# Chapter 6 - Electronic Structure of Atoms - Exercises: 6.76c

The condensed electron configuration of $Br$ is $$Br:[Ar]4s^23d^{10}4p^5$$ There is one unpaired electron in $Br$.

#### Work Step by Step

*Strategy: 1) Find the nearest noble gas element of the lowest atomic number. 2) Find out which shell is the outer shell and how many electrons there are in the outer shell (by looking at the periodic table). 3) Put the outer-shell electron in the orbitals and subshells according to Hund's rule. 1) The nearest noble gas element of the lower atomic number of $Br$ is $Ar$. Therefore, we would use $Ar$ in the condensed electron configuration of $Br$. 2) Looking at the periodic table, - $Br$ is on the 4th row, so the outer shell is the 4th shell. - The atomic number of $Br$ is 35, so the element has 35 electrons. The inner-shell electrons are 18 (the number of electrons in the first three shells). So, there are $35-18=17$ outer-shell electrons. 3) Procedure: - The s-subshell of the 4th shell is occupied first by 2 electrons ($4s^2$) - Then 10 electrons would occupy the d-subshell of the 3th shell ($3d^{10}$) - The last 5 electrons occupy the $4p$ subshell ($4p^5$) In conclusion, the condensed electron configuration of $Br$ is $$Br:[Ar]4s^23d^{10}4p^5$$ 4) Here we notice that all inner-shell electrons and the electrons in the $4s$ and $3d$ subshells are paired. Subshell $4p$ has 3 orbitals, with each orbital being able to contain at most 2 electrons. 5 electrons occupy subshell $4p$. So, 2 orbitals would be filled with 2 electrons each, leaving 1 orbital having only 1 electron. Therefore there is 1 unpaired electron.

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