The p Block Elements TEXTBOOK Solution

7.1. Why are pentahalides more covalent than trihalidcs?

Answer : 

The group 15 elements have 5 e-1 s in their valence shell. It is difficult to lose 3e-1s to form E3+ and even more difficult to lose 5e-1 s to form E5+. Thus, they have very little tendency to form ionic compounds.

Further, since the elements in +5 state have less tendency to lose e-1s than in the +3 state, elements in +5 state have more tendency to share e-1 s and hence pentahalides are more covalent than trihalides.

7.2. Why is BiH3 the strongest reducing agent amongst all the hydrides of group 15 elements? (C.B.S.E. 2013)

Answer : 

Down the group, the atomic size of the element (E) increases and the bond length of the corresponding E—H bond also increases. This adversely affects the bond dissociation enthalpy.

This means that amongst the trihydrides of the members of nitrogen family, the bond dissociation enthalpy of Bi—H bond is the least. Therefore, BiH3 is the strongest reducing agent among the hydrides of group 15 elements.

7.3. Why is N2 less reactive at room temperature?

Answer : 

Due to presence of triple bond between two N-atoms (N = N), the bond dissociation energy of N2 is very high. As a result, N2 becomes less reactive at room temperature.

7.4. Mention the conditions required to maximise the yield of ammonia.

Answer : 

Ammonia is prepared by Haber’s process as given below:

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q4

7.5. How does ammonia react with a solution of Cu2+?

Answer : 

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q5

7.6. What is the covalence of nitrogen in N2O5 ?

Answer : 

In N2O5 , each N-atom has four shared pairs of e-1 s as shown:

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q6

7.7. Why is bond angle in PH+4 ion higher than in PH3 ? (Pb. Board 2009)

Answer : 

In both PH3 and PH+4 ion, the phosphorus atom is sp3 hybridised. However, in PH3 the central atom has apyramidal structure due to the presence of lone electron pair on the phosphorus atom.

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q7

Because of lone pair : shared pair repulsion which is more than that of shared pair : shared pair repulsion, the bond angle in PH3 is nearly 93-6°. In PH+4 ion, there is no lone electron pair on the phosphorus atom. It has a tetrahedral structure with bond angle of 109°-28′. Thus, the bond angle in PH+4 ion is higher than in PH3.

7.8. What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2?

Answer : 

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q8

7.9. What happens when PCl5 is heated?

Answer : 

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q9

7.10. Write a balanced equation for the hydrolytic reaction of PC is in heavy water.

Answer : 

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q10

7.11. What is the basicity of H3PO4?

Answer : 

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q11

7.12. What happens when H3PO4 is heated?

Answer : 

On heating, H3POdisproportionates to form PH3 and H3PO4 with O.S. of-3and + 5.

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q12

7.13. List the important sources of sulphur.

Answer : 

Sulphur mainly occurs in the combined states in earth’s crust in the form of sulphates and sulphides.

Sulphates : gypsum (CaSO4.2H2O); epsom (MgSO4.7H2O); baryte (BaSO4), etc.

Sulphides : Galena (PbS); zinc blende (ZnS); copper pyrites (CuFeS2); iron pyrites (FeS2), etc. Traces of sulphur occur’as H2S and in organic materials such as eggs, proteins, garlic, onion, mustard, hair and wool.

7.14. Write the order of thermal stability of the – hydrides of Group 16 elements.

Answer : 

The thermal stability of hydrides of group 16 elements decreases down the group. This is because down the group, size of the element (M) increases, M-H bond length increases and thus, stability of M-H bond decreases so that it can be broken down easily. Hence, we have order of thermal stability as H2O > H2S > H2Se > H2Te > H2PQ

7.15. Why is H2O a liquid and H2S a gas?

Answer : 

Due to high electronegativity of O than S, H2O undergoes extensive intermolecular H-bonding. As a result, H2O exists as an associated molecule in which each O is tetrahedrally surrounded by four H2O molecules. Therefore, H2O is a liquid at room temperature.

On the other hand,H2S does not undergo H- bonding. It exists as discrete molecules which are held together by weak van der waals forces of attraction. A small amount of energy is required to break these forces of attraction. Therefore, H2S is a gas at room temperature.

7.16. Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe

Answer : 

Platinum (Pt) is a noble metal and does not react with oxygen directly.

7.17. Complete the following reactions:

  • (i)C2H2 + O2 ->
  • (ii) 4Al + 3 O2 ->

Answer : 

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q17

7.18. Why does O3 act as a powerful oxidising agent?

Answer : 

On heating, O3 readily decomposes to give O2 and nascent oxygen.

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q18

Since nascent oxygen is very reactive, therefore, O3 acts as a powerful oxidising agent.

7.19. How is O3 estimated quantitatively?

Answer : 

When O3 is treated with excess of KI solution buffered with borate buffer (pH = 9.2), I2 is liberated quantitatively.

