Question:
What are the four quantum numbers of 19th electron in Sc (\(Z = 21\))?
What are the four quantum numbers of 19th electron in Sc (\(Z = 21\))?
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Remember: 4s fills before 3d due to lower energy (Aufbau principle).
Updated On: Apr 23, 2026
- \(n = 4, l = 0, m = 0, s = +\frac{1}{2}\)
- \(n = 3, l = 2, m = 0, s = +\frac{1}{2}\)
- \(n = 4, l = 1, m = 0, s = +\frac{1}{2}\)
- \(n = 3, l = 1, m = 0, s = +\frac{1}{2}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Electronic configuration of Sc (\(Z = 21\)): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\).
Step 2: Detailed Explanation:
Order of filling: 1s (2 e\(^-\)), 2s (2 e\(^-\)), 2p (6 e\(^-\)), 3s (2 e\(^-\)), 3p (6 e\(^-\)), 4s (2 e\(^-\)).
19th electron enters 4s orbital. For 4s: \(n = 4, l = 0, m = 0, s = +\frac{1}{2}\).
Step 3: Final Answer:
Thus, quantum numbers: \(n = 4, l = 0, m = 0, s = +\frac{1}{2}\).
Electronic configuration of Sc (\(Z = 21\)): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\).
Step 2: Detailed Explanation:
Order of filling: 1s (2 e\(^-\)), 2s (2 e\(^-\)), 2p (6 e\(^-\)), 3s (2 e\(^-\)), 3p (6 e\(^-\)), 4s (2 e\(^-\)).
19th electron enters 4s orbital. For 4s: \(n = 4, l = 0, m = 0, s = +\frac{1}{2}\).
Step 3: Final Answer:
Thus, quantum numbers: \(n = 4, l = 0, m = 0, s = +\frac{1}{2}\).
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