• # question_answer 27) How would you explain the fact that the first ionisation enthalpy of sodium is lower than that of magnesium but its second ionisation enthalpy is higher than that of magnesium?

The electronic configurations of Na and Mg are: $Na(11)\,\,\,\,1{{s}^{2}},2{{s}^{3}}2{{p}^{6}},3{{s}^{1}}$ $Mg(12)\,\,\,\,\,\,\,1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{1}}$ In both the atoms, the electron is to be removed from $3s$-orbital but nuclear charge in magnesium is more than sodium. The force of attraction towards nucleus is more in Mg than in Na. Therefore, the ${{\Delta }_{i}}{{H}_{1}}$ of Na is less than${{\Delta }_{i}}{{H}_{1}}$ of Mg. After the loss of one electron, $N{{a}^{+}}$and $M{{g}^{+}}$ ions have electronic configurations: $N{{a}^{+}}\,\,\,\,\,\,\,1{{s}^{2}},2{{s}^{2}}2{{p}^{6}}$ $M{{g}^{+}}\,\,\,\,\,\,1{{s}^{2}},2{{s}^{2}}2{{p}^{6}}3{{s}^{1}}$ To remove another electron from $N{{a}^{+}}$ion is rather difficult as the ion has stable neon configuration. However, to remove another electron from $M{{g}^{+}}$ion is much easier as it is present on $3s$-orbital. Hence, ${{\Delta }_{i}}{{H}_{2}}$ of Na is higher than ${{\Delta }_{i}}{{H}_{2}}$of Mg.
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