Chemistry: The Central Science (13th Edition)

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

Chapter 7 - Periodic Properties of the Elements - Exercises: 7.30e


$Pd$ has the same number of electrons and the same electron configuration as $Sn^{4+}$.

Work Step by Step

$Sn^{4+}$ A neutral $Sn$ atom has 50 electrons. Losing 4 electrons to become $Sn^{4+}$ cation, it now has 46 electrons. The neutral atom that also has 46 electrons is $Pd$. The electron configuration of $Sn$ is $[Kr]4d^{10}5s^25p^2$. Now we need to remove 4 electrons so that $Sn$ would become $Sn^{4+}$. Again, we would remove the electrons of the orbitals with the highest principal quantum number first, which is 5 here. The electrons in orbitals $5s$ and $5p$ are also 4, so all these 4 electrons will be removed. Then, the electron configuration of $Sn^{4+}$ is $[Kr]4d^{10}$, which is also the electron configuration of $Pd$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.