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Step-by-Step Solution
Step 1: Understand the Given Information
We are given the Crystal Field Stabilization Energy (CFSE) of the octahedral complex
[\mathrm{Ti}(\mathrm{H}_2O)_6]^{3+} as -96.0\,\mathrm{kJ\,mol}^{-1} .
In an octahedral field for a d^1 configuration, CFSE = -0.4\,\Delta_0 ,
where \Delta_0 is the octahedral splitting energy.
The complex absorbs light of wavelength corresponding to \Delta_0 .
Step 2: Relate CFSE to \Delta_0
Use the expression
\mathrm{CFSE} = -0.4\,\Delta_0.
Given
\mathrm{CFSE} = -96.0\,\mathrm{kJ\,mol}^{-1},
we have
-96.0\,\mathrm{kJ\,mol}^{-1} = -0.4\,\Delta_0.
Solving for \Delta_0 :
\Delta_0 = \frac{-96.0\,\mathrm{kJ\,mol}^{-1}}{-0.4} = 240\,\mathrm{kJ\,mol}^{-1}.
Step 3: Convert \Delta_0 to Energy per Photon
1 kJ = 10^3 J, so
\Delta_0 = 240 \times 10^3\,\mathrm{J\,mol}^{-1}.
To find energy per photon, divide by Avogadro's number
N_A = 6 \times 10^{23}\,\mathrm{mol}^{-1} :
E_{\text{photon}}
= \frac{240 \times 10^3\,\mathrm{J\,mol}^{-1}}{6 \times 10^{23}\,\mathrm{mol}^{-1}}
= 4.0 \times 10^{-19}\,\mathrm{J\,photon}^{-1}.
Step 4: Calculate the Wavelength
Use the relation
E_{\text{photon}} = \frac{hc}{\lambda},
where
h = 6.4 \times 10^{-34}\,\mathrm{J\,s} (Planck's constant) and
c = 3.0 \times 10^8\,\mathrm{m\,s}^{-1} (speed of light). Solving for \lambda :
\lambda
= \frac{hc}{E_{\text{photon}}}
= \frac{(6.4 \times 10^{-34}\,\mathrm{J\,s}) \times (3.0 \times 10^8\,\mathrm{m\,s}^{-1})}{4.0 \times 10^{-19}\,\mathrm{J}}
= 4.8 \times 10^{-7}\,\mathrm{m}
= 480\,\mathrm{nm}.
Step 5: State the Final Answer
Thus, the wavelength of maximum absorption is approximately
480 nm.
