`"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 `(E)` is.

Options

A. `B_(2)O_(3)`
B. `B`
C. `H_(3)BO_(3)`
D. None of these

Explanation

To solve the question step by step, we will analyze the reactions provided and identify the compounds A, B, C, D, and E sequentially.

### Step 1: Identify Compound A
The first reaction involves Colemanite (Ca2B6O11) reacting with sodium carbonate (Na2CO3) upon fusion. The products of this reaction are:
- A: Sodium metaborate (NaBO2)
- B: Sodium tetraborate (Na2B4O7)
- CO2

Thus, we can write:
\[ \text{Colemanite} + \text{Na}_2\text{CO}_3 \xrightarrow{\text{Fused}} \text{NaBO}_2 \, (A) + \text{Na}_2\text{B}_4\text{O}_7 \, (B) + \text{CO}_2 \]

### Step 2: Identify Compound B
Next, we take compound A (NaBO2) and react it with carbon dioxide (CO2). The products of this reaction are:
- B: Sodium tetraborate (Na2B4O7)
- Na2CO3

So, we can express this reaction as:
\[ \text{NaBO}_2 \, (A) + \text{CO}_2 \rightarrow \text{Na}_2\text{B}_4\text{O}_7 \, (B) + \text{Na}_2\text{CO}_3 \]

### Step 3: Identify Compound C
Now, we take compound B (Na2B4O7) and react it with concentrated hydrochloric acid (HCl). The products of this reaction are:
- NaCl
- C: Boric acid (H3BO3)

Thus, we have:
\[ \text{Na}_2\text{B}_4\text{O}_7 \, (B) + \text{conc HCl} \rightarrow \text{NaCl} + \text{H}_2\text{B}_4\text{O}_7 \, (C) \]

### Step 4: Identify Compound D
Next, we take compound C (H2B4O7) and react it with water (H2O). The product of this reaction is:
- D: Boric acid (H3BO3)

This can be represented as:
\[ \text{H}_2\text{B}_4\text{O}_7 \, (C) + \text{H}_2\text{O} \rightarrow \text{H}_3\text{BO}_3 \, (D) \]

### Step 5: Identify Compound E
Finally, we take compound D (H3BO3) and subject it to strong heating. The product of this reaction is:
- E: Boron trioxide (B2O3)

This can be expressed as:
\[ 2 \text{H}_3\text{BO}_3 \, (D) \xrightarrow{\text{Strong heating}} \text{B}_2\text{O}_3 \, (E) + 3 \text{H}_2\text{O} \]

### Conclusion
Thus, the compound E is:
\[ \text{B}_2\text{O}_3 \]