as a fire extinguisher under the name pyrone.
Answer:
Carbon tetrachloride (CCl₄).

Question 19.
What happens when aluminium metal is dipped in cone, nitric acid?
Answer:
One molecule thick layer of oxide is formed on the surface of A1 and further reaction does not proceed. Al is said to become passive.

Question 20.
What happens when CO is passed over heated nickel?
Answer:
Ni + 4CO → Ni(CO)₄.
Tetra carbonyl nickel (O) is formed.

Question 21.
How does boron react with dinitrogen?
Answer:
2B + N₂ → 2BN (boron nitride)

Question 22.
What is the chemical formula of Borazole or borazine? Why is it called inorganic benzene?
Answer:
Borazole is B₃N₃H₆. Because of its similarity to benzene, it is called inorganic benzene.

Question 23.
Write down the structure of Borazole.
Answer:
B₃N₃H₆.

Question 24.
Why thallium prefers to show an oxidation state of +1 rather than +3?
Answer:
Due to the inert-pair effect.

Question 25.
How do you explain that anhydrous AlCl₃ is covalent, but hydrated AlCl₃ is electrovalent?
Answer:
In the presence of H₂O, Al₂Cl₆ dissociates into hydrated Al³⁺ and Cl^– ions due to the high heat of hydration of these ions.

Question 26.
Why BBr₃ is a stronger Lewis acid than BF₃?
Answer:
This is because the back donation of electrons into empty 2p orbital of boron atom from filled p orbital of Br atom is much less than that by F atom due to larger size of Br atom than F atom.

Question 27.
Why trihalides of group 13 elements fume in the moist air?
Answer:
They are hydrolysed by water forming hydrogen halides.
MX₃ + 3H₂O → M(OH)₃ + 3HX.

Question 28.
Aluminium forms [AlF₆]^3-, but boron does not form [BF₆]^3- why?
Answer:
Boron does not have vacant d-orbitals. Therefore, it cannot expand its coordination number beyond four.

Question 29.
Why boron halides do not exist as dimers whereas AlCl₃ exists as Al₂Cl₆?
Answer:
Boron atom being small iff size is unable to accommodate large-sized halogens around it.

Question 30.
Gold has much higher first ionisation energy than boron, yet gold is metal while boron is non-metal. Explain.
Answer:
This is based on their crystal structure. Gold has a coordination number of 12 while boron has 6 or less than 6.

Question 31.
Why CO₂ is a gas whereas SiO₂ is solid. Explain why?
Answer:
CO₂ forms monomeric linear molecules, while SiO₂ exists as a giant-sized 3-dimensional network structure.

Question 32.
C and Si are almost tetravalent, but Ge, Sn, Pb show divalency. Why?
Answer:
The inert pair effect is shown by Ge, Sn, Pb.

Question 33.
Account for the fact that PbX2 is more stable than PbX₄. [X = Cl, Br]
Answer:
Due to the inert pair effect, Pb shows an oxidation state of +2.

Question 34.
PbCl₄ is less stable than SnCl₄, but PbCl₂ is more stable than SnCl₂. Why?
Answer:
Stability of + 4 oxidation state decreases down the group while that of + 2 oxidation state increases due to inert pair effect.

Question 35.
Why carbon shows maximum catenation in group 14 elements?
Answer:
Due to the strong C-C bond, its bond dissociation energy is the highest among group 14 members,

Question 36.
Pyrosilicates contain which anion?
Answer:
Si₂O₇^6-.

Question 37.
Mention one industrial application of silicones.
Answer:
Silicones are used for making water-proof papers by coating them with a thin layer of silicones.

Question 38.
What is the basic building unit of silicates?
Answer:
SiO₄^4- tetrahedra.

Question 39.
Which element of group 13 forms amphoteric hydroxide?
Answer:
Aluminium.

Question 40.
Which element of group 13 forms the most stable oxidation state of + 1?
Answer:
Thallium.

Question 41.
Explain why silicon shows a higher covalency than carbon?
Answer:
Due to the presence of d-orbitals, silicon shows a higher covalency of 6.

Question 42.
What are silicones?
Answer:
Silicones are synthetic organosilicon compounds containing repeated R₂SiO units held by Si—O—Si linkages.

Question 43.
What are the 2 main uses of zeolites?
Answer:

1. Softening of hard water.
2. Catalysts in petrochemical industries.

Question 44.
Why is boric acid considered a weak acid?
Answer:
Because it is not able to release H⁺ ions on its own. It receives OH^– ions from the water molecule to complete its octet & in turn releases H⁺ ions.

Question 45.
[SiF₆]^-2 is known whereas [SiCl₆]^2- not. Why?
Answer:
The reasons are:

1. Six large chlorine atoms cannot be accommodated around the Silicon atom due to the limitation of its size.
2. Interaction between lone pair of chlorine atom & silicon atom is not very strong.

Question 46.
Diamond is co-valent, yet it has a high M.P. Why?
Answer:
Diamond has a three-dimensional networked structure involving a strong C—C bond, which are very difficult to break & in turn it has a high melting point.

Question 47.
What is the common name of the recently developed allotrope of carbon i.e. C60 molecule?
Answer:
Fullerene.

Question 48.
The M.P. & B.P. of Carbon & Silicon is very high. Why?
Answer:
This is due to the tendency of these elements to form giant molecules.

Question 49.
Why is boron metalloid?
Answer:
Boron resembles both metals & non-metals therefore it is metalloid.

Question 50.
Why does boron resemble Si?
Answer:
Both have a similar charge to radius ratio, i.e. similar polarizing power.

The p-Block Elements Important Extra Questions Short Answer Type

Question 1.
Although boric acid B(OH)₃ contains three hydroxyl groups, yet it behaves as a monobasic acid. Explain.
Answer:

Hydrated species
B(OH)₃ is not a protonic acid.

It behaves as a Lewis acid because it abstracts a pair of electrons from hydroxyl ion.

Question 2.
SiCl₄ forms [SiCl₆]^2- while CCl₄ does not form [CCl₆]^2- Explain.
Answer:
Carbon does not have d-orbitals and hence C.Cl₄ does not combine with Cl^– ions to give [CCl₆]^2-, On the other hand, silicon has vacant 3d-orbitals and thus can expand its covalency from 4 to 6. Therefore SiCl₄ combines with CL ions to form [SiCl₆]^2-

SiCl₄ + 2Cl^– → [SiCl₆]^2-

Question 3.
Why does not silicon form an analogue of graphite?
Or
Why does elemental silicon not form a graphite-like structure as carbon does? Explain.
Answer:
In graphite, C is sp² hybridised and each C is linked to three other C atoms forming hexagonal rings. Thus graphite has a two-dimensional sheet-like structure.

Silicon, on the other hand, does not form an analogue of C because of the following two reasons:

1. Silicon has a much lesser tendency for catenation than C as Si-Si bonds are much weaker than C-C bonds.
2. Silicon because of its larger size than C undergoes sp³ hybridisation.

Question 4.
Why carbon forms covalent compounds whereas lead forms ionic compounds?
Answer:
Carbon cannot lose electrons to form C₄ because the sum of four ionisation enthalpies is very high. It cannot gain four electrons to form C₄ because energetically it is not favourable. Hence C forms only covalent compounds. Down the group 14, ionisation enthalpies decrease, Pb being the last element has so low I.E. that it can lose electrons to form ionic compounds.

Question 5.
How is borax prepared from
(i) Colemanite ore
Answer:
Borax is also called sodium tetraborate decahydrate (Na₂B₄O₇.10H₂O). It can be prepared as follows:

From colemanite: Powdered mineral is boiled with sodium carbonate solution and filtered. The filtrate is concentrated and then cooled when crystals of borax.
Ca₂B₆O₁₁ + 2Na₂CO₃ → Na₂B₄O₇ + 2NaBO₂ + 2CaCO₃.

The mother-liquor which contains sodium meta-borate is treated with a current of C02, to convert it into borax which separates out.
4NaBO₂ + CO₂ → Na₂B₄O₇ + Na₂CO₃

(ii) Tincal.
Answer:
From Tincal: Tincal obtained from dried up lakes is boiled with water. The solution is filtered to get rid of insoluble impurities of clay, sand etc. The filtrate is concentrated to get the crystals of borax.

(iii) Boric acid?
Answer:
From boric acid: Boric acid is neutralised with sodium carbonate and the resulting solution is concentrated and cooled to get the crystals of borax Na₂B₄O₇.10H₂O.
4H₃BO₃ + Na₂CO₃ → Na₂B₄O₇ + 6H₂O + CO₂ ↑

Question 6.
Mention three important uses of borax
Answer:
It is used:

• As a flux soldering and welding in industry.
• In the manufacture of borosilicate glass (or pyrex glass).
• In making enamels and glazes.
• In stiffening of candle wicks.
• In softening of water.
• In a qualitative analysis for borax bead test in the laboratory.

Question 7.
What happens when a borax solution is acidified? Write a balanced equation for the reaction.
Answer:
When acidified aqueous solution of borax (Na₂B₄O₇) is heated, boric acid is formed.

Question 8.
Mention important uses of boric acid.
Answer:
Boric acid is used:

• In the manufacture of enamels and glazes for pottery.
• In making heat-resisting and shock resisting glass called boro glass (or pyrex glass).
• As a mild antiseptic for washing eyes

Question 9.
Mention some important properties of carbon monoxide.
Answer:

1. It is a colourless, odourless gas, slightly soluble in water.
2. It is highly poisonous. It combines with haemoglobin in the red blood cells to form carboxy-haemoglobin which cannot absorb oxygen and thus the supply of oxygen to the body is reduced.
3. It burns with a pale blue flame forming CO₂.
   2CO + O₂ → 2CO₂.
4. It is a reducing agent. It reduces some metal oxides into metals.
   Fe₂O₃ + 3CO → 2Fe + 3CO₂.
5. It combines With transition metals like iron, cobalt, nickel to form their carbonyl compounds.
   

Question 10.
What are halides of carbon? Give few examples.
Answer:
Carbon combines with halogens to form both simple and mixed tetrahalides. In the case of simple halides, all the four expected tetrahalides (e.g. CF₄, CCl₄, CBr₄ and Cl₄) are known to exist. The stability of the simple tetrahalides decreases with the increasing atomic mass of the halogen.
(CF₄ > CCl₄ > CBr₄ > Cl₄)
Amongst the mixed halides the better-known compounds are (CFCl₃, CF₂Cl₂ and C Cl₃Br).

Question 11.
What is allotropy? Give examples of allotropes.
Answer:
Two or more forms of the same element in the same physical state which differ in their physical properties but have the same chemical properties are called allotropic forms or (allotropes) and the phenomenon is called allotropy.

Carbon, phosphorus and sulphur are some elements that exhibit allotropy.

1. Diamond and graphite are allotropic forms of carbon.
2. Red phosphorus and white .phosphorus are allotropes of phosphorus.
3. Rhombic sulphur, monoclinic sulphur and plastic sulphur are allotropic forms of sulphur.

Question 12.
Give the uses of different allotropic forms of carbon.
Answer:

Forms of carbon	Uses
Diamond	Gemstone, cutting, drilling, grinding, polishing, industry.
Graphite	Reducing agent, refractories, pencils, high-temperature crucibles, electrode making, the moderator in nuclear reactors, high strength composite materials/
Activated carbon	Rubber industry, pigments in ink, paints and plastics.
Coke	Fuel, strut manufacture.
Charcoal	Fuel, reducing agent.

Question 13.
Give reasons:
(i) Graphite is used as a lubricant.
Answer:
Graphite has sp3 hybridized carbon with a layer structure. Due to wide separation & weak interlayer bonds, the two adjacent layers can easily slide over each other. This makes graphite act as a lubricant.

(ii) Cone. HN03 can be transported in an aluminium container.
Answer:
Aluminium on coming in contact with the cone; HNO₃ becomes passive due to the coating of aluminium oxide & thus the cone. HNO₃ can be transported in an aluminium container.

(iii) Aluminium utensils should not be kept in water overnight.
Answer:
Aluminium utensils should not be kept in water overnight because aluminium is readily corroded by water.

Question 14.
Write balanced equations for the reaction of elemental boron with elemental chlorine, oxygen & nitrogen at a high temperature.
Answer:

Question 15.
Why are boron halides & diborane referred to as “Electron deficient compounds”?
Answer:
Boron in its halides has only six electrons in its valence shell, therefore it is short by two electrons to complete its octet. As a result, a molecule of boron halide can accept a pair of electrons from any electron-rich compound that is why boron halides are called electron-deficient compounds. In diborane, the total no. of valence electrons is not sufficient to completely fill the. available orbitals. This gives an electron-deficient character to diborane.

Question 16.
Give reasons:
(i) A mix. of dil. NaOH & aluminium pieces are used to open drains.
Answer:
Aluminium dissolves in dil. NaOH with the evolution of H₂. This H₂ helps to open the drain.

(ii) Diamond is used as an abrasive.
Answer:
Diamond is the hardest substance known & thus used as abrasive & for cutting glass.

(iii) Aluminium wires are used to make transmission cables.
Answer:
Aluminium is cheaply available on a weight to weight basis, the electrical conductivity of aluminium is twice that of copper. Hence Aluminium wires are used to make transmission cables.

Question 17.
Write balanced equations:
(i) B₂H₆ + NH₃ →?
Answer:

(ii) NaH+ B₂H₆ →?
Answer:

(iii) BF₃ + LiH →?
Answer:

Question 18.
Describe briefly three different methods of obtaining elemental boron.
Answer:

1. By reduction of oxides: The reduction of oxides is carried out by electropositive metals like Mg.
   B₂O₃(s) + 3Mg(s) → 2B(s) + 3MgO(s)
2. By reduction of halides: The reduction of the volatile boron halides is carried but by dihydrogen at high temperatures
   
3. By the thermal decomposition of boron hydride
   

Question 19.
All the elements of group 13 except thallium show a +3 oxidation state while it shows a +1 oxidation state. Give reasons.
Answer:
The valence shell of group 13-elements have two electrons in s-subshell & 1-electron in p-subshell Therefore, they are expected to show a +3 oxidation state. In thallium, the electrons of the s-subshell do not take part in bond formation due to the inert pair effect & only one electron of the p-orbital participate in bond formation, thus it shows +1 oxidation state only.

Question 20.
Why carbon does not form ionic compounds?
Answer:
The electronic configuration of carbon atom is 1s², 2s², 2px’, 2py’ & has four valence electrons. In order to form ionic compounds, it has to either lose four electrons or gain four electrons. Since very high energies are involved in doing so. Thus carbon does not form ionic compounds. It completes its octet as a result of electron sharing & forms covalent compounds.

Question 21.
Gallium has higher ionization enthalpy than Aluminium. Explain.
Answer:
Gallium has higher ionization energy than Aluminium because of the higher effective nuclear charge. This is due to additional 3d electrons which do not screen the nuclear charge effectively so that the outer electrons are more strongly held.

Question 22.
Boron forms no compounds in a unipositive state but thallium in a unipositive state is quite unstable. Why?
Answer:
Boron has electronic configuration 2s²2p¹ & therefore forms compounds in a trivalent state. However, thallium prefers to form compounds in the +1 oxidation state rather than in the +3 oxidation state as suggested by its group number. This is due to the inert pair effect. According to this effect, the 6s² electrons in the case of heavy metals preferably do not take part in bonding.

Question 23.
What are silicones? How are they manufactured?
Answer:
Silicones are rubber-like polymers having Si – O-Si linkage & general formula R₂SiO. They are manufactured by hydrolysis of Chloro silicones,

Where R is Me/Ph groups

Question 24.
Mention any two dissimilarities of boron with other elements of group-13.
Answer:

1. All the compounds of boron are covalent in nature because of the non-existence of the B³⁺ ion. It is because of its high charge density.
2. The maximum covalence shown by Boron is 4 while other members of this group show a covalence of six or more.

Question 25.
Why boron trihalides form tetrahedral complexes?
Answer:
A Boron trihalide is an electron-deficient compound having six electrons in its outermost shell. Therefore it has a tendency to form a BX₃L type of complex after accepting an electron pair in the un-hybridized vacant p-orbital from a ligand (L) molecule. Because of this reason the shape of BX₃ changes from planar Triangular to tetrahedral in the BX₃L complex.

Question 26.
Discuss the pattern of variation in the oxidation states of Al & Tl.
Answer:
Aluminium shows a +3 oxidation state only while thallium, the last element of group 3 shows +3 oxidation state & +1 oxidation states. Tl⁺¹ is more stable than Tl³⁺ as is evident by redox potential data
Tl³⁺(aq) + 2e^– → Tl⁺ (aq) E° = +1.25 V

It happens because of the decrease in bond energy with size down the group (i.e. from Al → Tl). Thus the energy required to break the pair of ns² electrons is not compensated by the energy released during the formation of two additional bonds. Thus +1 oxidation state is more stable in thallium.

Question 27.
C—C bond length is shorter in graphite than the C-C bond length of diamond. Explain.
Answer:
Graphite has sp² hybridization & the C-C bond involves sp²—sp² hybridized carbon. Diamond has sp³ hybridization & the C—C bond involves sp³-sp³ hybridization. Furthermore, the more the S character in a hybridized atom, the smaller is the size of the hybridized orbital, resulting in more overlapping which leads to bond length.

Question 28.
[SiF₆]^2- is known whereas [SiCl₆]^2- not. Give reasons.
Answer:

1. Six large chlorine atoms cannot be accommodated around silicon atom due to the limitation of their size.
2. Interaction between lone pair of chlorine atoms.& silicon atom is not very strong.

Question 29.
SiCl. forms [SiCl]^2- while CClL does not form [CCl]^2-. Explain
Answer:
Carbon does not have d-orbitals, hence CCl₄ does not combine with Cl ions to form [CCl₆]^2-. On the other hand, silicon has vacant 3-d orbitals & thus SiCl₄ combines with Cl^– ions to form
[Sicy2-
SiCl₄ + 2Cl^– → [SiCl₆]^2-

In other words, carbon shows a fixed covalency of 4 but silicon exhibits varying covalency from 4 to 6.

Question 30.
Borazine is more reactive than benzene. Why?
Answer:
Both Borazine & Benzene are isoelectronic. In benzene C = C bonds are non-polar while N=B bonds in borazine are polar in nature due to the presence of a co-ordinate bond between N & B atoms. As a result, addiction is quite frequent in borazine while it is less in benzene because of delocalisation of π-electron charge.

The p-Block Elements Important Extra Questions Long Answer Type

Question 1.
(i) What are the different oxidation states exhibited by the group 14 elements? Discuss the stability of their oxidation states.
Answer:
The group 14 elements have four electrons in the outermost shell. The common oxidation states exhibited by these elements are +4 and +2. Since the sum of the first four ionisation enthalpies is very high, compounds in the +4 oxidation state are generally covalent in nature. In heavier members such as Ge, Sn and Pb, the tendency to show +2 oxidation state increases. It is due to the inability of ns2 electrons of the valence shell to participate in bonding.

The relative stabilities of these two oxidation states vary down the group. C and Si mostly show a +4 oxidation state. Ge forms stable compounds in the +4 state and only a few compounds in the +2 state. Sn forms compounds in both oxidation states (Sn in +2 state is a reducing agent).

Lead compounds in the +2 state are stable and in the +4 state are strong oxidising agents. In the tetravalent state, the number of electrons around the central atom in a molecule (e.g., carbon in CCl₄) is eight. Being electron precise molecules, they are normally not expected to act as an electron acceptor or electron donor.

Although carbon cannot exceed its covalence of more than 4, other elements of the group can do so. It is because of the presence of d-orbital in them. Due to this, their halides undergo hydrolysis and have a tendency to form complexes by accepting electron pairs from donor species. For example, the species like SiF₅^–, SiF₆^–, [GeCl₆]^2-, [Sn(OH)₆]^2- exist where the hybridisation of the central atom is sp³d².

(ii) What type of oxides are formed by group 14 elements? Which of them are acidic, neutral or basic?
Answer:
All members when heated in oxygen form oxides. There are -mainly two types of oxides, i. e., monoxide and dioxide of formula MO and MO₂ respectively. SiO only exists at high temperature. Oxides in the higher oxidation state of elements are generally more acidic than those in the lower oxidation state. The dioxides-CO₂, SiO₂ and GeO₂ are acidic, whereas SnO₂ and PbO₂ are amphoteric in nature. Among monoxides, CO is neutral, GeO is distinctly acidic whereas SnO and PbO are amphoteric.

Question 2.
(a) [SiF₆]^2- is known whereas [SiCl₆]^2- is not known. Give reasons
Answer:
(i) [SiF₆]^2- is known whereas [SiCl₆]^2- does not exist.
The main reasons are (i) six large chlorine atoms cannot be accommodated around silicon atom due to the limitation of their size.
(ii) Interactions between lone pairs of a chlorine atom and silicon atom are not very strong.

(b) Select the member (s) of group 14 that
(i) forms the most acidic oxide
Answer:
The most acidic dioxide is formed by carbon (CO₂).

(ii) is commonly found in the +2 oxidation state
Answer:
Lead is mostly found in the +2 oxidation state in its compounds.

(iii) used as a semi-conductor.
Answer:
Silicon and germanium are used as semiconductors.

(c) Explain why a diamond that is covalent has a high melting point?
Answer:
Though diamond has covalent bonding in it, yet it has a high melting point, because a diamond has a 3-dimensional network involving strong C—C bond, which are very difficult to break and in turn, it has a high melting point.

(d) Discuss the reaction of silica with
(i) NaOH
Answer:
SiO₂ reacts with HF as follows:
SiO₂ + 2NaOH → Na₂SiO₃ + H₂O

(ii) HE
Answer:
SiO₂ reacts with HF as follows:
SiO₂ + 4HF → SiF₄ + 2H₂O.

Question 3.
(a) Carbon exhibits catenation, whereas silicon does not. Explain.
Answer:
Carbon shows catenation because of its smaller size, high bond energy of C – C bond, the possibility of sp, sp², sp³ hybridisation and formation of multiple bonds C-C (1σ), C = C (1σ + 1π),- C = C (1σ + 2π). On the other hand, silicon shows only limited catenation because of its large atomic radius, low bond energy of Si-Si bond and absence of multiple bonds between Si atoms.

(b) How does boron differ from aluminium.
Answer:
Difference between boron and aluminium:

1. Boron is a non-metal but aluminium is a metal.
2. Boron is a semi-conductor while aluminium is a good conductor of electricity.
3. Boron forms a number of hydrides called boranes, but Al forms a polymeric hydride.
4. Halides of boron (except BF₃) are readily hydrolysed by water whereas halides of A1 are only partially hydrolysed by water.
5. B₂O₃ is acidic, but Al₂O₃ is amphoteric.
6. Boron hydroxide B(OH)₃ is acidic, but Al(OH)₃ is amphoteric.

(c) Write the similarities between boron and silicon.
Answer:
Similarities between boron and silicon:

1. Both are non-metals.
2. Both are semi-conductors
3. Boron and silicon form a number of covalent hydrides which have similar properties. For example, they spontaneously catch fire on exposure to air and are readily hydrolysed by water.
4. The halides of boron and silicon are readily hydrolysed by water.
5. Boron trioxide (B₂O₃) and silicon dioxide (SiO₂) are acidic in nature. These dissolve in alkali solution forming borates and silicates.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 8 Redox Reactions. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 8 Important Extra Questions Redox Reactions

Redox Reactions Important Extra Questions Very Short Answer Type

Question 1.
What are redox reactions? Give an example.
Answer:
Redox reaction is a reaction in which oxidation and reduction take place simultaneously, e.g.
Zn (s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Question 2.
Define oxidation and reduction in terms of electrons.
Answer:
Oxidation involves loss and reduction involves the gain of electrons.

