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 find 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. **Write the general formula for the compound**:
- The compound is \( MH_2 \), which means there are two hydrogen atoms bonded to one metal atom.

3. **Set up the equation based on the neutrality of the compound**:
- The sum of the oxidation states in a neutral compound must equal zero. Therefore, we can write the equation:
\[ \text{Oxidation state of } M + 2 \times (\text{Oxidation state of } H) = 0 \]

4. **Substitute the known values into the equation**:
- We know the oxidation state of \( M \) is \( +2 \), so we can substitute this into the equation:
\[ +2 + 2x = 0 \]
- Here, \( x \) represents the oxidation state of hydrogen.

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

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

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

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