For an electron with principal quantum number n = 3 and magnetic quantum number ml = -1, how many orbitals are possible?

• A. 3
• B. 1
• C. 2
• D. 6

Answer: (B)

Verified step by step guidance

1. Recall that the principal quantum number \(n\) determines the energy level and the possible values of the azimuthal quantum number \(l\), where \(l\) can take integer values from \(0\) to \(n-1\).
2. For \(n = 3\), the possible values of \(l\) are \(0\), \(1\), and \(2\). Each value of \(l\) corresponds to a subshell: \(l=0\) (s), \(l=1\) (p), and \(l=2\) (d).
3. The magnetic quantum number \(m_l\) depends on \(l\) and can take integer values from \(-l\) to \(+l\), including zero. For each \(l\), the number of possible \(m_l\) values equals \$2l + 1$.
4. Given \(m_l = -1\), this value must be within the range of \(m_l\) values for a particular \(l\). Check which \(l\) values allow \(m_l = -1\): for \(l=0\), \(m_l=0\) only; for \(l=1\), \(m_l = -1, 0, +1\); for \(l=2\), \(m_l = -2, -1, 0, +1, +2\).
5. Since \(m_l = -1\) is valid for \(l=1\) and \(l=2\), but the problem only specifies \(n=3\) and \(m_l=-1\) without specifying \(l\), the number of orbitals possible corresponds to the number of orbitals with that specific \(m_l\) value, which is exactly 1 orbital per \(m_l\) value.