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Is Na2[Co(SCN)4] Laporte allowed
Accepted Answer
The complex Na2[Co(SCN)4] is not Laporte allowed. Here's why:
Laporte Rule: This rule states that transitions between orbitals of the same parity (both even or both odd) are forbidden. This is because the electric dipole moment operator is odd, and the integral for the transition probability involves the product of the initial and final wavefunctions and the dipole operator.
Parity in Na2[Co(SCN)4]: In this complex, the cobalt(II) ion (Co2+) is in a tetrahedral coordination environment. Tetrahedral complexes have no center of symmetry. This means that the complex has
odd
parity.
Conclusion: Since the complex has odd parity, transitions within its d-orbitals are Laporte
allowed
. However, this complex exhibits strong color due to charge transfer transitions, where an electron moves from a ligand orbital to a metal orbital. These transitions are typically allowed and responsible for the color observed.
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