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Question

(A) $(P)$ In $PCl_5 	o PCl_3 + Cl_2$: $P$ in $PCl_5$ is $sp^3d$ (Trigonal bipyramidal,$+5$ state) and $P$ in $PCl_3$ is $sp^3$ (Pyramidal,$+3$ state)

Vedclass · Accepted Answer

$(P) 	o i, ii, iii, iv; (Q) 	o ii, iii, iv; (R) 	o ii, iii; (S) 	o ii, iii$

Vedclass · Answer

(A) $(P)$ In $PCl_5 	o PCl_3 + Cl_2$: $P$ in $PCl_5$ is $sp^3d$ (Trigonal bipyramidal,$+5$ state) and $P$ in $PCl_3$ is $sp^3$ (Pyramidal,$+3$ state). Changes: $(i), (ii), (iii), (iv)$.$(Q)$ In $SO_3 	o SO_2 + \frac{1}{2}O_2$: $S$ in $SO_3$ is $sp^2$ (Trigonal planar,$+6$ state) and $S$ in $SO_2$ is $sp^2$ (Bent,$+4$ state). Changes: $(ii), (iii), (iv)$. (Hybridisation remains $sp^2$).$(R)$ In $NH_3 + H^{+} 	o NH_4^+$: $N$ in $NH_3$ is $sp^3$ (Pyramidal,$-3$ state) and $N$ in $NH_4^+$ is $sp^3$ (Tetrahedral,$-3$ state). Changes: $(ii), (iii)$.$(S)$ In $H_2O + H^{+} 	o H_3O^{+}$: $O$ in $H_2O$ is $sp^3$ (Bent,$-2$ state) and $O$ in $H_3O^{+}$ is $sp^3$ (Pyramidal,$-2$ state). Changes: $(ii), (iii)$.

Column-$I$	Column-$II$
$(P)$ $\underline PCl_5 \to PCl_3 + Cl_2$	$(i)$ Change in hybridisation
$(Q)$ $\underline SO_3 \to \underline SO_2 + \frac{1}{2}O_2$	$(ii)$ Change in bond angle
$(R)$ $\underline NH_3 + H^{+} \to \underline NH_4^+$	$(iii)$ Change in shape
$(S)$ $H_2\underline O + H^{+} \to H_3\underline O^{+}$	$(iv)$ Change in oxidation state

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