NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions:

 Section Name Topic Name
2 Solutions
2.1 Types of Solutions
2.2 Expressing Concentration of Solutions
2.3 Solubility
2.4 Vapour Pressure of Liquid Solutions
2.5 Ideal and Non-ideal Solutions
2.6 Colligative Properties and Determination of Molar Mass
2.7 Abnormal Molar Masses

NCERT TEXTBOOK QUESTIONS SOLVED

2.1. Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Ans: Mass of solution = Mass of C6H6 + Mass of CCl4
= 22 g+122 g= 144 g
Mass % of benzene = 22/144 x 100 =15.28 %
Mass % of CCl4 = 122/144 x 100 = 84.72 %

2.2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Ans: 30% by mass of C6H6 in CCl4 => 30 g C6H6 in 100 g solution
.’. no. of moles of C6H6,(nC6h6) = 30/78 = 0.385
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q2

2.3. Calculate the molarity of each of the following solutions
(a) 30 g of Co(NO3)26H2O in 4·3 L of solution
(b) 30 mL of 0-5 M H2SO4 diluted to 500 mL.
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q3

2.4. Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
Ans: 0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH2CONH2) = 60 g mol-1
.’. 0.25 mole of urea = 0.25 x 60=15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
.’. 2.5 kg of solution contains urea =15/1.015 x 2.5 = 37 g

2.5. Calculate
(a) molality
(b) molarity and
(c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI solution is 1·202 g mL-1.
Ans:
Step I. Calculation of molality of solution
Weight of KI in 100 g of the solution = 20 g
Weight of water in the solution = 100 – 20 = 80 g = 0-08 kg
Molar mass of KI = 39 + 127 = 166 g mol-1.
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q5
Step II. Calculation of molarity of solution
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q5.1
Step III. Calculation of mole fraction of Kl
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q5.2

2.6. HS, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.
Ans: Solubility of H2S gas = 0.195 m
= 0.195 mole in 1 kg of solvent
1 kg of solvent = 1000g
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q6

2.7. Henry’s law constant for CO2 in water is 1.67 x 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.
Ans.:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q7

2.8 The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg respectively at 350 K. Calculate the composition of the liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition in the vapour phase.
Ans:
Vapour pressure of pure liquid A (\({ P }_{ A }^{ \circ }\)) = 450 mm
Vapour pressure of pure liquid B (\({ P }_{ B }^{ \circ }\)) = 700 mm
Total vapour pressure of the solution (P) = 600 mm
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q8

2.9. Vapour pressure of pure water at 298 K is 23.8 m m Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q9

2.10. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q10

2.11 Calculate the mass of ascorbic acid (vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1·5°C. (Kf for CH3COOH) = 3·9 K kg mol-1)
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q11

2.12. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Textbook Questions Q12

NCERT EXERCISES

2.1. Define the terra solution. How many types of solutions are formed? Write briefly about each type with an example.
Sol: A solution is a homogeneous mixture of two or more chemically non-reacting substances. Types of solutions: There are nine types of solutions.
Types of Solution Examples
Gaseous solutions
(a) Gas in gas Air, mixture of 02 and N2, etc.
(b) Liquid in gas Water vapour
(c) Solid in gas Camphor vapours in N2 gas, smoke etc.
Liquid solutions
(a) Gas in liquid C02 dissolved in water (aerated water), and 02 dissolved in water, etc.
(b) Liquid in liquid Ethanol dissolved in water, etc.
(c) Solid in liquid Sugar dissolved in water, saline water, etc.
Solid solutions
(a) Gas in solid Solution of hydrogen in palladium
(b) Liquid in solid Amalgams, e.g., Na-Hg
(c) Solid in solid Gold ornaments (Cu/Ag with Au)

2.2. Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What type of solid solution is this likely to be ?
Sol: The solution likely to be formed is interstitial solid solution.

2.3 Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
Sol: (i) Mole fraction: It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q3
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q3.1
(ii) Molality: It is defined as die number of moles of a solute present in 1000g (1kg) of a solvent.
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q3.2
NOTE: Molality is considered better way of expressing concentration of solutions, as compared to molarity because molality does not change with change in temperature since the mass of solvent does not vary with temperature,
(iii) Molarity: It is defined as the number of moles of solute present in one litre of solution.
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q3.3
NOTE: Molarity is the most common way of expressing concentration of a solution in laboratory. However, it has one disadvantage. It changes with temperature because volume of a solution alters due to expansion and contraction of the liquid with temperature.
(iv) Mass percentage: It is the amount of solute in grams present in 100g of solution.
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q3.4

Free online chemical calculator. Molarity Calculate Chemistry, mass or volume.

2.4. Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of acid if the density of the solution is 1·504 g mL-1 ?
Sol: Mass of HNO3 in solution = 68 g
Molar mass of HNO3 = 63 g mol-1
Mass of solution = 100 g
Density of solution = 1·504 g mL-1
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q4

2.5. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1 .2 g m L-1, then what shall be the molarity of the solution?
Sol: 10 percent w/w solution of glucose in water means 10g glucose and 90g of water.
Molar mass of glucose = 180g mol-1 and molar mass of water = 18g mol-1
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q5

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q5.1

2.6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2C0and NaHCOcontaining equimolar amounts of both?
Sol: Calculation of no. of moles of components in the mixture.
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q6

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q6.1

2.7. Calculate the percentage composition in terms of mass of a solution obtained by mixing 300 g of a 25% and 400 g of a 40% solution by mass.
Sol:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q7

2.8. An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C2 H6O2 ) and200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL-1, then what shall be the molarity of the solution?
Sol:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q8

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q8.1

2.9. A sample of drinking water was found to be severely contaminated with chloroform (CHCl3), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass).
(i) express this in percent by mass.
(ii) determine the molality of chloroform in the water sample.
Sol: 15 ppm means 15 parts in million (106) by mass in the solution.
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q9

2.10. What role does the molecular interaction play in solution of alcohol in water?
Sol: In case of alcohol as well as water, the molecules are interlinked by intermolecular hydrogen bonding. However, the hydrogen bonding is also present in the molecules of alcohol and water in the solution but it is comparatively less than both alcohol and water. As a result, the magnitude of attractive forces tends to decrease and the solution shows positive deviation from Raoult’s Law. This will lead to increase in vapour pressure of the solution and also decrease in its boiling point.

2.11. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Sol: When gases are dissolved in water, it is accompanied by a release of heat energy, i.e., process is exothermic. When the temperature is increased, according to Lechatlier’s Principle, the equilibrium shifts in backward direction, and thus gases becomes less soluble in liquids.

2.12. State Henry’s law and mention some of its important applications.
Sol:
Henry’s law: The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.
or
The partial pressure of a gas in vapour phase is proportional to the mole fraction of the gas (x) in the solution. p = KHX
where KH is Henry’s law constant.
Applications of Henry’s law :
(i) In order to increase the solubility of CO2 gas in soft drinks and soda water, the bottles are normally sealed under high pressure. Increase in pressure increases the solubility of a gas in a solvent according to Henry’s Law. If the bottle is opened by removing the stopper or seal, the pressure on the surface of the gas will suddenly decrease. This will cause a decrease in the solubility of the gas in the liquid i.e. water. As a result, it will rush out of the bottle producing a hissing noise or with a fiz.
(ii) As pointed above, oxygen to be used by deep sea divers is generally diluted with helium inorder to reduce or minimise the painfril effects during decompression.
(iii) As the partial pressure of oxygen in air is high, in lungs it combines with haemoglobin to form oxyhaemoglobin. In tissues, the partial pressure of oxygen is comparatively low. Therefore, oxyhaemoglobin releases oxygen in order to carry out cellular activities.

2.13. The partial pressure of ethane over a solution containing 6.56 × 10-3 g of ethane is 1 bar. If the solution contains 5.00 × 10-2 g of ethane, then what shall be the partial pressure of the gas?
Sol:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q13

2.14. According to Raoult’s law, what is meant by positive and negative deviaitions and how is the sign of ∆solH related to positive and negative deviations from Raoult’s law?
Sol: Solutions having vapour pressures more than that expected from Raoult’s law are said to exhibit positive deviation. In these solutions solvent – solute interactions are weaker and ∆solH is positive because stronger A – A or B – B interactions are replaced by weaker A – B interactions. Breaking of the stronger interactions requires more energy & less energy is released on formation of weaker interactions. So overall ∆sol H is positive. Similarly ∆solV is positive i.e. the volume of solution is some what more than sum of volumes of solvent and solute.
So there is expansion in volume on solution formation.
Similarly in case of solutions exhibiting negative deviations, A – B interactions are stronger than A-A&B-B. So weaker interactions are replaced by stronger interactions so , there is release of energy i.e. ∆sol H is negative.

2.15. An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1·004 bar at the boiling point of the solvent. What is the molecular mass of the solute ?
Sol:
According to Raoult’s Law,
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q15

2.16  Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35.0 g of octane?
Sol.
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q16

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q16.1

2.17.  The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it
Sol: 1 molal solution of solute means 1 mole of solute in 1000g of the solvent.
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q17

2.18. Calculate the mass of a non-volatile solute (molecular mass 40 g mol-1) that should be dissolved in 114 g of octane to reduce its pressure to 80%. (C.B.S.E. Outside Delhi 2008)
Sol: According to Raoult’s Law,
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q18

2.19. A solution containing 30g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18g of water is then added to the solution and the new of vapour pressure becomes 2.9 kPa at 298 K. Calculate
(i) molar mass of the solute.
(ii) vapour pressure of water at 298 K.
Sol: Let the molar mass of solute = Mg mol-1
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q19

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q19.1

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q19.2

2.20. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Sol: Mass of sugar in 5% (by mass) solution means 5gin 100g of solvent (water)
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q20

2.21. Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol-1. Calculate atomic masses of A and B.
Sol:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q21

2.22. At 300 K, 36 g glucose present per litre in its solution has osmotic pressure of 4·98 bar. If the osmotic pressure of the solution is 1·52 bar at the same temperature, what would be its concentration?
Sol:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q22

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q22.1

2.23. Suggest the most important type of intermolecular attractive interaction in the following pairs:
(i) n-hexane and n-octane
(ii) I2 and CCl4.
(iii) NaCl04 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H60)
Sol: (i) Both w-hexane and n-octane are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(ii) Both I2 and CCl4 are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(iii) NaCl04 is an ionic compound and gives Na+ and Cl04 ions in the Solution. Water is a polar molecule. Thus, the intermolecular interactions will be ion-dipole interactions.
(iv) Both methanol and acetone are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.
(v) Both CH3CN and C3H6O are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.

2.24. Based on solute solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
Sol: n-octane (C8H18) is a non-polar liquid and solubility is governed by the principle that like dissolve like. Keeping this in view, the increasing order of solubility of different solutes is:
KCl < CH3OH < CH3C=N < C6H12 (cyclohexane).

2.25. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol
Sol: (i) Phenol (having polar – OH group) – Partially soluble.
(ii) Toluene (non-polar) – Insoluble.
(iii) Formic acid (form hydrogen bonds with water molecules) – Highly soluble.
(iv) Ethylene glycol (form hydrogen bonds with water molecules) Highly soluble.
(v) Chloroform (non-polar)- Insoluble.
(vi) Pentanol (having polar -OH) – Partially soluble.

2.26. If the density of lake water is 1·25 g mL-1, and it contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake. (C.B.S.E. Outside Delhi 2008)
Sol:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q26

2.27. If the solubility product of CuS is 6 x 10-16, calculate the maximum molarity of CuS in aqueous solution.
Sol:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q27

2.28. Calculate the mass percentage of aspirin (C9H8O4 in acetonitrile (CH3CN) when 6.5g of CHO is dissolved in 450 g of CH3CN.
Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-28

2.29. Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 x 10-3 m aqueous solution required for the above dose.
Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q29

2.30. Calculate the amount of benzoic acid (C5H5COOH) required for preparing 250 mL of 0· 15 M solution in methanol.
Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q30

2.31. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-31
Fluorine being more electronegative than chlorine has the highest electron withdrawing inductive effect. Thus, triflouroacetic acid is the strongest trichloroacetic acid is second most and acetic acid is the weakest acid due to absence of any electron withdrawing group. Thus, F3CCOOH ionizes to the largest extent while CH3COOH ionizes to minimum extent in water. Greater the extent of ionization greater is the depression in freezing point. Hence, the order of depression in freezing point will be CH3COOH < Cl3CCOOH < F3CCOOH.

2.32. Calculate the depression in the freezing point of water when 10g of CH3CH2CHClCOOH is added to 250g of water. Ka = 1.4 x 1o-3 Kg = 1.86 K kg mol-1.

Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-32
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-32.1

Calculate the Dilution Factor by dividing the total diluted sample volume by the original sample volume using this dilution factor formula.

2.33. 19.5g of CH2FCOOH is dissolved in 500g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’s Hoff factor and dissociation constant of fluoroacetic acid.

Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-33

2.34. Vapour pressure of water at 293 K is 17·535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Solution:
According to Raoult’s Law,
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q34

2.35. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-35

2.36. 100g of liquid A (molar mass 140 g mol-1) was dissolved in 1000g of liquid B (molar mass 180g mol-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-36

2.37. Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot Ptotal, Pchlroform and Pacetone as a function of χacetone. The experimental data observed for different compositions of mixtures is:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-37q
Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-37

2.38. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80g of benzene is mixed with 100g of toluene.

Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-38

2.39. The air is a mixture of a number of gases. The major components are oxygen and nitrogen with an approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if Henry’s law constants for oxygen and nitrogen are 3.30 x 107 mm and 6.51 x 107 mm respectively, calculate the composition of these gases in water.

Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q39
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-39.1

2.40. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.

Solution:
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-40
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions-40.1

2.41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated. (C.B.S.E. 2013)
Solution:
Step I. Calculation of Van’t Hoff factor (i)
K2SO4 dissociates in water as :
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Exercises Q41

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NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties

NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chemistry-chapter-3-classification-elements-periodicity-properties/

NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties

Multiple Choice Questions
Single Correct Answer Type

Q1. Consider the isoelectronic species, Na+, Mg2+, F and O2-. The correct order of increasing length of their radii is

Chemistry Class 11 NCERT Ch 3

Sol: (b) Amongst isoelectronic ions, ionic radii decrease with increase in nuclear charge:
Mg2+(12) < Na+(11) < F (10) < 02- (8)

Q2. Which of the following is not an actinoid?
(a) Curium (Z = 96)
(b)Californium (Z = 98)
(c) Uranium (Z = 92)                               
(d)    Terbium (Z = 65)
Sol: (d) Actinoids are elements with Z= 90 – 103. Therefore, Terbium (Z = 65) is not an actinoid. Terbium (Z = 65) is a lanthanoid. Tb: [Xe]4f95d06s2

Q3. The order of screening effect of electrons of s, p, d and/orbitals of a given shell of an atom on its outer shell electrons is
(a) s>p> d>f
(b) f> d> p> s
(c) p < d < s <f
(d) f> p> s> d
Sol: (a) The screening effect of the orbitals follows the order of s >p > d>f as screening effect decreases from s-orbital to f-orbital in an atom on account of shape of the orbital.

Free Enthalpy Calculator is helpful to find the chnage in enthalpy of a reaction.

Q4. The first ionization enthalpies of Na, Mg, A1 and Si are in the order
(a)    Na < Mg > A1 < Si                          
(b)     Na>Mg>Al>Si
(c)    Na < Mg < A1 < Si                          
(d)    Na > Mg > A1 < Si

Sol: (a) The electronic configurations of Na and Mg are:

Na (11): [Ne] 3s1 and Mg (12): [Ne] 3s2 In both the atoms, the electron is to be removed from 3s-orbital but nuclear charge in Na is less than Mg. Thus, ionisation energy of Na is less than Mg (Na < Mg).
The electronic configurations of Mg and Al are:
Mg: [Ne] 3s2; Al: [Ne] 3s2 3pl
In Mg, the electron is to be removed from 3s-orbital while in Al, it is to be removed from 3p-orbital. Since it is easier to remove an electron from 3p-orbital in comparison to 3s-orbital, the ionization enthalpy of Mg is higher than Al (Mg > Al).
The electronic configurations of Al and Si are:
Al (13): [Ne] 3s2 3p1 Si (14): [Ne] 3s2 3p2
In both the atoms, the electron is to be removed from 3p-orbital but nuclear charge in Si is more than Al. Thus, ionisation enthalpy of Al is less than Si

Q5. The electronic configuration of gadolinium (Atomic number 64) is

NCERT Exemplar Class 11 Chemistry Ch 3
Sol: (c) The electronic configuration of La (Z = 57) is [Xe] 5dl 6s2. Therefore, further addition of electrons occurs in the lower energy 4f-orbital till it is exactly half-filled at Eu (Z = 63) Thus, the electronic configuration of Eu is [Xe] 4f7 6s2. Thereafter, addition of next electron does not occur in the more stable exactly half-filled 4f7 shell but occurs in the little higher energy 5d-orbital. Thus, the electronic configuration of Gd (Z = 64) is [Xe] 4f7 5dl 6s2.

Q6. The statement that is not correct for periodic classification of elements is
(a) The properties of elements are periodic function of their atomic numbers.
(b) Non-metallic elements are less in number than metallic elements.
(c) For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.
(d) The first ionization enthalpies of elements generally increase with increase in atomic number as we go along a period.
Sol: (c) For transition elements, the 3d-orbitals are filled with electrons after 3p and 4s-orbitals and before 4p-orbitals. The order of filling the orbitals is:
Is 2s 2p 3s 3p 4s 3d4p 5s 4d 5p 6s …

Q7. Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is
(a) F > Cl > Br > I
(b) F < Cl < Br < I
(c) F < Cl > Br > I
(d) F < Cl < Br > I
Sol: (c) Chlorine has higher electron gain enthalpy than fluorine. This is due to small size of fluorine atom, i.e., the electron density is high which resists the addition of an electron (F < Cl).
In general, electron gain enthalpy decreases as atomic size increases. Thus, electron gain enthalpy follows the order:
Cl > Br > I

Q8. The period number in the long form of the periodic table is equal to
(a) magnetic quantum number of any element of the period.
(b) atomic number of any element of the period.
(c) maximum principal quantum number of any element of the period.
(d) maximum Azimuthal quantum number of any element of the period.
Sol: (c) Since each period starts with the filling of electrons in a new principal quantum number, therefore, the period number in the long form of the periodic table refers to the maximum principal quantum number of any element in the period.
Period number = maximum n of any element
(where, n = principal quantum number).

Q9. The elements in which electrons are progressively filled in 4f-orbital are called
(a) actinoids.
(b) transition elements.
(c) lanthanoids.
(d) halogens.
Sol: (c) In lanthanoids, the electrons are filled in 4f-orbitals.

Q10. Which of the following is the correct order of size of the given species?
(a) I>I>I+
(b) I+>I>I
(c) i>i+>i
(d) i >i>i+

Sol: Anion is bigger than the parent atom and cation is smaller than the parent atom.
Thus, I > I > I+

Q11.  The formation of the oxide ion, 02-(g), from oxygen atom requires first an exothermic and then an endothermic step as shown below:
O(g) + e→0 (g), ∆H= -141 kJ mol-1
0(g) + e→O2 (g), ∆H =
+780 kJ mol-1
Thus process of formation of O2- ion in gas phase is unfavourable even though O2- is isoelectronic with neon. It is due to the fact that

(a) Oxygen is more electronegative.
(b) Addition of electron in oxygen results in larger size of the ion.
(c) Electron repulsion outweighs the stability gained by achieving noble gas configuration.
(d) 0 ion has comparatively smaller size than oxygen atom.
Sol:(c) There is a lot of repulsion when similar charges approach each other as
O(g), and electron are both negatively charged. To add an electron under such situation, the force of repulsion is to be overcome by applying external energy.

Q12. Comprehension given below is followed by some multiple choice questions. Each question has one correct option.
In the modem periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, s,p, d and f The modem periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 18, 18, 32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids, are placed at the bottom of the main body of the periodic table.
(i) The element with atomic number 57 belongs to

(a) s-block
(b) p-block
(c)  d-block
(d)  f-block

(ii) The last element of the p-block in 6th period is represented by the outermost electronic configuration.

(a)    7s2 7p6
(b)     5f14 6d10 7s2 7p0
(c)     4f14 5di0 6s26
(d)     4f14 5di0     6s2 6p4

(iii) Which of the elements whose atomic numbers are given below, cannot be accommodated in the present set up of the long form of the periodic table?

(a) 107
(b) 118
(c) 126               .
(d) 102

(iv) The electronic configuration of the element which is just above the element with atomic number 43 in the same group is________ .

(a) 1s2 2s22p6 3s2 3p6 3d5 4s2
(b) 1s2 2s2 2p6 3 s2 3p6 3d5 4s3 4p6
(c) 1s2 2s2 2p6 3s2 3p6 3d5 4s2
(d) ls2 2s2 2p6 3s2 3p6 3d74s2

(v) The element with atomic number 35, 53 and 85 are all_____ .

(a) noble gases
(b) halogens
(c) heavy metals
(d) light metals
Periodic Classification Of Elements NCERT Exemplar Class 11

Q13. Electronic configurations of four elements A, B, C and D are given below:
(A) 1s2 2s12p6             
(B)  1 s2 2s2 2p4
(C)     1s2 2s2 2p6 3s1                                  

(D)    Is2 2s2 2p5

Which of the following is the correct order of increasing tendency to gain electron?

(a) A < C < B < D
(b)     A < B < C < D
(c)    D < B < C < A                                 
(d)     D < A< B < C

Sol. (a) A – Is2 2s2 2p6 – Noble gas configuration

B -1s2 2s2 2p4 – 2 electrons short of stable configuration

C – 1s2 2s2 2p6 3.?1 – Requires one electron to complete 5-orbital

D -1s2 2s2 2p5 – Requires one electron to attain noble gas configuration

  • Noble gases have no tendency to gain electrons since all their orbitals are completely filled. Thus, element A has the least electron gain enthalpy.
  • Since element D has one electron less and element B has two electrons less than the corresponding noble gas configuration, hence, element D has the highest electron gain enthalpy followed by element B.
  • Since, element C has one electron in the 5-orbital and hence needs one more electron to complete it, therefore, electron gain enthalpy of C is less than that of element B. Combining all the facts given above, the electron gain enthalpies of the four elements increase in the order A < C < B < D.

More than One Correct Answer Type
Q14. Which of the following elements can show covalency greater than 4?
(a) Be (b) P (c) S (d) B
Sol: (b, c) P and S have d-orbitals in their valence shell and therefore, can accommodate more than 8. electrons in their respective valence shells. Hence they show covalency more than 4.

Q15. Those elements impart colour to the flame on heating in it, the atoms of which require low energy for the ionization (i.e., absorb energy in the visible region of spectrum). The elements of which of the following groups will impart colour to the flame?
(a) 2 (b) 13 (c) 1 (d) 17
Sol: (a, c) The elements of group 1 (alkali metals) and group 2 (alkaline earth metals) have low ionization enthalpies. Therefore, they impart colour to flame.

Q16. Which of the following sequences contain atomic numbers of only representative elements?
(a) 3, 33, 53, 87
(b) 2, 10, 22, 36
(c) 7, 17,25,37,48
(d) 9,35,51,88
Sol: (a, d) Elements of 5 and p-block elements are called representative elements. Elements of f-block (Z=21 – 30; 39 – 48; 57 and 72 – 80; 89 and 104 – 112) are called transition elements while those of f-block (with Z = 58-71 and Z = 90 – 103) are called inner transition elements.
(a) 3 – Group 1, 33 – group 15, 53 – group 17 and 87 – group 1.
(d) 9 – Group 17, 35 – Group 17, 51 – Group 15, 88 – Group 2.

Q17. Which of the following elements will gain one electron more readily in comparison to other elements of their group?
(a) S (g) (b) Na (g) (c) O (g) (d) Cl(g)
Sol:(a, d) Chlorine has highest tendency to gain an electron and its electron gain enthalpy (-ve) is high. O and S belong to group 16 but S has larger tendency to accept electron.

Q18. Which of the following statements are correct?
(a) Helium has the highest first ionization enthalpy in the periodic table.
(b) Chlorine has less negative electron gain enthalpy than fluorine.
(c) Mercury and bromine are liquids at room temperature.
(d) In any period, atomic radius of alkali metal is the highest.
Sol: (a, c, d) Chlorine has more negative electron gain enthalpy than fluorine due to bigger size and lesser electronic repulsion.

Q19. Which of the following sets contain only isoelectronic ions?
(a) Zn2+, Ca2+, Ga3+, Al3+                       
(b) K+, Ca2+, Sc3+, Cl
(c) P3-, S2- Cl,K+                          

(d) Ti4+, Ar, Cl3+, V5+
Sol:
(b, c)
(a) Zn2+ (30 – 2 = 28), Ca2+ (20 – 2 = 18), Ga3+ (31-3= 28), Al3+ (13 – 3 = 10) are not isoelectronic.
(b) K+ (19 – 1 = 18), Ca2+ (20 – 2 = 18), Sc3+ (21 – 3 = 18), Cl (17 + 1 = 18) are isoelectronic.
(c) P3- (15 + 3 = 18), S2- (16 + 2 = 18), Cl (17 + 1 = 18), K+ (19 – 1 = 18) are isoelectronic.
(d) Ti4+ (22 – 4 = 18), Ar (18), Cr3+ (24 – 3 = 21), V5+ (23 – 5 = 18) are not isoelectronic.

Q20. In which of the following options order of arrangement does not agree with the variation of property indicated against it?
(a) Al3+ < Mg2+ < Na+ < F (increasing ionic size)
(b) B < C < N < O (increasing first ionization enthalpy)
(c) I < Br < Cl < F (increasing electron gain enthalpy)
(d) Li < Na < K < Rb (increasing metallic radius)
Sol: (b, c) For increasing first ionization enthalpy, the order should be:
B < C < O < N
For increasing electron gain enthalpy, the order should be:
I < Br < F < Cl

Q21. Which of the following have no unit?
(a) Electronegativity (b) Electron gain enthalpy
(c) Ionisation enthalpy (d) Metallic character
Sol: (a, d) Electron gain enthalpy and ionization enthalpy have units of enthalpy.

Q22. Ionic radii vary in
(a) inverse proportion to the effective nuclear charge.
(b) inverse proportion to the square of effective nuclear charge.
(c) direct proportion to the screening effect.
(d) direct proportion to the square of screening effect.
Sol: (a, c) Ionic radii decreases as the effective nuclear charge increases.
Class 11 Chemistry NCERT Ch 3

Q23. An element belongs to 3rd period and group-13 of the periodic table. Which of the following properties will be shown by the element?
(a) Good conductor of electricity
(b) Liquid, metallic
(c) Solid, metallic  
(d) Solid, non metallic
Sol:(a, c) The element belonging to 3rd period and 13th group is aluminium which is a metal. Hence, it is solid, metallic and good conductor of electricity.

