The first ionization enthalpy of Na and Mg, respectively, are: 496, 737 and 786 kJ mol$$^{-1}$$. The first ionization enthalpy (kJ mol$$^{-1}$$) of Al is

We are given the first ionization enthalpies: Na = 496 kJ/mol, Mg = 737 kJ/mol, and we need to find the first ionization enthalpy of Al. The question also mentions 786 kJ/mol, but the standard known values are Na = 496, Mg = 737, and Al = 577 kJ/mol.

In the periodic table, as we move from left to right across a period, the ionization enthalpy generally increases due to increasing nuclear charge. However, Al (electronic configuration: $$[Ne] 3s^2 3p^1$$) has a lower first ionization enthalpy than Mg ($$[Ne] 3s^2$$). This is because the outermost electron of Al is in the 3p orbital, which is higher in energy and better shielded by the 3s electrons compared to Mg's 3s electron. The 3p electron is easier to remove than a paired 3s electron.

Among the given options, 577 kJ/mol correctly reflects this trend — it is lower than Mg's 737 kJ/mol but higher than Na's 496 kJ/mol, consistent with the anomalous dip at Al in the ionization energy trend across Period 3.

Hence, the correct answer is Option C.