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

Hint: 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.