### Related Formula According to Crystal Field Theory (CFT): - A complex is diamagnetic if all electrons are paired up (unpaired electrons, n=0$n=0$). - Strong field ligands (like mathrmCN^-$\mathrm{CN}^-$, mathrmNH_3$\mathrm{NH}_3$ with Co^3+$Co^{3+}$) cause pairing of electrons if Delta_o > P$\Delta_o > P$. {{SOLUTION_IMG}} ### Core Logic Analyze each complex: - **A. [mathrmMn(mathrmCN)_6]^3-$[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$**: - Mn^3+$Mn^{3+}$ has d^4$d^4$ configuration. - Strong field ligand mathrmCN^-$\mathrm{CN}^-$ causes pairing in t_2g$t_{2g}$ orbitals: t_2g^4 e_g^0$t_{2g}^4 e_g^0$. - There are 2 unpaired electrons rightarrow$\rightarrow$ *Paramagnetic*. - **B. [mathrmCo(mathrmNH_3)_6]^3+$[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$**: - Co^3+$Co^{3+}$ has d^6$d^6$ configuration. - mathrmNH_3$\mathrm{NH}_3$ acts as strong field ligand with Co^3+$Co^{3+}$, causing complete pairing: t_2g^6 e_g^0$t_{2g}^6 e_g^0$. - No unpaired electrons rightarrow$\rightarrow$ *Diamagnetic*. ### Step 1: Analyze complexes C and D - **C. [mathrmFe(mathrmCN)_6]^4-$[\mathrm{Fe}(mathrm{CN})_6]^{4-}$**: - Fe^2+$Fe^{2+}$ has d^6$d^6$ configuration. - Strong field ligand mathrmCN^-$\mathrm{CN}^-$ causes complete pairing: t_2g^6 e_g^0$t_{2g}^6 e_g^0$. - No unpaired electrons rightarrow$\rightarrow$ *Diamagnetic*. - **D. [mathrmCo(mathrmH_2mathrmO)_3mathrmF_3]$[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_3\mathrm{F}_3]$**: - Co^3+$Co^{3+}$ has d^6$d^6$ configuration. - Weak field ligands (mathrmF^-$\mathrm{F}^-$, mathrmH_2mathrmO$\mathrm{H}_2\mathrm{O}$) do not cause pairing: t_2g^4 e_g^2$t_{2g}^4 e_g^2$. - There are 4 unpaired electrons rightarrow$\rightarrow$ *Paramagnetic*. ### Step 2: Conclusion Only complexes B and C are diamagnetic, matching Option (4). ### Pattern Recognition Octahedral d^6$d^6$ ions (such as Co^3+$Co^{3+}$ or Fe^2+$Fe^{2+}$) coupled with strong-field ligands are exceptionally stable and always form low-spin, fully paired, diamagnetic complexes (t_2g^6 e_g^0$t_{2g}^6 e_g^0$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Core Logic The complex molecular composition is mathrmCo cdot 5NH_3 cdot Cl cdot SO_4$\mathrm{Co \cdot 5NH_3 \cdot Cl \cdot SO_4}$. Let's formulate the formulas for the two isomers: 1. **Isomer A**: Gives a white precipitate of mathrmAgCl$\mathrm{AgCl}$ when reacted with mathrmAgNO_3$\mathrm{AgNO_3}$. This means free chloride ions (mathrmCl^-$\mathrm{Cl}^-$) are present in the outer ionization sphere: [mathrmCo(NH_3)_5(SO_4)]mathrmCl$[\mathrm{Co(NH_3)_5(SO_4)}]\mathrm{Cl}$ 2. **Isomer B**: Gives a white precipitate of mathrmBaSO_4$\mathrm{BaSO_4}$ when reacted with mathrmBaCl_2$\mathrm{BaCl_2}$. This means free sulphate ions (mathrmSO_4^2-$\mathrm{SO}_4^{2-}$) are present in the outer ionization sphere: [mathrmCo(NH_3)_5Cl]mathrmSO_4$[\mathrm{Co(NH_3)_5Cl}]\mathrm{SO_4}$ Since these two isomers yield different ions in solution due to exchange of ligands between the coordination sphere and the ionization sphere, they exhibit **Ionisation isomerism**. ### Pattern Recognition Test for ions: - mathrmAgNO_3$\mathrm{AgNO_3}$ PPT rightarrow$\rightarrow$ free halide ion in outer sphere. - mathrmBaCl_2$\mathrm{BaCl_2}$ PPT rightarrow$\rightarrow$ free sulphate ion in outer sphere. - Outer-inner ion exchanges are always called **Ionisation isomerism**. