`NaBH_(4)+l_(2)toXuarr+Yuarr+2Nal`
`X+C_(2)H_(5)OHtoY uarr+D`
`X+HCl to Y uarr +E`
`D` gives folowwing colour with flame.

Options

A. Red
B. Green
C. Blue
D. No colour

Explanation

To solve the problem step-by-step, let's analyze the reactions provided in the question:

### Step 1: Reaction of NaBH₄ with I₂
- **Reaction**: \( \text{NaBH}_4 + \text{I}_2 \rightarrow X + Y + 2 \text{NaI} \)
- **Explanation**: Sodium borohydride (\( \text{NaBH}_4 \)) acts as a reducing agent and reduces iodine (\( \text{I}_2 \)) to iodide ions (\( \text{NaI} \)). The by-products of this reaction include diborane (\( \text{B}_2\text{H}_6 \)) and hydrogen gas (\( \text{H}_2 \)).
- **Identifying X and Y**: Here, \( X \) is \( \text{B}_2\text{H}_6 \) and \( Y \) is \( \text{H}_2 \).

### Step 2: Reaction of X with Ethanol
- **Reaction**: \( X + \text{C}_2\text{H}_5\text{OH} \rightarrow Y + D \)
- **Explanation**: When diborane (\( \text{B}_2\text{H}_6 \)) reacts with ethanol (\( \text{C}_2\text{H}_5\text{OH} \)), it forms tetraethyl borate (\( \text{C}_2\text{H}_5\text{O})_4\text{B} \)) and releases hydrogen gas (\( \text{H}_2 \)).
- **Identifying D**: Here, \( D \) is tetraethyl borate, which has a characteristic green color.

### Step 3: Reaction of X with HCl
- **Reaction**: \( X + \text{HCl} \rightarrow Y + E \)
- **Explanation**: Diborane (\( \text{B}_2\text{H}_6 \)) reacts with hydrochloric acid (\( \text{HCl} \)) to produce hydrogen gas (\( \text{H}_2 \)) and \( \text{B}_2\text{H}_5\text{Cl} \) (dichloroborane).
- **Identifying Y and E**: Here, \( Y \) is \( \text{H}_2 \) and \( E \) is \( \text{B}_2\text{H}_5\text{Cl} \).

### Step 4: Color of D
- **Conclusion**: The color of tetraethyl borate (\( D \)) is green.

### Final Answer:
- The color of \( D \) is green.

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