Which of the following is formed when diborane reacts with sodium hydride in diethyl ether?

Options

A. Sodium boride
B. Sodium borate
C. Sodium metaborate
D. sodium tetra hydridoborate

Answer

To determine what is formed when diborane (B2H6) reacts with sodium hydride (NaH) in diethyl ether, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are diborane (B2H6) and sodium hydride (NaH).

### Step 2: Understand the Reaction Conditions
The reaction takes place in the presence of diethyl ether, which is a solvent that can stabilize the reactants and products.

### Step 3: Write the Balanced Chemical Equation
When diborane reacts with sodium hydride, the reaction can be represented as follows:

\[ \text{B}_2\text{H}_6 + 2 \text{NaH} \rightarrow 2 \text{NaBH}_4 \]

This indicates that one mole of diborane reacts with two moles of sodium hydride to produce two moles of sodium tetrahydride borate (NaBH4).

### Step 4: Identify the Product
The product of this reaction is sodium tetrahydride borate (NaBH4). In this compound, boron is in the +3 oxidation state.

### Step 5: Conclusion
Thus, the final answer to the question is that when diborane reacts with sodium hydride in diethyl ether, sodium tetrahydride borate (NaBH4) is formed.

### Summary of the Solution
- Reactants: B2H6 and NaH
- Reaction: B2H6 + 2 NaH → 2 NaBH4
- Product: Sodium tetrahydride borate (NaBH4)