Question

If the CFSE of ${\left[Ti{\left(H_{2}O\right)}_6\right]}^{3+}$ is $-96.0kJ/mol$, this complex will absorb maximum at wavelength ___$nm$. (nearest integer)
Assume Planck's constant $(h)=6.4\times 10^{-34}Js$, Speed of light $(c)=3.0\times 10^{8}m/s$ and Avogadro's Constant $\left(N_A\right)=6\times 10^{23}/mol$.

Answer 1

Answer: 480

${\left(T{i}^{+3}{\left(H_{2}O\right)}_6\right)}^{3+}$
$T{i}^{+3}:3d^1$
$\text{ C.F.S.E. }=-0.4\times {\Delta }_0$
$=-\frac{96\times 10^3}{N_0}J$
${\Delta }_0=\frac{96\times 10^3}{0.4\times 6\times 10^{23}}$
$\Rightarrow \frac{hc}{\lambda }=\frac{96\times 10^3}{0.4\times 6\times 10^{23}}$
$\lambda =\frac{0.4\times 6\times 10^{23}\times 6.4\times 10^{-34}\times 3\times 10^8}{96\times 10^3}$
$=0.48\times 10^{-6}m$
$=480\times 10^{-9}m$
$=480nm$