Keeping Quiet Poem Summary Line By Line Explanation by Pablo Neruda in English

We have decided to create the most comprehensive English Summary that will help students with learning and understanding.

Keeping Quiet Poem Summary in English by Pablo Neruda

Keeping Quiet Poem by Pablo Neruda About the Poet

Pablo Neruda, was a Nobel Prize-winning poet. He adopted this pen name in 1920, in the memory of Czechoslovak poet, Jan Neruda.

Neruda wrote in a variety of styles, such as love poems as in his collection, Twenty Poems of Love and , a Song of Despair, surrealist poems, historical epics, and overtly political manifestos. He travelled the ? world for many years, and returned to Chile in 1943. In 1945, he joined the Communist Party of Chile and became Senator of the Republic. In 1971, Neruda won the Nobel Prize for Literature.

Pablo Neruda I remained active in the literary and political arena even before his death. His poetry bears an impact on his political activities and expresses his experiences in repression during his exile. He wrote over 140 poems. Colombian novelist Gabriel Garda Marquez once called him “the greatest poet of the 20th § century in any language”.

Poet Name Pablo Neruda
Born 12 July 1904, Parral, Chile
Died 23 September 1973, Santiago, Chile
Full Name Ricardo Eliecer Neftali Reyes Basoalto
Awards Nobel Prize in Literature, Lenin Peace Prize
Spouse Matilde Urrutia (m. 1966–1973)
Keeping Quiet Poem Summary by Pablo Neruda
Keeping Quiet Poem Summary by Pablo Neruda

Keeping Quiet Poem Introduction to the Poem

In this poem, the poet aims to appeal to the readers to take some time out of their busy life for a little introspection and retrospection. The title, “Keeping Quiet” is symbolic of stopping all the activities and keep the mind quiet by not doing anything, but to question and understand the purpose of the world that humans have created around themselves.

Keeping Quiet Poem Theme

In the poem, the poet advocates to keep quiet and still for a while to introspect and understand ourselves and our relationships, and to build a peaceful and harmonious world over. He asks all the human beings to stop all worldly activities for a while and spend few quiet moments in introspection.

Keeping Quiet Poem Summary in English

In this poem, the poet has emphasised the need to introspect and bring in the spirit of brotherhood among the people of the world. He wants people to stop talking and stop all movements symbolising agitation and restlessness till he counts twelve, that is, a short period of time. These moments of silence would be strange and exotic because in our mundane life, we are working towards achieving selfish goals, regardless of the others’ requirements and emotions. Hence, this sudden silence would give us an opportunity to introspect. Since we would not speak for a while, barriers between communities would break and a sense of brotherhood would prevail.

Man would get an opportunity to realise how he is destroying nature and how he is harming himself. Futile wars against man and nature would be arrested and a new feeling of unity would be experienced. The poet does not want his desire for inactivity to be misunderstood as a state of uselessness or in action. He wants man to learn a lesson from the earth. The earth appears to be inactive, yet it is selflessly productive. Man too could be productive and progressive without any aggression, selfishness and the urge for destruction.

Short Summary Of Keeping Quiet Reference-to-Context Questions

Read the extracts given below and answer the questions that follow.

1. Now we will count to twelve
and we will all keep still.
For once on the face of the Earth
let’s not speak in any language,
let’s stop for one second,
and not move our arms so much.

a. What is the significance of the number ‘twelve’ ?
Answer:
The number ‘twelve’ signifies the twelve divisions of time. It is a measure of time.

b. Which two activities does the poet want us to stop?
Answer:
The poet wants us to keep quiet and not speak. He does not want us to move our arms.

c. What does the poet mean by ‘let’s not speak in any language’?
Answer:
The poet pleads us to be silent and introspect.

d. Describe the pun in the word, ‘arms’.
Answer:
‘Arms’ refers to weapons and the arms in the human body.

2. It would be an exotic moment
without rush, without engines,
we would all be together
in a sudden strangeness.

a. What does ‘It’ refer to?
Answer:
‘It’ refers to the moment when there is peace and quietness which the poet considers would be exotic.

b. Who is the poet addressing to?
Answer:
The poet is addressing his readers.

c. What would be the moment like?
Answer:
The absence of hustle and bustle of life would create a feeling of peace and quiet which would make us united.

d. What does ‘exotic’ mean?
Answer:
It means ‘unique’ as the moment would be mysteriously different from the usually prevailing situation.

3. Fishermen in the cold sea,
would not harm whales
and the man gathering salt
would look at his hurt hands.

a. What are ‘fishermen’ symbolic of?
Answer:
‘Fishermen’ symbolise proficient hunters of all kinds. These people have been indiscriminately exploiting nature and Mother Earth for their own vested interests.

b. What message does the poet seem to give in these lines?
Answer:
The message that Neruda wishes to convey is that the desire of man for more and more progress and advancement has done more destruction than development. Man seems to have no concern and care for his brethren.

c. What image does Neruda portray in the last lines?
Answer:
The image that Neruda seems to create in the last lines is one of the incessant sufferings. In order to make life more comfortable, he pays no heed to the sufferings he has to undergo. The men, who gather salt do not look at their hurt palms because they are more busy making life more comfortable for themselves and for their family.

d. What basic idea does the poet empahsise upon?
Answer:
It is the environmental protection through man’s introspection and action.

4. Those who prepare green wars,
wars with gas, wars with fire,
victory with no survivors,
would put on clean clothes
and-walk about with their
brothers
in the shade, doing nothing.

a. Who are ‘those’ here?
Answer:
Here, ‘those’ are the politicians, scientists, statesmen and those overambitious powers to be, who in their zeal for dominance are involved in initiating and aggravating wars.

b. What are ‘green wars, wars with gas, wars with fire’?
Answer:
These are the nuclear and chemical weapons which have been created by man and used in waging wars against countries. In fact, these wars have brought mankind to its fatal end. These wars are also against the environment.

c. What does Neruda mean by ‘victory with no survivors’?
Answer:
The victory that may be gained with the help of nuclear and chemical warfare would actually leave no one alive in that area. There is no point of winning such a war which cannot be savoured as there would be no survivors.

d. Which poetic device is used in the third line of the extract?
Answer:
It is Paradox as opposite ideas are explained by this phrase.

5. What I want should not be
confused
with total inactivity.
Life is what it is about;
I want no truck with death.

a. What does Neruda imply by ‘total inactivity’?
Answer:
‘Total inactivity’ would imply that mankind should channelise his activities towards construction and he should put an end to his destructive activities. It certainly does not mean no action because life is an on-going process.

b. What is life about?
Answer:
Life cannot be brought to a standstill under any circumstances. It is an on-going process and it has to move on.

c. What does Neruda mean when he says, ‘I want no truck with death’?
Answer:
Neruda says that when he talks of counting upto twelve, meditating and introspecting, he is not advocating death-like silence. To bring all activities to a halt is only to facilitate introspection, and not to bring life to a standstill. Life is an on-going process and humanity should move on.

d. What is the underlying thought of the poet here?
Answer:
We should not act physically but we must introspect.

6. If we were not so single-minded
about keeping our lives moving,
and for once could do nothing,
perhaps a huge silence
might interrupt this sadness
of never understanding ourselves
and of threatening ourselves with
death.

a. What is man ‘single-minded’ about?
Answer:
Man is single-minded about his own progress and advancement. He is so focussed on his own development that he forgets to visualise the pros and cons of reckless development. The consequences and impact of materialistic progress should be taken into consideration.

b. Explain, ‘sadness of never understanding ourselves’.
Answer:
Mankind, in its race for materialistic success has sacrificed his emotional needs at some point of time. People are so engrossed in fulfilling their ambitions and successfully completing their duties that they tend to forego their inner mental satisfaction. Thus, they tend to become sad and isolated.

c. How has mankind threatened itself with death?
Answer:
The race for more arms and ammunitions and the desire to overpower and dominate over more and more areas and countries is what has threatened mankind with death. The mass destruction of nature is also one main reason for large-scale death of humanity.

d. What does ‘keeping our life moving’ mean?
Answer:
It means making progress without a second thought.

7. Perhaps the Earth can teach us
as when everything seems dead
and later proves to be alive.
Now I’ll count up to twelve
and you keep quiet and I will go.

a. What does the earth teach us?
Answer:
The earth teaches us how new life springs from dead remains and how there is life under apparent stillness.

b. What does the poet mean to achieve by counting upto twelve?
Answer:
The poet wants to achieve peace by counting upto twelve. He wants us to introspect in a moment of silence.

c. What is the significance of ‘keeping quiet’?
Answer:
Keeping quiet doesn’t mean being idle. It means that we should avoid all such activities, which are hurting the nature, and in turn, us.

d. What is the poetic device used in the first line?
Answer:
Personification because earth is depicted like an instructor to us.

A Thing of Beauty Summary Line By Line Explanation by John Keats in English

We have decided to create the most comprehensive English Summary that will help students with learning and understanding.

A Thing of Beauty Summary in English by John Keats

A Thing of Beauty Poem by John Keats About the Poet

John Keats (31 October, 1795 – 23 February, 1821) was an English Romantic poet. He was one of the main figures of the second generation of Romantic poets along with Lord Byron and P. B. Shelley, ; despite his work having been in publication for only four years before his death.

Although his poems were not generally well received by critics during his lifetime, his reputation grew after his death. So by the end of the 19th century, he had become one of the most beloved of all English | poets. He had a significant influence on a diverse range of poets and writers.

The poetry of Keats is characterised by sensual imagery, most notably in the series of odes. Today, his poems and letters are some of the most popular and most analysed in English literature. His notable works are: To autumn; Ode to a nightingale; On first looking into Chapman’s Homer; Ode on a Grecian urn.

Poet Name John Keats
Born 31 October 1795, Moorgate, City of London, United Kingdom
Died 23 February 1821, Rome, Italy
Poems Ode to a Nightingale, Ode on a Grecian Urn, To Autumn
Movies Arterial
A Thing of Beauty Summary by John Keats
A Thing of Beauty Summary by John Keats

A Thing of Beauty Introduction to the Poem

A British Romantic poet, John Keats (1795-1821), published his first epic poem, Endymion in 1818. It is a narrative about the relationship between a Goddess and her human lover. The poem is based on the Greek myth of Endymion, the shepherd who falls in love with the moon Goddess, Selene, whom the poet renames ‘Cynthia’. This extract, ‘A Thing of Beauty’ talks of how beautiful things give us pleasure and alleviate suffering and sorrow. The epic poem, ‘Endymion’ begins famously with the line, ‘A thing of beauty is a joy forever’ (first stanza).

A Thing of Beauty Theme

The very opening line of the poem, ‘A Thing of Beauty…. ‘ openly recites the very theme of the poem. Any beautiful object is always treasured in our mind because it provides us eternal and everlasting joy. The happiness that anything beautiful provides, never fades into nothingness but multiplies manifold whenever it returns to our mind.

A Thing of Beauty Summary in English

This poem presents the reader with the views of Keats on beauty and its value, and its ‘ importance to humans. Beauty, in whatever form it may be found, is an eternal joy to humans, because it offers humans the constant opportunity to reflect on that beauty, which stands in such stark contrast to the dejection, monotony and ugliness of our everyday lives.

In spite of all the difficulties and the sufferings that humans face, beauty has the ability to produce happiness and temporarily shift the burdens that humans bear.

Keats therefore, establishes that in his view of the world, life for humans consists of unremitting struggles and difficulties, and it is only beauty in its various guises that is able to shift those trials and at least temporarily produce happiness. The poem concludes with a list of things that constitute “beauty” for Keats, which include both physical objects which are examples of natural beauty such as daffodils, and also beauty that can be found in art, such as “the lovely tales we have heard or read”.

All of these forms of beauty act as “an endless fountain of immortal drink”, allowing humans to forget bleak reality and experience joy. This poem therefore, focuses on the theme of happiness and how it can be experienced. It inspires us and gives us the courage to fight against all odds.

Short Summary Of A Thing Of Beauty Reference-to-Context Questions

Read the extracts given below and answer the questions that follow.

1. A thing of beauty is a joy forever
Its loveliness increases, it will never
Pass into nothingness; but will keep
A bower quiet for us, and a sleep
Full of sweet dreams, and health, and quiet breathing.

a. Describe the kind of joy a beautiful thing provides.
Answer:
Anything that is beautiful provides us an unending, everlasting and eternal joy. It leaves an indelible imprint on our heart and soul. We love to relive the memories of that joyous experience whenever it flashes back in our mind.

b. Explain, ‘its loveliness increases’.
Answer:
The joy that we experience on seeing a beautiful object, multiplies whenever we remember it. Similarly, the loveliness of anything beautiful multiplies each time we visualise it in our thoughts. Compared to this, things which do not give us happiness only remain for a short while, and then disappear forever.

c. Why does a thing of beauty not pass into ‘nothingness’?
Answer:
A thing, the memory of which provides us eternal joy, will never fade away. The loveliness of a beautiful object does not fade away or die out because it leaves an indelible imprint on our heart and soul.

d. How can we get refreshed amidst beauty?
Answer:
The beautiful things give us peace like a quiet bower, a pleasant shady place providing us peaceful sleep, sweet dreams and good health and therefore, refreshes us.

2. Therefore, on every morrow, are we wreathing
A flowery band to bind us to the earth,
Spite of despondence, of the inhuman dearth
Of noble natures, of the gloomy days
Of all the unhealthy and o’er-darkened ways
Made for our searching:

a. Explain, ‘wreathing a flowery band to bind us to the earth’.
Answer:
Keats feels that every morning we weave a beautiful string of flowers or memories which help us support ourselves and motivate us to live our life to the fullest, instead of burdening our lives with pain and suffering.

b. Why is there an ‘inhuman dearth of noble natures’?
Answer:
In these days, there are few people who are noble in character and who rise above petty differences by being magnanimous and generous. There is a dearth of such noble souls on our earth; as man is selfish and self-centred.

c. What are ‘unhealthy and o’er-darkened ways’?
Answer:
The unhealthy and o’er-darkened ways refer to the trials and tribulations of life, the selfish and jealous methods we adopt to achieve our goals.

d. Explain, ‘spite of despondence’.
Answer:
It means despite the existence of gloom and darkness around.