Question 3.
Define an oxidizing agent. Name the best oxidizing agent.
Answer:
The oxidizing agent is a substance that can gain electrons easily. F₂ is the best oxidizing agent.

Question 4.
What is meant by reducing agent? Name the best reducing agent.
Answer:
The reducing agent is a substance that can lose electrons easily. Li is the best reducing agent.

Question 5.
In the reaction MnO₂ + 4HCl → MnCl₂ + H₂O which species is oxidized?
Answer:
HCl is oxidized to Cl₂

Question 6.
What is the oxidation state of Ni in Ni (CO)₄?
Answer:
Zero.

Question 7.
What is a redox couple?
Answer:
The redox couple consists of the oxidized and reduced form of the same substance taking part in an oxidation or reduction half-reaction, for example.
Zn²⁺(aq) / Zn, Cl₂ / Cl^–(aq) etc.

Question 8.
Define oxidation and reduction in the term of oxidation numbers.
Answer:
Oxidation involves an increase in oxidation number while reduction involves a decrease in oxidation number.
Sn²⁺ + 2Hg²⁺ → Sn⁴⁺ + Hg²⁺

Here Sn²⁺ gets oxidised while Hg²⁺ gets reduced.

Question 9.
What is the sum of oxidation numbers of all atoms in HIO₄4?
Answer:
Zero.

Question 10.
What is the oxidation number of N in (NH₄)₂ SO₂?
Answer:

The oxidation number of N is (NH₄)₂ SO₄ is – 3.

Question 11.
What is the oxidation number of O in
(i) OF₂
(ii) O₂F₂
Answer:
The oxidation number of oxygen in OF₂ is +2 whereas the oxidation number of oxygen in O₂F₂ is +1

Question 12.
At what concentration of M+(aq) will its electrode potential become equal to its standard electrode potential?
Answer:
At the concentration of IMol / L.

Question 13.
Select the oxidant in the following reaction.
H₂O₂ + O₃ → H₂O + 2O₂
Answer:
O₃ (zero to – 2)

Question 14.
What is the electrode potential of a standard hydrogen electrode?
Answer:
Zero.

Question 15.
Set up an electrochemical cell for the redox reaction: Ni²⁺(aq) + Fe(s) → Ni(s) + Fe²⁺(aq)
Answer:
Fe(s) / Fe²⁺(aq) || Ni²⁺(aq) / Ni(s)

Question 16.
What is the oxidation numbers of
(i) C in CH₂O
Answer:
Zero

(ii) pt in [pt(C₂H₄)Cl₃]
Answer:
2

Question 17.
What is the oxidation number of Mn in KMnO₄?
Answer:
The oxidation number of Mn in KMnO₄ is
KMnO₄ = 1 + x + 4 (- 2) = 0
x = + 7

Question 18.
What happens to the oxidation number of an element in oxidation?
Answer:
It increases.

Question 19.
Which reaction occurs at the cathode in a galvanic cell?
Answer:
Reduction.

Question 20.
Name one compound in which the oxidation number of Cl is +4.
Answer:
ClO₂.

Question 21.
If the reduction potential of an electrode is 1.28 v. What will be its oxidation potential?
Answer:
– 1.28 v.

Question 22.
Can we store copper sulfate in an iron vessel?
Answer:
No, because iron is more reactive than copper and thus holes will be developed in an iron vessel.
Cu²⁺ (aq) + Fe(s) → Fe²⁺(aq) + Cu(s)

Question 23.
The E° of Cu²⁺ / Cu is + 0.34 v. What does it signify?
Answer:
Copper lies below hydrogen in the activity series.

Question 24.
Calculate the oxidation number of underlined elements in the followings:
Na₂B₄O₇,O₃O₄
Answer:
Na₂B₄O₇ = 2 + 4x – 14 = 0
4x = 12
x = 3 .
O₃O₄ = x – 8 – 0
x = + 8

Question 25.
What is the oxidation number of alkali metals in its compounds?
Answer:
+1.

Question 26.
Determine the change in the oxidation number of S in H2S and S02 in the following reaction:
2H₂S(g) + SO₂(g) → 3S(s) + 2H₂S(g)
Answer:
The oxidation number of S changes from – 2 in H₂S and +4 in SO₂ to zero in elemental sulfur.

Question 27.
Can we use KCl as an electrolyte in the following cell?
Cu(s) | Cu²⁺(aq) || Ag⁺(aq)| Ag(s)
Answer:
KCl cannot be used as an electrolyte as a salt bridge because Cl^– ions will combine with Ag⁺ ions to form a precipitate of AgCl.

Question 28.
Define the EMF of the cell.
Answer:
EMF of the cell is defined as the difference in the electrode potential of the two half cells when the cell is not sending current through the circuit.

Question 29.
A solution of Na₂SO₄ was electrolyzed by using some inert molecules. What are the products at the electrodes?
Answer:
O₂ & H₂.

Question 30.
The reduction potentials are:
Cl₂ + 2e^– → 2Cl – E° = 1.36V
F₂ + 2e^– → 2F- E° = 2.87 V
Which is the better oxidising agent?
Answer:
F.

Question 31.
What is the relationship between standard oxidation potential and standard reduction potential?
Answer:
Both are equal in magnitude but opposite in sign.

Question 32.
Determine the oxidation number of nitrogen in HNO₃.
Answer:
HNO₃
1 + x +. 3(- 2) = 0
x = + 5

Question 33.
What is the oxidation number erf Fe in Na₄[Fe(CN)₆]?
Answer:

x – 1 × 6 = – 4 ,
x = +2

Question 34.
What is the maximum and minimum oxidation number of Nitrogen?
Answer:
The highest oxidation number of N is +5 and its minimum oxidation number is -3.

Question 35.
What is the basic principle of balancing the redox reactions by the ion-electron method?
Answer:
A number of electrons lost during oxidation = number of electrons gained during reduction.

Question 36.
What is the oxidation state of S in H₂SO₄ & H₂SO₃?
Answer:
H₂SO₄ = 2 ( + 1) + x + 4 (- 2) = 0
x = 8 – 2 = +6 ‘ ,
H₂SO₃ = 2 ( + 1) + x + 3 ( – 2) = 0
x = +4

Question 37.
What is the oxidation number of Xe in Ba₂XeO₂?
Answer:
Ba₂XeO₂ = 2 × 2 + x + 2( – 2) = 0
4 + x – 4 = 0
x = 0

Question 38.
Define half cell.
Answer:
The combination of an electrode and the solution in which it is dipped is called a half-cell.

Question 39.
Indicate the oxidizing and reducing agent in the following reaction:
2Cu²⁺ + 4I^– ⇌ 2CuI + I₂
Answer:
Cu²⁺ is an oxidizing agent and I^– is the reducing agent.

Question 40.
Write correctly the balanced half-reaction and the overall reaction for the following skeletal equation.
Fe(OH)₂ + H₂OZ ⇌ Fe(OH)₃ + H₂O (in basic medium)
Answer:
[Fe(OH)₂ + OH^– ⇌ Fe(OH)₃ + e^–] × 2
[H₂O₂ + 2e^– ⇌ 2OH^–]
2Fe(OH)₂ + H₂O₂ ⇌ 2Fe(OH)₃

Question 41.
Calculate the oxidation number of P in MgP₂O₇
Answer:
MgP₂O₇ = 2 × 2 + 2x + 7(-2) = 0
2x = 10
x = + 5

Question 42.
Arrange the following in order of increasing oxidation number of Mn:
MnCl₂, MnO₂, Mn(OH)₃, KMnO₄
Answer:

Question 43.
Arrange the followings in order of decreasing the oxidation number of oxygen:
HXO₄, HXO₃, HXO₂, HXO
Answer:

Question 44.
Arrange the following in order of increasing oxidation number of iodine:
I₂, HI, HIO₄, ICl
Answer:

Question 45.
How do you arrive at the oxidation number of Cr in Cr₂O₇^2-?
Answer:
(Cr₂O₇ )² = 2x – 14 = – 2
x = + 6

Question 46.
What is the charge on I mole of electrons?
Answer:
96500 coulombs.

Question 47.
Set up an electrochemical cell for the redox reaction.
Ni²⁺(aq) + Fe(s) ⇌ Ni(s) + Fe²⁺(aq)
Answer:
Fe(s) / Fe²⁺(aq) || Ni²⁺(aq) / Ni(s)

Question 48.
Define oxidation number?
Answer:
The oxidation number of an element may be defined as the charge which an atom of the element has in its ions. It is also known as the oxidation state.

Question 49.
Define electrode potential.
Answer:
The tendency, of an electrode to lose or gain electrons is called electrode potential.

Question 50.
Identify the strongest and weakest reducing agent from the metal.
Zn, Cu, Ag, Na, Sn
Answer:
Strongest reducing agent = Na Weakest reducing agent = Ag

Redox Reactions Important Extra Questions Short Answer Type

Question 1.
HNO3 acts only as an oxidant whereas HNOz acts both as an oxidant and reductant. Why?
Answer:
Ox. No. of N in HNO₃ = +5
Ox No. of N in HNO₂ = +3
Maximum oxidation numbers which N can show is = + 5
(∴ It has only 5 valance electrons 2S²2P³)

The Ox. No. of N in HNO₃ is maximum and it can only decrease. Therefore HNO₃ can act only as an oxidant. Minimum Ox. No. of N is -3.

Thus HNO₂ in which Ox. No. of N is +3 Can decrease as well as increase. Thus HNO₂ can act as an oxidant as well as a reductant.

Question 2.
Balance the following equation by the ion-electron method.
Zn(s) + NO₃^– → Zn²⁺(aq) + NH⁺ (aq) + H₂O(l) (In acid solution)
Answer:
Oxidation half reaction

Multiply (1) equation by 4 to equalise the no. of electron in both. Add both half reaction
4 Zn(s) → Zn²⁺(aq) + 8e^–
Zn(s) + NO₃^– (aq) + 10H⁺ (aq) → 4Zn²⁺ (aq) +NH⁺(aq)+ 3H₂2O(1)
– 1 + 10 = + 8 + 1
+ 9 = + 9

Question 3.
Balance the following equation in acidic medium by oxidation number method.

Answer:

Question 4.
Indicate the oxidising and reducing agent in the following reactions:
Answer:
(i) 2Mg + SO₂ → 2MgO + S
Mg = Reducing agent
SO₂ = Oxidising agent

(ii)2Cu²⁺ + 41 → 2CuI + I₂
Cu²⁺ = Oxidising agent
I- = Reducing agent

(iii)SO₂ + 2H₂S → 2H₂O + 3S
SO₂ = Oxidising ageiit
H₂S = Reducing agent

(iv)Sn²⁺ + 2Hg²⁺ → Hg₂²⁺ + Sn⁴⁺
Sn²⁺ = Reducing agent
Hg²⁺ = Oxidising agent

Question 5.
Which of the following redox reaction is oxidation & which is reduction?
Answer:
(i) Zn → Zn²⁺ + 2e^–
Oxidation
(ii) Cl2 + 2e^– → 2Cl-
Reduction
(iii) Fe → Fe²⁺ + 2e^–
Oxidation
(tv) Sn⁴⁺ + 2e^– → Sn²⁺
Reduction

Question 6.
What are the minimum and maximum oxidation numbers shown by sulfur?
Answer:
The minimum oxidation number shown by S is – 2 since it can acquire 2 more electrons to achieve the nearest inert gas [Ar] configuration.

The maximum Ox. No. shown by S is +6 since it has 6 valance electrons. (3S² 3P⁴)

Question 7.
Arrange the molecule NH₃, NO₂, HN₃, NO₂ & N₂H₄ in the decreasing order of the oxidation state of nitrogen.
Answer:

Question 8.
Can the reaction.
Cr₂O₇ ^2- + H₂O ⇌ 2CrO₄^2- + 2H⁺ be regarded as redox reaction.
Answer:
Ox. No. of Cr in Cr₂O₇^2- = + 6
Since the ox. no if Cr in CrO₄^2- = + 6

There is no change in ox. no. in the reaction therefore this reaction cannot be regarded as a redox reaction.

Question 9.
Balance the oxidation—reduction reaction
FeS₂ + O₂ → Fe₂O₃ + SO₂
Answer:
Fe²⁺ → Fe³⁺ 1 ↑
S₂^-1 → 2S⁺⁴ 10 ↑
O₂ → 2O^2- 4↓ × 11
4[Fe⁺² + S₂^-1] + 11O₂ → 4Fe³⁺ + 8S⁺⁴ + 22O^-2
or
3FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

Question 10.
Write the following redox reactions using half cell reactions:
(i) Zn(s) + FeCl₂(aq) → ZnCl₂(aq) + Fe(s)
Answer:
(Zn(s) → Zn²⁺(aq) + 2e^–
Fe²⁺(aq) + 2e^– → Fe(s)

(ii) Mg(s) + Cl2(g) → MgCl₂(s)
Answer:
Mg(s) → Mg²⁺ + 2e^–
Cl₂(g) + 2e^– → 2Cl^– (aq)

(iii) Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)
Answer:
Mg(s) → Mg²⁺ + 2e^–
2H⁺(aq) + 2e^– → H₂(g)

Question 11.
How would you know whether a redox reaction is taking place in an acidic/alkaline or neutral medium?
Answer:
If H⁺ or any acid appears on either side of the chemical equation, the reactions take place in the acidic medium. If OH^– or any base appears on either side of a chemical equation, the solution is basic. If neither H⁺; OH^– nor any acid or base is present in the chemical equation, the solution is neutral.

Question 12.
Is it possible to store:
(i) Copper sulphate solution in a zinc vessel.
Answer:
we cannot place CuSO₄ solution in a zinc vessel, if the following redox reaction occur.
Zn + CuSO₄ → ZnSO₄ + Cu
or
Zn + Cu²⁺ → Zn²⁺ + Cu

By convention, the cell may be represented by
Zn / Zn²⁺ || Cu²⁺ / Cu
K°_cell = E°_cu²⁺ cu- E°_zn²⁺
Zn = 0.34 -(- 0.76) = + 1.10v.
Since EMF comes out positive, therefore CuSO₄ reacts with zinc. So CuSO₄ solution cannot be stored in a zinc vessel.

(ii) Copper sulphate solution in silver vessel.
Answer:
We cannot store CuSO₄ solution in a silver vessel if the following reaction occurs.
2Ag + Cu²⁺ → 2Ag⁺ + Cu

By convention, the cell of the above redox reaction may be represented by
Ag / Ag⁺ || Cu²⁺ / Cu and E°cell = E°_cu²⁺ , cu^–, E°_Ag⁺
Ag = 0.34 – 0.80 = – 0.56 v.

Since the EMF of the cell is -ve, therefore CuSO₄ does not react with silver or CuSO₄ solution cannot be stored in a silver vessel.

Question 14.
The electrode potential of four metallic elements (A, B, C, D) are +0.80, -0.76, +0.12 and +0.34v. respectively. Arrange them in order of decreasing electron positive character.
Answer:
Higher the electrode potential, (E°) lower is the tendency of the metal to lose electrons .and hence lower is the electropositive character. So the electrode potential increase in the order of

Question 15.
What are the maximum and minimum oxidation numbers of N?
Answer:
The highest ox. no. of N is +5 since it has five electrons in the valence shell (2S2 2P3) and its minimum o.no. is -3 since it can accept three more electrons to acquire the nearest inert gas (Ne) configuration.

Question 16.
Find out the ox. no. of Cl in HCl, HClO, ClO;, CaOCl₂ & ClO₂.
Answer:
ox. no. of Cl = -1 in HCl, +1 in HClO, + 7 in ClO₄^– 0, in CaOCl₂ and + 4 in ClO₂.

Question 17.
What will happen if a nickel spatula is used to stir a solution of copper sulfate?
Answer:
Since E° of Ni (-0.25v) is lower than that of copper (+ 0.34v), therefore nickel has a higher tendency to lose electrons & copper has a higher tendency to gain electrons. Cu will deposit over the nickel spatula.

Question 18. Justify the reaction:
2Cu₂O(S) + Cu₂S (s) → 6Cu (s) + SO₂ (g) is a redox reaction. Identify the species oxidized, reduced, which acts as an oxidant and which acts as reductant.
Answer:

In this reaction copper is reduced from + 1 oxidation state to zero oxidation State and sulfur is oxidized from – 2 to +4 state. Therefore the reaction is a redox reaction further Cu₂O helps sulfur in Cu₂S itself and Cu₂O to decrease its oxidation number hence sulfur of Cu₂S is a reducing agent.

Question 19.
Write two information about the reaction-
Mg + H₂SO₄ → MgSO₄ + H₂
Answer:

1. Magnesium reacts with sulphuric acid to form magnesium sulfate and hydrogen.
2. 24 grams of magnesium reacts with 98 gm of sulphuric acid-producing (24 + 32 + 64) = 120 gm of magnesium sulfate and 2 gm of hydrogen.

Question 20.
Calculate the oxidation number of N in H – C = N
Answer:
As nitrogen is more electronegative than carbon, therefore, each covalent bond contributes one unit negative value to the oxidation number to nitrogen. Thus o. no. of nitrogen in HCN = -3

Question 21.
I₂ & Br₂ are added to a solution containing Br- and I- ions. What reaction will occur if,
I₂ + 2e^– → 21, E° = 0.54 v
& Br₂ + 2e^– → 2Br^–, E° = + 1.09 v?
Answer:
Since E° of Br₂ is higher than that of I₂, therefore Br₂ has a higher tendency to accept electrons than I₂. Conversely I- ions have a higher tendency to lose electrons than Br^– ions. Therefore, the following reaction will occur.

In other words, I^– ions will be oxidized to I₂ while Br₂ will be reduced to Br^– ions.

Question 22.
Is it possible to store copper sulfate solution in a gold vessel?
Answer:
We can store CuSO₄ solution in a gold vessel if the following redox reaction occur
2Au + 3Cu²⁺ → 2Au³⁺ + 3Cu
The cell corresponding to the above redox reaction may be represented as:
Au / Au³⁺ || Cu²⁺ / Cu and E°cell = E°_Cu²⁺ , Cu E°_Au³⁺
Au = 0.34 – 1.50 = 1.16 v

since the EMF of the above cell reaction is – ve, therefore, CuSO₄ solution does not react with gold in other words, the CuSO₄ solution can be stored in a gold vessel.

Question 23.
When magnesium ribbon burns in air two products are formed, magnesium oxide and magnesium nitride point out the oxidizing and reducing agent.
Answer:

Question 24.
What is the basic difference between electrochemical and electrolytic cells?
Answer:
In electrochemical cells electrical energy is generated as a result of the redox reaction which occurs in an electrolytic cell, the chemical reaction takes place as a result of electrical energy supplied.

Question 25.
Why are articles made of iron coated with zinc to check their rusting?
Answer:
Coating a layer of zinc on iron articles is called galvanization articles made up of iron are coated with zinc since zinc forms a protective coating on iron. When such articles are exposed to air. Zinc is oxidized to Zn²⁺ ions in preference to iron. Therefore zinc sacrifices itself for the sake of iron. The rusting of iron is prevented.

Question 26.
Identify the oxidizing agent, reducing agent, and the substance undergoing oxidation & reduction in the following reaction.
H₂SO₄ + 2HBr → SO₂ + Br₂ + 2H₂O
Answer:

Oxidizing agent – H₂SO₄
Reducing agent – HBr
Substance oxidized – HBr
Substance reduced – H₂SO₄

Question 27.
Consider the following galvanic cell.
Cd / Cd²⁺ (1M) || H⁺(1M) / H₂(g / atm) pt
(i) Write the overall cell reaction
Answer:
The anodic reaction is
Cd(s) → Cd²⁺(aq) + 2e^–

The Cathodic reaction is
2H⁺(aq) + 2e^– → H₂(g)

The overall reaction is
Cd(s) + 2H⁺(aq) → Cd²⁺(aq) + H₂(g)

(ii) What do the double vertical lines denote?
Answer:
The double vertical lines denote the salt bridge which connects the oxidation & reduction half cells.

Question 28.
Write the following redox reaction in the oxidation & reduction half-reaction.
(a) 2K(s) + Cl₂(g) → 2KCl(s)
Answer:
K(s) → K⁺(aq) + e^– (Oxidation)
Cl₂(g) + 2e^– → 2Cl^– (Reduction)

(b) 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)
Answer:
Al(s) → Al³⁺ (aq) + 3e^– (Oxidation)
Cu²⁺ + 2e^– → Cu(s) (Reduction)

Question 29.
Find out the oxidation number of iron in
[Fe(H₂O)₅ (NO)+] SO₄^—
Answer:
[Fe(H₂O)₅(NO)⁺]²⁺
x + 0 + 1 = 2
x = 2 – 1
= +1

Question 30.
Calculate the oxidation number of all the atoms in- CrO₄^2-.
Answer:
If x is the ox. no. of Cr in CrO₄^2- ion then
x + 4 × (- 2) x = – 2
x = 8 – 2 = 6
x = 6
For O → 6 + 4(x) = – 2
4x = – 8
x = \(\frac{-8}{4}\) = – 2

Redox Reactions Important Extra Questions Long Answer Type

Question 1.
Give the rules on the basis of which oxidation numbers are assigned to various elements.
Answer:
Various atoms are assigned oxidation number on the basis of the following rules:

1. An element in the free state has an oxidation number equal to zero, e.g. H₂, He, K, Ag all have zero ox. no.
2. In a binary compound of a metal and a non-metal, the oxidation number of metal is positive while that of non-metal is negative. In NaCl the ox. no. of sodium +1 and ox. n. of chlorine is  -1.
3. In a covalent compound, the atom with higher electronegativity has a negative oxidation number while another atom has a positive oxidation number.
4. The oxidation number of the radical or ions is equal to the electrical charge on it. for e.g. the ox. no. of Na+ is +1.
5. In neutral molecules, the algebraic sum of the oxidation number of all the atoms is zero.

Question 2.
Starting with the correctly balanced half-reaction, write the overall net ionic equation for the following change:
Chloride ion is oxidised to Cl₂ by MnO₄^– (in acid solution)
Answer:
The skeletal equation is
MnO₄^–+ H⁺ + Cl^–(aq) → Mn²⁺ + Cl₂(g) + H₂O (l)

Ox. no. of Mn change from +7 in MnO₄^– to +2
Whereas ox. no. of chlorine change from -1 in Cl^– ions to 0 in Cl₂.

Question 3.
Write the method used for balancing redox reaction by oxidation number method.
Answer:
The following steps are used for balancing the reactions by this methods:

1. Writing the skeletal equation for all the reactants and products of the reaction.
2. Assignment of the oxidation number of all atoms in each compound in the skeletal equation. Identify the atoms undergoing a change in their oxidation number.
3. Calculating the increase or decrease in oxidation number per atom and then for the whole molecule in which it occurs. If these are not equal then multiplying by suitable coefficients such that these become equal.
4. Now balancing the chemical reaction with respect to all atoms except H & O.
5. Finally balancing with respect to H & O atom for balancing oxygen atoms add H20 molecules to the side deficient in it.

