Question

Which of the following is paramagnetic ?

• A. CO
• B. \(O^-_2\)
• C. CN^-
• D. NO⁺

Answer 1

The Correct Option is B

 To determine which of the given molecules is paramagnetic, we need to consider their electronic configurations and the presence of unpaired electrons. Paramagnetism arises from the presence of unpaired electrons, which align with a magnetic field, causing attraction. Let's analyze each option:

1. CO (Carbon Monoxide): The CO molecule has a total of 14 electrons. In molecular orbital (MO) theory, its electronic configuration can be represented as: \(\sigma(1s)^2\sigma^*(1s)^2\sigma(2s)^2\sigma^*(2s)^2\sigma(2p_z)^2\pi(2p_x, 2p_y)^4\). All electrons are paired, so CO is diamagnetic.
2. \(O_2^-\) (Superoxide ion): The neutral O₂ molecule has 16 electrons. The addition of one electron gives O₂^- a total of 17 electrons: \(\sigma(1s)^2\sigma^* (1s)^2\sigma(2s)^2\sigma^*(2s)^2\sigma(2p_z)^2\pi(2p_x, 2p_y)^4\) \\ \(\pi^*(2p_x, 2p_y)^3\). The π* orbitals have unpaired electrons, making O₂^- paramagnetic.
3. CN^- (Cyanide ion): The CN molecule has 14 electrons, and the addition of one electron to form CN^- gives a total of 15 electrons. Its electronic configuration is similar to that of the nitrogen molecule (N₂): \(\sigma(1s)^2\sigma^*(1s)^2\sigma(2s)^2\sigma^*(2s)^2\pi(2p_x, 2p_y)^4\sigma(2p_z)^2\). All electrons are paired, so CN^- is diamagnetic.
4. NO⁺ (Nitrosonium ion): The NO molecule has 15 electrons. The removal of one electron to form NO⁺ leaves 14 electrons, resulting in the electronic configuration: \(\sigma(1s)^2\sigma^*(1s)^2\sigma(2s)^2\sigma^*(2s)^2\pi(2p_x, 2p_y)^4\sigma(2p_z)^2\). All electrons are paired, making NO⁺ diamagnetic.

From the analysis above, we find that the \(O^-_2\) ion is the only paramagnetic species due to the presence of unpaired electrons in its molecular orbitals.

Conclusion: The correct answer is \(O^-_2\\)