- Option (a) is correct
- Option (b) is correct
- Option (c) is correct
- Option (d) is correct
Solution
(A) $\mathrm{ICl}_4^{-}:$Steric Number $=\frac{1}{8}(7+(4 \times 7)+1)=\frac{36}{8}$ $=4+4$; b.p. $=4$, e.p. $=\frac{4}{2}=2$
$\therefore \quad \mathrm{ICl}_4^{-} \rightarrow \mathrm{sp}^3 \mathrm{~d}^2$ hybridization
(B)
$
\begin{aligned}
& \mathrm{NO}_3^{-}: \frac{1}{8}(5+(3 \times 6)+1) \\
& =\frac{1}{8} \times 24=3 \text { b.p., } \mathrm{sp}^2 \text { hybridization }
\end{aligned}
$
(C)
$
\begin{aligned}
& \mathrm{PCl}_4^{+}: \frac{1}{8}(5+(4 \times 7)-1) \\
& =\frac{1}{8} \times 32=4 \text { b.p., } \mathrm{sp}^3 \text { hybridization }
\end{aligned}
$
(D)
$
\begin{aligned}
& \mathrm{SiF}_6^{2-}: \frac{1}{8}(4+(6 \times 7)+2) \\
& =\frac{1}{8} \times 48=6 \text { b.p., } \mathrm{sp}^3 \mathrm{~d}^2 \text { hybridization }
\end{aligned}
$
Asked in: AP EAMCET 2022 (06 Jul Shift 1)
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