Q.(i) Predict the nature of hybridization of the central atom of PCl3 and mention the shape of the compound. [1+1]

$PCl_3$: $sp^3$, trigonal pyramidal. Interhalogens = compounds of two different halogens. $XeF_4$ hydrolyses to Xe, $XeO_3$, HF, and $O_2$. $H_2S > H_2O$ in acidity because bond strength, not electronegativity, dominates down the group.

(i) $PCl_3$ hybridisation and shape: Phosphorus in $PCl_3$ has 5 valence electrons; 3 are used to form P-Cl bonds and the remaining 2 form a lone pair, giving 4 electron domains (3 bond pairs + 1 lone pair) around P. Four electron domains correspond to $sp^3$ hybridisation. With one lone pair pushing the three P-Cl bonds together, the molecular shape (based on atom positions) is trigonal pyramidal.

(ii) Interhalogen compounds and $XeF_4$ hydrolysis: Interhalogen compounds are compounds formed by the combination of two different halogen atoms (e.g. $ClF$, $BrF_3$, $ICl_5$, $IF_7$), generally following the pattern $XX'_n$ ($n = 1, 3, 5, 7$), where the larger, less electronegative halogen $X$ combines with a smaller, more electronegative halogen $X'$.

Hydrolysis of $XeF_4$ is a disproportionation reaction where Xe goes from +4 to both 0 and +6 oxidation states:

$6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$

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