Question

Bond order and magnetic moment of $C{O^ + }$ is:
A) 2.5 and paramagnetic moment.
B) 3.5 and diamagnetic moment.
C) 3.5 and paramagnetic moment.
D) 2.5 and diamagnetic moment.

Answer 1

Hint: We know that whenever there is an unpaired electron in the electronic configuration i.e. there is a presence of an unpaired electron then the substance is said to be paramagnetic. On the other hand, whenever there is a paired electron in the electronic configuration i.e. there is a presence of paired electrons then the substance is said to show diamagnetic properties. And the total number of electron pairs is known as its bond order.

Complete step by step answer:
Bond order can be calculated by using the formula shown below.
$BO = \dfrac{{BE - ABE}}{2}$
where BO is known as bond order, BE is bonding electrons and ABE is known as antibonding electrons.
The number of Bonding and anti-molecular orbital can be determined using a molecular orbital diagram.
We know that the electronic configuration of C is $1{s^2}2{s^2}2{p^2}$ and the electronic configuration of oxygen is $1{s^2}2{s^2}2{p^4}$.
When the atomic orbitals of oxygen and carbon combine in a linear way then there is a formation of molecular orbitals. In carbon monoxide, the bonding molecular orbitals will resemble the atomic orbitals of oxygen more than they resemble carbon due to the electronegativity difference between them.
The molecular orbital configuration of $C{O^ + }$ is:
${\left( {\sigma _{1s}^b} \right)^2} < {\left( {\sigma _{1s}^a} \right)^2} < {\left( {\sigma _{2s}^b} \right)^2} < {\left( {\pi _x^b} \right)^2} = {\left( {\pi _y^b} \right)^2} < {\left( {\sigma _{2pz}^b} \right)^2} < {\left( {\sigma _{2s}^a} \right)^1}$
BO =$ \dfrac{{BE - ABE}}{2} = \dfrac{{10 - 3}}{2}$
 = 3.5
Since there is a presence of unpaired electrons in antibonding 2s orbital, so it is paramagnetic in nature.

Hence, the option (C) is correct.

Note: The bond order is also related to stability, bond length and bond dissociation energy. The bond order of $C{O^ + }$ is 3.5 while that of CO is 3. So $C{O^ + }$ has stronger bonds than CO.