MCQ Questions for Class 11 Biology Chapter 9 Biomolecules with Answers

Biomolecules Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Biology Chapter 9 Biomolecules with Answers Pdf free download. MCQ Questions for Class 11 Biology with Answers were prepared based on the latest exam pattern. We have provided Biomolecules Class 11 Biology MCQs Questions with Answers to help students understand the concept very well.

Class 11 Biology Chapter 9 MCQ With Answers

Biology Class 11 Chapter 9 MCQs On Biomolecules

Biomolecules Class 11 MCQ Question 1.
Feedback inhibition of enzymes is affected by which of the following
(a) enzyme
(b) substrate
(c) end products
(d) intermediate end products

Answer

Answer: (c) end products


Biomolecules MCQs Class 11 Question 2.
An example of competitive inhibition of an enzyme is the inhibition of
(a) succinic dehydrogenase by malonic acid
(b) cytochrome oxidase by cyanide
(c) hexokinase by glucose-6-phosphate
(d) carbonic anhydrase by carbon dioxide

Answer

Answer: (a) succinic dehydrogenase by malonic acid


MCQ On Biomolecules Class 11 Question 3.
Insulin is made up of _______ and _______.
(a) glucose and fructose
(b) glucose and fructose
(c) fructose and mannose
(d) mannose and glucose

Answer

Answer: (b) glucose and fructose
Explanation:
Insulin is a polysaccharide made up of glucose and fructose.


Class 11 Biology Chapter 9 MCQ Question 4.
Enzymes increase the rate of reaction by
(a) lowering activation energy
(b) increasing activation energy
(c) increasing temperature and pH
(d) decreasing temperature and pH

Answer

Answer: (a) lowering activation energy
Explanation:
Enzymes increase the rate of reaction by decreasing the activation energy.


MCQ Of Biomolecules Class 11 Question 5.
Which of the following statements regarding enzyme inhibition is correct?
(a) Competitive inhibition is seen when a substrate competes with an enzyme for binding to an inhibitor protein
(b) Non-competitive inhibitors often bind to the enzyme irreversibly
(c) Competitive inhibition is seen when the substrate and the inhibitor compete for the active site on the enzyme
(d) Non-competitive inhibition of an enzyme can be overcome by adding large amount of substrate

Answer

Answer: (c) Competitive inhibition is seen when the substrate and the inhibitor compete for the active site on the enzyme


Class 11 Biomolecules MCQs Question 6.
Which of the following is not a pyrimidine?
(a) Uracil
(b) Cytosine
(c) Guanine
(d) Thymine

Answer

Answer: (c) Guanine
Explanation:
Guanine is a purine.


Biomolecules Class 11 MCQ With Answers Question 7.
Hydrolysis of starch occurs with the help of
(a) Peptidase
(b) Amylase
(c) Sucrose
(d) Lipase

Answer

Answer: (b) Amylase


MCQs Of Biology Class 11 Chapter 9 Question 8.
Assertion: Arachidic acid is an unsaturated fatty acid.
Reason: There are one or more double bonds between carbon atoms in unsaturated fatty acids.
(a) Both Assertion and Reason are true and Reason is the correct explanation of the Assertion.
(b) Both Assertion and Reason are true but the Reason is not the correct explanations of Assertion.
(c) Assertion is true, but Reason is false.
(d) Both Assertion and Reason are false

Answer

Answer: (d) Both Assertion and Reason are false


Biomolecules Class 11 MCQs Question 9.
Which of the following influence feedback inhibition of enzyme?
(a) End product
(b) External factors
(c) Enzyme
(d) Substrate

Answer

Answer: (a) End product


MCQ Questions For Class 11 Biology Chapter 9 Question 10.
Which of the following is not a polysaccharide?
(a) Lactose
(b) Starch
(c) Glycogen
(d) Dextrin

Answer

Answer: (a) Lactose
Explanation:
Lactose is a disaccharide made up of glucose and galactose.


Chapter 9 Biology Class 11 MCQs Question 11.
Inulin is made up of _______ and _______.
(a) glucose and fructose
(b) glucose and fructose
(c) fructose and mannose
(d) mannose and glucose

Answer

Answer: (b) glucose and fructose
Explanation:
Inulin is a polysaccharide made up of glucose and fructose.


Class 11 Biology Biomolecules MCQs Question 12.
MCQ Questions for Class 11 Biology Chapter 9 Biomolecules with Answers 1
The structure shown above is
(a) mannose
(b) fructose
(c) glucose
(d) galactose

Answer

Answer: (c) glucose
Explanation:
MCQ Questions for Class 11 Biology Chapter 9 Biomolecules with Answers 2
Glucose molecule


MCQ Biomolecules Class 11 Question 13.
A protein having both structural and enzymatic traits is
(a) Collagen
(b) Trypsin
(c) Myosin
(d) Actin

Answer

Answer: (c) Myosin


Ch 9 Bio Class 11 MCQ Question 14.
NADP contains vitamin ______.
(a) B1
(b) B2
(c) B3
(d) B12

Answer

Answer: (c) B3
Explanation:
B3 also named as niacin.


Biology Class 11 Chapter 9 MCQs Question 15.
With reference to enzymes, which one of the following statements is true?
(a) Apoenzyme = Holoenzyme + Coenzyme
(b) Holoenzyme = Apoenzyme + Coenzyme
(c) Coenzyme = Apoenzyme + Holoenzyme
(d) Holoenzyme = Coenzyme – Apoenzyme

Answer

Answer: (b) Holoenzyme = Apoenzyme + Coenzyme


Question 16.
Inhibitor binds to the active site of the enzyme. Hence blocking the reaction. This is an example of
(a) allosteric inhibition
(b) feedback inhibition
(c) uncompetitive inhibition
(d) competitive inhibition

Answer

Answer: (d) competitive inhibition
Explanation:
When substrate binds to the active site of the enzyme, it completes the reaction.
When inhibitor binds to the active site of the enzyme, it blocks the reaction.


Question 17.
The fastest enzyme known is
(a) DNA polymerase
(b) carbonic anhydrase
(c) carbonic dehydrogenase
(d) DNA ligase

Answer

Answer: (b) carbonic anhydrase
Explanation:
The fastest enzyme known is carbonic anhydrase. It converts 106 molecules of carbon dioxide molecules per second.


Question 18.
Lecithin is a
(a) polysaccharide
(b) protein
(c) nucleic acid
(d) lipid

Answer

Answer: (d) liquid
Explanation:
Lecithin is a phospholipid.


Question 19.
The minimum amount of energy required to initiate a chemical reaction is called
(a) enzymatic energy
(b) activation energy
(c) substrate energy
(d) initiation energy

Answer

Answer: (b) activation energy
Explanation:
The minimum amount of energy required to initiate a chemical reaction is called activation energy.


Question 20.
Enzymes, vitamins and hormones are common in
(a) Enhancing oxidative metabolism
(b) Being synthesised in the body of organisms
(c) Being proteinaceous
(d) Regulating metabolism

Answer

Answer: (d) Regulating metabolism


We hope the given NCERT MCQ Questions for Class 11 Biology Chapter 9 Biomolecules with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Biology Biomolecules MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Biology MCQ:

MCQ Questions for Class 11 Chemistry Chapter 8 Redox Reactions with Answers

Redox Reactions Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 8 Redox Reactions with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided Redox Reactions Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.

Class 11 Chemistry Chapter 8 MCQ With Answers

Chemistry Class 11 Chapter 8 MCQs On Redox Reactions

Redox Reaction Class 11 MCQ Question 1.
The oxidation number of Cl in Cl2O7 is
(a) + 7
(b) + 5
(c) + 3
(d) – 7

Answer

Answer: (a) + 7
Explanation:
Cl show different oxidation state as -1 to +7 due to vacant d orbital. As oxygen is more electronegative than Cl. Oxygen size is small hence its more electronegative and show -2 oxidation states.
Here Cl2O7 then equation is: 2x + 7 × (-2) = 0
x = +7 hence oxidation state of Cl is +7. I think you get your answer how to find oxidation state.


MCQ On Redox Reaction Class 11 Question 2.
What is known as Autooxidation?
(a) Formation of H2O by the oxidation of H2O2.
(b) Formation of H2O2 by the oxidation of H2O.
(c) Both (1) and (2) are true
(d) None of the above

Answer

Answer: (b) Formation of H2O2 by the oxidation of H2O.
Explanation:
Autoxidation is any oxidation that occurs in presence of oxygen. The term is usually used to describe the degradation of organic compounds in air (as a source of oxygen). Autoxidation produces hydroperoxides and cyclic organic peroxides. These species can react further to form many products. The process is relevant to many phenomena including aging, paint, and spoilage of foods, degradation of petrochemicals, and the industrial production of chemicals. Autoxidation is important because it is a useful reaction for converting compounds to oxygenated derivatives, and also because it occurs in situations where it is not desired (as in the destructive cracking of the rubber in automobile tires or in rancidification).

Water automatically gets oxidised to hydrogen peroxide.


MCQ Of Redox Reaction Class 11 Question 3.
The tendency of an electrode to lose electrons is known as
(a) Electrode Potential
(b) Reduction Potential
(c) Oxidation Potential
(d) E.M.F.

Answer

Answer: (c) Oxidation Potential
Explanation:
The magnitude of the electrode potential of a metal is a measure of its relative tendency to lose or gain electrons. i.e., it is a measure of the relative tendency to undergo oxidation (loss of electrons) or reduction (gain of electrons).
M → Mn+ + ne (oxidation potential)
Mn+ + ne → M (reduction potential)


Class 11 Chemistry Redox Reaction MCQ Question 4.
If equal volumes of 1M KMnO4 and 1M K2Cr2O7 solutions are allowed to oxidize Fe2+ in acidic medium. The amount of iron oxidized will be:
(a) More with KMnO2
(b) More with K2Cr2O7
(c) Equal with both oxidising agents
(d) Cannot be determined

Answer

Answer: (b) More with K2Cr2O7
Explanation:
The reason due to which the amount of Fe oxidised will be more with ​K2Cr2O7 is:
the change in the oxidation state (or number) or n factor is greater with KMnO4
Also, ​K2Cr2O7 is a very strong oxidising agent and holds the ability to take the electrons but ​KMnO4 is more stronger than ​K2Cr2O7.


Redox Reaction MCQ Class 11 Question 5.
Which of the following processes does not involve either oxidation or reduction?
(a) Formation of slaked lime from quick lime
(b) Heating Mercuric Oxide
(c) Formation of Manganese Chloride from Manganese oxide
(d) Formation of Zinc from Zinc blende

Answer

Answer: (a) Formation of slaked lime from quick lime
Explanation:
Here, in this reaction
CaO + H2​O →Ca(OH)2
Oxidation number doesn’t change so its not a redox reaction.


Class 11 Redox Reaction MCQ Question 6.
One mole of N2H4 loses ten moles of electrons to form a new compound A. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in A? (There is no change in the oxidation state of hydrogen.)
(a) -1
(b) -3
(c) +3
(d) +5

Answer

Answer: (c) +3
Explanation:
First to find oxidation number of Nitrogen in N2H4
Oxidation number of H = +1
Let oxidation number of nitrogen be x
2x + 4(1) = 0
2x = -4
x = -2
Each nitrogen atom has -2 oxidation number. So taken both nitrogen atoms in account gives oxidation number -4.
Change in oxidation number of nitrogen on losing 10 mol of electrons (considering no change in oxidation number of hydrogen atoms)
-4 – (-10) = +6
Therefore, oxidation number of 2 nitrogen atoms in compound Y is +6. Hence, oxidation number of each nitrogen atom will be +3 in new compound Y.


Class 11 Chemistry Chapter 8 MCQ Question 7.
How many millilitres of 0.5 M H2SO4 are needed to dissolve 0.5 g of copper(II)carbonate?
(a) 6.01
(b) 4.5
(c) 8.1
(d) 11.1

Answer

Answer: (c) 8.1
Explanation:
The volume can be calculated :
N1V1 = N2V2
N1 = Normality of H2SO4 = 0.5 × 2 = 1 N
V1 = Volume of H2SO4
Molar mass of copper(II) carbonate = 123.5 g
N2 = Normality of copper (II) carbonate = (0.5×2)/(123.5) N
V2 = Volume of copper (II) carbonate = 1000 mL
So, after applying the formula,
1 × V1 = (0.5×2)/(123.5)×1000
Hence, V1 = 8.09 mL
= approx. 8.1 mL


Redox Reaction Class 11 MCQ Pdf Question 8.
The oxidation state of Cr in Cr (CO)6 is
(a) 0
(b) 2
(c) 2
(d) 6

Answer

Answer: (a) 0
Explanation:
CO (carbonyl) is a neutral ligand, hence oxidation state of Cr in Cr (CO)6 is zero


Redox Reactions Class 11 MCQ Question 9.
Which of the following processes does not involve oxidation of iron?
(a) Formation of Fe(CO)5 from Fe.
(b) Liberation of H2 from steam by iron at high temperature.
(c) Rusting of iron sheets.
(d) Decolourisation of blue CuSO4 solution by iron.

Answer

Answer: (a) Formation of Fe(CO)5 from Fe.
Explanation:
Oxidation number of Fe in Fe(CO)5 is zero.
In both Fe and Fe(CO)5, the oxidation state of iron is zero.
3Fe + 4H2O → Fe3O4 + 4H2
team
rusting
Fe → Fe2O3.xH2O
(+3)
CuSO4(aq) + Fe (s) → FeSO4(aq) + Cu(s)
(0) (+2)


MCQs On Redox Reaction Class 11 Question 10.
The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is
(a) One
(b) Two
(c) Five
(d) One fifth.

