P Block Elements

All given oxide have nitrogen – nitrogen bond except N₂O₅ as ;

Let a moles of SO₂Cl₂ is taken

Then no. of moles of H₂SO₄ = a moles

No. of moles of HCl = 2a moles

No. of moles of NaOH required = 2a + 2a = 4a = 16

$C\mathcal{l}{F}_3$ ->T-shaped (sp³d)

IF₇ ->Pentagonal bipyramidal (sp³d³)

BrF₅ -> Square pyramidal (sp³d²)

BrF₃ ->T-Shaped (sp³d)

I₂Cl₆ -> Triangular bipyramidal (sp³d)

$I{F}_5\to$  Square Pyramidal (sp³d²)

ClF -> (sp³)

ClF₅ -> Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ ® Square Pyramidal

Reducing power for group-15 hydrides increases down the group, so BiH₃ is the strongest reducing agent.

High leaving tendency corresponds to high reactivity towards hydrolysis. Hence order is.

Anhydride of HNO₃ produced by the treatment of P₄O₁₀ (anhydring reagent) in ratio of 4 : 1 is N₂O₅ which is acidic in nature. $4HN{O}_3+P_4{O}_{10}\to \underset{Acidicnature}{\underset{?}{2N_2{O}_5}}+{\left(HP{O}_3\right)}_4$

$Cl{O}_2^{-},C{l}_2,M{n}^{3+}$ can show disproportionation reaction while $Mn{O}_4^{-}$ cannot show disproportionation reaction is Mn is in +7 oxidation state.

Na/H₂ can not reduce a functional group as Na does not behaves as catalyst here.

Stability order of oxides (X₂O) is,

l₂O > Cl₂O > Br₂O

Bonds of halogen & oxygen are covalent due to less EN difference.

Stability of (I - O) bond is higher due to less polarity and that of (Cl-O) bond is higher due to multiple bonding.