The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.
Manganese(II) has the electronic configuration of 3d⁵. In octahedral complexes, the five d-orbitals split into lower energy t₂g orbitals (three orbitals) and higher energy e_g orbitals (two orbitals).
In [Mn(H₂O)₆]²⁺, water acts as a weak field ligand. The crystal field splitting energy (Δ₀) is small. The five d-electrons do not pair up and occupy all five d-orbitals singly following Hund's rule. This forms a high spin complex with five unpaired electrons, making it strongly paramagnetic.
In [Mn(CN)₆]⁴⁻, cyanide ion acts as a strong field ligand. The crystal field splitting energy (Δ₀) is large. The five d-electrons pair up in the lower energy t₂g orbitals. This forms a low spin complex with only one unpaired electron, making it weakly paramagnetic.
Thus, Crystal Field Theory explains this difference based on the ligand strength and the resulting crystal field splitting.
Explanation
This question tests understanding of Crystal Field Theory and how ligand strength affects electron pairing in d-orbitals. The context provides similar examples with Fe³⁺ complexes showing that CN⁻ (strong ligand) causes electron pairing while H₂O (weak ligand) does not. For Mn²⁺ (d⁵ configuration), weak field ligands give high spin complexes with 5 unpaired electrons, while strong field ligands give low spin complexes with 1 unpaired electron.
Solution Steps
Step 1: Identify the electronic configuration of Mn(II) as 3d⁵
Step 2: Analyze the effect of weak field ligand (H₂O) - small Δ₀, electrons remain unpaired, high spin complex with 5 unpaired electrons
Step 3: Analyze the effect of strong field ligand (CN⁻) - large Δ₀, electrons pair up in t₂g orbitals, low spin complex with 1 unpaired electron
Step 4: Conclude that ligand strength determines the magnetic behavior through crystal field splitting