`H_(2)O and H_(2)O_(2)` resemble in

Options

A. Hybridisation of oxygen
B. Oxidation state of oxygen
C. Structure
D. Bond angle

Answer

To determine how \( H_2O \) (water) and \( H_2O_2 \) (hydrogen peroxide) resemble each other, we will analyze their properties step by step.

### Step 1: Identify the Hybridization of Oxygen in \( H_2O \)

1. **Valency of Oxygen (O)**: Oxygen has a valency of 6.
2. **Monovalent Atoms (H)**: There are 2 hydrogen atoms, each contributing 1 to the bond count, so \( M = 2 \).
3. **Anions (A)**: There are no anions in \( H_2O \), so \( A = 0 \).
4. **Cations (C)**: There are no cations in \( H_2O \), so \( C = 0 \).

Using the hybridization formula:
\[ \text{Hybridization} = \frac{1}{2} (B + M + A - C) \]
Substituting the values:
\[ \text{Hybridization} = \frac{1}{2} (6 + 2 + 0 - 0) = \frac{8}{2} = 4 \]
Since the result is 4, the hybridization of oxygen in \( H_2O \) is \( sp^3 \).

### Step 2: Identify the Hybridization of Oxygen in \( H_2O_2 \)

1. **Valency of Oxygen (O)**: Again, the valency of oxygen is 6.
2. **Monovalent Atoms (H)**: There are 2 hydrogen atoms, so \( M = 2 \).
3. **Anions (A)**: There are no anions in \( H_2O_2 \), so \( A = 0 \).
4. **Cations (C)**: There are no cations in \( H_2O_2 \), so \( C = 0 \).

Using the same hybridization formula:
\[ \text{Hybridization} = \frac{1}{2} (B + M + A - C) \]
Substituting the values:
\[ \text{Hybridization} = \frac{1}{2} (6 + 2 + 0 - 0) = \frac{8}{2} = 4 \]
Thus, the hybridization of oxygen in \( H_2O_2 \) is also \( sp^3 \).

### Step 3: Compare the Hybridization

Both \( H_2O \) and \( H_2O_2 \) have the same hybridization of \( sp^3 \).

### Step 4: Conclusion

The resemblance between \( H_2O \) and \( H_2O_2 \) lies in their hybridization of oxygen, which is \( sp^3 \) for both compounds.

### Summary of Resemblance
- **Hybridization of Oxygen**: Both \( H_2O \) and \( H_2O_2 \) have \( sp^3 \) hybridization.