In the reaction of sodium hydride and water:

Options

A. sodium is reduced
B. hydrogen is oxidised
C. hydrogen is reduced
D. No element is oxidised or reduced

Explanation

To analyze the reaction between sodium hydride (NaH) and water (H2O) and determine which species is oxidized and which is reduced, we can follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
The reaction between sodium hydride and water can be represented as follows:

\[ \text{NaH} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2 \]

### Step 2: Assign oxidation states to each element in the reaction.
- In sodium hydride (NaH):
- Sodium (Na) has an oxidation state of +1.
- Hydrogen (H) has an oxidation state of -1.

- In water (H2O):
- Each hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

- In sodium hydroxide (NaOH):
- Sodium (Na) has an oxidation state of +1.
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

- In hydrogen gas (H2):
- Each hydrogen (H) has an oxidation state of 0.

### Step 3: Identify changes in oxidation states.
- For sodium (Na):
- Oxidation state remains +1 in NaH and NaOH, so it is neither oxidized nor reduced.

- For hydrogen (H):
- In NaH, the oxidation state is -1.
- In NaOH, the oxidation state is +1.
- In H2, the oxidation state is 0.

Here, we see that:
- The hydrogen in NaH changes from -1 to +1 (loss of electrons, hence oxidation).
- The hydrogen in H2O changes from +1 to 0 (gain of electrons, hence reduction).

### Step 4: Determine which species is oxidized and which is reduced.
- The hydrogen from sodium hydride (NaH) is oxidized (loses electrons).
- The hydrogen from water (H2O) is reduced (gains electrons).

### Conclusion:
- In this reaction:
- Hydrogen from NaH is oxidized.
- Hydrogen from H2O is reduced.

Thus, the correct options are:
- Hydrogen is oxidized (Option B).
- Hydrogen is reduced (Option C).