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
$\begin{aligned}
M & =\text { number of monovalent atoms } \\
C & =\text { total positive charge } \\
A & =\text { negative charge }
\end{aligned}$
(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