Which of the following has $sp^{3}d$ hybridization ?

Question

Accepted Answer

Option 2, $SF_{4}$, is the correct answer because it exhibits $sp^{3}d$ hybridization. Here's a detailed explanation:

1. **Explanation for $SF_{4}$:**
   - Sulfur in $SF_{4}$ has six valence electrons. When it forms $SF_{4}$, it uses four of these electrons to form bonds with four fluorine atoms, resulting in four sigma bonds.
   - The remaining two electrons form a lone pair on the sulfur atom.
   - To accommodate these bonds and the lone pair, sulfur undergoes $sp^{3}d$ hybridization, which involves the mixing of one s orbital, three p orbitals, and one d orbital.
   - This hybridization results in a trigonal bipyramidal electron geometry. However, due to the presence of the lone pair, the molecular shape is a "see-saw" configuration.

2. **Explanation for why other options are incorrect:**

- **Option 1: $NH_{4}^{+}$**
     - The ammonium ion, $NH_{4}^{+}$, involves nitrogen forming four sigma bonds with hydrogen atoms.
     - Nitrogen undergoes $sp^{3}$ hybridization, not $sp^{3}d$, as it uses one s orbital and three p orbitals to form these bonds.
     - There are no d orbitals involved because nitrogen does not have available d orbitals in its valence shell.

- **Option 3: $XeF_{4}$**
     - Xenon in $XeF_{4}$ forms four bonds with fluorine atoms and has two lone pairs.
     - The hybridization of xenon in this molecule is $sp^{3}d^{2}$, which involves one s orbital, three p orbitals, and two d orbitals.
     - This results in a square planar molecular geometry, not $sp^{3}d$ hybridization.

- **Option 4: $XeOF_{4}$**
     - In $XeOF_{4}$, xenon forms bonds with one oxygen and four fluorine atoms, and it has one lone pair.
     - The hybridization here is $sp^{3}d^{2}$, similar to $XeF_{4}$, involving one s orbital, three p orbitals, and two d orbitals.
     - This results in a square pyramidal molecular geometry, not $sp^{3}d$ hybridization.

In conclusion, $SF_{4}$ is the only molecule among the options that exhibits $sp^{3}d$ hybridization, which justifies why Option 2 is the correct answer.