Answer:
(a) The electronic configurations
of Na and Mg are:
\[Na(11):[Ne]3{{s}^{1}}\] and \[Mg(12):[Ne]3{{s}^{2}}\]
In both the atoms, the electron
is to be removed from 3s- orbital but nuclear charge in Na is less than Mg.
Thus, ionisation energy of Na is less than Mg (Na < Mg).
The electronic configurations of
Mg and Al are:
\[Mg[Nc]3{{s}^{2}};\,\,\,\,\,Al[Ne]3{{s}^{2}}3{{p}^{1}}\]
In Mg, the electron is to be
removed from 3s while in Al, it is to be removed from 3p. Since, it is easier
to remove electron from 3p in comparison to 3s, the ionisation enthalpy of Mg
is higher than Al (Mg > Al).
The electronic configuration of
Al and Si are:
\[Al(13):[Ne]3{{s}^{2}}3{{p}^{1}};Si(14):[Ne]3{{s}^{2}}3{{p}^{2}}\]
In both the atoms, the electron
is to be removed from 3p-orbital but nuclear charge in Si is more than Al. Thus,
ionisation enthalpy of Al is less than Si (Al < Si).
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