Pick out the incorrect statement

One electron is lost from both ${\mathrm{N}}_2^{+}$ and ${\mathrm{O}}_2^{+}$, causing a drop in bond strength and an increase in bond dissociation energy compared to ${\mathrm{N}}_2$ and ${\mathrm{O}}_2.$
An organism's species' bond order and length are inversely related.

$N_2$

Total valence electrons are equal to 10 ((5+5)).
As a result, its electronic configuration is $\left({\sigma }_{2}s{)}^2\right({\sigma }_2{s}^{*}{)}^2\left({\sigma }_{2}px{)}^2\right({\pi }_{2}py,{\pi }_{2}pz{)}^2\left({\sigma }_{2}p{x}^{*}\right)({\pi }_{2}p{y}^{*},{\pi }_2^{*}pz)$
signifies anti bonding electrons.
Bond order formula is represented by;
Bond order is calculated as follows: Bond order (8-2)/2=3 (Number of Bonding Electrons - Number of Antibonding Electrons)

∗means anti bonding electrons

Bond order formula can be written as;

Bond order $=\frac{( \mathrm{Number} \mathrm{of} \mathrm{Bonding} \mathrm{electrons} - \mathrm{Number} \mathrm{of} \mathrm{antibonding} \mathrm{electrons} )}{2}$$N_2^{+}$

Total valence electron =(5+51)=9 is present.
This means that its electrical arrangement is

$\left({\sigma }_{2}s{)}^2\right({\sigma }_2{s}^{*}{)}^2\left({\sigma }_{2}px{)}^2\right({\pi }_{2}p{y}^2,{\pi }_{2}p{z}^1\left)\right({\sigma }_{2}p{x}^{*}\left)\right({\pi }_2^{*}py,{\pi }_{2}p{z}^{*})$

Bond order  $\frac{(7-2)}{2}=2.5$

Bond Order 3 and N 2
The bond order of $N_2^{+}$ is 2.5. Therefore, ${\mathrm{N}}_2$ will have a longer bond length than $N_2^{+}$.