Problem: For the complex $K_{3}$[Fe(CN)_{6}], determine: (a) oxidation state of Fe, (b) IUPAC name, (c) number of unpaired electrons and magnetic moment.

Step 1 — Oxidation state of Fe:

The complex is $K_{3}$[Fe(CN)_{6}].

• $K^{+}$ contributes: 3 × (+1) = +3 overall positive charge outside.
• [Fe(CN)_{6}] must carry −3 to balance.
• $CN^{-}$ contributes: 6 × (−1) = −6.
• Let OS of Fe = x: x + (−6) = −3 → x = +3.
• Fe is in +3 oxidation state ($d^{5}$ configuration).

Step 2 — IUPAC name:

• Ligands: $CN^{-}$ → cyanido (anionic: add -o to cyanide → cyanido); 6 ligands → hexacyanido
• Central metal: Fe in anionic complex → ferrate (latin name root + -ate)
• Oxidation state: +3 → (III)
• Outer cation: $K^{+}$ → potassium
• Full name: Potassium hexacyanidoferrate(III)
• (This is also known as potassium ferricyanide in common nomenclature.)

Step 3 — Magnetic moment:

$Fe^{3+}$ is $d^{5}$. $CN^{-}$ is a very strong field ligand (top of spectrochemical series).

• Strong field + $d^{5}$: $\Delta o$ >> P, electrons pair up → Low spin
• t_{2}g orbital filling: 5 electrons fill t_{2}g as: t_{2}$g^{5}$ $eg^{0}$
• Unpaired electrons: only 1 (t_{2}g has orbitals that hold: 2, 2, 1 = 1 unpaired)
• μ = √(n(n+2)) = √(1×3) = √3 ≈ 1.73 BM

Contrast: [$FeF_{6}$]^{3-} — same $d^{5}$ $Fe^{3+}$ but $F^{-}$ is weak field → high spin → t_{2}$g^{3}$ $eg^{2}$ → 5 unpaired → μ = √(5×7) = √35 ≈ 5.92 BM