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 solve the question regarding the products formed from the reaction of sodium borohydride (NaBH₄) with iodine (I₂) in the presence of polyether, we can follow these steps:

### Step 1: Write the Reactants
The reactants in the reaction are:
- Sodium borohydride (NaBH₄)
- Iodine (I₂)
- Polyether (which acts as a solvent)

### Step 2: Identify the Reaction
When sodium borohydride reacts with iodine, it undergoes a redox reaction. Sodium borohydride acts as a reducing agent, while iodine acts as an oxidizing agent.

### Step 3: Determine the Products
The products formed from this reaction are:
- Diborane (B₂H₆)
- Sodium iodide (NaI)
- Hydrogen gas (H₂)

### Step 4: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction is:
\[ 2 \text{NaBH}_4 + \text{I}_2 \rightarrow \text{B}_2\text{H}_6 + 2 \text{NaI} + \text{H}_2 \]

### Step 5: Analyze the Oxidation States
In sodium borohydride (NaBH₄), the oxidation state of boron is -4. In diborane (B₂H₆), the oxidation state of boron is -3. This indicates that boron is oxidized, confirming that iodine is acting as an oxidizing agent.

### Step 6: Conclusion
Thus, the products formed from the reaction of 2 NaBH₄ with I₂ in the presence of polyether are:
- Diborane (B₂H₆)
- Sodium iodide (NaI)
- Hydrogen gas (H₂)

### Final Answer
The correct products formed are B₂H₆, NaI, and H₂.

---