Redox Reactions Important Extra Questions Numerical Problems

Question 1.
Determine the oxidation number of O in the following: OF2, Na₂O₂ & CH₃COOH
Answer:
(i) OF₂
Let the ox. no. of O = x
The ox. no. of each F = — 1
x – 2 = 0
x = +2

(ii) Na₂O₂
Let the o. no. of O = x
ox. no. of each Na = + 1
2 + 2x = 0
2x = – 2
x = – 1

(iii) CH₃COOH
Let the ox. no. of O = x
The ox. no. of each carbon atom = — 1
The ox. no. of hydrogen = +1
– 2 + 4 + 2x = 0
2x + 2 = 0
x = – 1

Question 2.
Determine the volume 6f M/8 KMnO₄ solution required to react completely with 25.0 cm3 of M/4 FeSO₄ solution in an acidic medium.
Answer:
The balanced ionic equation for the reaction is
MnO₄^– + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
from the balanced equation, it is evident that-
1 mole of KMnO₄ = 5 moles of FeSO₄

Applying the molarity equation to the balanced redox equation.

Thus the volume of M/8 KMnO₄ solution required = 10.0 ml.

Question 3.
How many grams of K₂Cr₂O₇ is required to oxidize Fe²⁺ present in 15.2 gm of FeSO₄ to Fe³⁺ if the reaction is carried out in an acidic medium.
Answer:
The balanced chemical equation for the redox reaction is
K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O
from the balanced equation, it is clear that 6 moles of FeSO₄ = 1

a mole of K₂Cr₂O₇ or 6 × 152 gm of FeSO₄ are oxidized by
K₂Cr₂O₇ = 294 gm
or
15.2 gm of FeSO₄ are oxidized by K₂Cr₂O₇

Question 4.
15.0 cm3 of 0.12 M KMnO₄ solution are required to oxidize 20 ml of FeSO₄ solution in an acidic medium. What is the concentration of FeSO₄ solution?
Answer:
The balanced chemical equation for the redox reaction is
2KMnO₄ + 10FeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8 H₂O

Applying molarity equation to the above redox reaction

Question 5.
16.6 gm of pure KI was dissolved in water and the solution was made up to one liter, cm³ of this solution was acidified with 20 cm³ of 2 MHCl the resulting solution required 10 cm³ of decinormal KIO₃ for complete oxidation of I- ions to ICl. Find out the value of v.
Answer:
The chemical equation for the redox reaction is
IO- + 2I^– + 6HCl → 3ICl + 3Cl^– + 3H₂O

Molarity of KI solution = \(\frac{16.6}{166}\) = 0.1 m

Applying molarity equation
\(\frac{0.1 \times v}{2}\) (KI) = \(\frac{10 \times 0.1}{1}\)(KIO3)
v = 20 cm³

Question 6.
Calculate the cone of hypo (Na₂S₂O₃ 5H₂O) solution in g dm-3 if 10.0 of this solution decolorized 15 ml of M/40 iodine solution.
Answer:
The balanced equation for the redox reaction is
2S₂O₃^2- + I₂ → 2I^– + S₄O₆^2-
from the balanced equation, it is evident that
2 moles of Na₂S₂O₃ = 1 mole of I₂

Applying the molarity equation we have,

Thus, the molarity of the hypo solution = 3/40 M
mol. mass of Na₂S₂O₃.5H₂O = 248 g mol^-1
cone, of Na₂S₂O₃.5H₂O = \(\frac{248 \times 3}{40}\)
= 18.6 gdm^-3

Question 7.
How many millimoles of potassium dichromate is required to oxidize 24 cm³ of 0.5 M mohr’s salt solution in an acidic medium.
Answer:
No. of millimoles of K₂Cr₂O₇ present in 24 cm³ of 0.5 m solution = 24 × 0.5 = 12
The balanced chemical equation for the redox reaction is
K₂Cr₂O₇ + 6(NH₄)₂SO₄.FeSO₄.6H₂O + 7H₂SO₄ → K₂SO₄ + 6(NH₄)₂SO₄ + 3Fe₂(SO₄)₃ + Cr₂(SO₄)₃ + 43H₂O from the balanced equations.

6 moles mohr’s salt are oxidised by K₂Cr₂O₇ = 1 moles
∴ 12 millimoles of mohr’s salt will be oxidised by
K₂Cr₂O₇ = \(\frac{1}{6}\) × 12 = 2 millimoies.

Question 8.
2.48 gm of Na₂S₂O₃.xH₂O was dissolved per liter of the solution. 20 cm³ of this solution required 10 cm³ of 0.01 M iodine solution. Find out the value of x?
Answer:
The balanced equation for the redox reaction is
2Na₂S₂O₃ + I2 → Na₂S₄O₆ + 2NaI
Let the molarity of Na₂S₂O₃ .xH₂O solution = M₁.

Applying molarity equation to the above redox reaction, we have
\(\frac{\mathrm{M}_{1} \times 20}{2}\)(Na2S2O3) = \(\frac{10 \times .01}{1}\)(I2)
∴ M₁ = 0.01 M

mol wt. of Na₂S₂O₃.xH₂O
= 2 × 23 + 2 × 32 + 3 × 16 + x × 18
= 158 + 18x

Amount of Na₂S₂O₃.xH₂O present per litre
= (158 + 18x) × 0.01 g

But the actual amount dissolved = 2.48 g. equating these values, we have
(158 + 18x) × 0.01 = 2.48
or
x = 5.

Question 9.
The half cell reactions with their oxidation potentials are:
Pb(s) → Pb²⁺(aq) + 2e^–, E°_oxi = + 0.13 v.
Ag(s) → Ag⁺(aq) + e^–, E°_oxi = – 0.80 v.
Write the cell reaction and calculate its EMF.
Answer:
The equations are as
Pb²⁺(aq) + 2e^– → Pb(s), E°_oxi = – 0.13 v
Ag⁺(aq) + e^– → Ag(s), (E° = + 0.80 v) …(2)

To obtain the equation for the cell reaction, multiply equation (2) by 2 and subtract equation (1) from it we get
Pb(s) + 2Ag⁺(aq) → Pb²⁺(aq) + 2Ag(s)
E°_cell = + 0.80 – ( – 0.13) = + 0.93 v

Question 10.
Predict whether zinc or silver reacts with 1 M H₂SO₄ to give out hydrogen or not. Given that the standard potential of zinc & silver are – 0.76 v & + 0.80 v respectively.
Answer:
(a) To predict the reaction of zinc with H₂SO₄ If Zn reacts, the following reactions should take place.
Zn + H₂SO₄ → ZnSO₄ + H₂
i.e. Zn + 2H⁺ → Zn²⁺ + H₂

By conventions, the cell will be represented as-
Zn / Zn²⁺ ∥ H⁺ / H₂

standard EMF of the cell,
E°cell = E°_H⁺/H2 – E°_zn²⁺/zn
= 0 – (- 0.76) = + 0.76 v

Thus the EMF of the cells comes out to be positive. Hence the reaction takes place.

(b) To predict the reaction, of silver with H₂SO₄.
If Ag reacts, the following reactions should take place.
2Ag + H₂SO₄ → Ag₂SO₄ + H₂
2Ag + 2H⁺ → 2Ag⁺ + H₂

By convention, the cell may be written as
Ag / Ag⁺ ∥ H⁺ / H₂
E°_cell = E°_H⁺,H2 – E°_Ag⁺,Ag
= 0 – 0.80 = – 0.80 v.
The EMF of the cell is negative
Hence, this reaction does not take place.

Question 11.
Calculate the oxidation number of Sb atoms in Sb₂O₅
Answer:
Sb₂O₅: Oxygen in common oxide has an oxidation state of -2. Therefore if x is the oxidation number of Sb in Sb₂O₅ then.
2 × x × 5 × (-2) = 0
x = \(\frac{10}{2}\) = 5
The ox. no. of Sb in Sb₂O₅ = +5

Question 12.
Calculate the ox. no. of sulphur is Na₂S₂O₃
Answer:
Ox. no. of various atom in Na₂S₂O₃
Na = +1, S = x, 0 = – 2
2( + l) + 2x + 3(- 2) = 0
2 + 2x – 6 = 0
2x = 4
x = + 2

Question 13.
Determine the oxidation number of O. in CH₃COOH.
Answer:
Let the ox. no. of O = x
The oxidation no. of each carbon atom = – 1
– 2 + 4 + 2x = 0
2x + 2 = 0
x = – 1

Question 14.
Determine the ox. no. of all the atoms in KClO₄.
Answer:
The ox.no. of K = +1
The ox. no. of Cl = x
The ox. no. of O = — 2
1 + x – 8 = 0
x – 7 = 0 .
x = 7
(K = +1, O = -2, Cl = 7)

Question 15.
Calculate the ox. no. of metal atom in Fe((CN)₆)^3-
Answer:
Fe((CN)₆)^3-
The ox. no. of CN- = – 1

Let the ox. no. of Fe be x
x + 6 (- 1) = – 3
x = + 3

Question 16.
Find out the ox. no. of iron in
[Fe(H₂O)₅(NO)⁺]SO₄
Answer:
[Fe(H₂O)₅(NO)⁺]²⁺
x + 0 + 1 = 2
x = 2 – 1 = +1

Question 17.
Find out the ox. no. of chlorine in HCl & HClO.
Answer:
(i) Cl in HCl is
+1 + x = 0
x = – 1

(ii) Cl in HCIO is
+ 1 + x – 2 = 0
x = +1

Question 18.
Find the ox. state of S in S₂O₄^2- or HSO₃^–
Answer:
(i) S₂O₄^2-
2x – 8 = – 2
2x = 6
x = +3

(ii) HSO₃^–
+ l + x – 6 = – 1
x – 5 = – 1
x = + 4

Question 19.
Calculate the emf of the cell.
M₁ = 1.00 M, M₂ = 0.40 M, PH₂ = 1.00 atm
Answer:
[Pb²⁺(aq)]
LOOM (H⁺ (aq)) = 0.40 M
emf 0.00 – (- 0.13) + \(\frac{0.059}{2}\) log \(\left(\frac{0.40}{1.00}\right)^{2}\)
= 0.13 – 0.0295 × 0.7959
= 0.13 – 0.023 = 0.107 v

Question 20.
Calculate the number of coulombs required to deposit 40.5g of aluminum when the electrode potential is
Answer:
The atomic mass of Al is 27.
The number of coulombs required to deposit 27g of Al is
3 × faradays
or
3 × 96500
Thus, the charge required to deposit 27 g of Al
= 3 × 96500 coulombs

Charge required to deposit 40.5 g of Al
= 3 × \(\frac{96500}{27}\) × 40.5 = 434250 c.

Question 21.
Calculate the ox. number of oxygen in OF₂
Answer:
OF₂
x – 2 = 0
x = +2

Question 22.
Calculate the ox. no. of Pb in Pb₃O₄
Answer:
3x + 4( -2) = 0
3x = 8; x = \(\frac{8}{3}\)

Question 23.
Calculate the ox. no. of S in H₂SO₄.
Answer:
H = +1, S = x, O = – 2
2 × 1 + x + 4 (- 2) = 0
2 + x – 8 = 0
x = + 6

Question 24.
Calculate the ox. no. of carbon atom in C₆H₁₂O₆
Answer:
If x is the oxidation number of c then,
6 × x + 12 × (+1) + 6 × (-2) = 0
6x = 0
x = 0

Question 25.
Calculate the ox. no. of Cr atom in CrO₄^2-.
Answer:
CrO₄^2- if x is the ox. no. of Cr in CrO₄^2- then
x + 4 × (-2) = – 2
x = 8 – 2 = 6
x = + 6

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 4 Chemical Bonding and Molecular Structure. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 4 Important Extra Questions Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure Important Extra Questions Very Short Answer Type

Question 1.
What change in energy takes place when a molecule is formed from its atoms?
Answer:
There is a fall in energy.

Question 2.
Arrange the following in order of increasing bond strengths.
F₂, O₂, N₂, Cl₂
Answer:
F₂ < Cl₂ < O₂ < N₂

Question 3.
Name the shapes of the following molecules: CH₄, C₂H₂, CO₂.
Answer:
CH₄: Tetrahedral; C₂H₂: Cylindrical; CO₂: linear

Question 4.
Arrange the following in order of increasing strengths of hydrogen bonding O, F, S, Cl, N.
Answer:
Cl < S < N < O < F;

Question 5.
Identify the compound/compounds in the following in which S does not obey the Octet rule: SO₂, SF₂, SF₄, SF₆.
Answer:
SF₄, SF₆.

Question 6.
Name one compound each involving sp³, sp², sp hybridization.
Answer:
sp³: CH₄: sp²: C₂H₄: sp: C₂H₂

Question 7.
Which orbitals can overlap to form a cr-bond and which orbitals can do so to form a π bond?
Answer:
s-s, s-p, p-p form a bond, and only p-p form π bond.

Question 8.
Which electrons take part in bond formation?
Answer:
Valence electrons present in the outermost shell.

Question 9.
What are SI units of dipole moment?
Answer:
Coulomb meter (Cm).

Question 10.
Name the method generally used for the calculation of lattice energy or electron affinity.
Answer:
Borh-Haber cycle.

Question 11.
You are given the electronic configuration of five neutral atoms A, B, C, D, and E.
A – 1s², 2s² 2p⁶ 3s²;
B = 1s², 2s² 2p⁶, 3s¹
C = 1s², 2s² 2p¹;
D = 1s², 2s² 2p⁵,
E = 1s², 2s² 2p⁶.
Write the empirical formula for the substance containing
(i) A and D,
Answer:
AD₂

(ii) B and D
Answer:
BD

(iii) only D
Answer:
D₂

(iv) Only E.
Answer:
E.

Question 12.
Which type of forces holds the atoms together in an ionic compound?
Answer:
Electrostatic forces of attraction.

Question 13.
Choose the compounds containing ionic, covalent, and coordinate bonds out of the following: MgO, CH₄, CaCl₂, HCl,
NH⁺₄,O₃.
Answer:
IONIC: MgO, CaCl₂;
Covalent: CH₄, HCl;
Coordinate:NH⁺₄, O₃.

Question 14.
Write the Lewis structure of the polyatomic ions CN^–, SO₄^2-.
Answer:

Question 15.
Arrange the following in order of decreasing C-C bond length:
C₂H₆, C₂H₂, C₂H₄
Answer:
H₃C – CH₃ > H₂C = CH₂ > HC ≡ CH.

Question 16.
Arrange the following in order of decreasing boiling point.
H₂O, H₂S, H₂Se.
Answer:
H₂O > H₂S > H₂Se.

Question 17.
Which of the following has the maximum bond angle? Why?
H₂O, CO₂, NH₃ CH₄.
Answer:
CO₂. Because it is a linear molecule (bond angle is 180°).

Question 18.
Write down the resonance structures of nitrous oxide.
Answer:

Question 19.
Sodium metal vaporizes on heating and the vapors have diatomic molecules of sodium (Na₂). What type of bonding is present in these molecules?
Answer:
Covalent.

Question 20.
Predict the molecular geometries of (a) CIF₃ (b) ICI^–₄
Answer:
(a) T-shaped
(b) Square planar.

Question 21.
What kind of bond exists between two non-metallic elements?
Answer:
Covalent bonds because the difference of electronegativity between two non-metals is usually very small.

Question 22.
Which of the following bonds is most polar?
S – Cl, S – Br, Se – Cl, or Se – Br.
Answer:
Se – Cl.

Question 23.
Write the Lewis structure of PCl₃.
Answer:

Question 24.
Define valency.
Answer:
The number of electrons that an atom gains or loses or shares with other atoms to attain noble gas configuration is termed its valency.

Question 25.
Explain why O-O bond lengths in ozone molecule are equal?
Answer:
O-O bond lengths in ozone molecules are equal because of resonance between their structures.

Question 26.
MgCl₂ is linear, but SnCl₂ is angular. Why?
Answer:
MgCl₂ has sp hybridization, whereas SnCl₂ has sp² hybridization.

Question 27.
Why CCl₄ does not give white precipitates with silver nitrate?
Answer:
CCl₄ is a covalent molecule and does not give Cl^– ions in solution.

Question 28.
What is the approximate bond strength for an ionic bond?
Answer:
It is about 400 kJ mol^-1

Question 29.
What is the approximate bond strength for a covalent bond?
Answer:
It is about 100 kJ mol^-1.

Question 30.
Though chlorine has nearly the same electronegativity as nitrogen, yet there is no hydrogen bonding in HCl. Why?
Answer:
The size of the chlorine atom (3 orbits) is bigger than N (2 orbits).

Question 31.
Which will form a stronger ionic bond?
(a) Na and F
(b) Na and Cl.
Answer:
(a) Na and F.

Question 32.
Out of MgO and NaCl which has more lattice energy?
Answer:
MgO.

Question 33.
Draw Lewis structure of N^3- ion.
Answer:

Question 34.
On what factors the polarity of the bond depends?
Answer:
Difference of electronegativity of the two atoms.

Question 35.
Arrange NaCl, AlCl₃ MgCl₂ in increasing covalent character.
Answer:
NaCl < MgCl₂ < AlCl₃.

Question 36.
Arrange KCl, LiCl, NaCl in increasing ionic character.
Answer:
LiCl < NaCl < KCl.

Question 37.
BF₃ is non-polar. Why?
Answer:
The dipole moment of BF₃ is zero because the BF₃ molecule is symmetrical.

Question 38.
What type of bond is formed when atoms have
(i) zero difference of electronegativity
Answer:
Non-polar covalent

(ii) a little difference of electronegativity
Answer:
Polar covalent

(iii) high difference of electronegativity?
Answer:
Electrovalent.

Question 39.
What is the state of hybridization of C atoms in
(i) diamond
Answer:
It is sp³ in diamond

(ii) graphite?
Answer:
It is sp² in graphite.

Question 40.
What type of orbitals can overlap to form a covalent molecule?
Answer:
Half-filled atomic orbitals containing electrons with opposite spins.

Question 41.
Why glucose, fructose, sucrose, etc. are soluble in the water though they are covalent compounds?
Answer:
These compounds contain OH groups which can form H- bonds with water.

Question 42.
What is the shape of NH₄⁺ ion?
Answer:
Due to the sp³ hybridization of N in NH₄⁺, it is tetrahedral in shape.

Question 43.
Both FIF and H₂O are associated through hydrogen¬bonding, yet the boiling point of HF is lower than that of HzO. Why?
Answer:
H-bonds formed by H₂O are double in number than those formed by HF molecules.

Question 44.
Why ethyl alcohol is completely miscible with water?
Answer:
This is because the OH group present in ethyl alcohol forms H-bonds with water.

Question 45.
Benzene ring contains alternate single and double bonds, yet all the C-C bonds are of equal length. Why?
Answer:
This is due to resonance in benzene.

Question 46.
When is a molecule paramagnetic in nature?
Answer:
A molecule is paramagnetic when it contains half-filled molecular orbitals.

Question 47.
Arrange the following molecular species in increasing order of stability.
Answer:
N₂^2- < N^2- = N²⁺ < N₂

Question 48.
Explain on the basis of the molecular orbital diagram why O₂ should be paramagnetic?
Answer:
O₂ molecule contains one unpaired electron in each of one π2p_x and π2p_y orbitals.

Question 49.
Define antibonding molecular orbital.
Answer:
The molecular orbital formed by the subtractive effect of the electron waves of the combining atomic orbitals is called an antibonding molecular orbital.

Question 50.
What is the effect of the process C₂ → C₂⁺ + e^– on bond order of C₂? [A.IS.B. 2002]
Answer:

Chemical Bonding and Molecular Structure Important Extra Questions Short Answer Type

Question 1.
Which out of CH₃F and CH₃Cl has a higher dipole moment and why?
Answer:
The dipole moment of CH₃Cl is greater than that of CH₃F. The C-F bond length in CH₃F is smaller than the C-Cl bond length in CH₃Cl. The charge separation in the C-F bond is more than in the Cl-C bond- fluoride being more electronegative than chlorine. The bond length has a greater effect than the charge separation. Hence the dipole moment of CH₃C1 is greater than that of CH₃F.

Question 2.
Define the term chemical bond. What are its different types?
Answer:
The attractive forces which hold the constituent atoms in molecules or species in lattices etc. are called a chemical bond.

They are of the following types:

1. Electrovalent or ionic bond
2. Covalent bond
3. Coordinate or dative bond
4. metallic bond
5. hydrogen bond
6. van der Waals forces.

Question 3.
Why covalent bonds are called directional bonds whereas ionic bonds are called non-directional?
Answer:
A covalent bond is formed by the overlap of half-filled atomic orbitals which have definite directions. Hence covalent bond is directional. In ionic compounds, each ion is surrounded by a number of oppositely charged ions and hence there is no definite direction.

Question 4.
AlF₃ is a high melting solid whereas SiF₄ is a gas. Explain why?
Answer:
AlF₃ is an ionic solid due to the large difference in electronegativities of Al and F whereas SiF₄ is a covalent compound and hence there are only weak van der Waal’s forces among its molecules.

Question 5.
Using the VSEPR theory identifies the type of hybridization and draw the structure of OF₂ What are oxidation states of O and F?
Answer:
The electron dot structure of OF₂ is

Thus the central atom (O-atom) has 4 pairs of electrons (2 bond pairs and 2 lone pairs). Hence oxygen in OF₂ is sp³ hybridised and the molecule is v-shaped oxidation state of F = – 1, oxidation state of O = + 2.

Question 6.
Account for the following: The experimentally determined N-F bond length in NF₃ is greater than the sum of the single covalent radii of N and F.
Answer:
This is because both N and F are small and hence have high-electron density. So they repel the bond pairs thereby making the N-F bond length larger.

Question 7.
Explain why the bond angle of NH₃ is greater than that of NF₃ while the bond angle of PH₃ is less than that of. PF₃.
Answer:
Both NH₃ and NF₃ are pyramidal in shape with one lone pair on N. However as F has a higher electronegativity than H, the electron pair is attracted more towards F in NF₃, i.e., the bond pairs of electrons are away from N or in other words, the distance between bond pairs is more. Hence the repulsions between bond pairs in NF₃ are less than in NH₃. Thus the lone pair repels the bond pairs of NF₃ more than it does in NH₃ As a result the bond angles decrease to 102.4° whereas, in ammonia, it decreases to 107.5° only.

PH₃ and PF₃ are also pyramidal in shape with one lone pair of P. But PF₃ has a greater bond angle than PH₃ (opposite to NH₃ and NF₃). This is due to resonance in PF₃ leading to the partial double bond character as shown below

As a result, repulsions between P-F bonds are larger and hence the bond angle is large. There is no possibility for the formation of double bonds in PH₃.

Question 8.
Why HCl is polar whereas the Cl₂ molecule is non-polar?
Answer:
In Cl₂ both atoms have the same electronegativity. Hence the shared pair of electrons is attracted equally by both and remains exactly in the center. NO end acquires a negative or positive charge. In HCl, chlorine is more electronegative than H. Hence shared pair of electrons is more attracted towards chlorine, which, therefore acquires a negative charge while H acquires a positive charge.

Question 9.
Write the Lewis structure of the nitrate ion NO₂
Answer:
Total no. of valence electrons on N = 2s²2p³ = 5
Total no. of valence electrons on two oxygen atoms = 2s²2p⁴
= 2 × 6 = 12
add one electron for negative charge = 5 + 12 + 1 = 18.
The skeletal structure of NO₂^– is written as O N O^–

Draw a single bond (one shared electron pair) between the nitrogen atom and each of the two oxygen atoms completing the octets on oxygen atoms. This, however, does not complete the octet on Nitrogen if the remaining two electrons, constitute lone pair on it.

