Given below are the statements about diborane
(a) Diborane is prepared by the oxidation of NaBH$$_4$$ with 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 statement(s) is

We need to evaluate each statement about diborane ($$\text{B}_2\text{H}_6$$).

Statement (a): Diborane is prepared by the oxidation of $$\text{NaBH}_4$$ with $$\text{I}_2$$. The reaction is $$2\text{NaBH}_4 + \text{I}_2 \to \text{B}_2\text{H}_6 + 2\text{NaI} + \text{H}_2$$. Here iodine acts as the oxidising agent and $$\text{NaBH}_4$$ is oxidised to produce diborane. This statement is correct.

Statement (b): Each boron atom in diborane is $$sp^3$$ hybridised (not $$sp^2$$), since each boron forms four bonds — two terminal B-H bonds and two bridging 3-centre-2-electron bonds. This statement is incorrect.

Statement (c): Diborane has two banana-shaped 3-centre-2-electron (3c-2e) bridging bonds, not one. This statement is incorrect.

Statement (d): Diborane is not planar. The four terminal hydrogen atoms and the two boron atoms lie in one plane, but the two bridging hydrogen atoms lie above and below this plane, giving the molecule a non-planar structure. This statement is incorrect.

Since only statement (a) is correct, the answer is option (2): (a) only.