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: 6.61

Answer

(a) and (d) represent impossible combinations of $n$ and $l$.

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 value of $l$ cannot exceed or even be equal with the value of $n$ (a) Orbital designation: $1p$. Therefore, $n=1$ and $l=1$. Since in this case, $l$ is equal with $n$, so this is an impossible combination. (b) Orbital designation: $4s$. Which means, $n=4$ and $l=0$. The value of $l$ in this case is smaller than the value of $n$. Therefore, this is a possible combination. (c) Orbital designation: $5f$. Which means $n=5$ and $l=3$. The value of $l$ in this case is smaller than the value of $n$. Therefore, this is a possible combination. (d) Orbital designation: $2d$. Which means $n=2$ and $l=2$. The value of $l$ in this case is equal with the value of $n$. So this is an impossible combination.
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