$H=\frac{1}{2}[V+M-C+A]$

where, $V=$ number of valence electrons of central atom

$M=$ number of monovalent atoms

$C=$ total positive charge

$\mathrm{A}=$ negative charge

$H=\frac{1}{2}[4+4-0+0]=4$

thus $s p^{3}$ hybridization

For $S F_{4}$ $H=\frac{1}{2}[6+4-0+0]=5$

thus $s p^{3} d$ hybridization

For $B F_{4}$

$H=\frac{1}{2}[3+4-0+1]=4$

thuss $p^{3}$ hybridization

For $N H_{4}^{+}$ $H=\frac{1}{2}[5+4-1+0]=4$

thuss $s p^{3}$ hybridization Thus, only in $S F_{4}$

the central atom does not have $sp ^3$ hybridization.