Answer

Answer: (b) Two
Explanation:
In alkaline medium the reduction of KMnO4 with KI will takes place as
2 KMnO4 + H2O → 2 KOH + 2 MnO2
KI + 3[O] → KIO3
Hence the overall reaction is
KI + 2KMnO4 + H2O → KIO3 + 2 KOH + 2 MnO2
So, one mole of KI will reduced two moles of KMnO4


MCQ Questions For Class 11 Chemistry Chapter 8 Question 11.
Which of the following reactions does not involve either oxidation or reduction?
(a) VO2+ → V2O3
(b) Na → Na+
(c) CrO2-4 → Cr2O2-7
(d) Zn2+ →Zn

Answer

Answer: (c) CrO2-4 → Cr2O2-7
Explanation:
In VO2+ →V2​O3, V is reduced from +4 to +3 oxidation state.
In Na → Na2+, Na is oxidised from to +1 oxidation state.
In CrO4-2​ → Cr2O7-2, Cr remains in same oxidation state +6.
In Zn+2 → Zn, Zn is reduced from +2 to 0 oxidation state.


Redox Reaction Class 11 MCQs Question 12.
KMnO4​ reacts with oxalic acid according to the equation 2MnO4​ + 5C2​O42-​ + 16H+ → 2Mn2+​ +10CO2​ + 8H2​O Here 20 mL of 0.1 M KMnO4​ is equivalent to
(a) 50 mL of 0.5 M C2H2O4
(b) 20 mL of 0.1 M C2H2O4
(c) 20 mL of 0.5 M C2H2O4
(d) 50 mL of 0.1 M C2H2O4

Answer

Answer: (d) 50 mL of 0.1 M C2H2O4
Explanation:
2MnO4 + 5C2O42- + 16H+ → 2Mn2+ + 10CO2 + 8H2O
Therefore, 2 moles of MNO4 equivalent to 5 moles of C2O42-
20 mL of 0.1 M KMnO4 = 2 moles of KMnO4
Also, 50 mL of 0.1 M C2H2O4 equivalent to 5 mol of C2O42-
Therefore, these are equivalent.


Ch 8 Chemistry Class 11 MCQ Question 13.
One mole of N2H4 loses ten moles of electrons to form a new compound A. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in A? (There is no change in the oxidation state of hydrogen.)
(a) -1
(b) -3
(c) +3
(d) +5

Answer

Answer: (c) +3
Explanation:
First to find oxidation number of Nitrogen in N2H4
Oxidation number of H = +1
Let oxidation number of nitrogen be x
2x + 4(1) = 0
2x = -4
x = -2
Each nitrogen atom has -2 oxidation number. So taken both nitrogen atoms in account gives oxidation number -4.
Change in oxidation number of nitrogen on losing 10 mol of electrons (considering no change in oxidation number of hydrogen atoms)
-4 – (-10) = +6
Therefore, oxidation number of 2 nitrogen atoms in compound Y is +6. Hence, oxidation number of each nitrogen atom will be +3 in new compound Y.


MCQs Of Redox Reaction Class 11 Question 14.
What is n-factor?
(a) Equal to product of Number of moles of electrons when Lost or gained by one mole of reductant or oxidant
(b) When Number of moles of electrons Lost or gained by one mole of reductant or oxidant is not same.
(c) Equal to Number of moles of electrons Lost or gained by one mole of reductant or oxidant
(d) None of the above

Answer

Answer: (c) Equal to Number of moles of electrons Lost or gained by one mole of reductant or oxidant.
Explanation:
For redox reaction it is considered as change in their oxidation number or change in their reduction number in both side of a chemical reaction.


Class 11 Chemistry Chapter 8 MCQ With Answers Question 15.
The oxidation number of Mn is maximum in
(a) MnO2
(b) K2MnO4
(c) Mn3O4
(d) KMnO4.

Answer

Answer: (d) KMnO4.
Explanation:
The electronic configuration of Mn is:
Mn(25) = [Ar]3d5 4s2, 4p0
In excited state, it can lose its all 7 electrons.
Hence, maximum oxidation sate exhibited by Mn is +7 which is in KMnO4.


Question 16:
The oxidation process involves
(a) Increase in oxidation number
(b) Decrease in oxidation number
(c) No change in oxidation number
(d) none of the above

Answer

Answer: (a) Increase in oxidation number
Explanation:
Oxidation process Involves:-
Addition of O2 or electronegative element
Removal of H/electropositive element
Loss of electrons
Increase in oxidation number


Question 17.
Metals generally react with dilute acids to produce hydrogen gas. Which one of the following metals does not react with dilute hydrochloric acid?
(a) Copper
(b) Magnesium
(c) Iron
(d) Silver

Answer

Answer: (b) Magnesium
Explanation:
Most of the metals such as Al, Cu, Fe etc. reacts with dilute acids to produce hydrogen gas but magnesium is an exception. Magnesium being an active metal liberates dihydrogen gas as it is allowed to react with dilute HCl. Thus all the given metals react with dilute acids.
Mg + 2HCl → MgCl2 + H2


Question 18.
The oxidation number of Xe in BaXeO6 is
(a) 8
(b) 6
(c) 4
(d) 10

Answer

Answer: (d) 10
Explanation:
Oxidation state of Ba in general = +2 and of O = −2
Applying formula, Sum of total oxidation state of all atoms = Overall charge on the compound.
Let oxidation state of Xe in BaXeO6 be x.
2 + x + 6(−2) = 0,
x = 10
But oxidation state 10 is not possible for Xe. In this case the oxidation state of Xe is equal to maximum possible oxidation state for Xe = +8.


Question 19.
The colourless solution of silver nitrate slowly turns blue on adding copper chips to it because of
(a) Dissolution of Copper
(b) Oxidation of Ag+ → Ag
(c) Reduction of Cu2+ ions
(d) Oxidation of Cu atoms.

Answer

Answer: (d) Oxidation of Cu atoms.
Explanation:
When copper turnings are added to silver nitrate solution, the solution becomes brown in color after sometime because copper is more reactive than silver so it displaces silver from silver nitrate solution and form copper nitrate solution.


Question 20.
A standard reduction electrode potentials of four metals are A = -0.250 V, B = -0.140 V, C = -0.126 V, D = -0.402 V The metal that displaces A from its aqueous solution is:
(a) A
(b) B
(c) C
(d) D

Answer

Answer: (d) D
Explanation:
Reduction potential of D is minimum i.e. −0.402 V. Thus oxidation potential of D is maximum i.e. to +0.402 V. D can oxidise itself and reduce other.
The aqueous solution A will be present in its ionic form and can be reduced by D as its reduction potential is higher than D.
Thus D can replace A from its Aqueous solution.


We hope the given NCERT MCQ Questions for Class 11 Chemistry Chapter 8 Redox Reactions with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Chemistry Redox Reactions MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Chemistry MCQ:

MCQ Questions for Class 11 Biology Chapter 8 Cell: The Unit of Life with Answers

Cell: The Unit of Life Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Biology Chapter 8 Cell: The Unit of Life with Answers Pdf free download. MCQ Questions for Class 11 Biology with Answers were prepared based on the latest exam pattern. We have provided Cell: The Unit of Life Class 11 Biology MCQs Questions with Answers to help students understand the concept very well.

Class 11 Biology Chapter 8 MCQ With Answers

Biology Class 11 Chapter 8 MCQs On Cell: The Unit of Life

Cell The Unit Of Life Class 11 MCQ Question 1.
The cytoplasmic connections from cell to cell are known as
(a) middle lamella
(b) plasmodesmata
(c) cell membrane system
(d) endoplasmic reticulum

Answer

Answer: (b) plasmodesmata


Cell The Unit Of Life MCQ Question 2.
Bacterial flagella is made up of
(a) tubulin
(b) flagellin
(c) chitin
(d) None of these

Answer

Answer: (b) flagellin
Explanation:
Bacterial flagella is made up of protein flagellin.


MCQ On Cell The Unit Of Life Class 11 Question 3.
Plasmolysis occurs due to-
(a) Absorption
(b) Endosmosis
(c) Osmosis
(d) Exosmosis

Answer

Answer: (d) Exosmosis


Class 11 Biology Chapter 8 MCQ Question 4.
The term Cell was given by-
(a) Leeuwenhoek
(b) Robert hooke
(c) Flemming
(d) Robert Brown

Answer

Answer: (b) Robert hooke


Cell Unit Of Life Class 11 MCQ Question 5.
Which one of the following also acts as a catalyst in a bacterial cell?
(a) 23 sr RNA
(b) 5 sr RNA
(c) sn RNA
(d) hn RNA

Answer

Answer: (a) 23 sr RNA


MCQs Of Biology Class 11 Chapter 8 Question 6.
Centrosome is found in-
(a) Cytoplasm
(b) Nucleus
(c) Chromosomes
(d) Nucleolus

Answer

Answer: (a) Cytoplasm


MCQ Questions For Class 11 Biology Chapter 8 Question 7.
Plasma membrane is
(a) impermeable
(b) semi-permeable
(c) completely permeable
(d) differentially permeable

Answer

Answer: (b) semi-permeable
Explanation:
Plasma membrane is a selective permeable membrane that allows only selective molecules to pass through it.
The permeability depends on the electric charge and polarity of the molecules.


Cell The Unit Of Life Class 11 MCQ Pdf Question 8.
Middle lamella is made up of ___________.
(a) calcium sulphide
(b) calcium pectate
(c) calcium carbonate
(d) calcium chloride

Answer

Answer: (b) calcium pectate
Explanation:
Middle lamella is made up of calcium pectate.


MCQ Questions For Class 11 Biology Chapter Cell The Unit Of Life Question 9.
Match the columns.
1. Cytoskeleton – A. hair-like outgrowth
2. Flagella – B. proximal region of centriole
3. Hub – C. bristle-like structures
4. Fimbriae – D. filamentous protein structure
(a) 1-D, 2-A, 3-B, 4-C
(b) 1-D, 2-C, 3-B, 4-A
(c) 1-B, 2-D, 3-A, 4-C
(d) 1-D, 2-A, 3-C, 4-B

Answer

Answer: (a) 1-D, 2-A, 3-B, 4-C
Explanation:
A filamentous protein structure present in the cytoplasm is called cytoskeleton.
Flagella is a hair-like outgrowths of the cell membrane.
The central part of the proximal region of the centriole is called hub.
Fimbriae are small bristle like fibres sprouting out of the cell.


Chapter 8 Biology Class 11 MCQs Question 10.
Which of the following does not have cell wall?
(a) Mycoplasma
(b) Bacteria
(c) PPLO
(d) Blue green algae

Answer

Answer: (a) Mycoplasma
Explanation:
Mycoplasma does not have cell wall.
PPLO (Pleuro Pneumonia Like Organisms)


Class 11 Cell The Unit Of Life MCQs Question 11.
Centrosome is found in-
(a) Cytoplasm
(b) Nucleus
(c) Chromosomes
(d) Nucleolus

Answer

Answer: (a) Cytoplasm


MCQ Of Chapter 8 Biology Class 11 Question 12.
The longest cell in human body is
(a) red blood cells
(b) white blood cells
(c) columnar epithelial cells
(d) nerve cells

Answer

Answer: (d) nerve cells
Explanation:
Nerve cells are the longest cells.


Class 11 Biology Chapter 8 MCQ With Answers Question 13.
The main site for synthesis of lipids is
(a) vacuoles
(b) RER
(c) SER
(d) Golgi body

Answer

Answer: (c) SER
Explanation:
The main site for synthesis of lipid is smooth endoplasmic reticulum.


Cell The Unit Of Life MCQ Class 11 Question 14.
The function of ribosomes is
(a) metabolism
(b) lipid synthesis
(c) protein synthesis
(d) photosynthesis

Answer

Answer: (c) protein synthesis
Explanation:
Ribosomes are called protein factories because they synthesize proteins.


Class 11 Bio Ch 8 MCQ Question 15.
Which is called Suicidal Bag?
(a) Centrosome
(b) Lysosome
(c) Mesosome
(d) Chromosome

Answer

Answer: (b) Lysosome


Question 16.
A nucleosome is a portion of the chromonema containing ______.
(a) both DNA and histones
(b) only histones
(c) both DNA and RNA
(d) only DNA

Answer

Answer: (a) both DNA and histones


Question 17.
The largest cell in the human body is-
(a) Nerve cell
(b) Muscle cell
(c) Liver cell
(d) Kidney cell

Answer

Answer: (a) Nerve cell


Question 18.
Keeping in view the fluid mosaic model for the structure of cell membrane, which one of the following statements is correct with respect to the movement of lipids and proteins from one lipid monolayer to the other (described as flip-flop movement)?
(a) Neither lipids, nor proteins can flip-flop
(b) Both lipids and proteins can flip-flop
(c) While lipids can rarely flip-flop, proteins can not
(d) While proteins can flip-flop, lipids can not

Answer

Answer: (c) While lipids can rarely flip-flop, proteins can not


Question 19.
Cell secretion is done by-
(a) Plastids
(b) ER
(c) Golgi apparatus
(d) Nucleolus

Answer

Answer: (c) Golgi apparatus


Question 20.
___________ increases the surface area for mitochondrial activity.
(a) Inner membrane
(b) Inter membrane space
(c) Matrix
(d) Cristae

Answer

Answer: (d) Cristae
Explanation:
The inner membrane forms a number of foldings called cristae towards the matrix. These cristae increases the surface area.


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Class 11 Biology MCQ:

MCQ Questions for Class 11 Chemistry Chapter 14 Environmental Chemistry with Answers

Environmental Chemistry Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 14 Environmental Chemistry with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided Environmental Chemistry Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.

Class 11 Chemistry Chapter 14 MCQ With Answers

Chemistry Class 11 Chapter 14 MCQs On Environmental Chemistry

Environmental Chemistry MCQ Chapter 14 Class 11 Question 1.
The agricultural field that produces maximum methane gas into atmosphere is
(a) Wheat field
(b) Paddy field
(c) Cotton field
(d) Groundnut field

Answer

Answer: (c) Cotton field
Explanation:
Rice is grown in flooded fields, a situation that depletes the soil of oxygen. Soils that are anaerobic (lacking oxygen) allow the bacteria that produce methane from decomposing organic matter to thrive. Some of this methane then bubbles to the surface, but most of it is diffused back into the atmosphere through the rice plants themselves. Hence, rice Paddy field that produces maximum methane gas into the atmosphere.


MCQ On Environmental Chemistry Chapter 14 Class 11 Question 2.
Fluorosis, the bone disease, is caused by the presence of
(a) Pesticides in water
(b) Fluorides in water
(c) Carbon monoxide in air
(d) Sulphur dioxide in air.

