NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire

NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire.

BoardCBSE
TextbookNCERT
ClassClass 12
SubjectHistory
ChapterChapter 8
Chapter NamePeasants, Zamindars and the State Agrarian Society and the Mughal Empire
Number of Questions Solved9
CategoryNCERT Solutions

NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire

Question l.
What are the problems in using the Ain as a source for reconstructing agrarian history ? How do historians deal with this situation ?
Solution :
(a) Problems : The Ain was revised five times by the author so that it might become authentic. In all the quantitative sections, all numeric data were reproduced in words so as to minimise the chances of errors. But in spite this there are following problems in using Ain as a source for reconstructing agrarian history :

  1. Numerous errors in totalling have been detected. These are ascribed to simple slips of arithmetic or of transcription by Abul Fazl’s assistants.
  2. Another problem is about the somewhat skewed nature of the quantitative data. Data were not collected uniformly from all provinces. For example, detailed information about caste composition of the zamindars was collected from many subas. However, such information is not available for Bengal and Orissa.
  3. Prices and wages that have been documented in the Ain pertains to areas in or around the imperial capital of Agra. It has, therefore, limited relevance for the rest of the country.
  4. Ain was to present a vision of Akbar’s empire where social harmony was provided by a strong ruling class. There was no place for a successful revolt against the Mughal state. Thus, whatever we learn about the peasants from the Ain remains a view from the top.

(b)

  1. The historians deal with the above problems by supplementing the descriptions contained in sources emanating from regions away from the Mughal capital, e.g., revenue records from Gujarat, Maharashtra and Rajasthan from the seventeenth and eighteenth centuries.
  2. The extensive records of the East India Company too provide useful descriptions of agrarian relations in eastern India. These sources describe the instances of conflicts between peasants, zamindars and the state. Thus, they provide an insight into peasants’ perception of and their expectations of fairness from the state.

Question 2.
To what extent is it possible to characterise agricultural production in the sixteenth- seventeeth centuries as subsistence agriculture? Give reasons for your answer.
Solution :
(a) During Mughal, India was basically an agricultural country. In the Mughal state of India a different varieties of crops were produced. In Bengal two varieties of rices were produced. But the focus on the cultivation of basic crops does not mean that only subsistence agriculture existed in medieval India.
(b) The Mughal state encouraged peasants to cultivate varieties of crops which brought in revenue especially cotton and sugarcane.
(c) Cotton was mainly grown in vast area which was spread over central India and the deccan plateau, whereas in Bengal sugarcane was mainly produced.
(d) Many varieties of cash crops such as oilseeds including mustard and lentils.
(e) An average peasant of that time grew both commercial and subsistence crops.

Question 3.
Describe the role played by women in agricultural production.
Solution :
The role played by women in agricultural production was as mentioned below :

  • Men and women worked shoulder to shoulder in the fields.
  • Men tilled and ploughed, while women sowed, weeded, threshed and winnowed the harvest.
  • During the medieval period, with the growth of nucleated villages and expansion in individuated peasant farming, the basis of production was the labour and resources of the entire household.
  • Inspite of above, there were biases related to women’s biological functions. For example, menstruating women were not allowed to touch the plough or the potter’s wheel in western India, or enter the groves where betel-leaves (paan) were grown in Bengal.

Question 4.
Discuss, with examples, the significance of monetary transactions during the period under consideration.
Solution :
The significance of monetary transactions during the period (sixteenth and seventeenth centimes) was substantial because the Mughal Empire was among the greatest empires that had managed to consolidate its power and resources. There was stability in the Ming (China), Safavid (Iran) and Ottoman (Turkey) empires. This stability led to create vibrant networks of overland trade from China to the Mediterranean Sea. The discovery of New World resulted in massive expansion of trade of India with Europe.

With the expansion of trade the importance of monetary transactions increased. The expansion of trade brought huge amount of bullion and silver into India where there was no natural resource of silver. This led to remarkable stability in the availability of metal currency, particularly in silver rupya in India. This facilitated an unprecedented expansion of minting of coins and the circulation of money in the economy. At the same time, there was no exchange of goods or barter system during this period. The payments were made in gold or silver coins. According to Giovanni Careri, all the gold and silver which circulates throughout the world, ultimately, comes into India due to its overseas trade. Thus a large amounts of cash transactions took place in the sixteenth and seventeenth centuries.

Question 5.
Examine the evidence that suggests that land revenue was important for the Mughal fiscal system.

Solution :
(i) Agriculture was the mainstay of the economy. Land Revenue collected was used to pay salaries and to meet different kinds of administrative expenses. So it was considered important to establish an administrative apparatus to ensure control over agricultural production.

(ii) Thus, before fixing land revenue, Mughal state first acquired specific information about the extent of agricultural lands and their produce.

(iii) Land revenue collection arrangements was consisted of two stages of assessment. These were Jama and hasil. Cultivators were given the choice to pay land revenue either in cash or kind. The state preferred to collect land revenue as cash. Attempts were made to maximize profits from the land revenue collection.

(iv) Both cultivated and cultivable lands were measured in each province to fix land revenue. According to a decree of Akbar, it was the responsibility of malguzar to make cultivator pay land revenue in kind and it was also kept open. Thus, it is clear from the evidence that the monetary transactions were very important. To continue this policy efforts by subsequent emperors like Aurangzeb continued to measure land for collection of land revenue.

Question 6.
To what extent do you think caste was a factor in influencing social and economic relations in agrarian society ?
Solution :
Agricultural production involved the intensive participation and initiative of the peasantry. There were different social groups, on the basis of caste and other factors, that were involved in agricultural expansion. This affected their social and economic relations in the agrarian society in the following ways :

  1. Deep inequalities on the basis of caste and other caste like distinctions made the cultivators a highly heterogeneous group.
  2. Among those who tilled the land, there was a sizeable number who worked as menials or agricultural labourers (majur).
  3. There was abundance of cultivable land but inspite of this certain caste groups were assigned menial tasks and were relegated to poverty. Such groups comprised a large section of the village population. They had the least resources and were constrained by their position in the caste hierarchy like the Dalits of modem India.
  4. In Muslim communites too menials like halalkhoran (scavengers) were housed outside the boundaries of the village. The mallahzadas (sons of boatmen) in Bihar were comparable to slaves.

Thus, there was a direct correlation between caste, poverty and social status at the lower level of society. It was, however, not so at the intermediate levels. For example, in Marwar, Rajputs were considered peasants like Jats who were accorded a lower status in the caste hierarchy. The Gauravas, who cultivated land around Vrindavan sought Rajput status in the seventeenth century. Ahirs, Gujars and Malis rose in the hierarchy due to the profitability of cattle rearing and horticulture.

Question 7.
How were the lives of forest dwellers transformed in the sixteenth and seventeenth centuries ?
Solution :
The forest dwellers were people who earned their livelihood by gathering of forest produce, hunting and shifting agriculture. These activities changed according to seasons. For example, the Bhils collected forest produce in the spring. They did fishing in the summer, cultivated during the monsoon months. They did hunting in the winter and autumn. Such a division of activities presumed and perpetuated mobility which was a distinctive feature of the forest dwellers. However, their lives were transformed in the sixteenth and seventeenth centuries in the following ways :

  1. In the sixteenth and seventeenth centuries, the state required elephants for the army. So, the peshkash – a form of tribute collected by the Mughal state – in the form of elephants too.
  2. In the Mughal political ideology, the hunt symbolised the overwhelming concern of the state to ensure justice to all the subjects, rich and poor. Thus, during hunting expeditions, the emperor personally attended to the grievances of the people. Such hunting expeditions affected the lives of the forest dwellers.
  3. Forests were cleaned for agricultural settlements. The spread of commercial agriculture impinged on the lives of forest dwellers, who lived on forest products like honey, beeswax and gum lac. Gum lac was exported from India. Elephants were captured and sold. There was [ exchange of commodities through barter too. All this changed due to commercial agriculture and agricultural settelments.
  4. Social factors too transformed the lives of forest dwellers. For example, chieftains of many tribes had become zamindars, some even became kings. They built up their armies and demanded that their fraternity to provide military service. In Assam, the Ahom kings had their paiks. They were people who obliged to render military service in exchange for land. Not only this the capture of wild elephants was declared a royal monopoly by the Ahom kings.
  5. With the establishment of tribal kingdoms in the north-east, war became a common feature.
  6. The cultural influences as that of sufi saints encouraged the forest-dwellers particularly agricultural communities to accept Islam.

