The first ionization enthalpy of Na, Mg and Si, respectively, are : 496, 737 and 786 kJ mol⁻¹. The first ionization enthalpy (kJ mol⁻¹) of Al is :

Solution

Chemistry ma’am · 3 steps

Al breaks the rising trend: its lone 3p electron leaves more easily than the paired 3s electron of Mg.

1. General trend. Across period 3 the nuclear charge rises while the shell stays the same, so ionisation enthalpy usually increases from Na to Si.
2. The dip at Al. Mg loses an electron from a filled 3s² subshell. Al loses its single 3p electron, which is higher in energy and partly shielded by the 3s² pair. So Al needs less energy than Mg.

   Element	Configuration	Electron removed	IE (kJ mol⁻¹)
   Na	[Ne]3s¹	3s	496
   Mg	[Ne]3s²	3s (filled subshell)	737
   Al	[Ne]3s² 3p¹	3p (lone, shielded)	between 496 and 737
   Si	[Ne]3s² 3p²	3p	786

3. So. Al has a larger nuclear charge than Na, so its value stays above 496, and it falls below Mg at 737. Only 577 fits, which is option (c).

The idea to keep. Ionisation enthalpy rises across a period but dips at group 13 (first p electron) and group 16 (first paired p electron). In period 3 the order is Na < Al < Mg < Si.

Answer from the NTA final answer key. Solution by PrepWiser.