Question

The bond order and magnetic behaviour of O\(_2^-\) ion are, respectively :

• A. 1 and paramagnetic.
• B. 1.5 and paramagnetic.
• C. 2 and diamagnetic.
• D. 1.5 and diamagnetic.

Hint:

For diatomic species of second-period elements, knowing the MOT energy level order is key. For O\(_2\), F\(_2\), and Ne\(_2\), the \(\sigma2p_z\) orbital is lower in energy than the \(\pi2p\) orbitals. For B\(_2\), C\(_2\), and N\(_2\), the order is reversed.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to determine the bond order and magnetic properties of the superoxide ion (O\(_2^-\)) using Molecular Orbital Theory (MOT).
Step 2: Key Formula or Approach:
1. Write the molecular orbital (MO) configuration for the O\(_2^-\) ion. An oxygen atom has 8 electrons, so an O\(_2\) molecule has 16 electrons. The O\(_2^-\) ion has 16 + 1 = 17 electrons.
2. Calculate the bond order using the formula: Bond Order = \(\frac{1}{2}\) (N\(_b\) - N\(_a\)), where N\(_b\) is the number of electrons in bonding MOs and N\(_a\) is the number of electrons in antibonding MOs.
3. Determine the magnetic behavior by checking for unpaired electrons in the MO configuration. If there are unpaired electrons, the species is paramagnetic. If all electrons are paired, it is diamagnetic.
Step 3: Detailed Explanation:
The MO energy level order for O\(_2\) and its ions is:
\( \sigma1s, \sigma^*1s, \sigma2s, \sigma^*2s, \sigma2p_z, (\pi2p_x = \pi2p_y), (\pi^*2p_x = \pi^*2p_y), \sigma^*2p_z \)
Now, we fill the 17 electrons of O\(_2^-\) into these orbitals:
\( (\sigma1s)^2 (\sigma^*1s)^2 (\sigma2s)^2 (\sigma^*2s)^2 (\sigma2p_z)^2 (\pi2p_x)^2 (\pi2p_y)^2 (\pi^*2p_x)^2 (\pi^*2p_y)^1 \)
Bond Order Calculation:
Number of bonding electrons (N\(_b\)) = 2(\(\sigma1s\)) + 2(\(\sigma2s\)) + 2(\(\sigma2p_z\)) + 4(\(\pi2p\)) = 10
Number of antibonding electrons (N\(_a\)) = 2(\(\sigma^*1s\)) + 2(\(\sigma^*2s\)) + 3(\(\pi^*2p\)) = 7
Bond Order = \( \frac{1}{2} (10 - 7) = \frac{3}{2} = 1.5 \)
Magnetic Behaviour:
Looking at the configuration, the last orbital, \( \pi^*2p_y \), contains a single, unpaired electron. The presence of this unpaired electron makes the O\(_2^-\) ion paramagnetic.
Step 4: Final Answer:
The bond order is 1.5 and the ion is paramagnetic. This corresponds to option (B).