Chlorine has the smallest atomic radius in Period 3 and forms PCl₃ with phosphorus.

Steps:

• Atomic radii decrease across Period 3, so D (smallest radius) is chlorine (Cl).
• E is another Period 3 element; D and E form compound L with melting point lower than E.
• Options with Cl (D) are A (MgCl₂, E=Mg) and D (PCl₃, E=P).
• Compare melting points: PCl₃ (-93°C) < phosphorus (44°C); MgCl₂ (714°C) > magnesium (650°C).

Why D is correct:

• PCl₃ is a molecular compound with weak van der Waals forces, yielding low melting point per intermolecular force trends.

Why the others are wrong:

• A: MgCl₂ melting point (714°C) exceeds magnesium's (650°C).
• B: Contains no Cl, so D cannot be Cl.
• C: Contains no Cl, so D cannot be Cl.

Final answer: D