Question 8.
Examine the role played by zamindars in Mughal India.
Solution :
Zamindars played a significant role in Mughal India as mentioned below :

  1. The zamindars had landed properties and enjoyed certain social and economic privileges due to their superior status in the society and due performance of certain services (khidmat) for the state.
  2. They had milkiyat lands which were cultivated for the private use of zamindars with the help of hired or servile labour. The zamindars could sell, bequeath or mortgage these lands at will.
  3. The zamindars collected revenue on behalf of the state. They were compensated for this financially.
  4. They had military resources such as armed contingent and fortresses (qilachas).
  5. In the social hierarchy, the zamindars constituted its very narrow apex.
  6. Contemporary documents give an impression that conquest may have been source of the origin of some zamindaris. A powerful military chieftain often dispossessed weaker people and expanded his zamindari.
  7. Zamindars spearheaded the colonisation of agricultural land. They helped in settling cultivators by providing them with the means of cultivation, including cash loans. The buying and selling of zamindaris accelerated the process of monetisation in the countryside. Zamindars sold the produce from their milkiyat lands and established markets (haats) where peasants came to sell their produce too.
  8. Although the zamindars were considered as an exploitative class but their relationship with the peasants were based on reciprocity, paternalism and patronage because the bhakti saints did not portray them as exploiters or oppressors of peasantry. Not only this, in a large ‘ number of agrarian uprisings in north India in the seventeenth century, zamindars often received the support of the peasantry in their struggle against the state.

Question 9.
Discuss the ways in which panchayats and village headmen regulated rural , society.
Solution :
Panchayats and village headmen regulated the rural society in the following ways :

  1. The village panchayat was an assembly of elders. Its decisions were binding on the members.
  2. The panchayat was headed by a headman known as muqaddam or mandal. The headman supervised the preparation of accounts, assisted by the accountant or patwari of the panchayat.
  3. The panchayat ensured that the caste boundaries among the various communities inhabitating the village were upheld. In eastern India all marriages were held in the presence of the mandal or headman. Thus, the headman was to oversee the conduct of the members of the village community “chiefly to prevent any offence against their caste”.
  4. Panchayats had the authority to levy fines and inflict serious punishments such as expulsion from the community. Such punishment was given as a deterrant to violation of caste norms.
  5. There were Jati panchayats of each caste. In Rajasthan, Jati panchayats arbitrated civil disputes between members of different castes. They decided the disputes related to claims on lands and marriages. Generally, the state respected the decisions of Jati panchayats.
  6. Sometimes petitions were presented to the panchayat complaining about extortionate taxation or the demand for unpaid labour (begar) imposed by the “superior” castes or officials of the state. These were submitted by the lower classes because they regarded the village panchayat as the court of appeal that would ensure that the state carried out its moral obligations and guaranteed justice. In such cases, the panchayats often suggested compromise and reconciliation. In case of failure of compromise, the peasants generally deserted the village because there was abundance of uncultivated land available in the villages.

We hope the NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 12 Atoms

NCERT Solutions for Class 12 Physics Chapter 12 Atoms

NCERT Solutions for Class 12 Physics Chapter 12 Atoms are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 12 Atoms.

BoardCBSE
TextbookNCERT
ClassClass 12
SubjectPhysics
ChapterChapter 12
Chapter NameAtoms
Number of Questions Solved17
CategoryNCERT Solutions

Question 1.
Choose the correct alternative from clues given at end of each statement:

  1. The size of the atom in Thomson’s model is the atomic size in Rutherford’s model.(much greater than/no different from/much less than.)
  2. In the ground state of…………… electrons are in stable equilibrium, while in ……………..  electrons always experience a net force. (Thomson’s model/Rutherford’s model.)
  3. A classical atom based on…………. is doomed to collapse. (Thomson’s model/Rutherford’s model.)
  4. An atom has a nearly continuous mass distribution in a………….. but has a highly non­uniform mass distribution in (Thomson’s model/Rutherford’s model.)
  5. The positively charged part of the atom possesses most of the mass in…………. (Rutherford’s model/both the models.)

Answer:

  1. no different from
  2. Thomson’s model; Rutherford’s model.
  3. Rutherford’s model.
  4. Thomson’s model, Rutherford’s model.
  5. both the models.

Question 2.
Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at a temperature 14 K). What results do you expect?
Answer:
The nucleus of a hydrogen atom is a proton. The mass of it is 1.67 x 10-27 kg, whereas the mass of an incident α-particle is 6.64 x 10-27 kg. Because the scattering particle is more massive than the target nuclei (proton). the α-particle won’t bounce back in even in a head-on collision. It is similar to a football colliding with a tennis ball at rest. Thus, there would be no large-angle scattering.

Question 3.
What is the shortest wavelength present in the Paschen series of spectral lines?
Answer:
The wavelength of the spectral lines forming the Paschen series is given by

NCERT Solutions for Class 12 Physics Chapter 12 Atoms 1
Question 4.

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits from the upper level to the lower level?
Answer:
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 2
Question 5.
The ground state energy of the hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state?
Answer:
K.E. = -E (Total energy)
= -(-13.6) = 13.6 eV
P.E. = 2 X E = 2 X (-13.6)
= -27.2 eV

Question 6.
A hydrogen atom initially in the ground level absorbs a photon which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.
Answer:
We know, the energy of an electron in the nth orbit of a hydrogen atom is given by
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 3
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 4

Question 7.
(a) Using Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels.
(b) Calculate the orbital period in each of these levels.
Answer:
(a) Speed of an electron in nth orbit of a hydrogen atom is given by
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 5

NCERT Solutions for Class 12 Physics Chapter 12 Atoms 6

Question 8.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 x 10-11 m. What are the radii of the n = 2 and n = 3 orbits?
Answer:
We know, the radius of the nth orbit of a hydrogen atom is given by rn = r0n2, where r0 = 5.3 x 10-u m is the radius of the innermost orbit of the hydrogen atom.
When n = 2, r2 = 5.3 x 10-u x 4
= 2.12 x 10-10m
When n = 3,
= 5.3 x 10-11 x 9
= 4.77 x 10-10m.

Question 9.
A 12.75 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Answer:

NCERT Solutions for Class 12 Physics Chapter 12 Atoms 7
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 8

Question 10.
In accordance with Bohr’s model, And the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 x 1011 m with an orbital speed
3 x 1014 m s-1. (Mass of earth = 6.0 x 1024 kg.)
Answer:
According to Bohr’s postulate of quantization of angular momentum
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 9

Question 11.
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.
(a) Is the average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(b) Is the probability of backward scattering (i.e., scattering of a-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of a-particles scattered at moderate angles is proportional to What clue does this linear independence on t provide?
(d) In which model is it completely wrong to ignore multiple scattering for the calculation of the average angle of scattering of a-particles by a thin foil?
Answer:
(a) About the same
(b) Much less
(c) It suggests that the scattering is predominantly due to a single collision, because the chance of a single collision increases linearly with the number of target atoms, and hence linearly with thickness.
(d) In Thomson’s model, a single collision causes very little deflection. The observed average scattering angle can be explained only by considering multiple scattering. So it is wrong to ignore multiple scattering in the Thomson model. In Rutherford’s model, most of the scattering comes through a single collision and multiple scattering effects can be ignored as a first approximation.

