Bond order is defined as the number of covalent bonds that are present between a pair of atoms. The stability of the molecule is determined by the bond order. The formula to find the bond order is given by:
Bond order= ½ (Nb-Na)
Where Na is the electrons in the antibonding orbitals
Nb is the electrons in the bonding orbitals.
Bond order can be both positive and negative that will determine the stability of the molecule. If the electrons in the bonding orbitals are greater than the electrons present in the antibonding orbitals (Nb>Na), the bond order will be positive. The Molecule formed will be stable.
On the other hand if the electrons in the bonding orbitals is less than the electrons present in the antibonding orbitals (Nb<Na), the bond order will be negative. The Molecule formed will be unstable.
In the question, the bond order of CO is 3. CO has 14 electrons.
There are 10 electrons in the bonding orbitals and 4 electrons in the antibonding orbitals.
B.O = ½ (Nb-Na)
= ½ (10-4) = 3
The other given compound that is N₂, NO⁺ and CN⁻ are isoelectronic with CO. Isoelectronics are the molecules that have the same number of electrons. As all these compounds are isoelectronic they have the same bond order that is 3.
On the other hand, NO⁻, have 16 electrons and have the bond order of 2.
Covalent bonds can be characterized on the basis of several bond parameters such as bond length, bond angle, bond order, and bond energy (also known as bond enthalpy). These bond parameters offer insight into the stability of a chemical compound and the strength of the chemical bonds holding its atoms together.
For example, The H—H bond enthalpy in hydrogen is 435.8 kJ mol-1. \
Bond order of H2 (H —H) =1
Bond order of 02 (O = O) =2
Bond order of N2 (N = N) =3
Read More: Chemical Bonding and Molecular Structure