Question

A d1 octahedral complex is found to absorb visible light, with the absorption maximum occurring at 517 nm. Calculate the crystal-field splitting energy, Δ , in kJ/mol.

Answer 1

Answer on Question #76742 – Chemistry – General Chemistry

A d1 octahedral complex is found to absorb visible light, with the absorption maximum occurring at 517 nm. Calculate the crystal-field splitting energy, $\Delta$, in kJ/mol.

Solution:

The crystal-field splitting energy is equal to the energy of transition of the electron, that is linked to the wavelength of the emitted light $\lambda$ as follows:

$$\Delta = E = \frac{hc}{\lambda},$$

Planck constant $h = 6.62 \times 10^{-34} \, \text{m}^2 \cdot \text{kg} \cdot \text{s}^{-1}$

Speed of light $c = 3 \times 10^8 \, \text{m} \cdot \text{s}^{-1}$

$$\Delta = \frac{6.62 \times 10^{-34} \left(m^2 \cdot kg \cdot s^{-1}\right) \cdot 3 \times 10^8 \left(m \cdot s^{-1}\right)}{517 \times 10^{-9} \, m} = 3.84 \times 10^{-19} \left(m^2 \cdot kg \cdot s^{-2}\right) = 3.84 \times 10^{-19} \, J$$

This value is the splitting energy per ion. To convert it into J per mol, we should multiply it by Avogadro number, $6.02 \times 10^{23} \, \text{mol}^{-1}$:

$$\Delta = 3.84 \times 10^{-19} \cdot 6.02 \times 10^{23} = 231.1 \, \text{kJ} \cdot \text{mol}^{-1}$$

Answer: 231.1 kJ·mol⁻¹

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