Question

In which of the following molecule back bonding is present?
(A) $B{{F}_{3}}$
(B) ${{H}_{2}}O$
(C) $N{{H}_{3}}$
(D) $HF$

Answer 1

After a sigma bond is formed between two adjacent atoms in a molecule, further multiple bonds may be present if one of the atoms has a lone pair of electrons and the adjacent atom has vacant orbitals in it.

Complete step by step solution:
The sideways overlapping of the filled and empty orbitals from the atoms with lone-pairs and with vacant p- or d-orbitals respectively, thus forming a weaker $\pi$-bond. This is known as the back- bonding, with its direction towards the vacant orbital.
So, the atoms of the molecule are generally from Period 2 and 3 of the periodic table, with at least one of them from the second period.
-As the donor atom, that is, the ligand with lone pairs has localised electrons or filled $\pi$-orbital like in F, O, N atoms.
-The acceptor atom, that is, metal has the empty orbitals of low energy like the p and d-orbitals, with the overlapping between atoms with similar orbitals and size is favoured.
So, the order of the extent of overlapping is as follows:
$2p-2p>2p-3d>2p-3p$
Therefore, in ${{H}_{2}}O$, $N{{H}_{3}}$ and $HF$ with hydrogen as one of the atoms with only s-orbital. It does not have vacant orbitals p- or d-orbitals present in it.
Whereas, in case of $B{{F}_{3}}$ with a trigonal planar geometry. After the formation of the three sigma-bonds, that is, B-F. The boron is left with a vacant p-orbital and the fluorine having three lone pairs in the $s{{p}^{3}}$ hybridised orbital, whose overlapping cause back- bonding.

Therefore, the molecule in which back-bonding is present is the option (A)- $B{{F}_{3}}$ molecule.

Note: Due to the back-boding occurrence in the molecule, it affects the stability, bond length, bond angle and bond order of the molecule. As it leads to the increase in the electron-density in between the bond. Thus, increasing the bond angle. Also, the bond length decreases along with the increase in bond order and stability.