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Identify the correct statement for $B _{2} H _{6}$ from those given below.
$(A)$ In $B _{2} H _{6}$, all $B - H$ bonds are equivalent.
$(B)$ In $B _{2} H _{6}$ there are four $3-$centre$-2-$electron bonds.
$(C)$ $B _{2} H _{6}$ is a Lewis acid.
$(D)$ $B _{2} H _{6}$ can be synthesized form both $BF _{3}$ and $NaBH _{4}$.
$(E)$ $B _{2} H _{6}$ is a planar molecule.
Choose the most appropriate answer from the options given below..... .
$(A)$ and $(E)$ only
$(B), (C)$ and $(E)$ only
$(C)$ and $(D)$ only
$(C)$ and $(E)$ only
Solution
$(A)\, (B)$ Two $3$ centre $-2$-electron bonds
$(C)$ $B _{2} H _{6}$ is $e ^{-}$deficient species
$(E)$ $B _{2} H _{6}$ is non – Planar molecule
$(D)$ $BF _{3}+ LiAlH _{4} \rightarrow 2 B _{2} H _{6}+3 LiF +3 AlF _{3}$
$NaBH _{4}+ I _{2} \rightarrow B _{2} H _{6}+2 NaI + H _{2}$