Question 12.
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the Coulomb attraction by a factor of about 10-40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Answer:
If electron and proton were bound by gravitational attraction, then
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 10
It is astonishing this value of r is much greater than the size of the universe.

Question 13.
Obtain an expression for the frequency of radiation emitted when a hydrogen atom
de-excites from the level it to level (n – 1). For large it, shows that this frequency equals the classical frequency of revolution of the electron in the orbit.
Answer:
The energy of an electron in the nth orbit of a hydrogen atom is given by
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 11
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 12
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 13

i.e. frequencies are equal. This is called Bohr’s correspondence principle.

Question 14.
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, a thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learned in the text. To stimulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10-10m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.
(b) You will find that the length obtained in («) many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in a non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr discard c and look for something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognizing that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, mg, and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer:
(a) Here, the dimensional formula of e is A1T1, the dimensional formula of me is M1, dimensional formula
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 14
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 15

Question 15.
The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?
Answer:
(a) K.E. = -E = – (-3.4 eV) = 3.4 eV
E. = 2E = 2 x (-3.4 eV)
= -6.8 eV
(b) Kinetic energy does not depend upon the choice of zero potential energy. Therefore, its value remains unchanged. However, the potential energy gets changed with the change in the zero levels of potential energy.

Question 16.
If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantization of orbits of planets around the sun?
Answer:
Applying Bohr’s quantization postulate,
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 15.1
i.e., n is very large. Since n is very large, the difference between the two successive energy or angular momentum levels is very small and the levels may be considered continuous.
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 16
Question 17.
Obtain the first Bohr’s radius and the ground state energy of a ‘muonic hydrogen atom’ (i.e. an atom in which a negatively charged muon (μ-1) of mass about 207 me orbits around a proton).
Answer:
Here the mass of the particle revolving around the proton is
NCERT Solutions for Class 12 Physics Chapter 12 Atoms 17

We hope the NCERT Solutions for Class 12 Physics Chapter 12 Atoms, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 12 Atoms, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era

NCERT Solutions for Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era.

BoardCBSE
TextbookNCERT
ClassClass 12
SubjectHistory
ChapterChapter 15
Chapter NameFraming the Constitution The Beginning of a New Era
Number of Questions Solved8
CategoryNCERT Solutions

NCERT Solutions for Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era

Question 1.
What were the ideals expressed in the Objectives Resolution ?
Solution :
The “Objectives Resolution” outlined the ideals of the Constitution of Independent India as mentioned below :

  • Independent sovereign republic : India will be an independent state without any foreign or internal control. Its head will be elected by the people.
  • Justice, equality and freedom : It will guarantee its citizens justice, equality and freedom.
  • Safeguards for minorities : It assured that “adequate safeguards shall be provided for minorities, backward and tribal areas, and Depressed and Other Backward Classes”.

Question 2.
How was the term minority defined by different groups?
Solution :
N.G. Ranga, a socialist who had been a leader of the peasant movement, urged that the term minorities be interpreted in economic terms. The real minorities were the poor and the downtrodden. Some considered that the real minorities were the masses of our country who were so depressed and oppressed that they were ot even able to take advantage of the ordinary civil rights. Singh spoke eloquently on the need to protect the tribes, and ensure conditions that could help them come up to the level of the general population.

Question 3.
What were the arguments in favour of greater power to the provinces ?
Solution :
The following arguments were given in favour of greater power to the provinces :

  1. K. Santhanam stated that by giving more powers to the Centre, we could not make it strong because in such a situation, the Centre would be overburdened with responsibilities. It would not be able to function effectively. On the other hand, if the Centre was relieved of some functions or powers it could function effectively and become stronger.
  2. He argued that weak status would cripple them. Their financial position would be weaker and without finances they would not be able to take up any project for development. They would have to depend on the central aid for education, sanitation and other welfare work for the people. He added a strong centre might lead the states to “revolt against the Centre” in the future.
  3. A member from Orissa warned that “the Centre is likely to break” since powers had been excessively centralised under the Constitution.

Question 4.
Why did Mahatma Gandhi think Hindustani should be the national language?
Solution :
In view of Mahatma Gandhi Hindustani was a language that the common people could easily understand. Hindustani was a blend of Hindi and Urdu. It was also popular among a large section of the people. Moreover, it was a composite language enriched by the interaction of diverse cultures. Words and terms from many different languages got incorporated into this language over the years. It made this language easily understandable by people from various regions.
As per Mahatma Gandhi Hindustani would be the ideal language of communication between the communities. It would help to unify Hindus and Muslims and the people from north and south. Language came to be associated with the politics of religious identities from the end of the 19th century. But Mahatma Gandhi retained his faith in the composite character of Hindustani.

Question 5.
What historical forces shaped the vision of the Constitution?
Solution :
Historical forces that shaped the vision of the Constitution were as given below :

  1. Historic efforts in the past : Nehru in his famous speech of 13 December 1946 referred to the American and French Revolutions. He thus linked the making of the Indian Constitution with the revolutionary moments in the past. But at the same time he emphasised not to copy the west but to learn from their experiments, achievements and failures.
  2. The will of the people : Nehru stated that the source of the Constituent Assembly was its strength i.e., the will of the people. So, members always kept in mind that passions that lay in the hearts of the masses of the Indian people and tried to fulfil them. Thus, Constituent Assembly was expected to represent the people.
  3. India, a large country with diversities : India is a large country with different religions, castes, communities, languages and groups. It was necessary to keep all united. So, the Constitution was prepared keeping in mind these diversities.
  4. Protection of minorities : There were minorities and depressed classes. It was necessary to protect their interests. Gandhiji had already started movement for upliftment of the Harijan. Thus, there were debates in the Assembly and provisions were incorporated for their protection and upliftment.
  5. Period of violence : There were riots and violence and communal frenzy. Under these circumstances it was necessary to have a strong Centre. There were arguments in favour of and against it. But ultimately more powers were given to the Centre.
  6. Problem of princely states : There were more than five hundred princely states. To accommodate them, it was absolutely necessary to have a federal system of government.

Question 6.
Discuss the different arguments made in favour of protection of the oppressed groups.
Solution :
The different arguments made in favour of protection of the oppressed groups were as mentioned below :

  1. It was argued that the problem of the “Untouchables” could not be resolved through protection and safeguards alone. Their disabilities were caused by the social norms and the moral values of caste society that had used their services and labour but kept them at a social distance.
  2. J. Nagappa from Madras pointed out that the suffering of the Depressed Classes was i due to their systematic marginalisation and not due to their numerical insignificance. They had
    no access to education, or share in the administration.
  3. K.J. Khanderkar from the Central Provinces argued that the Depressed Classes had been suppressed for thousands of years to such an extent that their bodies and minds were not able to march forward.

Question 7.
What connection did some of the members of the Constituent Assembly make between the political situation of the time and the need for a strong Centre ?
Solution :
The Constitution of India was framed between December 1946 and December 1949. It was a trouble-some time. There were riots and violence. There was the rising of the ratings of the Royal India Navy in Bombay and other cities in the spring of 1946. The violence culminated in the massacres that accompanied the transfer of populations when the Partition of India was announced. Some members of the Constituent Assembly made connection between the above political situation of the time and the need for a strong Centre as mentioned below :

  1. Referring to riots and violence in the country, many members had repeatedly stated that the powers of the Centre had to be greatly strengthened to enable it to stop the communal frenzy.
  2. Gopalaswami Ayyangar declared that “the Centre should be made as strong as possible”.
  3. Balakrishna Sharma from the United Provinces reasoned at length that only a strong Centre could plan for the well-being of the country, mobilise the available economic resources and establish a proper administration.

Question 8.
How did the Constituent Assembly seek to resolve the language controversy ?
Solution :
There were two main views about the language of the nation as mentioned below :

  1. A plea for Hindi : R.V. Dhulekar, from the United Provinces made a strong plea that Hindi be used as the language of making the constitution. He wanted Hindi to be declared a National Language.
  2. The fear of domination : Shrimati Durgabai from Madras explained her worries. She informed the house that there was strong opposition against Hindi in the South. She stated
    that the erosion of inclusive and composite character of Hindustani was bound to create anxieties and fears amongst different language groups.

