Chemistry NCERT Exemplar Solutions Class 11th Chapter Eleven

This is a Short Answers Type Questions as classified in NCERT Exemplar

4BF₃ + 3LiAlH₄? 2 B₂H₆ + 3LiF + 3AlF₃

                   (Z)                             (X)

B₂H₆ + 6H₂O?2H₃BO₃ + 6H₂

(X)                          (Y)                                

B₂H₆ + 3O₂? ΔB₂O₃ + 3H₂O

(X)

This is a Short Answers Type Questions as classified in NCERT Exemplar

(i) When moving down the periodic table within a group, ionization enthalpy generally decreases due to increased atomic radius. However, in the case of gallium (Ga) and aluminum (Al),  gallium experiences a higher effective nuclear charge due to less shielding from its inner electrons, resulting in a higher ionization enthalpy compared to aluminum.

(ii) As boron is smaller in size and the sum of its first three ionization enthalpies i.e. ΔH₁+ΔH₂+ΔH₃ is very large so boron does not allow to lose its all three valence electrons and exist as +3 ion rather it shares electrons to make covalent bonds.

(iii) As compared to boron, aluminium is larger in size and has vacant d-orbital through which it can expand its coordination number from +4 to +6. Can easily acquire the sp³d² hybridisation State. On the other hand, boron is smaller in size with no d-orbitals and can expand its coordination number maximum up to +4 so boron can maximum the form BF₄⁻ but not BF₆²⁻.

(iv) Lead is extra stable in its +2 oxidation state than in +4 oxidation state due to inert pair effect.

(v) Due to the inert Pair effect, lead is more stable in its +2 oxidation state than in +4. So if it is present in +4 oxidation state in any compound it can readily reduce itself by gaining two electrons. Thus acting as an oxidising agent.

Reaction: Pb⁴⁺+2e⁻→Pb²⁺

While for Sn2+, +4 oxidation state is extra stable than +2 oxidation state so in any compound where acid is present in +2 oxidation state it loses two more electron and oxidises itself to +4 oxidation state. Thus acting as a reducing agent.

Reaction: Sn²⁺→Sn⁴⁺+2e⁻

(vi) Electronic configuration of F & Clare: 

In fluorine the new electron added will be accommodated in 2p sub-shell while in chlorine the new electron goes to 3p subshell. The size of the 2p subshell is smaller than the size of 3p sub shell. So, the added electron in 2p sub shell experiences strong inter-electronic repulsion which is comparatively less in the case of 3p sub shell of chlorine. The added electron doesn't feel much nuclear attraction in fluorine, that's why fluorine doesn't easily gain new electrons. Therefore, fluorine has less negative electron gain enthalpy than chlorine.

(vii) When we move down the group in group 13 due to the inert pair effect for Thallium, +3 oxidation state is less stable than its +1 oxidation state. In Tl (NO₃)₃ the central atom Tl exist in +3 oxidation state which is not much stable so it readily gains two electrons to form compounds in which Thallium is in +1 oxidation state that is TINO₃.

(viii) As we move down the group in group 14 size of element increases, electronegativity decreases which results in the decrease of M-M born strength hence the catenation decreases

(ix) Complete hydrolysis of BF₃ does not occur, rather it gets partially hydrolysed. BF₃ reacts with water to form boric acid and HF the HF further reacts with H₃BO₃ and prevents Hydrolysation of BF₃.

(x) In group 14, as we move down the size of elements increases and electronegativity decreases. Silicon does not have sp² state and does not form sp² double bond like carbon does in graphite. Rather silicon exists in sp³ hybridisation state which forms 3-D structure more like diamond.