The `H-O-H` angle in water molecule is about

Options

A. `90^(@)`
B. `180^(@)`
C. `102^(@)`
D. `105^(@)`

Explanation

To determine the H-O-H angle in a water molecule, we can follow these steps:

### Step 1: Understand the Hybridization
The water molecule (H₂O) has an oxygen atom bonded to two hydrogen atoms. The hybridization of the oxygen atom in water is sp³. This means that the oxygen atom has four electron pairs (two bonding pairs and two lone pairs).

**Hint:** Remember that hybridization helps us understand the geometry of molecules.

### Step 2: Determine the Geometry
In a perfect tetrahedral arrangement (which has a bond angle of 109.5°), all four electron pairs would be positioned as far apart as possible. However, in water, two of these electron pairs are lone pairs.

**Hint:** Tetrahedral geometry is the starting point for understanding bond angles in molecules with sp³ hybridization.

### Step 3: Consider the Effect of Lone Pairs
Lone pairs occupy more space than bonding pairs, which causes the bond angles to decrease. The presence of these two lone pairs in water pushes the hydrogen atoms closer together, reducing the H-O-H bond angle.

**Hint:** Remember that lone pairs exert greater repulsive forces than bonding pairs, affecting bond angles.

### Step 4: Calculate the Bond Angle
The ideal tetrahedral angle is about 109.5°. However, due to the presence of the two lone pairs, the H-O-H bond angle in water is approximately 104.5° to 105°.

**Hint:** Use the ideal tetrahedral angle as a reference point to understand how lone pairs affect actual bond angles.

### Conclusion
The H-O-H angle in a water molecule is about 104.5° to 105°.

**Final Answer:** The H-O-H angle in water is approximately 105 degrees.

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