[ORGMalta]

Hey im trying to figure out this question, can anybody help please

The Pairing Energy of Co(III) is about 24 % greater then in the isoelectronic Fe(II). Rationalize.

By any chance does it have to do with B, the inner repulsive forces due to electrones, as they have different molecular sizes??

[Alpha-Omega]

To address this go to Crystal Field Theory.  The answer lies in the geometries that Co and Fe adopt.

Electron configurations for the transition metals are not
always easy to predict because they are dependent upon both orbital energies and electron-electron repulsion.

d orbitals experience a steeper drop in energy with increasing Zeff than the s orbital of the same period, thus
the electrons in d orbitals have lower energy.

Atomic radii for the transition metals decrease from left to right because the added d electrons don’t shield each other very well from the increasing nuclear charge (↑ Zeff).  Atomic radii toward the end of a period increase due to thegreater electron-electron repulsion (↑ shielding) as electrons are added to occupied orbitals. Transition metal radii do not differ as greatly as the main group elements do.

In the case of the low-spin configuration (small Δ), the complex, due to greater stability (lower energy), will
tolerate a small increase in energy to avoid an even larger increase in energy when electrons are paired; greater
repulsion, higher spin-pairing energy, P.

If Δ is large, the complex will accept the energetic premium needed to force two electrons into the same
orbital. High-spin complexes thus exhibit greater paramagnetism.

If Δ is less than P, the high-spin arrangement is more stable or lower in energy.

For an octahedral complex with metal ion configurations of d4 through d7 high and low-spin configurations are
possible.

For tetrahedral complexes Δ is almost always smaller than the spin-pairing energy, P, and thus, exist in high-spin
configurations.

Cobalt forms tetrahedral complexes.

Iron forms octahedral complexes.