Given below are two statements: Statement I: Crystal Field Stabilization Energy (CFSE) of [textCr(textH_2textO)_6]^2+$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ is greater than that of [textMn(textH_2textO)_6]^2+$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$. Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium ferrocyanide. In the light of the above statements, choose the correct answer from the options given below:

### Core Logic - Statement I: [textMn(textH_2textO)_6]^2+$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$ has d^5$d^5$ configuration with weak field ligands (CFSE = 0), whereas [textCr(textH_2textO)_6]^2+$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ has d^4$d^4$ configuration (CFSE = -0.6Delta_0$= -0.6\Delta_0$). Thus CFSE of chromium complex is greater in magnitude. - Statement II: Potassium ferricyanide textK_3[textFe(textCN)_6]$\text{K}_3[\text{Fe}(\text{CN})_6]$ has Fe^3+$Fe^{3+}$ (d^5$d^5$, 1 unpaired electron, mu = sqrt3$\mu = \sqrt{3}$ B.M.), while sodium ferrocyanide textNa_4[textFe(textCN)_6]$\text{Na}_4[\text{Fe}(\text{CN})_6]$ has textFe^2+$\text{Fe}^{2+}$ (d^6$d^6$, 0 unpaired electrons, mu = 0$\mu = 0$). Thus statement II is true. ### Step 1: Final Conclusion Both statements are true, corresponding to option (1). ### Pattern Recognition Sees: CFSE calculations and magnetic moment comparisons for coordination complexes. Trap: Miscalculating d$d$-electron count or ligand field strength. ### Chapter Mix Class 12 Chemistry: Coordination Compounds