Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them.

To determine the increasing order of bond order, we apply the principles of Molecular Orbital Theory (MOT) to calculate the bond order (B.O.) for each species using the formula: $B.O. = \frac{1}{2}(N_{b} - N_{a})$, where $N_{b}$ is the number of bonding electrons and $N_{a}$ is the number of antibonding electrons .

1. $He_{2}^{+}$: This species has 3 electrons. The configuration is $(\sigma 1s)^{2} (\sigma^{*} 1s)^{1}$ . Bond Order = $\frac{1}{2}(2 - 1) = 0.5$.
2. $O_{2}^{-}$ (Superoxide ion): This species has 17 electrons. Starting from the $O_{2}$ configuration (16 electrons, B.O. = 2.0), the additional electron occupies an antibonding $\pi^{*}$ orbital . Bond Order = $\frac{1}{2}(10 - 7) = 1.5$ .
3. $NO$ (Nitric oxide): This species has 15 electrons. The configuration includes 10 bonding electrons and 5 antibonding electrons (with one electron in a $\pi^{*}$ orbital). Bond Order = $\frac{1}{2}(10 - 5) = 2.5$.
4. $C_{2}^{2-}$: This species has 14 electrons and is isoelectronic with $N_{2}$ . The configuration is $(\sigma 1s)^{2} (\sigma^{*} 1s)^{2} (\sigma 2s)^{2} (\sigma^{*} 2s)^{2} (\pi 2p_{x}^{2} = \pi 2p_{y}^{2}) (\sigma 2p_{z})^{2}$ . Bond Order = $\frac{1}{2}(10 - 4) = 3.0$ .

Comparing the values: $He_{2}^{+} (0.5) < O_{2}^{-} (1.5) < NO (2.5) < C_{2}^{2-} (3.0)$. Thus, the correct increasing order is given in Option 3.