In view of the above differences, some members appealed for a spirit of accommodation and asked the members not to push the cause of Hindi aggressively. Thus, the language controversy was solved in the following way :

  • Hindi in the Devanagari script would be the official language.
  • Transition to Hindi would be gradual.
  • For the first fifteen years, English would continue to be used for all official purposes.
  • Each province was allowed to choose one of the regional languages for official work within the province.

Thus, referring to Hindi as the official rather that of the national language it was hoped that it would be acceptable to all.

We hope the NCERT Solutions for Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements

NCERT Class 12 Chemistry Solutions for Chapter 8  d-and f-Block Elements provides solutions to the questions provided in the textbook. These solutions are provided by the subject experts and are helpful for the students appearing for boards or competitive exams. The students can refer to these and enhance their conceptual knowledge.

NCERT Solutions are provided in the CBSE, MP, UP, and Gujarat boards. The students appearing for these boards can practice through the NCERT Solutions to score well in the examination.

BoardCBSE
TextbookNCERT
ClassClass 12
SubjectChemistry
ChapterChapter 8
Chapter Named-and f-Block Elements
Number of Questions Solved48
CategoryNCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements

The chapter d and f block elements is very important from the examination perspective. It explains the elements in the groups 3-12. A detailed understanding of this chapter will help the students to differentiate between the characteristics of d and f block elements. It also explains a comparative account of lanthanoids and actinoids.

The NCERT Solutions for Class 12 Chemistry Chapter 8 provides in-depth details of d and f blocks elements. The students are advised to refer to these solutions for better results.

NCERT IN-TEXT QUESTIONS

Question 1.
Silver has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition metal ?
Answer:
Silver (Z = 47) belongs to group 11 of (7-block (Cu, Ag, Au) and its outer electronic configuration is 4d105s1. It shows + 1 oxidation state (4d10 configuration) in silver halides (e.g. AgCl). However, it can also exhibit + 2 oxidation state (4d9 configuration) in compounds like AgF2 and AgO. Due to the presence of half filled d-orbital, silver is a transition metal.

Question 2.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol-1. Why?
Answer:
In 3d series from Sc to Zn, all elements have one or more unpaired e-1 s except Zn which has no unpaired electron as its outer EC is 3d104s2. Hence, the intermetallic bonding is weakest in zinc. Therefore, enthalpy of atomisation is lowest.

Question 3.
Which of the 3d series of transitional metals exhibits largest oxidation states and why?
Answer:
Mn (Z = 25) with electronic configuration [Ar]3d54s2 shows maximum oxidation state (+ 7) in its compounds since it has the maximum number of unpaired five i.e., seven. It shows largest variable oxidation state from + 2 to + 7 ( + 2, + 3, + 4, + 5, + 6, + 7) in its compounds.

Question 4.
The E°(M2+/M) value for copper is positive (+ 0·34 V). What is possibly the reason for this ? (C.B.S.E. Outside Delhi 2012, Sample Paper 2012)
Answer:
E°(M2+/M) for any metal is based upon three factors which have been discussed in the text part.
M(s) + ∆aH → M(g) ; (∆aH = Enthalpy of atomisation)
M(g) + ∆fH → M2+(g) ; (∆fH = Ionisation enthalpy)
M2+(g) + aq → M2+(aq); (∆hydH = Hydration enthalpy)
Copper has very high enthalpy of atomisation (energy required) and low enthalpy of hydration (energy released). In nut shell, the ∆fH i.e. ionisation enthalpy needed is not compensated by the energy released. Therefore E°(Cu2+/Cu) is positive.

Question 5.
How would you account for the irregular variation in ionisation enthalpies (first and second) in first series of transition elements ?
Answer:
The ionisation enthalpies of the transition metals are higher than those of s-block elements and less than the elements of p-block. Thus, these are less electropositive than the elements of s-block and at the same time more electropositive than the elements belonging to p-block present in the same period. In a transition series, the ionisation enthalpies increase from left to right. However, the gaps in the values of the two successive elements are not regular.

Question 6.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer:
Oxygen and fluorine have small size and high electronegativity. Hence, they can oxidise the metal to the highest O.S.

Question 7.
Which is a stronger reducing agent Cr2+ or Fe2+ and why ? (SamplePaper 2011, C.B.S.E. Outside Delhi 2010, 2014)
Answer:
Cr2+ is a stronger reducing agent than Fe2+. This is quite evident from the E° values ;
E°Cr3+/Cr2+ = – 0-41 V and E°Fe3+/Fe2+ = 0-77 V.
Reason : d4 → d3 occurs when Cr2+ changes to Cr3+ ion while d6 → d5 takes place when Fe2+ gets converted to Fe3+ ion.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 1
Now, d4 → d3 transition is easier as compared to d6 → d5 transition because in the second case, an electron is removed from a paired orbital which is rather difficult. Therefore, Cr2+ is a stronger reducing agent than Fe2+.

Question 8.
Calculate the spin magnetic moment of M2+(aq) ion (Z = 27).
Answer:
Electronic configuration of element M(Z = 27) : [Ar] 3d74s2
Electronic configuration M2+ (aq) ion : 3d7 or NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 2
Magnetic moment of M2+ (aq) ion with n = 3 ; \(\mu =\sqrt { n\left( n+2 \right) } \)
\(=\sqrt { 3\left( 3+2 \right) } =\sqrt { 15 } =3\cdot 87BM.\)

Question 9.
Explain why Cu+ ion is not stable in an aqueous solution. (C.B.S.E. Delhi 2011)
Answer:
In aqueous solution Cu+ (aq) undergoes disproportionation to form Cu2+ (aq) ion and Cu.
2Cu+(aq) → Cu2+(aq) + Cu(s)
The higher stability of Cu2+ (aq) in an aqueous solution may be attributed to its greater negative ∆hydH than that of Cu+ (aq). It compensates for the second ionisation enthalpy of Cu involved in the formation of Cl2+ ion. Thus, Cu+ (aq) ion changes to Cu2+ (aq) ion which is more stable.

Question 10.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Answer:
This is due to poor shielding by 5f-electrons in the actinoids than that by 4f e-1s in lanthanoids.

NCERT EXERCISE

Question 1.
Write down the electronic configuration of :
(a) Cr3+
(b) Cu+
(c) Co2+
(d) Mn2+
(e) Pm3+
(f) Ce4+
(g) Lu2+
(h) Th4+
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 3

Question 2.
Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?
Answer:
A half-filled d orbital or a completely filled d orbital is more stable than any other state. Mn2+ already has a half-filled stable state hence would not undergo oxidation to form Mn+3. On the other hand, Fe2+ on oxidation to Fe3+ will have half-filled d orbitals which are more stable.

Question 2.
Why are Mn2+ compounds more stable than Fe2+ compounds towards oxidation to their +3 state?
Answer:
Electronic configuration of Mn2+ is 3d5 while that of Fe2+ is 3d6. This shows that the Fe2+ ion has an urge to change to Fe3+ ion by losing an electron whereas the Mn2+ ion has no such tendency. Thus, the + 2 oxidation state of Mn is more stable as compared to the + 2 oxidation state of Fe.

Question 3.
Explain briefly how + 2 oxidation state becomes more and more stable in the first half of the first-row transition elements with increasing atomic number.
Answer:
In all the elements listed, with the removal of valence 45 electrons (+2 oxidation state), the 3d-orbitals get gradually occupied. Since the number of empty d-orbitals decreases or the number of unpaired electrons in 3d orbitals increases, the stability of the cations (M2+) increases from Sc2+to Mn2+.

Question 4.
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with an example.
Answer:
In the transition series, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, the electronic configuration of Fe(Z = 26) is [Ar] 3d64s2. It shows various oxidation states but Fe (III) is most stable because it has the configuration [Ar]3d5.

