i.(A) `overset(Delta)to ` glassy traparent beat (B) on platinum wire
(B) `+ CuSO_(4) rarr ` coloured bead(C )
ii `underset((D))((A))+conc. H_(2)SO_(4)+CH_(3)CH_(2)OH overset("ignite")rarr` green flame
iv.Aqueous solution (A) is alkline
Identify D.

Options

A. `(CH_(3))_(3)BO_(3)`
B. `C_(2)H_(5))_(3)BO_(3)`
C. `H_(3)BO_(3)`
D. None of these

Explanation

To solve the problem step by step, let's analyze each part of the question systematically.

### Step 1: Identify Compound A
The first part of the question states that heating compound A produces a glassy transparent bead (B) on a platinum wire. This is characteristic of borax, which is sodium tetraborate decahydrate (Na2B4O7·10H2O).

**Conclusion:**
A = Borax (Na2B4O7·10H2O)

### Step 2: Reaction of B with CuSO4
Next, the bead B (which is formed from borax) is treated with copper sulfate (CuSO4) to yield a colored bead (C). When borax is heated and treated with CuSO4, it forms a colored compound due to the presence of boron.

**Conclusion:**
B + CuSO4 → C (colored bead)

### Step 3: Reaction of A with H2SO4 and Ethanol
The next part of the question states that when compound A (borax) is treated with concentrated sulfuric acid (H2SO4) and ethanol (C2H5OH), and ignited, it produces a green flame.

The reaction can be summarized as follows:
1. Borax reacts with concentrated H2SO4 to form sodium sulfate (Na2SO4) and boric acid (H3BO3).
2. Boric acid further reacts with ethanol to form triethyl borate (C2H5)3BO3, which is responsible for the green flame.

**Conclusion:**
A + H2SO4 + C2H5OH (ignite) → D (green flame)

### Step 4: Identify Compound D
From the above reactions, we can deduce that compound D is triethyl borate, which is known to produce a green flame when burned.

**Conclusion:**
D = Triethyl borate (B(OEt)3)

### Final Answer
D is identified as triethyl borate.

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