Question

Which one of the following pairs of species have this same bond order?
A. \[{N_2},O_2^ - \]
B. \[CO,NO\]
C. \[{O_2},N{O^ + }\]
D. \[C{N^ - },CO\]

Answer 1

Hint: We know that the bond order is the number of bonds present between the two atoms which combines to form molecules.

Complete step by step answer
The MOT diagrams will help to solve these types of questions as it will give a clear picture of the molecules which are bonding and one which are non-bonding that takes part in the bond formation via covalent bond. The number of bonding and non-bonding are the same. The two atoms will combine to form the molecule from their distribution of electrons the bond angle, bond length, bond order, magnetic property (paramagnetic and diamagnetic) will come to know.
The bonding and non-bonding orbitals will be denoted with sigma or pi\[(\sigma \;\pi )\] and sigma star and pi star\[(\sigma {^ * }\;{\pi ^ * })\].MOT diagram we can predict the bond order, bond length and magnetic properties of the new molecule formed. MOT diagrams will not help us to predict the shape of the molecule
Both \[C{N^ - },CO\] number of electrons in both are 14.
Bond order \[ = {\rm{ }}\dfrac{{\left( {Nb - Na} \right)}}{2}\]
For \[{{O_2}^{ - }}\]
Bond order\[ = \dfrac{{10 - 7}}{2} = 1.5\]
Hence, bond order of \[{O_2}\] will be two.
For \[C{N^ - }\]
Bond order\[ = \dfrac{{10 - 4}}{2} = 3\]
For \[CO\]
Bond order\[ = \dfrac{{8 - 2}}{2}\; = 3\]
For \[N{O^ + }\]
Bond order\[ = \dfrac{{10 - 4}}{2}\; = 3\]
Hence, the bond order of \[NO\] will be \[2.5\]

So we can say that the correct answer is (D) \[C{N^ - },CO\].

Note:
Here the question can be solved by using isoelectronic species theory, the number of electrons should be the same. If we count the number of electrons in the \[C{N^ - },CO\] molecule the number of electrons in both will be fourteen. The one or more species if compared having the same number of electrons but different atoms are called isoelectronic species. It may be having positive, negative or neutral charge.