Question
Question: Suggest the reason why the \( \text{B - F} \) bond length in \( \text{B}{{\text{F}}_{\text{3}}} \) i...
Suggest the reason why the B - F bond length in BF3 is (130 pm) while that in BF-4 (143 pm) differ?
Solution
When two or more than two orbitals hybridize with each other to form hybrid orbitals then the bond length and the bond strength depends on the nature of the hybrid orbitals and the percentage of the “s-character” and the “p-character” of the orbitals.
Complete stepwise Solution
The s and the p orbitals can hybridize with each other to form three types of hybrid orbitals: sp hybridization, sp2 hybridization, and sp3 hybridization.
In the sp hybrid orbitals, the percentage of the s-character is 50 50%]
In the sp2 hybrid orbitals, the percentage of the s-character is 33.33 66.67\text{ }%]
In the sp3 hybrid orbitals, the percentage of the s-character is 25 75%].
Now, the s-orbitals are spherical in shape while the p-orbitals are dumbbell shaped. Hence the s-orbitals, having equal distribution of electrons on all sides, form better covalent bonds that the p-orbitals which have the electron density located only on the top and the bottom of the molecular plane. Therefore, the sp hybrid orbitals make the stronger bonds due to higher s-character than the sp2 hybrid orbitals which make stronger bonds than sp3 hybrid orbitals.
In the BF3 molecules, the hybridization of the boron atom is sp2 while that in the BF-4 ions is sp3 .
Hence, the BF3 molecules have stronger B - F bonds and shorter bond lengths as compared to the B - F bonds in BF-4 molecules.
Note
The hybridization is the phenomenon that involves intermixing of the atomic orbitals to have mixed orbitals with the properties of the orbitals involved in the bond formation. The number of the hybrid orbitals formed is the same as the number of the orbitals involved in bonding.