# 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?

1. A

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

2. B

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

3. C

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

4. D

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

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### 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 $\begin{array}{r}1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}3{d}^{5}\end{array}$

Hence the option A is correct.

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