How many electrons could have the following set of quantum numbers: n = 3, l = 2, m_l = 1?

• A. 0
• B. 2
• C. 1
• D. 4

Answer: (B)

Verified step by step guidance

1. Identify the quantum numbers given: principal quantum number \(n = 3\), azimuthal quantum number \(l = 2\), and magnetic quantum number \(m_l = 1\).
2. Understand that \(n = 3\) corresponds to the third energy level, and \(l = 2\) corresponds to the d subshell within that energy level.
3. The magnetic quantum number \(m_l = 1\) specifies one particular orbital within the d subshell (since \(m_l\) can range from \(-l\) to \(+l\), i.e., from \(-2\) to \(+2\) for \(l=2\)).
4. Each orbital can hold a maximum of 2 electrons, which differ by their spin quantum number \(m_s\) (either \(+\frac{1}{2}\) or \(-\frac{1}{2}\)).
5. Therefore, the number of electrons that can have the quantum numbers \(n=3\), \(l=2\), and \(m_l=1\) is 2, corresponding to the two possible spin states in that specific orbital.