O.N. of hydrogen in KH, `MgH_(2)` and NaH respectively would be -

Options

A. `-1,-1` and -1
B. `+1,+1` and `+1`
C. `+2,+1` and -2
D. `-2,-3` and -1

Answer

To determine the oxidation number (O.N.) of hydrogen in the compounds KH (Potassium Hydride), MgH₂ (Magnesium Hydride), and NaH (Sodium Hydride), we can follow these steps:

### Step 1: Identify the oxidation states of the metals
- **Potassium (K)** is an alkali metal and has an oxidation state of +1.
- **Sodium (Na)** is also an alkali metal and has an oxidation state of +1.
- **Magnesium (Mg)** is an alkaline earth metal and has an oxidation state of +2.

### Step 2: Write the general formula for the oxidation state of hydrogen
In hydrides, hydrogen typically has an oxidation state of -1 when bonded to metals that are more electropositive than hydrogen (like alkali and alkaline earth metals).

### Step 3: Calculate the oxidation state of hydrogen in each compound
1. **For KH:**
- Let the oxidation state of hydrogen be \( x \).
- The equation based on the compound is:
\[ +1 + x = 0 \quad \text{(since the compound is neutral)} \]
- Solving for \( x \):
\[ x = -1 \]

2. **For MgH₂:**
- Let the oxidation state of hydrogen be \( x \).
- The equation based on the compound is:
\[ +2 + 2x = 0 \quad \text{(since the compound is neutral)} \]
- Solving for \( x \):
\[ 2x = -2 \implies x = -1 \]

3. **For NaH:**
- Let the oxidation state of hydrogen be \( x \).
- The equation based on the compound is:
\[ +1 + x = 0 \quad \text{(since the compound is neutral)} \]
- Solving for \( x \):
\[ x = -1 \]

### Conclusion
The oxidation number of hydrogen in all three compounds (KH, MgH₂, and NaH) is -1.

### Summary of Results:
- O.N. of hydrogen in KH = -1
- O.N. of hydrogen in MgH₂ = -1
- O.N. of hydrogen in NaH = -1