The `N_(2)H_(4)` molecule contains :

Options

A. 4 lone pairs of electrons
B. 3 lone pairs of electrons
C. 2 lone pairs of electrons
D. No lone pair of electrons

Explanation

To determine how many lone pairs of electrons are present in the N₂H₄ (hydrazine) molecule, we can follow these steps:

### Step 1: Determine the Lewis Structure of N₂H₄
- N₂H₄ consists of two nitrogen (N) atoms and four hydrogen (H) atoms.
- Each nitrogen atom can form three bonds and has one lone pair of electrons.
- Each hydrogen atom can form one bond.

### Step 2: Count the Valence Electrons
- Nitrogen has 5 valence electrons, and there are two nitrogen atoms: 2 × 5 = 10 electrons.
- Hydrogen has 1 valence electron, and there are four hydrogen atoms: 4 × 1 = 4 electrons.
- Total valence electrons = 10 (from N) + 4 (from H) = 14 valence electrons.

### Step 3: Draw the Bonds
- Connect the two nitrogen atoms with a single bond.
- Connect each nitrogen atom to two hydrogen atoms with single bonds.
- This gives us the basic structure: H₂N-NH₂.

### Step 4: Assign Lone Pairs
- After forming the bonds, we need to distribute the remaining electrons to satisfy the octet rule.
- Each nitrogen atom has used 3 of its 5 valence electrons to form bonds (2 with H and 1 with the other N).
- Therefore, each nitrogen atom has 2 electrons left, which form 1 lone pair on each nitrogen atom.

### Step 5: Count the Total Lone Pairs
- Since there are 2 nitrogen atoms and each has 1 lone pair, the total number of lone pairs in N₂H₄ is 2.

### Conclusion
- The N₂H₄ molecule contains **2 lone pairs of electrons**.