QUANTUM PHYSICS FUNDAMENTALS - Chapter 8, Exercise 1 Solution
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Filling the n = 3 Electron Shell

PROBLEM
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Using the same real quantum-number rules as the chapter's worked
example, calculate how many electrons the n = 3 shell can hold in
total (3s + 3p + 3d subshells).

SOLUTION
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For shell n = 3, the orbital quantum number l ranges from 0 to n-1,
so l = 0, 1, 2 (the s, p, and d subshells). For each l, the number of
orbitals is (2l + 1), and each orbital holds 2 electrons (spin up and
spin down, per the Pauli exclusion principle).

    3s subshell (l=0): 1 orbital x 2 = 2 electrons
    3p subshell (l=1): 3 orbitals x 2 = 6 electrons
    3d subshell (l=2): 5 orbitals x 2 = 10 electrons

    Total: 2 + 6 + 10 = 18 electrons

ANSWER: The n = 3 shell can hold a maximum of 18 electrons.

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WHY THIS WORKS AS AN ANSWER

This result - 18 - is exactly the third of the three real "magic"
closed-shell numbers (2, 8, 18) the chapter opened with as the
original real chemical pattern Pauli set out to explain in 1925. The
chapter's own worked example already derived the first two (2 for
n=1's own single 1s subshell, 8 for n=2's combined 2s+2p subshells)
directly from the same quantum-number counting method; this exercise
completes the pattern for n=3, showing that all three of Pauli's own
original real observed "stable" electron counts fall directly out of
the exclusion principle's own counting rules, rather than needing to
be treated as separate, unexplained coincidences.