Short Answer Type Questions
Q24. Explain why the electron gain enthalpy of fluorine is less negative than that of chlorine.
Sol: In fluorine, the new electron to be added goes to 2p-subshell while in chlorine, the added electron goes to 3p-subshell. Since the 2p-subshell is relatively small as compared to 3p-subshell, the added electron in small 2p-subshell experiences strong interelectronic repulsions in comparison to that in 3p-subshell in Cl. As a result, the incoming electron does not feel much attraction from the nucleus and therefore, the electron gain enthalpy of F is less negative than that of Cl.

Q25. All transition elements are d-block elements, but all d-block elements are not transition elements. Explain.
Sol: All the elements in between s- and p-block, i.e., between group 2 and 13 are called transition elements. Elements in which the last electron enters the d-orbitals of their respective penultimate shells are called d-block elements. According to this definition, Zn, Cd and Hg cannot be regarded as d-block elements because last electron in these elements enters the s-orbital of their outermost shells rather than d-orbital of their penultimate shells. Therefore, these elements should not be regarded as d-block elements. However, the properties of these elements resemble with cf-block elements. Therefore, to make the study of periodic classification of elements more rational, these are studied along with d-block elements. Thus, on the basis of properties, all the transition elements are d-block elements but on the basis of electronic configuration, all d-block elements are not transition elements.

Q26. Identify the group and valency of the element having atomic number 119. Also predict the outermost electronic configuration and write the general formula of its oxide.
Sol: The present set up of the Long Form of the Periodic Table can accommodate at the maximum 118 elements. After this, according to Aufbau principle, 8.y-orbital should be filled. Therefore, the outer electronic configuration of element having atomic number 119 will be 8s1. Since it has one electron in the outermost s-orbital, its valency will be 1 and it should belong to group 1 along with alkali metals. The general formula of its oxide will be M20, where M represents the element.

Q27. Ionisation enthalpies of elements of second period are given below:
Ionisation enthalpy/kJ mol-1: 520, 899, 801, 1086, 1402, 1314, 1681, 2080. Match the correct enthalpy with the elements and complete the graph given in figure. Also write symbols of elements with their atomic number.
Chemistry Class 11 NCERT Solutions Ch 3

Sol: To match the correct enthalpy with the elements and to complete the graph, the following points are taken into consideration. As we move from left to right across a period, the ionization enthalpy keeps on increasing due to increased nuclear charge and simultaneous decrease in atomic radius. However, there are some exceptions given below-                                                                                                                    –

  • In spite of increased nuclear charge, the first ionisation enthalpy of B is lower than that of Be. This is due to the presence of fully filled 2s-orbital of Be [1s22s2] which is a stable electronic arrangement. Thus, higher energy is required to knock out the electron from fully filled 2.v-orbital. While B [1s2 2s2 2p1] contains valence electrons in 2s and 2p-orbitals. It can easily lose its one e from 2p-orbital in order to achieve noble gas configuration. Thus, first ionisation enthalpy of B is lower than that of Be.
    Since the electrons in 2s-orbital are more tightly held by the nucleus than those present in 2p-orbital, therefore, ionisation enthalpy of B is lower than that of Be.
  • The first ionisation enthalpy of N is higher than that of O though the nuclear charge of O is higher than that of N. This is due to the reason that in case of N, the electron is to be removed from a more stable, exactly half-filled electronic configuration (1s2 2s2 21x2ply 2p1z) which is not present in O (1s2 2s2 2p2x 2p1y 2p1z).
    Therefore, the first ionisation enthalpy of N is higher than that of O. The symbols of elements along with their atomic numbers are given in the following graph

Solutions Of Class 11 Chemistry NCERT Ch 3

Q28. Among the elements B, Al, C and Si,
(a) which element has the highest first ionization enthalpy
(b) which element has the most metallic character?
Justify your answer in each case.
Sol: Arranging the elements into different groups and periods:

Group 13 14
Period 2 B C
Period 3 Al Si
  • Ionization enthalpy increases along a period and decreases down a group. Therefore, C has the highest first ionization enthalpy.
  • Metallic character decreases along a period and increases down a group. Therefore, Al has the most metallic character.

Q29. Write four characteristic properties of p-block elements.
Sol: The four important characteristic properties of p-block elements are the following:
(a) p-Block elements include both metals and non-metals but the number of non-metals is much higher than that of metals. Further, the metallic character increases from top to bottom within a group and non-metallic character increases from left to right along a period in this block.
(b) Their ionization enthalpies are relatively higher as compared to s-block elements.
(c) They mostly form covalent compounds.
(d) Some of them show more than one (variable) oxidation states in their compounds. Their oxidizing character increases from left to right in a period and reducing character increases from top to bottom in a group.

Q30. Choose the correct order of atomic radii of fluorine and neon (in pm) out of the options given below and justify your answer.
(i) 72,160 (b) 160,160 (c) 72,72 (d) 160,72
Sol:(i) Atomic radius decreases as we move from left to right in a period in the periodic table. Fluorine has the smallest atomic radius. As we move to neon
in the same period, the atomic radius increases as it has van der Waals radius and van der Waals radii are bigger than covalent radii.

Q31. Illustrate by taking examples of transition elements and non-transition elements that oxidation states of elements are largely based on electronic configuration.
Sol: The oxidation state of an element is based on its electronic configuration. The various oxidation states of a transition metal are due to the involvement of (n-1)d and outer ns electrons in bonding.
For example, Ti (22, electronic configuration [Ar]3d24s2) can show three oxidation states (+2, +3 and +4) in various compounds like Ti02 (+4), Ti203 (+3) and TiO (+2).
The non-transition elements, mainly thep-block elements can show a number of oxidation states from +n to (n – 8) where, n is the number of electrons present in the outermost shell. For example, phosphorus can show -3, +3 and +5 oxidation states.
Lower oxidation states are ionic as the atom accepts the electron or electrons to achieve stable configuration while higher oxidation states are achieved by unpairing the paired electrons and shifting the electrons to vacant d-orbital.

Q32. Nitrogen has positive electron gain enthalpy whereas oxygen has negative. However, oxygen has lower ionization enthalpy than nitrogen explain.
Sol: The outermost electronic configuration of nitrogen is 2s22p1x2p1y2p1z. It is stable because it has exactly half filled 2p-subshell. Therefore, it has no tendency to accept extra electron and energy has to be supplied to add additional electron. Thus, electron gain enthalpy of nitrogen is slightly positive. On the other hand, the outermost electronic configuration of O is 2s22p2x2p1y2p1z. It has higher positive charge (+8) than nitrogen (+7) and lower atomic size than N. Therefore, it has a tendency to accept an extra electron. Thus, electron gain enthalpy of O is negative. However, oxygen has four electrons in the 2p subshell and can lose one electron to acquire stable half filled configuration and therefore, it has low ionization enthalpy. Because of stable configuration of N, it cannot readily lose electron and therefore, its ionization enthalpy is higher than that of O.

Q33. First member of each group of representative elements (i.e., s and p-block elements) shows anomalous behaviour. Illustrate with two examples.
Sol:First member of each group of s- and p-block elements shows anomalous behaviour due to the following reasons:
(i) Small size
(ii) High ionization enthalpy
(iii) High electronegativity
(iv) Absence of d-orbitals
Examples: Li in the first group shows different properties from the rest of elements like covalent nature of its compounds, formation of nitrides.

Similarly, beryllium, the first element of second group differs from its own group in the following ways:

  • Beryllium carbide reacts With water to produce methane gas while carbides of other elements give acetylene.
  • Beryllium shows a coordination number of four while other elements show a coordination number of six.

Q34. p-Block elements form acidic, basic and amphoteric oxides. Explain each property by giving two examples and also write the reactions of these oxides with wa

NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties

NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties-1

 

Q35. How would you explain the fact that first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Sol: Electronic configuration of Na is 1s22s22p63s1. After losing one electron from its outermost shell, sodium easily attains stable electronic configuration (1s22s22p6 ), while magnesium does not lose its electron easily due to presence of two electrons in s-orbital (ls22s22p63s2). Hence first ionisation energy of sodium is less than magnesium.
When one electron is removed from Na and Mg, their configurations become ls22s22p6 and 1s22s22p63s1 respectively. Now it is easier to remove one electron from 3s of Mg+ than 2p6 of Na+. Hence, second ionisation energy of Mg is less than Na.

Q36. What do you understand by exothermic reaction and endothermic reaction? Give one example of each type.
Sol:Exothermic reaction: The reaction which is accompanied with evolution of heat is known as exothermic reaction.
CaO + C02 → CaC03, ΔH= -178.3 kJ mol-1
Endothermic reaction: The reaction which is accompanied with absorption of energy is known as endothermic reaction, e.g.,
2NH3 → N2 + 3H2; ΔH = +91.8 kJ mol-1

Q37. Arrange the elements N, P, O and S in the order of
(i) increasing first ionisation enthalpy.
(ii) increasing non-metallic character.
Give reason for the arrangement assigned.
Sol: (i) S < P < O < N
Ionisation enthalpy increases from left to right in a period and decreases down the group. N has higher ionisation enthalpy than O due to extra stability of half-filled orbitals. Similarly, P has higher ionisation enthalpy than S due to half-filled orbitals.
(ii) P < S < N < O
Non-metallic character decreases down the group and increases from left to right in a period.

Q38. Explain the deviation in ionization enthalpy of some elements from the general trend by using the given figure.
NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties-2

Sol: Ionisation enthalpy increases in a period with increase in atomic number. The graph shows few exceptions and not the linear relationship. Ionisation enthalpy of Be is greater than that of B due to filled s-orbital in Be (Be – l.s2 1s2, B – 1s2 2s2 2pl).
Ionisation enthalpy of N is greater than that of O due to half-filled/?-orbitals in nitrogen.
(N – Is2 2s 22p3 O – ls 2 2s2 2p4).

Q39. Explain the following:
(a) Electronegativity of elements increases on moving from left to right in the periodic table.
(b) Ionisation enthalpy decreases in a group from top to bottom.
Sol: (a) The electronegativity generally increases on moving across a period from left to right (e.g., from Li to F in the second period). This is due to decrease in atomic size and increase in effective nuclear charge. As a result of increase in effective nuclear charge, the attraction for the outer electron and the nucleus increases in a period and therefore, electronegativity also increases.
(b) On moving down a group there is a gradual decrease in ionisation enthalpy. The decrease in ionisation enthalpy down a group can be explained in terms of net effect of the following factors:
(i) In going from top to bottom in a group, the nuclear charge increases.
(ii) There is a gradual increase in atomic size due to an additional main energy shell (n).
(iii) There is increase in shielding effect on the outermost electron due to increase in the number of inner electrons.
The effect of increase in atomic size and the shielding effect is much more than the effect of increase in nuclear charge. As a result, the electron becomes less and less firmly held to the nucleus as we move down the group. Hence, there is a gradual decrease in the ionization enthalpies in a group.

Q40. How does the metallic and non metallic character vary on moving from left to right in a period?
Sol: As we move from left to right in a period, the number of valence electrons increases by one at each succeeding element but number of shells remains same. Due to this, effective nuclear charge increases. More is the effective nuclear charge, more is the attraction between the nucleus and electron.
Hence, the tendency of the element to lose electrons decreases. This results in decrease in metallic character. Furthermore, the tendency of an element to gain electrons increases with increase in effective nuclear charge, so non- metallic character increases on moving from left to right in a period.

Q41. The radius of Na+ cation is less than that of Na atom. Give reason.
Sol: The radius of Na+ is less than Na atom because Na+ is formed by losing one energy shell.
Na – 1s2 2s2 2p6 3s1
Na+– 1s22s2 2p6

Q42. Among alkali metals which element do you expect to be least electronegative and why?  
Sol: Cs is the most electropositive or least electronegative element among the alkali metals since electronegative character decreases in a group.

Matching Column Type Questions
Q43. Match the correct atomic radius with the element.

Column I (Element) Column II (Atomic radius (pm)
Be 74
C 88
0 111
B 77
N 66

Sol: All the given elements are of same period and along period, atomic radii decrease because effective nuclear charge increases. Thus, the order of atomic radii is
O < N < C < B < Be or, Be = 111 pm, O = 66 pm, C = 77 pm, B = 88 pm, N = 74 pm.

Q44. Match the correct ionization enthalpies and electron gain enthalpies of the following elements.
NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties-3
Solu:
NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties-4

Q45. Electronic configurations of some elements are given in column I and their electron gain enthalpies are given in column II. Match the electronic configuration with electron gain enthalpy.

Column I

(Electronic configuration)

Column II ’

(Electron gain enthalpy/kJ moL-1

(i) 1s2 2s2 2p6 A. -53
(ii) ls22s22p63sl B. -328
(iii) ls22s22p5 C. -141
(iv) 1s2 2S2 2p4 D. +48

NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties-5

Assertion and Reason Type Questions
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Q46. Assertion (A): Generally, ionization enthalpy increases from left to right in a period.
Reason (R): When successive electrons are added to the orbitals in the same principal quantum level, the shielding effect of inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus.
(a) Assertion is correct statement and reason is wrong statement.
(b) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(c) Assertion and reason both are wrong statements.
(d) Assertion is wrong statement and reason is correct statement.
Sol: (b) Ionisation enthalpy increases from left to right across the period due to decrease in atomic size.
The electrons present within the subshell has almost same effective nuclear charge.

