Concept Explanation: Magnetic Moments in Octahedral Complexes

This question involves calculating the spin-only magnetic moments for Fe³⁺ complexes with different ligands (SCN^– and CN^–) and finding the difference between them. The magnetic moment depends on the number of unpaired electrons, which is determined by whether the complex is high-spin or low-spin. This, in turn, depends on the strength of the ligand field.

Step 1: Determine the Electron Configuration of Fe³⁺

Atomic number of Fe is 26. Its electron configuration is [Ar] 4s² 3d⁶.

For Fe³⁺, three electrons are removed. Typically, electrons are removed from the 4s orbital first, followed by the 3d orbital.

Therefore, the configuration of Fe³⁺ is [Ar] 3d⁵.

This means there are 5 electrons in the d orbitals.

Step 2: Understand Ligand Field Strength and Spin State

In an octahedral field, the five d orbitals split into two sets: the lower-energy t_2g set and the higher-energy e_g set. The energy difference between them is called Δ_o (crystal field splitting energy).

Whether a complex is high-spin (maximum unpaired electrons) or low-spin (minimum unpaired electrons) depends on the strength of the ligand:

• Weak Field Ligands: Small Δ_o. It is energetically favorable to place electrons in the higher e_g orbitals rather than pair them in the t_2g orbitals. This results in a high-spin complex.
• Strong Field Ligands: Large Δ_o. The energy cost to promote an electron is greater than the electron pairing energy. Therefore, electrons pair up in the t_2g orbitals. This results in a low-spin complex.

From the spectrochemical series:

SCN^– (bonding through S) is a weak field ligand.

CN^– is a strong field ligand.

Therefore:

• [Fe(SCN)₆]^3– will be a high-spin complex.
• [Fe(CN)₆]^3– will be a low-spin complex.

Step 3: Find the Number of Unpaired Electrons

For a d⁵ system:

High-spin complex (weak field, e.g., SCN^–): Electrons will remain unpaired to maximize spin. All five electrons are unpaired.

Number of unpaired electrons, $n$ = 5.

Low-spin complex (strong field, e.g., CN^–): Electrons will pair up. The t_2g set can hold 6 electrons. All 5 electrons will pair up as much as possible, leaving only one unpaired electron.

Number of unpaired electrons, $n$ = 1.

Step 4: Calculate the Spin-Only Magnetic Moment

The spin-only magnetic moment ($\mu$) is given by the formula:

Where $n$ is the number of unpaired electrons and BM stands for Bohr Magnetons.

For [Fe(SCN)₆]^3– (High-spin, n=5):

Approximated to the nearest integer: $6$ BM.

For [Fe(CN)₆]^3– (Low-spin, n=1):

Approximated to the nearest integer: $2$ BM.

Step 5: Find the Difference

Difference = |μ_high - μ_low| = |6 - 2| = $4$ BM.

Final Answer

The difference between the spin-only magnetic moments is $4$ Bohr Magnetons.

Related Topics & Formulae

Key Topics:

• Crystal Field Theory (CFT)
• Octahedral Crystal Field Splitting
• Spectrochemical Series
• High-spin vs. Low-spin Complexes
• Magnetic Properties of Coordination Compounds

Important Formulae:

Spin-only magnetic moment: $\mu =\sqrt{n(n+2)}\text{BM}$

Theory: The spin-only formula is derived from the quantum mechanical treatment of electron spin. It provides a good approximation for the magnetic moment of transition metal complexes, especially for 3d metals where orbital contributions are often quenched. The actual magnetic moment can be slightly higher than the spin-only value if there is a contribution from the orbital angular momentum.