Question

NO$^+$ has bond order

• A. 2
• B. 2.5
• C. 3
• D. 3.5
• E. 4

Hint:

Molecules with 14 electrons (like $N_2$, $CO$, $CN^-$, and $NO^+$) are isoelectronic and all have a stable triple bond with a bond order of 3.

Answer 1

The Correct Option is C

Concept: The bond order of a diatomic molecule or ion can be determined using Molecular Orbital (MO) theory. It is defined as half the difference between the number of electrons in bonding orbitals ($N_b$) and antibonding orbitals ($N_a$). $$\text{Bond Order} = \frac{1}{2}(N_b - N_a)$$

Step 1: {Calculate the total number of electrons in NO$^+$.}
Nitrogen (N) has 7 electrons and Oxygen (O) has 8 electrons. The positive charge indicates the loss of one electron. $$\text{Total electrons} = 7 + 8 - 1 = 14 \text{ electrons}$$

Step 2: {Determine the electronic configuration based on MO theory.}
For a 14-electron system (isoelectronic with $N_2$), the configuration is: $$\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, (\pi_{2p_x}^2 = \pi_{2p_y}^2), \sigma_{2p_z}^2$$

Step 3: {Calculate bonding and antibonding electrons.}
Bonding electrons ($N_b$) = 2 + 2 + 4 + 2 = 10
Antibonding electrons ($N_a$) = 2 + 2 = 4
$$\text{Bond Order} = \frac{10 - 4}{2} = \frac{6}{2} = 3$$