Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 6 - Electronic Structure of Atoms - Exercises: 6.75c


The condensed electron configuration of $Se$ is $$Se: [Ar]4s^23d^{10}4p^4$$

Work Step by Step

*Strategy: 1) Find the nearest noble gas element of the lower 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 selenium ($Se$) is argon ($Ar$). Therefore, we would use $Ar$ in the condensed electron configuration of $Se$. 2) Looking at the periodic table, - $Se$ is on the 4th row. The outer shell is the 4th shell. - The atomic number of $Se$ is 34, which means it has 34 electrons. Since it has 4 shells, there are 18 inner-shell electrons, and 34 - 18 = 16 outer-shell electrons. (The maximum electron number in the 1st shell is 2, 2nd shell is 8, 3rd shell is 8. Since the electrons would fill up the shells with lower energy first, all inner shells are fully occupied. In other words, $2+8+8=18$ electrons are inner-shell ones). 3) As we mentioned in the previous part, the first 2 outer-shell electrons would occupy the s-subshell of the 4th shell ($4s^2$). Then, the electrons would occupy the d-subshell of the 3rd shell. There are 5 orbitals in the d-subshell, each orbital can carry at most 2 electrons. So there are maximum 10 electrons that can occupy the d-subshell of the 3rd shell ($3d^{10}$) After 10 electrons occupy the d-subshell, still 4 electrons are left. These 4 electrons would then occupy the next lowest-energy subshell, which is the p-subshell of the 4th shell ($4p^4$). In conclusion, the condensed electron configuration of $Se$ is $$Se: [Ar]4s^23d^{10}4p^4$$
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