Bond Order of Kr2, B2, C2, N2 respectively are
A+0,+3,+1,+2
B+2,+3,+1,0
C+3,+2,+1,0
D+1, +3, +1,0
E+2, +2, +1,0
FQ,+1,+2,+3
The Correct Answer and Explanation is :
To calculate the bond order of diatomic molecules like ( \text{Kr}_2, \text{B}_2, \text{C}_2, \text{N}_2 ), we need to consider the molecular orbital theory. The bond order is given by the formula:
[
\text{Bond Order} = \frac{1}{2} \left( \text{Number of bonding electrons} – \text{Number of antibonding electrons} \right)
]
Let’s go over the molecular orbital theory for each of the given molecules, calculating the bond order step by step:
1. Krypton ((\text{Kr}_2))
Krypton has an atomic number of 36, so for the (\text{Kr}_2) molecule, we have a total of (36 + 36 = 72) electrons. In this case, krypton is a noble gas, and the electron configuration will place electrons in both bonding and antibonding molecular orbitals. However, the bond order for (\text{Kr}_2) is typically found to be 0, since the number of bonding electrons is equal to the number of antibonding electrons.
Bond Order of (\text{Kr}_2 = 0).
2. Boron ((\text{B}_2))
Boron has an atomic number of 5, so for (\text{B}_2), we have a total of (5 + 5 = 10) electrons. The molecular orbital filling for (\text{B}_2) will have electrons in both bonding and antibonding orbitals, and the bond order comes out to be 1.
Bond Order of (\text{B}_2 = 1).
3. Carbon ((\text{C}_2))
Carbon has an atomic number of 6, so for (\text{C}_2), we have a total of (6 + 6 = 12) electrons. Carbon follows similar molecular orbital filling principles as boron, but with slightly more bonding electrons. The bond order comes out to be 2.
Bond Order of (\text{C}_2 = 2).
4. Nitrogen ((\text{N}_2))
Nitrogen has an atomic number of 7, so for (\text{N}_2), we have a total of (7 + 7 = 14) electrons. The molecular orbital filling of (\text{N}_2) results in a bond order of 3.
Bond Order of (\text{N}_2 = 3).
Answer Choices:
- A: 0, 3, 1, 2
- B: 2, 3, 1, 0
- C: 3, 2, 1, 0
- D: 1, 3, 1, 0
- E: 2, 2, 1, 0
- F: 1, 2, 3
From the analysis, the correct answer is C (0, 3, 1, 2). This corresponds to the bond order of Krypton, Boron, Carbon, and Nitrogen molecules, respectively.
Conclusion:
The bond order depends on the number of bonding and antibonding electrons in the molecular orbitals. For (\text{Kr}_2), the bond order is 0 due to equal bonding and antibonding electrons. For (\text{B}_2), (\text{C}_2), and (\text{N}_2), the bond orders are 1, 2, and 3, respectively, based on their electron configurations and molecular orbital theory.
Let me generate the image now.
Here is the diagram illustrating the molecular orbital theory bond order calculations for (\text{Kr}_2, \text{B}_2, \text{C}_2,) and (\text{N}_2). It shows the bonding and antibonding electron configurations, helping to visualize how bond order is derived.