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

H_{3} B O_{3}

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[B H_{2} (N H_{3})_{2}]^{+} [B H_{4}]^{-}

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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 $B_{2} (C H_{3})_{6}$
(C) $C H_{3}$ group being larger can not form a bridge between two small sized boron atoms.

(A) $B_{2} H_{6} + 6 H_{2} O \to 2 H_{3} B O_{3} + 6 H_{2} ↑$

(B)
$B_{2} H_{6} + 2 N H_{3} \to [B H_{2} (N H_{3})_{2}]^{+} [B H_{4}]^{-}$
This is formed by unsymmetrical cleavage.
(D) $2 N a H + B_{2} H_{6} \to N a B H_{4}$

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