The first ionization energy of magnesium is smaller, as compared to that of elements X and Y, but higher than that of…

🧠 Mg sits in the middle of Period 3 — to find X, Y (higher IE) and Z (lower IE), look for elements right of Mg in Period 3 vs Group 1 below Mg $IE_1(\ce{Mg}) = 737\,kJ/mol$. Across Period 3, IE rises (with two anomalies). Down Group 1, IE falls. So elements ABOVE Mg in IE are roughly Si, P, S, Cl, Ar (further right in Period 3) — and below Mg in IE are alkali metals like Na, K.

🗺️ Identify Z (lower than Mg) Z must have $IE_1 < 737$. Na has $IE_1 = 496$ — yes, lower than Mg. So Z = Na.

Identify X and Y (higher than Mg) Both must exceed 737. Among the options: Cl has $IE_1 = 1251$ — higher. Ar has $IE_1 = 1521$ — higher (and the highest in Period 3). Li has $IE_1 = 520$ — LOWER than Mg, so Li cannot be X or Y. This rules out (a) and (b). Ne is in Period 2, not Period 3, but could still be considered. Ne has $IE_1 = 2080$ — higher than Mg. But option (d) makes Z = Cl, which contradicts Z having LOWER IE than Mg.

So only option (c): X = Ar, Y = Cl, Z = Na fits all three conditions.

⚠️ The trap Option (b) places Li with Mg — Li does have a low $IE$ but the question says X, Y are HIGHER than Mg, and Z is LOWER. Li has $IE_1 = 520 < 737$, so Li belongs in the Z slot, not the X/Y slot. Always check whether each candidate is above or below the reference before assigning.

$\boxed{\text{Answer: (c)}}$