3. … yes, in spite of all,
Some shape of beauty moves away the pall
From our dark spirits. Such the sun, the moon,
Trees old, and young, sprouting a shady boon
For simple sheep; and such are daffodils
With the green world they live in; and clear rills
That for themselves a cooling covert make
‘Gainst the hot season; the mid forest brake,
Rich with a sprinkling of fair musk-rose blooms;
And such too is the grandeur of the dooms
We have imagined for the mighty dead;

a. What images of beauty has the poet referred to?
Answer:
The poet appreciates the simplicity and serenity of beauty through the image of the sun, the moon, the trees, the sheep, the daffodils, the green pastures, the livid streams, and a fair bloom of musk-roses.

b. How is ‘tree’ a perfect example of a beautiful thing?
Answer:
The tree bears the heat of the sun to give shade and coolness to anyone who seeks shade and rest under its branches. Its greatness is a degree greater because it gives shade not only to the mighty ones, but also to the simplest of the animals, such as sheep.

c. What does ‘simple sheep’ symbolise?
Answer:
Sheep and lamb are envisioned as symbols of innocence and serene beauty. Jesus Christ was a shepherd and was surrounded by his flock of sheep, his followers. Keats has made special reference to the sheep as symbols of divine beauty.

d. Explain, ‘the green world they live in’.
Answer:
Beauty of nature is at its best in the lush green surroundings of meadows and pastures which provide support to all plants and animals. It is in this green world that all living creatures find true happiness and joy.

4. All lovely tales that we have heard or read;
An endless fountain of immortal drink,
Pouring unto us from the heaven’s brink.

a. What does Keats mean by ‘an endless fountain of immortal drink’?
Answer:
Beauty, according to Keats, is a perennial source of motivation and inspiration. He considers it an endless fountain from which mankind can drink the elixir of life.

b. Explain, ‘pouring unto us from the heaven’s brink’.
Answer:
Beauty, Keats says, has been showered upon us from the heaven by God as the greatest gift to man. This eternal and everlasting beauty is a perennial source of gaiety, which is eternal and everlasting.

c. Which poetic device is used in the 2nd line of the extract?
Answer:
hyperbole e.g., ‘endless fountain’

d. Which lovely tales does the poet speak of?
Answer:
The motivational stories of martyrs are the ‘lovely tales’ that the poet speaks of.

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

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

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

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

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

(ii) linear molecules?
Answer:
sp.

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

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

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

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

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

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

Question 8.
Give the I.U.P.A.C. names of the following compounds
Organic Chemistry Some Basic Principles and Techniques 1
Answer:
2-Bromo-4 – methyl pent-3- one
Organic Chemistry Some Basic Principles and Techniques 2
Answer:
4-Methyl-2-nitro pent – 3 – one
Organic Chemistry Some Basic Principles and Techniques 3
Answer:
2 – Ethoxy – 4 – methoxypent – 3 – one
Organic Chemistry Some Basic Principles and Techniques 4
Answer:
2-Bromo-4-nitro pent-3-one

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

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

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

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

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

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

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

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

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

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

Question 17.
Draw the structure of the tautomer of phenol and write its I.U.P.A.C name.
Answer:
Organic Chemistry Some Basic Principles and Techniques 5

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

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

  1. CH3O CH3 CH2 CH3 Its I.U.P.A.C. name is 1 -Methoxypropane.
  2. Organic Chemistry Some Basic Principles and Techniques 6

Question 20.
Give the I.U.P.A.C. name of CH2 = CH-CH (CH3)2
Answer:
Organic Chemistry Some Basic Principles and Techniques 7

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

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

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

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

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

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

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

Question 23.
Write the T.U.P.A.C. name of
Organic Chemistry Some Basic Principles and Techniques 8
Answer:
Compound is
Organic Chemistry Some Basic Principles and Techniques 9

Question 24.
Give the condensed and bond-line structural formula for
(a) Penta-1, 4-dien
Answer:
CH2 = CH – CH2 – CH = CH2
Organic Chemistry Some Basic Principles and Techniques 10

(b) Hexa-1, 3, 5 triene.
Answer:
CH2 = CH – CH = CH – CH = CH
Its bond line structure is
Organic Chemistry Some Basic Principles and Techniques 11

Question 25.
Write the I.U.P.A.C. names of HOOC-C ≡ C-COOH and
Organic Chemistry Some Basic Principles and Techniques 12
Answer:
But-2-yne – 1,4-dioic acid and 3-Methyl pentanenitrile.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 1.
Expand each of the following bond-line formulae to show all the atoms including carbon and hydrogen.
Organic Chemistry Some Basic Principles and Techniques 13
Answer:
Organic Chemistry Some Basic Principles and Techniques 14

Organic Chemistry Some Basic Principles and Techniques 15
Answer:
Organic Chemistry Some Basic Principles and Techniques 16

Organic Chemistry Some Basic Principles and Techniques 17
Answer:
Organic Chemistry Some Basic Principles and Techniques 18

Organic Chemistry Some Basic Principles and Techniques 19
Answer:
Organic Chemistry Some Basic Principles and Techniques 20

Question 2.
For each of the following compounds, write a more condensed and also their bond line formulae.
Organic Chemistry Some Basic Principles and Techniques 21
Answer:
Condensed formulae are
(CH3)2CH CH2 OH
Organic Chemistry Some Basic Principles and Techniques 25

Organic Chemistry Some Basic Principles and Techniques 22
Answer:
CH3(CH2)5 CHBr CH2 CHO
Organic Chemistry Some Basic Principles and Techniques 26

Organic Chemistry Some Basic Principles and Techniques 23
Answer:
HO(CH2)3 CH(CH3) CH(CH3)2
Organic Chemistry Some Basic Principles and Techniques 27

Organic Chemistry Some Basic Principles and Techniques 24
Answer:
HOCH(CN)2
Organic Chemistry Some Basic Principles and Techniques 28

Question 3.
What is the type of hybridisation of each carbon in the following compounds?
(a) CH3Cl
Answer:
Organic Chemistry Some Basic Principles and Techniques 29

(b) (CH3)2CO
Answer:
Organic Chemistry Some Basic Principles and Techniques 30

(c) CH3CN
Answer:
Organic Chemistry Some Basic Principles and Techniques 31

(d) HCONH2
Answer:
Organic Chemistry Some Basic Principles and Techniques 32

(e) CH3CH = CHCN.
Answer:
Organic Chemistry Some Basic Principles and Techniques 33

Question 4.
What is the shape of the following molecules:
(a) H2C = O
Answer:
In H2C = O; C is sp2 hybridised, hence its shape is H trigonal planar
Organic Chemistry Some Basic Principles and Techniques 34

(b) CH3F
Answer:
In CH3 -F; C is sp3 hybridized
∴ it is tetrahedral
Organic Chemistry Some Basic Principles and Techniques 35

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

Question 5.
Give the I.U.P. A.C. names of the following compounds:
Organic Chemistry Some Basic Principles and Techniques 36
Answer:
2-Ethylprop-2-en-l-ol

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

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

Organic Chemistry Some Basic Principles and Techniques 37
Answer:
3-Chloropropanal

Organic Chemistry Some Basic Principles and Techniques 38
Answer:
3-Methylbutane-l-al

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

Question 6.
Write the I.U.P.A.C. names of
Organic Chemistry Some Basic Principles and Techniques 39
Answer:
3-Ethyl-4-methylhept-5-en-2-one

Organic Chemistry Some Basic Principles and Techniques 40
Answer:
2-Ethyl-3-methylpent-2-en-1 -one.

Question 7.
Write 1.U.P.A.C. names of
Organic Chemistry Some Basic Principles and Techniques 41
Answer:
2-[2-methylclo but-l-enyl] ethanal

Organic Chemistry Some Basic Principles and Techniques 42
Answer:
2-(3-Oxobutyl) cyclohexane-1 one

Organic Chemistry Some Basic Principles and Techniques 43
Answer:
Methyl (2-oxo cyclopentane-1-carboxylate

Organic Chemistry Some Basic Principles and Techniques 44
Answer:
Cyclohex-2-en-l-ol

Organic Chemistry Some Basic Principles and Techniques 45
Answer:
2-Ethenyl-3-methyl-cyclohexa-l, 3-diene

Organic Chemistry Some Basic Principles and Techniques 46
Answer:
4-Formyl-2-oxo-cyclohexane-l carboxylic acid.

Question 8.
Draw the structures of
(i) Methyl t-butyl ether
Answer:
Organic Chemistry Some Basic Principles and Techniques 47

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

(iii) 2-Methyl buta, 1, 3-diene
Answer:
Organic Chemistry Some Basic Principles and Techniques 48

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

(v) Cyclo hex-2-en-l-ol
Answer:
Organic Chemistry Some Basic Principles and Techniques 49

(vi) l-Bromo-3-chloro cyclohex-1-ene
Answer:
Organic Chemistry Some Basic Principles and Techniques 50

Question 9.
Write I.U.P.A.C. names of
Organic Chemistry Some Basic Principles and Techniques 51
Answer:
(2-Isopropyl) benzene

Organic Chemistry Some Basic Principles and Techniques 52
Answer:
1-Phenylpropane-l-one

Organic Chemistry Some Basic Principles and Techniques 53
Answer:
1 -Phenylethan-1 -ol

Organic Chemistry Some Basic Principles and Techniques 54
Answer:
1, 3-Benzene dicarboxylic acid

Organic Chemistry Some Basic Principles and Techniques 55
Answer:
3-Phenylpropanal

Organic Chemistry Some Basic Principles and Techniques 56
Answer:
3-Boromo-4-hydroxybenzoic acid

Organic Chemistry Some Basic Principles and Techniques 57
Answer:
4-Hydroxy-3-methoxy benzaldehyde.

Question 10.
Write the structure of
(i) O-Ethyl anisole
Answer:
Organic Chemistry Some Basic Principles and Techniques 58

(ii) p— nitroaniline
Answer:
Organic Chemistry Some Basic Principles and Techniques 59

(iii) 4-Ethyl-I-fluoro-2-nitrobenzene.
Answer:
Organic Chemistry Some Basic Principles and Techniques 60

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

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

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

Question 12.
In which C-C bond of CH3CH2CH2Br, the inductive effect is expected to be the least.
Answer:
The magnitude of the inductive effect decreases with distance from the active centre and hence this effect is least in C2—C3 bond.
Organic Chemistry Some Basic Principles and Techniques 61

Question 13.
Write the resonance structures of
(a) CH3COO- and
Answer:
Organic Chemistry Some Basic Principles and Techniques 62

(b) CH6H5NH2. Show the movement of electrons by curved arrows.
Answer:
Organic Chemistry Some Basic Principles and Techniques 63

Question 14.
Which of the following pairs of structures do not constitute resonance structures?
Organic Chemistry Some Basic Principles and Techniques 64
Answer:
They differ in the position of atoms and hence are not resonance structures.
Organic Chemistry Some Basic Principles and Techniques 65
Answer:
They are a pair of resonance structures as they differ in the position of electrons.

Organic Chemistry Some Basic Principles and Techniques 66
Answer:
They differ in the position of atoms and so are not resonance structures. They are tautomers.

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

Question 15.
Write the resonance structures of CH2 = CH-CHO and arrange them in decreasing order of stability.
Answer:
Organic Chemistry Some Basic Principles and Techniques 67
The structure I is most stable as each C & O have octets completed and there is no charge on either of them.
II & III involve charge separation hence both are less stable than I. However II is more stable than III because in II more electronegative oxygen carries a negative charge.

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

Question 16.
Using curved arrow notation show the formation, of reactive intermediates when the following covalent bonds undergo heterolytic fission.
(a) CH3-SH3
Answer:
Organic Chemistry Some Basic Principles and Techniques 68

(b) CH3-CN
Answer:
Organic Chemistry Some Basic Principles and Techniques 69

(c) CH3-CU
Answer:
Organic Chemistry Some Basic Principles and Techniques 70

Question 17.
Giving proper justification categorise the following molecules/ ions as nucleophiles or electrophiles:
Organic Chemistry Some Basic Principles and Techniques 71
Answer:
Organic Chemistry Some Basic Principles and Techniques 72
are all nucleophiles as each one of them has one or more lone pairs of electrons to donate.

Organic Chemistry Some Basic Principles and Techniques 73
are all electrophiles. All these species have a sextet of electrons around positive centres.

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

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

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

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

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

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

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

Question 23.
Compare inductive & mesomeric effects.
Answer:

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

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

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

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

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

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

Question 27.
For each of the following compounds, write a more condensed formula & also their bond-line formula.
Organic Chemistry Some Basic Principles and Techniques 74
(b) HOCH2CH2CH2CHCH3CHCH3CH3
Organic Chemistry Some Basic Principles and Techniques 75
Answer:
Organic Chemistry Some Basic Principles and Techniques 76

Question 28.
Expand each of the following bond line formulae to show all the atoms including carbon & hydrogen.
Organic Chemistry Some Basic Principles and Techniques 77
Answer:
Organic Chemistry Some Basic Principles and Techniques 78

Organic Chemistry Some Basic Principles and Techniques 79
Answer:
Organic Chemistry Some Basic Principles and Techniques 80

Organic Chemistry Some Basic Principles and Techniques 81
Answer:
Organic Chemistry Some Basic Principles and Techniques 82

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

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

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

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

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

This method is based on the facts that

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

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

Obviously, this happens at a temperature that is lower than the boiling point of the substance or that of water. Thus an organic compound boils below its boiling points and chances of decomposition avoided. For example, a mixture of aniline (b.p 453 K) with decomposition and water (b.p. 373 K) under normal atmospheric pressure boils at 371K. At this temperature the
Organic Chemistry Some Basic Principles and Techniques 83
Steam Distillation

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

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

Question 2.
Dehydrobromination of compounds (A) and (B) yield the same alkene (c) Alkene (c) Can regenerate (A) and (B) by the addition of HBr in the presence and absence of peroxide respectively. Hydrolysis of A and B give isomeric products (D) and (E) respectively. 1, 1-Diphenyl ethane is obtained on the reaction of (C) of benzene in the presence of H+ ions. Give structures of A to E with reactions.
Answer:
Alkene (C) on reaction with benzene in the presence of H+ ions gives 1, 1-Diphenyl ethane. Therefore C must be styrene as depicted below
Organic Chemistry Some Basic Principles and Techniques 84
Now dehydrobromination of A and B give the same alkene C, i.e.,
styrene.
∴ A and B must be isomeric alkyl bromide.