Hence we have to resort to multiple bonding between N and one of the oxygen atoms (in this ease a double bond). This leads to the following Lewis dot structures.

Question 10.
Draw the Lewis dot structure of CO₃ ion.
Answer:
Total no. of valence electrons of CO₃ = 4 + 3 × 6 = 22
total no. of valence electrons on CO₃ = 22 + 2 = 24
The skeletal structure of CO₃ is
Putting one shared pair of electrons between C and O and completing the Octets of oxygen, we have

In the above structure, an octet of oxygen is not complete. Hence multiple bonding is required between C and O.

Question 11.
Calculate the formal charge on each oxygen atom of O₃ molecules and write its structure with formal charges.
Answer:
Lewis structure of O₃ is

Formal charge on O₁ = 6 – 4 – \(\frac{1}{2}\)(4) = 0
Formal charge on 0(2) = 6 – 2 – \(\frac{1}{2}\) (6) = + 1
Formal charge on 0(3) = 6 – 6 – \(\frac{1}{2}\) (2) = – 1.

Hence we represent O3 along with formal charges as follows.

Question 12.
Describe briefly the limitations of the Octet rule.
Answer:

1. In some molecules like LiCl, BeCl₂, BF₃, AlCl₃, central atoms like Li, Be, B, Al have not completed their octets Li: Cl,
   
2. In some molecules, there are an odd number of electrons like
   
3. In some molecules like PF₅, SF₆, IF₇, the central atoms like P, S, I have 10,12,14 electrons respectively.
4. Some noble gases also combine with oxygen and F to form a no. of compounds like XeOF₂, XeF₂, XeF_6, etc.
5. Octet theory does not explain the shape of the molecules.
6. It does not explain the relative stability of the molecules being totally silent about the energy of a molecule.

Question 13.
Explain the structure of the CO₂ molecule on the basis of resonance.
Answer:
The experimentally determined C to O bond length in CO₂ is 115 pm. The lengths of a normal C to oxygen double bond (C = 0) and carbon, to oxygen triple bond (C ≡ O), are 121 pm and 110 pm respectively. The C-O bond lengths in CO₂ (115 pm) lie between the values of C =0 and C-O. Obviously, a single Lewis structure cannot depict this position and it becomes necessary to write more than one Lewis structure.

∴ CO₂ is best described as the resonance hybrid of the canonical or resonance forms I, II, and III

Question 14.
How would the bond lengths vary in the dicarbon species
C₂, C₂^–, C₂^2-?
Answer:


As bond lengths are inversely proportional to bond order
∴ Bond lengths will be in order.
C₂ > C₂^– > C₂^2-.

Question 15.
What is the effect of the following ionization process on the bond order of O₂?
O₂ → O₂⁺ + e^–
Answer:
O₂ → O₂ + e^–

Thus bond order of O₂ which is = 2
increases to 2.5 in O₂⁺.

Question 16.
Draw the Lewis dot diagram of nitric acid, sulphuric acid, phosphorus acid, and hypophosphorous acid.
Answer:


Question 17.
Out of sigma and pi bonds, which one is stronger and why?
Answer:
Sigma (a) bond is stronger This is because a bond is formed by head-on overlapping of atomic orbitals and such overlapping being on the internuclear axis is large, n bond is formed by sidewise overlapping which is small and so a Pi bond is weaker.

Question 18. Arrange the following in order of decreasing bond angles
(i) CH₄, NH₃ H₂O, BF₃, C₂H₂
Answer:
C₂H₂ (180°) > BF₃ (120°) > CH₄ (109°, 28′) > NH₃ (107°) > H₂O (104.5°)

(ii) NH₃ NH₂, NH₄
Answer:
NH₄⁺ > NH₃ > NH₂^– [because no. of lone pairs of electrons present on N are 0, 1, 2 respectively].

Question 19.
Predict which out of the following molecules will have a higher dipole moment and why? CS₂ and OCS.
Answer:

both are linear molecules, but bond moments in CS₂ cancel out so that the net dipole moment = O. But in OCS bond moment of C= O is not equal to that of C = S. Hence it has a net dipole moment. Thus dipole moment of OCS is higher.

Question 20.
Out of p-orbital and sp-hybrid orbital which has greater directional character and why?
Answer:
sp orbital has a greater directional character than p-orbital. This is because the p-orbital has equal-sized lobes with equal electron density in both the lobes whereas the sp-hybrid orbital has greater electron density on one side.

Question 21.
Interpret the non-linear shape of H2S and non-planar shape of PCl₃ using VSEPR theory.
Answer:
H₂S
No. of electron pairs around S = \(\frac{6+2}{2}=\frac{8}{2}\) = 4

Question 22.
Why mobility of H⁺ ions is greater in ice as compared to liquid water? [CBSE. PMT 2005]
Answer:
H⁺ ions (obtained from ionization of water H₂O) in liquid water get easily hydrated. As a result, the size of the hydrated H⁺ ions is much larger than in ice. Therefore the mobility of H⁺ ions in water is less than in ice.

Question 23.
What is the hybrid state of BeCl₂? What will be the change in the hybrid state of BeCl₂ in the solid-state? [CBSE PMT 2005]
Answer:
In the vapor state at high temperature, BeCl₂ exists as a linear molecule. Cl – Be – Cl due to sp hybridization of Be.

In the solid-state, it has a polymeric structure as follows:

It is possible only due to the sp³ hybrid state of Be. Two half-filled hybrid orbitals will form normal covalent bonds with two Cl atoms. The other two Cl atoms are coordinated to Be atom by donating electron pairs into the empty hybrid orbitals of Be.

Question 24.
In both water and diethyl ether, the central atom viz. O- atom has the same hybridization. Then why do they have different bond angles? Which one has a greater bond angle?
Answer:
Both water and diethyl ether have the central atom O in an sp³ hybrid state with two lone pairs of electrons on O.

But due to the greater repulsions between two ethyl (C₂H₅) groups in diethyl ether than between two H-atoms in H₂O result in a greater bond angle (110°) in diethyl ether than 104.5° in that of water (H₂O).

Question 25.
BCl₃ is planar but anhydrous AlCl₃ is tetrahedral. Explain.
Answer:
B atom in BCl₃ is sp² hybridized due to which the shape of BCl₃ is triangular planar. Anhydrous AlCl₃ exists as a dimer with the formula Al₂Cl₆.

Thus, each atom of A1 in Al₂Cl₆ is surrounded by four bond pairs and hence it assumes a tetrahedral structure.

Question 26.
(a) How bond energy varies from N₂^– to N₂⁺ and why?
Answer:
Both N₂^– and N₂⁺ is, almost the same [though N₂^– is slightly less stable and hence has less bond energy than N₂⁺ due to the presence of a greater number of electrons in the antibonding molecular orbitals]

(b) On the basis of molecular orbital theory, what is a similarity between
(i) F₂, O₂^––
Answer:
Both F₂ and O₂^–– have the same bond order = 1 and so the same bond length

(ii) CO, N₂, NO⁺? [CBSE PMT 2004]
Answer:
All of them have the same bond order and bond length.

Question 27.
Phosphorus pentachloride dissociates in the vapor phase as
PCl₅ ⇌ PCl₄⁺ + Cl^–
The hybridization of P in PCl₅ is sp³d. What is it in PCl₄⁺?
Answer:
Phosphorus is sp³ hybridized in PCl₄ since it is isostructural with NH₄ which also has sp³ hybridization of Nitrogen.

Question 28.
Explain giving reasons whether BH₄ and H₃O⁺ will have the same/different geometry?
Answer:
The central atom in both the ions is surrounded by the same number of pairs of electrons, that is, 4. Hence they have the same tetrahedral geometry.

Question 29.
State with reasons, which is more polar CO₂ or N₂O?
Answer:
N₂O is more polar than CO₂ which is a linear molecule and thus symmetrical. Its net dipole moment is zero.

N₂O is linear but unsymmetrical. It is a resonance hybrid of the following canonical structures:

It has a net dipole moment of 0.116 D.

Question 30.
Out of peroxide ion (O₂) and superoxide ion (O₂ ) which has larger bond length and why?
Answer:
The bond order of O₂^– is 1.5 while that of O₂^2- is 1.0.
The lesser the bond order, the greater is the bond length as the bond order is inversely proportional to bond length. ( Hence O₂^2- has a larger bond length than O₂^2-.

Chemical Bonding and Molecular Structure Important Extra Questions Long Answer Type

Question 1.
Explain the formation of the following molecules according to the orbital concept, F₂, HF, O₂, H₂O, N₂, NH₃ molecules.
Answer:
1. Formation of F₂ molecule. Atomic number (Z) of fluorine is 9 and its orbital electronic configuration is 1s² 2s² 2p²_x, 2p²_y, 2p¹_z. Thus, a fluorine atom has one half-filled atomic orbital. Therefore, two atoms of fluorine combine to form the fluorine molecule as a result of the combination for their half-filled atomic orbitals shown in Fig. The two atoms get linked by a single covalent bond.

Formation of F₂ molecule

2. Formation of HF molecule. Fluorine atom, as stated above, has one half-filled atomic orbital. Hydrogen atom (Z = 1) has only one electron in Is orbital. Thus, the hydrogen fluoride (HF) molecule. is formed as a result of the combination (or overlap) of the half-filled orbitals belonging to the participating atoms.

Formation of HF molecule

3. Formation of O₂ molecules. The atomic number (Z) of oxygen is 8 and its orbital electronic configuration is 1s² 2s² 2p²_x 2p¹_x 2p¹_z. This means that an oxygen atom has two half-filled orbitals with one electron each. Two such atoms will combine to form a molecule of oxygen as a result of the overlap of the half-filled orbitals with opposite spins of electrons.

Formation of O₂ molecule
Thus, the two atoms of oxygen are bonded to each other by two covalent bonds or double bonds (O = O).

4. Formation of H₂O molecule. In the formation of the H₂O molecule, the two half-filled orbitals of the oxygen atom combine with the half-filled orbitals (1s) of the hydrogen atoms. Thus, the oxygen atom gets linked to the two hydrogen atoms by single covalent bonds as shown in

Formation of H₂O molecule

5. Formation of N₂ molecule. The atomic number of nitrogen is 7 and its orbital electronic configuration is 1s² 2s² 2p¹_x 2p¹_y 2p¹ _z. This shows that the nitrogen atom has three half-filled atomic orbitals. Two such atoms combine as a result of the overlap of the three half-filled orbitals and a triple bond gets formed (N = N).

Formation of N₂ molecule

6. Formation of NH₃ molecule. In the formation of ammonia (NH₃) molecule, three half-filled orbitals present in the valence shell of nitrogen atom combine with 1s orbital of three hydrogen atoms with one electron each. As a result, the nitrogen atom completes its octet and a molecule of NH₃ is formed in which the nitrogen atom is linked to three hydrogen atoms by covalent bonds.

Formation of NH₃ molecule

Question 2.
What is a hydrogen bond, what are its causes, and give the conditions for hydrogen bonding? What is the strength of hydrogen bonding? Describe the two types of hydrogen bonding.
Answer:
When hydrogen is connected to small highly electronegative atoms such as F, O, and N in such cases hydrogen forms an electrostatic weak bond with an electronegative atom of the second molecule, this type of bond binds the hydrogen atom of one molecule and the electronegative atom of the 2nd molecule is called as hydrogen bond. It is a weak bond and it is denoted by dotted lines e.g., in HF, hydrogen forms a weak bond with the electronegative F atom of the 2nd molecule neighboring HF.

So it means hydrogen is acting as a bridge between two molecules by one covalent bond and the other by a hydrogen bond. Due to this hydrogen bonding, HF will not exist as a single molecule but it will exist as an associated molecule (HF)_n. So hydrogen bond may be defined as a weak electrostatic bond that binds the hydrogen atom of one molecule and electronegative bond atoms (F, O, N) of the second neighboring molecule.

Cause of hydrogen bonding: When a hydrogen atom is bonded to an electronegative atom (say F, O, N) through a covalent bond, due to electronegativity difference, the electronegative atom attracts the shared pair of electrons towards its side with a great force as a result of which the shared pair of electrons will be displaced toward electronegative atom and away from a hydrogen atom.

Due to which hydrogen atom will acquire a slightly negative charge and if another molecule is brought nearer to it in such a way that electronegative atom of the second molecule faces hydrogen atom of the 1st molecule, due to opposite charges present on the atoms, an electrostatic bond will be formed between the hydrogen atom of one molecule and electronegative atom of 2nd molecule and this is called as hydrogen bond.

Conditions for hydrogen bonding. The following two necessary conditions for hydrogen bonding are:

1. Hydrogen atom should be connected to highly electronegative atom say F, O, or N.
2. The electronegative atom of which the hydrogen atom is connected should be the same in size.

The smaller the size of the electronegative atom greater will be the attraction of that atom for shared pair of electrons and hence that pair will be displaced more nearer to that atom and hence that atom will develop greater negative charge and the hydrogen atom will develop a greater positive charge and hence hydrogen atom of this molecule will easily attract negative atom of the IInd molecule and hence a hydrogen bond will be easily formed.

As both these conditions are satisfied only by F, O, N atoms so only three atoms show hydrogen bond.

Strength of Hydrogen Bond: A hydrogen bond is a very weak bond. It is weaker than an ionic or a covalent bond. Its strength range from 13 kJ mol-1 to 42 kJ mol^-1. The strength of the hydrogen bond for some of the molecules is the order H-F H (40 kJ mol^-1) > O-H…… O (28 kJ mol^-1) > H-N…… H (13 kJ mol^-1) whereas the strength of a covalent bond is quite high. For example, the bond strength of the H-H bond in H₂ is 433 kJ mol^-1.

Types of H-bonding
There are two types of hydrogen-bonds

1. Intermolecular hydrogen bond. It is formed between two different molecules of the same or different compounds. For example H-bond in case of HF, alcohol, or water.
2. Intramolecular Hydrogen bond. It is formed when a hydrogen atom is in between the two highly electronegative (F, O, N) atoms present within the same molecule. For example, in o-nitrophenol, hydrogen is in between the two oxygen atoms.
   

Question 3.
Compare the properties of electrovalent and covalent compounds.
Answer:
The main points of difference between electrovalent and covalent compounds are summed up in the table below:

Electrovalent/Ionic compounds	Covalent compounds
1. They are formed by sharing of electrons between two atoms.	1. They are formed by sharing of electrons between two atoms.
2. These compounds may be solids, liquids, or gases.	2. These compounds may be solids, liquids, or gases.
3. They are made up of molecules held together by weak van der Waals forces	3. They are made up of molecules held together by weak van der Waals forces of attraction.
4. They have generally high M and B points	4.            They have generally low melting and boiling points.
5. They are generally soluble in polar solvents like water and insoluble in organic solvents.	5. Covalent compounds are generally soluble in non-polar solvents such as benzene and insoluble in polar solvents.
6. Ionic compounds conduct electricity in the molten or dissolved state.	6. Covalent compounds are generally bad conductors of electricity.
7. Ionic compounds are non-directional and do not show isomerism.	7. Covalent compounds are rigid and directional. They show isomerism.
8. These compounds undergo ionic reactions which are very fast, almost instantaneous.	8. These compounds undergo molecular reactions which are very slow.

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 2 Structure of Atom Chemistry. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 2 Important Extra Questions Structure of Atom Chemistry

Structure of Atom Chemistry Important Extra Questions Very Short Answer Type

Question 1.
How many total electrons are present in nitrate ion?
Answer:
No. of electrons in NO₃^– ion
= No. of electrons on N + No. of electrons on 3 oxygen atom + one ē
= 7 + 3 × 8 + 1 = 32 electrons.

Question 2.
The nucleus of the atom of an element does not contain a neutron. Name the element and what does its nucleus consists of.
Answer:
The nucleus of hydrogen. It contains only one proton.

Question 3.
What are nucleons?
Answer:
The neutrons and protons present in the nucleus of an atom are collectively called nucleons.

Question 4.
Write electronic configurations of Chromium (At. Np. = 24).
Answer:
Cr = 24 = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁵, 4s¹.

Question 5.
Which of the following has the smallest de-Broglie wavelength? O₂, H₂, a proton, an electron.
Answer:
According to the de-Broglie equation λ = \(\frac{h}{m \times v}\) for same value of velocity λ ∝\(\frac{1}{m}\)
∴ O₂ molecule has shortest wavelength.

Question 6.
How many unpaired electrons are there is a carbon atom in the ground state?
Answer:
C = 6 = 1s², 2s², 2p¹_x 2p¹_y. There are only two unpaired electrons.

Question 7.
What type of spectrum is obtained when light emitted from the discharge tube containing hydrogen gas is analyzed?
Answer:
Emission line spectrum.

Question 8.
What is the maximum number of electrons in an atom having n = 3, l = 1 and s = + \(\frac{1}{2}\)?
Answer:
Three electrons (one each in 3p_x’ 3p_y’, 3p_z’ ).

Question 9.
Name the spectral line in the spectrum of H-atom obtained when an electron jumps from n = 4 to n = 2.
Answer:
Balmer Series.

Question 10.
Give some examples of electromagnetic radiation.
Answer:
Y-rays, X-rays, UV-rays, visible rays, radio waves, etc.

Question 11.
State two properties of electromagnetic radiation.
Answer:
Electromagnetic radiation shows the phenomenon of:

1. Interference,
2. Diffraction.

Question 12.
What is meant by the quantization of electron energy?
Answer:
It means that an electron in an atom has a certain, specific, discrete amount of energy.

Question 13.
What does a principal quantity number denote?
Answer:
It denotes a specific stationary state.

Question 14.
Why Bohr’s orbits are also called ‘energy levels’?
Answer:
Because they are associated with a certain definite amount of energy.

Question 15.
How many spherical nodal surfaces are there in a 3s- sub-shell?
Answer:
Two.

Question 16.
Out of 6s and 4f orbitals, which has higher energy and why?
Answer:
4forbital has higher energy, ((n + l) value of 4f is 7 while that of 6s is 6). The higher the (n + l) value of an orbital higher is the energy.

Question 17.
List the value/values of quantum numbers n and l for 4f electrons.
Answer:
n = 4, l = 3.

Question 18.
Out of 4s and 3d orbitals, which will have higher energy and why?
Answer:
3d orbital has higher energy as it has a higher value of (n + l).

Question 19.
Which of the following orbitals are not possible?
1p, 2p, 2d, 3f, 4f?
Answer:
1p, 2d, 3f is not possible.

Question 20.
Which orbital does not have directional characteristics?
Answer:
s-orbital.

Question 21.
An electron is in 3p-orbital. What are the permitted values of n, l, and m?
Answer:
n = 3, l = 1, m = – 1, 0, + 1.

Question 22.
Write designation of an orbital having n = 5,1 = 3.
Answer:
5f -orbital.

Question 23.
Consider the electronic configuration 1s° 2s° 2p° 3s¹.
Name the element having this configuration. Is it in an excited state or ground state?
Answer:
It is the configuration of the H-atom. It is in an excited state.

Question 24.
Which quantum number determines the
(i) size,
Answer:
Principal

(ii) orientation,
Answer:
Magnetic

(in) the shape of orbital?
Answer:
Azimuthal quantum number.

Question 25.
Which energy level does not have a p-orbital?
Answer:
First energy level (i.e., n – 1, K-shell).

Question 26.
Name an element that has only one d-electron.
Answer:
Scandium (atomic no. = 21).

Question 27.
Given an isotone of C-13 atom.
Answer:
Isotones are the atoms of different elements which have the same number of neutrons. N-14 is an isotone of a C-13 atom.

Question 28.
Which of the following orbitals does not make sense?
5s, 4f, 3p, 2d.
Answer:
2d orbital does not exist and thus makes no sense.

Question 29.
Name the famous experiment which showed for the first time that an atom has a nucleus.
Answer:
Rutherford experiment of scattering of a-particles.

Question 30.
Write the value of four quantum numbers for the valence electron of the sodium atom.
Answer:
Sodium atom has 11 electrons and its valence electron is 3s¹ (as configuration is 1s², 2s²p⁶, 3s¹)
∴ The value are: n = 3, l = 0, m = 0, s = + \(\frac{1}{2}\)

Question 31.
What is the relationship between velocity, wavelength, and frequency of radiation?
Answer:
These three characteristics of wave motion are related to each other as frequency = \(\frac{\text { Velocity }}{\text { Wavelength }}\)
In terms of symbols v = \(\frac{c}{λ}\).

Question 32.
How wave number (\(\vec{v}\) ) and wave length (λ) are related?
Answer:
\(\vec{v}\) = \(\frac{1}{λ}\) . Wave number is the reciprocal of wavelength.

Question 33.
Which series of hydrogen spectrum lies in the visible spectrum?
Answer:
Balmer series.

Question 34.
How do you distinguish the two electrons present in the same orbital?
Answer:
By their spin quantum no. s which has + \(\frac{1}{2}\) and – \(\frac{1}{2}\) values.

Question 35.
Name the principle which establishes that two electrons cannot have the same values for all the 4 quantum numbers.
Answer:
Pauli exclusion principle.

Question 36.
Write down the electronic configuration of Cu (= 29) in the ground state.
Answer:
Cu = 29 = 1s², 2s² 2p⁶, 3s² 3p⁶, 3d¹⁰ 4s¹.

Question 37.
What is the lowest value of n that allows g orbitals to exists?
Answer:
n = 5.

Question 38.
Why do an atom M and its ion M²⁺ have the same mass?
Answer:
Both have the same no. of neutrons and protons which are responsible for the mass of an atom.

Question 39.
The nucleus of an atom has 6 protons and 8 neutrons. What are its atomic number and mass number? What is this element1?
Answer:
At. No. = 6; Mass No. = 6 + 8 = 14; Element is Carbon.

Question 40.
What is the number of electrons having l = 0 in an atom with an atomic number 29?
Answer:
7. [1s², 2s², 3s², 4s¹, i.e., 2 + 2 + 1 = 7]

Question 41.
Express s, p, d, f for a shell in increasing order of energy.
Answer:
s < p < d < f. .

Question 42.
For n = 5; what are the possible values of l?
Answer:
When n = 5; l = 0,1, 2, 3, 4.

Question 43.
For l = 3, what are the possible values of m?
Answer:
m = – 3, – 2, – 1, 0, + 1, + 2, + 3.

Question 44.
The ionization potential of an atom is 13.6 V. How much energy is required to ionize it?
Answer:
13.6 eV.

Question 45.
The threshold wavelength for a metal surface is λ₀. How is it related to the work function of the metal?
Answer:
W₀ = \(\frac{h c}{\lambda_{0}}\)

Question 46.
Write the energy E of a photon in terms of frequency.
Answer:
E = hv where h is called Planck’ constant.

Question 47.
How much energy is required for the removal of the only electron present in the hydrogen atom?
Answer:
ΔE = E_∞ – E₁ = 0 – (- 1312 kJ mol^-1) – 1312 kJ mol^-1.

Question 48.
Which quantum number determines the
(a) shape
Answer:
Azimuthal

(b) orientation and
Answer:
Magnetic

(c) size of the orbital?
Answer:
Principal.

Question 49.
Write down the values of n, l, m, s of the last electron in potassium (Z = 19)?
Answer:
The last electron in potassium is present in 4s.
Its n = 4; l = 0; m = 0, s = + \(\frac{1}{2}\) or – \(\frac{1}{2}\).

