Question:

Given below are the statements about diborane
(a) Diborane is prepared by the oxidation of \(\mathrm{NaBH_4}\) with \(\mathrm{I_2}\)
(b) Each boron atom is in \(sp^2\) hybridized state
(c) Diborane has one bridged 3 centre-2-electron bond
(d) Diborane is a planar molecule
The option with correct statements is :

Show Hint

In \(\mathrm{B_2H_6}\) each B is \(sp^3\), there are two 3c-2e bridge bonds and the molecule is non-planar; it is made by \(\mathrm{NaBH_4}\) + \(\mathrm{I_2}\).
Updated On: Nov 4, 2026
  • (c) and (d) only
  • (a) only
  • (c) only
  • (a) and (b) only
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The Correct Option is B

Solution and Explanation

(a) \(2\mathrm{NaBH_4} + \mathrm{I_2} \rightarrow \mathrm{B_2H_6} + 2\mathrm{NaI} + \mathrm{H_2}\): correct (BH4- is oxidised).
(b) In diborane each boron is \(sp^3\) hybridised: incorrect.
(c) Diborane has two bridging 3-centre-2-electron (B-H-B) bonds, not one: incorrect.
(d) Four terminal B-H bonds lie in one plane, but the two bridge H atoms lie above and below it, so the molecule is non-planar: incorrect.
Only (a) is correct.
Answer: (2) (a) only
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