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

Step 1: Understanding the Question:
The question asks us to analyze the relationship between the three chemical species \(\text{N}_2\), \(\text{CO}\), and \(\text{NO}^+\) in terms of their electronic count (isoelectronic nature) and their bond orders.

Step 2: Key Formula or Approach:
1. Isoelectronic species: Chemical species that contain the exact same number of total electrons.
2. Bond Order: Calculated using Molecular Orbital (MO) theory. For homonuclear and heteronuclear diatomic molecules with 14 electrons, the bond order can be determined using the standard MO configuration:
\[ \text{Bond Order} = \frac{N_b - N_a}{2} \]
where \(N_b\) is the number of bonding electrons and \(N_a\) is the number of antibonding electrons.

Step 3: Detailed Explanation:
Let's first calculate the total number of electrons in each species:
- For \(\text{N}_2\): Each Nitrogen atom has 7 electrons.
\[ \text{Total electrons} = 7 + 7 = 14\text{ electrons} \]
- For \(\text{CO}\): Carbon has 6 electrons, and Oxygen has 8 electrons.
\[ \text{Total electrons} = 6 + 8 = 14\text{ electrons} \]
- For \(\text{NO}^+\): Nitrogen has 7 electrons, Oxygen has 8 electrons, and the positive charge indicates the loss of 1 electron.
\[ \text{Total electrons} = 7 + 8 - 1 = 14\text{ electrons} \]
Since all three species have exactly 14 electrons, they are isoelectronic.
Now let's determine their bond order:
- For any 14-electron diatomic system, the molecular orbital filling configuration is:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) \sigma_{2p_z}^2 \]
- The number of bonding electrons (\(N_b\)) is 10 (from \(\sigma_{1s}\), \(\sigma_{2s}\), \(\pi_{2p_x}\), \(\pi_{2p_y}\), and \(\sigma_{2p_z}\)).
- The number of antibonding electrons (\(N_a\)) is 4 (from \(\sigma^*_{1s}\) and \(\sigma^*_{2s}\)).
- Calculating the bond order:
\[ \text{Bond Order} = \frac{10 - 4}{2} = 3 \]
Since they are isoelectronic and share analogous molecular orbital configurations, they all have an identical bond order of 3.

Step 4: Final Answer:
Therefore, \(\text{N}_2\), \(\text{CO}\), and \(\text{NO}^+\) are isoelectronic and have identical bond order, which corresponds to option (A).