If the CFSE of [Ti(H_2O)_6]^{3+} is -96.0 kJ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For $[Ti(H_2O)_6]^{3+}$,the central metal ion is $Ti^{3+}$,which has a $3d^1$ electronic configuration.The $CFSE$ for a $d^1$ octahedral complex i

Vedclass · Accepted Answer

$480$

Vedclass · Answer

(C) For $[Ti(H_2O)_6]^{3+}$,the central metal ion is $Ti^{3+}$,which has a $3d^1$ electronic configuration.The $CFSE$ for a $d^1$ octahedral complex is given by $CFSE = -0.4 	imes \Delta_0$.Given $CFSE = -96.0 \ kJ / mol = -96000 \ J / mol$.Therefore,$\Delta_0 = \frac{96000}{0.4} = 240000 \ J / mol$.To find the wavelength $(\lambda)$ in $nm$,we use the relation $\Delta_0 = \frac{N_A 	imes h 	imes c}{\lambda}$.$\lambda = \frac{N_A 	imes h 	imes c}{\Delta_0} = \frac{6 	imes 10^{23} 	imes 6.4 	imes 10^{-34} 	imes 3.0 	imes 10^8}{240000}$.$\lambda = \frac{115.2 	imes 10^{-3}}{240000} = 0.48 	imes 10^{-6} \ m = 480 \ nm$.

If the $CFSE$ of $[Ti(H_2O)_6]^{3+}$ 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 $(N_A) = 6 \times 10^{23} / mol$.

Similar Questions

Which of the following is an incorrectly matched complex?

Why do transition elements form coloured compounds?

The transition metal complex with the highest value of crystal field splitting $(\Delta_{0})$ is ........ .

Which among the following ligands has the highest field strength?

In which of the following octahedral complex species the magnitude of $\Delta_0$ will be maximum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If the $CFSE$ of $[Ti(H_2O)_6]^{3+}$ 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 $(N_A) = 6 \times 10^{23} / mol$.