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q19

The I2 thus liberated is titrated against a standard solution of sodium thiosulphate using starch as an indicator.

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q19.1

7.20. What happens when sulp’hur dioxide is passed through an aqueous solution of Fe(III) salt?

Answer : 

SO2 acts as a reducing agent and reduces aqueous solution of Fe (III)salt to Fe (II) salt.

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q20

7.21. Comment on the nature of two S-O bonds formed in S02 molecule. Are the two S-O bonds in this molecule equal ?

Answer : 

SO2 exists as an angular molecule with OSO bond angle of 119.5°. It a resonance hybrid of two canonical-forms:

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q21

7.22. How is the presence of SO2 detected?

Answer : 

SOis a pungent smelling gas. It can be detected by two test:

7.23. Mention three areas in which H2SO4 plays an important role.

Answer : 

(i) Sulphuric acid is used for the manufacture of a number of chemicals like hydrochloric acid, phosphoric acid, nitric acid along with a large number of organic compounds.

(ii) A mixture of concentrated nitric acid and concentrated sulphuric acid is used in the manufacture of explosives like picric acid, T.N.T, dynamite etc.

(iii) Dilute solution of acid is employed in petroleum refining in order to remove the unwanted impurities of sulphur.

7.24. Write the conditions to maximize the yield of H2SO4 by Contact process.

Answer : 

The key step in the manufacture of sulphuric acid is oxidation of SO2 to SO3 in presence of V2O5 catalyst.

NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements Textbook Questions Q22

The reaction is exothermic and reversible. Hence, low temperature and high pressure are the favourable conditions for maximum yield of SO3. In practice a pressure of 2 bar and temperature of 720 K is maintained.

7.25. Why is Ka2 « Ka1 for H2SO4 in water?

Answer : 

H2SO4 is a very strong acid in water largely because of its first ionisation to H3O+ and HSO4– The ionisation of HSO4 to H3O+ and SO42- is very very small. That is why, Ka2« Ka1.

7.26. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising powers of F2 and Cl2.

Answer : 

The oxidising powers of both the members of halogen family are expressed in terms of their electron accepting tendency and can be compared as their standard reduction potential values.

F2 + 2e → 2F; E° = 2-87 V, Cl2 + 2e → 2Cl ; E° = 1-36 V

Since the E° of fluorine is more than that of chlorine, it is a stronger oxidising agent.

Explanation : Three factors contribute towards the oxidation potentials of both the halogens. These are :

(i) Bond dissociation enthalpy: Bond dissociation enthalpy of F2 (158 kJ mol-1) is less compared to that of Cl2 (242·6 kJ mol-1).

(ii) Electron gain enthalpy: The negative electron gain enthalpy of F (- 332·6 kJ mol-1) is slightly less than of Cl (-348·5 kJ mol-1).

(iii) Hydration enthalpy: The hydration enthalpy of F- ion (515 kJ mol-1) is much higher than that of Cl- ion (381 kJ mol-1) due to its smaller size.

From the available data, we may conclude that lesser bond dissociation enthalpy and higher hydration enthalpy compensate lower negative electron gain enthalpy of fluorine as compared to chlorine. Consequently, F2 is a more powerful oxidising agent than Cl2.

7.27. Give two examples to show the anomalous behaviour of fluorine.

Answer :

  1. Ionisation enthalpy, electro-negativity and electrode potential are higher for fluorine than the expected trends of other halogen.
  2. Fluorine does not show any positive oxidation state except in HOF.

7.28. Sea is the greatest source of some halogens. Comment.

Answer :

Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium but sodium chloride being the maximum makes sea water saline. Various sea weeds contain upto 0.5% iodine.

7.29. Give the reason for bleaching action of Cl2.

Answer :

Chlorine bleaches by oxidation Cl2 + H2O → HCl + HOCl → HCl + [O]

The nascent oxygen reacts with dye to make it colourless.

7.30. Name two poisonous gases which can be prepared from chlorine gas.

Answer :

COCl2 (phosgene), CCl3NO2 (tear gas)

7.31. Why is ICI more reactive than l2?

Answer :

In general, interhalogen compounds are more reactive than halogens due to weaker X-X’ bonding than X-X bond. Thus, ICI is more reactive than I2.

7.32. Why is helium used in diving apparatus?

Answer :

Helium along with oxygen is used in the diving apparatus by the sea divers. Since it is very little soluble in blood, it reduces decompression and causes less discomfort to the diver in breathing. A mixture of helium and oxygen does not cause pain due to very low solubility of helium in blood as compared to nitrogen.

7.33. Balance the following equation :

XeF6 + H2O → XeO2F2 + 4HF

Answer :

Do YourSelf …………………………

7.34. Why has it been difficult to study the chemistry of radon?

Answer :

Radon is radioactive with very short half-life which makes the study of chemistry of radon difficult.

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