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 A .

Options

A. `NaNH_(4)HPO_(4).4H_(2)O`
B. `Na_(2)B_(4)O_(7).10H_(2)O`
C. `CuSO_(4).SH_(2)O`
D. None of these

Explanation

To identify the compound A based on the given information, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formation of Glassy Transparent Bead:**
- When compound A is heated on a platinum wire, it forms a glassy transparent bead. This is characteristic of borax (sodium tetraborate), which can be represented by the formula \( \text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O} \).
- Upon heating, borax decomposes to form sodium metaborate (\( \text{NaBO}_2 \)) and boron trioxide (\( \text{B}_2\text{O}_3 \)), which are components of the glassy bead.

2. **Reaction with Copper Sulfate:**
- The glassy bead formed from A reacts with copper sulfate (\( \text{CuSO}_4 \)) to produce a colored bead. This indicates that the bead contains boron, which can react with copper ions to form colored complexes.

3. **Reaction with Concentrated Sulfuric Acid and Ethanol:**
- When A is treated with concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) and ethanol (\( \text{C}_2\text{H}_5\text{OH} \)), it produces a green flame. This is indicative of the formation of boron compounds that can emit a green flame when ignited, such as boric acid (\( \text{H}_3\text{BO}_3 \)).

4. **Alkaline Nature of Aqueous Solution:**
- The aqueous solution of A is alkaline. When borax dissolves in water, it hydrolyzes to form sodium hydroxide (\( \text{NaOH} \)) and boric acid, which makes the solution alkaline.

5. **Conclusion:**
- Based on the above observations, we can conclude that compound A is borax, which has the formula \( \text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O} \).

### Final Answer:
The compound A is **Borax** (sodium tetraborate), with the formula \( \text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O} \).

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