Q.Write the oxidation state, coordination number, nature of ligand, magnetic property and electronic configuration in octahedral crystal field for the complex K₄[Mn(CN)₆].

Step 1. K₄[Mn(CN)₆] has 4 K⁺ counter ions, so the complex anion carries -4 charge: x + 6(-1) = -4 gives x = +2. Manganese's oxidation state is +2, i.e. Mn²⁺.

Step 2. Coordination number: 6 monodentate CN⁻ ligands, each contributing one donor carbon atom -- coordination number 6, octahedral coordination polyhedron.

Step 3. Nature of ligand: CN⁻ sits at the strong-field end of the spectrochemical series (only CO ranks higher), so it is a strong-field ligand and forces pairing of the metal's d electrons.

Step 4. Mn²⁺ has outer configuration 3d⁵. With CN⁻ forcing pairing (Δ₀ > P), the electronic configuration in the octahedral crystal field is LOW spin: t2g⁵eg⁰ (all 5 electrons occupy the three t2g orbitals, giving 2 pairs and 1 unpaired electron, rather than the free-ion/high-spin t2g³eg² spread across all five orbitals).

Step 5. Magnetic property: with 1 unpaired electron, the complex is paramagnetic, μs = √[1×3] = 1.73 BM. …

Find manganese's oxidation state from the K₄ charge balance, place CN⁻ on the spectrochemical series to decide high spin vs low spin, write the resulting t2g/eg electron configuration for Mn²⁺ (d⁵), …

• Defaulting to the high-spin, free-ion-like configuration t2g³eg² for Mn²⁺ (d⁵) without checking that CN⁻ is a strong-field ligand here, which actually forces the low-spin t2g⁵eg⁰ configuration instead. …