Given below are the statements about diborane
(a) Diborane is prepared by the oxidation of NaBH₄ with I₂
(b) Each boron atom is in sp² hybridized state
(c) Diborane has one bridged 3 centre-2-electron bond
(d) Diborane is a planar molecule
The option with correct statements is :

Solution

Chemistry ma’am · 3 steps

Picture the diborane structure: two hydrogen bridges, a tetrahedral boron, and it is not flat.

1. Preparation. Iodine oxidises the hydride in sodium borohydride: 2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂. This is the standard laboratory method, so statement (a) is true.
2. The structure. Each boron forms two normal bonds to terminal H atoms and shares two bridging H atoms with the other boron. That makes four bonds per boron, arranged roughly tetrahedrally, so boron is sp³, not sp².

   Statement	Verdict	Reason
   (a) made from NaBH₄ and I₂	True	standard laboratory preparation
   (b) boron is sp²	False	four bonds on each B, so sp³
   (c) one 3c-2e bond	False	there are two B–H–B bridges
   (d) planar molecule	False	bridging H atoms sit above and below the plane

3. So. Only statement (a) is correct, which matches option (b).

The idea to keep. Remember diborane as two BH₂ units held by two banana bonds: 4 terminal B–H bonds, 2 three-centre two-electron bridges, sp³ boron, non-planar.

Answer from the NTA final answer key. Solution by PrepWiser.