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The value of IE₁, IE₂, IE₃ and IE₄ of an atom are respectively 7.5 eV 25.6 eV 48.6 eV and 170. eV. The electronic configuration of the atom will be

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$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$

answer is B.

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Detailed Solution

There is a high jump from IE₃ to IE₄. Therefore, it is difficult to remove the 4th valence electron. So, electronic configuration of element must have three valence electrons. Hence, the answer is (2).

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