The number of 2-centre-2-electron and 3-centre-2-electron bonds in $$B_2H_6$$, respectively, are:

We first recall the general idea of an ordinary covalent bond. A normal single bond involves two atoms sharing two electrons and is therefore called a “two-centre two-electron” bond, abbreviated as $$2c\!-\!2e$$.

Sometimes, however, two electrons are shared by more than two atoms. When exactly two electrons are shared by three atoms we call the bond a “three-centre two-electron” bond, abbreviated as $$3c\!-\!2e$$. Such bonds are also known by the popular name “banana bonds.”

Now we examine the molecule $$\mathrm{B_2H_6}$$, diborane. Its experimentally determined skeletal framework is

H     H

\    /

B — B

/    \

H     H

\   /

H   H

In words, each boron atom is attached to two “terminal” hydrogens (those that point outward) and is simultaneously bridged by two additional hydrogens that sit between the two boron atoms.

Let us count the two kinds of bonds one by one.

Step 1: Counting the ordinary $$2c\!-\!2e$$ bonds. Each boron has two terminal B-H bonds. Therefore

Number of terminal B-H bonds $$= 2 \text{ (per B)} \times 2 \text{ (B atoms)} = 4.$$

Because every terminal B-H linkage involves only one boron and one hydrogen and uses two electrons, all four of these are regular $$2c\!-\!2e$$ bonds.

Step 2: Identifying the $$3c\!-\!2e$$ bonds. The two bridging hydrogens each form a bond that involves the following three centres: B1, Hbridge, and B2. Experimental evidence (electron-diffraction, infrared spectra, etc.) shows that only two electrons are shared among these three centres. Thus every bridge corresponds to one $$3c\!-\!2e$$ bond.

There are exactly two such bridging hydrogens, hence

Number of $$3c\!-\!2e$$ bonds $$= 2.$$

Step 3: Summarising the count.

$$\text{Number of }2c\!-\!2e\text{ bonds} = 4$$

$$\text{Number of }3c\!-\!2e\text{ bonds} = 2$$

Comparing these numbers with the options given, we see that the pair “4 and 2” exactly matches Option A.

Hence, the correct answer is Option A.