In `NH_(4)^+` and `OF_(2)` th hybridisation of central atom respectively are .

Options

A. `sp^(3),sp^(2)`
B. `sp^(3),sp^(3)`
C. `sp^(3) d,sp^(3) d`
D. `sp^(3) d ,sp^(3)`

Explanation

To determine the hybridization of the central atoms in the molecules \( NH_4^+ \) (ammonium ion) and \( OF_2 \) (oxygen difluoride), we can follow these steps:

### Step 1: Analyze the structure of \( NH_4^+ \)

1. **Identify the central atom**: In \( NH_4^+ \), nitrogen (N) is the central atom.
2. **Count the valence electrons**: Nitrogen has 5 valence electrons. The ammonium ion has a +1 charge, which means it has lost one electron, leaving it with 4 valence electrons.
3. **Determine the number of bonds**: Nitrogen forms 4 single bonds with 4 hydrogen atoms.
4. **Determine the hybridization**: Since nitrogen forms 4 bonds and has no lone pairs, the hybridization is \( sp^3 \). The geometry is tetrahedral.

### Step 2: Analyze the structure of \( OF_2 \)

1. **Identify the central atom**: In \( OF_2 \), oxygen (O) is the central atom.
2. **Count the valence electrons**: Oxygen has 6 valence electrons. In \( OF_2 \), it forms 2 bonds with fluorine atoms, using 2 of its valence electrons.
3. **Determine the number of lone pairs**: After forming 2 bonds, oxygen has 4 remaining electrons, which means it has 2 lone pairs.
4. **Determine the hybridization**: With 2 bonding pairs and 2 lone pairs, the hybridization is also \( sp^3 \). The molecular geometry is bent or V-shaped due to the presence of lone pairs.

### Final Answer

The hybridization of the central atom in \( NH_4^+ \) is \( sp^3 \) and in \( OF_2 \) is also \( sp^3 \).

### Summary of Hybridization

- \( NH_4^+ \): \( sp^3 \)
- \( OF_2 \): \( sp^3 \)