The number of lone pair present in N-atom in `NH_(4)^(+)` ion is

Options

A. zero
B. 1
C. 2
D. 3

Answer

To determine the number of lone pairs present in the nitrogen atom in the ammonium ion (NH₄⁺), we can follow these steps:

### Step 1: Identify the Valence Electrons
Nitrogen (N) has 5 valence electrons. Each hydrogen (H) atom contributes 1 valence electron. Since there are 4 hydrogen atoms, they contribute a total of 4 valence electrons.

### Step 2: Calculate the Total Valence Electrons
For the ammonium ion (NH₄⁺):
- Valence electrons from nitrogen = 5
- Valence electrons from 4 hydrogen atoms = 4
- Total = 5 + 4 = 9 valence electrons

### Step 3: Adjust for the Charge
The ammonium ion has a positive charge (+1). This means we need to subtract 1 from the total number of valence electrons:
- Adjusted total = 9 - 1 = 8 valence electrons

### Step 4: Determine the Hybridization
Using the formula for steric number:
\[ \text{Steric Number} = \frac{1}{2} \left( V + M + \text{Charge (anion)} - \text{Charge (cation)} \right) \]
Where:
- V = number of valence electrons (5 for N)
- M = number of monovalent atoms (4 for H)
- Charge (cation) = +1

Substituting the values:
\[ \text{Steric Number} = \frac{1}{2} \left( 5 + 4 - 1 \right) = \frac{1}{2} \times 8 = 4 \]

### Step 5: Determine Bond Pairs and Lone Pairs
The steric number of 4 indicates that there are 4 regions of electron density around the nitrogen atom. In the case of NH₄⁺, nitrogen forms 4 single bonds with the 4 hydrogen atoms. Therefore:
- Bond pairs = 4
- Lone pairs = Steric number - Bond pairs = 4 - 4 = 0

### Conclusion
The number of lone pairs present in the nitrogen atom in the ammonium ion (NH₄⁺) is **0**.

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