0
$\begingroup$

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: $$\frac{\Delta_{\text o,\ce{H2O}}}{\Delta_{\text o,\ce{NH3}}}=\frac{\lambda_{\ce{NH3}}}{\lambda_{\ce{H2O}}}$$

$\endgroup$
2
$\begingroup$

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 $\ce{NH3}$: $$\lambda=\frac{600\ \mathrm{nm}}{1.25}=480\ \mathrm{nm}$$

$\endgroup$

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.