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 on the nitrogen atom in the \( NH_4^+ \) ion, we can follow these steps:

### Step 1: Identify the Valence Electrons
Nitrogen (N) has 5 valence electrons. Hydrogen (H) has 1 valence electron.

### Step 2: Analyze the Ion
The \( NH_4^+ \) ion is formed from ammonia (\( NH_3 \)) by removing one proton (H\(^+\)). In \( NH_3 \), nitrogen forms three bonds with three hydrogen atoms.

### Step 3: Count the Electrons Used in Bonding
In \( NH_3 \):
- Nitrogen uses 3 of its 5 valence electrons to form bonds with 3 hydrogen atoms.
- This leaves 2 valence electrons, which form a lone pair on the nitrogen atom.

### Step 4: Consider the Charge of the Ion
When we convert \( NH_3 \) to \( NH_4^+ \), we remove one electron (due to the positive charge). Therefore, the lone pair on nitrogen is lost when forming the \( NH_4^+ \) ion.

### Step 5: Determine the Number of Lone Pairs
After the removal of one electron, nitrogen in \( NH_4^+ \) has:
- 0 lone pairs (since the lone pair was lost when the positive charge was introduced).

### Conclusion
The number of lone pairs present on the nitrogen atom in the \( NH_4^+ \) ion is **0**.

---