All the products formed in the oxidation of $N a B H_{4}$ by $I_{2}$ , are:

B_{2} H_{6}

and

N a I

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B_{2} H_{6}, H_{2}

and

N a I

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B I_{3}

and

N a H

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N a B I_{4}

and

H I

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Solution

The correct option is B $B_{2} H_{6}, H_{2}$ and $N a I$
The reaction is as follows:

$2 N a B H_{4} (s) + I_{2} (s) \to B_{2} H_{6} (g) + 2 N a I (s) + H_{2} (g)$

Hence, the correct option is $B$

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