Q47. Assertion (A): Boron has a smaller first ionization enthalpy than beryllium. Reason (R): The penetration of a 2s electron to the nucleus is more than the 2p electron, hence, 2p electron is more shielded by the inner core of electrons that the 2s electrons.
(a) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
(b) Assertion is correct statement but reason is wrong statement.
(c) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(d) Assertion and reason both are wrong statement.

Sol: (c) Boron (1s22s22p1) has less first ionisation enthalpy than beryllium (1s22s2) because beryllium has fully filled .s-subshell.
2s-electrons are nearer to the nucleus as compared to 2s-electrons.

Q48. Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.
Reason (R): Size of the atom increases on going down the group and the added electron would be farther from the nucleus.
(a) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
(b) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(c) Assertion and reason both are wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
Sol:(b) On moving down the group, the electron gain enthalpy becomes less negative because on moving down the group the atomic size increases and the added electron lies away from the nucleus.

Long Answer Type Questions
Q49. Discuss the factors affecting electron gain enthalpy and the trend in its variation in the periodic table.
Sol: Factors affecting electron gain enthalpy:
(i) Nuclear charge: The electron gain enthalpy become more negative as the nuclear charge increases. This is due to greater attraction for the incoming electron if nuclear charge is high.
(ii) Size of the atom: With the increase in size of the atom, the distance between the nucleus and the incoming electron increases and this results in lesser attraction. Consequently, the electron gain enthalpy become less negative with increase in size of the atom of the element.
(iii) Electronic configuration: The elements having stable electronic configurations of half filled and completely filled valence subshells show very small tendency to accept additional electron and thus electron gain enthalpies are less negative.

Variation of electron gain enthalpies in periodic table:
Electron gain enthalpy, in general, becomes more negative from left to right in a period and becomes less negative as we go from top to bottom in a group.
(a) Variation down a group: On moving down a group, the size and nuclear charge increases. But the effect of increase in atomic size is much more pronounced than that of nuclear charge and thus the additional electron feels less attraction by the large atom. Consequently, electron gain enthalpy becomes less negative. This is clear from decrease of electron gain enthalpy in going from chlorine to bromine and to iodine.
(b) Variation along a period: On moving across a period, the size of the atom decreases and nuclear charge increases. Both these factors result in greater attraction for the incoming electron, therefore, electron gain enthalpy, in general, becomes more negative in a period from left to right. However, certain irregularities are observed in the general trend. These are mainly due to the stable electronic configurations of certain atoms.
Important Trends in Electron Gain Enthalpies:
There are some important features of electron gain enthalpies of elements. These are:
(i) Halogens have the highest negative electron gain enthalpies.
(ii) Electron gain enthalpy values of noble gases are positive while those of Be, Mg, N and P are almost zero.
(iii) Electron gain enthalpy of fluorine is unexpectedly less negative than that of chlorine.

Q50. Define ionization enthalpy. Discuss the factors affecting ionisation enthalpy of the elements and its trends in the periodic table.
Sol: Ionisation enthalpy: It is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom.

M(g) →M+(g) + eI1  = First ionisation enthalpy

Similarly, second and third electrons are also removed by providing successive ionisation enthalpies.

Factors on which Ionisation Enthalpy Depends:
(i) Size of the atom: The larger the atomic size, smaller is the value of ionisation enthalpy. In a larger atom, the outer electrons are far away from the nucleus and thus force of attraction with which they are attracted by the nucleus is less and hence can be easily removed.

Ionisation enthalpy ∝ 1/ Atomic size

(ii) Screening effect: Higher the screening effect, the lesser is the value of ionisation enthalpy as the screening effect reduces the force of attraction towards nucleus and hence the outer electrons can be easily removed.

Ionisation enthalpy ∝  1/ Screening effect

(iii) Nuclear charge: As the nuclear charge increases among atoms having same number of energy shells, the ionisation enthalpy increases because the force of attraction towards nucleus increases.

Ionisation enthalpy ∝ Nuclear charge

(iv) Half filled and fully filled orbitals: The atoms having half filled and fully filled orbitals are comparatively more stable, hence more energy is required to remove the electron from such atoms. The ionization enthalpy is rather higher than the expected value in case of such an atom.

Ionisation enthalpy ∝ Stable electronic configuration

(v) Shape of orbital: The s-orbital is more close to nucleus than the p-orbital of the same orbit. Thus, it is easier to remove electron from a p-orbital in comparison to s-orbital. In general, the ionisation enthalpy follows the following order

(s>p> d>f) orbitals of the same orbit.

Variation of ionisation enthalpy in the periodic table
ln general, the ionisation energy decreases down the group due to increase in atomic size. On the other hand, the ionisation energy increases across the period from left to right, again due to decrease in atomic size from left to right.

Q51. Justify the given statement with suitable examples “the properties of the elements are a periodic function of their atomic numbers”.
Sol. The similarities in the properties arise due to the same distribution of electrons in the outermost orbitals or electronic configuration which depends upon the atomic number. The elements present in a group or period exhibit similar chemical properties which depend upon the atomic number.

Q52. Write down the outermost electronic configuration of alkali metals. How will you justify their placement in group 1 of the periodic table?
NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties-6

Q53. Write the drawbacks in Mendeleev’s periodic table that led to its modification.
Sol: Drawbacks of Mendeleev’s periodic table:
1. Position of hydrogen: Hydrogen is placed in group I. However, it resembles the elements of group I (alkali metals) as well as the elements of group VILA, (halogens). Therefore, the position of hydrogen in the periodic table is not correctly defined.
2. Anomalous pairs: In certain pairs of elements, the increasing order of atomic masses was not obeyed. In these cases, Mendeleev placed elements according to similarities in their properties and not in increasing order of their atomic masses. For example, argon (Ar, atomic mass 39.9) is placed before potassium (K, atomic mass 39.1). Similarly, cobalt (Co, atomic mass 58.9) is placed before nickel (Ni, atomic mass 58.6) and tellurium (Te, atomic mass 127.6) is placed before iodine (I, atomic mass 126.9). These positions were not justified.
3. Position of isotopes: Isotopes are the atoms of the same element having different atomic masses but same atomic number. Therefore, according to Mendeleev’s classification, these should be placed at different places depending upon their atomic masses. For example, isotopes of hydrogen with atomic masses 1,2 and 3 should be placed at three places. However, isotopes have not been given separate places in the periodic table.
4. Some similar elements are separated and dissimilar elements are grouped together: In the Mendeleev’s periodic table, some similar elements were placed in different groups while some dissimilar elements
had been grouped together. For example, copper and mercury resembled
in their properties but they had been placed in different groups. At the same time, elements of group IA such as Li, Na and K were grouped with copper (Cu), silver (Ag) and gold (Au), though their properties are quite different.
5. Cause of periodicity: Mendeleev did not explain the cause of periodicity among the elements.
6. Position of lanthanoids (or lanthanides) and actinoids (or actinides):
The fourteen elements following lanthanum (known as lanthanoids, from atomic number 58-71) and the fourteen elements following actinium (known as actinoids, from atomic number 90 – 103) have not been given separate places in Mendeleev’s table.
In order to cover more elements, Mendeleev modified his periodic table.

Q54. In what manner is the long form of periodic table better than Mendeleev’s
periodic table? Explain with examples.
Sol: Superiority of the Long form of the Table over Mendeleev’s Table:
(i) This table is based on a more fundamental property, i.e., atomic number.
(ii) It correlates the position of the elements with their electronic configurations more clearly.
(iii) The completion of each period is more logical. In a period as the atomic number increases, the energy shells are gradually filled up until an inert gas configuration is reached. It eliminates the even and odd series of IV, V, VI and VII periods of Mendeleev’s periodic table.
(iv) The position of VIIIth group  is appropriate in this table. All the transition elements have been brought in the middle as the properties of transition elements are intermediate between s- and p-block elements.
(v) Due to separation of two subgroups, dissimilar elements do not fall together. One vertical column accommodates elements with same electronic configuration thereby showing same properties.
(vi) In this table, a complete separation between metals and non-metals has been achieved. The non-metals are present in the upper right comer of the periodic table.
(vii) There is a gradual change in properties of the elements with increase in their atomic numbers, i.e., periodicity of properties can be easily visualized. The same properties o recurrence in properties occur after the intervals of 2, 8, 8, 18, 18 and 32 elements which indicates the capacity of various periods of the table.
(viii) This arrangement of elements is easy to remember and reproduce.

Q55. Discuss and compare the trend in ionization enthalpy of the elements of group 1 with those of group 17 elements.
Sol: The ionization enthalpies decrease regularly as we move down a group from one element to the other. This is evident from the values of the first ionisation enthalpies of the elements of group 1 (alkali metals) and group 17 elements as given in table and figure.

Group ‘ 1 First ionisation enthalpies (kJ moL-1) Group

17

First ionisation enthalpies (kJ mol-1)
H 1312 F 1681
Li 520 Cl 1255
Na 496 Br 1142
K 419 I 1009
Rb 403 At 917
Cs 374

NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties-7
Given trend can be easily explaihed on the basis of increasing atomic size and screening effect as follows
(i) On moving down the group, the atomic size increases gradually due to the addition of one new principal energy shell at each succeeding element. Hence, the distance of the valence electrons from the nucleus increases.
Consequently, the force of attraction by the nucleus for the valence electrons decreases and hence the ionisation enthalpy should decrease.
(ii) With the addition of new shells, the shielding or the screening effect increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionisation enthalpy should decrease.
(iii) Nuclear charge increases with increase in atomic number. As a result, the force of attraction by the nucleus for the valence electrons should increase and accordingly the ionisation enthalpy should increase.
The combined effect of the increase in the atomic size and the screening effect more than compensates the effect of the increased nuclear charges. Consequently, the valence electrons become less and less firmly held by the nuclear and hence the ionisation enthalpies gradually decrease as move down the group.

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NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds

NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds:

Section Name Topic Name
9 Coordination Compounds
9.1 Werner’s Theory of Coordination Compounds
9.2 Definitions of Some Important Terms Pertaining to Coordination Compounds
9.3 Nomenclature of Coordination Compounds
9.4 Isomerism in Coordination Compounds
9.5 Bonding in Coordination Compounds
9.6 Bonding in Metal Carbonyls
9.7 Stability of Coordination Compounds
9.8 Importance and Applications of Coordination Compounds

NCERT Solutions CBSE Sample Papers ChemistryClass 12 Chemistry

NCERT INTEXT QUESTIONS

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9.1. Write the formulas for the following coordination compounds:
(i)Tetraamminediaquacobalt(IlI) chloride
(ii)Potassium tetracyanidonickelate(II)
(iii)Tris(ethanp-1,2-diamine) chromium(III) chloride
(iv)Amminebromidochloridonitrito-N- platinatc(II)
(v)Dichloridobis(ethane-l ,2-diamine) platinum (IV) nitrate
(vi)Iron(III)hexacyanidoferrate(II)
Ans: (i) [CO(NH3)4(H2O)2]Cl3.
(ii)K2[Ni(CN)4]
(iii)[Cr(en)3]Cl3
(iv)[Pt (NH3) Br Cl (N02)]
(v)[PtCl2(en)2](N03)2
(vi)Fe4[Fe(CN)6]3

9.2. Write IUPAC names of following co-ordination compounds :
(a) [CO(NH3)6]Cl3
(b) [CO(NH3)Cl]Cl2
(C) K3[Fe(CN)6]
(d) [K3[Fe(C2O4)3]
(e) K2[PdCl4]
(f) [Pt(NH3)2ClNH2CH3]Cl. (C. B. S. E. Delhi2013)
Ans:
(a) hexaamminecobalt (III) chloride
(b) pentaamminechloridocobalt (III) chloride
(c) potassium hexacyanoferrate (III)
(d) potassium trioxalatoferrate (III)
(e) potassium tetrachloridoplatinum (II)
(f) diamminechlorido (methylamine) platinum(II) chloride.

9.3. Indicate the types of isomerism exhibited by the . . following complexes and draw the structures for these isomers:
(i)K[Cr(H2O)2(C2O4)2]
(ii)[CO(en)3]Cl3
(iii)[CO(NH3)5(NO2)(NO3)2], .
(iv)[Pt(NH3)(H2O)Cl2]
Ans: (i)(a) geometrical isomerism (cis and tram)
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q3

NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q3.1

NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q3.2

9.4. Give evidence that [Co(NH3)5Cl]S0and [Co(NH3)5S04]Cl  are ionisation isomers.
Ans: When dissolved in water, they give different ions in solution which can be tested by adding AgN03 solution and BaCl2 solution, i.e.,
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q4

9.5. Explain on the basis of valence bond theory that [Ni(CN)4]2- ion with square planar structure is diamagnetic and [NiCl4]2- ion with tetrahedral geometry is paramagnetic. (Rajasthan Board 2012)
Ans: Outer electronic configuration of nickel (Z = 28) in ground state is 3d84s2. Nickel in this complex is in + 2 oxidation state. It achieves + 2 oxidation state by the loss of the two 4s-electrons. The resulting Ni2+ ion has outer electronic configuration of 3d8. Since CN ion is a strong field, under its attacking influence, two unpaired electrons in the 3d orbitals pair up.
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q5
Outer electronic configuration of nickel (Z = 28) in ground state is 3d84s2 Nickel in this complex is in + 2 oxidation state. Nickel achieves + 2 oxidation state by the loss of two 4s-electrons. The resulting Ni2+ ion has outer electronic configuration of 3d8. Since CP ion is a weak field ligand, it is not in a position to cause electron pairing.
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q5.1

9.6. [NiCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedraL Why?
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q6

9.7. [Fe(H2O)6]3+is strongly paramagnetic whereas [Fe(CN)6]3-is weakly paramagnetic. Explain.
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q7

NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q7.1

9.8. Explain[CO(NH3)6]2+ is an inner orbital complex.whereas [Ni(NH3)6]2+ is an outer orbital complex.
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q8

9.9. Predict the number of unpaired electrons in the square planar [Pt(CN)4]2- ion.
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q9

9.10. The hexaaquamanganese (II) ion contains five unpaired electrons while the hexacyano ion contains only one unpaired electron. Explain using crystal field theory.
Ans: Mn(II) ion has 3d5 configuration. In the presence of H2O molecules acting as weak field ligands, the distribution of these five electrons is \({ t }_{ 2 }^{ 3 }\)ge2 i. e., all the electrons remain unpaired to form a high spin complex. However, in the presence of CN acting as strong field ligands, the distribution of these electrons is \({ t }_{ 2g }^{ 5 }{ e }_{ g }^{ 0 }\) i.e., two t2g orbitals contain paired electrons while the third t2g orbital contains one unpaired electron. The complex formed is a low spin complex.

