Explain the following :

(i) Gallium has higher ionisation enthalpy than aluminium       (ii) Boron does not exist as B³⁺ ion         (iii) Aluminium forms [AlF₆] ^3– ion but boron does not form [BF₆] ^3– ion                                                                        (iv) PbX₂ is more stable than PbX_{4.            (v) Pb⁴⁺ acts as an oxidising agent but Sn²⁺ acts as a reducing agent.}

<p>This is a Short Answers Type Questions as classified in NCERT Exemplar</p><p><strong> (i)&nbsp;</strong>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.</p><p>&nbsp;</p><p><strong> (ii)</strong> 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.</p><p>&nbsp;</p><p><strong> (iii)</strong> 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⁻.</p><p>&nbsp;</p><p><strong> (iv)</strong> Lead is extra stable in its +2 oxidation state than in +4 oxidation state due to inert pair effect.</p><p>&nbsp;</p><p><strong> (v)</strong> 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.</p><p><strong>Reaction: </strong>Pb⁴⁺+2e⁻→Pb²⁺</p><p>&nbsp;</p><p>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.</p><p><strong>Reaction: </strong>Sn²⁺→Sn⁴⁺+2e⁻</p>

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