The lowest first ionization energy would be associated with which of the following configurations.

Let us see below the electronic configuration of elements and their group.

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^1\to \mathrm{Na}(\mathrm{Z}=11);\mathrm{Group} 1\mathrm{A}$

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^5\to \mathrm{F}(\mathrm{Z}=9); \mathrm{Group} \mathrm{VIIA}$

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^6\to \mathrm{Ne}(\mathrm{Z}=10);\hspace{0.17em}\mathrm{Group} \mathrm{VIIIA}$

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^2\to \mathrm{Si}(\mathrm{Z}=14);\mathrm{group} \mathrm{IVA}$

In the periodic table, IA group elements have lowest first ionization (I.E) values. Because the single valence electron is less strongly attracted by the nucleus force.