Organic Chemistry Some Basic Principles and Techniques 85
A and B can be obtained by the addition of HBr in the presence and absence of peroxide to styrene.
Organic Chemistry Some Basic Principles and Techniques 86
Hydrolysis of A and B give isomeric alcohols (D) & (E) as
Organic Chemistry Some Basic Principles and Techniques 87

Question 3.
What are reaction intermediates? How are they generated by bond fission?
Answer:
The species which are generated as a result of bond fission are called reaction intermediates. The important reaction intermediates are:
1. Free Radicals: A free radical may be defined as an atom or group of atoms having an impaired electron. These are obtained as a result of homolytic fission of covalent bonds.
Organic Chemistry Some Basic Principles and Techniques 88
These free radicals are neutral particles, extremely transient, (short-lived) and highly reactive. They get consumed as soon as they are formed. They pair up their electron with another electron from wherever it is available. They occur only as a reaction intermediate. Their presence is felt in reactions, but cannot be isolated in a free state. For example dissociation of Cl2 gas in the presence of Ultraviolet light produces free radicals.
Organic Chemistry Some Basic Principles and Techniques 89
The alkyl free radicals are obtained when free radical: Cl reacts with alkanes.
Organic Chemistry Some Basic Principles and Techniques 90
Free radical may be primary, secondary, tertiary depending upon whether, one, two or three carbon atom attached to the carbon atoms carrying the odd electron.
Organic Chemistry Some Basic Principles and Techniques 91
The stability is CH3 < 1° < 2° < 3°.

2. Carbocation or carbonium ion: It is defined as a group of atoms that contain positively charged carbon having only six electrons. It is obtained by heterolytic fission of a covalent bond involving a carbon atom.
Organic Chemistry Some Basic Principles and Techniques 92
They are also classified as primary, secondary and tertiary depending upon whether one, two or three carbon atoms are attached to the carbon bearing the positive charge as:
Organic Chemistry Some Basic Principles and Techniques 93
Thus the order of stability if CH3+ < 1° < 2° < 3°.

3. Carbanion: A carbanion may be defined as a species containing a carbon atom carrying a negative charge. These are generated by the atom in which the atom linked to carbon goes without the bonding electrons. As a result of this carbon acquires a negative charge. For example, the removal of hydrogen of methyl part of acetaldehyde molecule as H+ ion leaving both the electron on carbon.
Organic Chemistry Some Basic Principles and Techniques 94
They are also very reactive species. They are also classified as primary, secondary and tertiary depending upon whether one, two or three carbon atoms are attached to the carbon atom bearing negative
charge.
Organic Chemistry Some Basic Principles and Techniques 95
The order of stability is the reverse of free radicals and carbocations
CH3 > 1° > 2° > 3°.

(iv) Carbenes: The carbenes are reactive neutral species in which carbon atom has six electrons in the valency shell out of which two are shared. The simplest carbene is methylene (CH2). It is formed when diazomethane is decomposed by the action of light.
Organic Chemistry Some Basic Principles and Techniques 96
It is very reactive. It reacts with alkenes by adding to the double bond forming cyclopropane.
Organic Chemistry Some Basic Principles and Techniques 97

 

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

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

233g of BaSO4 contain sulphur = 32g
0. 582 g of BaSO4 contains sulphur
Organic Chemistry Some Basic Principles and Techniques 98

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

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

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

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

Now, 188 g of AgBr contains bromine = 80g.
Organic Chemistry Some Basic Principles and Techniques 99

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

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

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

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

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

Therefore 0.35g of BaSO4 will contain Sulphur
Organic Chemistry Some Basic Principles and Techniques 100

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

An Elementary School Classroom in a Slum Summary Line By Line Explanation by Stephen Spender in English

We have decided to create the most comprehensive English Summary that will help students with learning and understanding. In this article, we are Discussed the Summary Of An Elementary School Classroom In A Slum.

An Elementary School Classroom in a Slum Summary in English by Stephen Spender

An Elementary School Classroom in a Slum Poem by Stephen Spender About the Poet

Stephen Spender (1909-1995), an English poet and essayist, was one of the pioneers of poetic movement in the 1930s.. Spender took a keen interest in the social and political problems of his times. As a social reformer and pacifist of his time, he questions the value of education and the morals and ethics of individuals.

The trials and tribulations of the post-World War, Europe finds expression in Spender’s verse and prose. Books by Spender include ‘Poem of Dedication’, ‘The Edge of Being’, ‘The Creative Element’, etc.

Poet Name Stephen Spender
Born 28 February 1909, Kensington
Died 16 July 1995, Westminster, London, United Kingdom
Spouse Natasha Spender (m. 1941–1995), Inez Pearn (m. 1936–1939)
Education Gresham’s, University College, University College School Junior Branch, University of Oxford
Nominations Lambda Literary Award for Gay Fiction
An Elementary School Classroom in a Slum Summary by Stephen Spender
An Elementary School Classroom in a Slum Summary by Stephen Spender

An Elementary School Classroom in a Slum Introduction to the Poem

In this poem, Spender depicts the injustice which prevails in society. He talks of social injustice and class inequalities, denial of opportunities to slum dwellers, and also expresses his thoughts on the widening gap between the rich and the poor. He somewhere or the other seems to be hitting hard at the capitalist economy, which helps the rich to get richer and the poor poorer.

The Civil Rights Movement in America had gained pace and Spender’s poem supports it when he exposes the condition of the ‘haves’ and the ‘have notes’. The poem lucidly brings forth the miserable condition of the children of the slum and the inadequate educational facilities provided to them. The poem talks of racial discrimination and is a socialist proclamation that a country can prosper only if education reaches to the downtrodden in the society.

An Elementary School Classroom in a Slum Theme

  • In ‘An Elementary School Classroom in a Slum’, Spender beautifully brings out the themes of social injustice and class inequalities.
  • Poverty is also another theme of the poem. The poet creates an image of children in poverty. It is poverty that has caused the children to be weighed down, diseased and twisted. The poet believes that poverty is created through the oppressive power of capitalism.

An Elementary School Classroom in a Slum Stanza Wise Explanation Summary in English

The poem describes a primary school in a slum. Children studying in the slum classroom depict the social injustice and perpetual poverty, prevailing among the slum dwellers.

In the first stanza, Spender describes the miserable condition of the children. The faces of the children are unlike the usual children of schools. Instead of being exuberant and energetic, they are like rootless weeds, withered and worn out. They are unclean and untidy, as they are malnourished, sick and hungry. Just as weeds are not wanted in the garden, so are these children of the slum unwanted in the society. They have pale faces. Their hair is uncombed. A tall slim girl has her head bowed down as though she is exhausted physically because of malnutrition and emotionally because of poverty. The other students of the class are also in the same situation. There is a boy, who is as thin as paper, again because of malnutrition and lack of civic amenities.

He has eyes like that of a rat, searching for food and betterment. Another child in the class, who is a victim of genetic disorder, has gnarled bones and stunted growth. He has inherited this debilitating disease from his father and recites his lesson from his desk in a mellow and weak voice. In one corner of this poorly-lit and ill-equipped class, is a sweet, unnoticed young child lost in the world of his dreams. The dull and monotonous classroom does not interest him and hence, his mind deviates towards the squirrel in her tree room. He too, dreams of fun and frolic in an open space.

In the second stanza, the poet describes the dirty classroom. On the walls are displayed the names of people who have given donations. The bust of Shakespeare is displayed in the clear background of the sky. Walls have pictures of the beautiful Tylorese Valley as well as a map of the world. The children’s eyes can only view a narrow road enclosed with a dull sky. It is quite a dreary and depressing place for children.

In the third stanza, the pensive poet suddenly turns belligerent (aggressive) and feels that Shakespeare is ‘wicked’. This is because he misleads the children. He shows them a beautiful world of ships, sun and love which is not only unreal for them but has a corrupting influence on these children and instigates them to steal and try to escape from their cramped holes. Their existence is indeed very sad.

These emaciated children are so thin that it appears that they are ‘wearing’ only skins. The spectacles they are wearing have glass which has been broken and mended. Their entire appearance reeks of their deprivation. The poet shows his outrage by suggesting that the maps on their walls should show huge slums instead of beautiful scenic graphics.

Finally, in the last stanza, the poet reveals the appalling truth that there can be no change for the better unless a governor, a school inspector or an educationist or a visitor comes to the school. The map in their classroom is the only medium for the children to view the world outside their slums. The windows of their classroom shut them and confine them to their world of poverty and helplessness.

Next, poet appeals to those people who are in power to liberate these slum children from the horrendous life that they are leading. He also exhorts the people themselves, to break open these windows which appear to have sealed the fate of these children. He would like to see these children bask in the educational facilities in this world, and run carefree on the golden sands and enjoy a new lease of life and freedom. The poet earnestly desires that each and every child should be able to enjoy the fundamental right to freedom. They should have access to all kinds of books, new as well as old. They should also be able to learn from nature around them.

Spender ends the poem on a positive note as he expresses his belief that people who are ignited by the spirit of knowledge and learning are the ones who create history. It is the moral liability of everyone to break barriers between the haves and the have-nots, and give a meaningful education to all. For history remembers only those people who are educated and have enlightened themselves for a better world.

An Elementary School Classroom in a Slum Summary Reference-to-Context Questions

Read the extracts given below and answer the questions that follow.

1. Far far from gusty waves these children’s faces.
Like rootless weeds, the hair torn round their pallor:
The tall girl with her weighed-down head. The paper-
seeming boy, with rat’s eyes.

a. What are the children compared to?
Answer:
The children have pale faces and torn and scattered hair all over their faces like rootless weeds. A thin boy is compared to a paper.

b. Why do you think the tall girl is sitting with a weighed-down head?
Answer:
The tall girl is sitting with a weighed-down head because she is depressed of being poor. She also feels humiliated and embarrassed because of the lack of education.

c. Give two phrases which tell us that the children are undernourished.
Answer:
(i) rootless weeds
(ii) rat’s eyes

d. Name the poetic device used in the second line.
Answer:
Simile

2. The stunted, unlucky heir
Of twisted bones, reciting a father’s gnarled disease,
His lesson, from his desk. At back of the dim class
One unnoted, sweet and young. His eyes live in a dream,
Of squirrel’s game, in tree room, other than this.

a. Who is the ‘unlucky heir’ and what has he inherited?
Answer:
The stunted boy with twisted bones sitting in the slum classroom is the unlucky heir. He has inherited the deformity of gnarled disease.

b. What is the stunted boy reciting?
Answer:
The stunted boy has inherited the disease and despair of his parents and has become a carrier of his father’s disease and poverty. He is reciting his lesson in the class from his desk.

c. Who is sitting at the back of the dim class?
Answer:
One unnoted sweet and young dreamer, who dreams about a squirrel’s game, is sitting at the back of the dim class. He is dreaming of his future.

d. What is the ‘tree room’?
Answer:
It is the squirrel’s nest or hole from which the squirrel is moving in and out.

3. On sour cream walls, donations, Shakespeare’s head,
Cloudless at dawn, civilized dome riding all cities.
Belled, flowery, Tyrolese valley. Open-handed map
Awarding the world its world.

a. Which walls have been referred to in these lines?
Answer:
They are the walls of a classroom in an elementary school in a slum.

b. What is meant by ‘sour cream walls’?
Answer:
The walls are damp, unpleasant and dirty. They have not been painted freshly and is pealing off the surface.

c. What donations are there on the walls?
Answer:
On this wall many donated items have been put up that represent different world. A bust of Shakespeare and domes of huge buildings in the cities. There is also a reflection of the early morning cloudless sky on the wall. There is the beautiful picture of the valley of Tyrolese and the world map.

d. Explain, ‘Awarding the world its world’.
Answer
The rich people who have drawn these maps have depicted these places that are unreachable for these children. They are giving the world its world which however, does not belong to them.

4. And yet, for these
Children, these windows, not this map, their world.
Where all their future’s painted with a fog,
A narrow street sealed in with a lead sky
Far far from rivers, capes, and stars of words.

a. Who are ‘these children’? What do ‘these windows’ refer to?
Answer:
‘These children’ are the poor children living in a slum. ‘These windows’ are the windows of the classroom where the children are now sitting.

b. What has been said to be the world for these children?
Answer:
The narrow street under the dull sky has been said to be their world. Their fate is sealed by the windows in the classroom, and does not go beyond as the map suggests.

c. What has been said about their future?
Answer:
Their future is painted with fog. It means that the poor children have no bright hopes about their future. There is no one to guide them, their future is not clear and is sealed by the darkness of the sky that is above the narrow street they live in.

d. Explain the importance of the last line.
Answer:
These children have no access to the beauties of nature, they cannot see the rivers or the capes, or the stars of words. The highly literate people of the world will never be known to them.

5. Surely, Shakespeare is wicked, the map a bad example,
With ships and sun and love tempting them to steal—
For lives that slyly turn in their cramped holes
From fog to endless night? On their slag heap, these children
Wear skins peeped through by bones and spectacles of steel
With mended glass, like bottle bits on stones.

a. Why is Shakespeare described as wicked?
Answer:
Shakespeare is an epitome of high literary excellences, but in the slum schools, where hardly any learning takes place and the children are troubled by disease and despair, literary training is a far cry. That is why, Shakespeare has been described as wicked.

b. Explain: ‘from fog to endless night’.
Answer:
The expression describes the miserable and pathetic lives of the slum children. From foggy mornings till late nights, these children make desperate attempts to live their life, sustaining it despite all odds. Their life is full of misery, hopelessness and suffering.

c. What does the reference of ‘slag heap’ mean?
Answer:
The poet is comparing the extremely starved and malnourished bodies to the large pile of waste metal remains. When they sit within their classrooms, they appear to be the heaps of untidy piles of bones, or like a slag heap.

d. How do the slum children look like?
Answer:
Slum children look like skeletons wearing broken glasses as spectacles.

