Question

The sum of the bond orders of O$_2^+$, O$_2^-$, O$_2$, O$_2^{2-$, and the sum of the unpaired electrons in them respectively are

• A. 10, 4
• B. 10, 6
• C. 8, 4
• D. 8, 6

Hint:

Bond order in diatomic molecules can be calculated using molecular orbital theory, and unpaired electrons are determined by the occupancy of degenerate orbitals.

Answer 1

The Correct Option is A

Let’s break this down step by step to calculate the bond orders and unpaired electrons for the O$_2$ species and determine why option (1) is the correct answer.
Step 1: Determine the molecular orbital configuration and bond order
O$_2$ has 12 valence electrons. Using molecular orbital (MO) theory:

• O$_2$ (12 valence electrons): $\sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^4 \pi^*_{2p}^2$
  Bonding electrons = 8, antibonding = 4, bond order = $\frac{8 - 4}{2} = 2$, unpaired electrons = 2.
• O$_2^+$ (11 valence electrons): $\sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^4 \pi^*_{2p}^1$
  Bonding electrons = 8, antibonding = 3, bond order = $\frac{8 - 3}{2} = 2.5$, unpaired electrons = 1.
• O$_2^-$ (13 valence electrons): $\sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^4 \pi^*_{2p}^3$
  Bonding electrons = 8, antibonding = 5, bond order = $\frac{8 - 5}{2} = 1.5$, unpaired electrons = 1.
• O$_2^{2-}$ (14 valence electrons): $\sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^4 \pi^*_{2p}^4$
  Bonding electrons = 8, antibonding = 6, bond order = $\frac{8 - 6}{2} = 1$, unpaired electrons = 0.

Assuming O$_2^{2+}$ is included:

• O$_2^{2+}$ (10 valence electrons): $\sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p}^2 \pi_{2p}^4$
  Bonding electrons = 8, antibonding = 2, bond order = $\frac{8 - 2}{2} = 3$, unpaired electrons = 0.

Step 2: Sum the bond orders and unpaired electrons

• Bond orders: $3 + 2.5 + 2 + 1.5 + 1 = 10$
• Unpaired electrons: $0 + 1 + 2 + 1 + 0 = 4$

Step 3: Confirm the correct answer
The sum of bond orders is 10, and the sum of unpaired electrons is 4, matching option (1).
Thus, the correct answer is (1) 10, 4.