Animated Solution for Chemistry - s and p-Block Elements: The bond dissociation energy of B—F in ${\text{BF}}_3$ is $646{\text{ kJ mol}}^{-1}$ whereas that of C—F in ${\text{CF}}_4$ is $515{\text{ kJ mol}}^{-1}$. The correct reason for higher B—F bond dissociation energy as compared to that of C—F is

Imagine you are an atomic architect, tasked with building two molecules: Boron trifluoride (${\text{BF}}_3$) and Carbon tetrafluoride (${\text{CF}}_4$). Both molecules feature a central atom bonded to highly electronegative Fluorine atoms. On paper, you might expect the bonds to behave similarly. However, experimental data throws a massive curveball: the bond dissociation energy of the $\text{B—F}$ bond is a staggering $646{\text{ kJ mol}}^{-1}$, while the $\text{C—F}$ bond sits much lower at $515{\text{ kJ mol}}^{-1}$.

Why is the $\text{B—F}$ bond so incredibly robust? To solve this mystery, we must look beyond the simple lines we draw on paper and dive deep into the quantum mechanical reality of atomic orbitals.

Let's start by putting Boron under the microscope. Boron has an atomic number of $5$. In its ground state, its electronic configuration is $1s^{2}2s^{2}2p^1$. When it prepares to bond with three Fluorine atoms, it promotes an electron to achieve an excited state configuration of $1s^{2}2s^{1}2p_x^{1}2p_y^{1}2p_z^0$.

Boron then undergoes $s{p}^2$ hybridization, mixing its $2s$ and two of its $2p$ orbitals to form three identical $s{p}^2$ hybrid orbitals. These orbitals form the strong $\sigma$-bonds with the Fluorine atoms, creating a flat, trigonal planar geometry. But here is the critical detail: Boron is left with one completely empty, unhybridized $2p_z$ orbital sitting perpendicular to the molecular plane.

Now, let's look at the neighbors. Fluorine is a halogen with an atomic number of $9$ and a valence configuration of $2s^{2}2p^5$. After forming a single $\sigma$-bond with Boron, each Fluorine atom still possesses three lone pairs of electrons, which reside in fully occupied $2p$ orbitals.

We now have a fascinating setup: an electron-deficient Boron atom with an empty $2p$ orbital right next to an electron-rich Fluorine atom with filled $2p$ orbitals. Because Boron and Fluorine are both in the second period of the periodic table, their $2p$ orbitals are of identical size and energy.

This perfect energetic and spatial match allows Fluorine to donate one of its lone pairs back into Boron's empty $2p$ orbital. This sidewise overlap creates an additional $\pi$-bond. Because this interaction occurs between two $p$-orbitals, it is famously known as $p\pi -p\pi$ back bonding.

What does this back bonding actually do? It fundamentally changes the nature of the $\text{B—F}$ connection. The bond is no longer just a simple single $\sigma$-bond; it acquires a partial double bond character.

As a universal rule in chemistry, double bonds are shorter and significantly stronger than single bonds. The extra electron density pulling the nuclei together makes the bond much harder to break. This is the exact reason why the bond dissociation energy of $\text{B—F}$ skyrockets to $646{\text{ kJ mol}}^{-1}$.

So, why doesn't Carbon pull off the same trick in ${\text{CF}}_4$? Carbon has an atomic number of $6$. In its excited state, its configuration is $1s^{2}2s^{1}2p_x^{1}2p_y^{1}2p_z^1$. To bond with four Fluorine atoms, Carbon undergoes $s{p}^3$ hybridization, utilizing all of its valence orbitals to form four $\sigma$-bonds.

Unlike Boron, Carbon in ${\text{CF}}_4$ has absolutely no empty $p$-orbitals left. Even though the surrounding Fluorine atoms are practically begging to donate their lone pairs, Carbon simply has no room to accept them. Without an empty orbital, back bonding is impossible. The $\text{C—F}$ bond remains a pure single bond, which explains its lower bond dissociation energy of $515{\text{ kJ mol}}^{-1}$.

The dramatic difference in bond energies boils down to the availability of an empty orbital. The significant $p\pi -p\pi$ interaction between Boron and Fluorine in ${\text{BF}}_3$ fortifies the bond, whereas the lack of such interaction in ${\text{CF}}_4$ leaves it comparatively weaker. Therefore, the correct explanation is the significant $p\pi -p\pi$ interaction present in ${\text{BF}}_3$.