The difference between bond order of $\mathrm{CO}$ and ${\mathrm{NO}}^{+}$is $\frac{x}{2}$, where ' x ' =____________________
(Round off to the nearest integer)

Given data: It is given to identify the difference between bond order of CO and NO.

Explanation:

electronic configuration of CO=${\mathrm{\sigma }}_{1\mathrm{s}}^2{\mathrm{\sigma }}_{1\mathrm{s}}^{*2}{\mathrm{\sigma }}_{2\mathrm{s}}^2{\mathrm{\sigma }}_{2\mathrm{s}}^{*2}({\mathrm{\pi }}_{2\mathrm{px}}^2={\mathrm{\pi }}_{2\mathrm{py}}^2){\mathrm{\sigma }}_{2\mathrm{pz}}^2$

Bond order =1/2(bonding - anti-bonding) Bond order of $\mathrm{CO}=3$

electronic configuration of ${\mathrm{NO}}^{+}$=${\mathrm{\sigma }}_{1\mathrm{s}}^2{\mathrm{\sigma }}_{1\mathrm{s}}^{*2}{\mathrm{\sigma }}_{2\mathrm{s}}^2{\mathrm{\sigma }}_{2\mathrm{s}}^{*2}({\mathrm{\pi }}_{2\mathrm{px}}^2={\mathrm{\pi }}_{2\mathrm{py}}^2){\mathrm{\sigma }}_{2\mathrm{pz}}^2$
Bond order of ${\mathrm{NO}}^{+}=3$
Difference $=0=\frac{x}{2}$
x=0

Hence, correct option is: 0