`Na_(2) B_(4) O_(7) overset(740^(@) C)underset(Delta)(to) underset("Transparent")ubrace(underset(X" " + " "Y)(2Na BO_(2) + B_(2) O_(2)))`
`Z + CuO(s) to underset("Blue Bead")ubrace(Cu(BO_(2))_(2))`
The Z will be

Options

A. X
B. Y
C. Mixture of X & Y in 2: 1 ratio
D. Mixture of X & Y in 1 : 2 ratio

Answer

To solve the given question, we need to identify the compound Z that reacts with CuO to form a blue bead of Cu(BO2)2. The process involves understanding the decomposition of Na2B4O7 and the subsequent reactions.

### Step-by-Step Solution:

1. **Identify the decomposition of Na2B4O7**:
- When sodium tetraborate (Na2B4O7) is heated, it decomposes into sodium metaborate (NaBO2) and diboron dioxide (B2O2).
- The reaction can be written as:
\[ \text{Na}_2\text{B}_4\text{O}_7 \xrightarrow{740^\circ C} 2\text{NaBO}_2 + \text{B}_2\text{O}_2 \]
- Here, we denote sodium metaborate as X and diboron dioxide as Y.

2. **Identify the compound Z**:
- The problem states that Z will react with CuO to form a blue bead, which is Cu(BO2)2.
- The blue bead is formed when copper oxide reacts with a boron compound. The appropriate compound that reacts with CuO to form this blue bead is diboron dioxide (B2O2).

3. **Write the reaction for Z**:
- The reaction can be represented as:
\[ \text{B}_2\text{O}_2 + \text{CuO} \rightarrow \text{Cu(BO}_2\text{)}_2 \]
- Thus, we conclude that Z is B2O2.

4. **Final Answer**:
- Therefore, the compound Z is:
\[ \text{Z} = \text{B}_2\text{O}_2 \]

### Summary of the Solution:
- The compound Z that reacts with CuO to form the blue bead Cu(BO2)2 is diboron dioxide (B2O2).