Question

question 26 of 30
what are most likely the values of the quantum numbers ( l ) and ( m_l ) for an electron in a 2s orbital?

a. ( l = 0; m_l = 0 )

b. ( l = -\frac{1}{2}, +\frac{1}{2}; m_l = -2, -1, 0, 1, 2 )

c. ( l = 0, 1, 2; m_l = -\frac{1}{2}, +\frac{1}{2} )

d. ( l = 2; m_l = 0 )

Explanation:

1. Recall the rules for quantum numbers: The principal quantum number \( n \) determines the energy level. For a \( 2s \) orbital, \( n = 2 \). The azimuthal quantum number \( l \) (angular momentum quantum number) has values from \( 0 \) to \( n - 1 \). For \( s \)-orbitals (regardless of the principal quantum number \( n \)), \( l = 0 \) (since \( s \)-orbitals correspond to \( l = 0 \), \( p \)-orbitals to \( l = 1 \), \( d \)-orbitals to \( l = 2 \), etc.).
2. The magnetic quantum number \( m_l \) has values from \( -l \) to \( +l \), including \( 0 \). Since \( l = 0 \) for an \( s \)-orbital, \( m_l \) can only be \( 0 \) (because the range from \( -0 \) to \( +0 \) is just \( 0 \)).
3. Analyze the options:

• Option A: \( l = 0 \) (correct for \( s \)-orbital) and \( m_l = 0 \) (correct as \( l = 0 \) implies \( m_l = 0 \)).
• Option B: The values given for \( l \) are spin quantum number values (not azimuthal), and the \( m_l \) values are incorrect for \( l = 0 \).
• Option C: The \( l \) values are incorrect for an \( s \)-orbital (should be \( 0 \) only), and \( m_l \) values here are spin quantum numbers.
• Option D: \( l = 2 \) corresponds to \( d \)-orbitals, not \( s \)-orbitals, so this is incorrect.

Answer:

A. \( l = 0; m_l = 0 \)