6. On their slag heap, these children
Wear skins peeped through by bones and spectacles of steel
With mended glass, like bottle bits on stones.
All of their time and space are foggy slum.
So blot their maps with slums as big as doom.

a. Which two images are used to describe these slums?
Answer:
The two images used to describe slums are:
(i) foggy slums
(ii) slums as big as doom

b. What sort of life do these children lead?
Answer:
These children lead miserable lives. They are physically weak.

c. Which figure of speech is used in the last line?
Answer:
In the last line, ‘Simile’ is used as the figure of speech.

d. What request does the poet make here?
Answer:
The poet wants the maps to mention clearly the slums that they live in.

7. Unless, governor, inspector, visitor,
This map becomes their window and these windows
That shut upon their lives like catacombs,
Break O break open till they break the town

a. What is meant by ‘this map’?
Answer:
It is a map of the world which is hung on the wall in the classroom.

b. What are ‘these windows’ which the poet talks of?
Answer:
They are the classroom windows from where the children could see only a narrow street and a dull sky.

c. What has been referred to as ‘catacombs’?
Answer:
The little narrow homes of the slum-dwellers are referred to as ‘catacombs’. They lead a life that is shut inside the slum.

d. Why is there a mention of three categories of people?
Answer:
The three categories of people are the ones who can improve the conditions of the slum.

8. Break O break open till they break the town
And show the children to green fields, and make their world
Run azure on gold sands, and let their tongues
Run naked into books the white and green leaves open
History theirs whose language is the sun.

a. To whom does ‘they’ refer?
Answer:
‘They’ refers to the children sitting in the classroom of an elementary school in a slum.

b. What would they break?
Answer:
The poet hopes that one day they would break free from the chains of the slum. They will rise above all deprivations and create a beautiful world for themselves.

c. What other freedom should they enjoy?
Answer:
They should enjoy equal rights as citizens and get education entitled to them. They should have a bright future like all others.

d. Explain the last line of the extract.
Answer:
The poet feels that history will be changed if the people are educated. And according to him, history is created by those who have the warmth and exposure to the sun and the brightness of the wide world.

9. The stunted, unlucky heir
Of twisted bones, reciting a father’s gnarled disease,
His lesson, from his desk. At the back of the dim class
One unnoted, sweet and young.

a. Who is the unlucky heir?
Answer:
The boy with stunted growth is the unlucky heir of his father’s gnarled disease of twisted bones.

b. What will he inherit?
Answer:
All that he will inherit is his father’s gnarled disease of twisted bones.

c. Who is sitting at the back of the dim class?
Answer:
A young unnoted, sweet boy is sitting unnoticed at the back of the dim class.

d. Why is the disease referred to as ‘gnarled’?
Answer:
The disease has been referred to as ‘gnarled’ because it has restricted his growth and gave him twisted bones.

CBSE Class 6 Sanskrit Sample Paper Set 3

We have given detailed NCERT Solutions for Class 6 Sanskrit come in handy for quickly completing your homework.

CBSE Class 6 Sanskrit Sample Paper Set 3

निर्धारित समय: 3 घंटे
पूर्णांक: 80

खण्ड: – क
अपठित-अवबोधनम्

प्रश्न 1.
अधोलिखितं अनुच्छेदम् पठित्वा प्रश्नानाम् उत्तराणि लिखत- (10)

एकस्य वृक्षस्य शाखासु अनेके काकाः वसन्ति स्म। तस्य वृक्षस्य कोटरे एक सर्पः अपि अवसत्। काकानाम् अनुपस्थितौ सर्पः काकानाम् शिशून् खादति स्म। काकाः दुःखिताः आसन्। तेषु एकः वृद्धः काकः उपायम् अचिन्तयत्। राजसेवका: वृक्षस्य समीपे जलाशयम् आगच्छति स्म। शिलायां स्थितं तस्याः आभरणम् आदाय एकः काकः वृक्षस्य उपरि अस्थापयत्। राजसेवकाः काकम् अनुसृत्य वृक्षस्य समीपम् अगच्छन्। ते तत्र सर्प अमारयन्। अतः एव उक्तम्-उपायेन सर्वं सिद्धयति।

प्रश्ना:
I. एकपदेन उत्तरत- (1 × 4 = 4)
(i) के दुःखिताः आसन्?
(ii) कः उपायं अचिन्तयत्?
(iii) वृक्षस्य कोटरे कः अवसत्?
(iv) केन सर्वं सिद्धयति?

II. पूर्णवाक्येन उत्तरत- (2 × 2 = 4)
(i) काकाः कुत्र वसन्ति स्म?
(ii) सर्पः कान् खादति स्म?

III. यथा निर्देशं लिखत- (1 × 2 = 2)

(i) ‘शिलायां’ इत्यत्र का विभक्तिः?
(क) षष्ठी
(ख) सप्तमी
(ग) प्रथमा
(घ) द्वितीया

(ii) ‘अधः’ इति पदस्य विलोमपदं किम्?
(क) उपरि
(ख) समीपम्
(ग) अपि
(घ) सर्वं

खण्डः – ख
रचनात्मकं कार्यम्

प्रश्न 2.
मञ्जूषातः उचितपदानि चित्वा चित्राणाम् संस्कृतेन लिखत- (1 × 4 = 4)

मञ्जूषा – उल्लूकः, कपोतः, गजः, मयूरः, मण्डूकः, मत्स्यः

CBSE Class 6 Sanskrit Sample Paper Set 3 Q2

प्रश्न 3.
अधोलिखितशब्दान् आधृत्य सार्थकवाक्यानि रचयत- (4)

1. छात्राः कन्दुकेन कर्षन्ति
2. कृषकाः पुस्तकम् पचति
3. बालौ क्षेत्राणि खेलतः
4. महिला भोजनम् पठन्ति
  1. ______________
  2. ______________
  3. ______________
  4. ______________

प्रश्न 4.
चित्रम् दृष्ट्वा प्रश्नस्य उत्तरं लिखत- (1 × 5 = 5)

प्रश्न: 1.
नरस्य समीपे का तिष्ठति?
CBSE Class 6 Sanskrit Sample Paper Set 3 Q4

प्रश्नः 2.
जनकस्य अङ्के का अस्ति?
CBSE Class 6 Sanskrit Sample Paper Set 3 Q4.1

प्रश्न: 3.
कः हलेन कर्षति?
CBSE Class 6 Sanskrit Sample Paper Set 3 Q4.2

प्रश्न: 4.
एतौ कौ स्तः?
CBSE Class 6 Sanskrit Sample Paper Set 3 Q4.3

प्रश्नः 5.
एते के स्तः?
CBSE Class 6 Sanskrit Sample Paper Set 3 Q4.4

खण्ड: – ग
अनुप्रयुक्त व्याकरणम्

प्रश्न 5.
वर्णसंयोजनं कृत्वा कोष्ठके लिखत- (4)
यथा – त् + अ + इ + आ + ग् + अः = तडागः

  1. भ् + आ + र् + अ + त् + अ + स् + य् + अ = ______________
  2. व् + इ + ह् + आ + र् + अ + म् = ______________
  3. ब् + आ + ल् + उ + क् + आ = ______________
  4. स् + आ + ग् + अ + र् + आ + त् = ______________

प्रश्न 6.
(अ) उचितशब्दरूपैः रिक्तस्थानानि पूरयत- (½ × 4 = 2)
CBSE Class 6 Sanskrit Sample Paper Set 3 Q6

(ब) उचित धातुरूपैः रिक्तस्थानानि पूरयत- (½ × 4 = 2)
CBSE Class 6 Sanskrit Sample Paper Set 3 Q6.1

प्रश्न 7.
कोष्ठकात् उचितं पदं चित्वा वाक्यं पूरयत- (1 × 4 = 4)

  1. __________ कूजतः। (कोकिले / कोकिला)
  2. एतानि __________ सन्ति। (भवन / भवनानि)
  3. __________ लिखति। (छात्रा / छात्राः)
  4. __________ पतति। (नारिकेलं / नारिकेले)

प्रश्न 8.
मञ्जूषातः अव्ययपदानाम् प्रयोगेण रिक्तस्थानानि पूरयत- (1 × 4 = 4)

कुतः, कदा, कुत्र, कथं, किम्

  1. जगन्नाथपुरी __________ अस्ति?
  2. गङ्गानदी __________ प्रवहति?
  3. त्वं __________ गमिष्यसि?
  4. तव स्वास्थ्यं __________ अस्ति?

प्रश्न 9.
मञ्जूषातः शब्दान् चित्वा रिक्तस्थानानि पूरयत- (5)

कृत्वा, दृष्ट्वा, गृहीत्वा, तीर्खा, श्रुत्वा

  1. ते बालकाः __________ नद्यां उत्तीर्णाः।
  2. पथिकः बालकान् दुःखितान् __________ अपृच्छत्।
  3. पुस्तकानि __________ विद्यालयं गच्छ।
  4. ‘मोहनः कार्यं __________ गृहं गच्छति।
  5. ‘पथिकस्य वचनं __________ बालकाः प्रसन्नाः अभवन्।

प्रश्न 10.
कोष्ठकात् उचितं पदं चित्वा वाक्यानि पूरयत- (1 × 4 = 4)

  1. बालकाः __________ सह. पठन्ति। (मित्रैः/मित्र)
  2. नराः __________ वदन्ति। (मुखं/मुखने)
  3. फलानि __________ पतन्ति। (वृक्षात्/वृक्षस्य)
  4. __________ सैनिकाः सन्ति। (शिविरे/शिविरः)

खण्डः – घ
पठित-अवबोधनम्

प्रश्न 11.
अधोलिखितं गद्यांशं पठित्वा प्रश्नान् उत्तरत- (5)

एषः समुद्रतटः। अत्र जनाः पर्यटनाय आगच्छन्ति। केचन तरङ्गैः क्रीडन्ति। केचन च नौकाभिः जलविहारं कुर्वन्ति। तेषु केचन कन्दुकेन क्रीडन्ति। बालिकाः बालकाः च बालुकाभिः बालुकागृहं रचयन्ति। मध्ये-मध्ये तरङ्गाः बालुकागृहं प्रवाहयन्ति। एषा क्रीडा प्रचलति एव। समुद्रतटाः न केवलं पर्यटनस्थानानि। अत्र मत्स्यजीविनः अपि स्वजीविकां चालयन्ति।

प्रश्नाः
I. एकपदेन उत्तरत- (1 × 2 = 2)
(i) जनाः काभिः जलविहारं कुर्वन्ति?
(ii) समुद्रतटे जनाः किमर्थम् आगच्छन्ति?

II. पूर्णवाक्येन उत्तरत- (1 × 2 = 2)
(i) बालकाः केन क्रीडन्ति?
(ii) समुद्रतटे के स्वजीविकां चालयन्ति?

III. निर्देशानुसारं लिखत- (½ × 2 = 1)
(i) ‘तरङ्गः’ इति पदे किम् वचनम्?
(क) एकवचनम्
(ख) द्विवचनम्
(ग) बहुवचनम्

(ii) रिक्तस्थानं पूरयत-
तरङ्गाः __________ प्रवाहयन्ति।
(क) गृहम्
(ख) बालुका
(ग) बालुकागृहम्
(घ) मध्ये

प्रश्न 12.
अधोलिखितं पद्यांशं पठित्वा प्रश्नानाम् उत्तराणि लिखत- (5)

स्पृशन्ति पादैः पातालं च।
नभः शिरस्सु वहन्ति वृक्षाः।।
पयोदर्पणे स्वप्रतिबिम्बम्।
कौतुकेन पश्यन्ति वृक्षाः।।

प्रश्नाः
I. एकपदेन उत्तरत- (1 × 2 = 2)
(i) शिरस्सु नभः के वहन्ति?
(ii) वृक्षाः कैः पातालं स्पृशन्ति?

II. पूर्णवाक्येन उत्तरत- (1 × 2 = 2)
(i) वृक्षाः पयोदर्पणे कौतुकेन किम् पश्यन्ति?

III. निर्देशानुसारं उत्तरत- (½ × 2 = 1)
(i) ‘आकाशः’ इति पदस्य पर्यायः कः?
(क) पातालः
(ख) पयः
(ग) नभः
(घ) वृक्षः

(ii) ‘दर्पणे’ इति पदे का विभक्तिः?
(क) प्रथमा
(ख) तृतीया
(ग) चतुर्थी
(घ) सप्तमी

प्रश्न 13.
मञ्जूषातः उचितशब्दान् चित्वा रिक्तस्थानानि पूरयत- (4)

अगच्छन्, मानवाः, प्रियं, कृष्णः

  1. प्रियवाक्यप्रदानेन सर्वे __________ तुष्यन्ति।
  2. __________ वैनतेयः एकं पदं न गच्छति।
  3. पिकस्य वर्णः __________ भवति।
  4. __________ हि वक्तव्यं।

प्रश्न 14.
निम्नलिखितेभ्यः पदेभ्यः भिन्नप्रकृतिकं पंद चिनुत- (2)

  1. रामः, कृष्णः, सीता, बलरामः। ______________
  2. क्रीडति, धावति, मतिः, खेलति। ______________
  3. क्रूरता, मधुरता, गीता, कटुता। ______________
  4. श्वेतः, पीतः, नीलः, नीतः। ______________

प्रश्न 15.
मञ्जूषातः समानार्थकपदानि चित्वा लिखत- (4)

उचितम्, राज्ञः, मीनः, वा

  1. मत्स्यः
  2. अथवा
  3. युक्तम्
  4. नृपस्य

प्रश्न 16.
मञ्जूषातः विलोमपदानि चित्वा लिखत- (2)

ग्रीष्मे, सरसा, अक्षमम्, जागरणे

  1. क्षमम्
  2. स्वपने
  3. शीते
  4. नीरसा

प्रश्न 17.
तत्समशब्दान् लिखत- (2)

  1. चिडिया
  2. सूरज
  3. चाँद
  4. किसान

प्रश्न 18.
पद्यांशान् योजयत- (5)

(क) (ख)
1. पुस्तके पठितः पाठः जीवने यो न सार्थकः।
2. किं भवेत् तेन पाठेन। जीवने नैव साधितः
3. उद्येमन हि प्रविशन्ति मुखे मृगाः
4: सर्वः कामानवाप्नोतु सर्वो भद्राणि पश्यतु।

प्रश्न 19.
घटनाक्रमानुसारं अधोलिखितानि वाक्यानि पुनः लेखनीयानि। (½ × 6 = 3)

  1. अजीजः सफलः आसीत्।
  2. स्वामी तस्मै अवकाशस्य वेतनस्य च पूर्णं धनं ददाति।
  3. अजीजं दृष्ट्वा स्वामी चकितः भवति।
  4. सहसा एका मधुमक्षिका निर्गच्छति तस्य च हस्तं दशति।
  5. पीडितः स्वामी अत्युच्चैः चीत्करोति।
  6. एका अन्या मधुमक्षिका तस्य ललाटे दशति।

The Third Level Summary Line By Line Explanation by Jack Finney in English

We have decided to create the most comprehensive English Summary that will help students with learning and understanding.

