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 - Page 252: 6.76d


The condensed electron configuration of $V$ is $$V:[Kr]4s^{2}3d^3$$ There are 3 unpaired electrons.

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 $V$ is $Ar$. Therefore, we would use $Ar$ in the condensed electron configuration of $V$. 2) Looking at the periodic table, - $V$ is on the 4th row, so the outer shell is the 4th shell. - The atomic number of $V$ is 23, so the element has 23 electrons. The inner-shell electrons are 18 (the number of electrons in the first three shells). So, there are $23-18=5$ outer-shell electrons. 3) Procedure: - The s-subshell of the 4th shell is occupied first ($4s^2$) - The last 3 electron occupy the orbitals in the $3d$ subshell ($3d^3$) In conclusion, the condensed electron configuration of $V$ is $$V:[Kr]4s^{2}3d^3$$ 4) Here we notice that all inner-shell electrons and the electrons in the $4s$ subshell are paired. Subshell $3d$ has 5 orbitals. 3 electrons occupy subshell $3d$. So, each would occupy a different orbital. In other words, there are 3 orbitals each with just 1 electron. Therefore there are 3 unpaired electrons.
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