Animated Solution for Chemistry - s and p-Block Elements: The number of 2-centre-2-electron and 3-centre-2-electron bonds in ${\text{B}}_2{\text{H}}_6$, respectively, are

Have you ever looked at a chemical formula and thought, 'Wait, that doesn't add up?' Diborane, with the formula $B_2{H}_6$, is exactly one of those molecules. If you try to draw its Lewis dot structure using the rules you learned in basic chemistry, you will quickly run into a frustrating wall. Let's embark on a journey to understand how nature bends the rules to create this fascinating molecule.

To truly appreciate the structure of diborane, we first need to look at a molecule that looks very similar on paper: ethane, or $C_2{H}_6$. In ethane, the two carbon atoms are bonded to each other, and each carbon is bonded to three hydrogen atoms. This requires a total of seven bonds. Since each normal covalent bond requires two electrons, ethane needs $7\times 2=14$ valence electrons. Carbon provides 4 each (total 8), and hydrogen provides 1 each (total 6), giving exactly 14 electrons. Perfect!

Now, let's apply this logic to diborane ($B_2{H}_6$). Boron is in Group 13 of the periodic table, meaning each boron atom brings only 3 valence electrons to the table. Two boron atoms give us 6 electrons. The six hydrogen atoms provide another 6 electrons.

Do you see the problem? If diborane were to have the same structure as ethane, it would need 14 electrons. But it only has 12! It is exactly two electrons short of being able to form a complete set of normal two-center, two-electron ($2c-2e^{-}$) bonds. This is why diborane is famously known as an electron-deficient molecule.

Business as Usual

Nature is incredibly resourceful. Since there aren't enough electrons to go around, the molecule adopts a unique geometry to maximize the sharing of the electrons it does have.

Imagine the two boron atoms sitting in space. Four of the six hydrogen atoms position themselves on the outside of the molecule. We call these the terminal hydrogens.

Each of these four terminal hydrogens forms a standard, everyday covalent bond with a boron atom. In these bonds, one electron comes from the hydrogen and one comes from the boron, creating a localized bond between two nuclei.

In chemistry terminology, this is a two-center, two-electron ($2c-2e^{-}$) bond.

Since there are four terminal hydrogens, there are exactly four $2c-2e^{-}$ bonds. Let's do some quick math: $4\text{ bonds}\times 2\text{ electrons/bond}=8\text{ electrons}$.

We started with 12 valence electrons, and we have just used 8 of them for the terminal bonds. This leaves us with exactly 4 electrons to hold the rest of the molecule together.

The Magic of Banana Bonds

We have two boron atoms, two remaining hydrogen atoms, and only 4 electrons left. If the two boron atoms formed a bond with each other, that would use 2 electrons, leaving only 2 electrons for the two remaining hydrogens. That simply wouldn't work.

Instead, the two remaining hydrogen atoms move into the space between the two boron atoms. We call these the bridging hydrogens.

Here is where the magic happens. Instead of a bond existing between just two atoms, the electron cloud stretches out to encompass three atoms: Boron, Hydrogen, and Boron.

Two electrons are shared across these three nuclei. This creates a three-center, two-electron ($3c-2e^{-}$) bond.

Because the electron cloud has to bend around the space between the two boron atoms to include the hydrogen atom, the electron density takes on a curved shape. This is why these bonds are affectionately referred to as banana bonds.

There is one bridging hydrogen above the plane of the terminal atoms, and one bridging hydrogen below the plane. Therefore, there are exactly two $3c-2e^{-}$ banana bonds in the diborane molecule. These two bonds use up the remaining 4 electrons ($2\text{ bonds}\times 2\text{ electrons/bond}=4\text{ electrons}$), perfectly accounting for all 12 valence electrons!

Let's bring it all together and answer the question. We have meticulously broken down the structure of diborane and accounted for every single valence electron.

1. Terminal Bonds: There are 4 terminal B-H bonds. Each of these is a standard bond involving two atoms and two electrons. Thus, there are four $2c-2e^{-}$ bonds. 2. Bridging Bonds: There are 2 bridging B-H-B bonds. Each of these involves three atoms sharing two electrons. Thus, there are two $3c-2e^{-}$ bonds.

The question asks for the number of $2c-2e^{-}$ and $3c-2e^{-}$ bonds, respectively. Based on our structural analysis, the numbers are 4 and 2.

This elegant solution by nature not only solves the electron deficiency problem but also creates a molecule with fascinating chemical properties. Whenever you see a Group 13 element, always be on the lookout for these clever multicenter bonding strategies!