Calculate the CFSE for [Mn(H₂O)₆]⁴⁺ in terms of Δ_o?

Question

Quizplus · Accepted Answer

To calculate the CFSE for a coordination compound, we first need to determine the oxidation state of the metal ion in the complex. The overall charge of the complex is +1, and we know that the sulfate ligand (₆) has a charge of -2. The oxidation state of manganese (Mn) can be calculated as follows: (+1) + (2 x (-2)) + (-1) = -4 So Mn must have an oxidation state of +4 to balance out the charge of the complex. Next, we need to determine the number of d electrons in the Mn(IV) ion. Since Mn is in its +4 oxidation state, it has lost four electrons compared to its neutral state. The electron configuration of neutral Mn is [Ar] 3d5 4s2, so Mn(IV) has the configuration [Ar] 3d3. There are three d electrons in the Mn(IV) ion. Now we can calculate the CFSE using the formula: CFSE = -0.4 x (number of d electrons)^2 CFSE = -0.4 x (3)^2 = -1.2Δ_o Therefore, the best choice is (C).

Calculate the CFSE for [Mn(H2O)6]4+ in Terms of Δo

Calculate the CFSE for [Mn(H₂O)₆]⁴⁺ in Terms of Δ_o