Question 5.
What must be the stable oxidation state of the transition elements with the following electronic configuration in the ground states of their atoms: 3d3, 3d5, 3d8, 3d4?
Answer:
The maximum oxidation states of reasonable stability in the transition metals of 3d series correspond to the sum of s and d-electrons upto Mn. However, after Mn there is an abrupt decrease in oxidation states. In the light of this, most stable oxidation states of the elements are:
3d3 : 3d3s2 (+ 5);3 : 3d54s1 (+ 6) and 3d54s2 (+ 7)
3d8 : 3d84s2 (+ 2); 3d4 : 3d44s2 or 3d54s1 (+6)

Question 6.
Name the oxometal anions in the first transition series of transition metals in which the metal exhibits an oxidation state equal to its group number.
Answer:
Vanadate: VO3
chromate: CrO42-
permanganate: MnO4

Question 7.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction? (C.B.S.E. Delhi 2013)
Answer:
One common property associated with the elements in the periodic table is the variation in their atomic and ionic radii down the group and along a period. In general, these increase down the group due to the increase in the number of shells and decrease along a period considerably because of the increase in the magnitude of the effective nuclear charge.
Consequences of Lanthanoid Contraction
(a) Separation of Lanthanoids: Separation of lanthanoids is possible only due to lanthanoid contraction. All the lanthanoids have quite similar properties and due to this reason, they are difficult to separate. However, because of lanthanoid contraction, their properties (such as ionic size, ability to form complexes etc.,) vary slightly.

(b) Variation in basic strength of hydroxides: The basic strength of oxides and hydroxides decreases from La(OH)3 to Lu(OH)2. Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. The acidic strength which involves the cleavage of the O—H bond follows the reverse trend i.e. it increases along the series.

(c) Similarly in the atomic sizes of the elements of the second and third transition series present in the same group: We know that the atomic sizes of the elements generally increase appreciably down a group. Similar trend is also expected in the elements present in the different groups of d-block.

(d) Variation in standard reduction potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.
M3+ (aq) + 3e → M (aq)

(e) Variation in physical properties like melting point, boiling point, hardness etc: Various physical properties like m.pt., b.pt., hardness etc., increase with the increase in atomic number. This is because the attraction forces between the atoms increase as the size decreases.

Question 8.
What are the characteristics of transition elements and why are they called transition elements? Which of the d- block elements may not be regarded as the transition elements?
Answer:
In the transition elements, d-orbitais are successively filled. The general electronic configuration of transition elements is (n – 1) d1-10 ns1-2. There are three transition series. The first transition series involves the filling of 3d-orbitais. It starts from scandium (Z = 21) and goes upto zinc (Z = 30).

The second transition series involves the filling of 4 d-orbitais and includes 10 elements from yttrium (Z = 39) to cadmium (Z = 48). The third transition series invokes the filling of 5d-orbitals. The first element of this series lanthanum (Z = 57). It is followed by fourteen elements called lanthanides which involve the filling of 4f-orbitais. The net nine elements from hafnium (Z = 72) to mercury (Z = 80) belong to the third transition series.

There is an incomplete fourth transition series. it involves the filling of 6d- subshell starting from actinium (Z = 89) followed by elements with atomic number 104 onwards.

Question 9.
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Answer:
Transition elements contain partially filled d-orbitals whereas non-transition elements have no d-orbitals or have completely filled d -orbitals.

Question 10.
What are the different oxidation states exhibited by lanthanoids?
Answer:
+3 is the common oxidation state of the lanthanoids.
In addition to +3, oxidation states +2 and +4 are also exhibited by some of the lanthanoids.

Question 11.
Explain giving the reason:
(a) Transition metals and many of their compounds show paramagnetic behaviour. (H.P. Board 2014)
(b) The enthalpies of atomisation of transition metals are high. (Jharkhand Board 2010, C.B.S.E. Outside Delhi 2008, 2012, H.P. Board 2014)
(c) The transition metals generally form coloured compounds. (C.B.S.E. 2010. 2012)
(d) The transition metals and their compounds act as good catalysts. (C.B.S.E.Outside Delhi 2010) (C.B.S.E. Delhi 2008, Sample Paper 2010, H.P. Board 2017)
Answer:
(i) Paramagnetic arises due to the presence of unpaired electrons, each such electron has a magnetic moment associated with it due to its spin angular momentum. Transition metals have in its ground state or ionized state has a number of unpaired d-electrons which gives them a paramagnetic behaviour.

(ii) Transition metals have very high interatomic metallic interaction due to the involvement of greater number of electrons from (n – l)d in addition to the ns electrons. The greater the number of valence electrons, the stronger is the resultant bonding due to the greater overlapping of half-filled orbitals. Hence, more amount of energy is required to break these metallic bonds. Thus enthalpy of atomisation of transition metal is very high.

(iii) Colour of transition metal – compounds is due to the excitation of an electron from a lower energy d-orbital to a higher energy d orbital. The energy of excitation corresponds to the frequency of light absorbed and the colour observed corresponds to the complementary colour of the light absorbed (whose frequency lies on the visible region). The frequency of the light absorbed depends on the nature of ligand. Transition metals form coloured compounds due to the presence of vacant d-orbitals for the d-d transition of e’ which causes the colour.

(iv) The catalytic activity of transition metals is ascribed to their ability to adopt multiple oxidation states and to form complexes. Catalyst at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst this has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (activation energy is lowered. Transition metals have d and s orbitals to from these bonds.

Question 12.
What are interstitial compounds? Why are such compounds well known for transition metals?
Answer:
Interstitial compounds are the compounds formed as a result of the trapping of atoms of small elements like H, N, C, B etc. in the crystal lattices of certain metals. These are non-stoichiometric in nature and are neither ionic nor covalent. In fact, no proper bonds exist in the atoms of metals and non-metals involved in these compounds. Transition metals have a tendency to form such compounds. A few examples are: TiC,
Mn4N, Fe3H, VH0·56, Vse0·98 and Fe0·94O etc.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 4

  1. These are generally non-stoichiometric in nature. Therefore, they cannot be represented by a definite structure or formula.
  2. The compounds are neither covalent nor ionic and they don’t represent the normal oxidation states of the metals.
  3. Since the strengths of the metallic bonds in these compounds increase due to greater electronic interactions, they show high melting points and high metallic conductivity. However, these compounds are chemically inert.
  4. The conductivity of the metals remains unaffected in the corresponding interstitial compounds.

Question 13.
How is the variability in oxidation states of transition metals different from that of non-transition metals? Illustrate with examples.
Answer:
The variability of oxidation states, characteristics of transition elements arises out of incomplete filling of d orbitals in such a way that their oxidation states differ from each other by unity This is in contrast with the variability of oxidation states of non-transition elements where oxidation states normally differ by a unit of two.
eg: Vanadium: V+2, V+3, V+4, V+5
Chromium : Cr+2, Cr+3, Cr+4, Cr+5, C+6
Nitrogen: +5, +3, +1, -1, -3.

