Chemistry - CHEMICAL BONDING Question with Solution | TestHub

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 $\begin{array}{rl}& \mathrm{X}=\text{ no. of monovalent atoms }\\ & \mathrm{C}=\text{ charge on cation }\\ & \mathrm{A}=\text{ Charge on anion. }\end{array}$ (a) In ${\mathrm{SF}}_4$ $\mathrm{H}=\frac{1}{2}[6+4-0+0)=5$ (b) ${In\; I}_3^{-}$ $\mathrm{H}=\frac{1}{2}[7+2+1]=5$ (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 ${\mathrm{sp}}^3\mathrm{d}$).