`"Colemanite" + Na_(2)CO_(3) overset("Fused")rarr (A) + (B) + CO_(2)`
`underset("Solution")((A)) + CO_(2) rarr(B) + Na_(2) CO_(3)`
`(B) + conc HCl rarr NaCl + underset("Acid")((C))`
`underset("Acid")((C)) +H_(2)O rarr underset("Acid")((D))`
`(D) overset("Strong heating")rarr (E)`
`(E) +CuSO_(4) overset("Heated") underset("inflame")rarr underset("Blue-coloured compound")((F))`.
Compound `(A)` is.

Options

A. `NaBO_(2)`
B. `Na_(2)B_(4)O_(7)`
C. `Na_(3)BO_(3)`
D. `NaOH`

Explanation

To solve the problem step by step, we will analyze the reactions provided in the question and identify the compounds involved.

### Step 1: Identify the Reactants
The first reactant is "Colemanite," which has the chemical formula \( \text{Ca}_2\text{B}_6\text{O}_{11} \). The second reactant is sodium carbonate, \( \text{Na}_2\text{CO}_3 \).

### Step 2: Reaction of Colemanite with Sodium Carbonate
When Colemanite reacts with sodium carbonate upon fusion, the products formed are sodium metaborate and another compound. The balanced reaction can be represented as:
\[ \text{Ca}_2\text{B}_6\text{O}_{11} + \text{Na}_2\text{CO}_3 \xrightarrow{\text{Fused}} \text{NaB}_2\text{O}_4 + \text{Na}_2\text{CO}_3 + \text{CO}_2 \]
Here, the product (A) is sodium metaborate, which has the formula \( \text{NaB}_2\text{O}_4 \).

### Step 3: Identify Product (A)
From the reaction, we can conclude that product (A) is sodium metaborate, \( \text{NaB}_2\text{O}_4 \).

### Step 4: Further Reactions
The next part of the question involves the reaction of (A) with carbon dioxide and sodium carbonate, leading to another compound (B) and more carbon dioxide. This is a cyclic process that eventually leads to the formation of an acid (C) when (B) reacts with concentrated hydrochloric acid.

### Step 5: Identify the Final Products
The final products from the reactions lead to the formation of a blue-colored compound (F) when (E) reacts with \( \text{CuSO}_4 \) upon heating.

### Conclusion
Thus, the compound (A) is sodium metaborate, \( \text{NaB}_2\text{O}_4 \).

### Final Answer
**Compound (A) is sodium metaborate, \( \text{NaB}_2\text{O}_4 \).**

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