Question

The magnitude of crystal field stabilization energy (CFSE) of octahedral \( \text{[Ti(H}_2\text{O)}_6]^{3+} \) complex is 7680 cm\(^{-1}\). The wavelength at the maximum absorption (\( \lambda_{\text{max}} \)) of this complex is ............ nm (rounded up to the nearest integer).
 

Hint:

Use the relationship \( E = \frac{hc}{\lambda} \) to convert between the energy in cm\(^{-1}\) and the wavelength of maximum absorption.

Answer 1

Correct Answer: 520 - 521

For octahedral \([\mathrm{Ti(H_2O)6}]^{3+}), Ti(^{3+}\) is (d^1). The crystal field stabilization energy (CFSE) for a (d^1) ion in an octahedral field is
\[\mathrm{CFSE} = -0.4,\Delta{\text{oct}}.\]
Given the magnitude \(|\mathrm{CFSE}| = 7680\ \mathrm{cm^{-1}}\), we have
\[0.4,\Delta_{\text{oct}} = 7680\ \mathrm{cm^{-1}}\]
so
\[\Delta_{\text{oct}}=\frac{7680}{0.4}.\]
Compute exactly:
\[\frac{7680}{0.4}=\frac{7680}{4/10}=7680\times\frac{10}{4}=\frac{76800}{4}=19200\ \mathrm{cm^{-1}}.\]

Wavenumber \(\tilde{\nu}=\Delta_{\text{oct}}=19200\ \mathrm{cm^{-1}}\). The corresponding wavelength is
\[\lambda=\frac{1}{\tilde{\nu}}\ \text{(in cm)}=\frac{1}{19200}\ \text{cm}.\]
Convert to nm: \(1\ \text{cm}=10^{7}\ \text{nm}\), so
\[\lambda(\text{nm})=\frac{10^{7}}{19200}=\frac{100000}{192}=520.833\ldots\ \text{nm}.\]
Rounded to the nearest integer,
\[\boxed{\lambda_{\max}=521\ \text{nm}.}\]