I was looking at some quick revision notes for d&f block elements and it had this random fact: Out of $\ce{Mn^3+}$, $\ce{Cr^3+}$, $\ce{V^3+}$, and $\ce{Ti^3+}$, chromium readily forms complexes. I did not understand which properties of chromium make it advantageous for it to form complexes, was it due to its size? Its electronic configuration? Were there any other properties at play?


The premise of the statement is skewed. All four of those have the same (rather strong) tendency to form complexes since naked metal cations in aquaeous solution simply do not exist. However, a point may be made that the complexes of chromium(III) are the least labile (or most inert) ones of the series.

Th reason for this lies in the typical octahedral arrangement observed in transition metal aqua complexes. This arrangement causes splitting of the $\mathrm d$ orbitals (which are degenerate in ground state) into two sets of $\mathrm d$ orbitals: $\mathrm{t_{2g}}$ and $\mathrm{e_g}$. In very basic crystal field theory, this can be understood as the $\mathrm{e_g}$ orbitals being those pointing towards the ligands approaching along the coordinate axes and bearing a point-shaped negative charge; thus, these orbitals are destabilised (raised in energy). This affects the orbitals $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2 - y^2}$. On the contrary, the other three $\mathrm{d}$ orbitals — $\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ — are considered stabilised since the entire transformation should preserve the total energy. (This concept is not physico-chemically accurate in the way it is presented here but close enough.)

The energy difference between the two energy levels is defined as (the relative value of) $\pu{10Dq}$. Again, using the ‘conservation of total energy’ principle, the higher-up orbitals are defined as having an energy of $\pu{+6Dq}$ each while the stabilised orbitals are defined as having $\pu{-4Dq}$ each. A ligand field stabilisation enthalpy or LFSE can be calculated by taking the electrons in each orbital and adding up the corresponding $\pu{Dq}$ value to give a total energy; the higher this is, the more stable the configuration is relatively.

All complexes in question are high-spin complexes, meaning that even though the $\mathrm d$ orbitals are no longer degenerate they are still filled according to Hund’s rule. Since the order of the elements in the periodic table is $\ce{Ti, V, Cr, Mn}$ and since each is in the same oxidation state ($\mathrm{+III}$), we can instantly see that they differ by the number of $\mathrm d$ electrons only: they have the configuration $\mathrm{d^1, d^2, d^3}$ and $\mathrm{d^4}$, respectively.

If we add between one and four electrons into the five $\mathrm d$ orbitals according to Hund’s rule and then calculate the LFSE, we will get the following result:

\begin{array}{cr}\hline \text{conf.} & \text{LFSE}\\\hline \mathrm{d^1} & \pu{-4Dq}\\ \mathrm{d^2} & \pu{-8Dq}\\ \color{red}{\mathrm{d^3}} & \color{red}{\pu{-12Dq}}\\ \mathrm{d^4} & \pu{-6Dq}\\ \hline\end{array}

The highlighted value is the configuration with the lowest energy, i.e. the one which is most stabilised during complex formation. Therefore, chromium(III) complexes are the most inert complexes of the series.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.