`B_(2)H_(6)` reacts with `O_(2)` and `H_(2)O` respectively to form

Options

A. `HBO_2 and H_3BO_3`
B. `B_2O_3 and H_3BO_3`
C. `B_2O_3and H_3BO_3`
D. `B_2O_3 and [BH_4]^(-)`

Explanation

To solve the question of what products are formed when \( B_2H_6 \) (diborane) reacts with \( O_2 \) and \( H_2O \), we can break it down step by step.

### Step 1: Reaction of \( B_2H_6 \) with \( O_2 \)

1. **Write the reactants**: The reactants are \( B_2H_6 \) and \( O_2 \).
2. **Identify the products**: When diborane reacts with oxygen, it typically forms boron trioxide (\( B_2O_3 \)) and water (\( H_2O \)).
3. **Write the balanced equation**:
\[ 2 B_2H_6 + 3 O_2 \rightarrow 2 B_2O_3 + 6 H_2O \]
This shows that diborane reacts with oxygen to produce boron trioxide and water.

### Step 2: Reaction of \( B_2H_6 \) with \( H_2O \)

1. **Write the reactants**: The reactants are \( B_2H_6 \) and \( H_2O \).
2. **Identify the products**: When diborane reacts with water, it forms boric acid (\( H_3BO_3 \)) and hydrogen gas (\( H_2 \)).
3. **Write the balanced equation**:
\[ B_2H_6 + 6 H_2O \rightarrow 2 H_3BO_3 + 3 H_2 \]
This indicates that diborane reacts with water to produce boric acid and hydrogen gas.

### Final Products

From the reactions above, we can summarize the products formed:
- From the reaction with \( O_2 \): \( B_2O_3 \) and \( H_2O \)
- From the reaction with \( H_2O \): \( H_3BO_3 \) and \( H_2 \)

### Conclusion

The products formed when \( B_2H_6 \) reacts with \( O_2 \) and \( H_2O \) respectively are:
- \( B_2O_3 \) and \( H_3BO_3 \)