Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 41 - Atomic Physics - Exercises and Problems - Page 1245: 9

Answer

Aluminum (Al): $ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^1 $ Gallium (Ga): $ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^1 $ Indium (In): $ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^1 $

Work Step by Step

aluminum (Al), gallium (Ga), and indium (In). Let's determine the ground-state electron configurations for $ \bf Aluminum (Al)$: The atomic number is 13. Electron Configuration Process: The first 2 electrons fill the 1s orbital, the next 2 fill the 2s orbital, followed by 6 electrons in the 2p orbital, then 2 in the 3s orbital. The last electron goes into the 3p orbital. Thus, the ground-state electron configuration of Al is $$ \boxed{\;\;\text{Al}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^1\;\;} $$ Let's determine the ground-state electron configurations for $ \bf Gallium (Ga)$: The atomic number is 31. Electron Configuration Process: The electrons fill the orbitals in the order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, and 4p. First, 2 electrons fill the 1s orbital, followed by 2 in the 2s, 6 in the 2p, 2 in the 3s, and 6 in the 3p. The 4s orbital then gets 2 electrons, followed by the 3d orbital with 10 electrons, and finally, 1 electron in the 4p orbital. Thus, the ground-state electron configuration of Ga is $$ \boxed{\;\; \text{Ga}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^1\;\;} $$ Let's determine the ground-state electron configurations for $ \bf Indium (In)$: The atomic number is 49. Electron Configuration Process: The electrons fill the orbitals in the order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, and 5p. The first 2 electrons go into the 1s orbital, followed by 2 in the 2s, 6 in the 2p, 2 in the 3s, and 6 in the 3p. Then, 2 electrons fill the 4s orbital, followed by 10 electrons in the 3d, 6 in the 4p, 2 in the 5s, 10 in the 4d, and finally, 1 electron in the 5p orbital. Thus, the ground-state electron configuration of In is $$ \boxed{\;\; \text{In}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^1\;\;} $$
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