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 Evaluating Statement I: - [Cu(NH_3)_4]^2+$[Cu(NH_3)_4]^{2+}$: Cu^2+$Cu^{2+}$ is 3d^9$3d^9$, 1 unpaired electron. - [Ni(en)_3]^2+$[Ni(en)_3]^{2+}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, in octahedral field, 2 unpaired electrons. - [Ni(NH_3)_6]^2+$[Ni(NH_3)_6]^{2+}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, 2 unpaired electrons. - [Mn(H_2O)_6]^2+$[Mn(H_2O)_6]^{2+}$: Mn^2+$Mn^{2+}$ is 3d^5$3d^5$, weak field ligand H_2O$H_2O$ leads to high spin, 5 unpaired electrons. So [Mn(H_2O)_6]^2+$[Mn(H_2O)_6]^{2+}$ has the maximum number of unpaired electrons. Statement I is true. Evaluating Statement II: - [Ni(CO)_4]$[Ni(CO)_4]$: Ni(0)$Ni(0)$ is 3d^8 4s^2$3d^8 4s^2$, strong field CO pairs electrons to 3d^10$3d^{10}$, diamagnetic (0 unpaired). - [Ni(CN)_4]^2-$[Ni(CN)_4]^{2-}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, strong field CN^-$CN^-$ forces pairing rightarrow dsp^2$\rightarrow dsp^2$ square planar, diamagnetic (0 unpaired). - [NiCl_4]^2-$[NiCl_4]^{2-}$: Ni^2+$Ni^{2+}$ is 3d^8$3d^8$, weak field Cl^-$Cl^-$ does not pair rightarrow sp^3$\rightarrow sp^3$ tetrahedral, paramagnetic (2 unpaired). The pairs containing ONLY diamagnetic species: - \[NiCl_4]^2-, [Ni(CO)_4]\$\{[NiCl_4]^{2-}, [Ni(CO)_4]\}$ rightarrow$\rightarrow$ 1 para, 1 dia (No) - \[NiCl_4]^2-, [Ni(CN)_4]^2-\$\{[NiCl_4]^{2-}, [Ni(CN)_4]^{2-}\}$ rightarrow$\rightarrow$ 1 para, 1 dia (No) - \[Ni(CO)_4], [Ni(CN)_4]^2-\$\{[Ni(CO)_4], [Ni(CN)_4]^{2-}\}$ rightarrow$\rightarrow$ Both dia (Yes) The number of such pairs is exactly ONE. Statement II says two, so it is false. ### Step 1: Final Conclusion Statement I is true, Statement II is false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### 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

### Core Logic Let's check the d$d$-electron configurations and number of unpaired electrons (n$n$) in high spin octahedral complexes: - textV^3+ rightarrow 3d^2$\text{V}^{3+} \rightarrow 3d^2$ (n = 2, mu = sqrt8 approx 2.83 text BM$n = 2, \mu = \sqrt{8} \approx 2.83 \text{ BM}$) - textTi^2+ rightarrow 3d^2$\text{Ti}^{2+} \rightarrow 3d^2$ (n = 2, mu = 2.83 text BM$n = 2, \mu = 2.83 \text{ BM}$) - textNi^2+ rightarrow 3d^8$\text{Ni}^{2+} \rightarrow 3d^8$ (n = 2, mu = 2.83 text BM$n = 2, \mu = 2.83 \text{ BM}$) - textFe^2+ rightarrow 3d^6$\text{Fe}^{2+} \rightarrow 3d^6$ (n = 4, mu = sqrt24 approx 4.9 text BM > 3 text BM$n = 4, \mu = \sqrt{24} \approx 4.9 \text{ BM} > 3 \text{ BM}$) - textCo^2+ rightarrow 3d^7$\text{Co}^{2+} \rightarrow 3d^7$ (n = 3, mu = sqrt15 approx 3.87 text BM > 3 text BM$n = 3, \mu = \sqrt{15} \approx 3.87 \text{ BM} > 3 \text{ BM}$) ### Step 1: Summing Unpaired Electrons Only textFe^2+$\text{Fe}^{2+}$ (n = 4$n = 4$) and textCo^2+$\text{Co}^{2+}$ (n = 3$n = 3$) have magnetic moments > 3.0 text BM$> 3.0 \text{ BM}$. Sum of unpaired electrons = 4 + 3 = 7$= 4 + 3 = 7$. ### Pattern Recognition Sees: high spin octahedral complexes and spin-only magnetic moment threshold checks. Trap: Misidentifying high-spin versus low-spin electron pairing in d^6$d^6$ or d^7$d^7$ configurations. ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Core Logic Let's check the d$d$-electron configurations and number of unpaired electrons (n$n$) in high spin octahedral complexes: - textV^3+ rightarrow 3d^2$\text{V}^{3+} \rightarrow 3d^2$ (n = 2, mu = sqrt8 approx 2.83 text BM$n = 2, \mu = \sqrt{8} \approx 2.83 \text{ BM}$) - textTi^2+ rightarrow 3d^2$\text{Ti}^{2+} \rightarrow 3d^2$ (n = 2, mu = 2.83 text BM$n = 2, \mu = 2.83 \text{ BM}$) - textNi^2+ rightarrow 3d^8$\text{Ni}^{2+} \rightarrow 3d^8$ (n = 2, mu = 2.83 text BM$n = 2, \mu = 2.83 \text{ BM}$) - textFe^2+ rightarrow 3d^6$\text{Fe}^{2+} \rightarrow 3d^6$ (n = 4, mu = sqrt24 approx 4.9 text BM > 3 text BM$n = 4, \mu = \sqrt{24} \approx 4.9 \text{ BM} > 3 \text{ BM}$) - textCo^2+ rightarrow 3d^7$\text{Co}^{2+} \rightarrow 3d^7$ (n = 3, mu = sqrt15 approx 3.87 text BM > 3 text BM$n = 3, \mu = \sqrt{15} \approx 3.87 \text{ BM} > 3 \text{ BM}$) ### Step 1: Summing Unpaired Electrons Only textFe^2+$\text{Fe}^{2+}$ (n = 4$n = 4$) and textCo^2+$\text{Co}^{2+}$ (n = 3$n = 3$) have magnetic moments > 3.0 text BM$> 3.0 \text{ BM}$. Sum of unpaired electrons = 4 + 3 = 7$= 4 + 3 = 7$. ### Pattern Recognition Sees: high spin octahedral complexes and spin-only magnetic moment threshold checks. Trap: Misidentifying high-spin versus low-spin electron pairing in d^6$d^6$ or d^7$d^7$ configurations. ### Chapter Mix Class 12 Chemistry: Coordination Compounds