Question

Which of the following molecules(s) show/s paramagnetic behavior?
$\mathrm{O}_{2}$
$\mathrm{N}_{2}$
$\mathrm{F}_{2}$
$\mathrm{S}_{2}$

• A. A and B
• B. B and C
• C. D and B
• D. A and D

Hint:

Molecules with unpaired electrons are paramagnetic.

Answer 1

The Correct Option is D

The question asks which of the given molecules show paramagnetic behavior. To determine this, we need to understand the magnetic properties of molecules, which can be predicted based on their electronic configuration. Specifically, we look at the molecular orbital (MO) theory and Hund's rule which help us identify whether a molecule has unpaired electrons—indicative of paramagnetism.

1. Paramagnetism and Diamagnetism:
   • Paramagnetic substances contain unpaired electrons and are attracted by a magnetic field.
   • Diamagnetic substances have all paired electrons and are weakly repelled by a magnetic field.
2. Analyzing Options:
   • \(\mathrm{O}_{2}\): According to the MO theory, the oxygen molecule has the following electron configuration in its molecular orbitals: \((\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi_{2p_x}^*)^1 (\pi_{2p_y}^*)^1\). The presence of two unpaired electrons in the \(\pi^*\\) anti-bonding orbitals makes \(\mathrm{O}_2\) paramagnetic.
   • \(\mathrm{N}_{2}\): The electronic configuration for nitrogen molecule 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\\). All electrons are paired; thus, it is diamagnetic.
   • \(\mathrm{F}_{2}\): Its molecular orbitals are filled as \((\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi_{2p_x}^*)^2 (\pi_{2p_y}^*)^2\). All electrons are paired; hence it is diamagnetic.
   • \(\mathrm{S}_{2}\): The electronic configuration for the sulfur molecule (analogous to oxygen) in its molecular orbitals is similar to \(\mathrm{O}_{2}\) but at a higher energy level with unpaired electrons in the \(\pi\) anti-bonding orbitals, making it paramagnetic.
3. Conclusion:

Among the options given, both \(\mathrm{O}_{2}\) and \(\mathrm{S}_{2}\) have unpaired electrons and thus exhibit paramagnetic behavior.

Therefore, the correct answer for the given question, which specifically states \(\mathrm{S}_{2}\) as the paramagnetic molecule, aligns with our analysis.

Answer 2

1. Number of unpaired electrons: 
- (A) $\mathrm{O}_{2}$: 2 
- (B) $\mathrm{N}_{2}$: 0 
- (C) $\mathrm{F}_{2}$: 0 
- (D) $\mathrm{S}_{2}$: 2 
- (E) $\mathrm{Cl}_{2}$: 0 
2. Paramagnetic behavior: - Molecules with unpaired electrons exhibit paramagnetic behavior. 
- Therefore, $\mathrm{O}_{2}$ and $\mathrm{S}_{2}$ are paramagnetic. 
Therefore, the correct answer is (4) A & D only.