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# Bond order and magnetic property of CO is: Verified
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Hint: The molecular orbital theory helps us find out the bond order on the basis of molecular order diagrams. The Atomic no, of C$=6$ and O$=8$

The electronic configuration of C is $\text{1}{{\text{s}}^{2}}2{{s}^{2}}2{{p}^{2}}$and that of ${{O}^{+}}$is $1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$.
Steps for drawing the Molecular Orbital Diagram
Since $CO$ is a heteronuclear atom, the atom with more electronegativity will be placed lower in the energy level.
So
For the 1s degenerate orbital, $2$ electrons will go to σ$1s$ and $2$ electrons will go to $σ*1s$
For the 2s degenerate orbital, $2$ electrons will go to σ$2s$ and $2$ electrons will go to $σ*2s$
For the 2p degenerate orbital , $2$electrons go to $\pi 2{{p}_{x}}$ ,$2$electrons will go to $\pi 2{{p}_{y}}$ and $2$ electrons will go to σ$2{{p}_{z}}$.
Formula to calculate bond order is:$\dfrac{1}{2}(\text{No}\text{. of }{{\text{e}}^{-}}in\text{ bonding subshell - No}\text{. of }{{\text{e}}^{\text{-}}}\text{in antibonding subshell)}$
Acc. To the diagram 10 ${{e}^{-}}$s are in bonding subshell and 4 in antibonding subshell.
Putting the values in the formula:
$\dfrac{1}{2}(10-4)$
$=3$
Hence the Bond order of $CO$ is $=3$
Since all the electrons are paired in the molecular orbital diagram the $CO$ molecule is diamagnetic with the bond order $=3$

Note:
$C{{O}^{+}}$ does not have a symmetric Molecular diagram because it is a heteronuclear molecule. The more electronegative atom i.e., O is placed lower on the energy level. Due to this discrepancy in energies $σ*2s$ is placed higher than $\pi 2{{p}_{x}}$, $\pi 2{{p}_{y}}$, $σ2{{p}_{z}}$.
If all the electrons of the molecule are paired in the molecular orbital diagram the molecule is said to be diamagnetic.
If there are unpaired electrons present in the molecular orbital diagram the molecule is said to be paramagnetic.