Question

If the CFSE of $\left[ Ti \left( H _2 O \right)_6\right]^{3+}$ is $-960 kJ / mol$, this complex will absorb maximum at wavelength ___$nm$ (nearest integer) Assume Planck's constant $( h )=64 \times 10^{-34} Js$, Speed of light $( c )=30 \times 10^8 m / s$ and Avogadro's Constant $\left( N _{ A }\right)=6 \times 10^{23} / mol$

Hint:

The relationship between the energy of absorbed light and the wavelength is given by \( \frac{hc}{\lambda} = \Delta_0 \), where \( \Delta_0 \) is the crystal field splitting energy.

Answer 1

Correct Answer: 480

(Ti⁺³(H₂​O)_6​)³⁺ 
Ti⁺³:3d¹ 
C.F.S.E. =−0.4×Δ_0​ 
\(=−\frac{​96×103}{N_0}​J \)
\(Δ_0​=\frac{96×10^3​ }{0.4×6×10^{23}}\)
\(⇒\frac{hc}{λ}​=\frac{96×10^3​ }{0.4×6×10^{23}}\)
\(λ=\frac{0.4×6×10^{23}×6.4×10^{−34}×3×10^8​ }{96×10^3}\)
\(=0.48×10^{−6}m \)
\(=480×10^{−9}m \)
=480nm
So , the correct answer is 480.

Answer 2

For the complex \( [ \text{Ti}^{3+} (\text{H}_2\text{O})_6 ]^{3+} \), the electronic configuration is \( \text{Ti}^{3+}: 3d^1 \). The CFSE (Crystal Field Stabilization Energy) is given as -96.0 kJ/mol.

The formula for the CFSE is:

\[ \text{CFSE} = -0.4 \Delta_0 \]

Where \( \Delta_0 \) is the crystal field splitting energy. Using the given data:

\[ \text{CFSE} = -96 \times 10^3 \, \text{J/mol} \]

Now, solving for \( \Delta_0 \):

\[ \Delta_0 = \frac{96 \times 10^3}{6 \times 10^{23}} \quad \Rightarrow \quad \Delta_0 = 1.6 \times 10^{-19} \, \text{J} \]

Now, using the formula for the wavelength of absorption:

\[ \frac{hc}{\lambda} = \Delta_0 \]

Substitute the known values:

\[ \frac{6.4 \times 10^{-34} \times 3.0 \times 10^8}{\lambda} = 1.6 \times 10^{-19} \]

Solving for \( \lambda \):

\[ \lambda = \frac{6.4 \times 10^{-34} \times 3.0 \times 10^8}{1.6 \times 10^{-19}} = 480 \times 10^{-9} \, \text{m} \]

Thus, the wavelength is \( 480 \, \text{nm} \).