Coordination Compounds Question 17
Question 17 - 31 January - Shift 2
If the CFSE of $[Ti(H_2 O)_6]^{3+}$ is $-96.0 kJ / mol$, 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 $(N_A)=6 \times 10^{23} / mol$.
Show Answer
Answer: (480)
Solution:
Formula: CFSE
$ \begin{aligned} & (Ti^{+3}(H_2 O)_6)^{3+} \\ & Ti^{+3}: 3 d^{\prime} \\ & \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 \quad \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 \\ & =480 nm \end{aligned} $
