Match the species given in List-$I$ with their bond orders given in List-$II$.

List-$I$	List-$II$
$(1)$ ${\rm{NO}}$	$(A)$ $1.5$
$(2)$ ${\rm{CO}}$	$(B)$ $2.0$
$(3)$ ${\rm{O}}_2^-$	$(C)$ $2.5$
$(4)$ ${\rm{O}}_2$	$(D)$ $3.0$

(C) The bond order is calculated using the formula: $\text{Bond Order} = \frac{N_b - N_a}{2}$.
$(1)$ ${\rm{NO}}$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $\frac{10 - 5}{2} = 2.5$.
$(2)$ ${\rm{CO}}$ ($14$ electrons): Bond order = $3.0$.
$(3)$ ${\rm{O}}_2^-$ ($17$ electrons): Bond order = $\frac{10 - 7}{2} = 1.5$.
$(4)$ ${\rm{O}}_2$ ($16$ electrons): Bond order = $\frac{10 - 6}{2} = 2.0$.
Therefore,the correct matching is: $(1-C, 2-D, 3-A, 4-B)$.