Question

Consider the following changes I and II
Change I: ${O2 → O2^{2-}}$
Change II: ${O2 → O2^{-}}$
The correct statements about these changes (I) and (II) in accordance with MO theory are

• A. A, B & C only
• B. A & C only
• C. A & D only
• D. B & C only

Hint:

When analyzing MO theory questions, calculate the bond order and check for unpaired electrons to determine magnetic properties. Practice MO configurations for common molecules like $\ce{O2}$.

Answer 1

The Correct Option is D

Let’s analyze the changes using Molecular Orbital (MO) theory. For \( {O2} \), the MO configuration is \[ (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^*_{2p})^2, \] with 8 valence electrons in bonding orbitals and 4 in antibonding orbitals. Bond order = \( \frac{8 - 4}{2} = 2 \). \( {O2} \) is paramagnetic due to two unpaired electrons in the \( \pi^*_{2p} \) orbitals.

- Change I: \( {O2 \rightarrow O2^{2-}} \) (adds 2 electrons): The electrons go to the \( \pi^*_{2p} \) orbitals, making the configuration \( (\pi^*_{2p})^4 \). Bond order = \( \frac{8 - 6}{2} = 1 \). The bond order decreases by 1 (not 0.5), so (A) is incorrect, but (B) is correct. \( {O2^{2-}} \) becomes diamagnetic (no unpaired electrons), so the magnetic property changes.

- Change II: \( {O2 \rightarrow O2^{-}} \) (adds 1 electron): The configuration becomes \( (\pi^*_{2p})^3 \). Bond order = \( \frac{8 - 5}{2} = 1.5 \), a decrease of 0.5, so (C) is correct. \( {O2^{-}} \) remains paramagnetic (one unpaired electron), so the magnetic property does not change.

- Option (D): Since the magnetic property changes in (I) but not in (II), (D) is incorrect.

Thus, the correct statements are (B) and (C).