Question

The electronic spectrum of ${\left[\mathrm{Ti}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+3}$ shows a single broad peak with a maximum at $20,300{\mathrm{cm}}^{-1}$. The crystal field stabilization energy (CFSE) of the complex ion, in $\mathrm{kJ}\mathrm{mol}-1$, is:

• A. $83·7$
• B. $242·5$
• C. $145·5$
• D. $97$

Answer 1

The correct option is D

$97$

Step 1: Given data

Wavenumber $\left(\overline{v}\right)$=$20,300c{m}^{-1}$

Step 2: Formula used

The relation between energy and frequency is as follows:

$\mathrm{Change}\mathrm{in}\mathrm{energy}\left({∆}_{\circ }\right)=\mathrm{planck}\mathrm{constant}\left(\mathrm{h}\right)\times \mathrm{frequency}\left(\mathrm{v}\right)$

We know that $frequency\left(v\right)=\frac{speedoflight\left(c\right)}{wavelength\left(\lambda \right)}$. So,

${∆}_{\circ }=h\times \frac{c}{\lambda }$

It is understood that the relation between wavelength and wavenumber is, $\lambda =\frac{1}{wavenumber\left(\overline{v}\right)}$.

${∆}_{\circ }=hc\overline{v}·····\left(1\right)$

Step 3: Compute the crystal field stabilization energy

Substitute the known values in the equation $\left(1\right)$ to get the energy of one titanium ion,

$\begin{array}{rcl}{∆}_{\circ }& =& \left(6.6\times {10}^{-34}\right)\left(3.0\times {10}^{10}cm/s\right)\times 20300\\ & =& 4.019\times {10}^{-19}J\end{array}$

So, the energy of one-mole titanium is,

$$\begin{gathered} {∆}_{\circ }=6.02\times {10}^{23}\times 4.019\times {10}^{-19} \\ {∆}_{\circ }\simeq 243kJ/mol \end{gathered}$$

So, the crystal field stabilization energy is,

$\begin{array}{rcl}CFSE& =& 0.4{∆}_{\circ }\\ & =& 0.4\times 243\\ & =& 97kJ/mol\end{array}$ (Because titanium $\mathrm{Ti}\left[\mathrm{III}\right]$ contains one unpaired electron)

Thus the crystal field stabilization energy (CFSE) of the complex ion will be $97$$kJmo{l}^{-1}$.

Hence, option D is the correct answer.