If the CFSE of [Ti(H₂ O)₆]³⁺ is -96.0 kJ / mol, this complex will absorb maximum at wavelength ___________ nm. (nearest integer)
Assume Planck's constant (h) =6.4 × 10⁻³⁴ Js, Speed of light (c)=3.0 × 10⁸ m / s and Avogadro's
Constant (N_A)=6 × 10²³ / mol

Solution

Chemistry ma’am · 3 steps

CFSE is not the gap itself. For a d¹ ion, first divide by -0.4 to get Δₒ.

1. Find Δo. Ti³⁺ is d¹, and its one electron sits in a t_(2g) orbital. So CFSE = -0.4 Δₒ, which gives Δₒ = 96.0/0.4 = 240 kJ mol⁻¹.
2. Energy per ion. Each ion absorbs one photon to lift its electron from t_(2g) to e_g. So divide by N_A: E = (240 × 10³)/(6 × 10²³) = 4 × 10⁻¹⁹ J.

   λ = hc/E = (6.4 × 10⁻³⁴ × 3.0 × 10⁸)/(4 × 10⁻¹⁹) = 4.8 × 10⁻⁷ m

3. So. λ = 4.8 × 10⁻⁷ m, which is 480 nm.

The common slip: 1200. Taking 96.0 kJ mol⁻¹ as Δₒ itself gives 1200 nm. The CFSE of a d¹ ion is only -0.4 Δₒ.

The idea to keep. The absorbed photon matches Δₒ, not the CFSE. Convert CFSE to Δₒ using the electron configuration, then divide by N_A to get energy per ion.

Answer from the NTA final answer key. Solution by PrepWiser.