Question 50.
What is the sequence of energies of 3s, 3p, and 3d orbitals in
(i) H-atom
Answer:
H-atom: 3s = 3p = 3d

(ii) a multi-electron atom?
Answer:
3s < 3p < 3d.

Question 51.
Name the cations which do not have any electrons.
Answer:
H⁺, He²⁺.

Question 52.
How many quantum numbers are required to specify an orbital? Name them.
Answer:
Three quantum numbers.
These are

1. Principal quantum no. (n)
2. Azimuthal quantum No. (l)
3. Magnetic quantum no. (m).

Question 53.
What is observed when an opaque object is placed in the path of cathode rays?
Answer:
A shadow of an opaque object is observed on the wall opposite to cathode.

Question 54.
What happens when a very light paddle wheel is placed. in the path of cathode rays?
Answer:
It begins to undergo a rotatory motion.

Question 55.
What name was given to the particles which constitute cathode rays?
Answer:
Electron.

Question 56.
Arrange the following orbitals in the order in which electrons may be normally expected to fill them.
3s, 2p, 3p, 2s, 3d, 4s.
Answer:
According to the Aufbau principle, the given orbitals will be filled in the order: 2s, 2p, 3s, 3p, 4s, 3d.

Question 57.
Which fundamental property of an atom is not understood if we assume that an atom consists of a nucleus containing protons only and an extranuclear part containing an equal number of electrons?
Answer:
The mass number of atoms and stability of the nucleus cannot be explained.

Question 58.
The following ions are isoelectronic: F^–, Mg²⁺, O^2-. Write the common electronic configuration.
Answer:
Each given ion contains 10 electrons. The common configuration is that of the Neon atom, i.e., 1s², 2s² p⁶.

Question 59.
What is the atomic number of an element whose mass number is 23 and contains 12 neutrons in its nucleus? What is the symbol of an element?
Answer:
Atomic number = No. of protons in the nucleus
= Mass no. – No. of neutrons
= 23 – 12 = 11
The element is sodium and its symbol is Na.

Question 60.
An atom has 2 K, 8 L, and 5 M electrons. Write the electronic configuration of the atom. How many unpaired electrons are there in the atom?
Answer:
Electronic configuration of atom is: 1s², 2s²p⁶, 3s², 3p¹_x, 3p¹_y, 3p¹_z. It has three unpaired electrons.

Question 61.
Write the various possible quantum numbers for unpaired electron of Aluminium atom?
Answer:
Al = 13 = 1s², 2s² 2p⁶ 3s² 3p¹_x
n = 3; l = 1; m = – 1, 0, + 1; s = \(\frac{1}{2}\) or – \(\frac{1}{2}\) .

Question 62.
Give the values of quantum numbers for the electron with the highest energy in sodium atom.
Answer:
n = 3; l = 0; m = 0; s = + \(\frac{1}{2}\) or – \(\frac{1}{2}\) .

Question 63.
What do you observe in the spectrum of NaCl?
Answer:
Two yellow lines with a wavelength of 5890 Å and 5896 Å

Question 64.
What do you mean by saying that the energy of the electron is quantized?
Answer:
This means that the electrons in an atom have only definite values of energy.

Question 65.
Why are Bohr’s orbits called stationary states?
Answer:
This is because the energies of the orbits in which the electrons revolve are fixed.

Question 66.
What is the difference between a quantum and a photon?
Answer:
The smallest packet of energy of any radiation is called a quantum whereas that of light is called a photon.

Question 67.
Which quantum number does not follow from the solution of the Schrodinger wave equation?
Answer:
Spin quantum number.

Question 68.
How many orbitals will be possible in a g-subshell?
Answer:
For g-subsheil l = 4; m = 2l + 1; – 4, – 3, – 2, – 1, 0, + 1, + 2, + 3, + 4; 9 orbitals.

Structure of Atom Chemistry Important Extra Questions Short Answer Type

Question 1.
Enumerate the important characteristics of anode-rays (or positive rays). How this study led to the discovery of proton?
Answer:

1. The mass of positive particles which constitute these rays depend upon the nature of the gas in the tube.
2. The charge/mass (e/m) ratio of anode-rays is not constant but depends upon the nature of gas in the tube. The value of e/m is greatest for the lightest gas, hydrogen the electric charge on a lightest positively charged particle from hydrogen gas was found to be exactly equal in magnitude but opposite in sign to that of the electron. This lightest positively charged particle from hydrogen gas was named the proton. The mass of a proton is almost 1836 times that of the electron.

Question 2.
What are anode-rays? Illustrate their formation by a diagram.
Answer:
Anode-rays. If a perforated cathode is used in the discharge tube experiment, it is found that a certain type of radiation also travels from anode to cathode. These are called anode rays or positive rays.

Production of anode rays

Question 3.
Describe the important properties of cathode-rays. What is concluded about the nature of these rays?
Answer:
The cathode rays possess the following properties:

1. Travel in straight lines perpendicular to the surface of the cathode.
2. Consists of material particles.
3. Have got the heating effect.
4. Consists of negatively charged particles.
5. Produce X-rays when they strike against hard metals like copper, tungsten, platinum, etc.
6. Produce fluorescence when they strike glass or certain other materials like zinc sulfide.
7. Penetrate through thin aluminum foils and other metals.
8. Affect the photographic plates.

Question 4.
What are the main features of Rutherford’s model of an atom?
Answer:
The main features of this model are:

1. Atom is spherical and consists of two parts: Nucleus and extra-nuclear part.
2. The entire mass and entire positive charge are concentrated in a very small region at the center known as the nucleus.
3. The space surrounding the nucleus known as the extra-nuclear part is negatively charged so an atom as a whole is neutral.
4. Most of the extra-nuclear part is empty.
5. The electrons are not stationary but are revolving around the nucleus at very high speeds like planets revolving around the Sun.

Question 5.
What is meant by the dual nature of radiation?
Answer:
The fact that light energy is carried in terms of packets of energy (i.e., photons) as suggested by Planck’s theory means that light has a particle character. At the same time, the fact light has a wave character. These experimental facts led Einstein to suggest that light has a dual character, i.e., it behaves both like a wave and like a particle.

Question 6.
Describe the drawback to Rutherford’s model of the atom.
Answer:
The main drawback is that it could not explain the stability of an atom. Maxwell has shown that when electric charge is subjected to acceleration, it emits energy in the form of radiations. In Rutherford’s model of the atom, electrons are orbiting the nucleus and hence the direction of their velocity is constantly changing, i.e., electrons are accelerating.

This will cause the electrons will have lesser and lesser energy and will get closer and closer to the nucleus until at last, it spirals into the nucleus and thus does not provide a stable model of the atom.

Question 7.
What is the value of
(i) charge to mass ratio (e/m) of electrons,
Answer:
J. Thomson determined the value of e/m for electron by the study of deflection of electron beam under the simultaneous influence of electric and magnetic field perpendicular to each other, the e/m value is 1.76 × 10⁸ coulomb per gram of electrons.

(ii) charge of electrons,
Answer:
The charge of electrons was measured by Millikan in 1909 by his famous ‘oil drop’ experiment. It was found to be 1.60 × 10^-19 coulombs.

(iii) mass of an electron?
Answer:
The mass of electrons is 9.1 × 10^-28 g.

Question 8.
How is it concluded that electrons are a universal constituent of all matter?
Answer:
The charge/mass (e/m) ratio for the particles in the cathode rays (i.e., electron) is found to be the same irrespective of the nature of the cathode or the nature of the gas taken in the discharge tube. This shows that electrons are universal constituents of all matter.

Question 9.
Distinguish between an Emission spectrum and an Absorption spectrum.
Answer:
The important differences between the emission and absorption spectra are given in the following table:

Emission Spectrum	Absorption Spectrum
1. Emission spectrum is obtained when radiations emitted by the excited substance are analyzed in a spectroscope.	1. Absorption spectrum is obtained when the white light is first passed through the substance (in a gaseous state or in solution) and the transmitted light is analyzed in a spectroscope.
2. Emission spectrum consists of bright colored lines separated by dark spaces.	2. Absorption spectrum consists of dark lines in an otherwise continuous spectrum.

Question 10.
What are the shortcomings of Bohr’s atomic model?
Answer:

1. It couldn’t explain the spectra of multi-electron atoms.
2. It fails to explain the splitting of spectral lines when subjected to the electrostatic or magnetic fields (Stark or Zeeman’s effect).
3. It does not account for the fine splitting of spectral lines.
4. It affords a two-dimensional picture of the revolution of electrons while actually electron revolves around the nucleus in three dimensions.
5. It does not account for the shapes of molecules.

According to it, this is possible to determine simultaneously both the position and momentum of the electron accurately. But this is contrary to Heisenberg’s Uncertainty Principle.

Question 11.
Account for the stability of the atom with the help of Bohr’s theory.
Answer:
According to Bohr’s theory, an electron revolves around the nucleus only in a definite orbit and cannot lose energy continuously. It can lose energy only if it jumps from a higher orbit to a lower orbit but this is possible only if the electron has already acquired a higher energy level by absorbing a certain amount of energy. If no lower level is available, the electron cannot lose energy at all, i.e., an atom does not collapse. In other words, it is quite stable.

Question 12.
What are the main achievements of Bohr’s theory of the atom?
Answer:
The main achievements of Bohr’s theory of atom are:

1. It can explain the stability of the atom.
2. It successfully explains the line spectrum of hydrogen.
3. It explains the line spectra of single-electron ions like He⁺ and Li²⁺.

Question 13.
Write a short note on de-Broglie relation (or de-Borglie) equation.
Answer:
A moving material particle, like an electron, proton, etc. having mass m and velocity v is associated with wavelength X related by:
λ = \(\frac{h}{m v}=\frac{h}{p}\)
where p stands for the momentum of the particle and h is Planck’s constant. This relation is known as a de-Broglie equation.

Question 14.
Do atomic orbitals have sharp boundaries? Explain why or why not?
Or
Why don’t we draw a boundary surface diagram within the probability of finding the electron is 100%?
Answer:
No, atomic orbitals do not have sharp boundaries because the probability of finding the electron even at large distances may become very small, but not equal to zero.

Question 15.
What is the difference between the angular momentum of an electron present in 3p and that present in 4p?
Answer:
No difference, because angular momentum is given by = \(\frac{h}{2 \pi} \sqrt{l(l+1)}\), i.e., it depends only on the value of l and not the principal quantum no. n.

Question 16.
Why splitting of spectral lines takes place when the source giving the spectrum is placed in a magnetic field?
Answer:
In the presence of a magnetic field, the orbitals present in a subshell (which were degenerate) take up different orientations.

Question 17.
How many electrons in Sulphur (Z = 16) can have n + l = 3?
Answer:
16S = 1s², 2s² 2p⁶ 3s² 3p⁴

• For 1s², n + l = l+ 0 = l
• For 1s², n + l = 2 + 0 = 2
• For 2p⁶, n + 1 = 2 + 1= 3
• For 3s², n + l = 3 + 0 = 3
• For 3p⁴, n + l = 3 + l = 4.

Thus n + l = 3 for 2p6 and 3s2 electrons, i.e. for 8 electrons.

Question 18.
Why Pauli Exclusion principle is called the exclusion principle?
Answer:
This is because according to the principle, if one electron in an atom has some particular values for the four quantum numbers, then all the other electrons in that atom are excluded from having the same set of values.

Question 19.
Why Hund’s rule is called the rule of maximum multiplicity?
Answer:
This is because out of the various possible electronic configurations, only that Configuration is correct for which the total spin value is maximum.

Question 20.
How many orbitals are present in the M-shell?
Answer:

Structure of Atom Chemistry Important Extra Questions Long Answer Type

Question 1.
Describe the shape of s – and p – orbitals What do you mean by node or nodal surface?
Answer:
Shapes of Orbitals:
s-orbitals: These are spherically symmetrical and non-directional. Shapes of 1s and 2s orbitals are shown in Fig. The effective volume of 2s orbital is larger than Is orbital. Another important feature of 2s orbital is that there is a spherical shell within 2s (region without dots) where the probability of finding the electron is zero. This is called a node or a nodal surface. There are (n – 1) nodes in an s-orbital (where n is the energy level).

p-orbitals: There are three p-orbitals designated as p_x’, p_y’ or p_z’ which are oriented along the three mutually perpendicular axis x, y, and 2. Each, orbital consists of two lobes symmetrical about the particular axis and has a dumbbell shape. The two lobes are separated by a nodal plane.

Shapes of three 2p orbitals

The two lobes of each orbital are separated by a plane having x zero electron density. This plane is known as a nodal plane.

Question 2.
How does the Schrodinger wave equation help to understand the probability of finding the electron near the nucleus? What do you mean by an orbital?
Answer:
Probability Picture of Electrons:
Schrodinger incorporated the requirements of the uncertainty principle and de Broglie’s concept of matter waves and proposed a mathematical equation to describe the behavior of an electron in an atom. The equation was known as the Schrodinger wave equation.

The Schrodinger wave equation is

where x, y, and z are three space coordinates,
m is the mass of the electron,
h is Planck’s, constant

E is the total energy and V is the potential energy of the electron, φ (Greek letter psi) is the amplitude of the wave, called wave function, \(\frac{\partial^{2} \psi}{\partial x^{2}}\) refers to the second derivative of φ with respect to x only and so on.

The solution of this equation gave the mathematical expression which gives information about the various energy states and other measurable properties such as the radiation frequencies emitted or absorbed for the hydrogen atom. The solutions of the Schrodinger wave equation are called wave functions and are denoted by the symbol φ.

The physical significance of wave function: In the physical sense φ gives the amplitude of the wave associated with the electron. We know that in the case of light waves, the square of the amplitude, of the wave at a point is proportional to the intensity of light. Extending the same concept of electron wave motion, the square of the wave function, φ² may be taken as the intensity of electrons at any point.

In other words, φ² determines the probability density. Thus, φ² has been called the probability density and φ the probability amplitude. Thus, the solutions of the Schrodinger wave equation replace the discrete energy levels or orbits proposed by Bohr and led to the concept of the most probable regions in space in terms of φ2².

A large value of φ² means a high probability of finding the electron at that place and a small value of φ² means low probability. If φ² is almost zero at a particular point, it means that the probability of finding the electron at that point is negligible. Therefore, the wave mechanics approach gives meaningful wave functions which describe the position and energy levels of electrons in an atom.

Concept of Orbital: An orbital is a region in space around the nucleus where the probability of finding the electrons is maximum.

Structure of Atom Chemistry Important Extra Questions Numerical Problems

Question 1.
How many nucleons are present in an atom Nobelium, No? How many electrons are present in the atom? How many nucleons may be considered neutrons?
Answer:
Nucleons = 254, electrons = 102 and neutrons 254 – 102 = 152.

Question 2.
Complete the following table:


Answer:

Question 3.
Complete the following table:

Answer:

Question 4.
Find the number of protons, electrons an-d neutrons in
(a) ₁₃²⁷A³⁺
Answer:
p = 13, ē = 10, n = 14

(b) ₈¹⁵O^2-
Answer:
p = 8, ē = 10, n = 7.

Question 5.
Name the element whose atomic nucleus does not contain any neutrons.
Answer:
The nucleus of the hydrogen atom does not contain any neutron.

Question 6.
Calculate the total no. of electrons present in one mole of methane.
Answer:
1 Molecule of methane (CH₄) contains electrons = 6 + 4 = 10
1 Mole, i.e., 6.022 × 10²³ molecules will contain electrons = 6.022 × 10²⁴

Question 7.
Find (a) the total number and
(b) the total mass of neutrons in 7 mg of 14C
(Assume that the mass of a neutron = 1.675 × 10^-27 kg).
Answer:
1 mol of 14C = 6,022 × 10²³ atoms
i. e. 14 g of 14C = 6.022 × 10²³ atoms

∴ 14 g of it = 6.022 × 10²³ × 0.008 neutrons [∵ one atom ofuC contains = 14 – 6 = 8 neutrons]
and now 0.007 g of it will have
= 6.022 × 10²³ × 8 × 0.007/13
= 2.409 × 10²¹ neutrons .

(b) Mass of 1 neutrons = 1.675 × 10^-27 kg
∴ Mass of 2.409 × 10²¹ neutrons = 4.035 × 10^-6 kg.

Question 8.
A particular radio station broadcasts at a frequency of 1120 kHz (kilohertz). Another radio station broadcasts at a frequency of 98.7 MHz (megahertz). What are the wavelengths of the radiations from each station?
Answer:

Question 9.
Calculate the wave number of radiations having a frequency of 4 × 10¹⁴ Hz.
Answer:
v = 4 × 10¹⁴ Hz = 4 × 10¹⁴ sec^-1
Wave lemgth v̅ = \(\frac{1}{\lambda}=\frac{v}{c}\)
= \(\frac{4 \times 10^{14} \sec ^{-1}}{3 \times 10^{8} \mathrm{msec}^{-1}}\)
= 1.33 × 10⁶ m^-1
= 1.33 × 10⁴ cm^-1.

Question 10.
A photon of wave length 4 × 10^-7 m strikes on metal surface, the work function of the metal being 2.13 eV.
Calculate
(i) the energy of the photon (eV)
Answer:
Energy of the photon = E = hv

(ii) the kinetic energy of the emission and
Answer:
Kinetic energy of emission = \(\frac{1}{2}\) mv² = hv – hv₀
= 3.10 – 2.13 = 0.97 eV

(iii) the velocity of the photoelectron [leV = 1.602 × 10^-19 J].
Answer:

Question 11.
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. Calculate the Ionisation energy of solution in kJ mol^-1
Answer:

Question 12.
The wavelength of the first line in the Balmer Series is 656 nm. Calculate the wavelength of the second line and limiting line in the Balmer series.
Answer:
According to Rydberg’s formula

For the Balmer series n₁ = 2 and for the first line n₂ = 3

Divide (i) by (ii)

Question 13.
Calculate the wavelength of the spherical line in the Lyman series corresponding to n₂ = 3.
Answer:
For the Lyman series n₁ = 1
∴ v̅ = R\(\left[\frac{1}{1^{2}}-\frac{1}{3^{2}}\right]\)
= 109577 × \(\frac{8}{9}\) = 97490.7 m^-1
λ = \(\frac{1}{v}=\frac{1}{97490.7}\) cm^-1
= 102.6 × 10^-7cm
= 102.6 nm

Question 14.
Calculate the velocity of electrons in the first Bohr orbit of the hydrogen atom. Given that Bohr’s radius = 0,529 A. Planck’s constant h = 6.626 × 10^-34 Js mass of electron = 9.11 × 10^-31 kg and 1 J = 1 kg m² s^-1.
Answer:

Question 15.
The electron energy in a hydrogen atom is given by EH = (- 2.18 × 10^-18)/n² J. Calculate the energy required to, remove the electron completely from n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:

Question 16.
Give the values of the quantum numbers for the electron with the highest energy in the sodium atom.
Answer:
Electronic configuration of riNa = 1s², 2s²2p⁶ 3s¹
The electron with the highest energy is 3s¹ for which n = 3 and l = 0

Question 17.
Find the number of unpaired electrons in Fe²⁺ and Fe³⁺. At.No. of Fe = 26.
Answer:
Electronic configuration of 26Fe = 1s², 2s²2p⁶, 3s² 3p⁶ 3d⁶ 4s²
Fe²⁺ = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁶ It has 4 unpaired electrons
Fe³⁺ = 1s², 1s² 2p⁶, 3s² 3p⁶ 3d⁵ It has 5 unpaired electrons.

Question 18.
What atoms are indicated by the following electronic cofigurations?
(i) 1s², 2s²2p¹,
Answer:
Total no. of electrons = 2 + 2 + 1 = 5
∴ At. no. of element is 5.
Hence the element is Boron (B).

(ii) [Ar] 4s² 3d¹
Answer:
Total no. of electrons in [Ar] 4s² 3d¹ = 18 + 2 + 1 = 21
∴ At. no. of element is 21.
Hence the element is Scandium (Sc).

Question 19.
Give the electronic configuration of N^-3, K⁺, P^-3, O^2-.
Answer:

• N = 7 = 1s²,2s² 2p³
• N^3- = 10e = 1s², 2s² 2p⁶
• K = 19 = 1s², 2s² 2p⁶, 3s² 3p⁶ 4s¹
• K+ = 18e = 1s², 2s² 2p⁶, 3s² 3p⁶
• P = 15 = 1s², 2s² 2p⁶, 3s² 3p³
• p^3- = 18e^– = 1s², 2s² 2p⁶, 3s² 3p⁶
• O = 8 = 1s², 2s² 2p⁴
• O^2- = 10e^– = 1s², 2s² 2p⁶

Question 20.
Which out of Cu²⁺, Fe²⁺, Cr³⁺ has the highest paramagnetism and why?
Answer:
Paramagnetism is a property of unpaired electrons. More the no. of unpaired electrons, more the paramagnetism.
Cu²⁺ = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁹
∴ It has only one unpaired electron.
Fe²⁺ = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁶

∴ It has 4 unpaired electrons.
Cr³⁺ = 1s², 2s² 2p⁶, 3s² 3p⁶ 3d³

∴ It has 3 unpaired electrons.
Hence Fe²⁺ with 4 unpaired electrons is having the highest paramagnetism.

Question 24.
Find the velocity (in ms^-1) of the electron in the first Bohr orbit of radius a0. Also, find the de Broglie wavelength in m. Find the orbital angular momentum of 2p orbital of hydrogen in units of \(\frac{h}{2 \pi}\)
Answer:
For H-like particle, velocity in the nth orbit
V_n = 2.188 × 10⁶ × \(\frac{Z}{n}\) ms^-1

For H-atom, Z = 1; For 1st orbit n = 1
∴ v = 2.188 × 10⁶ ms^-1

∴ de Broglie wavelength λ = \(\frac{h}{mv}\)
= \(\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.188 \times 10^{6}}\) m
= 3.33 × 10^-10 m

∴ Orbital angular momentum = \(\sqrt{l(l+1)} \frac{h}{2 \pi}\)
For 2p orbital, l = 1
∴ Orbital angular momentum = \(\sqrt{l(l+1)} \frac{h}{2 \pi}\) = –\(\sqrt{2} \frac{h}{2 \pi}\) .

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 12 Mineral Nutrition. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 12 Important Extra Questions Mineral Nutrition

Mineral Nutrition Important Extra Questions Very Short Answer Type

Question 1.
What is tank farming?
Answer:
It is growing plants in water or solution culture.

Question 2.
Name any nitrogen-fixing symbiotic bacteria.
Answer:
Rhizobium.

Question 3.
What is necrosis?
Answer:
The death of tissues and cells and usually results in holes in the leaves is called necrosis.

Question 4.
Name the bacteria which convert ammonia into nitrite.
Answer:
Nitrosomonas.

Question 5.
What is the major role of calcium?
Answer:
It is a constituent of calcium pectate of the middle lamella of a cell wall.

Question 6.
What is chlorosis?
Answer:
Yellowing of leaves in a distinctive pattern due to lack of one or two other elements is called chlorosis.

Question 7.
From where do the plants get hydrogen?
Answer:
From the water absorbed by the plants.

Question 8.
What are hunger signs?
Answer:
Morphological abnormalities caused due to the deficiency of one or- the other essential elements.

Question 9.
What is premature abscission?
Answer:
Fall of leaves, flowers, or fruits before their maturation is called premature abscission.

Question 10.
Name two micronutrients.
Answer:
Boron, Copper,

Question 11.
Define hydroponics.
Answer:
The technique of growing plants in a nutrient solution is known as hydroponics.

