Molecular Orbital Theory - Bond Order

In Molecular Orbital Theory, a bond order of zero indicates that bonding electrons equal antibonding electrons, meaning no net attractive bonding exists and the molecule is completely unstable/does not exist (Statement I is false). As bond order increases, atoms are pulled closer by stronger shared electron density, so bond length decreases (Statement II is false). Hence, both statements are false.

$He_2$: electronic configuration $\sigma 1s^2\,\sigma^*1s^2$ Bond order $= \frac{1}{2}(N_b - N_a) = \frac{1}{2}(2 - 2) = 0$ Since the bond order is zero, the $He_2$ molecule does not exist.

NO: $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_z)^2(\pi2p_x)^2 = (\pi2p_y)^2(\pi^*2p_x)^1$; BO $= \frac{10-5}{2} = 2.5$ $CN^-$: $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_x)^2 = (\pi2p_y)^2(\sigma2p_z)^2$; BO $= \frac{10-4}{2} = 3$ CN: $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_x)^2 = (\pi2p_y)^2(\sigma2p_z)^1$; BO $= \frac{9-4}{2} = 2.5$ $CN^+$: $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_x)^2 = (\pi2p_y)^2$; BO $= \frac{8-4}{2} = 2$ So $CN^-$ has the highest bond order (3).

$CN^-$ and CO have bond order 3 each (both are isoelectronic with $N_2$, 14 electrons).

$N_2$ (14 electrons): $\sigma 1s^2\,\sigma^* 1s^2\,\sigma 2s^2\,\sigma^* 2s^2\,\pi 2p_x^2\,\pi 2p_y^2\,\sigma 2p_z^2$. Bond order $= \dfrac{10 - 4}{2} = 3$.
$N_2^+$ (13 electrons): one electron is removed from the bonding $\sigma 2p_z$ orbital. Bond order $= \dfrac{9 - 4}{2} = 2.5$.
Higher bond order means a stronger bond and greater dissociation energy.
So the dissociation energy of $N_2 \gt N_2^+$.