Question:medium

Among the following species, which one has the highest bond order according to Molecular Orbital Theory?

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Removing electrons from antibonding orbitals increases bond order, while adding electrons into antibonding orbitals decreases bond order.
Updated On: Jun 17, 2026
  • \(O_2\)
  • \(O_2^+\)
  • \(O_2^-\)
  • \(O_2^{2-}\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Molecular Orbital Theory (MOT) determines bond order using the number of electrons in bonding (\(N_b\)) and anti-bonding (\(N_a\)) orbitals: \( \text{Bond Order} = \frac{N_b - N_a}{2} \).
Step 2: Key Formula or Approach:
The molecular orbital configuration for the \(\text{O}_2\) series is:
\(\sigma 1s^2, \sigma^ 1s^2, \sigma 2s^2, \sigma^ 2s^2, \sigma 2p_z^2, (\pi 2p_x^2 = \pi 2p_y^2), (\pi^ 2p_x^1 = \pi^ 2p_y^1)\).
Step 3: Detailed Explanation:
- \(\text{O}_2\) (16 electrons): Bond Order = \(\frac{10-6}{2} = 2.0\).
- \(\text{O}_2^+\) (15 electrons): One electron is removed from an anti-bonding orbital, Bond Order = \(\frac{10-5}{2} = 2.5\).
- \(\text{O}_2^-\) (17 electrons): One electron is added to an anti-bonding orbital, Bond Order = \(\frac{10-7}{2} = 1.5\).
- \(\text{O}_2^{2-}\) (18 electrons): Two electrons are added to anti-bonding orbitals, Bond Order = \(\frac{10-8}{2} = 1.0\).
Step 4: Final Answer:
The species with the highest bond order is \(\text{O}_2^+\).
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