Question 14.
Describe the preparation of potassium dichromate from chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Answer:
Preparation from chromite:
Potassium dichromate is generally prepared from chromite ore (FeCr2O4). It is in fact, a mixed oxide Fe0.Cr2O3 of iron and chrome also called ferrochrome or chrome iron.
(i) Conversion of chromite ore into sodium chromate: Chromite ore is fused with sodium hydroxide or sodium carbonate in the presence of air.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 5
(ii) Conversion of sodium chromate into sodium dichromate :
The fused mass obtained above is extracted with water. Sodium chromate which is soluble in water goes into the solution leaving behind the insoluble ferric oxide (Fe2O3). The yellow solution of sodium chromate obtained above is treated with concentrated H2SO4 to form sodium dichromate which has an orange colour.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 6
(iii) Conversion of sodium dichromate into potassium dichromate:
Sodium dichromate is more soluble and less stable than potassium dichromate.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 7
Effect of increasing pH : The solution of potassium dichromate (K2Cr2O7) in water is orange in colour. On increasing the pH i.e. on adding the base, the potassium dichromate changes to potassium chromate (K2CrO4) which is yellow in colour. Thus, on increasing the pH, the colour of the solution changes from orange to yellow.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 8

Question 15.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
(a) iodide,
(b) iron (II) solution
(c) H2S.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 9
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 10
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 11

Question 16.
Describe the preparation of potassium permanganate. How does acidified permanganate solution react with:
(a) iron (II) solution
(b) SO2
(c) oxalic acid?
Write the ionic equations for the reactions.
Answer:
Potassium permanganate is prepared on a large scale from the mineral pyrolusite, MnO2.
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 12

Question 17.
For M2+/M and M3+/M2+ systems, the E° values of some metals are given :
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 13
Use this data to comment upon :
(a) The stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+.
(b) The ease with which iron can be oxidised as compared to the similar process for either chromium or manganese metal.
Answer:
As \({ E }_{ { Cr }^{ 3+ }/{ Cr }^{ 2+ } }^{ \circ }\) is negative (-0·4 V), this means that Cr3+ ions in solution cannot be reduced to Cr2+ ions or Cr3+ ions are very stable. As farther comparison of E° values shows that Mn3+ ions can be reduced to Mn2+ ion more readily than Fe3+ ions. Thus, in the light of this, the order of relative stabilities of different ions is Mn3+ < Fe3+ < Cr3+.
(b) From the E° values, the order of oxidation of the metal to the divalent cation is Mn > Cr > Fe.

Question 18.
Predict which of the following will be coloured in an aqueous solution? Ti3+, V3+,Cu+, Sc3+, Mn2+, Fe3+ and Co2+ Give reasons for each.
Answer:
Among the above mentioned, ions Ti3+, V3+ Mn2+, Fe3+, CO2+ will be coloured in its aqueous solution due to the ability of e’ to jump from a lower energy d orbital to a higher energy d orbital. In case of the ions Cu+, Sc3+ this d-d transition cannot take place either due to the absence of any e- in 3d orbital or due to complete filling of d orbital.

Question 19.
Compare the stability of the +2 oxidation state for the elements of the first transition series.
Answer:
The common oxidation state of 3d series elements is + 2 which arises due to the participation of only 4s electrons. The tendency to show the highest oxidation state increases from Sc to Mn then decreases due to the pairing of electrons in 3d subshell. Thus in the series Sc(II) does not exist, Ti(II) is less stable than Ti(IV). At the other end of the series, the oxidation state of Zn is +2 only.

Question 20.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration
(ii) atomic and ionic sizes and
(iii) oxidation state
(iv) chemical reactivity.
Answer:
(i) Electronic configuration:
The general electronic configuration for lanthanoids is [Xe] 0-14 5d 0-1 6s2 and that of actinoids is [Rn] 5f 0-14 6d0-1 7s2. Unlike 4f orbitals, 5 f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Oxidation states:
The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.

(iii) Atomic and ionic size Similar to lanthanoids, actinoids also exhibit actinoid contraction overall decrease in atomic and ionic radii. The contraction is greater due to the poor shielding effect of 5f orbitals. Hence there is an unexpected in the atomic and ionic sizes of actinoids.

(iv) Chemical activity
In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In the case of acids, they are slightly affected by nitric acid (because of the -formation of a protective oxide layer).

Question 21.
How would you account for the following :
(a) Of the d4 species, Cr2+ is strongly reducing while Mn3+ is strongly oxidising in nature.
(b) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents, it is easily oxidised.
(c) d1 configuration is very unstable in ions.
Answer:
(a) E° value of Cr3+/Cr2+ is negative (-0·41 V) while that of Mn3+/Mn2+ is positive (+ 1·57 V). This means Cr2+ ions can lose electrons to form Cr3+ ions and act as a reducing agent while Mn3+ ions can accept electrons and can act as oxidising agent.
(b) Cobalt (II) is stable in aqueous solution but in the presence of complexing agent, it undergoes change in oxidation state from +2 to +3 and is easily oxidised.
(c) The ion with d1 configuration is expected to be extremely unstable and has a great urge to acquire d° configuration (very stable) by losing the only electron present in the d-subshell.

Question 22.
What is meant by ‘disproportionation’ ? Give two examples of disproportionation reaction in aqueous solution.
Answer:
In a disproportionation reaction, an element undergoes an increase as well as decrease in its oxidation state forming
two different compounds. In other words, we can say that it can act both as reducing agent as well as oxidising agent.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 14

Question 23.
Which metal in the first series of transition metals exhibits + 1 oxidation state most frequently and why?
Answer:
Copper, because with +1 oxidation state an extra stable configuration, 3d10 results.

Question 24.
Calculate the number of unpaired electrons in the following gaseous ions :
Mn3+, Cr3+, V3+.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 15

Question 25.
Give examples and suggest reasons for the following features of the transition metal chemistry:
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits the highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
Answer:
(i) In case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atoms are not involved in bonding. As a result, it can donate electrons and behave as a base.

On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding. As a result, it can accept electrons and behave as an acid. For example, MnIIO is basic and MnVIIO is acidic.

(ii) Oxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits oxidation states in oxides and fluorides. For example, in OsF6 and V2O5, the oxidation states of Os and V are + 6 and respectively.

(iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size so, oxoanions of metal have the highest oxidation state. For example, in MnO4, the oxidation state of Mn is +7.

Question 26.
Give the steps in the preparation of (C.B.S.E. Delhi 2009 comptt.)
(a) K2Cr2O7 from chromite ore
(b) KMnO4 from pyrolusite ore.
Answer:
(a)
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 16
(b)
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 17

Question 27.
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Answer:
An alloy is a homogeneous mixture of two or more metals or metals and non-metals. An important alloy that contains lanthanoid metal is mischmetal which contains 50% Cerium and 25 % Lanthanum, with small amounts of Nd (Neodymium) and Pr (Praseody-mium). It is used in Mg-based alloy to produce bullets, shells, and lighter flints.

Question 28.
What are inner transition elements? Decide which of the following atomic numbers belong to inner transition elements :
29, 59, 74, 95, 102, 104.
Answer:
The inner transition elements also called/-block elements include the series of lanthanoids (Z = 58 to 71) and actinoids (Z = 90 to 103). This means that the elements with atomic numbers 59, 95, and 102 belong to inner transition elements.

Question 29.
The chemistry of actinoid elements is not so smooth as that of lanthanoids. Justify this statement by giving some examples of the oxidation states of these elements.
Answer:
Lanthanoids show a limited number of oxidation states, such as +2, +3 and +4 (+3 is the principal oxidation state). This is because of large energy gap between 5d and 4f- subshells. On the other hand, actinoids also show a principal oxidation state of +3 but show a number of other oxidation state also. For example, Uranium (Z = 92) exhibits oxidation states of +3, +4, +5, +6 and Neptunium (Z = 94) shows oxidation states of +3, +4, +5, +6 and +7. This is because of small energy difference between 5f and 6d orbitals.

Question 30.
Which is the last element in the series of actinoids? Write the electronic configuration of the element. Comment upon its possible oxidation state.
Answer:
Lawrencium (Lr = 103); [Rn] 5f146d17s2 oxidation state = +3.

Question 31.
Use Hund’s rule to derive the electronic configuration of Ce3+ ion and calculate its magnetic moment on the basis of the spin-spin formula.
Answer:
Cerium electronic configuration = [Xe]4f1 5d1 16s2
Ce+3 ion = [Xe]4f1
i.e., one unpaired electron is present
Magnetic moment, µ = \(\sqrt{n(n+2)}\) = \(\sqrt { 3 }\) = 1.73BM.