Answer

Answer: (b) Fluorides in water
Explanation:
Skeletal fluorosis is a bone disease caused by excessive accumulation of fluoride in the bones. In advanced cases, skeletal fluorosis causes painful damage to bones and joints.


Environmental Chemistry Class 11 MCQ Chapter 14 Question 3.
The most harmful air pollutant produced by automobiles is
(a) HNO3
(b) NO
(c) SO2
(d) CO.

Answer

Answer: (d) CO.
Explanation:
Motor vehicle emissions contribute to air pollution and are a major ingredient in the creation of smog in some large cities. Transportation contributes to more than half of the carbon monoxide, sulphur dioxide and nitrogen oxides, and almost a quarter of the hydrocarbons emitted into our air. Carbon monoxide (CO) is a colourless, odourless, toxic non-irritating gas. It is a product by incomplete combustion of fuel such as natural gas, coal or wood. Vehicular exhaust is a major source of carbon monoxide.


MCQ On Environmental Chemistry Class 11 Chapter 14 Question 4.
Which of the following is responsible for depletion of ozone layer in the upper strata of the atmosphere?
(a) Polyhalogens
(b) Ferrocene
(c) Fullerencs
(d) Freons.

Answer

Answer: (d) Freons.
Explanation:
Fluorocarbons such as freon-1 (CFCl3) and freon-12 (CF2Cl2) emitted as propellants in aerosol spray cans, refrigerators, fire fighting reagents etc. are stable compounds and chemically inert. They do not react with any substance with which they come in contact and thus float through the atmosphere unchanged and eventually enter the stratosphere. There they absorb UV radiations and break down liberating free atomic chlorine which causes decomposition of ozone. This results in the depletion of the ozone layer.


MCQs On Environmental Chemistry Chapter 14 Class 11 Question 5.
Photochemical smog is related to pollution if
(a) Air
(b) Water
(c) Soil
(d) None of these

Answer

Answer: (c) Soil
Explanation:
Photochemical smog is produced when pollutants from the combustion of fossil fuels react with sunlight. The energy in the sunlight converts the pollutants into other toxic chemicals. For photochemical smog to form, there must be other pollutants in the air, specifically nitrous oxides and other volatile organic compounds (VOCs).


Environmental Chemistry MCQ And Answers Pdf Class 11 Question 6.
Which of the following acts as a sink for CO?
(a) Plants
(b) Haemoglobin
(c) Microorganisms present in the soil
(d) Oceans

Answer

Answer: (c) Microorganisms present in the soil
Explanation:
CO is converted into CO2 by microorganism present in soil.


Environmental Chemistry MCQs Chapter 14 Class 11 Question 7.
Domestic waste mostly constitutes
(a) Non-biodegradable pollution
(b) Biodegradable pollution
(c) Effluents
(d) Air pollution

Answer

Answer: (b) Biodegradable pollution
Explanation:
Domestic waste mostly constitutes biodegradable pollution.


Environmental Chemistry MCQ Questions Chapter 14 Class 11 Question 8.
Which is most harmful for life on globe?
(a) Deforestation
(b) Soil Erosion
(c) Increasing Desert
(d) Nuclear Fallout.

Answer

Answer: (d) Nuclear Fallout.
Explanation:
Nuclear fallout, or simply fallout, is the residual radioactive material propelled into the upper atmosphere following a nuclear blast or a nuclear reaction conducted in an unshielded facility, so called because it “falls out” of the sky after the explosion and the shock wave have passed. While deforestation, soil erosion and increasing deserts can be controlled, nuclear fallout causes genetic mutation in humans and destruction that cannot be controlled or cured.


Environmental Chemistry MCQ With Answers Chapter 14 Class 11 Question 9.
One of the following is the chief source of water and soil pollution
(a) Agro industry
(b) Mining
(c) Thermal power stations
(d) All of these

Answer

Answer: (c) Thermal power stations
Explanation:
Thermal power plant, agro industry and mining all are responsible for soil and water pollution. (Due to extraction of chemicals which cause pollution)


MCQ Environmental Chemistry Pdf Chapter 14 Class 11 Question 10.
What is the indicator of pollution in water?
(a) Amount of oxygen
(b) Amount of hydrogen
(c) Amount of BOD
(d) Amount of nitrogen

Answer

Answer: (c) Amount of BOD
Explanation:
The amount of oxygen required to break down a certain amount of organic matter is called the biological oxygen demand (BOD). The amount of BOD in water is an indicator of the level of pollution.


Class 11 Chemistry MCQ With Answers Chapter 14  Question 11.
Which of the following acts as a sink for CO?
(a) Plants
(b) Haemoglobin
(c) Microorganisms present in the soil
(d) Oceans

Answer

Answer: (c) Microorganisms present in the soil
Explanation:
CO is converted into CO2 by microorganism present in soil.


Chemistry Class 11 MCQ Chapter 14 Question 12.
The false statement among the followings:
(a) The average residence time of NO is one month.
(b) Limestone acts as a sink for SOx.
(c) SOx can be removed from flue gases by passing through a solution of citrate ions.
(d) Ammonia acts as a sink for NOx.

Answer

Answer: (d) Ammonia acts as a sink for NOx.
Explanation:
The average residence time of NO is 4 days.


Environmental Chemistry MCQs Pdf Chapter 14 Class 11 Question 13.
A disease caused by eating fish contaminated by industrial waste, containing mercury compounds, is called as
(a) Minamata disease
(b) Brights disease
(c) Hashimotos disease
(d) Osteosclerosis

Answer

Answer: (a) Minamata disease
Explanation:
Disease caused by eating fish found in water contaminated with industrial waste having mercury is minamata disease.


Class 11 Chemistry MCQ With Answers Pdf Chapter 14 Question 14.
Which of the following is responsible for depletion of ozone layer in the upper strata of the atmosphere?
(a) Polyhalogens
(b) Ferrocene
(c) Fullerencs
(d) Freons.

Answer

Answer: (d) Freons.
Explanation:
Fluorocarbons such as freon-1 (CFCl3) and freon-12 (CF2Cl2) emitted as propellants in aerosol spray cans, refrigerators, fire fighting reagents etc. are stable compounds and chemically inert. They do not react with any substance with which they come in contact and thus float through the atmosphere unchanged and eventually enter the stratosphere. There they absorb UV radiations and break down liberating free atomic chlorine which causes decomposition of ozone. This results in the depletion of the ozone layer.


Environmental Chemistry Questions Chapter 14 Class 11 Question 15.
Photochemical smog is related to pollution if
(a) Air
(b) Water
(c) Soil
(d) None of these

Answer

Answer: (c) Soil
Explanation:
Photochemical smog is produced when pollutants from the combustion of fossil fuels react with sunlight. The energy in the sunlight converts the pollutants into other toxic chemicals. For photochemical smog to form, there must be other pollutants in the air, specifically nitrous oxides and other volatile organic compounds (VOCs).


Question 16.
Identify the wrong statement in the following:
(a) Acid rain is mostly because of oxides of nitrogen and sulphur
(b) Chlorofluorocarbons are responsible for ozone layer depltion
(c) Greenhouse effect is responsible for global warming
(d) Ozone layer does not permit infrared radiation from the sun to reach the earth.

Answer

Answer: (d) Ozone layer does not permit infrared radiation from the sun to reach the earth.
Explanation:
Allotropes of Oxygen-
O2 (Dioxygen) and O3 (Ozone)
Infrared radiation is most important for heating up the atmosphere and the surface of the earth. These rays have high wavelength and low frequency and are not very harmful. Ozone does not hinder infrared radiation.


Question 17.
Synthesis of ethanal commercially from which of the following reagent is the part of green chemistry?
(a) CH3CH2OH
(b) CH2 = CH2
(c) HC ≡ CH
(d) All of these

Answer

Answer: (b) CH2 = CH2


Question 18.
Which of the following is produced by a reaction of ultraviolet light?
(a) CO2
(b) SO2
(c) O3
(d) NO3

Answer

Answer: (c) O3
Explanation:
Ultraviolet radiations cause photochemical splitting of oxygen molecules in the stratosphere. The nascent oxygen combines with molecular oxygen to form ozone. In troposphere nascent oxygen comes from nitrogen dioxide.


Question 19.
Why certain parts of water bodies contain more number of algae?
(a) Due to favorable environment
(b) Due to run-off excess fertilizers
(c) Due to lack of fertilizers
(d) Due to water pollution

Answer

Answer: (b) Due to run-off excess fertilizers
Explanation:
Inorganic plant nutrients found in fertilizers. Run-off carries excess fertilizers into nearby water bodies, causing high levels of plant nutrients which promote the excessive growth of algae and other aquatic plants.


Question 20.
The substance which is a primary pollutant?
(a) H2SO4
(b) CO
(c) PAN
(d) Aldehydes

Answer

Answer: (b) CO
Explanation:
Important Air pollutants
Oxides of carbon (CO,CO2)
Oxides of sulphur (SO2 & SO3)
Oxides of nitrogen
Chlorofluoro carbons (CFC)
Hydrocarbons Like CH4
​Photochemical and Industrial smog.
Dust & other particulate matter from industries.


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Class 11 Chemistry MCQ:

MCQ Questions for Class 11 Biology Chapter 1 The Living World with Answers

The Living World Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Biology Chapter 1 The Living World with Answers Pdf free download. MCQ Questions for Class 11 Biology with Answers were prepared based on the latest exam pattern. We have provided The Living World Class 11 Biology MCQs Questions with Answers to help students understand the concept very well.

Class 11 Biology Chapter 1 MCQ With Answers

Biology Class 11 Chapter 1 MCQs On The Living World

The Living World Class 11 MCQ Question 1.
Identify the incorrect match.
(a) Physiology – Study of functions and processes of life
(b) Pedology – Soil science
(c) Limnology – Study of fresh water
(d) Kinesiology – Fossil study

Answer

Answer: (d) Kinesiology – Fossil study


MCQ Questions For Class 11 Biology Chapter 1 Question 2.
The Phylogenetic system of classification was put forth by
(a) Theophrastus
(b) George Bentham and Joseph Dalton Hooker
(c) Carolus Linnaeus
(d) Adolf Engler and Karl Prantl

Answer

Answer: (d) Adolf Engler and Karl Prantl


Class 11 Biology Chapter 1 MCQ Question 3.
Two taxonomic species are distinguished from each other by
(a) their failure to interbreed
(b) their ability to exchange gene freely
(c) their similarity in morphological characters
(d) discontinuity in a set of correlated characters

Answer

Answer: (a) their failure to interbreed
Explanation:
Two taxonomic species are distinguished from each other by their failure to interbreed.


Living World Class 11 MCQs Question 4.
Musca domestica is common name of
(a) Housefly
(b) Mosquito
(c) Snail
(d) Ant

Answer

Answer: (a) Housefly
Explanation:
Musca domestica is the common name of housefly.


Biology Class 11 Chapter 1 MCQs Question 5.
Keystone species are
(a) important for ecosystem
(b) important for plants
(c) endangered species
(d) extinct species

Answer

Answer: (a) important for ecosystem
Explanation:
Keystone species are plants or animals species that play a unique and crucial role in the way an ecosystem functions.
They are very important for ecosystem to function properly.


The Living World MCQ Class 11 Question 6.
The study of fish culture is called
(a) ophiology
(b) herpetology
(c) ichthyology
(d) pisciculture

Answer

Answer: (d) pisciculture


Living World Class 11 MCQ Question 7.
National Botanical Research Institute is located in
(a) Shimla
(b) Dehradun
(c) Howrah
(d) Lucknow

Answer

Answer: (d) Lucknow
Explanation:
National Botanical Research Institute is located in Lucknow.


Chapter 1 Biology Class 11 MCQs Question 8.
Biosystematics aims at
(a) Identification and arrangement of organisms on the basis of cytological characteristics
(b) The classification of organisms based on broad morphological characters
(c) Delimiting various taxa of organism and establishing their relationships
(d) The classification of organisms based on their evolutionary history and establishing their phylogeny on the totality of various parameters from all fields of studies

Answer

Answer: (d) The classification of organisms based on their evolutionary history and establishing their phylogeny on the totality of various parameters from all fields of studies


Class 11 Bio Chapter 1 MCQ Question 9.
In angiosperm, characters of flowers are used in classification because
(a) Flowers are attractive
(b) Flowers are large
(c) Characters of flowers are conservative
(d) None of the above

Answer

Answer: (c) Characters of flowers are conservative


MCQ Class 11 Biology Chapter 1 Question 10.
Scientific name of plants are given by
(a) International code for Botanical nomenclature
(b) International code for biological plants
(c) Indian code for Botanical nomenclature
(d) International code for zoological nomenclature

Answer

Answer: (a) International code for Botanical nomenclature
Explanation:
The scientific names of plants are given by International code for Botanical nomenclature.


Ncert Biology Class 11 Chapter 1 MCQ Question 11.
A system which gets a continuous flow of energy is called
(a) micropropagated system
(b) closed system
(c) open system
(d) steady state

Answer

Answer: (c) open system


Ch 1 Bio Class 11 MCQ Question 12.
Solanum, Panthera, Homo are examples of
(a) Family
(b) Division
(c) Genera
(d) Species epithet

Answer

Answer: (c) Genera
Explanation:
Solanum, Panthera, Homo are examples of generic names.


MCQ On Living World Class 11 Question 13.
Which of the following is correct?
(a) Musca domestica Order – Diptera Family – Muscidae Phylum – Arthropoda
(b) Mangifera indica Order – Sapindales Family – Anacardiaceae Phylum – Gymnospermae
(c) Triticum aestivum Order – Poles Family – Monocotyledons Phylum – Angiospermae
(d) Panthera leo Order – leo Family – Felidae Phylum – Mammalia

Answer

Answer: (a) Musca domestica Order – Diptera Family – Muscidae Phylum – Arthropoda
Explanation:
Mangifera indica belongs to Phylum Angiospermae.
Triticum aestivum belongs to family Poaceae.
Panthera leo belongs to order Carnivora.


