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.