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

1. For ${\mathrm{SO}}_4{}^{2-}$:

• Central atom: Sulfur (S)
• Valence electrons of S = 6
• Valence electrons of O = 6
• Total valence electrons = $6+4\times 6+2$ (for charge) = 32
• In ${\mathrm{SO}}_4{}^{2-}$, sulfur forms four sigma bonds with four oxygen atoms. There are no lone pairs on the sulfur atom.
• Steric Number (SN) = 4 (sigma bonds) + 0 (lone pairs) = 4
• A steric number of 4 corresponds to $s{p}^3$ hybridization.

2. For ${\mathrm{SF}}_4$:

• Central atom: Sulfur (S)
• Valence electrons of S = 6
• Valence electrons of F = 7
• Total valence electrons = $6+4\times 7=34$
• In ${\mathrm{SF}}_4$, sulfur forms four sigma bonds with four fluorine atoms.
• Number of lone pairs on S = $\frac{\text{Total valence electrons}-(\text{Number of bonding electrons})}{2}=\frac{34-(4\times 2)}{2}=\frac{34-8}{2}=\frac{26}{2}=13$ (This is incorrect for central atom lone pairs).
• A simpler method for central atom lone pairs: $\frac{1}{2}$ (Valence electrons of central atom + Number of monovalent atoms - charge) = $\frac{1}{2}(6+4-0)=\frac{10}{2}=5$. This is the total number of electron pairs.
• Number of bonding pairs = 4 (for 4 F atoms).
• Number of lone pairs = Total electron pairs - Bonding pairs = $5-4=1$.
• Steric Number (SN) = 4 (sigma bonds) + 1 (lone pair) = 5
• A steric number of 5 corresponds to $s{p}^{3}d$ hybridization.

3. For ${\mathrm{SF}}_2$:

• Central atom: Sulfur (S)
• Valence electrons of S = 6
• Valence electrons of F = 7
• Total valence electrons = $6+2\times 7=20$
• In ${\mathrm{SF}}_2$, sulfur forms two sigma bonds with two fluorine atoms.
• Number of lone pairs on S = $\frac{1}{2}$ (Valence electrons of central atom + Number of monovalent atoms - charge) = $\frac{1}{2}(6+2-0)=\frac{8}{2}=4$. This is the total number of electron pairs.
• Number of bonding pairs = 2 (for 2 F atoms).
• Number of lone pairs = Total electron pairs - Bonding pairs = $4-2=2$.
• Steric Number (SN) = 2 (sigma bonds) + 2 (lone pairs) = 4
• A steric number of 4 corresponds to $s{p}^3$ hybridization.

Based on the analysis:

• ${\mathrm{SO}}_4{}^{2-}$ has $s{p}^3$ hybridization.
• ${\mathrm{SF}}_4$ has $s{p}^{3}d$ hybridization.
• ${\mathrm{SF}}_2$ has $s{p}^3$ hybridization.

Therefore, ${\mathrm{SF}}_4$ does not undergo $s{p}^3$-hybridization.

Option Analysis:

• A) ${\mathrm{SO}}_4{}^{2-}$: Undergoes $s{p}^3$ hybridization. Incorrect.
• B) ${\mathrm{SF}}_2$ and ${\mathrm{SO}}_4{}^{2-}$: Both undergo $s{p}^3$ hybridization. Incorrect.
• C) ${\mathrm{SF}}_2$: Undergoes $s{p}^3$ hybridization. Incorrect.
• D) ${\mathrm{SF}}_4$: Undergoes $s{p}^{3}d$ hybridization. Correct.

Correct Answer: (D)