Identify the correct statement for $$B_2H_6$$ from those given below.
(A) In $$B_2H_6$$, all B $$-$$ H bonds are equivalent.
(B) In $$B_2H_6$$, there are four 3-centre-2-electron bonds.
(C) $$B_2H_6$$ is a Lewis acid.
(D) $$B_2H_6$$ can be synthesized from both $$BF_3$$ and $$NaBH_4$$.
(E) $$B_2H_6$$ is a planar molecule.
Choose the most appropriate answer from the options given below :

We need to identify the correct statements about $$B_2H_6$$ (diborane).

Analyzing each statement:

Statement (A): "All B-H bonds are equivalent" - This is incorrect. In $$B_2H_6$$, there are two types of B-H bonds:

- 4 terminal B-H bonds (2-centre-2-electron bonds, shorter, ~1.19 Å)

- 2 bridging B-H-B bonds (3-centre-2-electron bonds, longer, ~1.33 Å)

Statement (B): "There are four 3-centre-2-electron bonds" - This is incorrect. There are only two 3-centre-2-electron (3c-2e) bonds in $$B_2H_6$$, corresponding to the two bridging B-H-B bonds.

Statement (C): "$$B_2H_6$$ is a Lewis acid" - This is correct. Boron in $$B_2H_6$$ is electron-deficient (has an incomplete octet). It readily accepts electron pairs from Lewis bases like $$NH_3$$, $$CO$$, etc.

Statement (D): "$$B_2H_6$$ can be synthesized from both $$BF_3$$ and $$NaBH_4$$" - This is correct.

$$3NaBH_4 + 4BF_3 \rightarrow 2B_2H_6 + 3NaBF_4$$

This is a standard industrial synthesis of diborane.

Statement (E): "$$B_2H_6$$ is a planar molecule" - This is incorrect. $$B_2H_6$$ has a non-planar structure. The four terminal H atoms and the two B atoms are in one plane, while the two bridging H atoms lie above and below this plane.

Therefore, only statements (C) and (D) are correct.

The correct answer is Option C: (C) and (D) only.