The oxidation number of hydrogen in`MH_(2)` is : (where M `equiv` bivalent metal)

Options

A. `+1`
B. `-1`
C. `+2`
D. `-2`

Explanation

To determine the oxidation number of hydrogen in the compound \( MH_2 \), where \( M \) is a bivalent metal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the oxidation state of the metal \( M \)**:
- Since \( M \) is a bivalent metal, its oxidation state is \( +2 \).

2. **Assign a variable for the oxidation state of hydrogen**:
- Let the oxidation state of hydrogen be \( x \).

3. **Set up the equation based on the overall charge of the compound**:
- The compound \( MH_2 \) is neutral, meaning the sum of the oxidation states must equal zero.
- The equation can be set up as follows:
\[ \text{Oxidation state of } M + 2 \times \text{Oxidation state of } H = 0 \]
- Substituting the known values:
\[ +2 + 2x = 0 \]

4. **Solve for \( x \)**:
- Rearranging the equation gives:
\[ 2x = -2 \]
- Dividing both sides by 2:
\[ x = -1 \]

5. **Conclusion**:
- The oxidation number of hydrogen in \( MH_2 \) is \( -1 \).

### Final Answer:
The oxidation number of hydrogen in \( MH_2 \) is \( -1 \).