Question

give all the possible values of the four quantum numbers of an electron in the following orbital: 3d n : l : m_l : m_s : write your answer in a fractional form.

Explanation:

Step1: Determine the principal quantum number (n)

The orbital is \( 3d \), so the principal quantum number \( n \) is the coefficient, which is 3.

Step2: Determine the azimuthal quantum number (l)

For a \( d \) orbital, the azimuthal quantum number \( l \) is 2 (since \( s \): \( l = 0 \), \( p \): \( l = 1 \), \( d \): \( l = 2 \), \( f \): \( l = 3 \)).

Step3: Determine the magnetic quantum number (\( m_l \))

The magnetic quantum number \( m_l \) can take values from \( -l \) to \( +l \), so for \( l = 2 \), \( m_l \) can be \( -2, -1, 0, 1, 2 \).

Step4: Determine the spin quantum number (\( m_s \))

The spin quantum number \( m_s \) can be either \( \frac{1}{2} \) or \( -\frac{1}{2} \).

Answer:

\( n \): \( 3 \)
\( l \): \( 2 \)
\( m_l \): \( -2, -1, 0, 1, 2 \)
\( m_s \): \( \frac{1}{2} \) or \( -\frac{1}{2} \)