9.11. Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)42+ ion, given that β4 for this complex is 2.1 x 1013.
Ans: Overall stability constant (β4) = 2.1 x 1013.
Thus, the overall dissociation constant is
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Intext Questions Q11

NCERT EXERCISES

9.1. Explain the bonding in coordination compounds in terms of Werner’s postulates.
Ans: The main postulates of Werner’s theory of coordination compounds are as follows:
(a)Metals possess two types of valencies called
(i) primary valency which are ionisable; (ii) secondary valency which are non- ionisable
(b)Primary valency is satisfied by the negative ions and it is that which a metal exhibits in the formation of its simple salts.
(c)Secondary valencies are satisfied by neutral ligand or negative ligand and are those which metal exercises in the formation of its complex ions. Every cation has a fixed number of secondary valencies which are directed in space about central metal ion in certain fixed directions, e.g„ In CoCl3-6NH3, valencies between Co and Cl are primary valencies and valencies between Co and NH3 are secondary. In COCl3-6NH3 , six ammonia molecules linked to Co by secondary valencies are directed to six corners of a regular octahedron and thus account for structure of COCl3-6NH3 as follows:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q1
In modern theory, it is now referred as coordination number of central metal atom or ion.

9.2. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain why?
Ans: When FeSO4 and (NH4)2SO4 solutions are mixed in 1 : 1 molar ratio, a double salt known as Mohr’s salt is formed. It has the formula FeSO4.(NH4)2SO4.6H2O. In aqueous solution, the salt dissociates as :
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q2
The solution gives the tests for all the ions including Fe2+ ions. On the other hand, when CuSO4 and NH3 are mixed in the molar ratio of 1 : 4 in solution, a complex [Cu(NH3)4]SO4 is formed. Since the Cu2+ ions are a part of the complex entity (enclosed in square bracket), it will not give their characteristic tests as are given by Fe2+ ions.

9.3. Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Ans: Coordination entity: It constitutes of a central atom/ion bonded to fixed number of ions or molecules by coordinate bonds e.g. [COCl3(NH3)3], [Ni (CO)4] etc.
Ligand : The ions/molecules bound to central atom/ion in coordination entity are called ligands. Ligands in above examples are CL, NH3, CO Coordination number : This is the number of bond formed by central atom/ion with ligands. Coordination polyhedron : Spatial arrangement of ligands defining the shape of complex. In above cases Co and Ni polyhedron are octahedral and tetrahedral in [CoCl3 (NH3)3] and [Ni(CO)4] respectively.
Homoleptic : Metal is bound to only one kind of ligands eg Ni in[Ni(CO)4]
Heteroletric Metal is bound to more than one kind of ligandseg Coin [CoCl3(NH3)3]

9.4. What is meant by unidentate didentate and ambidentate ligands? Give two examples for each.
Ans: A molecule or an ion which has only one donor atom to form one coordinate bond with the central metal atom is called unidentate ligand, e.g,, Cl- and NH3.
A molecule or ion which contains two donor atoms and hence forms two coordinate bonds with the central metal atom is called adidentate
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q4
A molecule or an ion which contains two donor atoms but only one of them forms a coordinate bond at a time with the central metal atom is called
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q4.1

9.5. Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H2O)(CN)(en)2]2+ (ii) [CoBr2(en)2]+ (iii) [PtCl4]2- (iv) K3[Fe(CN)6] (v) [Cr(NH3)3CI3]
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q5

9.6. Using IUPAC norms, write the formulae for the following : (C.B.S.E. Foreign 2015)
(a) tetrahydroxozincate(II)
(b) hexaammineplatinum (TV)
(c) potassiumtetrachloridopalladate(II)
(d) tetrabromidocuprate (II)
(e) hexaaminecobalt(III) sulphate
(f) potassiumtetracyanonicklate (II)
(g) potassiumtrioxalatochromate(III)
(h) pentaamminenitrito-O-cobalt(III)
(i) diamminedichloridoplatinum(II)
(j) pentaamminenitrito-N-cobalt (III). (C.B.S.E. Delhi 2012)
Ans: (a) [Zn(OH)4]2-
(b) [Pt(NH3)6]4+
(c) K2[PdCl4]
(d) [Cu(Br)4]2-
(e) [CO(NH3)6]2 (SO4)3
(f) K2[Ni(CN)4]
(g) K3 [Cr(OX)3]
(h) [CO(NH3)5ONO]2+
(i) [Pt(NH3)2Cl2]
(j) [CO(NH3)5NO2]2+.

9.7. Using IUPAC norms write the systematic names of the following:
(i) [Co(NH3)6]CI3,
(ii)[Pt(NH3)2CI (NH2CH3)] Cl
(iii) [Ti(H20)6]3+
(iv) [Co(NH3)4Cl(N02)]CI
(v)|Mn(H20)6]2+
(vi)[NiCl4]2-
(vii)[Ni(NH3)6]CI2
(viii)[Co(en)3]3+
(ix) [Ni(CO)4]
Ans: (i) Hexaammine cobalt (III) chloride.
(ii) Diammine chlorido (methylamine) platinum (II) chloride.
(iii) Hexaaquatitanium (III) ion.
(iv) Tetraammine chlorido nitrito-N-cobalt (IV) chloride.
(v)Hexaaquamanganese (II) ion.
(vi)Tetrachloridonickelate (II) ion.
(vii)Hexaammine nickel (II) chloride.
(viii)Tris (ethane -1,2-diamine) cobalt (III) ion.
(ix) Tetra carbonyl nickel (0).

9.8. List various types of isomerism possible for coordination compounds, giving an example of each.
Ans: Coordination compounds exhibit stereo isomerism and structural isomerism.
Two types of stereoisomerism and their examples are as follows.
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q8

9.9. How many geometrical isomers are possible in . the following coordination entities?
(i) [Cr(C2O4)3]3- (ii) [CoCl3(NH3)3]
Ans: (i) [Cr(C2O4)3]3- => No geometrical isomers
are possible in this coordination entity.
(ii) [Co(NH3)3 Cl3] => Two geometrical isomers are possible (fac and mer) in this coordination entity.

9.10. Draw the structures of optical isomers of
(i) [Cr(C2O4)3]3-
(ii)[PtCI2(en)2]2+
(iii)[Cr(NH3)2CI2(en)]+
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q10

NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q10.1

9.11. Draw all the isomers (geometrical and optical) of
(i)[CoCl2(en)2]+
(ii)[Co(NH3) Cl (en)2]2+
(iii) [Co(NH3)2Cl2(en)]+
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q11

9.12. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl) (Py)] and how many of these will exhibit optical isomerism?
Ans: Three isomers of[Pt(NH3)(Br)(Cl)(Py)] are possible.
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q12
These are obtained by keeping the position of one of the ligand, say NH3 fixed and rotating the positions of others. This type of isomers do not show any optical isomerism. Optical isomerism only rarely occurs in square planar or tetrahedral complexes and that too when they contain unsymmetrical chelating ligand.

9.13.  Aqueous copper sulphate solution (blue in colour) gives: (i) a green precipitate with aqueous potassium fluoride and (ii)a bright green solution with aqueous potassium chloride. Explain these experimental results.
Ans: Aqueous CuS04 solution exists as [Cu(H20)4]S04 which has blue colour due to [Cu(H20)4]2+ ions.
(i) When KF is added, the weak H20 ligands are replaced by F ligands forming [CUF4]2- ions which is a green precipitate.
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q13
(ii)When KCl is added, Cl ligands replace the weak H20 ligands forming [CuCl4]2- ion which has bright green colour.
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q13.1

9.14. What is the coordination entity formed when excess of aqueons KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S (g) is passed through this solution?
Ans: First cupric cyanide is formed which decomposes to give cuprous cyanide and cyanogen gas. Cuprous cyanide dissolves in excess of potassium cyanide to form the complex, K3[Cu(CN)4],
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q14
Thus, coordination entity formed in the above reaction is [Cu(CN)4]3-. As CN is a strong ligand, the complex ion is highly stable and does not dissociate/ionize to give Cu2+ ions. Hence, no precipitate,with H2S is formed.

9.15.  Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)6]4-
(ii) [FeF6]3-
(iii) [Co(C2O4)3]3-
(iv) [CoF6]3-
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q15

NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q15.1

9.16. Draw figure to show the splitting of d-orbitals in an octahedral crystal field.
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q16

9.17. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Ans: The crystal field splitting, ∆0, depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong fields in which, the splitting will be large whereas others produce weak fields and consequently result in small splitting of d-orbitals. In general, ligands can be arranged in a series in the order of increasing field strength as given below :
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q17

9.18. What is crystal field splitting energy? How does the magnitude of Δ0 decide the actual configuration of d-orbitals in a coordination entity?
Ans: When the ligands approach a transition metal ion, the d-orbitals split into two sets, one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy (Δ0 for octahedral field). If Δ0 < P (pairing energy), the fourth electron enters
one of the e°g, orbitals giving the configuration t32ge1g, thus forming high spin complexes. Such ligands for which Δ0 < P are called weak field ligands. If Δ0 > P, the fourth electron pairs up in one of the t2g orbitals giving the configuration t42ge1g thereby forming low spin complexes. Such ligands for which Δ0> P are called strong field ligands.

9.19. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q19

9.20. A solution of [Ni(H20)6]2+ is green but a solution of [Ni(CN)4]2-is colourless. Explain.
Ans: In [Ni(H20)6]2+, Ni is in + 2 oxidation state and having 3d8 electronic configuration, in which there are two unpaired electrons which do not pair in the presence of the weak H20 ligand. Hence, it is coloured. The d-d transition absorbs red light and the complementary light emitted is green.
In [Ni(CN)4]2- Ni is also in + 2 oxidation state and having 3d8 electronic configuration. But in presence of strong ligand CN the two unpaired electrons in the 3d orbitals pair up. Thus, there is no unpaired electron present. Hence, it is colourless.

9.21. [Fe(CN)6]4- and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
Ans: In both the complexes, Fe is in + 2 oxidation state with d6 configuration. This means that it has four unpaired electrons.Both CN ion and H2O molecules which act as ligands occupy different relative positions in the spectrochemical series. They differ in crystal field splitting energy (∆0). Quite obviously, they absorb radiations corresponding to different wavelengths/frequencies from the visible region of light. (VIBGYOR) and the transmitted colours are also different. This means that the complexes have different colours in solutions.

9.22. Discuss the nature of bonding in metal carbonyls.
Ans: In metal carbonyl, the metal carbon bond (M – C) possess both the σ and π -bond character. The bond are formed by overlap of atomic orbital of metal with that of C-atom of carbon monoxide in following sequence:
(a)σ -bond is first formed between metal and carbon when a vacant d-orbital of metal atom overlaps with an orbital containing lone pair of electrons on C-atom of carbon monoxide (: C = O:)
(b)In addition to σ -bond in metal carbonyl, the electrons from filled d-orbitals of a transition metal atom/ ion are back donated into anti bonding π-orbitals of carbon monoxide. This stabilises the metal ligand bonding. The above two concepts are shown in following figure:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q22

9.23. Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:
(i)K3[CO(C2O4)3I (ii) cis-[Cr(en)2Cl2]Cl (iii) (NH4)2[CoF4] (iv) [Mn(H20)6]SO4
Ans: (i) K3[Co(C2O4)3] =>[CO(C204)3]3-. x + 3 (-2) = -3 . Oxidation state, x=+3 Coordination number is also 6 as C2042- is didentate. Co+3 is a case in which all electrons are paired
(ii) cis – [Cr(en)2Cl2]+ Cl
x + 0—2 =+1
Oxidation state, x =+3
Coordination number is 6 as ‘en’ is didentate. Cr3+ is a cfi case, paramagnetic.
(iii) (NH4)2[COF4] = (NH4)22+[COF4]2-
x —4 =—2.
Oxidation state, x = + 2
Coordination number=4.
Co2+ is a d5 case, paramagnetic
(iv)[Mn(H20)6]2+S042-
x+0f+2
.•. Oxidation state, x- + 2
Coordination number is 6.
Mn+2 is a dcase, paramagnetic

9.24. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H20)2(C204)2|-3H20 (ii) [Co(NH3)5CIlCl2 (iii) CrCI3(Py)3
(iv)Cs[FeCl4] (v)K4|Mn(CN)6|
Ans: (i) K[Cr(H20)2(C204)2|-3H20 IUPAC name is potassiumdiaquadioxalatochromate (III) hydrate.
Coordination number = 6
Oxidation state of Cr: x + 0 + 2 (-2) = – 1
.‘. x = + 3
Shape is octahedral Electronic configuration of Cr3+ = 3d3=t32gg .
Magnetic moment,
\(\mu =\sqrt { n(n+2 } =\sqrt { 3×5 } =\sqrt { 15 } BM\)
= 3-87 BM
(ii) [Co(NH3)5CIlCl2IUPAC name is pentaamminechloridocobalt (III) chloride Coordination number of Co = 6 Shape is octahedral.
Oxidation state of Co, x + 0 -1 = + 2 .’. x = + 3
Electronic configuration of Co3+ = 3d6 = t62gg   n=0, μ =0 .
(iii) CrCI3(Py)3. IUPAC name is
trichloridotripyridine chromium (III).
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q24

9.25. What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes.
Ans: Formation of a complex in solution is an equilibrium reaction. It may be represented as
\(M+4L\rightleftharpoons M{ \quad L }_{ 4 }\)
The equilibrium constant of this reaction is the measure of stability of the complex. Hence the equilibrium constant is also called as stability constant or Instability constant may be defined as equilibrium constant for reverse reaction. The formation of above complex may also be written in successive steps:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q25
Stability constant is written as
β4=K1K2K3K4.
Greater the stability constant, stronger is the metal-ligand bond.
The stability of complex will depend on
(a)nature of metal
(b)Oxidation state of metal
(c)Nature of ligand e g. chelating ligand form stabler complexes
(d)Greater the basic strength of the ligand, more will be the stability.

