The 1$$^{st}$$ ionization enthalpy for Mg is +737 kJ/mol. The most probable estimated value of the 2$$^{nd}$$ ionization enthalpy of Mg is ______.

The atomic configuration of neutral magnesium is $$1s^2\,2s^2\,2p^6\,3s^2$$.

• First ionization enthalpy ($$\Delta_iH_1$$) removes one $$3s$$ electron:
$$Mg(g) \;\xrightarrow{+\;737\;{\rm kJ\;mol^{-1}}}\; Mg^{+}(g) + e^-$$

• The resulting $$Mg^{+}$$ ion now has the configuration $$1s^2\,2s^2\,2p^6\,3s^1$$. The next (second) electron to be removed is the remaining $$3s$$ electron, still situated in the outermost $$n=3$$ shell. Because the effective nuclear charge experienced by this electron is higher (one electron fewer while the nuclear charge is unchanged), a larger energy input is needed, but no shell‐change is involved yet.

Hence the second ionization enthalpy ($$\Delta_iH_2$$) must be:

1. Positive (energy has to be supplied).
2. Greater than $$737\;{\rm kJ\;mol^{-1}}$$ but not astronomically large (huge jump occurs only for the third ionization when the noble-gas core begins to ionize).

Among the given choices, only $$+1450\;{\rm kJ\;mol^{-1}}$$ satisfies both conditions: it is larger than the first ionization enthalpy yet still far below the multi-thousand values typical after the noble-gas core is breached.

Therefore the most probable second ionization enthalpy of magnesium is

Option C which is: $$+1450\;{\rm kJ\;mol^{-1}}$$