The spin-only magnetic moment (\(\mu\)) of a complex is determined by the number of unpaired electrons (n) present in the central metal ion. The formula for the spin-only magnetic moment is given by:

\(\mu = \sqrt{n(n+2)}\) BM

where BM stands for Bohr Magneton, the unit of magnetic moment.

Spin Magnetic Moment of Complex B: [Fe(phen)₃]Cl₂

• The complex is [Fe(phen)₃]Cl₂. The chloride ions (Cl^-) are outside the coordination sphere, carrying a total charge of -2. Therefore, the complex cation [Fe(phen)₃] must have a charge of +2.
• phen (1,10-phenanthroline) is a neutral ligand. Let the oxidation state of Iron (Fe) be x. The charge balance gives: \(x + 3 \times (0) = +2\). Thus, \(x = +2\).
• The metal ion in complex B is Fe²⁺.
• The electronic configuration of neutral Fe is [Ar] 3d⁶ 4s². For Fe²⁺, it is [Ar] 3d⁶.
• phen (1,10-phenanthroline) is a strong field ligand. Strong field ligands cause a large crystal field splitting energy (\(\Delta_o\)), which leads to pairing of electrons in the lower energy orbitals before filling the higher energy orbitals (low spin complex) for d⁴ to d⁷ ions in octahedral complexes.
• For Fe²⁺ (d⁶) in an octahedral field with a strong field ligand like phen, the electrons are arranged in a low spin configuration: t_2g⁶ e_g⁰.
• The number of unpaired electrons (n) in this configuration is 0.
• Calculating the spin-only magnetic moment for complex B: \(\mu_B = \sqrt{0(0+2)} = \sqrt{0} = 0\) BM.

Spin Magnetic Moment of Complex A: cis-[Fe(phen)₂(NCS-N)₂]

• The complex is cis-[Fe(phen)₂(NCS-N)₂]. This is a neutral complex.
• phen (1,10-phenanthroline) is a neutral ligand. NCS-N (thiocyanate bonded through Nitrogen) is an anionic ligand (NCS^-) with a charge of -1. There are two NCS-N ligands, contributing a total charge of 2 \(\times\) (-1) = -2.
• Let the oxidation state of Iron (Fe) be y. The charge balance gives: \(y + 2 \times (0) + 2 \times (-1) = 0\). Thus, \(y - 2 = 0\), which means \(y = +2\).
• The metal ion in complex A is Fe²⁺, which has a d⁶ electronic configuration.
• Complex A contains both phen (strong field ligand) and NCS-N (moderately strong field ligand, stronger when bonded via N than S). In mixed ligand complexes, the overall ligand field strength determines the spin state. While phen is a strong field ligand favoring low spin, the presence of two NCS-N ligands might result in an overall ligand field strength that is not strong enough to force all the d⁶ electrons to pair up, especially compared to the complex with three phen ligands (Complex B).
• For Fe²⁺ (d⁶), if the ligand field is relatively weak or moderate, it forms a high spin complex. In a high spin configuration for an octahedral field, the electrons are distributed according to Hund's rule before pairing: t_2g⁴ e_g².
• The number of unpaired electrons (n) in this high spin configuration is 4 (two in t_2g and two in e_g).
• Calculating the spin-only magnetic moment for complex A: \(\mu_A = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24}\).
• \(\sqrt{24}\) is approximately 4.899. So, \(\mu_A \approx 4.89\) BM.

Based on the analysis:

• Complex A (cis-[Fe(phen)₂(NCS-N)₂]) is likely high spin with 4 unpaired electrons, resulting in a spin-only magnetic moment of approximately 4.89 BM.
• Complex B ([Fe(phen)₃]Cl₂) is low spin with 0 unpaired electrons, resulting in a spin-only magnetic moment of 0 BM.

Comparing these values with the given options, the correct match is 4.89 BM for A and 0 BM for B.