Question:
Which quantum number combination is impossible for a hydrogenic atom orbital?
Which quantum number combination is impossible for a hydrogenic atom orbital?
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Always check \(l < n\) first — most errors come from this rule.
Updated On: Jun 10, 2026
- \(n = 3, l = 2, m_l = -2, m_s = +\frac{1}{2}\)
- \(n = 4, l = 0, m_l = 0, m_s = -\frac{1}{2}\)
- \(n = 3, l = 3, m_l = +1, m_s = +\frac{1}{2}\)
- \(n = 2, l = 1, m_l = 0, m_s = -\frac{1}{2}\)
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The Correct Option is C
Solution and Explanation
Concept:
Quantum numbers follow strict rules:
• \(n = 1, 2, 3, ...\)
• \(l = 0 \text{ to } (n-1)\)
• \(m_l = -l \text{ to } +l\)
• \(m_s = \pm \frac{1}{2}\)
Step 1: Check condition for l For \(n = 3\), maximum value of \(l = 2\)
Step 2: Evaluate options Option (C) has \(l = 3\), which violates \(l \le n-1\) Thus, it is impossible.
• \(n = 1, 2, 3, ...\)
• \(l = 0 \text{ to } (n-1)\)
• \(m_l = -l \text{ to } +l\)
• \(m_s = \pm \frac{1}{2}\)
Step 1: Check condition for l For \(n = 3\), maximum value of \(l = 2\)
Step 2: Evaluate options Option (C) has \(l = 3\), which violates \(l \le n-1\) Thus, it is impossible.
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