Class 11 Bio Ch 1 MCQ Question 14.
Binomial Nomenclature was given by
(a) Ernst Mayr
(b) Alexander Agassiz
(c) Carolus Linnaeus
(d) Alexander Agassiz and Carlous Linnaeus

Answer

Answer: (c) Carolus Linnaeus
Explanation:
The system of providing scientific names with two components i.e. generic name and specific epithet is called Binomial Nomenclature.
Binomial Nomenclature was given by Carolus Linnaeus.


MCQ Questions For Class 11 Biology Chapter 1 With Answers Question 15.
Systematics deals with
(a) Nomenclature of new organisms
(b) Evolutionary relationship between organisms
(c) Identification of newly discovered organisms
(d) All of these

Answer

Answer: (d) All of these
Explanation:
Linnaeus used Systema Naturae as the title of his publication.
The scope of systematics includes identification, nomenclature and classification.
Systematics takes into account evolutionary relationship between organisms.


MCQ Chapter 1 Biology Class 11 Question 16.
Which of the following shows maximum diversity?
(a) Plantas
(b) Animalia
(c) Monerans
(d) Protistans

Answer

Answer: (b) Animalia


MCQs Of Biology Class 11 Chapter 1 Question 17.
Which of the following does not come under taxon?
(a) Species
(b) Kingdom
(c) Division
(d) Key

Answer

Answer: (d) Key
Explanation:
Taxon is a group of one or more population of organisms.
Kingdom, division and species come under taxon but key is taxonomical aid.


MCQ Of Chapter 1 Biology Class 11 Question 18.
The ozone layer is found in the
(a) hydrosphere
(b) stratosphere
(c) lithosphere
(d) troposphere

Answer

Answer: (b) stratosphere


Class 11 Biology Chapter 1 MCQs Question 19.
Taxonomic studies depends on
(a) Ecological information of organisms.
(b) Structure of cell and development process of organisms.
(c) External and internal structure of organisms.
(d) All of the above.

Answer

Answer: (d) All of the above.
Explanation:
Taxonomic studies depends on ecological information of organisms, structure of cell and development process of organism and external and internal structure of organisms.


Biology MCQs For Class 11 Chapter 1 With Answers Pdf Question 20.
Wheat belongs to division ______.
(a) Angiospermae
(b) Gymnospermae
(c) Poacae
(d) None of the above

Answer

Answer: (a) Angiospermae
Explanation:
Wheat belongs to Angiospermae division.


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Class 11 Biology MCQ:

MCQ Questions for Class 11 Chemistry Chapter 13 Hydrocarbons with Answers

Hydrocarbons Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 13 Hydrocarbons with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided Hydrocarbons Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.

Class 11 Chemistry Chapter 13 MCQ With Answers

Chemistry Class 11 Chapter 13 MCQs On Hydrocarbons

Hydrocarbons Class 11 MCQ Question 1.
A dibromo derivative of an alkane reacts with sodium metal to form an alicyclic hydrocarbon. The derivative is ______.
(a) 2, 2-dibromobutane
(b) 1, 1-dibromopropane
(c) 1, 4-dibromobutane
(d) 1, 2-dibromoethane

Answer

Answer: (c) 1, 4-dibromobutane
Explanation:
The derivative is 1, 4 dibromobutane. This on reaction with sodium metal gives cyclobutane.
MCQ Questions for Class 11 Chemistry Chapter 13 Hydrocarbons with Answers 1
This reaction is an example of internal Wurtz reaction.


MCQ On Hydrocarbons Class 11 Question 2.
The position of double bond in alkenes can be located by :
(a) Hydrogenation of oil
(b) Ozonolysis
(c) Photolysis
(d) Hydration

Answer

Answer: (b) Ozonolysis
Explanation:
Ozonolysis is the cleavage of an alkene or alkyne with ozone to form organic compounds in which the multiple carbon-carbon bonds have been replaced by a double bond to oxygen. The outcome of the reaction depends on the type of multiple bonds being oxidized.

Bromine water can be also used to identify the position of a double bond. In this reaction, red-brown colour of bromine gets turned into coulorless indicating that there is a double bond.


Hydrocarbons Class 11 MCQ Pdf Question 3.
Nitrobenzene on reaction with conc. HNO3/H2SO4 at 80 – 100°C forms which one of the following products?
(a) 1, 2-Dinitrobenzene
(b) 1, 3-Dinitrobenzene
(c) 1, 4-Dinitrobenzene
(d) 1, 2, 4-Trinitrobenzene

Answer

Answer: (b) 1, 3-Dinitrobenzene
Explanation:
NO2 is an m-directing group and hence, 1, 3-dinitrobenzene is formed.
MCQ Questions for Class 11 Chemistry Chapter 13 Hydrocarbons with Answers 2


MCQ Of Hydrocarbons Class 11 Question 4.
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
(a) –C ≡ N
(b) -SO3H
(c) -COOH
(d) -NO2

Answer

Answer: (d) -NO2
Explanation:
The correct order of deactivation is
−NO2 > −C ≡ N > −SO32 H > −COOH


Hydrocarbon MCQ Class 11 Question 5.
Which of the following compounds will exhibit geometrical isomerism?
(a) 1 – Phenyl – 2 – butene
(b) 3 – Phenyl – 1 – butene
(c) 2 – Phenyl – 1 butene
(d) 1, 1 – Diphenyl – propene .

Answer

Answer: (a) 1 – Phenyl – 2 – butene
Explanation:
C6H5CH2−CH = CH−CH3 (1-phenyl-2-butene) exhibits the phenomenon of geometrical isomerism due to Cab = Cad structure, so its two isomers are possible which are given are follow :
MCQ Questions for Class 11 Chemistry Chapter 13 Hydrocarbons with Answers 3


Hydrocarbons MCQs With Answers Question 6.
The order of decreasing reactivity towards an electrophilic reagent, for the following : (i) Benzene (ii) Toluene (iii) Chlorobenzene (iv) Phenol would be:
(a) (i) > (ii) > (iii) > (iv)
(b) (ii) > (iv) > (i) > (iii)
(c) (iv) > (iii) > (ii) > (i)
(d) (iv) > (ii) > (i) > (iii)

Answer

Answer: (d) (iv) > (ii) > (i) > (iii)
Explanation:
Benzene having any activating group i.e., OH, R etc., undergoes electrophilic substitution very easily as compared to benzene itself. Thus toluene (C6H5CH3), phenol (C6H5OH) undergo electrophilic substitution very readily than benzene. Chlorine with +E and +M effect deactivates the ring due to strong -I effect. So, it is difficult to carry out the substitution in chlorobenzene than in benzene, so correct order is Phenol (iv) > Toluene (ii) > Benzene (i) > Chlorobenzene (iii)


Hydrocarbons MCQs Class 11 Question 7.
Pure methane can be prepared by
(a) Soda lime decarboxylation
(b) Kolbes electrolytic method
(c) Wurtz reaction
(d) Reduction with H2

Answer

Answer: (a) Soda lime decarboxylation
Explanation:
Methane cannot be prepared by either Wurtz reaction, Kolbes electrolytic method or by reduction of alkenes with H2. While acetic acid salt on heating with soda lime gives methane.
CH3COONa → CH4 + Na2CO3


Hydrocarbon Class 11 MCQ Question 8.
Hydrocarbon containing following bond is most reactive
(a) C ≡ C
(b) C = C
(c) C-C
(d) All of these

Answer

Answer: (a) C ≡ C
Explanation:
−C ≡ C−is most reactive because sp-hybridization.


MCQs On Hydrocarbons Class 11 Question 9.
The compound C3H4 has a triple bond, which is indicated by its reaction with
(a) Bromine water
(b) Bayers reagent
(c) Fehling solution
(d) Ammonical silver nitrate

Answer

Answer: (d) Ammonical silver nitrate
Explanation:
CH3−C ≡ C−H + AgNO3 → CH3 −C ≡ C−Ag
Propyne Ammonical Siver salt of Propyne


Hydrocarbons MCQs Question 10.
Isopropyl bromide on Wurtz reaction gives
(a) Hexane
(b) Propane
(c) 2, 3 – Dimethylbutane
(d) Neohexane

Answer

Answer: (c) 2, 3 – Dimethylbutane
Explanation:
CHMe2 − Br + 2Na + Br − CHMe2 + Br − CHMe2 \(\underrightarrow { dryEther } \) Me2CH − CHMe2 + 2NaBr


Hydrocarbons Class 11 MCQs Question 11.
Nitrobenzene on reaction with conc. HNO3/H2SO4 at 80 – 100°C forms which one of the following products?
(a) 1, 2-Dinitrobenzene
(b) 1, 3-Dinitrobenzene
(c) 1, 4-Dinitrobenzene
(d) 1, 2, 4-Trinitrobenzene

Answer

Answer: (b) 1, 3-Dinitrobenzene
Explanation:
NO2 is an m-directing group and hence, 1, 3-dinitrobenzene is formed.


Class 11 Hydrocarbons MCQs Question 12.
The angle strain in cyclobutane is
(a) 24°44
(b) 29°16
(c) 19°22
(d) 9°44

Answer

Answer: (d) 9°44
Explanation:
According to Baeyers strain theory, the amount of the strain is directly proportional to the angle through which a valency bond has deviated from its normal position . i.e., Amount of deviation) in cyclobutane (d) = (109∘28 − 90∘)/(2) = 9∘44


MCQ On Chapter Hydrocarbons Class 11 Question 13.
The first fraction obtained during the fractionation of petroleum is:
(a) Gasoline
(b) Diesel Oil
(c) Hydrocarbon Gases
(d) Kerosene Oil

Answer

Answer: (c) Hydrocarbon Gases
Explanation:
During fractionation or distillation of petroleum, gases light hydrocarbons in the form of gases are obtained from first fraction.


Hydrocarbons Class 11 Important Questions With Answers Pdf Question 14.
2 – Phenylpropene on acidic hydration gives
(a) 2 – Phenyl – 2 – propanol
(b) 2 – Phenyl – 1 – propanol
(c) 3 – Phenyl – 1 – propanol
(d) 1 – Phenyl – 2 – propanol

Answer

Answer: (a) 2 – Phenyl – 2 – propanol
Explanation:
Acidic hydration of 2-phenyl propene follows electrophilic reaction mechanism forming an intermediate 3° carbocation (more stable), there by forming 2-phenyl-2-propanol.
MCQ Questions for Class 11 Chemistry Chapter 13 Hydrocarbons with Answers 4


Class 11 Chemistry Chapter 13 MCQ Question 15.
The coal tar fraction which contains phenol is:
(a) Heavy Oil
(b) Light Oil
(c) Middle Oil
(d) Green Oil

Answer

Answer: (c) Middle Oil
Explanation:
The second fraction of coal-tar distillation which distills at 170°−230°C is called middle-oil or olic oil. It contains mainly phenol, naphthalene etc.


Class 11 Chemistry Hydrocarbons MCQs Question 16.
Which one of these is not true for benzene?
(a) There are three carbon-carbon single bonds and three carbon-carbon double bonds.
(b) Heat of hydrogenation of benzene is less than the theoretical value
(c) It forms only one type of mono substituted product
(d) The bond angle between carbon-carbon bonds is 120 Degree

Answer

Answer: (a) There are three carbon-carbon single bonds and three carbon-carbon double bonds.
Explanation:
There are three carbon-carbon single bonds and three carbon-carbon double bonds.


Hydrocarbons Class 11 MCQ Pdf Downloa Question 17.
The catalyst used in Friedel – Crafts reaction is
(a) Aluminium Chloride
(b) Anhydrous Aluminium Chloride
(c) Ferric Chloride
(d) Copper .

Answer

Answer: (b) Anhydrous Aluminium Chloride
Explanation:
It is a catalyst-based electrophilic substitution reaction. Lewis acids like anhydrous AlCl3, anhydrous FeCl3, BF3 and BiCl3 can be used as catalysts. The catalyst facilitates more effective attack by the electrophile (R+, RCO+ etc.) on the benzene ring.


Question 18.
Alkyl halides react with dialkyl copper reagents to give?
(a) Alkanes
(b) Alkenes
(c) Hydrogen
(d) Carbon

Answer

Answer: (a) Alkanes
Explanation:
Alkyl halides react with dialkyl copper reagents to give Alkanes.


Question 19.
The lowest alkene, that is capable of exhibiting geometrical isomerism is
(a) Ethene
(b) But – 1- ene
(c) But – 2 – ene
(d) Propene.

Answer

Answer: (c) But – 2 – ene
Explanation:
The lowest alkene which is capable of exhibiting geometrical isomerism is 2-Butene


Question 20.
Presence of a nitro group in a benzene ring
(a) Activates the ring towards electrophilic substitution
(b) Renders the ring basic
(c) Deactivates the ring towards nucleophilic substitution
(d) Deactivates the ring towards electrophilic substitution

Answer

Answer: (d) Deactivates the ring towards electrophilic substitution
Explanation:
−NO2 group shows −M effect, so withdraws the electron density from the ring and hence deactivate the ring towards electrophilic aromatic substitution.


We hope the given NCERT MCQ Questions for Class 11 Chemistry Chapter 13 Hydrocarbons with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Chemistry Hydrocarbons MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Chemistry MCQ:

MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics with Answers

Thermodynamics Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided Thermodynamics Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.

Class 11 Chemistry Chapter 6 MCQ With Answers

Chemistry Class 11 Chapter 6 MCQs On Thermodynamics

Thermodynamics Class 11 MCQ Question 1.
Third law of thermodynamics provides a method to evaluate which property?
(a) Absolute Energy
(b) Absolute Enthalpy
(c) Absolute Entropy
(d) Absolute Free Energy

Answer

Answer: (c) Absolute Entropy
Explanation:
The Third Law of Thermodynamics is concerned with the limiting behavior of systems as the temperature approaches absolute zero. Most thermodynamics calculations use only entropy differences, so the zero point of the entropy scale is often not important. However, the Third Law tells us about the completeness as it describes the condition of zero entropy.


