The four quantum numbers (n, ℓ, m_ℓ, m_s) for the last electron of sodium (Z = 11) in its ground state are:
A(3, 0, 0, +½)
B(2, 1, 0, +½)
C(3, 1, 0, +½)
D(2, 0, 0, +½)
Answer & Solution
Correct answer: A. (3, 0, 0, +½)
1. Ground-state configuration of Na: 1s² 2s² 2p⁶ 3s¹.
2. The 3s¹ electron has principal quantum number n = 3.
3. For an s orbital ℓ = 0 and m_ℓ = 0; choose m_s = +½ by convention.
4. So the set is (3, 0, 0, +½).
_Source: NCERT Class 11 Chem Ch 2 §2.6 Quantum Numbers + Electronic Configuration_
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