Question 12.
What is the toxicity of micronutrients?
Answer:
Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10 percent is considered toxic.

Question 13.
What is nitrogen?
Answer:
Nitrogen is a constituent of amino acids, proteins, hormones, chlorophylls, and many vitamins.

Question 14.
Define nitrogen fixation.
Answer:
The process of conversion of nitrogen (N ) to ammonia is termed nitrogen fixation.

Question 15.
Define the biological nitrogen fixation.
Answer:
The reduction of nitrogen to ammonia by living organisms is called biological nitrogen fixation.

Question 16.
Write the two deficiency symptoms of molybdenum.
Answer:

1. Premature falling of flowers.
2. Low levels of as carbic acid arid reduced organic nitrogen content.

Mineral Nutrition Important Extra Questions Short Answer Type

Question 1.
What are chelators or chelating agents?
Answer:
These are usually organic chemicals that hold or bind iron in the form of soluble complexes to make available iron to the plant. The chelator itself is not taken up by the plant. EDAA (Ethylene Diaminotetra Acetic Acid) is a commonly used chelator in water culture experiments.

Question 2.
What type of condition is created by leghaemoglobin in the root nodules of the legume?
Answer:
Leghaemolobin is an oxygen scavenger, it creates anaerobic conditions in the cells of root nodules and protects the nitrogen-fixing enzyme nitrogenase of the bacteroids.

Question 3.
What are micronutrients? Give examples.
Answer:
The elements that are required by the plants in fewer amounts or traces are called micronutrients e.g. B(Boron). Mo (Molybdenum). Mn (Manganese). Cl (chlorine). These elements are present in plant tissues. They mostly act as cofactor or activator of enzymes.

Question 4.
Some bags of fertilizers are labeled 15 – 15 – 15. What does it mean?
Answer:
The number 15 – 15 – 15 on the bags indicates the percentage by weight of nitrogen, phosphorus, and potassium in the chemical fertilizer. The majority of the fertilizers contain these elements in bulk such fertilizers are called complete fertilizers. Common fertilizers consist of chemicals either singly or in various compositions like urea, nitrate of soda, ammonium sulfate, etc.

Question 5.
What do you mean by mineral nutrition?
Answer:
The utilization of minerals by plants for growth and development is called mineral nutrition. The minerals are obtained from the soil for their growth. Plant analysis reveals the presence of a large number of minerals and nutrients in the soil. The number and amount of the mineral elements present varies from plants to plant/

Question 6.
Differentiate between micronutrients and macronutrients:
Answer:
Difference between micronutrients and macronutrients:

Micronutrients	Macronutrients
1. These are present in plant tissues in larger concentrations and are known as major elements.	1. These are present in plant tissues in relatively smaller concentrations and are thus known as trace elements.
2. Their concentration in plant tissues is 1mg or more per gram of dry matter of plant.	2. Their Concentration in plant tissues is less than 1mg per gram of dry matter of plant.
3. These do not have any toxic effect if present in slight excess.	3. These produce toxic effect if present m slight excess.
4. These act as structural constituents of cell organs and reserve food materials or functional biomolecules.	4. There are mostly involved in the functioning of enzymes, either by acting as cofactor or activators of enzymes.

Question 7.
Mention symptoms of any four mineral deficiencies in plants.
Answer:

1. Chlorosis: Nondevelopment or loss of chlorophyll that leads to yellowing the entire leaf or part of it is termed chlorosis.
2. Mottling: It is the appearance of patches of green and non-green areas on leaves.
3. Necrosis: It is the localized death of tissue of the leaf.
4. Curling: It is caused due to unequal growth of the leaf.

Question 8.
Write a short note on Industrial Apological Nitrogen fixation.
Answer:
Ammonia is produced industrially by a direct combination of nitrogen and hydrogen (obtained from H₂O) at high temperatures and pressure. Subsequently, NH₃ is converted into various types of fertilizers such as urea, potash, etc which are used for plant growth and protein synthesis. ’

Question 9.
What is ion-exchange absorption?
Answer:
Ions, both cations and anions, have a tendency to get absorbed on the surfaces of the cell walls and exchange with ions present in the soil solution. The process of exchange between absorbed ions and ions in solution is known as ion exchange.

Mode of Iron- absorption carbonic acid exchange

Question 10.
What are the deficiency symptoms of calcium?
Answer:

1. Termination of meristematic activity due to the disintegration of meristematic tissues.
2. Chlorosis on margins of leaves followed by curling and necrosis.
3. Thickening of roots.
4. Premature fall of flowers reduced seed formation.
5. Blossom end rot disease in tomato.

Question 11.
Give three criteria of essentiality of an element.
Answer:
The elements which are required for plant growth and to complete their life cycle, are known as essential elements.

The criteria for the essentiality of an element are given below:

1. The element is necessary for growth and reproduction.
2. The element is required specifically and not replaced by another element.
3. The element must be involved in the metabolism of the plant.

Out of 105 elements only 20 or so have been found essential for plant growth and metabolism.

Question 12.
Write an explanatory note on the Biological fixation of nitrogen.
Answer:
Biological nitrogen fixation is the conversion of molecular nitrogen into nitrogenous compounds through living organisms.

Nitrogen-fixing organisms are of two types:

1. Symbiotic and
2. Non-Symbiotic.

Symbiotic nitrogen-fixing bacteria or Rhizobium Leguminaris occurs in the roots of leguminous plants such as gram. lea. soybean, groundnut, and various pules. It grows to form root nodules containing a pigment leghaemoglobin. It contains enzymes that catalyze biological nitrogen fixation.

Molecular nitrogen is converted into an ammonia molecule in 3 steps. Each step of reduction requires 5 ATP molecules. Ammonia is not liberated but combines with the keto group to form amino acids.

Representation of Biological Nitrogen Fixation

Question 13.
Write short notes on reductive amination and transmutation.
Answer:
Reductive amination: In this process, ammonia reacts with a ketoglutaric acid and forms glutamic acid as indicated below a ketoglutaric acid+.
NH₄⁺ + NAD (P) H — Glutamata + H₂O + NAD (P)

Transmination: It involves the transfer of amino groups from one amino acid. Glutamic acid is the main amino acid from which other 17 amino acids are formed through transamination. The enzyme responsible for such a reaction is termed transaminase.

Question 14.
Complete the following table.

Nutrient	Deficiency symptoms
1. Nitrogen
2. Sulphur
3. Phosphorus
4. Iron
5. Manganese
6. Magnesium
7. Copper

Answer:

Nutrient	Deficiency symptoms
1. Nitrogen	Stunted growth, yellowing of leaves.
2. Sulphur	Chlorosis, stunted growth, and premature leaf fall.
3. Phosphorus	Stunted growth.
4. Iron	Chlorosis, decrease in meristematic activity.
5. Manganese	Chlorosis, Mottling.
6. Magnesium	Chlorosis, stunted growth.
7. Copper	Necrosis, dieback of shoots.

Question 15.
Write two different micronutrients and Macronutrient.
Answer:

Micronutrient	Macronutrient
1. Needed in very small quantity.	1. Needed in large quantity.
2. Needed for functional compounds	2. Needed both for structural and functional compounds.

Question 16.
What is active mineral uptake?
Answer:
The absorption of minerals by the roots at the expense of metabolic energy against the concentration gradient is called active uptake. Certain carriers are active in the cell membrane that takes the minerals from the outside to the inside of the cell.

Question 17.
What is the role of endodermis in roots?
Answer:
In roots, the endodermis is well marked due to the presence of Casparian strips on its anticlinal wall. It prevents the plasmolysis of cells of endodermis and allows the soil water to pass through.

Question 18.
Write a brief note on mineral nutrition in plants.
Answer:
Plants have an autotrophic mode of nutrition. There synthesize organic matter by utilizing raw materials from outside. All inorganic materials except carbon, hydrogen, and oxygen are obtained from the soil. The source of these inorganic materials is the minerals present in the soil.

The process of absorption, distribution, and utilization of these mineral substances by the plants for their growth and development is termed mineral nutrition. Woodward (1699) was the first person to show that plants obtain minerals from the soil for their growth and development.

These findings were confirmed by De Saussure (1804). Plant analysis reveals the presence of a large number of mineral nutrients in the soil. The number and amount of mineral elements present vary from plant to plant. In the same plant, these parameters differ from place to place. Plants absorb all mineral elements present in the soil whether required for their growth or not.

Mineral Nutrition Important Extra Questions Long Answer Type

Question 1.
What is water culture? How will you determine the essentiality of mineral elements experimentally?
Answer:
The technique of growing the plants by placing their roots in different nutrient solutions instead of growing in soil is termed hydroponics or water culture. For determining the essence of the mineral element, seedlings are grown in a balanced nutrient solution, are taken as control solution, lacking one or another element.

The growth of the plants grown in containers containing nutrient solution deficient in one or the other mineral element is compared to that of seedling grown in the balanced nutrient solution. If the plant shows some deficiency symptoms, it implies that the mineral element which was lacking in that culture set is an essential element.

However, the mineral element is a non-essential element if by eliminating that element from the nutrient solution, the growth is comparable to that of the control set.

An experiment set up to determine the essentiality of minerals by water-culture technique.

Question 2.
What is the role of micronutrients in the plant’s life? What are the deficiencies symptoms of iron?
Answer:

Micronutrient	Role and Function
1. Boron (B)	Pectin formation in cell wall translocation of sugar and absorption of water.
2. Molybedenurn (Mo)	Reduction of nitrates constituent of nitrate reductase and play important role in photophosphorylation.
3. Manganese (Mn)	Helps in nitrogen metabolism. chlorophyll synthesis and activation of enzymes.
4. Copper (Cu)	It is the main component of enzymes and plastocyanin.
5. Chloride (Cl.)	Chloride helps in the transfer of electrons.
6. Zinc (Zn)	Synthesis of auxins and acts as an activator,

Deficiency symptoms of iron (Fe):

1. The appearance of interveinal chlorosis, symptoms appearing first in younger leaves. Veins remain green.
2. Chlorosis is followed by necrosis.
3. Reduction in meristematic activity.
4. Retardation of growth.

Question 3.
Make a list of macronutrients and mention their major function.
Answer:
The macronutrients are carbon, hydrogen. oxygen, nitrogen. phosphorus. sulfur, potassium. calcium, magnesium, and silicon.
Carbon: It regulates the metabolic activities required by meristematic and differentiatiñg tissues.

Carbon, hydrogen, and oxygen: These elements are absolutely essential for plant growth. These enter into all chemical compositions of all types of organic compounds like carbohydrates, proteins, lipids, organic acids, amino acids. enzymes, nucleic acids, hormones, etc. These are protoplasmic and formwork elements.

Nitrogen: Nitrogen is essential for all metabolic activities as various biochemical reactions occur in presence of enzymes, It plays an important role in cell division, vegetative, and reproduction growth.

Phosphorus: It is the structural component of nucleic acids. Phospho-lipids, nucleoproteins, ATP, NADP+, sugar phosphates, and a number of co-enzymes. Phosphorus plays an indispensable role in energy, metabolism. It plays an active role in metabolic processes like photosynthesis, respiration, and protein synthesis.

Potassium: It is essential for the functioning of a large number of enzymes taking part in different metabolic activities like photosynthesis, respiration, starch synthesis, synthesis of nucleic acids. It controls the closing and opening of stomata.

Calcium: It is essential for the control of carbohydrate metabolism. It plays some role in binding nucleic acids and proteins in chromosomes.

Magnesium: It is essential for binding together two subunits of ribosomes. It is essential for fat metabolism, carbohydrate metabolism. It is also an activator of enzymes involved in the synthesis of nucleic acids.

Silicon: It plays an important role in the Biological activities of the plants.

Question 4.
Define the following:
(i) Nutrients,
Answer:
Nutrients: The chemical substances used by living organisms as raw materials for metabolic activities are termed nutrients.

(ii) Nutrition,
Answer:
Nutrition: The uptake and utilization of both inorganic and organic raw materials by a living organism for their growth, various metabolic activities, and development is called nutrition.

(iii) Micronutrients
Answer:
Micronutrients: Micronutrients are the essential elements present in plant tissues in relatively lesser amounts i.e. less than 1 mg per gram of dry matter. These mostly act as cofactor or activator of enzymes. These are iron, copper, zinc, manganese, molybdenum, boron, and chlorine.

(iv) Macronutrients,
Answer:
Macronutrients: Macronutrients are the essential elements present in plant tissues in relatively larger concentrations, i.e. at least 1 mg per gram of dry matter. These are carbon, hydrogen, oxygen, nitrogen, sulfur, phosphorus, calcium, magnesium, and potassium.

(v) Active absorption,
Answer:
Active absorption: It is observed that the concentration of K’ions in vacuolar sap was found to be 1000 times more than the pond water. This can occur by utilization of metabolic energy only. The absorption of minerals by the plant against the concentration gradient involving the expenditure of energy is termed active absorption. Inactive absorption, the minerals move from the soil water from low concentration to higher concentration within the cell.

(vi) Passive absorption,
Answer:
Passive absorption: Passive absorption is the absorption of minerals by physical processes not involving the direct expenditure of metabolic energy. A substance moves passively from higher concentration to lower concentration. Ions can also be absorbed and accumulated against an F.CP (Electro Chemical Potential) gradient without the use of metabolic activities. Several theories have been proposed to explain the movement of ions such as ion exchange. Donnan equilibrium and mass flow of ions.

(vii) Symplastic movement and
Answer:
Symplastic Movement: It is the type of movement in which, ions entering the cell wall of the epidermis move across the cell wall of the cortex, cytoplasm of endodermis, the cell wall of the pericycle, and finally in the xylem vessels.

(viii) Apoplastic movement.
Answer:
In apoplastic transport, water and minerals flow in an upward direction via the apoplast to the xylem in the root. The concentration of solutes transported in aboveground organs is established through a combination of import from the xylem, absorption by cells, and export by the phloem

Question 5.
What do you understand by heterotrophic mode of nutrition? Elaborate your answer with suitable examples.
Answer:
This is the type of nutrition, in which organisms obtain readymade organic food materials from some other source and are not capable of synthesizing these from inorganic raw materials of their own. The organ¬isms, which show this type of mode is known as Heterotrophs.

The heterotrophs are divided into two main types:

1. parasites
2. saprophytes.

Parasites obtain readymade organic food material from other living plants or animals. The plant or animal which provides food to the parasite is termed the host. Many bacteria and fungi are parasites. They cause various diseases in their hosts.

Total Stemparasite

Some flowering plants also show parasitic modes of nutrition. These plants send haustorial or parasitic roots into the host to draw nutrients from it. Depending upon the organ of the host on which parasite is attached, it may be a stem parasite or root parasite.

Saprophytic plants such as Morotropa, bacteria, fungi grow on decaying animal and vegetable matter and absorb the organic food from it.

Heterotrophic plants could be symbiotic and insectivorous also.

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 11 Transport in Plants. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 11 Important Extra Questions Transport in Plants

Transport in Plants Important Extra Questions Very Short Answer Type

Question 1.
Define heat of wetting or hydration.
Answer:
Imbibition of water is always associated with heat generation is called wetting or hydration.

Question 2.
What phenomenon is always associated with imbibition?
Answer:
Heating of wetting or hydration.

Question 3.
Which solution de-plasmolyse the plasmolyse solution?
Answer:
Hypotonic solution.

Question 4.
What is the full form of D.P.D.?
Answer:
Diffusion pressure deficit.

Question 5.
Name the element that regulates turgidity in guard cells.
Answer:
Potassium.

Question 6.
Which type of guard cells are found in grasses?
Answer:
Dumb bell-shaped.

Question 7.
Which is the important factor that affects water potential?
Answer:
Solute concentration.

Question 8.
Why do plants growing in arid regions bear small leaves with sunken stomata?
Answer:
To reduce transpiration.

Question 9.
Name two antitranspirants.
Answer:
Abscisic acid (ABA), Phenyl Mercuric Acetate (PMA).

Question 10.
When does wilting occur?
Answer:
When the rate of evaporation of water exceeds the rate of uptake of water by roots.

Question 11.
What is water potential?
Answer:
The capacity of a solution to hold the maximum amount of water is called water potential.

Question 12.
What will be the water potential of distilled water at normal pressure and temperature?
Answer:
Zero.

Question 13.
What are the porins?
Answer:
The porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria allowing molecules up to the size of small proteins to pass through.

Question 14.
Define the water potential.
Answer:
Water potential is a concept fundamental to understanding water movement.

Question 15.
What is antiport?
Answer:
In an antiport, both molecules move in opposite directions.

Question 16.
What is flaccid?
Answer:
When the water flows into the cell and out of the cell are in equilibrium the cells are said to be flaccid.

Question 17.
Define isotonic.
Answer:
If the external solution balances tire osmotic pressure cytoplasm it is said to be isotonic.

Question 18.
Define translocation.
Answer:
The bulk movement of substances through the conducting or vascular tissues of plants is called translocation.

Question 19.
Define Casparian strip.
Answer:
The inner boundary of the cortex, the endodermis, is impervious to water because of a bond of suberised matrix called the Casparian strip.

Question 20.
Define cohesion.
Answer:
The mutual attraction between water molecules.

Question 21.
What is transpiration?
Answer:
Transpiration is the evaporative loss of water by the plants. It occurs mainly through the stomata in the leaves.

Transport in Plants Important Extra Questions Short Answer Type

Question 1.
What is osmosis?
Answer:
It is the movement of water molecules from a region of its higher concentration to low concentration through the plasma membrane. It is a vital process. Various physiological process in plants takes through osmosis.

Question 2.
What is the active transport of water?
Answer:
It is the transport of water molecules against the concentration gradient with the utilization of energy (ATP). The molecules from low to high concentration move through the active transport of water.

Question 3.
How does transpiration differ from guttation?
Answer:

Transpiration	Guttation
1. It is the process of loss of water in the form of water vapour from the aerial part of the plant body.	1. It is the process of elimination of water in the form of water droplets from leaf margins.
2. It mostly takes place through stomata.	2. It takes place only through hydathodes.

Question 4.
Define permanent wilting co-efficient or permanent wilting percentage.
Answer:
The percentage of water on a dry weight basis of the soil that still remains at the time when the plant shows permanent wilting is termed as permanent wilting co-efficient. It various between 1-1.5% depending upon the texture of the soil. It is higher in clayey soil than sandy soil.

Question 5.
What are the conditions for imbibition to take place?
Answer:
There are two conditions necessary for imbibition to take place:

1. Water potential gradient between the surface of the adsorbent and the liquid imbibed.
2. The affinity between the adsorbent and the imbibed liquid. It is a type of diffusion by which the movement of water takes place.

Question 6.
Give an account of the water relations of a plant cell when it is placed in
(a) Hypertonic solution,
Answer:
When the cell is placed in hypertonic solution exosmosis occurs, the protoplast contracts and the cell membrane detaches from the cell wall and contracted protoplast. This contraction of the protoplast by ex¬osmosis is termed plasmolysis.

(b) Hypotonic solution.
Answer:
When a plasmolysed cell is placed in water or hypotonic solution, endosmosis occurs and the protoplast regains the original position. This phenomenon is termed de-plasmolysis.

Question 7.
What is the ascent of sap?
Answer:
Water and minerals absorbed by the plants through roots from the soil are transported to different parts of the plant as these play a vital role in their growth. Water along with various dissolved inorganic substances in it are termed sap. The upward translocation of sap (water and dissolved inorganic substances) from the roots to the aerial parts of the plant is termed ascent of sap.

Question 8.
What are the advantages of transpiration?
Answer:

1. Ascent of sap: Transpiration pull created in leaves is responsible for ascent of sap.
2. Absorption of water: Transpiration pull is also responsible for passive absorption of water.
3. The distribution of minerals in different parts of the plant is done by transpiration.
4. Cooling effect: Transpiration lowers the temperature of the leaf and causes a cooling effect.
5. The increased rate of transpiration favours the development of tissue, which provides strength to the plant.
6. Excessive transpiration induces hardness which imparts resistance of plants to drought.

Question 9.
What is the diffusion pressure? What are the factors which affect the rate of diffusion?
Answer:
The pressure exerted by the particles due to their tendency to diffuse from the region of higher concentration to the region of lower concentration.

Various factors affect the rate of diffusion:

• Diffusion pressure gradient
• Temperature
• The density of diffusing substance
• The density of the medium.

Question 10.
Define water holding capacity or field capacity of the soil.
Answer:
After heavy rainfall or irrigation, the amount of water actually retained by soil even against the force of gravity is termed as water holding capacity or field capacity of the soil. It is expressed in terms of the percentage of water present per unit dry weight of soil.

Question 11.
Describe osmosis as a special case of diffusion.
Answer:
Uptake and distribution of water, solutes, and gases occur in a plant as a result of diffusion. The diffusion of water through a semipermeable membrane is known as osmosis. The diffusion of water molecules continues across the membrane until an equilibrium is attained. Osmosis can be demonstrated by a simple experiment as follows:

A thistle funnel is taken and tied with a semipermeable membrane (parchment paper) to the wide mouth of the thistle funnel and made tight.

The thistle funnel is filled with concentrated sugar solution and its wide mouth is dipped into water contained in a beaker. The membrane allows water molecules to pass through and not the sugar molecules. The level of sugar solution will rise in the funnel from ‘A’ to ‘B’. This demonstrates osmosis.

A demonstration of osmosis. A thistle funnel is filled with sucrose solution and kept inverted in a beaker containing water,
(a) Water will diffuse across the membrane (as shown by arrows) to raise the level of the solution in the funnel
(b) Pressure can be applied as shown to stop the water movement into the funnel.

Question 12.
What is the factor which affects the rate of transpiration?
Answer:
There are two factors.
A. External (Environmental) factors.
B. Internal (Living factors).

A. External factors:

1. Light: Causes stomatal opening
2. Temperature: High temperature decreases relative humidity increasing transpiration.
3. Humidity: It directly affects the rate of transpiration that is related to the vapour pressure of the atmosphere.
4. Wind: High velocity of wind causes closure of stomata.
5. Soil moisture: The rate of transpiration is directly proportional to the quantity of available moisture in the soil.

B. Internal factor:

1. Root Shoot Ratio: Roots absorbs water, should transpire water, hence their ratio affects transpiration.
2. Leaf area: Smaller plants tend to transpire more rapidly per unit area than larger plants.
3. Leaf Anatomy: Modification of leaves affects transpiration.

Question 13.
In what way does the concept of water potential help in explaining water movement?
Answer:
The additional pressure (more than osmotic pressure) is applied so that water can be made to flow out of the solution.
There are two factors

1. The concentration of dissolved solutes in a solution,
2. Pressure difference determines the chemical potential of water, which is the driving force for water movement in plants.

This chemical potential is the water potential. It is the difference in the free energy of water molecules in solution and in the pure state at the same temperature and pressure.

Question 14.
Of what importance is potassium in the opening and closing of stomata?
Answer:
The opening and closing of stomata are controlled by the accumulation of solutes in the guard cells. In the guard cell solutes are taken from neighbouring epidermal and mesophyll cells. The major solute which is taken in by the guard cells is potassium.

The rise in potassium level causes the stomatal opening and a decrease in level causes stomatal closing. The uptake of potassium controls the water potential. The extent of K⁺ accumulation in guard cells determines the size of the stomatal opening.

Question 15.
Does transpiration serve any useful function in the plants?
Answer:
Transpiration helps in the movement of xylem sap. It increases the absorption of mineral nutrients by the roots from the soil. Solar radiation absorbed by the leaves is used in photosynthesis but some radiations will cause heating of leaves. Transpiration, however, reduces the heating of the leaves.

