Of the species, NO, NO$$^+$$, NO$$^{2+}$$ and NO$$^-$$, the one with minimum bond strength is:

We have to compare the bond strengths of four related species. In Molecular Orbital (MO) theory, bond strength is directly connected with the bond order. A larger bond order means more bonding electrons (or fewer antibonding electrons), the bond becomes shorter and stronger; a smaller bond order means the bond is weaker.

The numerical relation is stated first:

$$$\text{Bond order} \; (B.O.) \;=\; \dfrac{N_b - N_a}{2}$$$

where $$N_b$$ is the total number of electrons present in bonding molecular orbitals and $$N_a$$ is the total number of electrons present in antibonding molecular orbitals.

For all second-period diatomic species (B$$_2$$ to N$$_2$$, O$$_2$$, F$$_2$$ and their ions) the order of the valence MOs that we must fill is the same:

$$$\sigma(2s)\,\lt \,\sigma^\*(2s)\,\lt \,\sigma(2p_z)\,\lt \,\pi(2p_x)=\pi(2p_y)\,\lt \,\pi^\*(2p_x)=\pi^\*(2p_y)\,\lt \,\sigma^\*(2p_z)$$$

Now we count the valence electrons in each species.

For atomic nitrogen, $$Z_N = 7$$, so $$N(2s^2 2p^3)$$ contributes $$5$$ valence electrons.
For atomic oxygen, $$Z_O = 8$$, so $$O(2s^2 2p^4)$$ contributes $$6$$ valence electrons.

Hence the neutral molecule NO has

$$$5 + 6 = 11 \text{ valence electrons.}$$$

Starting from this neutral 11-electron case, we can reach the ions simply by adding or removing electrons:

NO $$\; :\; 11 \text{ e}^-$$
NO$$^+$$ $$\; :\; 10 \text{ e}^-$$ (one electron removed)
NO$$^{2+}$$ $$: \; 9 \text{ e}^-$$ (two electrons removed)
NO$$^-$$ $$: \; 12 \text{ e}^-$$ (one electron added)

Now we place these electrons into the MO ladder one by one and calculate $$N_b$$ and $$N_a$$ every time. Because the first four electrons (two in $$\sigma(2s)$$ and two in $$\sigma^\*(2s)$$) cancel each other’s bonding effect, we can do the bookkeeping efficiently from the 5th electron onward, but for clarity every single filling step is shown.

(1) The neutral molecule NO   -  11 electrons

$$$\sigma(2s)^2\;\; \sigma^\*(2s)^2\;\; \sigma(2p_z)^2\;\; \pi(2p_x)^2\;\; \pi(2p_y)^2\;\; \pi^\*(2p_x)^1$$$

Bonding electrons: $$N_b = 2 + 2 + 4 = 8$$
Antibonding electrons: $$N_a = 2 + 1 = 3$$

$$B.O._{\text{NO}} = \dfrac{8 - 3}{2} = \dfrac{5}{2} = 2.5$$

(2) The cation NO$$^+$$   -  10 electrons

The electron removed comes from the highest-energy occupied orbital of NO, i.e. the antibonding $$\pi^\*(2p)$$. So after removal:

$$$\sigma(2s)^2\;\; \sigma^\*(2s)^2\;\; \sigma(2p_z)^2\;\; \pi(2p_x)^2\;\; \pi(2p_y)^2$$$

Bonding electrons: $$N_b = 2 + 2 + 4 = 8$$
Antibonding electrons: $$N_a = 2$$

$$B.O._{\text{NO}^+} = \dfrac{8 - 2}{2} = 3$$

The bond order has increased, so the bond in NO$$^+$$ is stronger than in NO.

(3) The dication NO$$^{2+}$$   -  9 electrons

The second electron must now be removed from the next highest occupied MO, which is a bonding $$\pi(2p)$$ orbital. We therefore have:

$$$\sigma(2s)^2\;\; \sigma^\*(2s)^2\;\; \sigma(2p_z)^2\;\; \pi(2p_x)^2\;\; \pi(2p_y)^1$$$

Bonding electrons: $$N_b = 2 + 2 + (2+1) = 7$$
Antibonding electrons: $$N_a = 2$$

$$B.O._{\text{NO}^{2+}} = \dfrac{7 - 2}{2} = \dfrac{5}{2} = 2.5$$

The bond order comes back down to 2.5, the same as the neutral molecule.

(4) The anion NO$$^-$$   -  12 electrons

The extra electron has nowhere to go except the partially filled antibonding $$\pi^\*(2p)$$ level, giving:

$$$\sigma(2s)^2\;\; \sigma^\*(2s)^2\;\; \sigma(2p_z)^2\;\; \pi(2p_x)^2\;\; \pi(2p_y)^2\;\; \pi^\*(2p_x)^2$$$

Bonding electrons: $$N_b = 2 + 2 + 4 = 8$$
Antibonding electrons: $$N_a = 2 + 2 = 4$$

$$B.O._{\text{NO}^-} = \dfrac{8 - 4}{2} = 2$$

Now we compile the bond orders:

$$$\text{NO}^{2+} : 2.5,\quad \text{NO}^+ : 3,\quad \text{NO} : 2.5,\quad \text{NO}^- : 2$$$

The smallest bond order is $$2$$, which belongs to NO$$^-$$. Because bond strength decreases as bond order decreases, the species with the minimum bond strength is the anion NO$$^-$$.

Hence, the correct answer is Option D.