Chemistry - CHEMICAL BONDING Question with Solution | TestHub

When the number of hybrid orbitals, $H$ is 4 , the hybridisation is ${\mathrm{sp}}^3$. $\mathrm{H}=\frac{1}{2}[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}]$ where, $V=$ number of valence electrons of central atom $\begin{array}{rl}M& =\text{ number of monovalent atoms }\\ C& =\text{ total positive charge }\\ A& =\text{ negative charge }\end{array}$ (a) For ${\mathrm{CH}}_4$, $\mathrm{H}=\frac{1}{2}[4+4-0+0]=4\text{, thus }{\mathrm{sp}}^3$ hybridisation (b) For ${\mathrm{SF}}_4$, $\mathrm{H}=\frac{1}{2}[6+4-0+0]=5\text{, thus }{\mathrm{sp}}^3\mathrm{d}$ hybridisation (c) For ${\mathrm{BF}}_4^{-}$, $\mathrm{H}=\frac{1}{2}[3+4-0+1]=4,\text{ thus }{\mathrm{sp}}^3$ hybridisation (d) For ${\mathrm{NH}}_4^{+}$, $H=\frac{1}{2}[5+4-1+0]=4$, thus $s{p}^3$ hybridisation Thus, only in ${\mathrm{SF}}_4$, the central atom does not have ${\mathrm{sp}}^3$ hybridisation