The first ionization enthalpy of Na, Mg and Si, respectively, are 496, 737 and 786 kJ mol-1. The first ionization…

🧠 Al has lower $IE_1$ than Mg — the $3p^1$ electron of Al is easier to pull than the $3s^2$ pair of Mg This is the famous Group 2 → Group 13 anomaly. Mg has a stable $3s^2$ closed subshell ($IE_1 = 737$). Al removes its lone $3p^1$ electron, which sits in a higher-energy orbital and is shielded by the $3s^2$ pair. So $IE_1$(Al) drops below $IE_1$(Mg).

🗺️ Narrow down using the data Given: Na = 496, Mg = 737, Si = 786. Al sits between Na and Mg: $496 < IE_1(\ce{Al}) < 737$.

Scan options: (a) 487 — too low, smaller than Na. (b) 768 — too high, between Mg and Si. (d) 856 — way above Mg. (c) 577 — fits cleanly in the $496$–$737$ window.

Standard tabulated value: $IE_1(\ce{Al}) = 577\,kJ/mol$. ✓

⚠️ The trap Option (b) 768 looks "natural" because it sits neatly between Mg (737) and Si (786) — a smooth left-to-right rise. But ionization enthalpy doesn't rise smoothly through Period 3; the Mg → Al step DROPS due to the anomaly. Always check for the Group 2 → 13 dip before assuming a smooth trend.

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