The user wants me to identify the species from the given options that has a bond order different from that of carbon monoxide (CO).

Step-by-step Solution:

1. Determine the bond order of CO. First, we need to find the total number of electrons in the CO molecule.

   • Carbon (C) has 6 electrons.
   • Oxygen (O) has 8 electrons.
   • Total electrons in CO = $6 + 8 = 14$ electrons.

   Species with 14 electrons have the following Molecular Orbital (MO) configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$

   The bond order is calculated using the formula: $\text{Bond Order} = \frac{1}{2} (N_b - N_a)$ where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.

   • $N_b$ = 2 (in $\sigma 1s$) + 2 (in $\sigma 2s$) + 2 (in $\pi 2p_x$) + 2 (in $\pi 2p_y$) + 2 (in $\sigma 2p_z$) = 10

   • $N_a$ = 2 (in $\sigma^* 1s$) + 2 (in $\sigma^* 2s$) = 4

   • Bond Order of CO = $\frac{1}{2} (10 - 4) = \frac{6}{2} = 3$.

2. Apply the isoelectronic principle. Species with the same number of total electrons (isoelectronic species) have the same bond order. We can check which of the options are isoelectronic with CO (i.e., have 14 electrons).

3. Analyze the options:

   • A: NO⁻

     • Nitrogen (N) has 7 electrons.
     • Oxygen (O) has 8 electrons.
     • The negative charge (-1) adds 1 electron.
     • Total electrons in NO⁻ = $7 + 8 + 1 = 16$ electrons.
     • Since NO⁻ has 16 electrons, it is not isoelectronic with CO. Its bond order will be different.
     • Let's calculate the bond order for a 16-electron species (like O₂): MO configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$ $N_b = 10$, $N_a = 6$ Bond Order of NO⁻ = $\frac{1}{2} (10 - 6) = \frac{4}{2} = 2$.
     • The bond order of NO⁻ (2) is different from the bond order of CO (3).

   • B: NO⁺

     • Nitrogen (N) has 7 electrons.
     • Oxygen (O) has 8 electrons.
     • The positive charge (+1) removes 1 electron.
     • Total electrons in NO⁺ = $7 + 8 - 1 = 14$ electrons.
     • NO⁺ is isoelectronic with CO. Therefore, its bond order is 3.

   • C: CN⁻

     • Carbon (C) has 6 electrons.
     • Nitrogen (N) has 7 electrons.
     • The negative charge (-1) adds 1 electron.
     • Total electrons in CN⁻ = $6 + 7 + 1 = 14$ electrons.
     • CN⁻ is isoelectronic with CO. Therefore, its bond order is 3.

   • D: N₂

     • Nitrogen (N) has 7 electrons.
     • Total electrons in N₂ = $7 + 7 = 14$ electrons.
     • N₂ is isoelectronic with CO. Therefore, its bond order is 3.

4. Conclusion: The bond order of CO, NO⁺, CN⁻, and N₂ is 3. The bond order of NO⁻ is 2. The question asks for the species with a bond order different from that in CO. Therefore, NO⁻ is the correct answer.