Question 16.
Describe the role of osmotic potential in regulating the water potential of plant cells.
Answer:
Osmotic potential is the amount by which the water potential of pure water is reduced by the presence of the solute. The osmotic potential has a negative value.

If we apply additional pressure, the water can be flown out of the solution.

Osmosis is driven by two factors:

1. The concentration of dissolved solutes in solution,
2. Pressure difference.

Water potential is the driving force for water movement in plants. Water potential represents the free energy associated with water. Osmotic potential regulates the flow of water molecules through the membrane.

Question 17.
Explain the facilitated diffusion.
Answer:
In facilitated diffusion special proteins help move substances across membranes without the expenditure of ATP energy. Facilitated diffusion cannot cause net transport of molecules from a low to high concentration-this would require the input of energy.

The transport rate reaches a maximum when all of the protein transporters are being used (saturation). Facilitated diffusion is very specific: it allows the cell to select substances for uptake. It is sensitive to inhibitors that react with protein side chains.

Question 18.
What is active transport? Explain with an appropriate example.
Answer:
Active transport is carried out by membrane proteins. Hence different proteins in the membrane play a major role in both active as well as passive transport. Pumps are proteins that use energy to carry substances across the cell membrane. These pumps can transport substances from a low concentration to a high concentration.

The transport rate reaches a maximum when all the protein transporters are being used or are saturated. Like enzymes, the carrier protein is very specific in what it carries across the membrane. These proteins are sensitive to inhibitors that react with protein side chains.

Question 19.
What is imbibition?
Answer:
Imbibition is a special type of diffusion when water is absorbed by solids-colloids-causing them to enormously increase in volume. The classical examples of imbibition are absorption of water by seeds and by dry wood. The pressure that is produced by the swelling of wood has been used by prehistoric man to split rocks and boulders.

If it were not for the pressure due to imbibition, seedlings would not have been able to emerge out of the soil into the open; they probably would not have been able to establish.

Question 20.
What is turgor pressure? Give its two roles in plants.
Answer:
The pressure exerted by the cell sap on its wall when it has absorbed the maximum amount of water is called turgor pressure.

1. Leaves stand erect and look fresh due to turgor pressure.
2. The movement of soluble food in phloem is due to turgor pressure.

Question 21.
What is guttation?
Answer:
Oozing of droplets along the leaf margin on the vein endings at night is called guttation. In the morning when water evaporates, a layer of salts remains on the leaf which may cause burning.

Guttation occurs through the hydathodes. Education occurs when absorption exceeds transpiration and water pressure builds up in the xylem vessels. It forces the water outward through the hydathode.

Transport in Plants Important Extra Questions Long Answer Type

Question 1.
Describe the theories related to the translocation of water.
Answer:
There are three most important theories related to the translocation of water.

1. Root pressure theory
2. Capillarity
3. Cohesion theory.

1. Root pressure theory: Water flows from higher water potential to low water potential. Water from the soil is absorbed by root hairs and conducted through xylem vessels. Mineral ions from the soil are taken up by roots and get deposited in the xylem vessels.

When the stem of a plant is cut transversely above the soil surface, a drop of the xylem sap will exude from the cut surface. This indicates the presence of positive pressure in the xylem. This pressure is known as Root Pressure.

2. Capillarity: Capillarity means a rise in water in tubes of small diameter kept in a water vessel. The uptake of water through xylem vessel is possible in small size plants through capillarity. This is due to the forces of adhesion and cohesion.

Adhesive forces attract molecules of different kinds whereas cohesive forces attract molecules of the same kind to each other. According to this theory, water is taken due to the force of adhesion and flows upward due to the force of cohesion.

3. Cohesion: This is the most important theory of water movement through plants. It is based on the force of cohesion between water molecules. This sets up a continuous water column from the top to the root tip of the plant. According to this theory water evaporates from the leaf to the atmosphere, results in a decrease in the water potential of epidermal cells.

This loss of water is balanced by water moving from adjacent cells along a water potential gradient. The movement of the water occurs from the soil to the root. Uptake of water is termed as cohesion theory and also known as transpirational pull.

Question 2.
Is there a general mechanism to explain the opening and closing of stomata? Justify your answer.
Answer:
There is no general mechanism to explain the opening and closing of stomata. Because opening and closing of stomata are regulated by the accumulation of solute in the guard cells. Solutes are taken in the guard cells, as a result, osmotic potential and water potential of guard cells are lowered, the guard cells become turgid and swell size, resulting in the stomatal opening. With a decline in guard cell solutes, water moves out, resulting in the stomatal opening.

There are two theories to explain the mechanism of opening and closing of stomata.

1. Classical starch sugar conversion theory: According to this theory, the change in osmotic concentration is brought about due to the conversion of starch into glucose and vice-versa.
2. K+ Influx and Efflux theory: According to this theory when the leaf is exposed to light, the pH of the guard ceils rises due to the active transfer of H+ ions from the cytoplasm into chloroplast’s utilization of CO. in photosynthesis. In the majority of plants, stomata remain open during the day and close at night.

Hence, there is no general mechanism to explain the stomatal opening and closing.

Question 3.
Mention some factors that influence stomatal opening and closing. How are these factors involved in regulating stomatal behaviour?
Answer:
Factors affecting stomatal movements:

1. Light: In most of the plant’s stomata open during the day. The effect of light causes the opening of stomata or it may be either due to the hydrolysis of starch into glucose.
2. The water content of leaves: A decrease of water content in stomatal cells results in an increase in their D.P.D. Water from guard cells moves into these cells and stomata close.
3. CO₂ concentration: Low CO₂ concentration in guard cells causes the opening of stomata.
4. pH: High pH stimulates the opening of stomata and low pH causes closure of stomata and high concentration of CO₂ causes closure of stomata.
5. Temperature: High temperature stimulates the opening of the stomata.
6. Atmospheric Humidity: Humid environment favours opening and dryness causes closure of stomata.
7. Minerals: Minerals like P, Mg, Ca etc. affect the stomatal opening. A high concentration of K+ ions causes the opening of stomata.
8. Growth Hormones: Cytokinins stimulates the opening of stomata. Abscisic acid induces the closure of stomata.

Question 4.
Write short notes on:
(i) Cohesion-Tension and Transpiration pull theory.
Answer:
Transpiration pull theory: Ascent of sap has been explained satisfactorily by Dixon with the help of a theory called Transpiration pull theory. According to this theory water continuously evaporates from the turgid and moist cell walls of mesophyll cells in the leaves.

It makes the mesophyll air saturated. The air outside the leaf is dry. So a gradient is set up which allows the water vapours to go out from the interior of the leaf to the outside through the stomata. The mesophyll cells draw water from the deeper tissue, which in turn take water from the xylem of the leaf. It creates a kind of pull in the leaf called transpiration pull.

The xylem of the leaf is connected to the xylem of the stem and further to the xylem of the roots. Since there is a continuous column of water in the plant, water is virtually lifted up due to transpiration pull a situation similar to one like drawing a bucket of water from a well. The column of water does not break because of the great force of cohesion among the water molecules. This theory is also called the cohesion of water molecules theory.

(ii) Mass flow hypothesis.
Answer:
Mass flow hypothesis: The carbohydrates prepared in the leaves are translocated to other parts of the plant in the form of sucrose through phloem at the expense of metabolic energy. Munch’s mass flow hypothesis is the most accepted theory for the translocation of organic food.

According to this hypothesis, organic substances move from the region of high osmotic pressure to the region of low osmotic pressure due to the development of a gradient of turgor pressure. This can be proved by taking two interconnected osmometers. One of the osmometers has a high solute cone than the other. The whole apparatus is placed in water.

Water enters the osmometer with a high solute cone. It creates high turgor pressure in it. High turgor pressure forces the solution to move through the tube to the other osmometer. It is called mass flow. If somehow, the solute is continuously added to the donor osmometer and converted into the osmotically inactive compound in the other osmometer, this system can work indefinitely.

Munch’s mass flow apparatus.

By going through these CBSE Class 11 Maths Notes Chapter 16 Probability Class 11 Notes, students can recall all the concepts quickly.

Probability Notes Class 11 Maths Chapter 16

Coin, Die and Playing cards
(i) Coin : It has two sides, viz. head and tail. If we have more than one coin, coins are regarded as distinct, if not otherwise stated.

(ii) Die : A die has six faces marked 1, 2, 3, 4, 5 and 6. If we have more than one die, all dice are regarded as different, if not otherwise stated.

(iii) Playing cards : Its pack has 52 cards. There are four suits, viz., spade, heart, diamond and club, each having 13 cards. There are two colours red (heart and diamond) and black (spade and club) each having 26 cards. In 13 cards of each suit, there are 3 face cards viz. king, queen and jack. So, there are in all 12 face cards in a pack of playing cards. Also, there are 16 honour cards, 4 of each suit viz., ace, king, queen and jack.

The set of all possible outcomes of an experiment is called the sample space of that experiment. It is usually denoted by S. The elements of S are called sample points and the subset of S is called an event.

Note : Elements of sample space are known as sample points.

If an experiment conducted repeatedly under the identical conditions does not give necessarily the same result, then the experiment is called random experiment. The result of the experiment is called outcome.

Different types of events :

1. Simple event: If an event has only one sample point of the sample space, it is called a simple (or elementary) event.
2. Compound event : When an event is composed of a number of simple events, then it is called a compound event.
3. Null event: An event having no sample point is called null event. It is denoted by Φ. It is also known as impossible event.
4. Sure event: The event which is certain to occur is said to be the sure event.
5. Equally likely events : Events are called equally likely, when we do not expect the happening of one event in preference to the other.
6. Mutually exclusive events: A set of events is said to be mutually exclusive, if the happening of one event A excludes the happening of the other event B, i.e., A ∩ B = Φ
7. Exhaustive events : A set of events is said to be exhaustive, if the performance of the experiment always results in the occurrence of at least one of them.

If A is an event of an experiment whose sample space is S, then its probability P(A) is given by
P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{\text { Number of favourable cases }}{\text { Total number of exhaustive cases }}\)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B), where A and B are any two events.

P(A ∪ B) = P(A) + P(B), where A and B are mutually exclusive events.

0 ≤ P(A) ≤ 1, for an event A.

P(\(\overline{\mathrm{A}}\)) = 1 – P(A), where P(\(\overline{\mathrm{A}}\)) denotes the probability of not happening the event A.

P(A ∪ \(\overline{\mathrm{A}}\)) = P(S) = 1.

If the event A implies the event B, then P(A) ≤ P(B).

P(A ∪ B ∪ C) = PC A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C).

For any two events A and B,
(i) P(A ∩ \(\overline{\mathrm{B}}\)) = P(A) – P(A ∩ B)
(ii) P(\(\overline{\mathrm{A}}\) ∩ B) = P(B) – P(A ∩ B).

Two events A and B associated with the same random experiment are independent, if and only if, P(AB) = P(A).P(B).

Let p₁ p₂, …, p_n be the probabilities of r independent events A₁ A₂,…, A_n respectively. Then, the probability that at least one of r events happen = 1 – (1 -P₁)(1 – p₂) (1 – p₃)… (1 – P_r)
= \(\frac{\text { No. of favourable cases }}{\text { No. of cases which are not favourable }}\)

Odds in favour of an event
= \(\frac{\text { No. of cases which are not favourable }}{\text { No. of cases which are favourable }}\)

By going through these CBSE Class 11 Maths Notes Chapter 14 Mathematical Reasoning Class 11 Notes, students can recall all the concepts quickly.

Mathematical Reasoning Notes Class 11 Maths Chapter 14

Statement : A sentence is called mathematically acceptable statement, if it is either true of false but not both.

Negation of a Statement: The denial of a statement is called negation of the statement e.g.
If p : Diagonals of a rectangle are equal.

So, ~ p : Diagonals of a rectangle are not equal. This may also be written as
~ p : It is false that diagonals of a rectangle are equal. Further, it may also be written as
~ p : There is at least one rectangle whose diagonals are not equal.

Compound Statement: A compound statement is a statement which is made up of two or more statements.
Each statement is called a component statement.

The Connecting Word “And” : We can connect the two statements by the word “AND”, e.g. ‘
p : 48 is divisible by 4.
q : 48 is divisible by 6.
p and q : 48 is divisible by 4 and 6.

Truth Value of p and q : The statementp and q are both true, otherwise it is false, i.e., it is false, when
(i) p is true and q is false.
(ii) p is false and q is true.
(iii) p is false and q is false.

The Connecting Word “OR” : The statement p and q may be connected by the connecting word ‘OR’: i.e., p or q, e.g.
p : Ice cream is available at dinner.
q : Coffee is available at dinner.
p or q : Ice cream or coffee is available at dinner.

Exclusive “OR” : In a statement p or q, if one of the statements either p or q is true, then the statement is true. Thus, connecting of word “OR” is exclusive.

Inclusive “OR” : In a statement either both are true, then connecting word ‘OR’ is inclusive, e.g.
At plus one level, a student can either opt for Biology or Mathematics or both.
Here, the connecting word “OR” is inclusive.

Truth value of p or q : When p and q statements both are false, then p or q is also false, otherwise it is true.
Thus, p or q is true, when
(i) p is true, q is false.
(ii) p is false, q is true.
(iii.) p and q both are true.

Quantifier “There Exists” : There exists, is used for at least one e.g.
p : There exist, a quadrilateral whose all sides are equal.
The statement is equivalent to
There is at least one quadrilateral whose all sides are equal.

Quantifier “For All” : In mathematical statements “for all” is commonly used e.g. For all n ∈ N implies that each n is a natural rumber.

Implications : “If then”, “only if’ and “if and only if’ are known as imphcations. .
If p then q : The statement if “p then q” says that in the event if p is true, the i q must be true e.g.
‘ I a numbei ” a multiple of 4, then it is a multiple of 2.
Here p : A ni her is a multiple of 4.
and q : The m. aber is a multiple of 2.
When p is true i.e., a number is a multiple of 4, then q is true
i. e., the number is a multiple of 2.

“If p then q” may be used as : l.p implies q : It is denoted by p ⇒ q. The symbol ⇒ stands for implies.
2. p is sufficient condition for q : A number is a multiple of 4 is sufficient to conclude that the number is a multiple of
3. p only if q. The number is a multiple of 4 only if it is a multiple of 2.
4. q is a necessary condition for p. When the number is a multiple of 4, it is necessarily a multiple of 2.
5. – q implies ~ p (~ stands for negation or not)

Converse Statement : The converse of “if p then q” is if q then p e.g.
If a number x is odd, then x² is also odd.
Its converse is : If x² is odd, then x is also odd.

Contrapositive Statement: Contrapositive of “if p then q” is if not q then not p. e.g. contrapositive of
“If a number is divisible by 4, it is divisible by 2”, is
“If a number is not divisible by 2, then it is not divisible by 4”.

Truth Value of “If p then q” : Truth value of the statement “if p then q” is false, when p is true and q is false, otherwise it is true, ue., It is true, when
(i) p is true, q is true.
(it) p is false, q is true.
(iii) p is false, q is false.

However, three methods are adopted to test the truth value of this statement.
(1) Assuming that p is true, prove that q must be true. (Direct method).
(2) Assuming that q is false, prove thatp must be false.
(3) Assuming thatp is true and q is false, obtain a contradiction. (Contradiction method).

Statements with “If and Only If’: “If and only if’, represented by the symbol ⇔, has the following equivalent forms :
(i) p if and only if q.
(it) q if and only p.
(iii) p is necessary and sufficient condition for q and vice-verse.
(iv) p ⇔ q.

Truth Value of “If and only If” : Truth value of statements with “if and only iP is true when p and q are both true or false, otherwise it is false, ue.., statement if and only if is true, when
(i) p is true, q is true.
(ii) p is false, q is false.

Statement with if and one if is false, when
(i) p is true, q is false
(ii) p is false, q is true.

To, Prove A Statement By Contradiction
Example : To prove that √3 is irrational.
Let √3 be rational. So, √3 = p/q, where p and q are co-prime.
p² = 3q² ⇒ 3 divides p or 3 is a factor of p.
Let p = 3k. ∴p² = 9k².
∴ p² = 3q²becomes 9k² = 3q² or 3k² = q².
This means 3 is a factor or q.
i. e., 3 is a factor of p and q both. This is a contradiction.
∴ √3 is irrational.

Counter Example : To prove that a statement is false, we find an example (called counter example) by which the given statement is not valid. Then, we say that the statement is false.

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 8 Cell: The Unit of Life. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 8 Important Extra Questions Cell: The Unit of Life

Cell: The Unit of Life Important Extra Questions Very Short Answer Type

Question 1. Who gave the term chromosome?
Answer:
Waldayer.

Question 2.
Define a cell coat.
Answer:
A clear layer of oligosaccharide outside the cell membrane in some animal cells.

Question 3.
Where is dynein present?
Answer:
In microtubules of flagella.

Question 4.
Define sarcoplasmic reticulum.
Answer:
It is defined as the ER found in striated muscles.

Question 5.
What are pili?
Answer:
Pili are elongated, tubular structure in Gram-ve bacteria.

Question 6.
What are fimbriae?
Answer:
Fimbriae are small, bristle-like fibers sprouting out of the cell in bacteria.

Question 7.
Name the organelle of the cell called ‘Suicidal-bag’?
Answer:
Lysosomes are called suicidal bags of the cell.

Question 8.
Define cytoplasm.
Answer:
The cytoplasm is a jelly fluid of protoplasm composed of inorganic and organic matters containing many organelles like ribosomes, ER, vacuole, etc.

Question 9.
Define plasmodesmata.
Answer:
Plasmodesmata are connections between two cell walls that are interrupted by small pores having fine threads of cytoplasm.

Question 10.
What are the three parts of a flagellum?
Answer:

1. filament
2. hook and
3. basal body.

Question 11.
What is an asymmetric karyotype?
Answer:
A karyotype that shows the larger size, the difference between smaller and larger chromosome of the set, and have few metacentric chromosomes.

Question 12.
What is the origin of the Golgi complex?
Answer:
Smooth endoplasmic reticulum.

Question 13.
What are lysosomal enzymes?
Answer:
Hydrolases.

Question 14.
What are tunnel proteins?
Answer:
Integral proteins of plasma membrane functioning as channels.

Question 15.
Which organelle is called the “engine of the cell”?
Answer:
Ribosomes where protein synthesis occurs.

Question 16.
Define symplasm.?
Answer:
It is a plasmodesma that helps to maintain the continuity of living matter and cytoplasm.

Question 17.
What is desmotubule?
Answer:
A fine cytoplasmic canal lined by a plasma membrane and has ER.

Question 18.
What are permeases?
Answer:
Enzymes facilitate the entry of substances through the plasma membrane.

Question 19.
Who gave the unit membrane concept?
Answer:
Robertson.

Question 20.
Expand PPLO.
Answer:
Pleuro Pneumonia likes organisms.

Question 21.
What holds the ribosome together in a polyribosome?
Answer:
The mRNA chain.

Question 22.
Who discovered the cell?
Answer:
The cell was discovered by Robert Hooke.

Question 23.
What is totipotency?
Answer:
The capacity of every plant cell to develop into a whole plant is totipotency.

Question 24.
Who discovered the nucleus?
Answer:
Nucleus was discovered by Robert Brown in the cells of roots of orchids.

Question 25.
What is tonoplast?
Answer:
Tonoplast is the membrane around vacuole.

Question 26.
What are receptor molecules?
Answer:
Receptor molecules are specific proteins in the cell membrane that enter into the cells like hormones.

Question 27.
What is absent in erythrocytes?
Answer:
Nucleus, Aerobic respiration, DNA.

Question 28.
Who proposed the cell theory.
Answer:
Cell theory was proposed by M.J. Schleiden and Theodore Schwann.

Question 29.
Name two types of cells that retain their mitotic ability but seldom divide.
Answer:
Liver and muscle cells.

Question 30.
Who concluded, “Cells are the ultimate units forming the structure of all plant tissues”?
Answer:
Mathias Jacob Schleiden, a German botanist concluded that the cells were the ultimate units forming the structure of all plant tissues.

Question 31.
What is the contribution of Leeuwenhoek to cell biology?
Answer:
The invention of the microscope.

Question 32.
Name the site for protein synthesis in the cell.
Answer:
Ribosome.

Question 33.
What is differentiation?
Answer:
The process by which cells lose their specialization is called differentiation.

Question 34.
Name some unicellular organisms.
Answer:
Amoeba, Paramecium, Euglena and Acetabularia.

Question 35.
Name two organelles, other than the nucleus, which contain DNA.
Answer:
Answer:
Mitochondria or chloroplast.

Cell: The Unit of Life Important Extra Questions Short Answer Type

Question 1.
Give the fundamental similarities in all cells:
Answer:
Fundamental similarities in all cells are:

1. Hereditary characters are transmitted through nucleic acids.
2. The basic structure of membranes of all cell organelles is the same.
3. Method of aerobic respiration.
4. Mechanism of synthesis of nucleic acids and proteins within the cells.

Question 2.
What are Cytoskeletal Structures?
Answer:
Cytoskeletal Structures: The ability of eukaryotic cells to adopt different types of shapes and to perform directed movement depends on the cytoskeleton.

There are three principal types of protein filaments

1. Microfilaments,
2. Microtubules, and
3. Intermediate filaments.

These constitute the cytoskeleton. The microfilaments are 8 nm in diameter, either scattered or organized into the network or parallel arrays within the matrix. They play a major role in cell motion (changes in shape). Such cellular movements associated with the microfibres are movements of pigment granules, amoeboid movements, and protoplasmic streaming. These microfilaments consist of actin-like proteins.

Question 3.
What are the main functions of the cell wall?
Answer:
The main functions of the cell wall are:

1. It gives a definite shape to the cell and protects the internal organelles.
2. It provides a framework and lends support to the plasma membrane.
3. It prevents the cell from desiccation.
4. It counteracts physically the osmotic pressure produced by the cell contents.
5. It helps in the transport of materials and metabolites in and out of the cell.

Question 4.
List the functions of Golgi bodies.
Answer:
The functions of Golgi bodies are:

1. Storage, condensation, and packaging of the material.
2. Several enzymes are localized in Golgi bodies.
3. During spermatogenesis, Golgi apparatus form the euro some.
4. Mucilage and gums are secreted in plant cells due to the action of the Golgi apparatus.

Question 5.
Name different types of the endoplasmic reticulum.
Answer:
There are two types of EM.

1. Smooth ER (Agranular) and
2. Rough ER.

Question 6.
Describe the functions of the three organelles, viz Golgi bodies, chloroplasts, and mitochondria.
Answer:
(a) Functions of Golgi bodies:

1. Carbohydrate synthesis of mucopolysaccharides
2. Formation of acrosome
3. Formation of the lysosome.
4. Formation of the plasma membrane.
5. Formation of the cell wall.
6. Absorption of compounds.
7. Production of hormones.
8. Formation of pigments.
9. Yolk deposition.

(b) Functions of chloroplast:

1. Their main function is to trap the sun’s energy and to convert it into the chemical energy of food by photosynthesis.
2. Storage of starch,
3. Chloroplasts in fruits and flowers change into chromoplasts.

(c) Functions of Mitochondria:

1. Powerhouses the cell and stores energy as ATP.
2. Several respiratory enzymes are found in mitochondria.
3. DNA is also contained in mitochondria.
4. They regulate the concentration of calcium ions in the cells.

