JEE Mains · Chemistry · STD 11 -p-Block elements - I
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 \((A)\) and \((E)\) only
- B \((B), (C)\) and \((E)\) only
- C \((C)\) and \((D)\) only
- D \((C)\) and \((E)\) only
Answer & Solution
Correct Answer
(C) \((C)\) and \((D)\) only
Step-by-step Solution
Detailed explanation
\((A)\, (B)\) Two \(3\) centre \(-2\)-electron bonds \((C)\) \(B _{2} H _{6}\) is \(e ^{-}\)deficient species \((E)\) \(B _{2} H _{6}\) is non - Planar molecule \((D)\) \(BF _{3}+ LiAlH _{4} \rightarrow 2 B _{2} H _{6}+3 LiF +3 AlF _{3}\)…
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