Question

In which of the following molecules, vacant orbital cannot participate in bonding?
A. ${B_2}{H_6}$
B. $A{l_2}C{l_6}$
C. $\left[ {{H_3}N.{\text{ }}B{F_3}} \right]$
D. $S{i_2}{H_6}$

Answer 1

Back bonding is the bonding between two atoms in which one atom can have one vacant orbital and the other have a lone pair of electrons.

Complete step by step answer:
Back bonding is a type of resonance. It is the sharing of electrons between the atomic orbital on one atom with the antibonding orbital on another atom.
Back bonding is only possible In $\left[ {{H_3}N.{\text{ }}B{F_3}} \right]$ due to the presence of an empty p-orbital on B and lone pairs of electrons on the p orbital of F. In this case B act as a Lewis acid and F act as a Lewis base. Lone pair on the F is donated to the p orbital of B.
We know that F is highly electronegative in nature. So, it has a tendency to take back the electrons which are donated to B. So there will be a continued jumping of electrons.

Hence the correct answer is (C) i.e $\left[ {{H_3}N.{\text{ }}B{F_3}} \right]$ . Due to back bonding its orbital is not available for bonding.

Note:
Back bonding increases the stability of molecules. This bonding occurs mostly in organometallic compounds. In metal carbonyls the presence of back bonding will decreases the bond order and $C - O$ bond length will increase. This will decrease the $C - O$ stretching frequency and increase the $M - C$ Bond order. Hence $M - C$ bond length decreases and $M - C$ stretching frequency increases. Back bonding can affect the properties such as dipole moment, hybridisation etc. CO, $N{\left( {Si{H_3}} \right)_{3,}}P{F_3}$ are some molecules showing back bonding.