Multiple Choice
What is the maximum number of electrons in an atom that can have the quantum numbers n = 3, l = 2, m_l = 0?
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9. Quantum Mechanics
Quantum Numbers: Number of Electrons
Multiple Choice
Which set of quantum numbers is possible for an electron in a 3d orbital?
A
n = 3, l = 0, m_l = 0, m_s = +1/2
B
n = 3, l = 1, m_l = 0, m_s = -1/2
C
n = 3, l = 2, m_l = 1, m_s = +1/2
D
n = 2, l = 2, m_l = -2, m_s = +1/2
Verified step by step guidance1
Recall the meaning and allowed values of each quantum number: the principal quantum number \(n\) determines the energy level and must be a positive integer (\(n = 1, 2, 3, \ldots\)).
The azimuthal (angular momentum) quantum number \(l\) defines the subshell and can take integer values from \$0\( to \)n-1\(. For a 3d orbital, \)n = 3\( and \)l = 2\( because \)l = 2$ corresponds to the d subshell.
The magnetic quantum number \(m_l\) can take integer values from \(-l\) to \(+l\), including zero. For \(l = 2\), \(m_l\) can be \(-2, -1, 0, 1,\) or \$2$.
The spin quantum number \(m_s\) can only be \(+\frac{1}{2}\) or \(-\frac{1}{2}\), representing the two possible spin states of an electron.
Check each given set of quantum numbers against these rules: the set must have \(n = 3\), \(l = 2\), \(m_l\) within \(-2\) to \$2$, and \(m_s = \pm \frac{1}{2}\) to be valid for a 3d electron.
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