Products formed are
`2 Na BH_(4) + l_(2) overset("Polyether")(to)`

Options

A. Hl , Nal and `H_(2)`
B. `B_(2) H_(6), Nal ` and Hl
C. `B_(2) H_(6) , Nal` and `H_(2)`
D. `H_(3) BO_(3) + H_(2)`

Answer

To determine the products formed when sodium borohydride (NaBH₄) reacts with iodine (I₂), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are sodium borohydride (NaBH₄) and iodine (I₂).

2. **Write the Balanced Chemical Equation**:
The balanced chemical equation for the reaction of sodium borohydride with iodine can be written as:
\[ 2 \text{NaBH}_4 + \text{I}_2 \rightarrow \text{B}_2\text{H}_6 + 2 \text{NaI} + \text{H}_2 \]
This equation indicates that two moles of sodium borohydride react with one mole of iodine to produce diborane (B₂H₆), sodium iodide (NaI), and hydrogen gas (H₂).

3. **Identify the Products**:
From the balanced equation, we can identify the products formed:
- Diborane (B₂H₆)
- Sodium iodide (NaI)
- Hydrogen gas (H₂)

4. **Select the Correct Option**:
Now, we need to match the products with the given options:
- Option 1: HI, NaI, H₂ (Incorrect)
- Option 2: B₂H₆, NaI, HI (Incorrect)
- Option 3: B₂H₆, NaI, H₂ (Correct)
- Option 4: H₃BO₃, H₂ (Incorrect)

The correct answer is Option 3: B₂H₆, NaI, H₂.

### Final Answer:
The products formed when 2 moles of NaBH₄ react with I₂ are B₂H₆ (diborane), NaI (sodium iodide), and H₂ (hydrogen gas).

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