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What will happen if there are electrons with opposite spins present in the orbitals of ligand and metal atom during the splitting of orbitals in crystal field splitting?

Magnetic fields of two electrons with opposite spins should cancel out each other, thus its net magnetic field would be zero, there would be no repulsion between orbitals of the ligand and metal atom, and splitting of the orbitals $\mathrm{e_g}$ and $\mathrm{t_{2g}}$ would not happen.

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    $\begingroup$ It's not the magnetic field due to spin that causes the splitting. It is the electrostatic field, coupled with the lowering of the symmetry from spherical to octahedral or whatever, that is behind the splitting of the d orbitals. $\endgroup$
    – Ian Bush
    Commented Dec 13, 2022 at 7:09

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The repulsion between two electrons due to being in the same spin-state is only a part of the net repulsion between the electrons. Coulombic repulsion is present between pairs of electrons with same and opposite spins. The Hamiltonian includes the Exchange integral, the Coulombic integral, etc. In certain cases, it is even favorable for electrons to occupy parallel spins (see magnetism). The only condition in which Pauli exclusion principle applies is when the electrons are present in the same orbital, having all other quantum numbers identical.

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