The oxidation number of H in `LiAlH_4` is

Options

A. `-1`
B. `+1`
C. 0
D. `+3`

Answer

To find the oxidation number of hydrogen in lithium aluminium hydride (LiAlH₄), we can use the X method. Here’s the step-by-step solution:

### Step 1: Identify the compound and its components
The compound is lithium aluminium hydride (LiAlH₄). It consists of lithium (Li), aluminium (Al), and hydrogen (H).

### Step 2: Assign known oxidation states
- Lithium (Li) is an alkali metal and typically has an oxidation state of +1.
- Aluminium (Al) usually has an oxidation state of +3.

### Step 3: Let the oxidation number of hydrogen be X
We need to find the oxidation number of hydrogen (H) in this compound. We will denote the oxidation number of hydrogen as X.

### Step 4: Write the equation based on the sum of oxidation states
The overall charge of the compound LiAlH₄ is neutral (0). Therefore, we can set up the equation based on the sum of the oxidation states:

\[ \text{Oxidation state of Li} + \text{Oxidation state of Al} + \text{4} \times \text{Oxidation state of H} = 0 \]

Substituting the known values:

\[ (+1) + (+3) + 4X = 0 \]

### Step 5: Solve for X
Now, we simplify the equation:

\[ 1 + 3 + 4X = 0 \]

This simplifies to:

\[ 4 + 4X = 0 \]

Now, isolate X:

\[ 4X = -4 \]

\[ X = -1 \]

### Conclusion
The oxidation number of hydrogen in lithium aluminium hydride (LiAlH₄) is -1.

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