`H_(3)BO_(3)(s) + aq. NaOH overset(Delta)rarr(X)`,
`H_(3)BO_(3)(s) + "molten" NaOH overset(Delta)rarr (Y)`
Compound (X) & (Y) are respectively,

Options

A. `Na_(3)BO_(3), Na_(3)B`
B. `Na_(3)BO_(3), NaBO_(2)`
C. `Na[B(OH)_(4)], Na_(3)BO_(3)`
D. `Na_(3)BO_(3), Na[B(OH)_(4)`

Explanation

To solve the question, we need to analyze the two reactions involving boric acid (H₃BO₃) and sodium hydroxide (NaOH) under different conditions.

### Step-by-Step Solution:

1. **Identify the first reaction:**
- The first reaction involves solid boric acid (H₃BO₃) reacting with aqueous sodium hydroxide (NaOH) when heated.
- The balanced equation for this reaction is:
\[ H₃BO₃(s) + aq. NaOH \overset{\Delta}{\rightarrow} NaBO₂H₄ \]
- Here, the product formed is sodium tetrahydroborate (NaBO₂H₄).

2. **Identify the second reaction:**
- The second reaction involves solid boric acid (H₃BO₃) reacting with molten sodium hydroxide (NaOH) when heated.
- The balanced equation for this reaction is:
\[ H₃BO₃(s) + \text{molten } NaOH \overset{\Delta}{\rightarrow} Na₃BO₃ + 3H₂O \]
- Here, the product formed is sodium borate (Na₃BO₃) along with water (H₂O).

3. **Determine the compounds (X) and (Y):**
- From the first reaction, we found that the product (X) is sodium tetrahydroborate (NaBO₂H₄).
- From the second reaction, we found that the product (Y) is sodium borate (Na₃BO₃).

4. **Final Answer:**
- Therefore, the compounds (X) and (Y) are:
- (X) = NaBO₂H₄
- (Y) = Na₃BO₃

### Conclusion:
The correct answer for the compounds (X) and (Y) is:
- **X = NaBO₂H₄**
- **Y = Na₃BO₃**