9.26. What is meant by the chelate effect? Give an example.
Ans: When a didentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a five or a six membered ring is formed, the effect is called chelate effect. For example,
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q26

9.27. Discuss briefly giving an example in each case the role of coordination compounds in :
(a) biological systems,
(b) analytical chemistry,
(c) medicinal chemistry, and
(d) extraction/metallurgy of metals.
Ans:
(i) Coordination compounds are of great importance in biological systems. The pigment responsible for photosynthesis, chlorophyll, is a coordination compound of magnesium. Haemoglobin, the red pigment 1 of blood which acts as oxygen carrier is a coordination compound of iron. Vitamin B12, cyanocobalamine, the anti- pernicious anaemia factor, is a coordination compound of cobalt. Among the other compounds of biological importance with coordinated metal ions are the enzymes like, carboxypeptidase A and carbonic anhydrase (catalysts of biological systems).

(ii) There is growing Interest in the use of chelate therapy in medicinal chemistry. An example is the treatment of problems caused by the presence of metals in toxic proportions in plant/animal systems. Thus, excess of copper and iron are removed by the chelating ligands D-penicillamine and desferrioxime B via the formation of coordination compounds.

EDTA is used in the treatment of lead poisoning. Some coordination compounds of platinum effectively inhibit the growth of tumours. Examples are: ds-platin and related compounds.

(iii) Coordination compounds find use in many qualitative and quantitative chemical analysis. The familiar colour reactions given by metal ions with a number of ligands (especially chelating ligands), as a result of formation of coordination entities, form the basis for their detection and estimation by classical and instrumental methods of analysis. Examples of such reagents include EDTA, DMG (dimethylglyoxime), a-nitroso- β-naphthol, cupron, etc.

(iv) Some important extraction processes of metals, like those of silver and gold, make use of complex formation. Gold, for example, combines with cyanide in the presence of oxygen and water to form the coordination entity [Au(CN)2]- in aqueous solution. Gold can be separated in metallic form from this solution by the addition of zinc.

9.28. How many ions are produced from the complex Co(NH3)6Cl2 in solution?
(i) 6
(ii) 4
(iii)3
(iv)2
Ans: Coordination number of cobalt = 6
Hence, the complex is [Co (NH3)6] Cl2. It ionizes in the solution as follows :
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q28
Thus, three ions are produced. Hence, the correct option is (iii)

9.29. Amongst the following ions? Which one has the highest magnetic moment value:
(i) [Cr(H2O)6]3+
(ii) [Fe(H20)6]2+ (iii) [Zn(H20)6]2+
Ans: The oxidation states are: Cr (III), Fe (II) and Zn (II).
Electronic configuration of Cr3+ = 3d3, unpaired electron = 3
Electronic configuration of Fe2+ = 3d6, unpaired electron = 4
Electronic configuration ofZn2+ = 3d10, unpaired electrons = 0
\(\mu =\sqrt { n\quad (n+2) }\)
where V is number of unpaiared electrons Hence, (ii) has highest value of magnetic moment.

9.30. The oxidation number of cobalt in K[Co(CO)4] is
(i)+1
(ii)+3
(iii)-1
(iv)-3
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q30

9.31. Amongst the following, the most stable complex is:
(i) [Fe(H2O)6] (ii) [Fe(NH3)6]3+
(iii) [Fe(C2O4)3]3- (iv) [FeCl6]3-
Ans: In each of the given complex, Fe is in + 3 oxidation state. As C2042-is didentate chelating ligand, it forms chelate rings and hence (iii) out of complexes given above is the most stable complex.

9.32. What will be the correct order for the wavelengths of absorption in the visible region for the following:[Ni(NO2)6]4-, [Ni(NH3)6]2+, [Ni(H20)6]2+?
Ans: As metal ion is fixed, the increasing field strengths, i.e., the CFSE values of the ligands from the spectro-chemical series are in the order: H20<NH3< NO2;
Thus, the energies absorbed for excitation will be in the order:
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds Exercises Q32

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NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chemistry-chapter-4-chemical-bonding-molecular-structure/

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Multiple Choice Questions
Single Correct Answer Type

Q1. Isostructural species are those which have the same shape and hybridization. Among the given species identify the isostructural pairs.
(a) [NF3 and BF3]
(b) [BF4andNH+4]
(c) [BC13 and BrCl3]                             
(d) [NH3 and N03 ]
Sol: (b) NF3 is pyramidal whereas BF3 is planar triangular.
BF4 and NH+4 ions both are tetrahedral.
BC13 is triangular planar whereas BrCl3 is pyramidal.
NH3 is pyramidal whereas N03 is triangular planar.

Q2. Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?
(a) C02
(b) HI                      
(c) H20                     
(d) S02

chemical bonding NCERT exemplar pdf Chapter 4

Q3. The types of hybrid orbitals of nitrogen in N02, N03 and NH4 respectively are expected to be
(a) sp, sp3 and sp2                                  
(b) sp, sp2 and sp3
(c) sp2,
sp and sp3                                   
(d) sp2, sp3 and sp
Sol: (b) The number of orbitals involved in hybridization can be determined by the application of formula:

chemical bonding NCERT exemplar Chapter 4

where H = number of orbitals involved in hybridization
V= valence electrons of central atom
M= number of monovalent atoms linked with central atom
C = charge on the cation
A = charge on the anion
NCERT exemplar class 11 chemistry chapter 4

Q4. Hydrogen bonds are formed in many compounds e.g., H20, HF, NH3. The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds: The correct decreasing order of the boiling points of above compounds is
(a) HF>H20>NH3                        
(b) H20>HF>NH3
(c) NH3>HF>H20                       

(d) NH3>H20>HF
Sol: (b) Strength of H-bonding depends on the electronegativity of the atom which follows the order: F > O > N .
Strength of H-bond is in the order:
H……. F > H…….. O > H…….. N
But each H20 molecule is linked to 4 other H20 molecules through H-bonds whereas each HF molecule is linked only to two other HF molecules.
Hence, correct decreasing order of the boiling points is HzO > HF > NH3.

Q5. In PO43- ion the formal charge on the oxygen atom of P – O bond is
(a) +1                      (b) -1                        (c) -0.75                    (d) +0.75
Sol: (b) Formal charge of the atom in the molecule or ion = (Number of valence electrons in free atom) – (Number of lone pair electrons + 1/2 Number of bonding electrons)

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Q6. In N03 ion, the number of bond pairs and lone pairs of electrons on nitrogen atom are
(a) 2, 2                    (b) 3, 1                      (c) 1,3                      (d) 4, 0
Sol: (d) In N-atom, number of valence electrons = 5
Due to the presence of one negative charge, number of valence electrons = 5 + 1 = 6. One O-atom forms two bonds (= bond) and two O-atom are shared with two electrons of N-atom.
Thus, 3 O-atoms are shared with 8 electrons of N-atom.
Number of bond pairs (or shared pairs) = 4
Number of lone pairs = 0

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-1
Q7. Which of the following species has tetrahedral geometry?

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-2

Q9. Which molecule/ion out of the following does not contain unpaired electrons?
(a) N+2
(b) 02                        
(c) O22-                    
(d) B2

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-3

Q10. In which of the following molecule/ion all the bonds are not equal?
(a) XeF4                 
(b) BF4                    
(c) C2H4                  
(d) SiF4
Sol:
(c) C2H4 has one double bond and four single bonds. Bond length of double bond (C = C) is smaller than single bond (C – H).

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-4

Q11. In which of the following substances will hydrogen bond be strongest?
(a) HCl
(b) H20                     
(c) HI                      
(d) H2S
Sol: (b) HC1, HI and H2S do not from H-bonds. Only H20 forms hydrogen bonds. One H20 molecule forms four H-bonding.

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-5

Q12. If the electronic configuration of an element is 1s2 2s2 2p6 3s2 3p6 3d2 4s2 , the four electrons involved in chemical bond formation will be .
(a) 3p6
(b 3p6, 4s2          
(c)  3p6, 3d2         
(d) 3d1,4s2

Sol:(d) In transition elements (n -1 )d and ns orbitals take part in bond formation.

Q13. Which of the following angle corresponds to sp2 hybridisation?
(a) 90° (b)     120°                (c)     180°               (d) 109°
Sol: (b) sp2 hybridisation gives three sp2 hybrid orbitals which are planar triangular forming an angle of 120° with each other.

The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A ls22s22p6
B ls22s22p63s23p3
C ls22s22p63s23ps

Q14. Stable form of A may be represented by the formula
(a) A
(b) A2                    
(c) A3                   
(d)  A4
Sol:
(a) The given electronic configuration shows that A represents noble gas because the octet is complete. A is neon which has 10 atomic number.

Q15. Stable form of C may be represented by the formula
(a) C
(b) C2                     
(c) C3                   
(d)  C4
Sol:
(b) The electronic configuration of C represent chlorine. Its stable form is dichlorine (Cl2), i.e., C2.

Q16. The molecular formula of the compound formed from B and C will be
(a) BC
(b) B2C                 
(c)  BC2                
(d)  BC3
Sol:
(d) B represent P and C represents Cl. The stable compound is PC13 i.e., BC3.

Q17. The bond between B and C will be
(a) ionic
(b) covalent         
(c) hydrogen
(d) coordinate
Sol: (b) Both B and C are non-metals and therefore, bond formed between them will be covalent.

Q18. Which of the following order of energies of molecular orbitals of N2 is correct?
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-6

Q19. Which of the following statements is not correct from the view point of molecular orbital theory?
(a) Be2 is not a stable molecule.
(b) He2 is not stable but He+2 is expected to exist.
(c) Bond strength of N2 is maximum amongst the homonuclear diatomic molecules belonging to the second period.
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-7
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-8

Thus, dinitrogen (N2) molecule contains triple bond and no other molecule of second period have more than double bond. Hence, bond strength of N2 is maximum amongst the homonuclear diatomic molecules belonging to the second period.
(d) It is incorrect. The correct order of energies of molecular orbitals in N2 molecule is

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-9

Q20. Which of the following options represents the correct bond order?
(a) o2>o2>o+2
(b) o2<o2<o2
(c) o2>o2<o2

(d) o2<o2>o+2

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-10

Q21. The electronic configuration of the outermost shell of the most electronegative element is
(a) 2s22p5
(b) 3s23p5                               

(c) 4s24p5                
(d) 5s25ps
Sol:
(a) The electronic configuration represents
2s22p5= fluorine = most electronegative element
3s23p5 = chlorine
4s24p5 = bromine
5s25p5 = iodine

Q22. Amongst the following elements, whose electronic configurations are given below, the one having the highest ionization enthalpy is
(a) [Ne]3s23p1
(b) [Ne]3s23p3
(c) [Ne]3s23p2

(d) [Ar]3d104s24p3
Sol:
(b) (b) and (d) have exactly half-filled p-orbitals but (b) is smaller in size than Hence, (b) has highest ionization enthalpy.

More than One Correct Answer Type

Q23. Which of the following have identical bond order?
(a) CN (b) NO+                      (c) 02                        (d) 022-
Sol:
(a, b) CN (number of electrons = 6 + 7 + 1 = 14)
NO+ (number of electrons = 7 + 8 – 1 = 14)
02 (number of electrons = 8 + 8 + 1 = 17)
022- (number of electrons = 8 + 8 + 2 = 18)
Thus, CN and NO+ because of the presence of same number of electrons, have same bond order.

Q24. Which of the following attain the linear structure?
(a) BeCl2
(b) NCO+                  
(c) N02                     
(d) CS2
Sol:
(a, d) BeCl2 (Cl – Be – Cl) and CS2 (S = C = S) both are linear, NCO+ is non-linear. However, remember that NCO(N = C = O) is linear because it is isoelectronic with C02.
N02 is angular with bond angled 132° and each O – N bond length of 1.20Ao (intermediate between single and double bond).

