Multiple Choice
Which of the following combinations of principal quantum number n and angular momentum quantum number l represent possible atomic orbitals?
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9. Quantum Mechanics
Quantum Numbers: Principal Quantum Number
Multiple Choice
Which one of the following sets of quantum numbers is not possible for an electron in an atom?
A
n = 2, l = 1, m_l = 0, m_s = +1/2
B
n = 3, l = 2, m_l = -2, m_s = -1/2
C
n = 1, l = 1, m_l = 0, m_s = +1/2
D
n = 4, l = 0, m_l = 0, m_s = -1/2
Verified step by step guidance1
Recall the allowed ranges for each quantum number: the principal quantum number \(n\) must be a positive integer (\(n = 1, 2, 3, \ldots\)), the azimuthal quantum number \(l\) can take integer values from \$0\( up to \)n-1\(, the magnetic quantum number \)m_l\( ranges from \)-l\( to \)+l\( in integer steps, and the spin quantum number \)m_s$ can be either \(+\frac{1}{2}\) or \(-\frac{1}{2}\).
Check the set \(n = 2, l = 1, m_l = 0, m_s = +\frac{1}{2}\): since \(l\) must be less than \(n\), and \$1 < 2\(, this set is possible. Also, \)m_l = 0\( is within \)-1\( to \)+1\(, and \)m_s$ is valid.
Check the set \(n = 3, l = 2, m_l = -2, m_s = -\frac{1}{2}\): here, \(l = 2\) is less than \(n = 3\), \(m_l = -2\) is within \(-2\) to \(+2\), and \(m_s\) is valid, so this set is possible.
Check the set \(n = 1, l = 1, m_l = 0, m_s = +\frac{1}{2}\): since \(l\) must be less than \(n\), but here \(l = 1\) is not less than \(n = 1\), this set violates the rule and is not possible.
Check the set \(n = 4, l = 0, m_l = 0, m_s = -\frac{1}{2}\): \(l = 0\) is less than \(n = 4\), \(m_l = 0\) is valid for \(l = 0\), and \(m_s\) is valid, so this set is possible.
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