A + 2Cl^{-} rightarrow BWhere A is PCl_4^+ an | vedclass

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(A) $1$. Analyze the hybridization of $A$ $(PCl_4^+)$: The central atom $P$ has $5 + 4 - 1 = 8$ valence electrons,forming $4$ bonds. Hybridization is

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Only $I, II$ and $III$

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(A) $1$. Analyze the hybridization of $A$ $(PCl_4^+)$: The central atom $P$ has $5 + 4 - 1 = 8$ valence electrons,forming $4$ bonds. Hybridization is $sp^3$ ($0$ $d$-orbitals).$2$. Analyze the hybridization of $B$ $(PCl_6^-)$: The central atom $P$ has $5 + 6 + 1 = 12$ valence electrons,forming $6$ bonds. Hybridization is $sp^3d^2$ ($2$ $d$-orbitals).$3$. Evaluate $(I)$: From $B$ $(sp^3d^2)$ to $A$ $(sp^3)$,the number of $d$-orbitals involved decreases from $2$ to $0$. Thus,statement $(I)$ is $INCORRECT$.$4$. Evaluate $(II)$: In $A$ $(sp^3)$,$p$-character is $75\%$. In $B$ $(sp^3d^2)$,$p$-character is $3/6 = 50\%$. Thus,from $A$ to $B$,$p$-character decreases. Statement $(II)$ is $INCORRECT$.$5$. Evaluate $(III)$: $A$ has $4$ covalent bonds; $B$ has $6$ covalent bonds. They are not the same. Statement $(III)$ is $INCORRECT$.$6$. Evaluate $(IV)$: $PCl_6^-$ is formed by $PCl_5 + Cl^- \rightarrow PCl_6^-$. $PCl_5$ acts as a Lewis acid. This is a super-octet molecule. Statement $(IV)$ is $CORRECT$.$7$. Since $(I), (II),$ and $(III)$ are incorrect,the answer is $A$.

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