MCQ On Thermodynamics Chemistry Class 11 Question 2.
One mole of which of the following has the highest entropy?
(a) Liquid Nitrogen
(b) Hydrogen Gas
(c) Mercury
(d) Diamond

Answer

Answer: (b) Hydrogen Gas
Explanation:
The measure of randomness of a substance is called entropy. Greater the randomness of molecules of a substance greater is the entropy. Here hydrogen gas has more entropy as it shows more randomness/disorderliness due to less molar mass than all the given substances and also in the gas phase.


Thermodynamics MCQ Class 11 Question 3.
The enthalpy of vaporisation of a substance is 8400 J mol-1 and its boiling point is –173°C. The entropy change for vaporisation is :
(a) 84 J mol-1K-1
(b) 21 J mol-1K-1
(c) 49 J mol-1K-1
(d) 12 J mol-1K-1

Answer

Answer: (a) 84 J mol-1K-1
Explanation:
∆S = (qrev)/ (T)
= (\(\frac {8400}{100}\))
= 84 J mol-1K-1


Thermodynamics Class 11 Chemistry MCQ Question 4.
The species which by definition has ZERO standard molar enthalpy of formation at 298 K is
(a) Br2(g)
(b) Cl2(g)
(c) H2O(g)
(d) CH4(g)

Answer

Answer: (b) Cl2(g)
Explanation:
This is possible only for elements, chlorine is a gas at this temperature, but bromine is a liquid, so it is possible only for chlorine.


Thermodynamics Chemistry Class 11 MCQ Pdf Question 5.
In a reversible process the system absorbs 600 kJ heat and performs 250 kJ work on the surroundings. What is the increase in the internal energy of the system?
(a) 850 kJ
(b) 600 kJ
(c) 350 kJ
(d) 250 kJ

Answer

Answer: (c) 350 kJ
Explanation:
∆E = q + w
= (600 – 250)
∆E = 350 J


MCQ Of Thermodynamics Class 11 Question 6.
Which of the following is true for the reaction? H2O (l) ↔ H2O (g) at 100° C and 1 atm pressure
(a) ∆S = 0
(b) ∆H = T ∆S
(c) ∆H = ∆U
(d) ∆H = 0

Answer

Answer: (a) ∆S = 0
Explanation:
Equilibrium
Therefore, ∆ G = 0 = ∆ H – T ∆ S
Or T∆ S = ∆H


Thermodynamics Chemistry MCQs With Answers Pdf Question 7.
Calculate the heat required to make 6.4 Kg CaC2 from CaO(s) and C(s) from the reaction: CaO(s) + 3 C(s) → CaC2(s) + CO (g) given that ∆f Ho (CaC2) = -14.2 kcal. ∆f H° (CO) = -26.4 kcal.
(a) 5624 kca
(b) 1.11 × 104 kcal
(c) 86.24 × 10³
(d) 1100 kcal

Answer

Answer: (b) 1.11 × 104 kcal
Explanation:
n = (Mass)/ (Molecular weight)
= (6.4 × 10³)/ (64)
= 100
For 1 mole of CaC2
∆ H = ∆Hf (CaC) + Hf (CO) – Hf (CaO)
= -14.2 – 26.4 + 151.6 = 111.1 kcal
For 100 moles, ∆H = 1.11 × 104 Kcal


Class 11 Chemistry Chapter 6 MCQ Question 8.
In a system where ∆E = -51.0 kJ, a piston expanded against a pext of 1.2 atm giving a change in volume of 32.0 L. What was the change in heat of this system?
(a) -36 kJ
(b) -13 kJ
(c) -47 kJ
(d) 24 kJ

Answer

Answer: (c) -47 kJ
Explanation:
w = -1.2 (32) × 101.3
= – 3.89 KJ
= -4 (approx.)
= ∆ E = – 51.0 KJ
Therefore, E = q + w
– 51 = q – 4
Therefore, q = – 47 KJ


MCQ Questions For Class 11 Chemistry Chapter 6 Question 9.
A system absorb 10 kJ of heat at constant volume and its temperature rises from 270 C to 370 C. The value of ∆ U is
(a) 100 kJ
(b) 10 kJ
(c) 0 kJ
(d) 1 kJ

Answer

Answer: (b) 10 kJ
Explanation:
At constant volume w = 0
Therefore, ∆ U = q = 10 kJ


Class 11 Chemistry Chapter 6 MCQ With Answers Question 10.
An ideal gas is taken around the cycle ABCA as shown in P-V diagram The next work done by the gas during the cycle is equal to:
MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics with Answers 1
(a) 12P1V1
(b) 6P1V1
(c) 5P1V1
(d) P1V1

Answer

Answer: (c) 5P1V1
Explanation:
Work done = Area under P-V graph = (\(\frac {1}{2}\)) (5P1) (2V1) = 5P1 V1


Class 11 Thermodynamics MCQ Question 11.
In which of the following process, a maximum increase in entropy is observed?
(a) Dissolution of Salt in Water
(b) Condensation of Water
(c) Sublimation of Naphthalene
(d) Melting of Ice

Answer

Answer: (c) Sublimation of Naphthalene
Explanation:
The order of entropy in solid, liquid and gas is gas > liquid > solid .Hence, in sublimation of naphthalene, maximum increase in entropy is observed.


Class 11 Chemistry Thermodynamics MCQ Question 12.
The bond energy (in kcal mol-1) of a C-C single bond is approximately
(a) 1
(b) 10
(c) 83-85
(d) 1000

Answer

Answer: (c) 83-85
Explanation:
C–C bond 83–85 kcal/mol

It is the energy required to break the bond .It is defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolysis reaction at 0 K (absolute zero)


Thermodynamics MCQs With Answers Pdf Class 11 Question 13.
Which thermodynamic function accounts automatically for enthalpy and entropy both?
(a) Helmholtz Free Energy (A)
(b) Internal Energy (E)
(c) Work Function
(d) Gibbs Free Energy

Answer

Answer: (d) Gibbs Free Energy
Explanation:
Gibbs free energy combines the effect of both enthalpy and entropy. The change in free energy (ΔG) is equal to the sum of the change of enthalpy (∆H) minus the product of the temperature and the change of entropy (∆S) of the system.
∆G = ∆H – T∆S
ΔG predicts the direction in which a chemical reaction will go under two conditions: (1) constant temperature and (2) constant pressure.
If ΔG is positive, then the reaction is not spontaneous (it requires the input of external energy to occur) and if it is negative, then it is spontaneous (occurs without the input of any external energy).


Thermodynamics Class 11 MCQ Questions Question 14.
The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283.0 kJ mol-1 respectively. The enthalpy of formation of carbon monoxide is:
(a) -676 kJ
(b) 110.5 kJ
(c) -110.5 kJ
(d) 676.5 kJ

Answer

Answer: (c) -110.5 kJ
Explanation:
C(s) + O2 à CO2 ∆H1 = -393.5
CO + (\(\frac {1}{2}\)) O2 à CO2 ∆H2 = -283.0
C(s) + (\(\frac {1}{2}\)) O2 à CO ∆H = ∆H1 – ∆H2
= -393.5 + 283 = -110.5 KJ


MCQ On Thermodynamics Class 11 Chemistry Question 15.
The amount of the heat released when 20 ml 0.5 M NaOH is mixed with 100 ml 0.1 M HCl is x kJ. The heat of neutralization is
(a) -100 × kJ/mol
(b) -50 × kJ/mol
(c) 100 × KJ/mol
(d) 50 × kJ/mol

Answer

Answer: (a) -100 × kJ/mol
Explanation:
Normality of NaOH = Molarity × acidity
= 0.5 × 1 = 0.5 N
Total heat q produced = x kJ
Heat of neutralisation
= [(q)/ (Volume of acid or base)] ×1000× (1/normality of acid or base)
= (\(\frac {x}{20}\)) × 1000 × (\(\frac {1}{0.5}\))
= 100 x kJmol-1
Since heat is liberated, heat of neutralisation = −100 x kJmol-1


MCQs On Thermodynamics Chemistry Question 16.
Based on the first law of thermodynamics, which one of the following is correct?
(a) For an isothermal process, q = +w
(b) For an isochoric process, ΔU = -q
(c) For an adiabatic process, ΔU = -w
(d) For a cyclic process, q = -w

Answer

Answer: (d) For a cyclic process, q = -w
Explanation:
(1) ΔU = q + w. For an isochoric process, w = −PΔV = 0. Hence, ΔU = qv
(2) For an adiabatic process, q = 0. Hence, ΔU = w
(3 ) For an isothermal process, ΔU = 0 Hence, q = −w
(4) For a cyclic process , ΔU = 0. Hence, q = −w


Thermodynamics Chemistry Class 11 MCQ Question 17.
A system absorb 10 kJ of heat at constant volume and its temperature rises from 270°C to 370°C. The value of ∆ U is
(a) 100 kJ
(b) 10 kJ
(c) 0 kJ
(d) 1 kJ

Answer

Answer: (b) 10 kJ
Explanation:
At constant volume w = 0
Therefore, ∆ U = q = 10 kJ


Chemical Thermodynamics MCQ Question 18.
The temperature of the system decreases in an ______.
(a) Adiabatic Compression
(b) Isothermal Expansion
(c) Isothermal Compression
(d) Adiabatic Expansion

Answer

Answer: (d) Adiabatic Expansion
Explanation:
In adiabatic process heat is neither added nor removed from system. So the work done by the system (expansion) in adiabatic process will result in decrease of internal energy of that system (from first law).
As internal energy is directly proportional to the change in temperature there will be temperature drop in an adiabatic process.


Thermodynamics MCQ Chemistry Question 19.
Which of the following salts will have maximum cooling effect when 0.5 mole of the salt is dissolved in same amount of water. Integral heat of solution at 298 K is given for each salt?
(a) KNO3 (∆H = 35.4 kJ mol-1)
(b) NaCl (∆H = 5.35 kJ mol-1)
(c) HBr (∆H = -83.3 kJ mol-1)
(d) KOH ( ∆H = -55.6 kJ mol-1)

Answer

Answer: (a) KNO3 (∆H = 35.4 kJ mol-1)
Explanation:
More the heat absorbed, more will be the cooling effect. Hence, more the positive value of ∆ H, more the cooling effect.


Question 20.
Standard enthalpy of vapourisation ΔHvap for water at 100oC is 40.66 kJmol-1. The internal energy of vapourisation of water at 100°C (in kJmol-1) is (Assume water vapour to behave like an ideal gas)
(a) 43.76
(b) 40.66
(c) 37.56
(d) -43.76

Answer

Answer: (c) 37.56
Explanation:
For gaseous reactants and products, we have a relation between standard enthalpy of vaporization (ΔHvap) and standard internal energy (ΔE) as-
ΔHvap = ΔE+ Δng​ RT whereas,
Δng = n2​ − n1, i.e., difference between no. of moles of reactant and product.
For vaporization of water,
H2O (l) → H2O(g)
​Therefore, Δng = 1 − 0 = 1
Therefore, ΔHvap = ΔE+ Δng RT
⇒ ΔE = ΔHvap − Δng
​RT = 40.63 − (1×8.314×10-3 × 373) = 37.53 KJ/mol
Hence the value ΔE for this process will be 37.53KJ/mol.


We hope the given NCERT MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Chemistry Thermodynamics MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Chemistry MCQ:

MCQ Questions for Class 11 Chemistry Chapter 5 States of Matter with Answers

States of Matter Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 5 States of Matter with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided States of Matter Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.

Class 11 Chemistry Chapter 5 MCQ With Answers

Chemistry Class 11 Chapter 5 MCQs On States of Matter

States Of Matter Class 11 MCQ Question 1.
The vapour pressure of water at 300 K in a closed container is 0.4 atm. If the volume of container is doubled, its vapour pressure at 300 K will be
(a) 0.8 atm
(b) 0.2 atm
(c) 0.4 atm
(d) 0.6 atm

Answer

Answer: (c) 0.4 atm
Explanation:
Vapour pressure depends on T only and it does not depend on container volume.


States Of Matter MCQ Question 2.
Name the liquid with higher vapour pressure in the following pairs:
(a) Alcohol, glycerine (b) Petrol, kerosene (c) mercury, water.
(a) Alcohol, Water, Petrol
(b) Petrol, Water, Alcohol
(c) Alcohol, Petrol, Water
(d) None of these

Answer

Answer: (c) Alcohol, Petrol, Water
Explanation:
The vapour pressure of the liquid is inversely proportional to the magnitude of the intermolecular forces of attraction present. Based on this, the liquid with higher vapour pressure in the different pairs is: (a) Alcohol, (b) Petrol, (c) Water.


States Of Matter MCQ With Answers Question 3.
The rate of diffusion methane is twice that of X. The molecular mass of X is
(a) 64.0
(b) 32.0
(c) 40
(d) 80

Answer

Answer: (a) 64.0
Explanation:
Let rate of diffusion of gas x, r1 = a
Therefore, rate of diffusion of methane, r2 = 2 a
According to Grahams Law of Diffusion
(\(\frac {r_1}{r_2}\)) = (\(\sqrt{\frac {M_2}{M_1}}\))
M1 = Molecular mass of gas x
M2 = Molecular mass of Methane = 16 g
Therefore, (\(\frac {a}{2a}\)) = (\(\sqrt{\frac {16}{M_2}}\))
Squaring both the sides, (\(\frac {1}{4}\)) = (\(\frac {16}{M_2}\))
or, M2 = 16 × 4 = 64 g


Class 11 Chemistry Chapter 5 MCQ With Answers Question 4.
The state of matter that shows the uniformity of behavior :
(a) Solid Liquid
(b) Liquid
(c) Gas
(d) None of the Above

Answer

Answer: (c) Gas
Explanation:
Of the three states of matter, the gaseous state is the simplest and shows greatest uniformity in behaviour. Gases show almost similar behaviour irrespective of their chemical nature. This state is characterized by:
Gases maintain neither the volume nor the shape. They completely fill the container in which they are placed.

They expand appreciably on heating. Gases are highly compressible. The volume of the gas decreases when the pressure increases. They diffuse rapidly into space. Gases exert equal pressure in all directions.
All gases are colourless except a few e.g. chlorine (greenish yellow) bromine (reddish brown), nitrogen dioxide (reddish brown)

The behaviour of gases can be described by certain quantitative relationships called gas laws. They give the relationship between mass, pressure, volume and temperature.