The Third Level Summary in English by Jack Finney

The Third Level by Jack Finney About the Author

Jack Finney (2 October 1911-16 November 1995) was born in Milwaukee, Wisconsin, and given the name John Finney. His father died when he was three years old and he was renamed Walter Braden Finney in honour of his father. Yet the nickname Jack remained with him throughout his life. He attended Knox College in Galesburg, Illinois. His best-known works are science fiction and thrillers. Two of his novels, ‘The Body Snatchers’ and ‘Good Neighbor Sam’ became the basis of popular films.

Poet Name Jack Finney
Born 2 October 1911, Milwaukee, Wisconsin, United States
Died 14 November 1995, Greenbrae, California, United States
Awards World Fantasy Award—Life Achievement
Movies Invasion of the Body Snatchers, The Invasion
The Third Level Summary by Jack Finney
The Third Level Summary by Jack Finney

The Third Level Introduction to the Chapter

The Grand Central Station of New York has two levels. But Charley, the thirty-one year old protagonist of the story, a city dweller, declares that there are three and claims to have been there as well.

The Third Level Theme

The story, ‘The Third Level’ clearly explores the science fiction genre of ‘time travel’. Jack Finney, the recipient of the World Fantasy Award, interweaves fantasy with reality in his projection of time travel. Charley, the protagonist wishes to be transported to the third level, the world of Galesburg, Illinois, 1894, which is supposed to be a much happier and quieter place to be in.

The story also dwells on the theme of escapism as a psychological refuge from the grim realities of the present day world along with a desire to stay with the past—a desire that Charley’s wife Louisa does not contest. Sam has also happily escaped, with no desire to return to his old profession.

The story exposes the vulnerable side of the common man. Surrounded by myriad problems, we humans, sometimes experience a craving for peace and serenity, and look for possible escapes. This story is about time intersection, an illusion, a kind of long dream that we do not experience during our sleep.

The Third Level Summary in English

“The Third Level”, written by Jack Finney blends fiction with reality. It is also about a man’s wish to escape from the harsh realities of present life. Charley, though he does not admit it, wants to go into the past as he is unhappy. He is unhappy with his wife. In fact, he is as unhappy as he finds the world in which he lives full of hurry, tension and war. His psychiatrist friend, Sam tells his stamp collecting also as an escape into the past.

There were only two levels at Grand Central Station. However, Charley found a third one. It was by chance. Many a times, he was lost there. He was always discovering new doorways, new corridors and new tunnels. He had begun to think that the Grand Central was always pushing out tunnels and new corridors like roots of a huge tree.

There he lost his way and found himself on the third level. This level was entirely different and olcl-fashioned. The locomotive, the brass spittoons and the naked gaslights belonged to the previous century. He desired to escape to Galesburg, the town of his dreams. However, he was nearly arrested. The money he gave to pay the fare was different from that in use those days. The booking clerk thought that he was cheating. Charley, thus, ran into the present. He never found the third level again. However, his psychiatrist friend, who did not believe that the third level existed, found it and escaped to Galesburg of 1894.

The Third Level Main Characters in the Chapter

Charley

Charley is a thirty-one year old man in a tan gabardine suit and a straw hat. One night, on his way back from work, he decides to take the subway at the Grand Central Station, which as everyone knows, has two levels. Preoccupied and in a hurry, Charley discovers an unknown exit that takes him through a long corridor, into the third level.

Here, there were fewer ticket windows, the man at the booth wore green eye-shades, the lights were open-flame gas lights, and women wore old-fashioned, fully covered dresses. The newspaper, ‘The World’, was dated June 11,1894. Charley knows that from there, the third level of the Grand Central, he could go to anywhere in the United States, 1984. He decides to buy two tickets to Galesburg, Illinois, for his wife Louisa and himself from the ticket window in the third level.

Galesburg, with its big old houses, huge lawns and tremendous trees represents an idyllic world to Charley, with the World War II still forty years into the future. However, the clerk at the window refuses the currency Charley offers. Charley leaves, deciding to return the next day, after converting all his savings into old-style currency. But Charley has never again found the third level.

When Charley tells his psychiatrist friend, Sam Weiner about this, Sam tells him that it was “a waking dream wish fulfilment” as Charley was “unhappy” in the modern world with its insecurities and fears, and just wanted to escape. Charley never again found the corridor that led to the third level at the Grand Central. Ironically, his friend Sam, the psychiatrist, disappeared, only to reach Galesburg, Illinois, in 1894.

Sam

Sam Weiner is Charley’s friend, and psychiatrist, and the next most important character in the story. He concludes that the third level is a figment of Charley’s imagination, induced by the pressures of modern living.

When Charley fails to find the third level of the Grand Central Station, his wife Louisa is worried for him and tells him to stop looking for it. But after sometime, both start looking for it because they find proof that the third level exists. Charley’s friend, Sam Weiner disappears. A first-day cover that Charley discovers in his collection, is signed by Sam and is from Galesburg, Illinois, dated July 18,1894. Charley subsequently discovers that Sam had bought eight- hundred dollars worth of old-style currency and moved to Galesburg, Illinois, in 1894. He had . set up a hay, feed and grain business as he had always said that it is what he really wished to do. Clearly, he could not go back to his old business—psychiatry—in Galesburg, Illinois, in 1894.

Louisa

Charley’s wife Louisa was initially angry with Sam’s suggestion that Charley was unhappy, ‘ when Charley tells him about his sojourn to the third level of the Grand Central Station in New York. Then Sam explains that it is not marital unhappiness, but dissatisfaction and discontent with modern day living with its insecurity, fear, war and worry. To escape from these pressures, Charley’s mind had sought refuge in the idyllic world of the third level. Louisa’s conviction in the existence of the third level is affirmed only when she sees the note sent by Sam himself, from Galesburg, Illinois, dated July 18, 1894. Since then, Louisa has been actively involved in looking for the third level, along with her husband, Charley.

The Third Level Summary Questions and Answers

Question 1.
Why did Charley meet a psychiatrist?
Answer:
Charley met a psychiatrist, since he was in a dilemma. He felt sure that he had been on the third level of the Grand Central Station, which everyone knows has only two levels. Even the Presidents of the railroads would swear on a stack of timetables that there were only two levels.

Question 2.
What was the psychiatrist’s diagnosis?
Answer:
The psychiatrist said that Charlie was unhappy. The modern world, full of insecurity, fear, war and worry oppressed him, and he just wanted to escape.

Question 3.
What proof did the psychiatrist provide?
Answer:
Charley’s psychiatrist and his friends said that his stamp-collecting was an indication of his desire to seek “a temporary refuge from reality”, as was his collection of first-day covers.

Question 4.
What was Charley’s argument when the psychiatrist told him that the stamp collection was a temporary refuge from reality?
Answer:
Charley argued that his grandfather lived in nice and peaceful times, yet he was the one . who had started the stamp collection. He did not need any “temporary refuge from reality”. He added that President Roosevelt collected stamps too.

Question 5.
How does Charley describe Galesburg, Illinois, 1894?
Answer:
Charley describes it as a wonderful town with a leisurely way of life with big old frame houses, huge lawns, tremendous trees and a peaceful and tranquil world. During summer evenings, people sat in their lawns, with men smoking cigars and women waving palm-leaf fans. The first World War was twenty years away and the second World War was forty years into the future.

Question 6.
What is a first-day cover?
Answer:
When a new stamp is issued, stamp collectors buy some and use them to mail envelopes to themselves on the very first day of sale and the postmark proves the date. The envelope is called a first-day cover. They are never opened. You just put a blank paper inside the envelope.

Question 7.
What role does the first-day cover play in the story?
Answer:
One night,while fussing with his stamp collection, Charley comes across a first-day cover that should not have been there. It had been mailed to his grandfather at his home in Galesburg in July 18, 1894. However, instead of a blank paper, it contained a letter for Charley from Sam. It urged him to come back to the third level with Louisa, and keep looking for it till he found it.

Question 8.
What was the content of the note that Sam wrote to Charley?
Answer:
Sam said that he had found the third level, that he had already been there for two weeks, that life was peaceful, calm and tranquil. He urged Charley and Louisa to go back to the third level and keep looking for it till they found it.

Question 9.
How was Charley often lost on the Grand Central Station?
Answer:
Cllarley had went to the Grand Central Station hundreds of times. However, at times, he was always lost in new doorways and corridors. Once, he entered a tunnel and came out in the lobby of a hotel. Another time, he reached in an office building.

Question 10.
How did Charley compare the Grand Central to a huge tree? Why?
Answer:
Charley always found new tunnels and staircase at the Grand Central. He began to suspect that Grand Central was like a huge tree. It used to push out new corridors and tunnels like the roots of a tree.

The Laburnum Top Poem Summary Line By Line Explanation by Ted Hughes in English

We have decided to create the most comprehensive English Summary that will help students with learning and understanding.

The Laburnum Top Poem Summary in English by Ted Hughes

The Laburnum Top Poem by Ted Hughes About the Poet

Name Ted Hughes
Born 17 August 1930, Mytholmroyd, United Kingdom
Died 28 October 1998, North Tawton, United Kingdom
Education Pembroke College, University of Cambridge
Spouse Carol Orchard (m. 1970–1998), Sylvia Plath (m. 1956–1963)
Awards Costa Book of the Year, Guggenheim Fellowship for Creative Arts, US & Canada, Guardian Children’s Fiction Prize
The Laburnum Top Poem Summary by Ted Hughes
The Laburnum Top Poem Summary by Ted Hughes

The Laburnum Top Poem Summary in English

Laburnum is known as amaltas in Hindi.

On a bright September afternoon, the laburnum, a short tree with hanging branches and yellow flowers, stood soundless and motionless. A few leaves that had yellowed and some seeds lay scattered around it. Just then, all of a sudden a small singing bird with yellow feathers on its wings, the goldfinch, arrived chirping. She entered the foliage like a lizard—smooth, watchful and hasty.

As she entered, the tree suddenly seemed to come alive. It started up like a machine. There were shrill sounds of twittering arid the tree seemed to quiver with joy. The mother bird was like the engine of her family. Like an engine she added life to the tree and flitted from branch to branch, showing her striped face, with yellow and black markings that were peculiar to her. Then with a mysterious, low whistle she flew off into the sky. Once again the laburnum quietened down as it was before her arrival.

The Laburnum Top Poem Summary Questions and Answers

1. The Laburnum top is silent, quite still
In the afternoon yellow September sunlight,
A few leaves yellowing, all its seeds fallen.

a. Mention the poetic device used in the ‘The Laburnum top is silent, quite still’. Why has it been used? Alteration. The Laburnum top is silent, quite still. The repeated sound creates a musical effect.

b. Mention the colour suggested by the poet. Also mention the words that suggest colour.
The colour is yellow. The colour is suggested by the yellow September sunlight and the yellowing leaves.

c. What is the stillness disturbed by?
It is disturbed by the arrival of the goldfinch.

d. What season is the poet talking about? Mention the words that suggest the season.
Autumn. The words that suggest the season are September; leaves yellowing and all the seeds fallen.

2. Till the goldfinch comes, with a twitching chirrup
A suddenness, a startlement, at a branch end.
Then sleek as a lizard, and alert, and abrupt,
She enters the thickness, and a machine starts up
Of chitterings, and a tremor of wings, and trillings –
The whole tree trembles and thrills.

a. What changes the scene completely?
Answer:
The arrival of the goldfinch changes the scene completely.

b. What is the affect on the tree?
Answer:
There is a lot of movement and sound.

c. How does the bird enter the tree? Mention the poetic device.
Answer:
Like a lizard, watchfully and unexpectedly. Simile.

d. ‘… and a machine starts up.’ What is compared to a machine? What is the poetic device?
Answer:
The tree becomes noisy and trembles like a machine. Metaphor.

e. What is the mood of the poem now? How is it suggested?
Answer:
Happy and excited as if trembling with delight. The word that suggest it are ‘The whole tree trembles and thrills’. .

f. Mention the words that show movement?
Answer:
enters, machine starts up, tremor of wings, and trillings, trembles, thrills.

g. Mention the words that show sound. Name the poetic device,
Answer:
twitching chirrup, machine starts up; chitterings. Onomatopoeia.

h. Mention the adjectives used for the goldfinch.
Answer:
sleek, alert, and abrupt.

3. It is the engine of her family.
She stokes it full, then flirts out to a branch-end
Showing her barredface identity mask

a. Who is the‘she’in these lines?
Answer:
The goldfinch.

b. What is her arrival compared to?
Answer:
The starting of an engine.

c. What are the similarities in an engine and a tree then?
Answer:
Both are noisy and pulsating.

d. What does she do to the engine?
Answer:
She ‘stokes it full’ implying she adds fuel to it and causes it to come to life.

e. What is the ‘barred face identity mask’?
Answer:
It’s distinctive face with yellow and black markings.

f. What does she do on the tree?
Answer:
She moves about playfully from branch to branch.

