Answer:
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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