To determine the bond order of diatomic species, we use the Molecular Orbital Theory (MOT). The bond order is calculated as half the difference between the number of electrons in bonding molecular orbitals ($N_b$) and antibonding molecular orbitals ($N_a$).

Formula: Bond Order $=\frac{1}{2}(N_b-N_a)$

Let's calculate the bond order for each species in the given options:

Option A:

• NO: Total electrons = 7 (N) + 8 (O) = 15.
• Electronic configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p}{)}^2({\pi }_{2p}{)}^4({\pi }_{2p}^{\ast }{)}^1$
• $N_b=2+2+2+4=10$, $N_a=2+2+1=5$
• Bond Order $=\frac{1}{2}(10-5)=2.5$
• CO: Total electrons = 6 (C) + 8 (O) = 14.
• Electronic configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p}{)}^4({\sigma }_{2p}{)}^2$
• $N_b=2+2+4+2=10$, $N_a=2+2=4$
• Bond Order $=\frac{1}{2}(10-4)=3.0$
• Bond orders are different.

Option B:

• N${}_2$: Total electrons = 7 (N) + 7 (N) = 14.
• Electronic configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p}{)}^4({\sigma }_{2p}{)}^2$
• $N_b=2+2+4+2=10$, $N_a=2+2=4$
• Bond Order $=\frac{1}{2}(10-4)=3.0$
• O${}_2$: Total electrons = 8 (O) + 8 (O) = 16.
• Electronic configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p}{)}^2({\pi }_{2p}{)}^4({\pi }_{2p}^{\ast }{)}^2$
• $N_b=2+2+2+4=10$, $N_a=2+2+2=6$
• Bond Order $=\frac{1}{2}(10-6)=2.0$
• Bond orders are different.

Option C:

• O${}_2^{2-}$: Total electrons = 8 (O) + 8 (O) + 2 (charge) = 18.
• Electronic configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p}{)}^2({\pi }_{2p}{)}^4({\pi }_{2p}^{\ast }{)}^4$
• $N_b=2+2+2+4=10$, $N_a=2+2+4=8$
• Bond Order $=\frac{1}{2}(10-8)=1.0$
• B${}_2$: Total electrons = 5 (B) + 5 (B) = 10.
• Electronic configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p}{)}^2$
• $N_b=2+2=4$, $N_a=2+2=4$
• Bond Order $=\frac{1}{2}(4-4)=0$ (This is incorrect. For B${}_2$, the order of MOs is ${\sigma }_{2s}<{\sigma }_{2s}^{\ast }<{\pi }_{2p}<{\sigma }_{2p}$. So, $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\pi }_{2p}{)}^2$. $N_b=2+2=4$, $N_a=2$. Bond Order $=\frac{1}{2}(4-2)=1.0$)
• Bond orders are the same (1.0).

Option D:

• O${}_2^{+}$: Total electrons = 8 (O) + 8 (O) - 1 (charge) = 15.
• Electronic configuration: $({\sigma }_{1s}{)}^2({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2({\sigma }_{2s}^{\ast }{)}^2({\sigma }_{2p}{)}^2({\pi }_{2p}{)}^4({\pi }_{2p}^{\ast }{)}^1$
• $N_b=2+2+2+4=10$, $N_a=2+2+1=5$
• Bond Order $=\frac{1}{2}(10-5)=2.5$
• NO${}_{+}$: Total electrons = 7 (N) + 8 (O) - 1 (