Question

$[\mathrm{Ti}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{3+}$ absorbs light of wavelength 498 nm during a d–d transition. The octahedral splitting energy for the above complex is ___________ $\times {10}^{–19}\mathrm{J}$. (Round off to the nearest integer). $\mathrm{h}=6.626\times {10}^{–34}\mathrm{Js};\mathrm{c}=3\times {10}^8{\mathrm{ms}}^{–1}.$

Answer 1

Step 1: Given data

Wavelength $\mathrm{\lambda }=498\mathrm{nm}$

Speed of light $\mathrm{c}=3\times {10}^8{\mathrm{ms}}^{-1}$

Planks constant $\mathrm{h}=6.626\times {10}^{–34}\mathrm{Js}$

Step 2: Calculating the energy

We know that the octahedral splitting energy is:

$$\begin{gathered} {∆}_0\mathrm{or}\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ \Rightarrow \mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ \Rightarrow \mathrm{E}=\frac{6.626\times {10}^{-34}\mathrm{Js}\times 3\times {10}^8{\mathrm{ms}}^{-1}}{498\times {10}^{-9}\mathrm{m}} \\ \Rightarrow \mathrm{E}=0.0399\times {10}^{-17}\mathrm{J} \\ \Rightarrow \mathrm{E}=3.99\times {10}^{-19}\mathrm{J} \\ \Rightarrow \mathrm{E}\approx 4\times {10}^{-19}\mathrm{J} \end{gathered}$$

Therefore, $[\mathrm{Ti}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{3+}$ absorbs light of wavelength 498 nm during a d–d transition.

The octahedral splitting energy for the above complex is $4\times {10}^{–19}\mathrm{J}$.