$\ce{NH3}$ is known to be a strong field ligand, while $\ce{Cl}$ is known to be a weak field ligand.
Is $\ce{[Co(NH3)4Cl2]Cl}$ a high spin complex or a low spin complex?
I assumed this to be a high spin complex. Oxidation state of $\ce{Co}$ is $+3$. And electronic configuration of $\ce{Co(III)}$ is $\mathrm{[Ar]~3d^6}$.
Since I assumed this to be a high spin complex — pairing of electrons of $\mathrm{3d}$ orbitals will not happen. The $6$ pairs of electrons from the ligands must be included in the $\mathrm{4s}$, $\mathrm{4p}$, $\mathrm{4d}$ orbitals — which leaves electrons unpaired in the $\mathrm{3d}$ orbital, making this a paramagnetic complex, with hybridisation $\mathrm{sp^3d^2}$.
Answer given to this question says otherwise. Hints are welcome!
This question is an excerpt of a question that appeared in the JEE(A) - 2016
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