`"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 `(F)` is.

Options

A. `CuS`
B. `Cu_(2)O`
C. `CuSO_(3)`
D. `Cu(BO_(2))_(3)`

Explanation

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

### Step 1: Reaction of Colemanite with Sodium Carbonate
The first reaction involves Colemanite (Ca2B6O11) reacting with sodium carbonate (Na2CO3) when fused. The products formed are:
- (A): Sodium Metaborate (NaBO2)
- (B): Sodium Tetraborate (Na2B4O7)
- Byproduct: Carbon Dioxide (CO2)

**Products:**
- A = NaBO2
- B = Na2B4O7
- CO2 is also produced.

### Step 2: Reaction of A with CO2
Next, sodium metaborate (A) reacts with carbon dioxide (CO2) to form:
- (B): Sodium Tetraborate (Na2B4O7)
- Byproduct: Sodium Carbonate (Na2CO3)

**Products:**
- A + CO2 → B + Na2CO3

### Step 3: Reaction of B with Concentrated HCl
Now, sodium tetraborate (B) reacts with concentrated hydrochloric acid (HCl) to produce:
- Sodium Chloride (NaCl)
- (C): Tetraboric Acid (H2B4O7)

**Products:**
- B + HCl → NaCl + C (where C = H2B4O7)

### Step 4: Reaction of C with Water
The tetraboric acid (C) then reacts with water (H2O) to yield:
- (D): Tetraboric Acid remains the same (H2B4O7)

**Products:**
- C + H2O → D (where D = H2B4O7)

### Step 5: Strong Heating of D
Upon strong heating of (D), we get:
- (E): Boron Trioxide (B2O3)
- Byproduct: Water (H2O)

**Products:**
- D (H2B4O7) → E (B2O3) + H2O

### Step 6: Reaction of E with Copper Sulfate
Finally, boron trioxide (E) reacts with copper sulfate (CuSO4) when heated in a flame to produce:
- (F): Copper(II) Borate (CuB2O4), which is a blue-colored compound.

**Products:**
- E + CuSO4 → F (where F = CuB2O4)

### Conclusion
The blue-colored compound (F) formed in the last reaction is **Copper(II) Borate (CuB2O4)**.

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