A $\ce{[M(H2O)6]^2+}$ complex typically absorbs at around $600\ \mathrm{nm}$. It is allowed to form a new complex $\ce{[M(NH3)6]^2+}$ that should have absorbtion at?

The absorbtion is because of transition of $\mathrm{e_2g}$ electron to $\mathrm{t_{2g}}$ and vice versa which depend on crystal field stabilization energy as: $$\Delta = \frac{hc}{\lambda=600\ \mathrm{nm}}\approx2.0\ \mathrm{eV}$$

But I don't know how to convert it in terms of $\Delta_\text o$ so that I can know the actual configuration.

In $\ce{H2O}$ there will be no pairing of electrons as it is weak and reverse with $\ce{NH3}$.

This looks useful but isn't without the metal configuration: $$\frac{\Delta_{\text o,\ce{H2O}}}{\Delta_{\text o,\ce{NH3}}}=\frac{\lambda_{\ce{NH3}}}{\lambda_{\ce{H2O}}}$$


1 Answer 1


The given wavelength of $\lambda=600\ \mathrm{nm}$ may be directly converted to various other quantities, namely wavenumber $\tilde\nu$, frequency $\nu$, photon energy $E_\text{p}$, and molar energy $E_\text{m}$:

$$\tilde\nu=\frac{1}{\lambda}=16\,700\ \mathrm{cm^{-1}}$$ $$\nu=\tilde\nu c=\frac{c}{\lambda}=5.00\times 10^{14}\ \mathrm{s^{-1}}=500\ \mathrm{THz}$$ $$E_\text{p}=h\nu=\frac{hc}{\lambda}=3.31\times 10^{-19}\ \mathrm{J}=2.07\ \mathrm{eV}$$ $$E_\text{m}=N_\text{A}h\nu=\frac{N_\text{A}hc}{\lambda}=199\ \mathrm{kJ\ mol^{-1}}$$

With regard to electronic spectra of coordination compounds, usually wavenumbers $\tilde\nu$ expressed in $\mathrm{cm^{-1}}$ are used.

The electronic configuration of $\ce{M^2+}$ in the given complex $\ce{[M(H2O)6]^2+}$ is unknown. Nevertheless, we may assume that the complex has octahedral geometry and that the observed absorption maximum probably corresponds to promoting one electron from the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$) to an $\mathrm{e_g}$ orbital ($\mathrm{d}_{x^2{-}y^2}$ or $\mathrm{d}_{z^2}$). Thus, it directly gives an estimate for $\Delta_\text{o}=10\ \mathrm{Dq}$ as $16\,700\ \mathrm{cm^{-1}}$.

The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{NH3}$, the value of $\Delta_\text{o}=10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are increased.

An empirical equation for estimating the value of $\Delta_\text{o}$ for any pair of metal and ligands is $$\Delta_\text{o}=f \cdot g$$ where $f$ is the ligand parameter and $g$ the contribution from the central atom. The parameter $f$ describes the ligand strength relative to water; i.e. the value for $\ce{H2O}$ is $f=1.00$. The empirical value for $\ce{NH3}$ is $f=1.25$.

Since the ligand field splitting for $\ce{[M(H2O)6]^2+}$ is $\Delta_\text{o} = 16\,700\ \mathrm{cm^{-1}}$, the corresponding ligand field splitting for $\ce{[M(NH3)6]^2+}$ may be estimated as $\Delta_\text{o} = 1.25 \times 16\,700\ \mathrm{cm^{-1}}= 20\,900\ \mathrm{cm^{-1}}$.

The resulting wavenumber $\tilde\nu=20\,900\ \mathrm{cm^{-1}}$ may be converted to a wavelength of approximately $\lambda=480\ \mathrm{nm}$.

Since $\lambda=\frac{1}{\tilde\nu}$, the new wavelength could have been directly calculated from the given original value of $600\ \mathrm{nm}$ using the parameter $f=1.25$ for $\ce{NH3}$: $$\lambda=\frac{600\ \mathrm{nm}}{1.25}=480\ \mathrm{nm}$$


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