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In which one of the following species the central atom has the type of hybridisation which is not the same as that present in the other three?
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The correct answer is:
$\mathrm{SbCl}_5^{2-}$
Key Idea:- Molecules having the same number of hybrid orbitals, have same hybridisation and number of hybrid orbitals,
$\mathrm{H}=\frac{1}{2}[\mathrm{~V}+\mathrm{X}-\mathrm{C}+\mathrm{A}]$
where, $\mathrm{V}=$ no. of valence electrons of central atom
$\begin{aligned}
& \mathrm{X}=\text { no. of monovalent atoms } \\
& \mathrm{C}=\text { charge on cation } \\
& \mathrm{A}=\text { Charge on anion. }
\end{aligned}$
(a) In $\mathrm{SF}_4$
$\mathrm{H}=\frac{1}{2}[6+4-0+0)=5$
(b) $\operatorname{In~I}_3^{-}$
$\mathrm{H}=\frac{1}{2}[7+2+1]=5$
(c) In $\mathrm{SbCl}_5^{2-}$,
$\mathrm{H}=\frac{1}{2}[5+5+2)=6$
(d) In $\mathrm{PCl}_5$,
$H=\frac{1}{2}[5+5+0-0]=5$
Since, only $\mathrm{SbCl}_5^{2-}$ has different number of hybrid orbitals (ie, 6) from the other given species, its hybridisation is different from the others, ie, $s p^3 d^2$. (The hybridisation of other species is $\operatorname{sp}^3 \mathrm{~d}$).
$\mathrm{H}=\frac{1}{2}[\mathrm{~V}+\mathrm{X}-\mathrm{C}+\mathrm{A}]$
where, $\mathrm{V}=$ no. of valence electrons of central atom
$\begin{aligned}
& \mathrm{X}=\text { no. of monovalent atoms } \\
& \mathrm{C}=\text { charge on cation } \\
& \mathrm{A}=\text { Charge on anion. }
\end{aligned}$
(a) In $\mathrm{SF}_4$
$\mathrm{H}=\frac{1}{2}[6+4-0+0)=5$
(b) $\operatorname{In~I}_3^{-}$
$\mathrm{H}=\frac{1}{2}[7+2+1]=5$
(c) In $\mathrm{SbCl}_5^{2-}$,
$\mathrm{H}=\frac{1}{2}[5+5+2)=6$
(d) In $\mathrm{PCl}_5$,
$H=\frac{1}{2}[5+5+0-0]=5$
Since, only $\mathrm{SbCl}_5^{2-}$ has different number of hybrid orbitals (ie, 6) from the other given species, its hybridisation is different from the others, ie, $s p^3 d^2$. (The hybridisation of other species is $\operatorname{sp}^3 \mathrm{~d}$).
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