MCQ On States Of Matter Class 11 Question 5.
A gas deviates from ideal behavior at a high pressure because its molecules:
(a) Attract one another
(b) Show the Tyndall Effect
(c) Have kinetic energy
(d) Are bound by covalent bonds

Answer

Answer: (a) Attract one another
Explanation:
The basic concept of the kinetic-molecular theory give us the information why real gases deviate from ideal behavior. The molecules of an ideal gas are assumed to occupy no space and have no attractions for one another. Real molecules, however, do have finite volumes, and they do attract one another. So, a gas deviates from ideal behavior at a high pressure because its molecules attract one another.


MCQ Questions For Class 11 Chemistry Chapter 5 Question 6.
The value of universal gas constant R depends on
(a) Temperature of Gas
(b) Volume of Gas
(c) Number of Moles of Gas
(d) Units of Volume,Temperature and Pressure

Answer

Answer: (d) Units of Volume,Temperature and Pressure
Explanation:
The value of the gas constant R depends on the units used for pressure, volume and temperature.


Class 11 Chemistry Chapter 5 MCQ Question 7.
The critical volumes of four gases A, B, C, D are respectively 0.025 L, 0.312 L, 0.245 L, 0.432 L, the gas with highest value of van der Wall constant b is
(a) A
(b) B
(c) C
(d) D

Answer

Answer: (d) D
Explanation:
Vc = 3b = 3 × 4N × (4/3) πr³


MCQ Of States Of Matter Class 11 Question 8.
Which of the following statement is wrong for gases?
(a) Gases do not have definite shape and volume
(b) Volume of the gas is equal to the volume of the container confining the gas
(c) Confined gas exert uniform pressure on the wall of the container in all directions
(d) Mass of the gas cannot be determined by weighing a container in which it is contained

Answer

Answer: (d) Mass of the gas cannot be determined by weighing a container in which it is contained
Explanation:
Mass of the gas = mass of the cylinder including gas – mass of empty cylinder. So mass of a gas can be determined by weighing the container in which it is enclosed. Thus, the statement (d) is wrong for gases.


MCQs On States Of Matter Class 11 Question 9.
In van der Waal equation of state of gas laws, the constant b is a measure of
(a) Intermolecular collisions per unit volume
(b) Intermolecular attraction
(c) Volume occupied by the molecules
(d) Intermolecular repulsions

Answer

Answer: (c) Volume occupied by the molecules
Explanation:
In van der Waals equation of state of the gas law, the constant b is a measure of the volume occupied by the molecules. It gives the effective size of the gas molecules. The greater value of b indicates a larger size of the molecules and smaller compressible volume.


State Of Matter Class 11 MCQ Question 10.
The volume of 2.8 g of carbon monoxide at 27°C and 0.0821 atm is
(a) 30 L
(b) 3 L
(c) 0.3 L
(d) 1.5 L

Answer

Answer: (a) 30 L
Explanation:
According to the ideal gas equation, we have
PV = nRT
​PV = (\(\frac {w}{M}\)) RT
​V = ​ (\(\frac {w}{M}\)) (\(\frac {RT}{P}\))
Given values are:
w = 2.8 g
M = Molar mass of CO = 28 g mol-1
T = 27°C = (273 + 27) = 300 K
P = 0.821 atm
R = 0.0821 L atm mol-1 K-1
Putting the values in the formula we get :
V = (2.8 g /28 g mol-1) × (0.0821 L atm mol-1 K-1) × (300 K)/(0.821 atm)
= 3 L


MCQ Questions On States Of Matter Class 11 Question 11.
If 20cm³ gas at 1 atm. is expanded to 50 cm³ at constant T, then what is the final pressure
(a) 20 × 150
(b) 50 × 120
(c) 1 × 120 × 50
(d) None of these

Answer

Answer: (a) 20 × 150
Explanation:
At constant T, P1V1 = P2V2
1 × 20 = P2 × 50;
P2 = (\(\frac {20}{ 50}\)) × 1


States Of Matter MCQ Class 11 Question 12.
How many of the know elements exist as gases at 25°C?
(a) 9
(b) 11
(c) 12
(d) 15

Answer

Answer: (b) 11


MCQ States Of Matter Class 11 Question 13.
The states of matter having no definite shape but definite volume:
(a) Gas
(b) Liquid
(c) Solid
(d) None of the Above

Answer

Answer: (b) Liquid
Explanation:
In a liquid, particles will flow or glide over one another, but stay toward the bottom of the container. The attractive forces between particles are strong enough to hold a specific volume but not strong enough to keep the molecules sliding over each other.


Class 11 States Of Matter MCQ Question 14.
The rise or fall of a liquid within a tube of small bore is called:
(a) Surface Tension
(b) Capillary Action
(c) Viscosity
(d) Formation of Curvature

Answer

Answer: (b) Capillary Action
Explanation:
Capillarity, rise or depression of a liquid in a small passage such as a tube of small cross-sectional area, like the spaces between the fibres of a towel or the openings in a porous material. Capillarity is not limited to the vertical direction. Water is drawn into the fibres of a towel, no matter how the towel is oriented.

Liquids that rise in small-bore tubes inserted into the liquid are said to wet the tube, whereas liquids that are depressed within thin tubes below the surface of the surrounding liquid do not wet the tube. Water is a liquid that wets glass capillary tubes; mercury is one that does not. When wetting does not occur, capillarity does not occur.

Capillarity is the result of surface, or interfacial, forces. The rise of water in a thin tube inserted in water is caused by forces of attraction between the molecules of water and the glass walls and among the molecules of water themselves. These attractive forces just balance the force of gravity of the column of water that has risen to a characteristic height. The narrower the bore of the capillary tube, the higher the water rises. Mercury, conversely, is depressed to a greater degree, the narrower the bore.


States Of Matter Class 11 MCQ Questions Question 15.
The theory which explains that gases consist of molecules, which are in rapid option is known as:
(a) Daltons Atomic Theory
(b) Bohrs Theory
(c) Rutherfords Atomic Theory
(d) Kinetic Molecular Theory

Answer

Answer: (d) Kinetic Molecular Theory
Explanation:
The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term “molecule” will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)

Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.
The molecules composing the gas are negligibly small compared to the distances between them.
The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.

Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).
The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.


Question 16.
Falling drop of water is spherical due to:
(a) Hydrogen Bonding
(b) Surface Tension
(c) Capillary Action
(d) Vlscosity

Answer

Answer: (b) Surface Tension
Explanation:
Raindrops take up the spherical shape due to the surface tension of water which is caused due to the tendency of water molecules to stick together. The spherical shape is having the least possible surface area due to which it can resist any of the external force in the atmosphere.


Question 17.
The rates of diffusion of gases are inversely proportional to square root of their densities . This statement refers to :
(a) Daltons Law
(b) Grahams Law
(c) Avogadros Law
(d) None of the Above

Answer

Answer: (b) Grahams Law
Explanation:
Grahams law states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight. In the same conditions of temperature and pressure, the molar mass is proportional to the mass density. Therefore the rate of diffusion of different gases is inversely proportional to the square root of their mass densities.
r α (\(\sqrt{\frac {1}{d}}\))
and r α (\(\sqrt{\frac {1}{M}}\))


Question 18.
The law, which states that at constant temperature, the volume of a given mass of gas is inversely proportional is pressure, is known as:
(a) Boyles law
(b) Charles law
(c) Combine gas law
(d) Avogadros law

Answer

Answer: (a) Boyles law
Explanation:
In 1662 Robert Boyle studied the relationship between volume and pressure of a gas of fixed amount at constant temperature. He observed that volume of a given mass of a gas is inversely proportional to its pressure at a constant temperature. Boyles law, published in 1662, states that, at constant temperature, the product of the pressure and volume of a given mass of an ideal gas in a closed system is always constant. It can be verified experimentally using a pressure gauge and a variable volume container. It can also be derived from the kinetic theory of gases: if a container, with a fixed number of molecules inside, is reduced in volume, more molecules will strike a given area of the sides of the container per unit time, causing a greater pressure.
A statement of Boyles law is as follows:
The volume of a given mass of a gas is inversely related to pressure when the temperature is constant.​
V ∝ (\(\frac {1}{P}\)) meaning “Volume is inversely proportional to Pressure”, or
P ∝ (\(\frac {1}{V}\)) meaning “Pressure is inversely proportional to Volume”, or
where P is the pressure, and V is the volume of a gas, and k1 is the constant in this equation.


Question 19.
If helium and methane are allowed to diffuse out of the container under the similar conditions of temperature and pressure, then the ratio of rate of diffusion of helium to methane is:
(a) 2 : 1
(b) 1 : 2
(c) 3 : 5
(d) 4 : 1

Answer

Answer: (a) 2 : 1
Explanation:
According to Grahams law
(\(\frac {r_1}{r_2}\)) = (\(\sqrt{\frac {M_1}{M_2}}\))
(rHe/rCH4) = (\(\sqrt{\frac {16}{4}}\))
= (\(\frac {1}{2}\))


Question 20.
When you heat a sample of gas, what happens to the particles that make up the gas?
(a) The particles move faster.
(b) The particles break apart
(c) The particles get smaller
(d) The particles become more dense

Answer

Answer: (a) The particles move faster.
Explanation:
There is a great deal of empty space between particles, which have a lot of kinetic energy. The particles move very fast and collide into one another when the gas is heated up, causing them to diffuse, or spread out, until they are evenly distributed throughout the volume of the container.


We hope the given NCERT MCQ Questions for Class 11 Chemistry Chapter 5 States of Matter with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Chemistry States of Matter MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Chemistry MCQ:

MCQ Questions for Class 11 Chemistry Chapter 10 The s-Block Elements with Answers

The s-Block Elements Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 10 The s-Block Elements with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided The s-Block Elements Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.

Class 11 Chemistry Chapter 10 MCQ With Answers

Chemistry Class 11 Chapter 10 MCQs On The s-Block Elements

S Block Elements Class 11 MCQ Question 1.
CsOH is
(a) Strongly basic
(b) Weakly basic
(c) Slightly acidic
(d) Amphoteric.

Answer

Answer: (a) Strongly basic
Explanation:
Caesium hydroxide or cesium hydroxide (CsOH) is a chemical compound consisting of caesium ions and hydroxide ions. It is a strong base (pkb = -1.76), much like the other alkali metal hydroxides such as sodium hydroxide and potassium hydroxide.


MCQ On S Block Elements Class 11 Question 2.
Solvays process is used for the manufacture of :
(a) NaOH
(b) (Na2CO3. 10H2O)
(c) K2CO3
(d) Na2O2

Answer

Answer: (b) (Na2CO3. 10H2O)
Explanation:
The Solvay process or ammonia soda process is used for the manufacture of sodium carbonate
(Na2CO3. 10H2O).


S-Block Elements MCQs With Answers Pdf Question 3.
Alkali metals give a _________________when dissolved in liquid ammonia
(a) Deep blue solution
(b) Colourless
(c) Red colour
(d) None of the Above

Answer

Answer: (a) Deep blue solution
Explanation:
When an alkali metal is dissolved in liquid ammonia, it results in the formation of a deep blue coloured solution. The ammoniated electrons absorb energy corresponding to a red region of visible light. Therefore, the transmitted light is blue in colour.


S Block Elements MCQ Question 4.
What are Oxo-Acids?
(a) Acid containing Oxygen
(b) Acid containing Sulphur
(c) Acid containing Carbon
(d) None of the Above

Answer

Answer: (a) Acid containing Oxygen
Explanation:
An oxyacid, oxoacid, or ternary acid is an acid that contains oxygen. Specifically, it is a compound that contains hydrogen, oxygen, and at least one other element, with at least one hydrogen atom bond to oxygen that can dissociate to produce the H+ cation and the anion of the acid. e.g., carbonic acid, H2CO3 (OC(OH)2; sulphuric acid, H2SO4 (O2S(OH)2).


MCQ Of S Block Elements Class 11 Question 5.
In curing cement plasters, water is sprinkled from time to time. This helps in
(a) Converting sand into silicic acid
(b) Keeping it cool
(c) Developing interlocking needle like crystals of hydrated silicates
(d) Hydrating sand and gravel mixed with cement.

Answer

Answer: (c) Developing interlocking needle like crystals of hydrated silicates
Explanation:
Water develops interlocking needle ­like crystals of hydrated silicates. The reactions involved are the hydration of calcium aluminates and calcium silicates which change into their colloidal gels.

At the same time, some calcium hydroxide and aluminium hydroxides are formed as precipitates due to hydrolysis. Calcium hydroxide binds the particles of calcium silicate together while aluminium hydroxide fills the interstices rendering the mass impervious


Class 11 Chemistry S Block Elements MCQ Question 6.
The substance not likely to contain CaCO3 is
(a) Dolomite
(b) A marble statue
(c) Calcined gypsum
(d) Sea shells.

Answer

Answer: (c) Calcined gypsum
Explanation:
The composition of gypsum is CaSO4 ∙2H2O. It does not have CaCO3


MCQs On S Block Elements Question 7.
Carnallite is the mineral of:
(a) Na
(b) Ca
(c) Mg
(d) None of the Above

Answer

Answer: (c) Mg
Explanation:
Carnallite is an evaporite mineral a hydrated potassium magnesium chloride. It is variably coloured yellow to white, reddish or blue. It occurs with a sequence of potassium and magnesium evaporite. It is an uncommon double chloride mineral that forms under specific conditions. It is an important source of potash.


MCQ On S Block Elements Question 8.
The wire of flash bulb is made up of:
(a) Mg
(b) Ag
(c) Cu
(d) Ba

Answer

Answer: (a) Mg
Explanation:
Magnesium metal is used for the preparation of the wire of flash bulb.


Class 11 Chemistry Chapter 10 MCQ Question 9.
The basic strength of which hydroxide is maximum
(a) LiOH
(b) NaOH
(c) Ca (OH)2
(d) KOH.

Answer

Answer: (d) KOH.
Explanation:
The basic strength increases down the group and decreases along a period.


