Animated Solution for Chemistry - s and p-Block Elements: In which one of the following molecules strongest back donation of an electron pair from halide to boron is expected?

Unveiling the Power of Back Bonding

Imagine you are Boron in a Boron Trihalide ($B{X}_3$) molecule. You have formed three covalent bonds with three halogen atoms, but there is a problem. You only have six electrons in your valence shell. You are electron-deficient, hungry for a full octet, and sitting there with an empty $2p$ orbital perpendicular to your molecular plane.

This is the classic setup for a Lewis acid—a species desperate to accept an electron pair. But before an external molecule can come to the rescue, help arrives from within.

The halogen atoms surrounding Boron are rich in electrons. They have fully filled $p$ orbitals containing lone pairs. Seeing Boron's empty $2p$ orbital, a halogen atom can generously donate one of its lone pairs into Boron's empty space.

This internal sharing of electrons is called back bonding (specifically, $p\pi -p\pi$ back donation). It acts like a partial double bond, temporarily satisfying Boron's electron deficiency and stabilizing the molecule.

The Overlap Principle

Now, the critical question arises: In which boron trihalide (${\text{BF}}_3$, ${\text{BCl}}_3$, ${\text{BBr}}_3$, or ${\text{BI}}_3$) is this back bonding the strongest? The answer lies in the geometry of the orbitals.

For a strong $\pi$ bond to form, the overlapping orbitals must be of similar size and energy. Think of it like two people trying to give each other a high-five; it works best if they are standing on the same level.

- In ${\text{BF}}_3$, Boron uses its $2p$ orbital, and Fluorine uses its $2p$ orbital. Because both are $2p$ orbitals, their sizes match perfectly. The $2p-2p$ overlap is highly effective, leading to strong back donation. - In ${\text{BCl}}_3$, Boron still uses its $2p$ orbital, but Chlorine uses a larger $3p$ orbital. The size mismatch makes the $2p-3p$ overlap much weaker. - As we move to ${\text{BBr}}_3$ ($2p-4p$) and ${\text{BI}}_3$ ($2p-5p$), the halogen orbitals become so large and diffused that the overlap with Boron's tiny $2p$ orbital is practically negligible.

Because the orbital size increases down the halogen group, the effectiveness of back bonding decreases in the order:

Therefore, the strongest back donation is expected in ${\text{BF}}_3$.

Bonus Insight for JEE: How does this affect their Lewis acidity? Since strong back bonding in ${\text{BF}}_3$ satisfies Boron's electron hunger internally, it doesn't need external electrons as much. Conversely, ${\text{BI}}_3$ gets almost no internal help, making it desperately hungry for electrons. Thus, the Lewis acidity trend is the exact opposite of the back bonding trend: ${\text{BI}}_3>{\text{BBr}}_3>{\text{BCl}}_3>{\text{BF}}_3$.