A gas phase atom with the electronic configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{6}$ loses three electrons. What is the electron configuration of the resulting gas phase ion?

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$\begin{array}{r} 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1} 3 d^{4} \end{array}$

$\begin{aligned} 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} \end{aligned}$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{3}$

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5} 4 s^{1} 3 d^{5}$

answer is A.

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Detailed Solution

Atomic number is iron is 26. Its electronic configuration is

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{6}$ .

So, after loosing three electrons , the electronics configuration becomes $\begin{aligned} 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} \end{aligned}$ .

Hence the option A is correct.

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