S-Block Elements MCQs With Answers Question 10.
The composition of Sorels cement is
(a) KCl × MgCl2× 6H2O
(b) MgCl2 × 5MgO × (xH2O)
(c) MgCO3 × CaCO3
(d) CaSO4 × 2H2O

Answer

Answer: (b) MgCl2 × 5MgO × (xH2O)
Explanation:
Mixture of MgCl2 and MgO is called Sorels cement. It is MgCl2 × 5MgO × (xH2O)


Class 11 S Block Elements MCQs Question 11.
Which of the following alkali metals has the least melting point?
(a) Na
(b) K
(c) Rb
(d) Cs.

Answer

Answer: (d) Cs.
Explanation:
Atomic size increases as we move down the alkali group. As a result, the binding energies of their atoms in the crystal lattice decrease. Also, the strength of metallic bonds decreases on moving down a group in the periodic table. This causes a decrease in the melting point. Among the given metals, Cs is the largest and has the least melting point.


MCQ Of S Block Elements Question 12.
The basic strength of which hydroxide is maximum
(a) LiOH
(b) NaOH
(c) Ca(OH)2
(d) KOH.

Answer

Answer: (d) KOH.
Explanation:
The basic strength increases down the group and decreases along a period.


S Block Elements MCQs With Answers Pdf Question 13.
The hydration energy of Mg2+ is larger than that of
(a) A13+
(b) Na+
(c) Be26
(d) Mg3+

Answer

Answer: (b) Na+
Explanation:
Hydration energy depends on charge of ion and ionic radius. Higher the charge, greater the hydration energy. On the other hand, smaller the size, greater the hydration energy. Charge is considered first for comparison. Hence, Mg2+ has higher hydration energy than Na+.


S Block Elements MCQ Class 11 Question 14.
Milk of lime reacts with chlorine to form ________, a constituent of bleaching powder.
(a) Ca(OCI)2
(b) Ca(CIO2)2
(c) Ca(CIO3)2
(d) Ca(CIO4)2

Answer

Answer: (a) Ca(OCI)2
Explanation:
Milk of lime reacts with chlorine to form bleaching powder.
2Ca(OH)2 + 2Cl2 → CaCl2 + Ca(OCl)2 + 2H2O


MCQ Questions On S Block Elements Question 15.
What happens when Calcium carbonate is heated to 1200 K?
(a) Carbon Monoxide
(b) Calcium hydroxide
(c) Calcium Oxide
(d) Both (2) and (3) are correct

Answer

Answer: (d) Both (2) and (3) are correct
Explanation:
Calcium carbonate is strongly heated until it undergoes thermal decomposition to form calcium oxide and carbon dioxide. The calcium oxide (unslaked lime) is dissolved in water to form calcium hydroxide (limewater).
CaCO3 → CaO + CO2


S Block Elements Class 11 MCQ Pdf Question 16.
The salt that is added to table salt to make it flow freely in rainy season is
(a) KCl
(b) KI
(c) Ca3(PO4)2
(d) Na3PO4

Answer

Answer: (c) Ca3(PO4)2
Explanation:
Ca3(PO4)2. Both Ca and P are needed by human beings. Also they prevent moisture absorbing power of other components such as MgCl2, CaCl2, CaSO4 and MgSO4 present in commercial sodium chloride.


MCQ Questions For Class 11 Chemistry Chapter 10 Question 17.
CsOH is
(a) Strongly basic
(b) Weakly basic
(c) Slightly acidic
(d) Amphoteric.

Answer

Answer: (a) Strongly basic
Explanation:
Caesium hydroxide or cesium hydroxide (CsOH) is a chemical compound consisting of caesium ions and hydroxide ions. It is a strong base (pkb = -1.76), much like the other alkali metal hydroxides such as sodium hydroxide and potassium hydroxide.


MCQ Questions On S-Block Elements Pdf Question 18.
Which of the following metals is not manufactured by electrolysis?
(a) Na
(b) Mg
(c) Al
(d) Fe

Answer

Answer: (d) Fe
Explanation:
Method of extraction of a metal depends on the reactivity of the metal. Iron (Fe) is not manufacture by electrolysis. Moderately reactive metals like zinc and iron are extracted by reduction of their oxides using carbon.


Question 19.
The wire of flash bulb is made up of:
(a) Mg
(b) Ag
(c) Cu
(d) Ba

Answer

Answer: (a) Mg
Explanation:
Magnesium metal is used for the preparation of the wire of flash bulb.


Question 20.
Usually CaCl2 is preferred over NaCl for cleaning snow on roads particularly in very cold countries this is because
(a) NaCl makes the road slippery but CaCl2 doesnot
(b) CaCl2 is hygroscopic but NaCl is not
(c) CaCl2 is less soluble in H2O than NaCl
(d) eutectic mixtue of CaCl/H2O freezes at -55°C while that of NaCl/H2O freezes at -18°C

Answer

Answer: (d) eutectic mixtue of CaCl/H2O freezes at -55°C while that of NaCl/H2O freezes at -18°C
Explanation:
Eutectic mixture of CaCl2/H2O freezes at -55°C while that of NaCl/H2O freezes at -18°C.

When added to ice. NaCl and CaCl2 ionise into respective ions (Na+, Ca2+ & Cl) this leads to increase in the number of particles (ions). As CaCl2 has greater number of ions, Increasing the number of particles /ions increases the vant hoff factor. Vant hoff factor is directly proportional to the depression in freezing point thus lowers the freezing point of ice causing preventing the formation of ice.


We hope the given NCERT MCQ Questions for Class 11 Chemistry Chapter 10 The s-Block Elements with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Chemistry The s-Block Elements MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Chemistry MCQ:

MCQ Questions for Class 11 Chemistry Chapter 2 Structure of Atom with Answers

Structure of Atom Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 2 Structure of Atom with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided Structure of Atom Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.

Class 11 Chemistry Chapter 2 MCQ With Answers

Chemistry Class 11 Chapter 2 MCQs On Structure of Atom

Structure Of Atom Class 11 MCQ Question 1.
The increasing order (lowest first) for the values of e/m (charge/mass) for
(a) e, p, n, α
(b) n, p, e, α
(c) n, p, α, e
(d) n, α, p, e

Answer

Answer: (d) n, α, p, e
Explanation:
(i) (e/m) for (i) neutron = (\(\frac{0}{1}\)) = 0
(ii) α− particle = (\(\frac{2}{4}\)) = 0.5
(iii) Proton = (\(\frac{1}{1}\)) = 1
(iv) electron = (\(\frac{1}{1837}\)) = 1837.


Structure Of Atom MCQ Class 11 Question 2.
The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol-1. The energy required to excite the electron in the atom from n = 1 to n = 2 is
(a) 8.51 × 105 J mol-1
(b) 6.56 × 105 J mol-1
(c) 7.56 × 105 J mol-1
(d) 9.84 × 105 J mol-1

Answer

Answer: (d) 9.84 × 105 J mol-1
Explanation:
Energy required when an electron makes transition from n = 1 to n = 2
E2=−(1.312 × 106 × (1)²)/(2²)
= −3.28 × 105 J mol-1
E1 = −1.312 × 106 J mol-1
ΔE = E2 − E1
=−3.28 × 105−(−13.2 × 106)
ΔE = 9.84×105 J mol-1


Class 11 Chemistry Chapter 2 MCQ Question 3.
or a given principal level n = 4, the energy of its subshells is in the order
(a) s < p < d < f
(b) s > p > d > f
(c) s < p < f < d
(d) f < p < d < s

Answer

Answer: (a) s < p < d < f
Explanation:
Order of energy is:
s < p < d < f
MCQ Questions for Class 11 Chemistry Chapter 2 Structure of Atom with Answers 1


MCQ Questions For Class 11 Chemistry Chapter 2 Question 4.
A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at:
(a) 518 nm
(b) 1035 nm
(c) 325 nm
(d) 743 nm

Answer

Answer: (d) 743 nm
Explanation:
From Law of Conservation of energy, energy of absorbed photon must be equal to combined energy of two emitted photons.
ET = E1 + E2 ….. (1)
Where E1 is Energy of first emitted photon emitted and E2is Energy of second emitted photon.
Energy E and wavelength λ of a photon are related by the equation
E= (hc)/ (λ)….. (2)
Where Plancks constant = h, c is velocity of light.
Substituting the values from (2) in (1) we get
(hc/λT) = (hc)/ (λ1) + (hc)/ (λ2)
Or (\(\frac{1}{λ_T}\)) = (\(\frac{1}{λ_1}\)) + (\(\frac{1}{λ_2}\)) …… (3)
Substituting given values in (3) we get
(\(\frac{1}{355}\)) = (\(\frac{1}{680}\)) + (\(\frac{1}{λ_2}\))
Or \(\frac{(1)}{(λ_T)}\) = (\(\frac{1}{355}\)) − (\(\frac{1}{680}\))
⇒ (\(\frac{1}{λ_2}\)) = (680 − 355)/ (355 × 680)
⇒ λ2 = 742.77nm


Class 11 Chemistry Chapter 2 MCQ With Answers Question 5.
Which of the following statements in relation to the hydrogen atom is correct?
(a) 3s orbital is lower in energy than 3p orbital
(b) 3p orbital is lower in energy than 3d orbital
(c) 3s and 3p orbitals are of lower energy than 3d orbital
(d) 3s, 3p and 3d orbitals all have the same energy

Answer

Answer: (d) 3s, 3p and 3d orbitals all have the same energy
Explanation:
A hydrogen atom has 1st configuration and these its, 3p and 3d orbitals will have same energy wrt 1s orbital.


MCQ Of Structure Of Atom Class 11 Question 6.
The magnetic quantum number specifies
(a) Size of orbitals
(b) Shape of orbitals
(c) Orientation of orbitals
(d) Nuclear Stability

Answer

Answer: (c) Orientation of orbitals
Explanation:
The magnetic quantum number specifies orientation of orbitals.


MCQ On Structure Of Atom Class 11 Question 7.
The electronic configuration of silver atom in ground state is
(a) [Kr]3d104s1
(b) [Xe]4f145d106s1
(c) [Kr]4d105s1
(d) [Kr]4d95s2

Answer

Answer: (c) [Kr]4d105s1
Explanation:
The electronic configuration of Ag in ground state is [Kr]4d105s1


Ch 2 Chemistry Class 11 MCQ Question 8.
Which of the following element has least number of electrons in its M-shell?
(a) K
(b) Mn
(c) Ni
(d) Sc

Answer

Answer: (a) K
Explanation:
K = 19 = 1s²2s22p63s23p6s1
3s23p6 = m-shell
= k has only 8 electrons in M shell


Structure Of Atom Class 11 MCQ Questions And Answers Pdf Question 9.
Which one of the following sets of ions represents a collection of isoelectronic species? (Atomic nos.: F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)
(a) K+, Ca2+, Sc3+, Cl
(b) Na+, Ca2+ , Sc3+, F
(c) K+, Cl, Mg2+, Sc3+
(d) Na+, Mg2+, Al3+, Cl

Answer

Answer: (a) K+, Ca2+, Sc3+, Cl
Explanation:
Isoelectronic species are those which have same number of electrons.
K+ = 19 – 1 = 18; Ca2+ = 20 – 2 = 18; Sc3+ = 21 – 3 = 18; Cl = 17 + 1 = 18
Thus all these ions have 18 electrons in them.


Atomic Structure Class 11 MCQ Questions Question 10.
In the ground state, an element has 13 electrons in its M-shell. The element is_____.
(a) Copper
(b) Chromium
(c) Nickel
(d) Iron

Answer

Answer: (b) Chromium
Explanation:
M shell means it is third shell ⇒ n = 3
Number of electrons in M shell = 13
⇒ 3s23p63d5
The electronic configuration is: (1s2) (2s2 2p6) (3s2 3p6 3d5) (4s1)
The element is chromium is Cr.


Chapter 2 Chemistry Class 11 MCQ Question 11.
The electrons of the same orbitals can be distinguished by
(a) Principal quantum number
(b) Azimuthal quantum number
(c) Spin quantum number
(d) Magnetic quantum number

Answer

Answer: (c) Spin quantum number
Explanation:
Electrons occupying the same orbital are distinguished by Spin quantum number.
For spin Quantum number it has two values +1/2 or -1/2,
Hence the value of n, l , m are same for the two electrons occupying in the same orbitals, but only the is different, which is
Therefore, Spin quantum number explains the direction through which the electron spins in an orbital. so obviously there are only 2 possible directions. Which is either clockwise or anticlockwise.
So the electron which are available in the same orbitals, must have opposite spins. Hence spin quantum number distinguished b/w the two electrons.


Class 11 Chemistry Ch 2 MCQ Question 12.
Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively:
(a) 12 and 4
(b) 12 and 5
(c) 16 and 4
(d) 16 and 5

Answer

Answer: (b) 12 and 5
Explanation:
24Cr → 1s2 2s22p6 3s2 3p6 3d5 4s1
As we know for p, l = 1 and d, l = 2
For l = 1, total number of electrons = 12 [2p6 and 3p6]
For l = 2, total number of electrons = 5 [3d5]


Chemistry Class 11 Chapter 2 MCQ Question 13.
A body of mass 10 mg is moving with a velocity of 100 ms-1. The wavelength of de-Broglie wave associated with it would be (Note: h = 6.63 × 10-34 Js)
(a) 6.63 × 10-37 m
(b) 6.63 × 10-31 m
(c) 6.63 × 10-34 m
(d) 6.63 × 10-35 m

Answer

Answer: (b) 6.63 × 10-31 m
Explanation:
m = 10 mg
= 10 × 10-6 kg
v = 100 ms-1
λ = \(\frac {(h)}{(mv)}\)
= (6.63×10-34)/ (10 × 10-6 × 100)
= 6.63 × 10-31 m


MCQ Of Chapter 2 Chemistry Class 11 Question 14.
The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol-1. The energy required to excite the electron in the atom from n = 1 to n = 2 is
(a) 8.51 × 105 J mol-1
(b) 6.56 × 105 J mol-1
(c) 7.56 × 105 J mol-1
(d) 9.84 × 105 J mol-1

Answer

Answer: (d) 9.84 × 105 J mol-1
Explanation:
Energy required when an electron makes transition from n = 1 to n = 2
E2 = −(1.312 × 106 × (1)²)/(2²)
= −3.28 × 105 J mol-1
E1 = −1.312 × 106 J mol-1
ΔE = E2 − E1
= −3.28 × 105−(−13.2 × 106)
ΔE = 9.84 × 105 J mol-1


Class 11 Chemistry Chapter 2 MCQ Questions With Answers Question 15.
In Hydrogen atom, energy of first excited state is – 3.4 eV. Then find out KE of same orbit of Hydrogen atom
(a) 3.4 eV
(b) 6.8 eV
(c) -13.6 eV
(d) +13.6 eV

Answer

Answer: (a) 3.4 eV
Explanation:
For hydrogen atom,
The kinetic energy is equal to the negative of the total energy.
And the potential energy is equal to the twice of the total energy.
The first excited state energy of orbital = -3.4 eV
and The kinetic energy of same orbital = -(-3.4 eV) = 3.4 eV
Therefore, the kinetic energy of same orbit of hydrogen atom is 3.4 eV.