Q25. CO is isoelectronic with
(a) NO+
(b) N2                       
(c) SnCl2                  
(d) N02
Sol:
(a, b) Number of electrons in CO =14
Number of electrons in NO+ =14
Number of electrons in N2 = 14
Number of electrons in SnCl2 = 84
Number of electrons in N02 = 24

Q26. Which of the following species have the same shape?
(a) C02
(b) CC14                 
(c) 03                        
(d) N02
Sol:
(c, d) C02 →Linear, CC14 → Tetrahedral, 03 →Angular (V-shaped), N02 →Angular (V-shaped)

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-11

Q27. Which of the following statements are correct about CO32- ?
(a) The hybridization of central atom is sp3.
(b) Its resonance structure has one C – O single bond and two C = O double bonds.
(c) The average formal charge on each oxygen atom is 0.67 units.
(d) All C – O bond lengths are equal.

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-12

Q28. Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic?
(a) N2                     
(b) N22- 
(c) 02                      
(d) o22-

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-13
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-14

Q29. Species having same bond order are
(a) N2                      
(b) N2                       
(C) F+2                      
(d) o2

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-15

Q30. Which of the following statements are not correct?
(a) NaCl being an ionic compound is a good conductor of electricity in the solid state.
(b) In canonical structures there is a difference in the arrangement of atoms.
(c) Hybrid orbitals form stronger bonds than pure orbitals.
(d) VSEPR theory can explain the square planar geometry of XeF4.
Sol: (a, b)
(a) Ionic compounds are good conductors only in molten state or aqueous solution since ions are not furnished in solid state.
(b) In canonical structures there is a difference in arrangement of electrons.

Short Answer Type Questions
Q31. Explain the non linear shape of H2S and non planar shape of PCl3 using valence shell electron pair repulsion theory.
Sol: The Lewis structure of H2S is:

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-16
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-17

Q32. Using molecular orbital theory, compare the bond energy and magnetic character of 0+2 and O2

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-18

Q33. Explain the shape of BrF5.
Sol: Br-atom has configuration:
1s2, 2s22p6 , 3s23p63d10, 4s24p5
To get pentavalency, two of the p-orbitals are unpaired and electrons are shifted to 4d-orbitals.

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-19
In this excited state, sp3d2-hybridisation occurs giving octahedral structure. Five positions are occupied by F atoms forming sigma bonds with hybrid bonds and one position occupied by lone pair, i.e., the molecule as a square pyramidal shape.
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-20

Q34. Structures of molecules of two compounds are given below:

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-21

(a) Which of the two compounds will have intermolccular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding?
(b) The melting point of a compound depends on. among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show higher melting point.
(c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with water easily and be more soluble in it?
Sol:(a) Compound I will form intramolecular hydrogen bond because N02 and OH groups are close together whereas it is not so in compound II. Compound II will have intermolecular hydrogen bonding. Bonding in both the cases is shown below:

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-22

(b) As a large number of molecules can be linked together through intermo­lecular hydrogen bonding, compound II will show higher melting point.

(c) Due to intramolecular hydrogen bonding, compound I will not be able to form hydrogen bonds with H20 molecules. Hence, it will be less soluble in water. However, compound II can form hydrogen bonds with H20 molecules easily and hence it will be more soluble in water.

Q35. Why does type of overlap given in the following figure not result in the bond formation?
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-23

Sol: In first figure, the ++ overlap is equal to +- overlap and therefore, these cancel out and net overlap is zero.
In second figure, no overlap is possible because the two orbitals are perpendicular to each other.

Q36. Explain why PC15 is trigonal bipyramidal whereas IF5 is square pyramidal.
Sol: PC15 – The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are represented below

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-24
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-25

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-26

Sol: Dimethyl ether has larger bond angle than water. This is because there is more repulsion between bond pairs of CH3 groups attached in ether than between bond pairs of hydrogen atoms attached to oxygen in water. The carbon of CH3 group in ether is attached to three hydrogen atoms through c bonds and electron pairs of these bonds add to the electron charge density on carbon atom. Hence, repulsion between two -CH3 groups will be more than that between two H atoms.

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-27

Q38. Write Lewis structure of the following compounds and show formal charge on each atom. HN03, No2, H2so4
Sol:
Formal charge on an atom in a Lewis structure
= [total number of valence electrons in free atom] – [total number of non-bonding (lone pairs) electrons]
—1/2 [total number of bonding or shared electrons]

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-28
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-29
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-30
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-31

Q39. The energy of σ2pz: molecular orbital is greater than 2px and 2pv molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species:
N2, N+2, N2, N22+

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-32
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-33

Q40. What is the effect of the following processes on the bond order in N-, and 02?
(i) N2 → N+2 + e (ii) 02 → O+2 + e

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-34

Q41. Give reasons for the following: ‘
(a) Covalent bonds are directional bonds while ionic bonds are non- directional.
(b) Water molecule has bent structure whereas carbon dioxide molecule is linear.
(c) Ethyne molecule is linear.
Sol: (i) Since the covalent bond depends on the overlapping of orbitals between different orbitals, the geometry of the molecule is different. The orientation of overlap is different. The orientation of overlap is the factor responsible for their directional nature.
(ii) Due to presence of two lone pairs of electrons on oxygen atom in HiO the repulsion between Ip-lp is more. C02 undergoes sp hybridization resulting in linear shape (O = C = O) while H20 undergoes .sp3 hybridisation resulting in distorted tetrahedral or bent structure.

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-35

(iii) In ethyne molecule carbon undergoes sp hybridization with two unhybridised orbitals. One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C – C sigma bond while the other hybridized orbital of each carbon atom overlaps axially with S orbitals of hydrogen atoms forming σ bonds. Unhybridised orbitals form π bonds

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-36

Q42. What is an ionic bond? With two suitable examples explain the difference between an ionic and covalent bond?
Sol: An ionic bond is formed as a result of the electrostatic attraction between the positive and negative ions formed by transfer of electrons from one atom to another.

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-37

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-38

Q43. Arrange the following bonds ‘in order of increasing ionic character giving reason.
N-H, F-H, C-H and O-H

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-39

Q44. Explain why CO2-3 ion cannot be represented by a single Lewis structure. How can it be best represented?
Sol: A single Lewis structure of CO2-3  ion cannot explain all the properties of this ion. It can be represented as a resonance hybrid of the following structures:

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-40
If it were represented only by one structure, there should be two types of bonds, i.e., C = O double bond and C – O single bonds but actually all bonds are found to be identical with same bond length and same bond strength.

Q45. Predict the hybridization of each carbon in the molecule of organic compound given below. Also indicate the total number of sigma and pi bonds in this molecule.

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-41

Q46. Group the following in linear and non-linear molecules: H20, HOC1, BeCl2 C120

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-42

Q47. Elements X, Y and Z have 4, 5 and 7 valence electrons respectively, (i) Write the molecular formula of the compounds formed by these elements individually with hydrogen, (ii) Which of these compounds will have the highest dipole moment?

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-43

Q48. Draw the resonating structure of (i) Ozone molecule (ii) Nitrate ion
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-44

Q49. Predict the shapes of the following molecules on the basis of hybridization. BC13, ch4, co2, nh3
Sol:
BCl3 – sp2 hybridisation – Trigonal planar
CH4 – sp3 hybridisation – Tetrahedral .
NH3 – sp3 hybridisation – Distorted tetrahedral or Pyramidal

Q50. All the C – O bonds in carbonate ion (CO2-3) are equal in length. Explain.
Sol: Carbonate ion is represented by resonating structures as given below:

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-44

Q51. What is meant by the term average bond enthalpy? Why there is difference in bond enthalpy of O – H bond in ethanol (C2H5OH) and water?

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-45

Matching Column Type Questions
Q52. Match the species in Column I with the type of hybrid orbitals in Column II.

Column I Column II
(i) SF4 (a) sp3cf
(ii) if5 (b) d2sp3
(iii) NO2+ (c) sp3 d
(iv) NH4 (d) sp3
(e) sp

 

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-46

Q53. Match the species in Column I with the geometry/shape in Column II.

Column I Column II
(i) H30+ (a) Linear
(ii) HC = CH (b) Angular
(iii) Cl02 (c) Tetrahedral
(iv) NH+4 (d) Trigonal bipyramidal
(e) Pyramidal

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-47

Q54.  Match the species in Column I with the bond order in Column II.

Column I , . Column II
(i) NO (a) 1.5
(ii) CO (b) 2.0
(iii) o2 (c) 2.5
(iv) 02 (d) 3.0

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-48

Q55. Match the items given in Column I with examples given in Column II.

Column I Column II
(i) Hydrogen bond (a) C
(ii) Resonance (b) LiF
(iii) Ionic solid (c) H2
(iv) Covalent solid (d) HF
(e) 03

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-49

Q56. Match the shape of molecules in Column I with the type of hybridization in Column II.

Column I Column II
(i) Tetrahedral (a) sp2
(ii) Trigonal (b) sp
(iii) Linear (c) sp3

Sol: (i) →c; (ii) → a; (iii) —> b

sp3 hybridisation – Tetrahedralshape
sp2 hybridisation – Trigonal shape
sp hybridization – Linear shape

Assertion and Reason Type Questions

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Q57. Assertion (A): Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.
Reason (R): This is because sodium and chloride ions acquire octet in sodium chloride formation.
(a) A and R both are correct, and R is the correct explanation of A.
(b) A and R both are correct, but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A and R both are false.
Sol: (a) Sodium chloride (Na+CL) is stable ionic compound because both Na+ and CL ions have complete octet in outermost shell.

Q58. Assertion (A): Though the central atom of both NH3 and H20 molecules are sp3 hybridised, yet H – N – H bond angle is greater than that of H – O – H.
Reason (R): This is because nitrogen atom has one lone pair and oxygen atom has two lone pairs.
(a) A and R both are correct, and R is the correct explanation of A.
(b) A and R both are correct, but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A and R both are false.

Sol:(a)

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-50
H20 has two lone pairs while NH3 has single lone pair, hence, H20 involves greater lone pair-bond pair repulsion.

Q59. Assertion (A): Among the two O – H bonds in H20 molecule, the energy required to break the first O – H bond and other O – H bond is the same.
Reason (R): This is because the electronic environment around oxygen is the same even after breakage of one O – H bond.
(a) A and R both are correct, and R is the correct explanation of A.

(b) A and R both are correct, but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A and R both are false.
Sol: (d) Bond energy of two (-O – H) bonds in H20 will be different.

Long Answer Type Questions
Q60. (i) Discuss the significance/applications of dipole moment.
(ii) Represent diagrammatically the bond moments and the resultant dipole moment in C02, NF3 and CHCl3.

Sol: (i) Dipole moment plays very important role in understanding the nature of chemical bond. A few applications are given below:
(a) Distinction between, polar and non-polar molecules. The measurement of dipole moment can help us to distinguish between polar and non-polar molecules. Non-polar molecules have zero dipole
moment while polar molecules have some value of dipole moment.
(b) Degree of polarity in a molecule. Dipole moment measurement also gives an idea about the degree of polarity specially in a diatomic molecule. The greater the dipole moment, the greater is the polarity in such a molecule.
(c) Shape of molecules. In case of molecules containing more than two atoms, the dipole moment not only depends upon the individual dipole moments of the bonds but also on the arrangement of bonds.
.Thus, dipole moment is used to find the shapes of molecules.
(d) Ionic character in a molecule. Knowing the electronegativities of atoms involved in a molecule, it is possible to predict the nature of chemical bond formed. If the difference in electronegativities of two atoms is large, the bond will be highly polar. As an extreme case, when the electron is completely transferred from one atom to another, an ionic bond is formed. Therefore, the ionic bond is regarded as an extreme case of covalent bond. The greater the difference in electronegativities of the bonded atoms, the higher is the ionic character
(e) Distinguish between cis- and trans- isomers. Dipole moment measurements help to distinguish between cis- and trans- isomers because ds-isomer has usually higher dipole moment than trans isomer.
(f) Distinguish between ortho, meta and para isomers. Dipole moment measurements help to distinguish between o-, m- and p-isomers because the dipole moment of p-isomer is zero and that of o-isomers is more than that of m-isomer.

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-51

Q61. Use the molecular orbital energy level diagram to show that N2 would be expected to have a triple bond, F2 a single bond and Ne2 no bond.NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-52
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-53
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-54
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-55
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-56

No bond is formed between two Ne atoms or in other words, Ne2 does not exist.
Bond Order = ½ (10 – 10) = 0 (No bond)

Q62. Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can you interpret energy changes taking place in the formation of dihydrogen?
Sol: The valence bond theory was put forward by Heitler and London in 1927. It was later improved and developed by L. Pauling and J.C. Slater in 1931. The valence bond theory is based on the knowledge of atomic orbitals and electronic configurations of elements, overlap criteria of atomic orbitals and stability of molecule.
The main points of valence bond theory are
(i) Atoms do not lose their identity even after the formation of the molecule.
(ii) The bond is formed due to the interaction of only the valence electrons as the two atoms come close to each other. The inner electrons do not participate in the bond formation.
(iii) During the formation of bond, only the valence electrons from each bonded atom lose their identity. The other electrons remain unaffected.
(iv) The stability of bond is accounted by the fact that the formation of bond is accompanied by release of energy. The molecule has minimum energy at a certain distance between the atoms known as intemuclear distance. Larger the decrease in energy, stronger will be the bond formed.