4. Then with eerie delicate whistle-chirrup whisperings
She launches away, towards the infinite
And the laburnum subsides to empty.

a. What does the goldfinch finally do?
Answer:
It flies away from the laburnum after producing a whistling sound.

b. Why has the sound been described as ‘eerie delicate whistle-chirrup whisperings’?
Answer:
The sound of a goldfinch is peculiar. It is a liquid, twittering song with trills.

c. Where does she go?
Answer:
She flies into the sky.

d. What is the effect on the tree?
Answer:
It once again reverts back into silence.

e. What do the words ‘eerie delicate’ suggest?
Answer:
They suggest an unusual and weak sound.

CBSE Class 6 Sanskrit Sample Paper Set 2

We have given detailed NCERT Solutions for Class 6 Sanskrit come in handy for quickly completing your homework.

CBSE Class 6 Sanskrit Sample Paper Set 2

निर्धारित समय : 3 घंटे
पूर्णांक : 80

खण्ड: – क
अपठित-अवबोधनम्

प्रश्न 1.
अधोलिखितं कथां पठित्वा प्रश्नानाम् उत्तराणि लिखत- (10)

एकदा एकः नृपः भोजनं वितरति स्म। राजपुरुषाः एकम् ऋषिकुमारं दृष्ट्वा अवदन्-भवान् अपि भोजनं करोतु। इयम् एव राजाज्ञा इति। सः तरूणः ऋषिकुमारः अवदत्-भोः! अहम् श्रमम् कृत्वा एव भोजनं करोमि। ये जनाः वृद्धाः असहायाः च तेभ्यः एव भोजनं यच्छत। एषः आसीत् रैक्वः ऋषिकुमारः। राजा तस्य समीपं गत्वा ज्ञानार्जनम् अकरोत्।

प्रश्नाः
I. एकपदेन उत्तरत- (1 × 4 = 4)
(i) ऋषिकुमारः किं कृत्वा भोजनं करोति स्म?
(ii) ऋषिकुमारस्य नाम किम् आसीत्?
(iii) नृपः किम् वितरति स्म?
(iv) नृपः कस्य समीपं गत्वा ज्ञानार्जनम् अकरोत्?

II. पूर्णवाक्येन उत्तरत- (2 × 2 = 4)
(i) नृपः ऋषिकुमारस्य समीपं गत्वा किम् अकरोत्?
(ii) केभ्यः भोजनं दातव्यम्?

III. भाषिककार्यम् कुरुत- (1 × 2 = 2)
(i) ‘अवदन्’ इति क्रियापदस्य कर्तृपदं किम्?
(क) राजपुरुषाः
(ख) एकम्
(ग) ऋषि
(घ) कुमारः

(ii) ‘करोमि’ इत्यत्र कः लकार:?
(क) लुट्
(ख) लट्
(ग) लोट
(घ) लङ्

खण्डः – ख
रचनात्मकं कार्यम्

प्रश्न 2.
मञ्जूषातः उचितपदानि चित्वा चित्राणाम् नामानि संस्कृतेन लिखत- (1 × 4 = 4)

मञ्जूषा – अश्वः, उष्ट्रः, सिंहः, कुक्कुरः, वृषभः, वानरः

CBSE Class 6 Sanskrit Sample Paper Set 2 Q2

प्रश्न 3.
अधोलिखितशब्दान् आधृत्य सार्थकवाक्यानि रचयत- (4)

1. वानराः जले गर्जन्ति
2. सिंहाः वनेषु कूर्दन्ति
3. खगाः आकाशे तरन्ति
4. मत्स्याः वृक्षेष उत्पतन्ति
  1. _______________
  2. _______________
  3. _______________
  4. _______________

प्रश्न 4.
मञ्जूषातः समुचितपदानि चित्वा अनुच्छेदं पूरयत- (10)

दुष्यन्तस्य, पराक्रमी, मातृभूमिः, भरतः, विशालः, राजधानी, सह, गङ्गा, पर्वतराजः, नाम

भारतदेशः अस्माकं ___(1)___ अस्ति। प्राचीनकाले ___(2)___ नाम राजा भारतम् अपालयत्। सः शकुन्तलायाः ___(3)___ च पुत्रः आसीत्। भरतः ___(4)___ आसीत्। निर्भीकः सः सिंहशावकैः ___(5)___ अक्रीडत्। भरतेन शासितम् देशः भारतम् इति ___(6)___ अभवत्। भारतदेशः अति ___(7)___ देशः अस्ति। नवदेहली भारतस्य ___(8)___ अस्ति। भारतस्य उत्तरदिशायाम् ___(9)___ हिमालयः अस्ति। पुण्यनदी ___(10)___ हिमालयात् एव प्रवहति।

खण्ड: – ग
अनुप्रयुक्त-व्याकरणम्

प्रश्न 5.
वर्णसंयोजनं कृत्वा कोष्ठके लिखत- (4)

यथा- प् + उ + ष् + प् + ए = पुष्पे

  1. प् + अ + क् + व् + आ + न् + इ = _______________
  2. भ् + अ + व् + अ + न् + त् + इ = _______________
  3. ध् + अ + र् + आ + य् + आ + म् = _______________
  4. ल् + इ + ख् + आ + म् + अः = _______________

प्रश्न 6.
(अ) उचितशब्दरूपैः रिक्तस्थानानि पूरयत- (½ × 4 = 2)
CBSE Class 6 Sanskrit Sample Paper Set 2 Q6

(ब) उचितधातुरूपैः रिक्तस्थानानि पूरयत- (½ × 4 = 2)
CBSE Class 6 Sanskrit Sample Paper Set 2 Q6.1

प्रश्न 7.
कोष्ठकात् उचितं पदं चित्वा वाक्यं रिक्तस्थानानि पूरयत- (1 × 4 = 4)

  1. _______________ कच्छपाः भ्रमन्ति। (तडागे / तडागाः)
  2. _______________ भ्रमराः गुञ्जन्ति (पुष्पं / पुष्पेषु)
  3. नपः _______________ पतति। (अश्वः / अश्वात)
  4. _______________ विद्यालयं गच्छामि। (अहम् / आवाम्)

प्रश्न 8.
मञ्जूषातः उचितं अव्ययपदं चित्वा रिक्तस्थानं पूरयत- (1 × 5 = 5)

अद्य, अधुना, प्रातः, कदा, सर्वदा

  1. _______________ भ्रमणं स्वास्थ्याय भवति।
  2. _______________ सत्यं वद।
  3. त्वं _______________ मातुलगृहं गमिष्यसि?
  4. _______________ विज्ञानस्य युगः अस्ति।
  5. _______________ रविवासरः अस्ति।

प्रश्न 9.
(अ) मञ्जूषातः कर्तृपदं चित्वा रिक्तस्थानानि पूरयत- (2)

अहम्, युवाम्, त्वम्, तौ

  1. _______________ लेखम् लिखसि।
  2. ______________ कथां पठसि।
  3. _______________ फलानि खादामि।
  4. _______________ बालौ स्तः।

(ब) मञ्जूषातः क्रियापदं चित्वा रिक्तस्थानानि पूरयत- (2)

गमिष्यावः, पिबतः, स्थ, पाठयति

  1. महिले शीतपेयं _______________।
  2. आवाम् पुस्तकालयं _______________।
  3. आचार्यः शिष्यान् _______________।
  4. यूयम् उत्तमाः बालकाः _______________।

प्रश्न 10.
कोष्ठकात् उचितं पदं चित्वा वाक्यानि पूरयत- (4)

  1. सुदामा _______________ मित्रम् आसीत्। (श्रीकृष्णः / श्रीकृष्णस्य)
  2. वयम् _______________ खादामः। (फलस्य / फलानि)
  3. _______________ आभूषण विद्या। (नरम् / नरस्य)
  4. बालकाः _______________ क्रीडन्ति। (क्रीडाक्षेत्र / क्रीडाक्षेत्रे)

खण्ड: – घ
पठित-अवबोधनम्

प्रश्न 11.
अधोलिखितं गद्यांशं पठित्वा प्रश्नान् उत्तरत- (5)

उत्सवप्रियः भारतेदशः। अत्र कुत्रचित् शस्योत्सवः भवति, कुत्रचित् पशूत्सवः भवति, कुत्रचित् धार्मिकोत्सवः भवति कुत्रचित् च यानोत्सवः। एतेषु एव अस्ति अन्यतमः पुष्पोत्सवः इति। अयं ‘फूलवालों की सैर’ इति नाम्ना प्रसिद्धः अस्ति।

देहल्याः मेहरौलीक्षेत्रे ऑक्टोबर्मासे अस्य आयोजनं भवति। अस्मिन् अवसरे तत्र बहुविधानि पुष्पाणि दृश्यन्ते। परं प्रमुखम् आकर्षणं तु अस्ति पुष्पनिर्मितानि व्यजनानि।

प्रश्ना:
I. एकपदेन उत्तरत- (1 × 2 = 2)
(i) पुष्पोत्सवः केन नाम्ना प्रसिद्धः अस्ति?
(ii) पुष्पोत्सवस्य आयोजनं कुत्र भवति?

II. पूर्णवाक्येन उत्तरत- (1 × 2 = 2)
(i) पुष्पोत्सवस्य आयोजनं कदा भवति?
(ii) पुष्पोत्सवस्य प्रमुख आकर्षणं कानि सन्ति?

III. निर्देशानुसारम् उत्तरत- (½ × 2 = 1)
(i) ‘एतेषु’ इत्यत्र का विभक्तिः?
(क) प्रथमा
(ख) सप्तमी
(ग) षष्ठी
(घ) पञ्चमी

(ii) ‘भवति’ इत्यत्र कः पुरुषः?
(क) प्रथमपुरुषः
(ख) मध्यम्पुरुषः
(ग) उत्तमपुरुषः

प्रश्न 12.
अधोलिखितं पद्यांशं पठित्वा प्रश्नानाम् उत्तराणि लिखत- (5)

1. उन्नतवृक्षं तुङ्गं भवनं
क्रान्त्वाकाशं खलु याम।
कृत्वा हिमवन्तं सोपानं
चन्दिरलोकं प्रविशाम॥

I. एकपदेन उत्तरत- (1 × 2 = 2)
(i) तुङ्ग किम् अस्ति?
(ii) वयं कुत्र प्रविशाम?

II. पूर्णवाक्येन उत्तरत- (1 × 2 = 2)
वयं किं क्रान्त्वा आकाशं याम?

III. यथानिर्देशम् उत्तरत- (½ × 2 = 1)
(i) ‘गगनं’ इति पदस्य समानार्थकम् पदं किम्?
(क) आकाशं
(ख) वृक्षं
(ग) भवनम्
(घ) सोपानं

(ii) ‘कृत्वा’ इत्यत्र कः प्रत्ययः?
(क) वन्त
(ख) वत्
(ग) क्त्वा
(घ) मन्त

प्रश्न 13.
मञ्जूषातः पर्यायवाचिपदानि चित्वा लिखत- (½ × 4 = 2)

धरित्री, तृषा, अधिकम्, वस्त्राणि

  1. पिपासा = _______________
  2. वसनानि = _______________
  3. पृथ्वी = _______________
  4. विपुलम् = _______________

प्रश्न 14.
मञ्जूषातः विलोम पदानि चित्वा लिखत- (½ × 4 = 2)

प्रातः, कृष्ण, निर्धनम्, दूरे

  1. श्वेतः = _______________
  2. धनिकम् = _______________
  3. पार्वे = _______________
  4. सायम् = _______________

प्रश्न 15.
उपयुक्तकथनानाम् समक्षम् ‘आम्’ अनुपयुक्तकथनानाम् समक्षं ‘न’ इति लिखत- (6)

  1. वसन्तसमये पिककाकयोः भेदः भवति (_______________)
  2. नव बालकाः एव स्नातुम् अगच्छन्। (_______________)
  3. एकः बालकः नद्यां मग्नः। (_______________)
  4. शक्रावतारस्य तीर्थस्य निवासी धीवरः आसीत्। (_______________)
  5. धीवरः मत्स्यविक्रयेण कुटुम्बस्य भरणं करोति स्म। (_______________)
  6. धीवरेण राजकीयं अगुलीयकम् प्राप्तम्। (_______________)

प्रश्न 16.
घटनाक्रमानुसारं अधोलिखितानि वाक्यानि पुनः लेखनीयानि- (½ × 6 = 3)

  1. स्वामी अजीजाय अवकाशस्य पूर्णं धनं ददाति।
  2. अजीजः सरलः परिश्रमी च आसीत्।
  3. अजीजः पेटिकाम् आनयति।
  4. एकदा सः गृहम् गन्तुम् अवकाशं वाञ्छति।
  5. पीडितः स्वामी अत्युच्चैः चीत्करोति।
  6. मक्षिके स्वामिनं दशतः।

प्रश्न 17.
श्लोकाशान् योजयत- (4)

(क) (ख)
1. गच्छन् पिपीलकः याति वर्षन्तु विपुलं जलम्।
2. सूर्यस्तपतु मेघाः वा योजनानां शतान्यपि।
3. काकः कृष्णः पिकः कृष्णः वचने का दरिद्रता।
4. तस्मात् प्रियं हि वक्तव्यम् को भेदः पिककाकयोः

CBSE Class 6 Sanskrit Sample Paper Set 1

We have given detailed NCERT Solutions for Class 6 Sanskrit come in handy for quickly completing your homework.

CBSE Class 6 Sanskrit Sample Paper Set 1

निर्धारित समय : 3 घंटे
पूर्णांक : 80

खण्ड: – क
अपठित-अवबोधनम्

प्रश्न 1.
अधोलिखितं अपठितं गद्याशं पठित्वा प्रश्नान् उत्तरत- (10)

पुरा अयोध्यायाः नृपः दशरथः आसीत्। तस्य त्रयः पुत्राः आसन्। तेषु श्रीरामः ज्येष्ठः पुत्रः आसीत्। श्रीरामस्य पत्नी सीता आसीत्। पितुः दशरथस्य वचनस्य पालनाय श्रीरामः लक्ष्मणसीताभ्याम् सह वनं अगच्छत्। वने रावणः सीताम् अहरत्। श्रीरामः रावणं हत्वा सीतां मोचितवान्।

प्रश्नाः
I. एकपदेन उत्तरत- (1 × 4 = 4)
(क) सीता कस्य पत्नी आसीत्?
(ख) अयोध्यायाः नृपः कः आसीत्?
(ग) कः सीताम् अहरत्?
(घ) रावणं कः अमारयत्?

