N- H Bond in Ammonia molecule is

Options

A. `sigma_(s-s)`
B. `sigma_(p - s)`
C. `sigma_(sp^3 -s)`
D. `sigma_(sp^3 - p)`

Answer

To determine the nature of the N-H bond in the ammonia molecule (NH₃), we can follow these steps:

### Step 1: Draw the Lewis Structure of Ammonia
- The nitrogen atom (N) has 5 valence electrons.
- Each hydrogen atom (H) has 1 valence electron.
- In ammonia, nitrogen forms three bonds with three hydrogen atoms and has one lone pair of electrons.

**Lewis Structure:**
```
H
|
H - N - H
|
(Lone Pair)
```

### Step 2: Count the Electron Pairs
- In ammonia, nitrogen has:
- 3 bond pairs (N-H bonds)
- 1 lone pair
- Total electron pairs = 3 (bond pairs) + 1 (lone pair) = 4 electron pairs.

### Step 3: Determine the Hybridization of Nitrogen
- To accommodate 4 electron pairs, nitrogen undergoes hybridization.
- The hybridization that accommodates 4 pairs is **sp³** hybridization.
- This is because 1 s orbital and 3 p orbitals combine to form 4 equivalent sp³ hybrid orbitals.

### Step 4: Identify the Type of Bond Formed
- The N-H bond is formed by the overlap of nitrogen's sp³ hybrid orbital with hydrogen's 1s orbital.
- The type of bond formed by this overlap is a **sigma (σ) bond** because the overlap occurs along the internuclear axis.

### Conclusion
- The N-H bond in the ammonia molecule is a **sigma bond** formed by the overlap of nitrogen's sp³ hybrid orbital and hydrogen's s orbital.

### Final Answer
The N-H bond in the ammonia molecule is a **sigma (σ) bond** formed by sp³-s overlap.

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