Charge Transfer Spectra and Types

Charge Transfer Spectra or Band

A charge transfer spectra or band may be defined as the peak arising from the transition in which an electron is transferred from one atom or group of atoms in the molecule to another one.

In other words, the transition occurs between molecular orbitals that are centered on different atoms or groups. Unlike $d\text{-}d$ transitions, charge transfer transitions are Laporte and spin-allowed and therefore, show very intense (strong) absorption. The charge transfer bands have extinction coefficients in the 500–2000 unit range (for $d\text{-}d$ transitions, the coefficients are typically below 1000 units).

These bands mostly occur near the ultraviolet region. When these transitions occur in the visible region, the compound shows intense color. In the spectra of some metal complexes, an overlap between the end of the charge transfer band and $d\text{-}d$ absorption occurs. This makes cleanly resolving the full $d\text{-}d$ spectrum of the complex impossible.

A charge transfer transition is an intramolecular transition, so it requires much more energy than $d\text{-}d$ transitions. It may be regarded as an internal redox process because the charge distribution in the excited state differs considerably from the ground state.

Nature of Charge Transfer Spectra

1. These are Laporte and spin-allowed transitions: $$\Delta l = \pm 1 \quad \text{and} \quad \Delta S = 0$$
2. Charge transfer transitions mostly occur in the UV and visible regions.
3. Transitions occurring exclusively in the visible region are responsible for the distinct color of the complexes.
4. It is observed almost universally across all complexes.

Types of Charge Transfer Spectra

Charge transfer spectra may be classified into four primary pathways:

1. Ligand to Metal Charge Transfer spectra (LMCT)
2. Metal to Ligand Charge Transfer spectra (MLCT)
3. Metal to Metal (Intervalence) Charge Transfer spectra (MMCT)
4. Ligand to Ligand (Intra-Ligand) Charge Transfer spectra (LLCT)

1. Ligand to Metal Charge Transfer Spectra (LMCT)

If the transfer of an electron takes place from the ligand to the metal, it is called ligand to metal charge transfer (LMCT). These transitions take place with lower energy as the metal becomes more easily reducible and the ligand gets readily oxidizable.

Conditions for LMCT

• Metals should be in a high oxidation state.
• Metals should have high ionization energy so that they possess empty orbitals at fairly low energies.
• Ligands should have lone pairs of electrons of relatively high energy and low electron affinity.

A majority of charge transfer complexes of transition metals involve $L \rightarrow M$ transfer as expected from the availability of nonbonding or antibonding orbitals on the metal.

2. Metal to Ligand Charge Transfer Spectra (MLCT)

If the transfer of an electron takes place from the metal to the ligand, it is called metal to ligand charge transfer (MLCT). Charge transfer processes in this opposite direction are favored in complexes that have occupied metal-centered orbitals and vacant, low-lying ligand-centered orbitals.

Conditions for MLCT

• Metals should be in a low oxidation state.
• Metal $d$-orbitals are heavily filled.
• Metal $d$-orbitals are relatively high in energy.
• Ligands must have empty $\pi^*$-antibonding orbitals.

MLCT mainly occurs with ligands having accessible antibonding $\pi$-orbitals such as $\text{CO}$, $\text{NO}$, $\text{CN}^-$, $\text{SCN}^-$, pyridine, bipyridine, pyrazine, dithioline, and o-phenanthroline.

In octahedral complexes, when both the $t_{2g}$ and $e_g^*$ orbitals belonging to the metal are occupied, two MLCT bands ($\pi^* \leftarrow t_{2g}$ and $\pi^* \leftarrow e_g^*$) are observed (where $\pi^*$ is the vacant ligand orbital). If only one of these sets is occupied, then only a single charge transfer band is observed.

3. Metal to Metal Charge Transfer Spectra (Intervalence Transition)

In these transitions, an electron gets excited from the valence shell of one metal atom to the valence shell of another. The electron transfer typically takes place from an atom of a lower oxidation state to an atom of a higher oxidation state.

