To determine the hybridization of the central atom, we can use the steric number (SN) method, which is the sum of the number of sigma bonds and lone pairs around the central atom.

Step 1: Determine the steric number for each molecule/ion.

• A) ${\mathrm{SF}}_4$: Sulfur (S) is the central atom. It has 6 valence electrons. It forms 4 sigma bonds with 4 fluorine atoms. Remaining electrons = $6-4=2$, which form 1 lone pair. Steric number = $4\text{ (sigma bonds)}+1\text{ (lone pair)}=5$. A steric number of 5 corresponds to ${\mathrm{sp}}^3\mathrm{d}$ hybridization.
• B) ${\mathrm{BF}}_4^{-}$: Boron (B) is the central atom. It has 3 valence electrons. The negative charge adds 1 electron, so total valence electrons for B = $3+1=4$. It forms 4 sigma bonds with 4 fluorine atoms. Remaining electrons = $4-4=0$, so 0 lone pairs. Steric number = $4\text{ (sigma bonds)}+0\text{ (lone pairs)}=4$. A steric number of 4 corresponds to ${\mathrm{sp}}^3$ hybridization.
• C) ${\mathrm{NH}}_4^{+}$: Nitrogen (N) is the central atom. It has 5 valence electrons. The positive charge removes 1 electron, so total valence electrons for N = $5-1=4$. It forms 4 sigma bonds with 4 hydrogen atoms. Remaining electrons = $4-4=0$, so 0 lone pairs. Steric number = $4\text{ (sigma bonds)}+0\text{ (lone pairs)}=4$. A steric number of 4 corresponds to ${\mathrm{sp}}^3$ hybridization.
• D) ${\mathrm{CH}}_4$: Carbon (C) is the central atom. It has 4 valence electrons. It forms 4 sigma bonds with 4 hydrogen atoms. Remaining electrons = $4-4=0$, so 0 lone pairs. Steric number = $4\text{ (sigma bonds)}+0\text{ (lone pairs)}=4$. A steric number of 4 corresponds to ${\mathrm{sp}}^3$ hybridization.

Step 2: Identify the molecule where the central atom does not have ${\mathrm{sp}}^3$ hybridization.

Based on the calculations, only ${\mathrm{SF}}_4$ has ${\mathrm{sp}}^3\mathrm{d}$ hybridization, while the others have ${\mathrm{sp}}^3$ hybridization.

Option Analysis:

• A) ${\mathrm{SF}}_4$: The central sulfur atom has 4 bond pairs and 1 lone pair, leading to a steric number of 5 and ${\mathrm{sp}}^3\mathrm{d}$ hybridization. This is the correct answer.
• B) ${\mathrm{BF}}_4^{-}$: The central boron atom has 4 bond pairs and 0 lone pairs, leading to a steric number of 4 and ${\mathrm{sp}}^3$ hybridization.
• C) ${\mathrm{NH}}_4^{+}$: The central nitrogen atom has 4 bond pairs and 0 lone pairs, leading to a steric number of 4 and ${\mathrm{sp}}^3$ hybridization.
• D) ${\mathrm{CH}}_4$: The central carbon atom has 4 bond pairs and 0 lone pairs, leading to a steric number of 4 and ${\mathrm{sp}}^3$ hybridization.

Correct Answer: (A)