Select the incorrect statement about $\mathrm{N}_2$ molecule.

Using Molecular Orbital Theory, we evaluate each statement for $\mathrm{N}_2$ and $\mathrm{O}_2$.

For $\mathrm{N}_2$ (14 electrons), the configuration is $(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\sigma_{2p_z})^2$, with bond order $\frac{1}{2}(10 - 4) = 3$ and diamagnetic behavior.

For $\mathrm{O}_2$ (16 electrons), the configuration is $(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi_{2p_x}^*)^1 (\pi_{2p_y}^*)^1$, with bond order $\frac{1}{2}(10 - 6) = 2$ and paramagnetic behavior.

Statement A: $\mathrm{N}_2$ has higher bond order (3) than $\mathrm{O}_2$ (2), so it is more stable, making this correct.

Statement B: Both molecules have 10 bonding electrons, so this is incorrect.

Statement C: $\mathrm{N}_2$ bond order is 3, so this is correct.

Statement D: $\mathrm{N}_2$ has all electrons paired, so it is diamagnetic, making this correct.

The incorrect statement is B.