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: 6.57b

Answer

$n=2$ and $l=0$.

Work Step by Step

*NOTES TO REMEMBER: In an orbital designation, - the number represents the value of $n$. - the letter represents the value of $l$, according to the following rule: $l=0$, the letter used is $s$. $l=1$, the letter used is $p$. $l=2$, the letter used is $d$. $l=3$, the letter used is $f$. The orbital designation is $2s$. Therefore, $n=2$ and $l=0$.
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