Valence bond Treatment of Hydrogen Molecule:
Consider two hydrogen atoms A and B approaching each other havingnuclei Ha and HB and the corresponding electrons eA and eB respectively.
When atoms come closer to form molecules new forces begin to operate.
(a) The force of attraction between nucleus of atom and electron of another atom.
(b) The force of repulsion between two nuclei of the atom and electron of two atoms.

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-57
NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-58
Fig. (a) Two hydrogen atoms at a large distance and hence, no interaction, (b) Two hydrogen atom closer to each other atomic orbitals begin to interact, (c) Attractive and repulsive forces in hydrogen atoms when interaction begins. In case of hydrogen: Figure ‘a’ shows that two hydrogen atoms are at farthest distances and their electron distribution is absolutely symmetrical.
(a) When two hydrogen atom start coming closer to each other, the electron cloud becomes distorted and new attractive and repulsive forces begin to operate as shown in figure ‘c’
(b) In figure ‘c’ dotted lines show attractive forces present in atom already and bold lines show the new attractive and repulsive forces.
(c) It has been found experimentally that the magnitude of net attractive forces is more than net repulsive forces. Thus stable hydrogen molecule is formed.

Potential energy diagram for formation of hydrogen molecules:
When two hydrogen atoms are at farther distance, there is no force operating between them, when they start coming closer to each other, force of attraction comes into play and their potential energy starts decreasing. As they come closer to each other potential goes on decreasing, but a point is reached, when potential energy acquires minimum value.
Note:
(a) This distance corresponding to this minimum energy value is called the distance of maximum possible approach, i.e. the point which corresponds to minimum energy and maximum stability.
(b) If atoms come further closer than this distance of maximum possible approach, then potential energy starts increasing and force of repulsion comes into play and molecules starts becoming unstable.

NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure-59

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NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers

NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers:

Section Name Topic Name
15 Polymers
15.1 Classification of Polymers
15.2 Types of Polymerisation Reactions
15.3 Molecular Mass of Polymers
15.4 Biodegradable Polymers
15.5 Polymers of Commercial Importance

NCERT INTEXT QUESTIONS

15.1. What are polymers?
Ans: Polymers are high molecular mass substances (103 — 107u) consisting of a very large number of simple repeating structural units joined together through covalent bonds in a linear fashion. They are also called macromolecules. Ex: polythene, nylon 6,6, bakelite, rubber, etc.

15.2. How are polymers classified on the basis of structure?
Ans: On the basis of structure, polymers are classified into three types. These are linear chain polymers, branched chain polymers and crossedlinkedpolymers.

1. Linear chain polymers: In this case, the monomer units are linked to one another to form long linear chains. These linear chains are placed one above the other and are closely packed in space. The close packing results in high densities, tensile strength and also high melting and boiling points. High density polyethene is a very common example of this type. Nylon, polyesters and PVC are also linear chain polymers.
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Intext Questions Q2

2. Branched chain polymers: In this type of polymers, the monomer units are linked to form long chains which have also side chains or branched chains of different Lengths attached to them. As a result of branching, these polymers are not closely packed in space. They have low densities, low tensile strength as well as low melting and boiling points. Some common examples of such polymers are ; low density polyethene, amylopectin, starch, glycogen etc.

3. Cross: linked polymers. In these polymers, also called net—work polymers, the monomer units are linked together to form three dimensionaL net—work as shown in the figure. These are expected to be quite hard, rigid and brittle. Examples of cross linked polymers are bakelite, glyptal. melamine formaldehyde polymer etc.

15.3. Write the names of the monomers of the following polymers:
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Intext Questions Q3

NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Intext Questions Q3.1
Ans: (i) Hexamethylene diamine NH2-(CH2)6NH2 and adipic acid HOOC – (CH2)4 – COOH
(ii) Caprolactum
(iii) Tetrafluoroethene F2C = CF2

15.4. Classify the following as addition and condensation polymers:
Terylene, Bakelite, Polyvinyl chloride,Polythene
Ans: Addition polymers: Polyvinyl chloride, Polythene
Condensation polymers : Terylene, bakelite.

15.5. Explain the difference between Buna-N and Buna-S.
Ans: Both Buna-N and Buna-S are synthetic rubber and are co-polymers in nature. They differ in their constituents.
Buna-N: Constituents are : buta-1, 3-diene and acrylonitrile.
Buna-S: Constituents are : buta-1, 3-diene, and styrene. They condense in the presence of Na.

Buna – S: It is a co—polymer of 1. 3 – butadiene and styrene and is prepared by the polymerisation of these components in the
ratio of 3 : 1 in the presence of sodium.
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Intext Questions Q5

Buna-N (Nitrile rubber):  h is a co-polymer of buta-1. 3-diene and acrylonitrile. It is formed as follows:
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Intext Questions Q5.1

15.6. Arrange the following polymers in increasing order of their intermolecuiar forces.
(i) Nylon 6,6, Buna-S, Polythene
(ii) Nylon 6, Neoprene, Polyvinyl chloride
Ans: On the basis of intermolecuiar forces, polymers
are classified as elastomers, fibres and plastics. The increasing order of intermolecuiar forces is: Elastomer < Plastic < fibre.
Thus, we have
(i)Buns-S < Polythene < Nylon 6,6
(ii)Neoprene < Polyvinyl chloride < Nylon 6.

NCERT EXERCISES

15.1. Explain the terms polymer and monomer.
Ans: Polymers are high molecular mass substances consisting of a very large number of simple repeating structural units joined together through covalent bonds in a regular fashion. Polymers are also called macromolecules. Some examples are polythene, nylon-66, bakelite, rubber, etc. Monomers are the. simple and reactive molecules from which the polymers are prepared either by addition or condensation polymerisation. Some examples are ethene, vinyl chloride, acrylonitrile, phenol and formaldehyde etc.

15.2. What are natural and synthetic polymers ? Give two examples of each.
Ans:
1. Natural polymers: The polymers which occur in nature mostly in plants and animals are called natural polymers. A few common examples are starch, cellulose, proteins, rubber nucleic acids, etc. Among them, starch and cellulose are the polymers of glucose molecules. Proteins are formed from amino acids which may be linked in different ways. These have been discussed in detail in unit 15 on biomolecules. Natural rubber is yet another useful polymer which is obtained from the latex of the rubber tree. The monomer units are of the unsaturated hydrocarbon 2-methyl-i, 3-butadiene, also called isoprene.
Example of natural polymers: Natural rubber, cellulose, nucleic acids, proteins etc.

2. Synthetic polymers: The polymers which are prepared in the laboraroiy are called synthetic polymers. These are also called man made polymers and have been developed in the present century to meet the ever increasing demand of the modem civilisation.
Example of synthetic polymers: Dacron (or terylene), Bakelite, PVC, Nylon-66, Nylon-6 etc.

15.3. Distinguish between the terms homopolymer and copolymer and give an example of each.
Ans: Polymers whose repeating structural units are derived from only one type of monomer units are called homopolymers, e.g., PVC polyethene, PAN, teflon, polystyrene, nylon- 6 etc.
Polymers whose repeating structural units are derived from two or more types of monomer molecules are copolymers, e.g., Buna-S, Buna-N, nylon-66, polyester, bakelite.

15.4. How do you explain the functionality of a monomer?
Ans: Functionality of a monomer implies the number of bonding sites present in it. For example, monomers like propene, styrene, acrylonitrile have functionality of one which means that have one bonding site.
Monomers such as ethylene glycol, hexamethylenediamine, adipic acid have functionality of two which means that they have two bonding sites.

15.5. Define the term polymerisation?
Ans: It is a process of formation of a high molecular Sol. mass polymer from one or more monomers by linking together a large number of repeating structural units through covalent bonds.

15.6. Is (-NH — CHR—CO-)n a homopolymer or copolymer?
Ans: It is a homopolymer because the repeating structural unit has only one type of monomer, i.e., NH2—CHR—COOH.

15.7. In which classes, are the polymers classified on the basis of molecular forces?
Ans: Polymers are classified into four classes on the basis of molecular forces. These are:
elastomers, fibres, thermoplastic polymers and thermosetting polymers.

1. Elastomers: In these polymers, the intermolecular forces are the weakest. As a result, they can be readily stretched by applying small stress and regain their original shape when the stress is removed. The elasticity can be further increased by introducing some cross – links in the polymer chains. Natural rubber is the most popular example of elastomers. A few more examples are of: buna-S, buna-N and neoprene.

2. Fibres: Fibres represent a class of polymers which are thread-like and can be woven into fabrics in a number of ways. These are widely used for making clothes, nets, ropes, gauzes etc. Fibres possess high tensile strength because the chains possess strong intermolecular forces such as hydrogen bonding. These forces are also responsible for close packing of the chains. As a result, the fibres are crystalline in nature and have aJso sharp melting points. A few common polymers belonging to this class are nylon – 66, terylene and polyacrylonitrile etc.

3. Thermoplastics: These are linear polymers and have weak van der Waals forces acting in the various chains and are intermediate of the forces present in the elastomers and in the fibres. When heated, they melt and form a fluid which sets into a hard mass on cooling, Thus, they can be cast into different shapes by using suitable moulds. A few common examples are polyethene and polystyrene polyvinyls etc. These can be used for making toys, buckets, telephone apparatus, television cabinets etc.

4. Thermosetting plastics: These are normally semifluid substances with low molecular masses. When heated, they become hard and infusible due to the cross-linking between the polymer chains. As a result, they also become three dimensional in nature. They do not melt when heated. A few common thermosetting polymers are bakelite, melamine-formaldehyde, urea-formaldehyde and polyurethane etc.

15.8. How can you differentiate between addition and condensation polymerisatiop?
Ans: In addition polymerization, the molecules of the same or different monomers simply add on to one another leading to the formation of a macromolecules without elimination of small molecules like H2O, NH3 etc. Addition polymerization generally occurs among molecules containing double and triple bonds. For example, formation of polythene from ethene and neoprene from chloroprene, etc. In condensation polymerisation, two or more bifunctional trifimctional molecules undergo a series of independent condensation reactions usually with the elimination of simple molecules like water, alcohol, ammonia, carbon dioxide and hydrogen chloride to form a macromolecule. For example, nylon-6,6 is a condensation polymer of hexamethylenediamine and adipic acid formed by elimination of water molecules.

15.9. Explain the term copolymerisation and give two examples.
Ans: When two or more different monomers are allowed to polymerise together the product formed is called a copolymer, and the process in called copolymerisation. Example, Buna-S and Buna-N. Buna- S is a copolymer of 1, 3- butadiene and styrene while Buna-N is a copolymer of 1,3-butadiene and acrylonitrile.

15.10. Write the free radical mechanism for the polymerisation of ethene.
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q10

NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q10.1

15.11. Define thermoplastics and thermo setting polymers with two examples of each
Ans: Thermoplastics polymers are linear polymer which can be repeatedly melted and moulded again and again on heating without any change in chemical composition and mechanical strength. Examples are polythene and polypropylene.
Thermosetting polymers, on the other hand, are permanently setting polymers. Once on heating in a mould, they get hardened and set, and then cannot be softened again. This hardening on heating is due to cross- linking between different polymeric chains to give a three dimensional network solid. Examples are bakelite, melamine-foimaldehyde polymer etc.

15.12. Write the monomers used for gettingThe following polymers:
(i) Polyvinylchloride
(ii) Teflon (iii) Bakelite
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q12

15.13. Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q13

15.14. How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Ans: Natural rubber is cis-polyisoprene and is obtained by 1, 4-polymerization of isoprene units. In this polymer, double bonds are located between C2 and C3 of each isoprene unit. These cis-double bonds do not allow the polymer chains to come closer for effective interactions and hence intermolecular forces are quite weak. As a result, natural rubber, i.e., cis-polyisoprene has a randomly coiled structure not the linear one and hence show elasticity.
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q14

15.15. Discuss the main purpose of vulcanisation of rubber.
Ans: Natural rubber has the following disadvantages:
(a) It is soft and sticky and becomes even more so at high temperatures and brittle at low temperatures. Therefore, rubber is generally used in a narrow temperature range (283-335 K) where its elasticity is maintained.
(b)It has large water absorption capacity, has low tensile strength and low resistance to abrasion.
(c)It is not resistant to the action of organic solvents.
(d)It is easily attacked by oxygen and other oxidising agents. .
To improve all these properties, natural rubber is vulcanised by heating it with about 5% sulphur at 373-415 K. The vulcanized rubber thus obtained has excellent elasticity over a larger range of temperature, has low water absorption tendency and is resistant to the action of organic solvents and oxidising agents.

15.16. What are the monomeric repeating units of Nylon-6 and Nylon 6,6?
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q16

15.17. Write the names and structures of the monomers of the following polymers:
(i) Buna-S (ii) Buna-N (iii) Dacron (iv) Neoprene
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q17

15.18. Identify the monomer in the following polymeric structures:
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q18
Ans:
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q18.1

15.19. How is dacron obtained from ethylene glycol and terephthalic acid?
Ans: Dacron is obtained by condensation polymerization of ethylene glycol and terephthalic acid with the elimination of water molecules. The reaction is carried out at 420 – 460 K in presence of a catalyst consisting of a mixture of zinc acetate and antimony trioxide.
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q19

15.20. What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester.
Ans: Polymers which disintegrate by themselves over a period of time due to environment degradation by bacteria, etc., are called biodegradable polymers. Example is PHBV, i. e., Poly-β-Hydroxybutyrate-co-β- Hydroxyvalerate.
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers Exercises Q20

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