II. पूर्णवाक्येन उत्तरत- (2 × 2 = 4)
(i) श्रीरामः किमर्थम् वनं अगच्छत्?
(ii) दशरथस्य कति पुत्राः आसन्?

III. निर्देशानुसारम् उत्तर- (1 × 2 = 2)
(i) ‘राजा’ इति पदस्य पर्यायः कः?
(क) पितुः
(ख) नृपः
(ग) ज्येष्ठः
(घ) त्रयः

(ii) ‘अगच्छत्’ इत्यत्र कः लकारः?
(क) लट्
(ख) लोट
(ग) लङ्
(घ) लृट्

खण्ड: – ख
रचनात्मक-कार्यम्

प्रश्न 2.
मञ्जूषातः उचितपदानि चित्वा चित्राणाम् नामानि संस्कृतेन लिखत- (1 × 4 = 4)

मञ्जूषा – कूपः, कृषकः, पितरौ, विद्युतव्यजनम्, शीतकम्, कमलम्

CBSE Class 6 Sanskrit Sample Paper Set 1 Q2

प्रश्न 3.
अधोलिखितशब्दान् आधृत्य सार्थकवाक्यानि रचयत- (4)

बालिकाः गायति
मयूराः गर्जतः
गायक: पठन्ति
सिंहौ नृत्यन्ति
  1. ____________
  2. ____________
  3. ____________
  4. ____________

प्रश्न 4.
मञ्जूषातः समुचितपदानि चित्वा कथां पूरयत- (10)

पिपासितः, उपायम्, स्वल्पम्, पाषाणस्य, कार्याणि, उपरि, सन्तुष्टः, पातुम्, इतस्ततः, कुत्रापि

एकदा एकः काकः ___(1)___ आसीत्। सः जलं पातुम् ___(2)___ अभ्रमत्। परं ___(3)___ जलं न प्राप्नोत्। अन्ते सः एकं घटम् अपश्यत्। घटे ___(4)___ जलम् आसीत्। अतः सः जलम् ___(5)___ असमर्थः अभवत्। सः एकम् ___(6)___ अचिन्तयत्। सः ___(7)___ खण्डानि घटे अक्षिपत्। एवं क्रमेण घटस्य जलम् ___(8)___ आगच्छत्। काकः जलं पीत्वा ___(9)___ अभवत्। परिश्रमेण एव ___(10)___ सिध्यन्ति न तु मनोरथैः।

खण्ड: – ग
अनुप्रयुक्त-व्याकरणं

प्रश्न 5.
वर्णसंयोजनं कृत्वा कोष्ठके पदं लिखत- (1 × 4 = 4)

यथा- ल् + इ + ख् + अ + त् + इ = लिखति

  1. ख + अ + न् + इ + त् + र् + अ + म् = _____________
  2. प् + उ + र् + आ + ण् + आ + न् + इ = _____________
  3. प् + अ + र् + ण् + अ + म् = _____________
  4. प् + उ + स् + त् + अ + क् + अ + म् = _____________

प्रश्न 6.
पदानां वर्णविच्छेदं कुरुत- (1 × 4 = 4)

यथा- शुनकः = श् + उ + न् + अ + क् + अः

  1. वातायानम् = _____________
  2. विमानम् = _____________
  3. पुस्तकम् = _____________
  4. लेखनी = _____________

प्रश्न 7.
कोष्ठकात् उचितं पदं चित्वा वाक्यं पूरयत- (1 × 4 = 4)

  1. _____________ समयं सूचयति। (घटिका / घटिके)
  2. _____________ पतन्ति। (पत्रे / पत्राणि)
  3. _____________ कूर्दति। (वानराः / वानरः)
  4. _____________ पठन्ति। (बालकाः / बालक:)

प्रश्न 8.
निर्देशानुसारं वाक्यानि रचयत- (1 × 4 = 4)

यथा- नारिकेलं पतति। (बहुवचने) = नारिकेलानि पतन्ति।

  1. पुष्पम्-विकसति। (द्विवचने) = _____________
  2. खगाः उत्पतन्ति। (एकवचने) = _____________
  3. अश्वः चलति। (द्विवचने) = _____________
  4. पेटिका अस्ति। (बहुवचने) = _____________

प्रश्न 9.
सः, तौ, ते इत्येतेभ्यः उचितं सर्वनामपदं चित्वा रिक्तस्थानानि पूरयत- (1 × 4 = 4)

  1. छात्रौ पठतः = _____________
  2. गजाः चलन्ति। = _____________
  3. बालकाः क्रीडन्ति। = _____________
  4. गायकः गायति। = _____________

प्रश्न 10.
मञ्जूषातः कर्तृपदं चित्वा रिक्तस्थानानि पूरयत- (1 × 5 = 5)

पश्यन्ति, चलति, लिखन्ति, विहरतः, नृत्यतः

  1. चटके ____________।
  2. मयरौ ____________।
  3. जनाः ____________।
  4. छात्राः ____________।
  5. पिपीलिका ____________।

खण्डः – घ
पठित-अवबोधनम्

प्रश्न 11.
अधोलिखितम् गद्यांशं पठित्वा प्रश्नान् उत्तरत- (5)

एकस्मिन् वने शृगालः बकः च निवसतः स्म। तयोः मित्रता आसीत्। एकदा प्रातः शृगालः बकम् अवदत्-“मित्र! श्वः त्वं मया सह भोजनं कुरु।” शृगालस्य निमन्त्रणेन बकः प्रसन्नः अभवत्।
अग्रिमे दिने सः भोजनाय शृगालस्य निवासम् अगच्छत्। कुटिलस्वभावः शृगालः स्थाल्यां बकाय क्षीरोदनम् अयच्छत्। बकम् अवदत् च-“मित्र! अस्मिन् पात्रे आवाम् अधुना सहैव खादावः।”

प्रश्ना:
I. एकपदेन उत्तरत- (1 × 2 = 2)
1. शृगालः बकः च कुत्र निवसतः स्म?
2. शृगालः बकाय किम् अयच्छत्?

II. पूर्णवाक्येन उत्तरत- (1 × 2 = 2)
1. कस्य निमन्त्रेण बकः प्रसन्नः अभवत्?
2. शृगालः कीदृशः आसीत्?

III. निर्देशानुसारं कुरुत- (½ × 2 = 1)
1. ‘निमन्त्रेण’ इत्यत्र का विभक्तिः?
2. ‘आवाम्’ इति पदे कः पुरुषः?

प्रश्न 12.
अधोलिखितं पद्यांशं पठित्वा प्रश्नान् उत्तरत- (5)

कुत आगच्छसि मातुलचन्द्रः!
कुत्र गमिष्यसि मातुलचन्द्रः!

अतिशयविस्तृतनीलाकाशः
नैव दृश्यते क्वचिदवकाशः।
कथं प्रयास्यसि मातुलचन्द्रः!
कुत आगच्छसि मातुलचन्द्रः!

कथमायासि न भो! मम गेहम्
मातुल! किरसि कथं न स्नेहम्।

I. एकपदेन उत्तरत- (1 × 2 = 2)
1. अतिशयविस्तृतः कः?
2. मातुलः कः?

II. पूर्णवाक्येन उत्तरत- (2)
1. क: नैव दृश्यते?

III. यथानिर्देशानुसारं कुरुत- (½ × 2 = 1)
1. ‘गमिष्यसि’ इति पदे कः लकार:?
(क) लट
(ख) लृट्
(ग) लङ्
(घ) लोट

2. ‘गृहम्’ इति पदस्य पर्यायपदं किम्?
(क) गेहम्
(ख) स्नेह
(ग) मम
(घ) कथं

प्रश्न 13.
मञ्जूषातः पर्यायवाचिपदानि चित्वा लिखत- (2)

तरू:, पयः, सततम्, खगाः

  1. निरंतरम्
  2. विहगाः
  3. वृक्षः
  4. जलम्

प्रश्न 14.
मञ्जूषातः विलोमपदानि चित्वा लिखत- (2)

दीर्घतमः, आकाशः, मित्रता, तमः

  1. पातालः
  2. शत्रुता
  3. लघुतमः
  4. प्रकाशः

प्रश्न 15.
उपकथनानाम् समक्षम् ‘आम्’ अनुपयुक्तकथनानाम् समक्षं ‘न’ इति लिखत- (6)

  1. कोच्चितटः नारिकेलफलेभ्यः ज्ञायते। ____________
  2. भारतस्य पूर्वदिशायाम् हिन्दमहासागरः अस्ति। ____________
  3. चेन्नईनगरस्य मेरीनातटः देशस्य सागरतटेषु दीर्घतमः। ____________
  4. वैनतेयः पशुः अस्ति। ____________
  5. अस्माभिः प्रियं न वक्तव्यम्। ____________
  6. पिकः श्वेतः भवति। ____________

प्रश्न 16.
घटनाक्रमानुसारं अधोलिखितानि वाक्यानि पुनः लेखनीयानि- (½ × 6 = 3)

  1. एकदाः सः गृहम् गन्तुम् अवकाशं वाञ्छति।
  2. एषः अवकाशस्य अपि वेतनं ग्रहीष्यति।
  3. परम् एतदर्थं त्वं वस्तुद्वयम् आनय।
  4. स्वामी चतुरः आसीत्।
  5. अजीजः स्वामिनः एव सेवायां लीनः आसीत्।
  6. अहम् तुभ्यम् अवकाशस्य वेतनस्य च सर्वं धनं दास्यामि।

प्रश्न 17.
श्लोकांशान् मेलयत- (4)

(क) (ख)
1. गृहं जीर्णं न वर्षासु तौ तु क्षेत्राणि कर्षतः।
2. हलेन च कुदालेन या शुष्का कण्टकावृता।
3. पादयोन पदत्राणे वृष्टिं वारयितुम् क्षमम्।
4. धरित्री सरसा जाता शरीरे वसनानि नो

CBSE Class 7 Sanskrit Sample Paper Set 5

We have given detailed NCERT Solutions for Class 7 Sanskrit come in handy for quickly completing your homework.

CBSE Class 7 Sanskrit Sample Paper Set 5

निर्धारित समय : 3 घंटे
अधिकतम अंक : 80

खण्ड: – क
अपठित-अवबोधनम्

प्रश्न 1.
अधोलिखितं अनुच्छेदं पठित्वा प्रदत्तप्रश्नानाम् उत्तराणि लिखत- (10)

पुरा छात्राः गुरुकुलं गत्वा विद्यां प्राप्नुवन्ति स्म। गुरुकुले गुरुशिष्य-परम्परा प्रचलिता आसीत्। आचार्यः स्वशिष्येषु पुत्रवत् स्निह्यति स्म। आचार्य कदापि स्वज्ञानस्य व्यापारम् न अकरोत्। गुरुदक्षिणारुपेण आचार्यः केवलं इदम् एव वाच्छति स्म यत् तस्य शिष्यः स्वज्ञानस्य सुप्रयोगम् एव कुर्यात्। शिष्याः आचार्यस्य सम्मानम् कुर्वन्ति स्म। छात्राः आचार्येण कथिते मार्गे एव चलन्ति स्म। धन्या एतादृशी गुरुपरम्परा।

प्रश्नाः
I. एकपदेन उत्तरत- (1 × 2 = 2)
(i) छात्राः कुत्र गत्वा विद्यां प्राप्नुवन्ति स्म?
(ii) कः शिष्येषु पुत्रवत् स्निह्यति स्म?

II. पूर्णवाक्येन उत्तरत- (2 × 2 = 4)
(i) गुरुशिष्य परम्परा कुत्र प्रचलिता आसीत्?
(ii) गुरुदक्षिणारूपेण गुरुः किम् वाञ्छति स्म?

