   Chapter 11, Problem 44QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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# Which of the following orbital designations is (are} not possible?msp;  a .  3 f                       c .  4 d b .  5 s                       d .  1 p

Interpretation Introduction

Interpretation:

We have to check the given orbital designations and determine the impossible ones.

Concept Introduction:

The values of ‘n’ (principal quantum number) and ‘l’ (azimuthal quantum number) for different energy levels can be used to determine the orbital designation.

3f and 1p orbitals are not possible.

Reason for correct option: In any orbital designation the first number indicates the principal energy level whereas next symbol indicates the orbital that is decided by value of ‘l’. For example;

• For 3f:.
 0 1 2 3s 3p 3d

n = 3

l = 0 to n-1 =

Hence for 3rd energy level; 3f is not possible.

• For 1p:.
 0 1s

n = 1

l = 0 to n-1 =

Hence for 1st energy level only 1s is possible, not 1p.

Reasons for incorrect options:

• 5s means the principal energy level (n) is 5. The values of ‘l’ for n=5 will be from 0 to n-1. Hence, it will be 0, 1, 2, 3 and 4. The zero value for ‘l’ is for s-orbital. So 5th energy level has 5s, 5p, 5d, 5f and 5g sub-levels.
• For 4d:.
 0 1 2 3 4s 4p 4d 4f

n = 4

l = 0 to n-1 =

Hence, 5s and 4d are possible.

Explanation

Reason for correct option: In any orbital designation the first number indicates the principal energy level whereas next symbol indicates the orbital that is decided by value of ‘l’. For example;

• For 3f:.
 0 1 2 3s 3p 3d

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