From $B_{2} H_{6}$ , all the following can be prepared except :

B_{2} O_{3}

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H_{3} B O_{3}

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B_{2} (C H_{3})_{6}

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N a B H_{4}

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Solution

The correct option is C $B_{2} (C H_{3})_{6}$
$B_{2} H_{6} + 3 O_{2} \to B_{2} O_{3} + 3 H_{2} O + H e a t$

$B_{2} H_{6} + 6 H_{2} O \to H_{3} B O_{3} + 6 H_{2}$

$2 N a H + B_{2} H_{6} \to e t h e r N a B H_{4}$

From $B_{2} H_{6}$ , all can be prepared except $B_{2} (C H_{3})_{6}$

Hence option $C$ is correct.

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