Which of the following molecules show/s paramagnetic behavior?
(A) $$O_2$$ (B) $$N_2$$ (C) $$F_2$$ (D) $$S_2$$ (E) $$Cl_2$$
Choose the correct answer:

For a molecule to be paramagnetic, it must contain at least one unpaired electron. The Molecular Orbital (MO) approach is the most reliable way to count unpaired electrons in homonuclear diatomic molecules.

Rule: Fill MOs in the order of increasing energy while obeying Hund’s rule (maximum multiplicity) and the Pauli exclusion principle. In period-2 diatomics the ordering changes after $$N_2$$:
• For $$B_2,\,C_2,\,N_2$$ the sequence is $$\sigma_{2p_z}$$ > $$\pi_{2p_x} = \pi_{2p_y}$$.
• For $$O_2,\,F_2$$ and beyond (including $$S_2,\,Cl_2$$) the sequence is $$\pi_{2p_x} = \pi_{2p_y}$$ > $$\sigma_{2p_z}$$.

Case A: $$O_2$$
Total valence electrons per atom = 6, so molecule has $$12$$.
Filling the energy-correct MO sequence (after $$N_2$$) gives:
$$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\pi_{2p_x}^{*1}\,\pi_{2p_y}^{*1}$$
Two electrons remain unpaired in the degenerate $$\pi_{2p}^{*}$$ orbitals, so $$O_2$$ is paramagnetic.

Case B: $$N_2$$
Total valence electrons = $$10$$.
Using the first energy sequence (before $$O_2$$):
$$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\sigma_{2p_z}^2$$
All electrons are paired ⇒ $$N_2$$ is diamagnetic.

Case C: $$F_2$$
Total valence electrons = $$14$$.
Filling after $$O_2$$ sequence:
$$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\pi_{2p_x}^{*2}\,\pi_{2p_y}^{*2}$$
All electrons are paired ⇒ $$F_2$$ is diamagnetic.

Case D: $$S_2$$
Sulphur is one period below oxygen but forms a similar $$S_2$$ molecule with valence 6 e⁻ per atom ⇒ $$12$$ electrons, and the same MO ordering used for $$O_2$$ applies (the 3p orbitals follow the same pattern).
Hence configuration of $$S_2$$ ends with $$\pi_{3p_x}^{*1}\,\pi_{3p_y}^{*1}$$.
Two unpaired electrons are present, so $$S_2$$ is paramagnetic.

Case E: $$Cl_2$$
Chlorine contributes 7 valence electrons; $$Cl_2$$ has $$14$$. Filling the 3p MO diagram analogous to $$F_2$$ gives all electrons paired ⇒ $$Cl_2$$ is diamagnetic.

Summary: The molecules with unpaired electrons are $$O_2$$ and $$S_2$$ only.

Therefore, the correct option is Option D: A & D only.