Question 32.
Name the members of the lanthanoid series which exhibit+4oxidatk>nstatesandthosewhichexhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
Answer:
+ 4 oxidation state in Ce (Z = 58), Pr (Z = 59), Tb (Z = 65).
+ 2 oxidation state in Nd (Z = 60), Sm(Z=62), Eu (Z = 63), Tm (Z=69), Yb (Z = 70).
+ 2 oxidation state is exhibited when the lanthanoid has the configuration 5cf 6s2 so that two electrons are-easily lost.
+ 4 oxidation state is exhibited by the elements which after losing four electrons acquire configuration 4f° or 4f1

Question 33.
Compare the chemistry of the actinoids with that of lanthanoids with reference to:
(i) electronic configuration
(ii) oxidation states and
(iii) chemical reactivity.
Answer:
(i) Electronic configuration:
The general electronic configuration for lanthanoids is [Xe] 0-14 5d 0-1 6s2 and that of actinoids is [Rn] 5f 0-14 6d0-1 7s2. Unlike 4f orbitals, 5 f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Oxidation states:
The principal oxidation states of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in the +3 state than in the +4 state.

(iii) Atomic and ionic size Similar to lanthanoids, actinoids also exhibit actinoid contraction overall decrease in atomic and ionic radii. The contraction is greater due to the poor shielding effect of 5f orbitals. Hence there is an unexpected in the atomic and ionic sizes of actinoids.

(iv) Chemical reactivity
In the lanthanoid series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In the case of acids, they are slightly affected by nitric acid (because of the -formation of a protective oxide layer).

Question 34.
Write the electronic configuration of the elements with atomic numbers 61, 91, 101, 109.
Answer:
Promethium or Pm (Z = 61) [Xe]544f55d06s2
Protactinium or Pa (Z = 91) [Rn] 4f26d17s2
Mendelevium or Md (Z = 101) [Rn] 5f16d07s2
Meitnerium or Mt (Z = 109) [Rn] 5f146d77s2

Question 35.
Compare the general characteristics of the first transition series of transition metals with those of the second and third transition series metals in the respective vertical columns. Give special emphasis on the following points :
(i) electronic configuration
(ii) oxidation states
(iii) ionisation enthalpies
(iv) atomic sizes.
Answer:
(i) Electronic configuration. There are some exceptions in the electronic configurations in all the three series.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 18
(ii) Oxidation state. The elements belonging to the different series but present in the same group have similar electronic configurations and therefore, exhibit almost same variable oxidation states. In general, these are maximum in the middle of the series while minimum towards the end. Transition elements show variable oxidation states due to the participation of ns and (n – 1) d electrons in bonding because the energies of ns and (n – 1) d-subshells are quite close. The stability of a particular oxidation state depends upon the nature of the element with which the transition metal forms the compound.

(iii) Ionisation enthalpies. In general, the ionisation enthalpies in all three transition series increase from left to the right. However, the gaps in the two successive elements in a particular series are small and are also not regular. The first three ionisation enthalpies of the elements present in the first transition series are given in the text part. The ∆iH1 [ values of the elements belonging to the 5d series and higher as compared to those belonging to 3d and Ad series in the same group because of poor shielding by intervening 4f electrons present.

(iv) Atomic size. In all three transition series, the atomic, as well as ionic radii of the elements, increase from left to right. The values for 3d series are given in the text part. However, the increase in their values is not as much as expected since the shielding by (n – 1 )d electrons is not as much as expected. In a particular group, the atomic radius of the elements belonging to Ad series is more than the elements in the 3d series. However, the gaps in the elements belonging to Ad and 5d series are negligible on account of lanthanoid contraction which the elements of the 5d experience.

Question 36.
Write down the number of 3d electrons in each of the following ions :
Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+.
Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
Answer:
The number of 3d electrons in the ions are :
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 19
For the explanation of the involvement of 3d orbitals in the hydrated ions (octahedral in nature) consult the next unit on coordination compounds.

Question 37.
Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
Answer:
The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways.
(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements of the 2nd and 3rd transition series). However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series this is due to lanthanoid configuration.

(ii) +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.

(iii) The enthalpies of atomization of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.

(iv) The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of strong metallic bonding (M – M bonding)

(v) The dements of the first transition series from low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.

Question 38.
What can be inferred from the magnetic moment values of the following complex species?
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 20
Answer:
The magnetic moment of a compound is given by the relation (µ) = \(\sqrt { n\left( n+2 \right) } \) B.M, where n is the number of unpaired electrons.
For one unpaired electron (n = 1) ; µ = \(\sqrt { 1\left( 1+2 \right) } =\sqrt { 3 } =1\cdot 73\quad B.M.\)
For two unpaired electrons (n – 2) ; µ = \(\sqrt { 2\left( 2+2 \right) } =\sqrt { 8 } =2\cdot 83\quad B.M.\)
For three unpaired electrons (n = 3); µ = \(\sqrt { 3\left( 3+2 \right) } =\sqrt { 15 } =3\cdot 87\quad B.M.\)
For four unpaired electrons (n = 4) ; µ = \(\sqrt { 4\left( 4+2 \right) } =\sqrt { 24 } =4\cdot 9\quad B.M.\)
For five unpaired electrons (n = 5) ; µ = \(\sqrt { 5\left( 5+2 \right) } =\sqrt { 35 } =5\cdot 92\quad B.M.\)
*In the light of the above value, let us gather the desired information about the complex species that are mentioned
(i) K4[Mn(CN)6]
Oxidation state of Mn : [Mn(CN)6]4- , x + 6(-l) = -4 or x = -4 + 6 = + 2
The magnetic value of 1·73 B.M. indicates the presence of one unpaired electron in the complex. When six, CN ions (or ligands) approach Mn2+ ion, electrons in 3d orbitals pair up to make available six vacant orbitals involving d2sp3 hybridisation.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 21
The complex is octahedral and is paramagnetic due to one unpaired electron.
(ii) [Fe(H2O)6]2+
Oxidation state of Fe : [Fe(H2O)6]2+ ; r + 6 (0) = +2
The magnetic moment value of 5·3 B.M. indicates that there are four unpaired electrons in the complex. This means that the electrons in Fe2+ ion do not pair up when six H20 molecules (or ligands) approach it. Since the desired number of vacant orbitals (six) are available, die complex formed is sp3d2 hybridised.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 22
The complex is octahedral and is paramagnetic due to four unpaired electrons. It is also called outer orbital complex because 4d (n = 4) orbitals are involved.
(iii) K2[MnCl4]
Oxidation state of Mn : [MnCl4]2-, x + 4(-1) = -2 or x = -2 + 4= + 2
The magnetic moment value of 5·9 B.M. indicates that there are five unpaired electrons in the complex. This means that all the five 3d orbitals in Mn2+ ion are involved in the bond formation. The complex is sp3 hybridized in which one vacant 4s and three vacant 4p orbitals participate.
NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements 23
The complex is therefore, tetrahedral in nature.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture

NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture.

BoardCBSE
TextbookNCERT
ClassClass 12
SubjectHistory
ChapterChapter 12
Chapter NameColonial Cities Urbanisation, Planning and Architecture
Number of Questions Solved9
CategoryNCERT Solutions

NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture

Question 1.
To what extent are census data useful in reconstructing patterns of urbanisation in the colonial context ?
Solution :
The census data are useful in reconstructing patterns of urbanisation in the colonial context in the following ways :

  1. The censuses reveal that after 1800 the urbanisation in India was slow. The proportion of the urban population to the total population in India was almost stagnant. Between 1900 and 1940 the urban population increased from about 10 per cent of the total population to about 13 per cent.
  2. Smaller towns did not grow economically but Bombay, Madras and Calcutta grew rapidly.
  3. The introduction of railways changed the centre of economic activity from traditional towns to towns connected with the railways.

The above facts provide us the patterns of urbanisation but the historians have found the census data misleading because the census operation was a means by which social data were converted into convenient statistics about the population. There were many shortcomings in it. For example, classification of different sections of population was arbitrary. There were overlapping identities of people. People w7ere too suspicious of census operations and did not cooperate with the officials. Thus, census data is invaluable but should be studied carefully in restructuring patterns of urbanisation in colonial India.

