### Related Formula textParamagnetic species: Complexes with unpaired electron count (n) > 0$\text{Paramagnetic species: Complexes with unpaired electron count } (n) > 0$ ### Core Logic Let's perform electron tracking across every entry using CFT parameters: 1. [textCo(textNH_3)_6]^3+$[\text{Co}(\text{NH}_3)_6]^{3+}$: textCo^3+$\text{Co}^{3+}$ (3d^6$3d^6$), textNH_3$\text{NH}_3$ is SFL implies t2g^6 e_g^0$\implies t{2g}^6 e_g^0$, unpaired electrons = 0$= 0$ (Diamagnetic). 2. [textCo(textC_2textO_4)_3]^3-$[\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$: textCo^3+$\text{Co}^{3+}$ (3d^6$3d^6$), Oxalate acts as SFL here implies t_2g^6 e_g^0$\implies t_{2g}^6 e_g^0$, unpaired electrons = 0$= 0$ (Diamagnetic). 3. [textMnCl_6]^3-$[\text{MnCl}_6]^{3-}$: textMn^3+$\text{Mn}^{3+}$ (3d^4$3d^4$), textCl^-$\text{Cl}^-$ is WFL implies t_2g^3 e_g^1$\implies t_{2g}^3 e_g^1$, unpaired electrons = 4$= 4$. 4. [textMn(textCN)_6]^3-$[\text{Mn}(\text{CN})_6]^{3-}$: textMn^3+$\text{Mn}^{3+}$ (3d^4$3d^4$), textCN^-$\text{CN}^-$ is SFL implies t_2g^4 e_g^0$\implies t_{2g}^4 e_g^0$, unpaired electrons = 2$= 2$. 5. [textCoF_6]^3-$[\text{CoF}_6]^{3-}$: textCo^3+$\text{Co}^{3+}$ (3d^6$3d^6$), textF^-$\text{F}^-$ is WFL implies t_2g^4 e_g^2$\implies t_{2g}^4 e_g^2$, unpaired electrons = 4$= 4$. 6. [textFe(textCN)_6]^3-$[\text{Fe}(\text{CN})_6]^{3-}$: textFe^3+$\text{Fe}^{3+}$ (3d^5$3d^5$), textCN^-$\text{CN}^-$ is SFL implies t_2g^5 e_g^0$\implies t_{2g}^5 e_g^0$, unpaired electrons = 1$= 1$. 7. [textFeF_6]^3-$[\text{FeF}_6]^{3-}$: textFe^3+$\text{Fe}^{3+}$ (3d^5$3d^5$), textF^-$\text{F}^-$ is WFL implies t_2g^3 e_g^2$\implies t_{2g}^3 e_g^2$, unpaired electrons = 5$= 5$. ### Step 1: Finding Common Electronic Counts Reviewing unpaired counts among paramagnetic entities: - n=1$n=1$: 1 complex ([textFe(textCN)_6]^3-$[\text{Fe}(\text{CN})_6]^{3-}$) - n=2$n=2$: 1 complex ([textMn(textCN)_6]^3-$[\text{Mn}(\text{CN})_6]^{3-}$) - n=4$n=4$: 2 complexes ([textMnCl_6]^3-$[\text{MnCl}_6]^{3-}$ and [textCoF_6]^3-$[\text{CoF}_6]^{3-}$) - n=5$n=5$: 1 complex ([textFeF_6]^3-$[\text{FeF}_6]^{3-}$) The highest matching sub-group frequency has a count of 2. ### Pattern Recognition CFT Shortcut tracking: For 3d^4$3d^4$ weak field and 3d^6$3d^6$ weak field systems, the unpaired counts identically match (n=4$n=4$). Spotting that textMn^3+text/WFL$\text{Mn}^{3+}\text{/WFL}$ and textCo^3+text/WFL$\text{Co}^{3+}\text{/WFL}$ both leave 4 electrons unpaired immediately provides the pair answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Core Logic This pathway corresponds to the standard confirmatory test for cobalt (mathrmCo^2+$\mathrm{Co}^{2+}$) ions in qualitative inorganic analysis: 1. mathrmCoS$\mathrm{CoS}$ dissolves in aqua regia to yield cobalt chloride (mathrmCoCl_2$\mathrm{CoCl}_2$): mathrmCoS + textaqua regia ightarrow mathrmCoCl_2$\mathrm{CoS} + \text{aqua regia} ightarrow \mathrm{CoCl}_{2}$ 2. Treating this solution with potassium nitrite (mathrmKNO_2$\mathrm{KNO}_{2}$) in the presence of acetic acid (mathrmAcOH$\mathrm{AcOH}$) oxidizes mathrmCo^2+$\mathrm{Co}^{2+}$ to mathrmCo^3+$\mathrm{Co}^{3+}$, precipitating potassium cobaltinitrite as a characteristic yellow solid: mathrmCoCl_2 + 7mathrmKNO_2 + 2mathrmCH_3mathrmCOOH ightarrow mathrmK_3[mathrmCo(mathrmNO_2)_6]downarrow (textyellow) + 2mathrmNaCl + mathrmNO + 2mathrmCH_3mathrmCOOK + mathrmH_2mathrmO$\mathrm{CoCl}_{2} + 7\mathrm{KNO}_{2} + 2\mathrm{CH}_{3}\mathrm{COOH} ightarrow \mathrm{K}_{3}[\mathrm{Co}(\mathrm{NO}_{2})_{6}]\downarrow (\text{yellow}) + 2\mathrm{NaCl} + \mathrm{NO} + 2\mathrm{CH}_{3}\mathrm{COOK} + \mathrm{H}_{2}\mathrm{O}$ ### Pattern Recognition A yellow precipitate formed specifically upon adding mathrmKNO_2$\mathrm{KNO}_2$ and acetic acid is a definitive signature of potassium cobaltinitrite, mathrmK_3[mathrmCo(mathrmNO_2)_6]$\mathrm{K}_3[\mathrm{Co}(\mathrm{NO}_2)_6]$. This confirms the starting sulfide was mathrmCoS$\mathrm{CoS}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Qualitative Analysis

### Core Logic An aqueous nickel (II) chloride solution contains the green hexaquarickel(II) complex, [mathrmNi(mathrmH_2mathrmO)_6]^2+$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$. Ethane-1,2-diamine ('en') is a bidentate ligand that binds more strongly than water, shifting the crystal field splitting parameter (Delta_o$\Delta_o$) to higher energies as it replaces water molecules: 1. Initial state: [mathrmNi(mathrmH_2mathrmO)_6]^2+text (Green)$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}\text{ (Green)}$ 2. Adding 1 equivalent of 'en' forms a mono-en complex: [mathrmNi(mathrmH_2mathrmO)_6]^2+ + mathrmen ightarrow [mathrmNi(mathrmH_2mathrmO)_4(mathrmen)]^2+text (Pale Blue) + 2mathrmH_2mathrmO$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+} + \mathrm{en} ightarrow [\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{4}(\mathrm{en})]^{2+}\text{ (Pale Blue)} + 2\mathrm{H}_{2}\mathrm{O}$ 3. Adding a 2nd equivalent forms a bis-en complex: [mathrmNi(mathrmH_2mathrmO)_4(mathrmen)]^2+ + mathrmen ightarrow [mathrmNi(mathrmH_2mathrmO)_2(mathrmen)_2]^2+text (Blue / Purple) + 2mathrmH_2mathrmO$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{4}(\mathrm{en})]^{2+} + \mathrm{en} ightarrow [\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{2}(\mathrm{en})_{2}]^{2+}\text{ (Blue / Purple)} + 2\mathrm{H}_{2}\mathrm{O}$ 4. Adding a 3rd equivalent forms the tris-en complex: [mathrmNi(mathrmH_2mathrmO)_2(mathrmen)_2]^2+ + mathrmen ightarrow [mathrmNi(mathrmen)_3]^2+text (Violet) + 2mathrmH_2mathrmO$[\mathrm{Ni}(\mathrm{H}_{2}\mathrm{O})_{2}(\mathrm{en})_{2}]^{2+} + \mathrm{en} ightarrow [\mathrm{Ni}(\mathrm{en})_{3}]^{2+}\text{ (Violet)} + 2\mathrm{H}_{2}\mathrm{O}$ This progressive ligand replacement shifts the absorption spectrum, changing the solution's visible color from Green ightarrow$ ightarrow$ Pale Blue ightarrow$ ightarrow$ Blue ightarrow$ ightarrow$ Violet. ### Pattern Recognition Replacing weak-field ligands (like mathrmH_2mathrmO$\mathrm{H}_2\mathrm{O}$) with stronger bidentate chelating ligands (like 'en') increases crystal field splitting. For mathrmNi^2+$\mathrm{Ni}^{2+}$, this ligand substitution always follows the specific chromatic progression: Green ightarrow$ ightarrow$ Pale Blue ightarrow$ ightarrow$ Blue ightarrow$ ightarrow$ Violet. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Core Logic Consider an octahedral coordination environment for a d^4$d^4$ transition metal ion configuration: * Weak Field Ligand (WFL): The crystal field splitting energy is smaller than the pairing energy (Delta_o < P$\Delta_o < P$). Consequently, electrons prefer to occupy the higher-energy e_g$e_g$ orbitals rather than pair up in the lower-energy t_2g$t_{2g}$ orbitals. This leads to a high spin complex with the configuration: t_2g^3 e_g^1$t_{2g}^{3} e_{g}^{1}$ * Strong Field Ligand (SFL): The splitting energy is larger than the pairing energy (Delta_o > P$\Delta_o > P$). Electrons pair up in the t_2g$t_{2g}$ orbitals before occupying the e_g$e_g$ subshell, resulting in a low spin complex with the configuration: t_2g^4 e_g^0$t_{2g}^{4} e_{g}^{0}$ ### Pattern Recognition An electron occupying an e_g$e_g$ orbital before the t_2g$t_{2g}$ orbitals are fully paired requires a weak-field ligand. This configuration maximizes the number of unpaired electrons, which is the defining characteristic of a high-spin complex. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Related Formula textMoles of AgCl text precipitated = textMoles of ionizable Cl^- text ions outside the coordination sphere$\text{Moles of } AgCl \text{ precipitated} = \text{Moles of ionizable } Cl^{-} \text{ ions outside the coordination sphere}$ ### Core Logic Since 1 mole of the complex yields 2 moles of AgCl_(s)$AgCl_{(s)}$, there must be exactly 2 chloride ions outside the coordination sphere to undergo precipitation: [Co(NH_3)_5Cl]Cl_2 rightarrow [Co(NH_3)_5Cl]^2+(aq) + 2Cl^-(aq)$[Co(NH_{3})_{5}Cl]Cl_{2} \rightarrow [Co(NH_{3})_{5}Cl]^{2+}(aq) + 2Cl^{-}(aq)$ This dissociation produces a total of 3 moles of ions per mole of the complex, perfectly consistent with the problem constraints. ### Pattern Recognition Number of precipitated AgCl$AgCl$ moles directly equates to the count of counter-anions located outside the square brackets. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds