Question

How is bond order calculated and why is Bond order of $C{O}^{+}$ $3.5$ ?

Answer 1

Hint :Bond order is the difference of bonding and anti bonding electrons divided by two. Draw the Molecular Orbital diagram for CO and remove an electron from oxygen to get the molecular orbital diagram for $C{O}^{+}$ . Atomic no, of C $=6$ and O $=8$ .

Complete Step By Step Answer:
The electronic configuration of C is $\text{1}{\text{s}}^{2}2s^{2}2p^2$ and that of $O^{+}$ is $1s^{2}2s^{2}2p^3$ .
Steps for drawing the Molecular Orbital Diagram
Since $C{O}^{+}$ is a heteronuclear atom, the atom with more electronegativity will be placed lower in the energy level.
For the 1s degenerate orbital, $2$ electrons will go to σ $1s$ and $2$ electrons will go to $$ σ $\ast 1s$
For the 2s degenerate orbital, $2$ electrons will go to σ $2s$ and $2$ electrons will go to σ $\ast 2s$
For the 2p degenerate orbital , $2$ electrons go to $\pi 2p_x$ , $2$ electrons will go to $\pi 2p_y$ and $2$ electrons will go to σ $2p_z$ .
Formula to calculate bond order is: ${\displaystyle \frac{1}{2}}(\text{No}\text{. of }{\text{e}}^{-}in\text{ bonding subshell - No}\text{. of }{\text{e}}^{\text{-}}\text{in antibonding subshell)}$
Acc. To the diagram 10 $e^{-}$ s are in bonding subshell and 4 in antibonding subshell.
But this is the Molecular diagram for $CO$ , to get Molecular orbital diagram for $C{O}^{+}$ , we will remove an $e^{-}$ for the Subshell with higher energy level.
In case of heteronuclear molecules like $C{O}^{+}$ the σ $\ast 2s$ is placed higher than $\pi 2p_x$ , $\pi 2p_y$ , σ $2p_z$ . So, the $e^{-}$ is removed from σ $\ast 2s$ subshell.
Therefore, $\text{No}\text{. of }{\text{e}}^{-}in\text{ bonding subshell }$ $=10$ and $\text{No}\text{. of }{\text{e}}^{\text{-}}\text{in antibonding subshell}$ $=3$
Putting the values in the formula:
 ${\displaystyle \frac{1}{2}}(10-3)$
 $=3.5$
Hence the Bond order of $C{O}^{+}$ is 3.5.

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 $\ast 2s$ is placed higher than $\pi 2p_x$ , $\pi 2p_y$ , σ $2p_z$ .