Question 2.
What do the terms “White” and “Black” Town signify?
Solution :
The British had white skin as they were often described ‘white’ and they considered themselves as superior to others. On the other hand, the blacks had brown or black skin. So they were known as the ‘black’. The White signified their superiority over the black due to the colour of their skin. The British symbolised the Black areas full of chaos and anarchy, filth and disease and on the other hand, the white areas stood for cleanliness and hygiene. In Black areas, epidemics like cholera and plague often broke out. So the British took stringent measures to ensure sanitation and public health to prevent diseases of the Black areas. They ensured underground piped water supply and introduced sewerage and drainage system in White areas. Thus, we can say, the White Towns were those parts of the colonial towns where the White people lived. These towns had wide roads, barracks, churches, paradeground, big bungalows and gardens, symbolised settled city life, whereas the Indian lived in Black Towns, were said to be unorganised and a source of filth and disease.

Question 3.
How did prominent Indian merchants establish themselves in the colonial city?
Solution :
The prominent Indian merchants established themselves in the colonial city in the following ways :

  1. With the expansion of railways, the countryside from where raw materials and labour were obtained was linked to the cities like Bombay, Madras and Calcutta. This gave an opportunity to the Indian merchants to set up modern factories. Thus, after the 1850s, cotton mills were set up by Indian merchants and entrepreneurs in Bombay.
  2. Kanpur specialised in leather, woollen, cotton textiles and Jamshedpur where steel factory was established by J. Tata, specialised in steel.
  3. The American Civil War started in 1861 gave another opportunity to the Indian merchants for earning huge profits. Bombay was the most important city of India. By the late nineteenth century, Indian merchants in Bombay established cotton mills.

Question 4.
Examine how concerns of defence and health gave shape to Calcutta.
Solution :
Sirajudaula, the Nawab of Bengal in 1756, sacked the small fort from Britisher. In this fort the British traders had built to house their goods. Consequently, when Sirajudaula was defeated in the Battle of Plassey, the British built a new fort, Fort William which could not be easily attacked. Around this a vast open space was left. This open space ‘ was called the Maidan or garermath. This was done for security reasons, because there would be no obstructions to a straight time of fire from the Fort against an advancing enemy army. Soon the British began to move out of the Fort. They built residences along the periphery of the Maidan. This indicates that how the English Settlement in Calcutta began to take shape. The vast open space around the Fort William became the significant town planning measure in Calcutta (Now Kolkata).

Lord Wellesley was more concerned about the conditions that existed in the cities. Cities were overcrowded, and had no sanitation facilities. He issued an administrative order in 1803 on the need for town planning and set up various committees for this purpose open places in the city would make the city healthier. As a result of this, many bazaars, ghats, burial ground and tanneries were cleared or removed. After Wellesley’s departure, the Lottery Committee carried on with the work of town planning in Calcutta.

Question 5.
What are the different colonial architectural styles which can be seen in Bombay city ?
Solution :
The different colonial architectural styles which can be seen in Bombay city are as mentioned below :

  1. European style : In mid nineteenth century, the buildings were constructed in the European style to create a familiar landscape in an alien country and to symbolise their superiority.
  2. Indian style : As the Indians used European architecture, the British adopted Indian style that can be seen in the construction of bungalows in Bombay and all over India.
  3. Neo-classical or new classical style : Its characteristics are construction of geometrical structures fronted with lofty pillars. Town Hall is an example of this style.
  4. Neo-Gothic style : Its characteristics are high-pitched roofs, pointed arches and detailed decoration: Secretariat, University of Bombay and High Court were made in this style. The most spectacular example of the neo-Gothic style is the Victoria Terminus.
  5. Indo-saracenic style : It was a hybrid architectural style that combined the Indian with the European. It was inspired by the medieval buildings in India with their domes, chhatris, jalis and arches. The Gateway of India and the Taj Mahal Hotel that was built by Jamsetji Tata belong to this style.

Question 6.
How were urban centres transformed during the eighteenth century?
Solution :
(i) The disintegration of the mughal empire after the death of Aurangzab paved the way of emergence of paverful regional powers. The capital cities of these regional kingdom likes Lucknow, Poona, Nagpur and Barda now become important. Taking the advantage of this opportunity many nobles and officials created new urban settlements such as the qasbah and ganj.

(ii) The European companies too had set up their bases in different parts of India during the sixteenth and seventeenth centuries. For example the Portugues (in Panaji in 1570) and the British in Madras in 1639. With the expansion in commercial activity, towns began to emerge as trading centres.

(iii) From the mid-eighteenth century trading. Centres like Surat and Dhakha which had grown in the seventeenth century now began to decline as trade shifted to other places. When the British acquired Bengal and the east Indian’s Company’s trade hereafter expanded the colonial port cities likes Madras and Calcutta. These new part cities began to emerge as the new economic capitals.

(iv) In these newly developed cities many new buildings were built and new occupations developed. People flocked to these cities in large numbers. By the nineteenth century, these newly developed cities become the biggest cities in India.

Question 7.
What were the new kinds of public places that emerged in the colonial city ? What functions did they serve ?
Solution :
New kinds of public places that emerged in the colonial city were as given below :

  1. Fort St. George (Madras), Fort William (Calcutta), the Fort George (Bombay). These were the fortified areas of British settlement.
  2. The Writers’ Building in Calcutta : It was the building where the servants of the East India Company in India stayed on arrival in the country. Later this building became a government office.
  3. Clubs, racecourses and theatres were built for the ruling elites exclusively on racial grounds.
  4. Cantonment places were developed. Here Indian troops under European command were stationed. These were considered safe enclaves for Europeans.
  5. Simla, founded during the course of Gurkha war, and Darjeeling were hill stations that became strategic places for billeting troops, guarding frontiers and launching campaigns against enemy rulers.
  6. Public places such as public parks, theatres, and cinema halls came into existence for providing new forms of entertainment and social interaction.
  7. Government House Calcutta : It was built by Lord Wellesley for himself in Calcutta.

Question 8.
What were the concerns that influenced town planning in the nineteenth century?
Solution :

  1. Two concerns which influenced the town planning in the nineteenth century were defence and health.
  2. In many towns British built forts to protect their factories. Around the fort, a vast open space was left open. This vast space was known as the Maidan.
  3. It was done so that there would be no obstructions to a straight line of fire from the Fort against an advancing enemy.
  4. Attempts were also made to improve the sanitation and cleanliness by creating open spaces in the city.
  5. For this purpose, in Calcutta many bazaars, ghats and burial grounds were cleared.

Question 9.
To what extent were social relations transformed in the new cities ?
Solution :
The social relations were transformed in the new cities in the following ways :

  1. New transport facilities as horse-drawn carriages, trams and buses meant that people could live at a distance from the city centre. This led to separation of the place of work from the place of residence. People travelled from home to office or factory.
  2. The sense of coherence and familiarity of the old towns disappeared. The public places such as parks, theatres and cinema halls provided new forms of entertainment and social interaction.
  3. New social groups came into existence. The “middle classes” increased due to the coming of all types of people i.e., clerks, doctors, teachers, lawyers and others. With the spread of education, people could put forward their views in newspapers and journals. People started questioning old customs and practices.
  4. Women entered new professions as factory workers, teachers, theatres and film actresses. However, their entry into public spaces remained the objects of social censure.
  5. There was a dramatic contrast between extreme wealth and poverty. The new cities were bewildering places where life seemed always in a flux. Paupers from the villages came to cities in search of employment. The male migrants left their families in the villages because jobs were uncertain and food was expensive. But yet the villagers participated in religious festivals, tamashas and swangs which mocked the pretensions of their masters, Indian and European.

We hope the NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture, drop a comment below and we will get back to you at the earliest.