The difference between bond orders of CO and NO$$^{\oplus}$$ is $$\frac{x}{2}$$ where x = _________ (Round off to the Nearest Integer)

Electrons in $$NO^+$$ = 14$$e^-$$
Electrons in $$CO$$ = 14$$e^-$$

Molecular Orbital Energy Sequence

For molecules with 14 or fewer electrons (such as $N_2$, $B_2$, or $C_2$), the energy level sequence for molecular orbitals is:

$$ \sigma_{1s} < \sigma_{1s}^{*} < \sigma_{2s} < \sigma_{2s}^{*} < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi_{2p_x}^{*} = \pi_{2p_y}^{*} < \sigma_{2p_z}^{*} $$

Electron Configuration (14 Electrons)

Filling all 14 electrons into this sequence yields the following configuration:

$$ \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^{2} \pi_{2p_y}^{2} \sigma_{2p_z}^{2} $$

Bond Order Calculation

The bond order is calculated using the formula:

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

Where:

• $N_b$ = Number of electrons in bonding orbitals (without the *).
• $N_a$ = Number of electrons in antibonding orbitals (with the *).

 Count the electrons:

• Bonding electrons ($N_b$): $\sigma_{1s}^{2}$, $\sigma_{2s}^{2}$, $\pi_{2p_x}^{2}$, $\pi_{2p_y}^{2}$, $\sigma_{2p_z}^{2}$ $\rightarrow 2 + 2 + 2 + 2 + 2 =$ 10 electrons
• Antibonding electrons ($N_a$): $\sigma_{1s}^{*2}$, $\sigma_{2s}^{*2}$ $\rightarrow 2 + 2 =$ 4 electrons

 Calculate the Bond Order:

$$ \text{Bond Order} = \frac{1}{2} (10 - 4) $$$$ \text{Bond Order} = \frac{1}{2} (6) $$$$ \text{Bond Order} = 3 $$

Therefore, any diatomic molecule or ion with exactly 14 electrons (like the $N_2$ molecule or the $CN^-$ ion) has a bond order of 3, indicating a stable triple bond.