Question

Which one of the following statements is correct about \(\text{N}_2\), \(\text{CO}\), and \(\text{NO}^+\)?

• A. These are isoelectronic and have identical bond order.
• B. These are isoelectronic and have different bond orders.
• C. These are not isoelectronic but have identical bond order.
• D. These are neither isoelectronic nor have identical bond order.

Hint:

Diatomic species with 14 electrons (like \(\text{N}_2\), \(\text{CO}\), \(\text{NO}^+\), \(\text{CN}^-\)) are extremely stable and have a bond order of 3.
For any diatomic species, you can quickly find the bond order using 14 electrons as the baseline (Bond Order = 3), and subtracting 0.5 for every electron added or removed from 14.

Answer 1

The Correct Option is A

The given question asks us to identify the correct statement about \(\text{N}_2\), \(\text{CO}\), and \(\text{NO}^+\). We will determine if these species are isoelectronic and whether they have identical bond orders. Let's examine their electronic configurations and bond orders.

1. Isoelectronic Species:
   • Two species are isoelectronic if they have the same number of electrons. Let's compare the electronic configuration of each:
   • \(\text{N}_2\): Nitrogen (\(Z = 7\)) has an atomic number of 7. Hence, the total number of electrons in \(\text{N}_2\) is \(7 + 7 = 14\).
   • \(\text{CO}\): Carbon (\(Z = 6\)) has an atomic number of 6, and Oxygen (\(Z = 8\)) has an atomic number of 8. Thus, the total number of electrons in \(\text{CO}\) is \(6 + 8 = 14\).
   • \(\text{NO}^+\): Nitrogen (\(Z = 7\)) has an atomic number of 7, and Oxygen (\(Z = 8\)) has an atomic number of 8. Being a cation with a charge of +1, it loses one electron, making the total \(7 + 8 - 1 = 14\).
   • Since all three species (\(\text{N}_2\), \(\text{CO}\), \(\text{NO}^+\)) have 14 electrons, they are isoelectronic.
2. Bond Order Calculation:
   • The bond order (BO) of a diatomic molecule can be determined using the molecular orbital theory formula: \text{BO} = \frac{\text{number of bonding electrons} - \text{number of antibonding electrons}}{2}.
   • For isoelectronic species with 14 electrons, the bond order calculation results in identical values because the molecular orbital configuration is the same.
   • Using the molecular orbital theory, the bond order for these species is 3.
3. Conclusion:
   • As shown above, \(\text{N}_2\), \(\text{CO}\), and \(\text{NO}^+\) are isoelectronic with identical bond orders of 3.
   • Therefore, the correct answer is: "These are isoelectronic and have identical bond order."