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 $\left( N _{ A }\right)=6 \times 10^{23} / mol$

Answer 1

(Ti+3(H2​O)6​)3+
Ti+3:3d1
C.F.S.E. =−0.4×Δ0​
$=−\frac{​96×103}{N_0}​J$
$Δ_0​=\frac{96×10^3​ }{0.4×6×10^{23}}$
$⇒\frac{hc}{λ}​=\frac{96×10^3​ }{0.4×6×10^{23}}$
$λ=\frac{0.4×6×10^{23}×6.4×10^{−34}×3×10^8​ }{96×10^3}$
$=0.48×10^{−6}m$
$=480×10^{−9}m$
=480nm
So , the correct answer is 480.