Magnetic Moment Calculation for f⁵ Ion

To calculate the magnetic moment for an f⁵ ion in its ground state, we first need to determine the ground state term symbol using Hund's rules. The f subshell has 7 orbitals and can hold up to 14 electrons. For an f⁵ configuration, we have 5 electrons.

Ground State Term Symbol for f⁵

We fill the 7 f orbitals (with m_l values +3, +2, +1, 0, -1, -2, -3) with 5 electrons following Hund's rules:

1. Maximize spin multiplicity (2S+1). Electrons will occupy different orbitals with parallel spins as much as possible. For 5 electrons in 7 orbitals, all 5 can have parallel spins.
2. Maximize orbital angular momentum (L). Place electrons in orbitals with the highest positive m_l values first.

Filling electrons:

• Electron 1: m_l = +3, m_s = +1/2
• Electron 2: m_l = +2, m_s = +1/2
• Electron 3: m_l = +1, m_s = +1/2
• Electron 4: m_l = 0, m_s = +1/2
• Electron 5: m_l = -1, m_s = +1/2

Total spin angular momentum (S) = 5 × (1/2) = 5/2.
Spin multiplicity (2S+1) = 2(5/2) + 1 = 6.

Total orbital angular momentum (L) = Sum of m_l values = +3 +2 +1 +0 -1 = 5.
The value L=5 corresponds to an H term.

Total angular momentum (J): For configurations less than half-filled (f⁵ is less than half-filled as 5 < 7), J = |L - S|.

\(J = |5 - 5/2| = |10/2 - 5/2| = 5/2\)

The ground state term symbol is ^2S+1L_J = ⁶H_5/2.

Calculating Magnetic Moment

For lanthanide ions (which have f electrons), the magnetic moment is generally calculated using the total angular momentum J, as the orbital contribution is significant and not quenched. The formula for the magnetic moment (\(\mu\)) in Bohr Magnetons (B.M.) is given by:

\(\mu = g\sqrt{J(J+1)}\) B.M.

where g is the Landé g-factor, calculated using the formula:

\(g = 1 + \frac{S(S+1) - L(L+1) + J(J+1)}{2J(J+1)}\)

We have L=5, S=5/2, and J=5/2.

First, calculate the g-factor:

\(g = 1 + \frac{(5/2)(5/2+1) - 5(5+1) + (5/2)(5/2+1)}{2(5/2)(5/2+1)}\)
\(g = 1 + \frac{(5/2)(7/2) - 5(6) + (5/2)(7/2)}{2(5/2)(7/2)}\)
\(g = 1 + \frac{35/4 - 30 + 35/4}{35/2}\)
\(g = 1 + \frac{70/4 - 30}{35/2}\)
\(g = 1 + \frac{35/2 - 60/2}{35/2}\)
\(g = 1 + \frac{-25/2}{35/2}\)
\(g = 1 - \frac{25}{35}\)
\(g = 1 - \frac{5}{7}\)
\(g = \frac{7}{7} - \frac{5}{7} = \frac{2}{7}\)

Now, calculate the magnetic moment using the g-factor and J value:

\(\mu = g\sqrt{J(J+1)}\)
\(\mu = \frac{2}{7}\sqrt{\frac{5}{2}(\frac{5}{2}+1)}\)
\(\mu = \frac{2}{7}\sqrt{\frac{5}{2}(\frac{7}{2})}\)
\(\mu = \frac{2}{7}\sqrt{\frac{35}{4}}\)
\(\mu = \frac{2}{7} \times \frac{\sqrt{35}}{\sqrt{4}}\)
\(\mu = \frac{2}{7} \times \frac{\sqrt{35}}{2}\)
\(\mu = \frac{\sqrt{35}}{7}\)

The calculated magnetic moment for the f⁵ ion in its ground state is \(\frac{\sqrt{35}}{7}\) B.M.

The final answer is \(\sqrt{35}\)/7.