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

Question

Accepted Answer

To determine the hybridization of the given molecules and ions, we need to consider the number of electron pairs around the central atom, which includes both bonding pairs and lone pairs. The hybridization can be deduced from the steric number, which is the sum of the number of atoms bonded to the central atom and the number of lone pairs on the central atom.

1. **Option 1: $NH_4^+$**
   - The ammonium ion ($NH_4^+$) consists of a nitrogen atom bonded to four hydrogen atoms. There are no lone pairs on the nitrogen atom.
   - The steric number is 4 (4 bonding pairs + 0 lone pairs).
   - Hybridization for a steric number of 4 is $sp^3$.
   - Therefore, $NH_4^+$ does not have $sp^3d$ hybridization.

2. **Option 2: $SF_4$**
   - Sulfur tetrafluoride ($SF_4$) consists of a sulfur atom bonded to four fluorine atoms, with one lone pair on the sulfur atom.
   - The steric number is 5 (4 bonding pairs + 1 lone pair).
   - Hybridization for a steric number of 5 is $sp^3d$.
   - Therefore, $SF_4$ has $sp^3d$ hybridization, making Option 2 the correct answer.

3. **Option 3: $XeF_4$**
   - Xenon tetrafluoride ($XeF_4$) consists of a xenon atom bonded to four fluorine atoms, with two lone pairs on the xenon atom.
   - The steric number is 6 (4 bonding pairs + 2 lone pairs).
   - Hybridization for a steric number of 6 is $sp^3d^2$.
   - Therefore, $XeF_4$ does not have $sp^3d$ hybridization.

4. **Option 4: $XeOF_4$**
   - Xenon oxytetrafluoride ($XeOF_4$) consists of a xenon atom bonded to one oxygen atom and four fluorine atoms, with one lone pair on the xenon atom.
   - The steric number is 6 (5 bonding pairs + 1 lone pair).
   - Hybridization for a steric number of 6 is $sp^3d^2$.
   - Therefore, $XeOF_4$ does not have $sp^3d$ hybridization.

In conclusion, $SF_4$ is the only molecule among the options that exhibits $sp^3d$ hybridization, which is why Option 2 is the correct answer.