III. यथानिर्देशम् उत्तरत- (2 × 2 = 4)
(क) ‘चलन्ति स्म’ इति क्रियापदस्य कर्तृपदम् किम्?
(i) शिष्याः
(ii) आचार्णण
(iii) कथिते
(iv) मार्गे।

(ख) ‘अकरोत्’ इति पदे कः लकार:?
(i) लट्
(ii) लङ्
(iii) लृट्
(iv) लोट्।

खण्ड: – ख
रचनात्मक कार्यम्

प्रश्न 2.
मञ्जूषातः उचितपदानि चित्वा अनुच्छेदस्य रिक्तस्थानानि पूरयत- (10)

मञ्जूषातः – श्लोकसंख्या, श्रीरामचन्द्रस्य, विदुषां, लिखितं, अवश्यमेव, वाल्मीकिः, प्रियपुस्तकम्, नारायणावताररूपेण, किष्किन्धा, सप्तकाण्डानि।

आदिकाव्यं, रामायणं मम ______(1)______ अस्ति। मर्यादापुरुषोतमस्य ______(2)______ पावनं चरित्रं रामायणे प्राप्यते। अस्य आदिकाव्यस्य लेखक: महर्षिः ______(3)______ आसीत्। अतएव तस्य रचना ‘आदिकाव्य’ इति ______(4)______ मतम्। रामायणम् अनुष्टुप् छन्दसि ______(5)______। रामायणे ______(6)______ चतुर्विशति सहस्त्रम् वर्तते। रामायणे ‘बाल अयोध्या-अरण्य ______(7)______ सुन्दर लंका-उत्तर इति’ सप्त ______(8)______ विद्यन्ते। रामायणे श्रीरामचन्द्रस्य ______(9)______ वर्णनं उपलभ्यते। मानवैः जीवने रामायणं ______(10)______ पठनीयम्।

प्रश्न 3.
वाक्यानि रचयत- (1 × 5 = 5)

  1. मुनिः ____________
  2. नरः ____________
  3. महिला ____________
  4. प्रतिदिनम् ____________
  5. ईश्वरः ____________

प्रश्न 4.
चित्रम् दृष्ट्वा मञ्जूषायाम् प्रदत्तशब्दानाम् सहायतया पञ्चवाक्यानि पूरयत- (1 × 5 = 5)

प्रसन्नाः, शीतपेयम्, समुद्रतटस्य, नौका, नारिकेलवृक्षाः

CBSE Class 7 Sanskrit Sample Paper Set 5 Q4

  1. इदम् चित्रम् ____________ अस्ति।
  2. समुद्रतटे ____________ सन्ति।
  3. समुद्रजले ____________ चलति।
  4. समुद्रतटे एकः नरः ____________ पिबति।
  5. सर्वे जनाः ____________ दृश्यन्ते।

खण्ड: – ग
अनुप्रयुक्त-व्याकरणं

प्रश्न 5.
अधोलिखितवाक्येषु रेखाङ्कितपदानां समुचितं सन्धि सन्धि-विच्छेदं वा प्रदत्तविकल्पेभ्यः चित्वा लिखत- (1 × 4 = 4)

(क) वार्ता अति + इव महत्त्वपूर्णा आसीत्।
(i) अतिव
(ii) अतीव
(iii) अतीइव
(iv) अत्यीव

(ख) अद्य सहसैव वृष्टिः आगता।
(i) सहसा + एव
(ii) सह + सैव
(iii) सहस + एव
(iv) सहसा + इव।

(ग) छात्राः पुस्तकालये पुस्तकानि पठन्ति।
(i) पुस्तक + आलये
(ii) पुस्तक + लये
(iii) पुस्त + कालये
(iv) पुस्तक + अलये

(घ) इति + उक्त्वा अर्जुन : तूष्णीं अभवत्।
(i) इतीक्त्वा
(ii) इत्युक्त्वा
(iii) इत्युक्त्वा
(iv) इतिउकत्वा

प्रश्न 6.
उदाहरणानुसारं शब्दरूपेषु रिक्तस्थानानि पूरयत- (½ × 6 = 3)

CBSE Class 7 Sanskrit Sample Paper Set 5 Q6

प्रश्न 7.
(अ) अधोलिखितान समयवाचकान् अङ्कान् पदेषु लिखत- (½ × 2 = 1)

  1. 10 : 30 ______
  2. 9 : 00 ______

(ब) मञ्जूषातः अङ्कान् कृते पदानि चिनुत। (½ × 2 = 1)
एकादश, नवतिः, पञ्चदश, विंशतिः
20, 15, 90, 11

प्रश्न 8.
कोष्ठकात् उचितं पदं चित्वा रिक्तस्थानानि पूरयत- (1 × 3 = 3)

(i) राघवः ____________ फलानि खादति। (मधुरं, मधुराः, मधुराणि)
(ii) वयम् ____________ ग्रहान् गणयाम। (सर्वं, सर्वान्, सर्वा)
(iii) ____________ गगने वायुविहारं करवाम। (नीलं, नीले, नीलः)

प्रश्न 9.
उचितं विभक्तिपदं चित्वा वाक्यपूर्तिः कुरुत- (1 × 4 = 4)

1. नभ: ____________ शोभते
(क) चन्द्रः
(ख) चन्द्रेण
(ग) चन्द्रम्
(घ) चन्द्राय

2. ____________ नमः।
(क) शिवं
(ख) शिवाय
(ग) शिवः
(घ) शिवात्

(3) अध्यापकः ____________ क्रुध्यति।
(क) शिष्यं
(ख) शिष्याय
(ग) शिष्ये
(घ) शिष्यः

(4) ____________ परितः वृक्षाः सन्ति।
(क) ग्राम्
(ख) ग्रामः
(ग) ग्रामाः
(घ) ग्रामात्

प्रश्न 10.
उदाहरणानुसारं धातुरूपेषु रिक्तस्थानानि पूरयत- (1 × 5 = 5)

CBSE Class 7 Sanskrit Sample Paper Set 5 Q10

प्रश्न 11.
मञ्जूषातः समुचित पदानि चित्वा वाक्यपूर्तिः क्रियताम्। (1 × 4 = 4)

अलम्, बहिः, मा, उपरि

  1. मञ्चस्य ____________ गायिकाः तिष्ठन्ति।
  2. ____________ कोलाहलेन।
  3. असत्यं ____________ वद।
  4. नगरात् ____________ नदी वहति।

खण्ड: – घ
पठित-अवबोधनम्

प्रश्न 12.
अधोलिखितं गद्यांशं पठित्वा प्रश्नानाम् उत्तराणि लिखत- (4)

(क) अधुना निखिले संसारे कलहस्य अशान्तेः च वातावरणम् अस्ति। मानवाः परस्परम् न विश्वसन्ति। ते परस्य कष्टम् स्वकीयं कष्टम् न गणयन्ति। अपि च समर्थाः देशाः असमर्थान् देशान् प्रति उपेक्षाभावं प्रदर्शयन्ति, तेषाम् उपरि स्वकीयं प्रभुत्वं स्थापयन्ति। संसारे सर्वत्र विद्वेषस्य, शत्रुतायाः, हिंसायाः च भावना दृश्यते। देशानाम् विकासः अपि अवरुद्धः भवति।

प्रश्नाः
I. एकपदेन उत्तरत- (½ × 2 = 1)
(i) के परस्परं न विश्वसन्ति?
(ii) केषाम् विकासः अवरुद्धः भवति?

II. पूर्णवाक्येन उत्तरत- (1 × 2 = 2)
(i) संसारे सर्वत्र कीदृशी भावना दृश्यते?
(ii) संसारे कस्य वातावरणम् अस्ति?

III. निर्देशानुसारं उत्तरत- (½ × 2 = 1)
(क) ‘सम्पूर्णे’ इत्यर्थे किम् पदम् प्रयुक्तम्?
(i) परस्य
(ii) निखिले
(iii) अशान्ते:
(iv) दृश्यते

(ख) ‘प्रदर्शयन्ति’ इति क्रियापदस्य कर्तृपदं किम्?
(i) समर्थाः
(ii) देशाः
(iii) देशान्
(iv) उपेक्षाभावं

अधोलिखितं पद्यांशं पठित्वा प्रश्नानाम् उत्तराणि लिखत- (4)

(ख) सर्वदा व्यवहारे स्यात् औदार्यं सत्यता तथा।
ऋजुता मृदुता चापि कौटिल्यं न कदाचन।।

प्रश्ना:
I. एकपदेन उत्तरत- (½ × 2 = 1)
(i) व्यवहारे कदाचन किम् न स्यात्?
(ii) सत्यता कदा व्यवहारे स्यात्?

II. पूर्णवाक्येन उत्तरत- (1 × 2 = 2)
(i) व्यवहारे सर्वदा का का स्यात्?

III. यथानिर्देशं कुरुत- (½ × 2 = 1)
(क) ‘सरलता’ इति पदस्य समानार्थकपदं किम्?
(i) मृदुता
(ii) ऋजुता
(iii) सत्यता
(iv) सर्वदा

(ख) ‘सर्वदा व्यवहारे स्यात् औदार्यं’ इत्यत्र क्रियापदं किम्?
(i) स्यात्
(ii) सर्वदा
(iii) व्यवहारे
(iv) औदार्य

प्रश्न 13.
श्लोकांशान् मेलयत- (4)

(क) (ख)
1. सत्येन धार्यते पृथ्वी विद्यायाः संग्रहेषु च।
2. आहारे व्यवहारे च सर्व सत्येन प्रतिष्ठितम्।
3. सत्येन वाति वायुश्च सत्येन तपते रविः।
4. धनधान्यप्रयोगेषु त्यक्तलज्जः सुखी भवेत्।

प्रश्न 14.
विलोमपदानि मेलयत- (½ × 6 = 3)

(क) (ख)
1 उन्नतः विर्कीय
2. गगने अवनतः
3. चित्वा पृथिव्याम्
4. दु:खी शोकः
5. हर्षः असुन्दरः
6 सुन्दरः सुखी

प्रश्न 15.
मञ्जूषातः समानार्थकपदानि चित्वा लिखत- (½ × 6 = 3)

पृथ्वी, अपायः, दृष्ट्वा, कूर्मः, भृत्यः, अधुना

  1. वसुन्धरा ____________
  2. कच्छपः ____________
  3. हानिः ____________
  4. अधुना ____________
  5. वीक्ष्य ____________
  6. सेवकः ____________

प्रश्न 16.
स्थूलपदानां स्थाने समुचितं प्रश्नवाचकं प्रदत्तविकल्पेभ्यः चित्वा प्रश्ननिर्माणं कुरुत- (1 × 4 = 4)

(क) पुष्पेषु चित्रपतङ्गाः डयन्ते।
(i) कासु
(ii) केषु
(iii) काः
(iv) कस्मिन्

(ख) सलिले नौका सेलति।
(i) का
(ii) कः
(iii) काः
(iv) के

(ग) पर्वतेभ्यः प्रस्तराणि आनीय सेतोः निर्माणं भवति।
(i) काभ्यः
(ii) केभ्यः
(iii) कः
(iv) काभिः

(घ) कर्मकराः सेतोः निर्माणम् कुर्वन्ति।
(i) कः
(ii) काः
(iii) के
(iv) किम्

प्रश्न 17.
आम् / न इति लिखत- (1 × 3 = 3)

  1. वने व्याघ्रः गर्जति। ____________
  2. सिंह: नीचैः क्रोशति। ____________
  3. अनारिका गृहम् आगत्य मातरम् अपृच्छत्। ____________

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

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

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

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

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

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

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

(b) HCN
Answer:
+ 2

(c) H2CO3
Answer:
+ 4

(d) CaC2.
Answer:
-1.

Question 4.
What is the state of hybridization of C in
(a) CO32-
Answer:
sp2

(b) CCl4
Answer:
sp3

(c) diamond
Answer:
sp3

(d) graphite?
Answer:
sp2

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

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

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

Question 8.
Write balanced equations to show hydrolysis reactions of CO32- and HCO3.
Answer:
CO32- + H2O ⇌ OH + HCO3
HCO3 + H2O ⇌ OH + H2CO3.

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

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

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

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

Question 13.
What happens when boric acid is heated?
Answer:
The p-Block Elements Class 11 Important Extra Questions Chemistry 1
Question 14.
What are boranes?
Answer:
Stable covalent hydrides of boron like B2H6, B4H10, B5H9 on analogy with alkanes are called boranes.

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

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

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

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

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

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

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

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

Question 23.
Write down the structure of Borazole.
Answer:
B3N3H6.
The p-Block Elements Class 11 Important Extra Questions Chemistry 2
Question 24.
Why thallium prefers to show an oxidation state of +1 rather than +3?
Answer:
Due to the inert-pair effect.

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

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

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

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

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

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

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

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

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

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

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

Question 36.
Pyrosilicates contain which anion?
Answer:
Si2O76-.

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

Question 38.
What is the basic building unit of silicates?
Answer:
SiO44- tetrahedra.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The p-Block Elements Important Extra Questions Short Answer Type

Question 1.
Although boric acid B(OH)3 contains three hydroxyl groups, yet it behaves as a monobasic acid. Explain.
Answer:
The p-Block Elements Class 11 Important Extra Questions Chemistry 3
Hydrated species
B(OH)3 is not a protonic acid.

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

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

SiCl4 + 2Cl → [SiCl6]2-

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

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

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

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

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

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

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

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

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

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

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

Question 7.
What happens when a borax solution is acidified? Write a balanced equation for the reaction.
Answer:
When acidified aqueous solution of borax (Na2B4O7) is heated, boric acid is formed.
The p-Block Elements Class 11 Important Extra Questions Chemistry 4
Question 8.
Mention important uses of boric acid.
Answer:
Boric acid is used:

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

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

  1. It is a colourless, odourless gas, slightly soluble in water.
  2. It is highly poisonous. It combines with haemoglobin in the red blood cells to form carboxy-haemoglobin which cannot absorb oxygen and thus the supply of oxygen to the body is reduced.
  3. It burns with a pale blue flame forming CO2.
    2CO + O2 → 2CO2.
  4. It is a reducing agent. It reduces some metal oxides into metals.
    Fe2O3 + 3CO → 2Fe + 3CO2.
  5. It combines With transition metals like iron, cobalt, nickel to form their carbonyl compounds.
    The p-Block Elements Class 11 Important Extra Questions Chemistry 5

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

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

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

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

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

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

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

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

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

Question 14.
Write balanced equations for the reaction of elemental boron with elemental chlorine, oxygen & nitrogen at a high temperature.
Answer:
The p-Block Elements Class 11 Important Extra Questions Chemistry 6

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

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

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

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

Question 17.
Write balanced equations:
(i) B2H6 + NH3 →?
Answer:
The p-Block Elements Class 11 Important Extra Questions Chemistry 7
(ii) NaH+ B2H6 →?
Answer:
The p-Block Elements Class 11 Important Extra Questions Chemistry 8
(iii) BF3 + LiH →?
Answer:
The p-Block Elements Class 11 Important Extra Questions Chemistry 9
Question 18.
Describe briefly three different methods of obtaining elemental boron.
Answer:

  1. By reduction of oxides: The reduction of oxides is carried out by electropositive metals like Mg.
    B2O3(s) + 3Mg(s) → 2B(s) + 3MgO(s)
  2. By reduction of halides: The reduction of the volatile boron halides is carried but by dihydrogen at high temperatures
    The p-Block Elements Class 11 Important Extra Questions Chemistry 10
  3. By the thermal decomposition of boron hydride
    The p-Block Elements Class 11 Important Extra Questions Chemistry 11

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

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

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

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

Question 23.
What are silicones? How are they manufactured?
Answer:
Silicones are rubber-like polymers having Si – O-Si linkage & general formula R2SiO. They are manufactured by hydrolysis of Chloro silicones,
The p-Block Elements Class 11 Important Extra Questions Chemistry 12
Where R is Me/Ph groups

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

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

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

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

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

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

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

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

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

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

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

The p-Block Elements Important Extra Questions Long Answer Type

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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