Which of the following is a feasible set of orbital quantum numbers (n, l, m_l) for an electron in an atom?

• A. n = 3, l = 2, m_l = -2
• B. n = 1, l = 1, m_l = 0
• C. n = 2, l = 2, m_l = 1
• D. n = 4, l = 3, m_l = 4

Answer: (A)

Verified step by step guidance

1. Recall the rules for quantum numbers: the principal quantum number \(n\) must be a positive integer (\(n = 1, 2, 3, \ldots\)).
2. The azimuthal (orbital) quantum number \(l\) can take integer values from \$0\( up to \)n-1\(, so \)l\( must satisfy \)0 \leq l \leq n-1$.
3. The magnetic quantum number \(m_l\) can take integer values from \(-l\) up to \(+l\), so \(m_l\) must satisfy \(-l \leq m_l \leq l\).
4. Check each set of quantum numbers against these rules: for example, for \(n=1\), \(l\) can only be \$0\(, so \)l=1\( is not allowed; for \)n=2\(, \)l\( can be \)0\( or \)1\(, so \)l=2\( is not allowed; for \)n=4\(, \)l\( can be \)0\( to \)3\(, so \)l=3\( is allowed but \)m_l=4\( is not because \)m_l\( must be between \)-3\( and \)3$.
5. Identify the set that satisfies all conditions: \(n=3\), \(l=2\), \(m_l=-2\) is valid because \(l=2\) is less than \(n=3\), and \(m_l=-2\) is within the range \(-2\) to \$2$.