Question

The bond order of $NO$ is 2.5 while that of $N{{O}^{+}}$ is 3. Which of the following statements is true for these two species?
(A) Bond length in $N{{O}^{+}}$ is equal to that in $NO$.
(B) Bond length in $NO$ is greater to that in $N{{O}^{+}}$.
(C) Bond length in $N{{O}^{+}}$ is greater to that in $NO$.
(D) Bond length is unpredictable.

Answer 1

Hint: Bond order can be calculated with the help of molecular orbital theory. Understand the molecular orbital theory and the MOT diagram for the molecule under consideration. The bond order can be calculated by knowing the number of electrons in bonding orbitals and number of electrons in antibonding orbitals. The formula depicting the relation is given below:
$B.O=\dfrac{({{e}^{-}}_{B}-{{e}^{-}}_{AB})}{2}$
Where,
B.O stands for bond order
${{e}^{-}}_{B}$ denotes number of electrons in bonding orbitals
${{e}^{-}}_{AB}$ denotes number of electrons in antibonding orbitals

Complete Solution :
Bond order, as given by Linus Pauling can be defined as the difference between the number of bonds and antibonds present in a molecule.
Bond order can also be defined as half the difference between the number of bonding electrons and the number of antibonding electrons. This is in accordance with molecular orbital theory.
This definition often yields results similar to the bonds reaching equilibrium lengths. However, it is not applicable for stretched bonds.
We know that bond order is inversely proportional to bond length. This means that greater the bond order, shorter the length of bond.
It is given to us that the bond order of $NO$ is 2.5 while that of $N{{O}^{+}}$ is 3.
Based on the above statements we can conclude that bond length in $NO$ is greater to that in $N{{O}^{+}}$.
So, the correct answer is “Option B”.

Note: It is important to know that in molecules exhibiting resonance, the bond order need not be an integer. For example, the bond order of carbon atoms in benzene is 1.5.