Answer:
a) [Fe(H2O)6]3+
b) [Fe(CN)6]3−
c) [Ru(CN)6]3-
Explanation:
. [Mn(H2O)6]2+ or [Fe(H2O)6]3+
The both complexes are d5 complexes with the same ligand , water. Water is a weak ligand and note that Mn^2+ often have a crystal field stabilization energy of zero hence
[Fe(H2O)6]3+ will possess a greater ∆o value.
The splitting of d orbitals according to the crystal field theory depends on the;
i)geometry of the complex
ii) nature of the metal ion,
iii)charge on the metal ion,
iv) ligands that surround the metal ion.
When the geometry and the ligands are held constant, the order of crystal field splitting is as follows;
Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+
[Fe(H2O)6]3+ or [Fe(CN)6]3−
[Fe(CN)6]3− will have a greater ∆o because the cyanide ion is a strong field ligand compared to water. A strong field ligand causes a greater splitting of the octahedral crystal field compared to a weak field ligand.
. [Fe(CN)6]3− or [Ru(CN)6]3-
[Ru(CN)6]3- will exhibit a greater crystal field splitting. Crystal field splitting increases with the second and third row transition elements when compared to the crystal field splitting of the first row transition elements. Note that, there is an increase of approximately 30%–50% in Δo on going from a first-row transition metal to a second-row metal and another 30%–50% increase on going from a second-row to a third-row metal when they have the same geometry and oxidation state.
, then the bond is non-polar.
, then the bond will be covalent.
, then the bond will be ionic.