Question

Draw orbital-energy splitting diagrams and use the spectrochemical series to show the orbital occupancy for each of the following (assuming that $\mathbf{\left(}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\right)}$is a weak-field ligand): role="math" localid="1663325015190" $\mathbf{\left(}\mathbf{a}\mathbf{\right)}{\mathbf{\left[}\mathbf{Cr}{\left(H_{2}O\right)}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathbf{b}\mathbf{\right)}{\mathbf{\left[}\mathbf{Cu}{\left(H_{2}O\right)}_{\mathbf{4}}\mathbf{\right]}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{c}\mathbf{\right)}{\mathbf{\left[}{\mathbf{FeF}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{-}}$

Short answer

1. 3 d electrons
2. 9 d electrons
3. 5 d electrons

Step-by-step solution

Crystal field splitting energy and spectrochemical series

The crystal field splitting energy $∆$represents the difference in energy between the two higher energy eg orbitals and the three lower energy orbitals ${\mathrm{T}}_2\mathrm{g}$;there are five orbitals which are splitted in two higher energy orbitals and three lower energy orbitals because of the electrostatic attractions between metal cation and ligands' negative charge.

Spectrochemical series (from weak-field to strong-field ligands):

${\mathrm{I}}^{-}<{\mathrm{Cl}}^{-}<{\mathrm{F}}^{-}<{\mathrm{OH}}^{-}<{\mathrm{H}}_2\mathrm{O}<{\mathrm{SCN}}^{-}<{\mathrm{NH}}_3<{\mathrm{en}}^{-}<{\mathrm{NO}}_2^{-}<{\mathrm{CN}}^{-}<\mathrm{CO}$

- the weaker the field is, the lower the splitting energy is$∆$

- the stronger the field is, the higher the splitting energy is$∆$

Orbital-energy splitting diagram

Orbital-energy splitting diagram

Orbital-energy splitting diagram