Question 16.
Which of the following sets of quantum numbers represents the highest energy of an atom?
(a) n = 3, l = 0, m = 0, s = + \(\frac {1}{2}\)
(b) n = 3, l = 1, m = 1, s = + \(\frac {1}{2}\)
(c) n = 3, l = 2, m = 1, s = + \(\frac {1}{2}\)
(d) n = 4, l = 0, m = 0, s = + \(\frac {1}{2}\)

Answer

Answer: (c) n = 3, l = 2, m = 1, s = + \(\frac {1}{2}\)
Explanation:
n = 3, l = 0 represents 3s orbital n = 3, l = 1 represents 3p orbital n = 3, l = 2 represents 3d orbital n = 4, l = 0 represents 4s orbital The order of increasing energy of the orbitals is 3s < 3p < 4s < 3d.


Question 17.
In the Bohrs model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in a quantum state n is:
(a) 1
(b) 2
(c) -1
(d) -2

Answer

Answer: (c) -1
Explanation:
As we know, in Bohr model KE of an electron in an orbit = + (\(\frac {1}{2}\)) (\(\frac {e^2}{r_n}\))
Total energy of electron in an orbit = -(\(\frac {e^2}{2r_n}\))
Therefore, \(\frac {(KE)}{(TE)}\) = -1


Question 18.
Which of the following statements does not form a part of Bohrs model of hydrogen atom?
(a) Energy of the electrons in the orbit is quantised
(b) The electron in the orbit nearest the nucleus has the lowest energy
(c) Electrons revolve in different orbits around the nucleus
(d) The position and velocity of the electrons in the orbit cannot be determined simultaneously

Answer

Answer: (d) The position and velocity of the electrons in the orbit cannot be determined simultaneously
Explanation:
The position and velocity of electrons cannot be determined simultaneously does not fit in with the Bohrs model of H atom. It is a part of Heisenbergs uncertainty principle


Question 19.
Which of the following statements in relation to the hydrogen atom is correct?
(a) 3s orbital is lower in energy than 3p orbital
(b) 3p orbital is lower in energy than 3d orbital
(c) 3s and 3p orbitals are of lower energy than 3d orbital
(d) 3s, 3p and 3d orbitals all have the same energy

Answer

Answer: (d) 3s, 3p and 3d orbitals all have the same energy
Explanation:
A hydrogen atom has 1s1 configuration and these its, 3p and 3d orbitals will have same energy wrt 1s orbital.


Question 20.
A sub-shell with n = 6 , l = 2 can accommodate a maximum of
(a) 12 electrons
(b) 36 electrons
(c) 10 electrons
(d) 72 electrons

Answer

Answer: (c) 10 electrons
Explanation:
n = 6, ℓ = 2 means 6d → will have 5 orbitals.
Therefore max 10 electrons can be accommodated as each orbital can have maximum of 2 electrons.


We hope the given NCERT MCQ Questions for Class 11 Chemistry Chapter 2 Structure of Atom with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Chemistry Structure of Atom MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Chemistry MCQ:

MCQ Questions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry with Answers

Some Basic Concepts of Chemistry Class 11 MCQ Online Test With Answers Questions

Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided Some Basic Concepts of Chemistry Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.

Class 11 Chemistry Chapter 1 MCQ With Answers

Chemistry Class 11 Chapter 1 MCQs On Some Basic Concepts of Chemistry

Some Basic Concepts Of Chemistry MCQ Question 1.
Approximate atomic weight of an element is 26.89. If its equivalent weight is 8.9, the exact atomic weight of element would be
(a) 26.89
(b) 8.9
(c) 17.8
(d) 26.7

Answer

Answer: (d) 26.7
Explanation:
Atomic weight = (Equivalent weight × Valency)
=(8.9 × 3) = 26.7
(Valency = (26.89)/(8.9) ≈ 3).


Class 11 Chemistry Chapter 1 MCQ Questions With Answers Question 2.
The number of moles present in 6 gms of carbon is:
(a) 2
(b) 0.5
(c) 5
(d) 1

Answer

Answer: (b) 0.5
Explanation:
The molar mass of
12C is 12.0 gmol-1.
NA (Avogadros number = 6.022×1023)
12C atoms have a mass of 12.0 g.
Given that: – 6.0 g.
Thus (6.0 g)/ (12.0 gmol-1) = 0.50 mol.


Class 11 Chemistry Chapter 1 MCQ Question 3.
What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO3 and 0.1 M NaCl are mixed together
(a) 0.1 N
(b) 0.2 M
(c) 0.05 M
(d) 0.25 M

Answer

Answer: (b) 0.2 M
Explanation:
0.1 M AgNOA will react with 0.1 M NaCl to form 0.1 M NaNOA.
But as the volume is doubled, conc. of NO3 = \(\frac { (0.1) }{ (2) } \)
= 0.05 M


Some Basic Concepts Of Chemistry Class 11 MCQ With Answers Question 4.
The -ve charged particles is called:
(a) Anion
(b) Cation
(c) Radical
(d) Atom

Answer

Answer: (a) Anion
Explanation:
A charged particle, also called an ion, is an atom with a positive or negative charge.
This happens whenever something called an ionic bond forms.
Two particles that have different numbers of electrons (the smallest particle in an atom which is negative) start reacting to each other.
The particle that has the greater amount of electrons takes the other particles electrons.
One becomes positive because it lost an electron, and the other negative because it got another electron.
The two particles become attracted to each other and mix together, making a new kind of particle.


MCQ Questions For Class 11 Chemistry Chapter 1 Question 5.
Which is not a unit of pressure:
(a) Bar
(b) N/m²
(c) Kg/m²
(d) Torr

Answer

Answer: (c) Kg/m²
Explanation:
Pressure is force per unit area so Kg/m² cannot be a unit of pressure.


Some Basic Concepts Of Chemistry Class 11 MCQ Question 6.
What is the normality of a 1 M solution of H3PO4
(a) 0.5 N
(b) 1.0 N
(c) 2.0 N
(d) 3.0 N

Answer

Answer: (d) 3.0 N
Explanation:
H3PO4 is tribasic
So N = 3M
= 3 × 1 = 3.


Chemistry Class 11 Chapter 1 MCQs Question 7.
The total number of ions present in 111 g of CaCl2 is
(a) One Mole
(b) Two Mole
(c) Three Mole
(d) Four Mole

Answer

Answer: (c) Three Mole
Explanation:
Molecular weight of
CaCl2 = 111g/mol
Ions in one calcium chloride molecule = Ca+2 + 2Cl = 3 ions
Now no. of molecules in 111g/mol of CaCl2 = Avogadros number
= 6.02 × 1023 molecules
So number of ions in 111g/mol of CaCl2
= 3 × 6.02 × 1023 ions
= 3 moles.


Ch 1 Chemistry Class 11 MCQ Question 8.
Which of the following weighs the most?
(a) One g – atom of nitrogen
(b) One mole of water
(c) One mole of sodium
(d) One molecule of H2SO4

Answer

Answer: (c) One mole of sodium


Chapter 1 Chemistry Class 11 MCQ Question 9.
Under similar conditions of pressure and temperature, 40 ml of slightly moist hydrogen chloride gas is mixed with 20 ml of ammonia gas, the final volume of gas at the same temperature and pressure will be
(a) 100 ml
(b) 20 ml
(c) 40 ml
(d) 60 ml

Answer

Answer: (b) 20 ml
Explanation:
NH3(g) + HCl(g)→ NH4Cl(s)
t = 0 40ml 0
t = t 20ml solid
Final volume = 20ml.


Chemistry Class 11 Chapter 1 MCQ Question 10.
An organic compound contains carbon , hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be
(a) CHO
(b) CH4O
(c) CH3O
(d) CH2O

Answer

Answer: (c) CH3O
Explanation:
MCQ Questions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry with Answers 1


Class 11 Chemistry Ch 1 MCQ Question 11.
Which of the following cannot give iodometric titrations
(a) Fe3+
(b) Cu2+
(c) Pb2+
(d) Ag+

Answer

Answer: (c) Pb2+
Explanation:
Atom in highest oxidation state can oxidize iodide to liberate I2 which is volumetrically measured by iodometric titration using hypo.
2I →I2
Pb+2 → Lowest oxidation state cannot oxidise iodide to I2


MCQ Class 11 Chemistry Chapter 1 Question 12.
What is the concentration of nitrate ions if equal volumes of 0.1 MAgNO3 and 0.1 M NaCl are mixed together
(a) 0.1 M
(b) 0.2 M
(c) 0.05 M
(d) 0.25 M

Answer

Answer: (c) 0.05 M
Explanation:
0.1M AgNO3 will react with 0.1M NaCl to form 0.1M NaNO3.
But as the volume doubled, conc. of NO2
= 0.12
= 0.05 M


MCQ Of Chapter 1 Chemistry Class 11 Question 13.
The number of moles present in 6 gms of carbon is:
(a) 2
(b) 0.5
(c) 5
(d) 1

Answer

Answer: (b) 0.5
Explanation:
The molar mass of
12C is 12.0 gmol-1.
NA (Avogadros number = 6.022 × 1023)
12C atoms have a mass of 12.0 g.
Given that: – 6.0 g.
Thus (6.0 g)/ (12.0 gmol-1) = 0.50 mol.


MCQ On Some Basic Concepts Of Chemistry Question 14.
Which of the following contains same number of carbon atoms as are in 6.0 g of carbon (C – 12)?
(a) 6.0 g Ethane
(b) 8.0 g Methane
(c) 21.0 g Propane
(d) 28.0 g CO

Answer

Answer: (b) 8.0 g Methane
Explanation:
6g carbon
Moles of carbon = (6/12) = 0.5 mol
Number of carbon atoms
= 0.5 × NA =0.5NA (NA is Avogadro number)
6g ethane (C2H6 two atoms of C per mole)
Moles = (6/30) = 0.2 mol
Number of carbon atoms = 0.2 × 2 × NA = 0.4 NA
(Number of carbon atoms = moles of compound × number of C atoms per mol × Avogadro number)
8g methane (CH4)
Moles = (8/16) = 0.5 mol
Number of carbon atoms = 0.5 × 1 × NA = 0.5 NA
21g propane (C3H8)
Moles = (21/44) =0.48 mol
Number of carbon atoms = 0.48 × 3 × NA = 1.44 NA
28g CO
Moles = (28/28) =1 mol
Number of carbon atoms = 1 × 1 × NA = NA


MCQ Of Ch 1 Chemistry Class 11 Question 15.
The significant figures in 3400 are
(a) 2
(b) 5
(c) 6
(d) 4

Answer

Answer: (b) 5
Explanation:
As we know that all non-zero unit are significant number.


Question 16.
A symbol not only represents the name of the element but also represents
(a) Atomic Mass
(b) Atomic Number
(c) Atomicity
(d) Atomic Volume

Answer

Answer: (c) Atomicity
Explanation:
Symbol of an element represents the name along with the number of same atoms in room condition, which is its atomicity, of that element.


Question 17.
What is the normality of a 1 M solution of H3PO4
(a) 0.5 N
(b) 1.0 N
(c) 2.0 N
(d) 3.0 N

Answer

Answer: (d) 3.0 N
Explanation:
H3PO4 is tribasic
So N = 3M
= 3 × 1 = 3.


Question 18.
Which of the following weighs the most?
(a) One g – atom of nitrogen
(b) One mole of water
(c) One mole of sodium
(d) One molecule of H2SO4

Answer

Answer: (c) One mole of sodium


Question 19.
The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with ZnSO4.7H2O. The atomic weight of M is
(a) 40.3
(b) 36.3
(c) 24.3
(d) 11.3

Answer

Answer: (c) 24.3
Explanation:
As the given sulphate is isomorphous with ZnSO4.7H2O its formula would be MSO4.7H2O.m is the atomic weight of M, molecular weight of MSO4.7H2O
= (m + 32 + 64 + 126) = (m + 222)
Hence % of M= [(m)/ (m+222)] × 100 = 9.87(given)
Or
100 m = (9.87 m + 222 × 9.87)
Or 90.13 m = (222 × 9.87)
Or m = (222 × 9.87)/(90.13)
= 24.3.


Question 20.
The S.I unit of temperature is :
(a) Kelvin
(b) Celsius
(c) Fahrenheit
(d) Centigrade

Answer

Answer: (a) Kelvin
Explanation:
The SI unit of temperature is the Kelvin. Although the Celsius temperature scale is also used, it is considered a derived SI unit and is generally used to measure everyday temperatures.

In the Kelvin scale, the lowest possible temperature is 0 K. This reading is called absolute zero; however, nothing could possibly ever reach this temperature. Absolute zero, according to the Celsius scale, is approximately minus 273 degrees Celsius. Water boils at 373 Kelvin and freezes at 273 Kelvin.

Scientists discovered absolute zero when they figured that as temperature decreases, the volume of a gas also gets smaller. They graphed this relationship and found that for each substance tested, zero volume for each substance would hypothetically occur around minus 273 degrees Celsius, the equivalent of absolute zero.


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