Coordination Compounds Part 4

Concept:Crystal Field Stabilization Energy (CFSE) depends on the $d$-electron configuration and the ligand field strength. Spin-only magnetic moment depends on the number of unpaired electrons.Explanation:Statement I: In $[Cr(H_2O)_6]^{2+}$, $Cr^{2+}$ is $d^4$ and $H_2O$ is a weak field ligand.This gives a high-spin configuration: $t_{2g}^3 e_g^1$.CFSE = $3(-0.4\Delta_o) + 1(+0.6\Delta_o) = -0.6\Delta_o$.In $[Mn(H_2O)_6]^{2+}$, $Mn^{2+}$ is $d^5$ and high-spin: $t_{2g}^3 e_g^2$.CFSE = $3(-0.4\Delta_o) + 2(+0.6\Delta_o) = 0$.Since $-0.6\Delta_o$ is more stabilizing than $0$, the CFSE of the chromium complex is greater. Statement I is true.Statement II: Potassium ferricyanide $K_3[Fe(CN)_6]$ contains $Fe^{3+}$ ($d^5$).$CN^-$ is a strong field ligand, so low-spin: $t_{2g}^5$, with 1 unpaired electron.Spin-only magnetic moment $\mu = \sqrt{1(1+2)} = \sqrt{3}$ BM.Sodium ferrocyanide $Na_4[Fe(CN)_6]$ contains $Fe^{2+}$ ($d^6$).Strong field gives low-spin $t_{2g}^6$, 0 unpaired electrons, $\mu = 0$.Thus ferricyanide has a greater magnetic moment. Statement II is true.Answer:Both statements are true. The correct option is B.