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Can a coordination complex be classified as high spin if it possesses the same number of unpaired electrons in both weak and strong field ligand environments? I am curious about the factors that determine the spin state of a complex and how it may or may not change in response to different ligand strengths. Any insights into the interplay between ligand field effects and the spin state of coordination complexes would be greatly appreciated.

Is the statement "With strong field ligand, Cr(III) octahedral complex will be high spin"(it is $d^3$ so there will be no effect of WFL or SFL on number of unpaired electrons)

correct or incorrect

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1 Answer 1

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High spin and low spin are the relative terms in splitting of electrons to the $t_{2g}$ and $e_g$ orbitals. According to the molecular orbital theory (MOT), the ligands approach the metal symmetrically along the vertices of an octahedron with the metal at center. Considering the axes of approach of ligand to be $\pm x$, $\pm y$ and $\pm z$, the ligands interact with the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (along with other valance orbitals), splitting them into $e_g$ bonding and $e_g^*$ antibonding orbitals. In this case, $e_g$ is the symmetry of the orbitals. Rest of the $d$-orbitals of the metal, $d_{xy}$, $d_{yz}$ and $d_{zx}$ does not interact with the ligand and stay as non-bonding orbitals of $t_{2g}$ symmetry.

Thus, according to MOT, the interactions of metals with the ligands make the d-orbitals split. The electrons fill up the orbitals one by one, and the number is such that the number of electrons in the $d$-orbitals of the metal exactly match up with that to be filled up in the $t_{2g}$ and $e_g^*$ orbitals, assuming the outermost $s$-orbital of the metal is filled and each ligand provides a contribution of two electrons.

When a "strong field ligand" couples strongly with the metal, the $e_g^*$ orbital has higher energy than the exchange energy (pairing energy) of electrons when compared to $t_{2g}$. In this case, the electrons, specifically the fourth and fifth electron to be filled, find it energetically more favourable to pair up in $t_{2g}$, thus reducing the spin quantum number and hence making a low spin complex.

On the other hand, a "weak field ligand" couples weakly with the metal orbitals, and thus the exchange energy is higher compared to the split between $t_{2g}$ and $e_g$. Thus, the fourth and fifth electrons prefer occupying the $e_g^*$ orbitsl in this case, thus making it a high spin complex.

A table with the electronic configuration for ideal strong and weak field ligands are given below:

No. of $d$-electrons $t_{2g}$ SF $e_g^*$ SF Unpaired $e^-$ in SF $t_{2g}$ WF $e_g^*$ WF Unpaired $e^-$ in WF
1 1 0 1 1 0 1
2 2 0 2 2 0 2
3 3 0 3 3 0 3
4 4 0 2 3 1 4
5 5 0 1 3 2 5
6 6 0 0 4 2 4
7 6 1 1 5 2 3
8 6 2 2 6 2 2
9 6 3 1 6 3 1
10 6 4 0 6 4 0

For a full diagram of the MO for octahedral metal-ligand interactions, please check this link.

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