Question 7.
Describe functions of flagella and cilia.
Answer:
Functions of flagella and cilia:

1. Flagella and cilia help the organisms in movement and loco¬motion.
2. They help the organism to swim in the water.
3. Social and flagella are associated with the motility of cells.

Question 8.
Distinguish between:
(a) Microtubules and microfilaments.
Answer:
Differences between Microtubules and Microfilaments:

Points	Microfilaments	Microtubules
(1) Structure	(1) Actin is the main component of microfilament and so they are contractile.	(1) (α) and (β) tubulin proteins are the main components of the microtubules so they are non-contractile.
(2) Diameter	(2) 5 – 6 nm	(2) 25 nm
(3) Sub-units	(3) Absent	(3) 13 protofilaments from a microtubule

(b) Primary wall and secondary wail.
Answer:
Difference between Primary and Secondary cell walls:

Character	Primary cell wall	Secondary cell wall
(1) Location	(1) Primary cell wall is found in plant cells only.	(1) Secondary cell wall is found in nature and non-dividing cells.
(2) Thickness	(2) 1 – 3 pm	(2) 5 – 10 pm
(3) Growth	(3) Growth internal	(3) Growth is accretionary.
(4) Chemical  Structure	(4) It is made of pectin cellulose or hemicellulose	(4) It is made of cellulose, hemicellulose suberin, cutin, or lignin.

(c) Leucoplast and chromoplast.
Answer:
Differences between Leueoplast and Chromoplast:

Leueoplast	Chromoplast
1. Colourless plastids devoid of any pigment.	1. May contain pigments other than chlorophyll.
2. They have the capacity to develop pigment when needed. Leucoplasts are of three types- amyloplast, proteinoplast, and elaioplast.	2. They synthesize and store other pigments such as carotenoids.

Question 9.
Describe the ultrastructure and functions of
(a) nucleus
Answer:
Ultrastructure and functions of Nucleus: Electron microscopic studies reveal that the nucleus is bounded by two membranes, which make the nuclear envelope. The outer and inner membranes are separated by a narrow space, perinuclear space.

The outer membrane remains in continuation with the endoplasmic reticulum (ER) and the inner one surrounds the nuclear contents. At some points, the nuclear envelope is interrupted by the presence of small structures called nuclear pores.

These pores are enclosed by circular pores. These pores are enclosed by circular structures called the annuli. The pores and annuli unite to form the pore complex. These pores help in the exchange of materials between nucleoplasm and cytoplasm. Nuclear membrane disappears during cell.-division. It reappears during nuclear recognization in the telophase stage.

The nucleoplasm contains chromatin and nucleolus The nucleolus is a rounded structure, it is not separated from the rest to the nucleoplasm by a membrane. It is associated with a specific nucleolar organizing region (NOR) of some chromosomes. The nucleolus is the “site for ribosomal RNA synthesis.” The cells which remain engaged in protein synthesis have larger and more numerous nucleoplasm their nucleoplasm.

(b) mitochondrion and
Answer:
Ultrastructure of Mitochondria: They are spherical or elongated or rod-like cell organelle. They are known as the “powerhouse of the cell.” Mitochondria was fust discovered by Hofmeister (1851) in the cells of a pteridophyte. Equisetum ( 1851). They were named ‘mitochondrion’ by Benda (1897).

Mitochondria is a double membranous organelle, the outer membrane is smooth while the space between the two membranes is called the outer chamber while the space surrounded by the inner membrane is an inner chamber that is filled with homogenous fluid.

The inner membrane has a large number of F particles or Oxysome. These are sites for oxidative mitochondrial phosphorylation. Each Oxysome has a head, stalk, and base. The number, of elementary particles in mitochondria on maybe 104-105.’ The head of F particle contains ATP’ ( synthetase enzyme hence they are said to be ATP profiles.

(c) plastid.
Answer:
Functions of Mitochondria:
Ultrastructure of plastid: Plastids are green-colored cell organelles in the form of plastids. They are bounded by two membranes about 300 A in total thickness. Each membrane is about 40 to 60 A’ thick. Both the membranes are separated by a clear space of about 25 – 27 A.

The inner membrane is very intricately elaborated to form a system of lamellae. Internally the chloroplasts are divisible into two parts, stroma – (colorless; ground substance) and the membrane system (made of closed flattened sacs called thylakoids).

The thylakoids are closely packed in certain areas. They appear as piles of coins placed one above the other. These structures are called grana. As many as 40 to 60 grana may be present in a single chloroplast and eish granum may contain 2 to 100 coin-like thylakoids.

Thylakoids can be seen in a variety of configurations in different species of plants. The arrangement can be in the form of simple parallel sacs running lengthwise, 4, or maybe in a complex interconnecting network of the sacs. The chloroplasts invariably have the same starch granules which often accumulate near a special region known as pyrenoid in algae.

Question 10.
Describe the fluid mosaic model of the plasma membrane.
Answer:
Structure of plasma membrane: Fluid mosaic model of the plasma membrane was suggested by, S, Singer, and G. Nicholson in the 1970s.

Fluid-mosaic model of membrane

According to this model, the lipids and proteins are arranged in a mosaic fashion. The matrix is a highly viscous fluid of two layers of phospholipids having two types of protein molecules – extrinsic and intrinsic proteins.

The phospholipid layer is bimolecular and their hydrophilic ends are pointed towards the top and bottom respectively. Peripheral (extrinsic) proteins are superficially arranged on either side and can be easily separated. They have enzymatic properties and also make membranes selectively permeable.

Integral (intrinsic) proteins are tightly held in place by strong hydrophilic or hydrophobic interactions or both and are difficult to remove from the membranes.

Question 11.
What is a cell envelope? Describe its chemical nature.
Answer:
Cell envelop and its chemical nature:
Cell envelope: The bacterial cells have a chemically complex cell envelope. The layers of cell envelope stack upon one another and are bonded together tightly.

Three basic layers are identified

1. The outermost glycocalyx
2. Cell wall and
3. The cell membrane (plasma membrane).

Each layer of the envelope performs a specific function. They act as a single protective unit as a whole. The cell envelope accounts for about 10 – 50% of the cell volume.

The glycocalyx is the outermost layer. It has a coating of macromolecules. These macromolecules protect cells. They help in adhesion. It differs in thickness and chemical composition in different bacteria. Some prokaryotes have a loose sheath called the slime layer.

This slimy layer protects the cells from loss of water and nutrients. In some bacteria, there may be a thick covering called a capsule. The capsule and slime layer consists of polysaccharides and proteins. The capsule is responsible for the gummy and sticky character of the cell. This layer may be highly specific and immunogenic in bacteria:

Question 12.
Give the difference between cell walls of Gram-positive and Gram-negative bacteria?
Answer:

Gram-positive bacteria	Gram-negative bacteria
1. Their cell wall is only single-layered and 100-200 A thick.	1. Their cell wall consists of two layers and is 70-120 A in thickness.
2. They are stained by gram stain.	2. They are not stained by gram stain.
3. They do not have pili.	3. They have pili
4. In gram-positive bacteria they are mesosomes.	4. In gram-negative bacteria they are not present. They are poorly developed.

Question 13.
What are the cell inclusions in a prokaryotic cell?
Answer:
Cell inclusions in prokaryotic cells are granules or inclusion bodies. They lie freely in the cytoplasm. For example, phosphate granule; glycogen granules, sulfur granules, gas vacuole, poly-(ii) hydroxybutyrate. There may be metachromatic granules.

Question 14.
What is the structure of the ribosome?
Answer:
Ribosome Palade (1855) coined the term ribosome and he dis-covered them in animal cells. The size of the ribosome varies from 150 A0 to 250 A° in diameter. The size of the ribosome can be determined by the speed with which they sediment in a centrifugal field. The sedimenting speed measured in a unit called Swedberg unit S, The size of ribosomes present in higher animal cells is about 80S whereas the size in bacteria is about 70S.

Structure of ribosomes: Each ribosome is made up of two sub-units. For example, the ribosome of SOS has the 60S and 40S sub-units whereas 70S ribosomes have 50S and 30S subunits. These sub-units are further formed of smaller sub-units. The ribosomes are formed in the nucleolus.

Each sub-unit of ribosomes consists of about an equal amount of RNA protein. The ribosomal proteins are formed somewhere in the cytoplasm but then it migrates to the nucleoli. The ribosomal RNA is formed by ribosomal genomes found in the nuclear DNA. The ribosomal protein joins the so formed ribosomal RNA and results in, the formation of ribosomes.

Question 15.
Mention the differences between Pilus and Fimbriae.
Answer:
The differences between Pilus and Fimbriae:

Pilus	Fimbriae
1. The pili are elongated and tribal structure.	1. The fimbriae are small bristle-like fibers.
2. Pili is made up of a protein called pilin.	2. Fimbriae are composed of helically arranged protein sub-units.
3. Pili are involved in the mating process.	3. Fimbriate attach bacteria to a solid surface.

Question 16.
Give point is the justification of the statement “cell is the basic unit of life.”
Answer:
The cell-The basic unit of life: All living organisms are composed of small, tiny structures or compartments i.e. cells. These cells are called the building blocks of life.

The cells in the true sense are considered as the basic unit of life because all the life processes i.e., metabolism, responsiveness, reproduction are carried out by the cells. The cell is the seat of all metabolic (anabolic and catabolic) processes. Respiration, nutrition release of energy for the body are earned out within the cells only. Even the animals and plants (organisms) reproduce because the cell reproduces individually. Growth occurs because cells grow and multiply.

Let us take the example of Amoeba, a unicellular organism. In Amoeba all the life processes are performed within the boundaries of the single cell. This is true for all other multicellular organisms. The only difference, in the multicellular organisms, is the body of these organisms is made up of many cells. In these organisms, the cells do not behave independently but get organized into tissues.

Each tissue is specialized to perform specific functions. Different tissues then get, organized into organs that perform certain specific functions. Different organs are finally organized to form organ systems. Now it must be very clear that the basic structure of tissues, organs, and organ system are the cells only. These tissues, organs, and organ systems of the .organisms work because the cell works.

On account of the above, it can be said that cells are the structural and functional unit of the living being, hence it is the basic unit of life.

Question 17.
“The function of an organism is the result of the sum total of activities and interactions of the constituent cells.” Comment.
Answer:
The cells are building blocks of bodies. They are the functional units of bodies -Schleiden and Schwann observed that all animals and ‘ plants are composed of cells. The cells make tissues, tissues make, organs and organs make organ system. Organ systems are interrelated to perform a specific function of that system. In the body of an individual, there are different types of cells that serve different types of functions.

The activities of an organism are sum total of co-ordinate activities and interactions of the constituent cells, the cells in the body have the same hereditary material. All new cells of an organism develop from the pre-existing cells, so all cells have DNA hereditary material or RNA which passes from one generation to other by division.

So each cell has the same genetic information and it has the potential to give rise to a new individual. This property of the cell is called totipotency.

Question 18.
Multicellular organisms have better survival chances than their unicellular counterparts. Why?
Answer:
Division of labor: Some organisms are made up of just one cell (Amoeba, Paramecium, Chlamydomonas, etc.) These are termed unicellular organisms. In these organisms, all the vital life activities are performed within a single cell. On the other hand, most organisms are formed by many cells.

The size of these organisms is also big. In these multicellular organisms, all the body cells do not perform all the vital activities of life. The multicellular organisms have an advantage over unicellular organisms because in these organisms cells play a more specialized role in life activities.

For example in multicellular animals some cells of the body perform the function of movement (muscle cells), some perform the function of digestion or respiration or the removal of the wastes from the body. A group of specific types of cells performs only some limited functions.

Similarly in higher plants, some of the cells perform protective functions (epidermal cells) some cells perform the function of transport of water, mineral, and food substance in the body (xylem and phloem).

These cells would perform other functions except for which they are specialized. The group of similar cells performing similar functions is termed tissues. This system is very advantageous to multicellular organisms. Because the work has been divided by the cells of the body. This is called the division of labor. Due to this division of labor among the cells all the vital activities in the body of an organism function in a coordinated way.

Question 19.
Write a note on the structure of the nucleus.
Answer:
Structure of Nucleus: The nucleus is one of the most important components of the cell. It is, therefore, called the control center of the cell as it controls the various metabolic activities of the cell. The nucleus is situated in the cytoplasm of the cell. Usually, it is round but many different shaped nuclei can be seen in some cells.

It is surrounded by two porous membranes called nuclear membranes which remain continuous with the ER. Within the nuclear membrane is present a liquid substance called nucleoplasm. The nucleoplasm contains chromatin material of two types heterochromatin and euchromatin.

Question 20.
Write a brief note on the bacterial cell wall.
Answer:
The cell wall of Bacteria: It is formed of murein or peptidoglycan. It consists of polysaccharides cross-linked with short amino acid chains. In Gram, -ve bacteria covering is formed of lipopolysaccharide and present around cell wall. It may provide. specific adhesion properties to these cells. It determines the shape of the cell.

Question 21.
“The function of an organism is the result of the sum total of activities and interactions of the constituent cells.” Comment
Answer:
The-cells are building blocks of bodies. They are the functional units of bodies -Schleiden and Schwann observed that all animals and plants are composed of cells. The cells make tissues, tissues make organs and organs make organ systems. Organ systems are interrelated to perform a specific function of that system.

In the body of an individual, there are different types of cells that serve different types of functions. The activities of an organism are sum total of co-ordinate activities and interactions of the constituent cells. The cells in the body have the same hereditary material All new cells of an organism develop from the pre-existing cells, so all cells have DNA hereditary material or RNA which passes from one generation to another by division.

So each cell has the same genetic information and it has the potential to give rise to a new individual. This property of the cell is called totipotency.

Question 22.
How is the multicellular organization of the body more advanced than unicellular organization?
Answer:
The multicellular organization is more advanced than the unicellular organization as:

1. Division of labor among cells increases efficiency. In a multicellular organism, cells are differentiated. Specialized cells perform specific functions. Their varied type of cells is more efficient than a single-celled organism.
2. Increase in survival capacity. In multicellular organisms, the death of few cells does affect the whole organism.

Question 23.
Give four examples of specific functions performed by varied cells.
Answer:

1. Nerve cells are specialized to transmit nerve impulses.
2. Muscle cells are specialized for contraction
3. Cells of the pancreas for secretion of insulin.
4. Red blood cells for the transport of oxygen,

Question 24.
“Cell theory has shortcomings”. Justify.
Answer:
There are some shortcomings of cell theory which are given below.

1. Schleiden and Schwann were not the first scientists to prove that living beings are made up of cells.
2. Rudolf Virchow stated that “all-new cells arise from the pre-existing cells”.
3. Now it is able to explain that all cells contain hereditary material which passes from one generation to the next by cell division. It ensures the transfer of characters from parents to their offsprings.

Question 25.
Why does the efficiency of a cell decrease with an increase in size in a unicellular organism?
Answer:
With the increase in size, volume increases but surface area exposed to the environment does not increase correspondingly. This limits the exchange of information and materials through the surface. As a result efficiency of the cell as an autonomous unit decreases.

Question 26.
The cell is “an open dynamic system” discuss.
Answer:
The cell is an open system because it allows the entry and exit of matter and energy. It takes up food, oxygen, water, and salts for its substance, growth, and division. The cell also takes up energy from- food and operates the metabolic processes. It gives out waste products, secretions, and energy.

Question 27.
On what basis can we consider the cell as an autonomous unit?
Answer:
The cell can be considered as an autonomous unit because:

1. Each cell-carries out all fundamental biological processes independently.
2. Each cell oxidizes food material and utilizes that energy and some nutrient molecules to synthesize complex molecules.
3. The cell uses these molecules to build new structures and to replace worn-out cells.
4. The cell respires and exchanges gases with the environment.
5. It reproduces to form new cells with similar hereditary characters,
6. It also maintains an internal physiochemical environment.

Question 28.
Distinguish between Extrinsic and intrinsic flow of information
Answer:

Extrinsic	Intrinsic
1. Flow of genetic, information within the cell.	1. Flow of functional information from outside the cell,
2. Regulates all activities of the cell.	2. Regulates some activities of the cell.
3. Information flows in the form of nucleic acids.	3. Information flows in the form of protein or other types of molecules.

Cell: The Unit of Life Important Extra Questions Long Answer Type

Question 1.
What is the role of the plasma membrane in the compartmentalization of the cell?
Answer:
Every cell is enclosed on all sides by the distinct covering called the plasma membrane.

1. It maintains the individuality of the cell by preventing the mixing of cell contents with extracellular materials.
2. The cell is not a sealed compartment and the exchange of materials is allowed by the cell membrane in a selective and regulated manner.
3. In animals, the cell membrane has on its surface certain chemical that can recognize cells of the same kind. This helps the cells to aggregate and defend themselves against microbes.
4. The cell membrane also receives messages from outside and passes them to adjacent cells.
5. The cell membrane selectively allows the passage of certain molecules and ions from the seawater to enter the cells of organisms living in the sea.

Question 2.
Compare a plant cell with an animal cell.
Answer:

Plant cell	Animal cell
1. Plant cells form all the amino acids, coenzymes, and vitamins.	1. Animal cells cannot form all the amino acids, coenzymes, and vitamins.
2. A plant cell consists of a cellulose cell wall surrounding the plasma membrane	2. The cell wall is absent. The limiting membrane of the cell is the plasma membrane.
3. On plasmolyzing, the cells do not burst due to the presence of a cell wall.	3. On plasmolyzing animal cells burst due to the absence of a cell wall.
4. A plant cell consists of plastids in general.	4. In an animal cell, the plastids are generally absent.
5. Plant cells consist Of large vacuoles.	5. Vacuoles are absent.
6. Crystal may be present in plant cells.	6. Crystals are usually absent in animal cells.
7. Centriole is absent in plant cells.	7. Centriole is present though absent in invertebrates.
8. Mitochondria are fewer.	8. Mitochondria are numerous.

Question 3.
The cells of unicellular organisms are usually spherical whereas those of multicellular tend to be many-sided. Why?
Answer:

• It is true that the cells of unicellular organisms tend to be spherical. It is because of the following reasons.
• Surface tension: Surface tension shapes the spherical way as in the case in air-borne soap bubbles.
• The free-floating cells with thin membranes tend to be spherical as it is the most economical shape that can confine a given mass of protoplasm.

However, the shape and the size of the cell depend upon the place where they are present and the functions they have to perform. In multicellular animals, the cells tend to become faceted as they come in contact with each other in the same way as the spherical soap bubbles become flattened when they are jammed together in a small space.

This phenomenon can be best seen in the early stages of the development of an embryo. The cell mass still remains spherical for some time but when the cells multiply the shape changes because of adjusting themselves to the available space. This is also true in plants, they have in addition cellulose also.

But the most important aspect of cell shape is the functions each cell has to perform.

Question 4.
Discuss how the method of science is reflected in the formulation t of cell theory?
Answer:
Methods of science and cell theory: Scientific method of solving a problem is a realistic approach and it helps us in finding out the truth.

The scientist after making observations of many samples proposes a hypothesis that is subject to be tested before it is taken as a theory.
1. In the case of cell theory Theodor Schwann after examining many types of tissues found that all these have cells and the cells have a nucleus and cytoplasm universally present and then after comparing its structure to that of plant cells proposed a hypothesis that bodies of animals and plants are composed of cells and products of cells.

2. This hypothesis was later confirmed by Schleiden who examined a large variety of plant tissues and found all of them to be composed of cells. Thus the hypothesis of Schwann become a theory of Schleiden and Schwann.

Further, as the scientific method says a theory can be changed, modified, or discarded on the basis of new discoveries. The cells theory of Schleiden and Schwann was modified by Rudolf Virchow who was the first to explain that, “cells divide and all-new cells must come from pre-existing cells.”

Thus we can say that method of science is fully reflected in the formulation. of the cell theory.

Question 5.
If, as the second law of thermodynamics states, “the free energy in any system tends to decrease”, how is it that the earth maintains so many living organisms, each in a highly organized, high free energy state?
Answer:

1. To maintain the organization of living organisms every system of energy, try to reduce entropy. Entropy is the degree of randomness. If the system is left on its own it increases in entropy.
2. Earth receives a continuous supply of energy from the sun in the form of photons of light.
3. 0.2 to 1% of the solar energy received by the earth enters the biosphere in the form of chemical energy through the process of photosynthesis. Heterotrophs depend upon autotrophs for food.
4. Flows of energy take place from photosynthesizers to heterotrophs „
   forming food chains and food webs.
5. Approximately 10% of the energy is conserved at each trophic level. This is how earth maintains so many living organisms in a highly organized state.

Question 6.
Distinguish between prokaryotes and eukaryotes.
Answer:

Features	Prokaryotes	Eukaryotes
1. Nucleus	Not enclosed membrane.	Enclosed is the nuclear membrane.
2. Cell wall	Present in all except smallest prokaryote, when present contain muramic acid.	Variable, if present does no contain muramic acid.
3. Endoplas mic reticulum	Absent	Universally present.
4. Organelles	Membrane-bound organelles absent.	Present.
5. Ribosomes	Present freely in the cytoplasm Only of 70S type               with	Present freely in cytoplasm or associated  E.R. (80S type).
6. Centrioles	Absent	Present in animal cells
7. Size range	100-2000 nm	10,000-100,000 nm.

Question 7.
Explain the structure and function of mitochondria.
Answer:
Structure of Mitochondria: They are spherical or elongated or rod-like cell organelle. Mitochondria were first discovered by Hofmeister in the cells of a pteridophyte, Equisetum. They were named ‘Mitochondrion’ by Benda.

Mitochondria is a double membranous organelle, the outer membrane is smooth while the inner one is folded into a number of cristae. The space between the two membranes is called the outer chamber while the space surrounded by the inner membrane is an inner chamber that is filled with homogenous fluid. The inner membrane has a large number of F, particles or exosomes.

These are the sites for oxidative phosphorylation. Each Oxysome has a head, stalk, base; The number of elementary particles in a mitochondrion maybe 104-105. The head of F, particle contain ATP synthetase enzyme hence they are said to be ATP particles.

Mitochondria from animal cell

Functions of mitochondria:

1. A powerhouse of the cell and store energy as ATP.
2. Several respiratory enzymes are found in mitochondria.
3. DNA is also contained in mitochondria.
4. They regulate the concentration of calcium ions in the cells.

Question 8.
Explain in detail the structure of a typical eukaryotic chloroplast.
Answer:
Chloroplast: Chloroplasts are relatively large organelles that are visible under the light microscope. The chloroplasts are green in color and vary in size and shape from species to species. In higher plants, chloroplasts measure 2 to 4 × 5 to 10 μ in size.

Details structure of chloroplast: They are bounded by two mem-branes about 300 Å in total thickness. Each membrane is about 40 to 60 Å thick. Both the membranes are separated by a clear space of about 25-27 Å. The inner membrane is very intricately elaborated to form a system of lamellae.

Internally the chloroplasts are divided into two parts
(a) Stroma and
(b) the membrane system.

The thylakoids are closely packed in certain areas. They appear as

Structure of chloroplast

piles of coins placed one above the other. These structures are called grana. As many as 40 to 60 grana may be present in a single chloroplast and each granum may contain 2 to 100 coins like thylakoids. Thylakoids can be seen in a variety of configurations in different species of plants.

The arrangement can be in the form of simple parallel Sacuniiing lengthwise or maybe in a complex interconnecting network of the seek. The chloroplasts invariably have some starch granules which often accumulate near a special region known as pyrenoid in algae.