Compounds containing metal-metal bonds are intensely colored. This color is due to allowed $\sigma \rightarrow \sigma^*$, $\pi \rightarrow \pi^*$, and $\delta \rightarrow \delta^*$ transitions. Metal carbonyls containing an $\text{M-M}$ single bond are generally intense in color due to $\sigma \rightarrow \sigma^*$ transitions, while the color of compounds containing a quadruple bond is driven by $\delta \rightarrow \delta^*$ transitions.

For example, $\text{Mn}_2(\text{CO})_{10}$ is bright yellow, $[\text{Re}_2\text{Cl}_8]^{2-}$ is deep blue, $\text{Co}_2(\text{CO})_8$ is purple-black, $[\text{Mo}_2\text{Cl}_8]^{4-}$ is cherry red, and $\text{Fe}_2(\text{CO})_9$ is gold.

Prussian blue, $\text{KFe}[\text{Fe}(\text{CN})_6]$, shows an intense blue color because of the transfer of an electron from $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$. In prussian blue, the $\text{Fe}^{2+}$ ion is octahedrally coordinated to the carbon atoms of the cyanide ligands, and $\text{Fe}^{3+}$ is octahedrally coordinated to the nitrogen atoms of the cyanide ligands. Thus, electron transfer occurs directly through the bridging cyanide ligands. Similarly, red lead ($\text{Pb}_3\text{O}_4$) contains both $\text{Pb(II)}$ and $\text{Pb(IV)}$, and electron transfer between them produces its intense red color.

4. Intra-Ligand Charge Transfer

Some organic ligands behave inherently as a chromophore, meaning another type of absorption band—an intraligand band—may be observed. There are four classic electronic transitions within a chromophore: $\sigma \rightarrow \sigma^*$, $\pi \rightarrow \pi^*$, $n \rightarrow \pi^*$, and $n \rightarrow \sigma^*$. When such a ligand coordinates with a metal ion, the characteristic energy of absorption shifts.

Why is the color of octahedral complexes less intense than tetrahedral complexes?

Solution:
The extinction coefficient is notably higher and the color of tetrahedral complexes is more intense relative to octahedral complexes due to $d\text{-}p$ orbital mixing in tetrahedral geometries.

Due to this mixing, some $p$-character is introduced into the $t_2$ subset of the $d$-orbitals. Consequently, the Laporte symmetry selection rule is relaxed to some extent, raising the extinction coefficient value higher than that of an octahedral system. Furthermore, the inversion center ($gerade$ property) is inherently lost in a tetrahedral geometry.

We know that: $$\Delta_{\text{tet}} = \frac{4}{9}\Delta_{\text{oct}}$$

Because the crystal field splitting energy value is smaller, transitions in tetrahedral complexes occur at lower energies that fall cleanly within the visible spectrum, yielding a more intense perceived color.

Practice Question: Which of the following has the lowest frequency?

A. $[\text{Co}(\text{NH}_3)_5\text{F}]^{2+}$
B. $[\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+}$
C. $[\text{Co}(\text{NH}_3)_5\text{Br}]^{2+}$
D. $[\text{Co}(\text{NH}_3)_5\text{I}]^{2+}$

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View Answer & Explanation

Option D is the correct answer.

Explanation: According to the spectrochemical series, a weaker field ligand causes a smaller crystal field splitting energy ($\Delta_o$). Therefore, the relative order of $\Delta_o$ values for these halopentammine complexes is:

$[\text{Co}(\text{NH}_3)_5\text{I}]^{2+} < [\text{Co}(\text{NH}_3)_5\text{Br}]^{2+} < [\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+} < [\text{Co}(\text{NH}_3)_5\text{F}]^{2+}$

Using Planck's relation equation: $$E = h\nu$$

The complex with the smallest crystal field splitting gap ($\Delta_o$) will absorb light of the longest wavelength, which inherently translates to the lowest energy transition and therefore the lowest frequency ($\nu$). Since iodine is the weakest field ligand here, $[\text{Co}(\text{NH}_3)_5\text{I}]^{2+}$ satisfies this condition.