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' constant ( h ) = 6 . 4 × 10 - 34 Js , Speed of light ( c ) = 3 . 0 × 10 8 m / s and Avogadro's constant N A = 6 × 10 23 / mol .

Step-by-step solution

In Ti H 2 O 6 3 + , oxidation state of Titanium is +3. Thus, electronic configuration: Ti + 3 : 3 d 1 ⇒ t 2 g 1 , 0 , 0 , e g 0 , 0 We know CFSE = - 0 . 4 n t 2 g + 0 . 6 n e g Δo ( Δo   =   Octahedral   splitting   energy ) Δo = 96 × 10 3 0 . 4 × 6 × 10 23 = 4 × 10 - 19   J Using planck's equation and fact that octahedral splitting energy per complex is: ⇒ Δo = hc λ = 4 × 10 - 19   J ⇒ λ = 6 . 4 × 10 - 3