Compound `(A)` on reaction with iodine in the solvent diglyme gives a hydride `(B)` and hydrogen gas. The product `(B)` is instantly hydrolysed by water or aqueous alkali forming compound `( C)` and liberating hydrogen gas. The compound `( C)` in aqueous solution behaves as a week mono basic acid. But in presence of certain organic polyhydroxy compound behaves as a strong monobasic acid. The hydride `(B)` in air catches fire spontaneously forming oxide which gives coloured beads with transition metal compounds.
Which of the following statement is correct for the product `(C)`?

Options

A. It is an odd electron molecule.
B. It in water acts as proton donor.
C. It in solid state have hydrogen bonding.
D. It is a useful primary standard for titrating against acids.

Explanation

To solve the question step by step, we will analyze the reactions and properties of the compounds involved:

### Step 1: Identify Compound A
- **Given**: Compound (A) reacts with iodine in diglyme to form a hydride (B) and hydrogen gas.
- **Assumption**: Compound (A) is likely sodium borohydride (NaBH₄), which reacts with iodine to form diborane (B₂H₆) and sodium iodide (NaI).

### Step 2: Reaction of Compound A with Iodine
- **Reaction**:
\[ 2 \text{NaBH}_4 + \text{I}_2 \rightarrow \text{B}_2\text{H}_6 + 2 \text{NaI} + \text{H}_2 \]
- **Conclusion**: The hydride (B) formed is diborane (B₂H₆).

### Step 3: Hydrolysis of Hydride B
- **Given**: Hydride (B) is hydrolyzed by water or aqueous alkali to form compound (C) and liberate hydrogen gas.
- **Reaction**:
\[ \text{B}_2\text{H}_6 + 6 \text{H}_2\text{O} \rightarrow 2 \text{H}_3\text{BO}_3 + 3 \text{H}_2 \]
- **Conclusion**: The compound (C) formed is orthoboric acid (H₃BO₃).

### Step 4: Properties of Compound C
- **Given**: Compound (C) behaves as a weak monobasic acid in aqueous solution.
- **Behavior**:
- In water, orthoboric acid does not dissociate completely, thus acting as a weak acid.
- It can also form the tetrahydroxyborate ion (BO₃(OH)₄⁻) in equilibrium with water.

### Step 5: Interaction with Organic Polyhydroxy Compounds
- **Given**: In the presence of certain organic polyhydroxy compounds, compound (C) behaves as a strong monobasic acid.
- **Explanation**: This behavior can be attributed to the formation of more stable complexes or enhanced ionization in the presence of these compounds.

### Step 6: Behavior of Hydride B in Air
- **Given**: Hydride (B) catches fire spontaneously in air, forming an oxide.
- **Conclusion**: The oxide formed is likely boron oxide (B₂O₃), which can react with transition metal compounds to form colored beads.

### Step 7: Correct Statement about Compound C
- **Given Options**:
1. It is an odd electron molecule.
2. In water, it acts as a proton donor.
3. In solid state, it has hydrogen bonding.
4. It is a useful primary standard for titrating against acids.

- **Analysis**:
- **Option 1**: Incorrect. Orthoboric acid has an even number of electrons.
- **Option 2**: Incorrect. It acts as a Lewis acid and does not donate protons directly.
- **Option 3**: Correct. In solid state, orthoboric acid forms a network of hydrogen bonds.
- **Option 4**: Incorrect. It is not a primary standard for titration.

### Final Answer
The correct statement regarding compound (C) is:
**In solid state, it has hydrogen bonding.**

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