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: 6.77d


The element mentioned here is tellurium ($Te$). It has 2 unpaired electrons.

Work Step by Step

*Strategy: 1) Find out the number of electrons it has by summing all the electrons in different subshells. 2) The number of electrons will be equal to the atomic number of the element. 3) Look up the periodic table to find out which element it is. 1) $[Kr]5s^24d^{10}5p^4$ - A $Kr$ atom has 36 electrons. - Subshell $5s$: 2 electrons - Subshell $4d$: 10 electrons - Subshell $5p$: 4 electrons. In total, the element has 52 electrons. 2) Since the number of electrons is equal to the atomic number of the element, the atomic number of the element is 52. 3) The element that has the atomic number 52 is tellurium ($Te$). 4) All the inner-shell electrons (those in the $Kr$) are paired. The s-subshell has 1 orbital and can carry at most 2 electrons and the d-subshell has 5 orbitals and can carry at most 10 electrons. Therefore, both subshell $5s$ and $4d$ are completely filled and the electrons are all paired. P-subshell has 3 orbitals, each can carry 2 electrons in maximum. Here there are 4 electrons that need to occupy. So the problem is like the b) part. According to Hund's rule, each of the first 3 electrons would occupy a different orbital of subshell $5p$. Then the last electron would randomly occupy 1 orbital out of 3. After all, there are 1 two-electron orbital and 2 one-electron orbitals. There are 2 unpaired electrons as a result.
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