Which of the following set of quantum numbers is possible for an electron in a set of 6f orbitals?

• A. n = 6, l = 3 , ml_l = 0, m_s = 0
• B. n = 6, l = 2, m_l = 1, m_s = –1/2
• C. n = 9, l = 1, m_l = –2, m_s = +1/2
• D. n = 6, l = 3, m_l = 0, m_s = +1/2
• E. n = 4, l = 2, m_l = –2, m_s = +1

Answer: (D)

Verified step by step guidance

1. Recall the four quantum numbers: the principal quantum number \(n\), the azimuthal (angular momentum) quantum number \(l\), the magnetic quantum number \(m_l\), and the spin quantum number \(m_s\).
2. Understand the allowed ranges for each quantum number: \(n\) is a positive integer (\(n = 1, 2, 3, \ldots\)), \(l\) ranges from \(0\) to \(n-1\), \(m_l\) ranges from \(-l\) to \(+l\) in integer steps, and \(m_s\) can only be \(+\frac{1}{2}\) or \(-\frac{1}{2}\).
3. Check the given quantum numbers for the 6f orbital: since \(f\) orbitals correspond to \(l = 3\), verify that \(n = 6\) and \(l = 3\) are consistent, which they are because \(l\) must be less than \(n\).
4. Verify that \(m_l\) is within the allowed range for \(l = 3\), so \(m_l\) must be one of \(-3, -2, -1, 0, 1, 2, 3\). The value \(m_l = 0\) is valid.
5. Confirm that \(m_s\) is either \(+\frac{1}{2}\) or \(-\frac{1}{2}\). The value \(m_s = 0\) or \(+1\) is not allowed, so only \(m_s = +\frac{1}{2}\) or \(-\frac{1}{2}\) are valid spin quantum numbers.