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Core Logic Let's systematically analyze the coordination, ligand strength, hybridization, and magnetic behavior of each complex: 1. **[mathrmFeF_6]^3-$[\mathrm{FeF}_{6}]^{3-}$**: - mathrmFe^3+$\mathrm{Fe}^{3+}$ (3mathrmd^5$3\mathrm{d}^5$). mathrmF^-$\mathrm{F}^-$ is a weak-field ligand (WFL). No pairing occurs. - Outer-orbital complex: mathrmsp^3mathrmd^2$\mathrm{sp}^3\mathrm{d}^2$. - Paramagnetic (5$5$ unpaired electrons). 2. **[mathrmFe(CN)_6]^3-$[\mathrm{Fe(CN)}_{6}]^{3-}$**: - mathrmFe^3+$\mathrm{Fe}^{3+}$ (3mathrmd^5$3\mathrm{d}^5$). mathrmCN^-$\mathrm{CN}^-$ is a strong-field ligand (SFL). Pairing occurs. - Config: mathrmt_2mathrmg^5\ mathrme_mathrmg^0$\mathrm{t}_{2\mathrm{g}}^5\ \mathrm{e}_{\mathrm{g}}^0$ (one unpaired electron remains implies$\implies$ **Paramagnetic**). - Inner-orbital complex: **mathrmd^2mathrmsp^3$\mathrm{d}^2\mathrm{sp}^3$**. 3. **[mathrmMn(CN)_6]^3-$[\mathrm{Mn(CN)}_{6}]^{3-}$**: - mathrmMn^3+$\mathrm{Mn}^{3+}$ (3mathrmd^4$3\mathrm{d}^4$). mathrmCN^-$\mathrm{CN}^-$ is an SFL. Pairing occurs. - Config: mathrmt_2mathrmg^4\ mathrme_mathrmg^0$\mathrm{t}_{2\mathrm{g}}^4\ \mathrm{e}_{\mathrm{g}}^0$ (two unpaired electrons remain implies$\implies$ **Paramagnetic**). - Inner-orbital complex: **mathrmd^2mathrmsp^3$\mathrm{d}^2\mathrm{sp}^3$**. 4. **[mathrmCo(C_2mathrmO_4)_3]^3-$[\mathrm{Co(C}_2\mathrm{O}_4)_3]^{3-}$**: - mathrmCo^3+$\mathrm{Co}^{3+}$ (3mathrmd^6$3\mathrm{d}^6$). Oxalate is a chelating SFL here. Full pairing occurs. - Config: mathrmt_2mathrmg^6\ mathrme_mathrmg^0$\mathrm{t}_{2\mathrm{g}}^6\ \mathrm{e}_{\mathrm{g}}^0$ (zero unpaired electrons implies$\implies$ Diamagnetic). - Inner-orbital complex: mathrmd^2mathrmsp^3$\mathrm{d}^2\mathrm{sp}^3$. 5. **[mathrmMnCl_6]^3-$[\mathrm{MnCl}_{6}]^{3-}$**: - mathrmMn^3+$\mathrm{Mn}^{3+}$ (3mathrmd^4$3\mathrm{d}^4$). mathrmCl^-$\mathrm{Cl}^-$ is a WFL. No pairing occurs. - Outer-orbital complex: mathrmsp^3mathrmd^2$\mathrm{sp}^3\mathrm{d}^2$. - Paramagnetic (4$4$ unpaired electrons). 6. **[mathrmCoF_6]^3-$[\mathrm{CoF}_{6}]^{3-}$**: - mathrmCo^3+$\mathrm{Co}^{3+}$ (3mathrmd^6$3\mathrm{d}^6$). mathrmF^-$\mathrm{F}^-$ is a WFL. No pairing occurs. - Outer-orbital complex: mathrmsp^3mathrmd^2$\mathrm{sp}^3\mathrm{d}^2$. - Paramagnetic (4$4$ unpaired electrons). Thus, only [mathrmFe(CN)_6]^3-$[\mathrm{Fe(CN)}_{6}]^{3-}$ and [mathrmMn(CN)_6]^3-$[\mathrm{Mn(CN)}_{6}]^{3-}$ are both **paramagnetic** and involve **mathrmd^2mathrmsp^3$\mathrm{d}^2\mathrm{sp}^3$** hybridization. ### Pattern Recognition VBT shortcut: - Strong-field ligand complexes with d^4text--d^6$d^4\text{--}d^6$ central ions form inner-orbital mathrmd^2mathrmsp^3$\mathrm{d}^2\mathrm{sp}^3$ complexes. - Of those, check the number of electrons: d^6$d^6$ is completely paired (diamagnetic), but d^5$d^5$ ([Fe(CN)_6]^3-$[Fe(CN)_6]^{3-}$) and d^4$d^4$ ([Mn(CN)_6]^3-$[Mn(CN)_6]^{3-}$) both leave unpaired electrons in the t_2g$t_{2g}$ orbitals (paramagnetic). ### Evaluation Rubric / Model Answer Detailed individual classification of each complex based on VBT/CFT to yield the correct count of 2$2$ inner-orbital paramagnetic complexes. ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Core Logic Let us evaluate both statements individually: * **Statement I**: A homoleptic complex contains only one type of ligand. For an octahedral complex using monodentate ligands, the general formula is [Ma_6]$[Ma_6]$. Since all coordination positions are populated identically by the exact same ligand, swapping spatial positions produces no structural difference, hence it cannot demonstrate geometrical or optical isomerism. **Statement I is true.**

* **Statement II**: Cis-platin and trans-platin have the chemical formula [Pt(NH_3)_2Cl_2]$[Pt(NH_3)_2Cl_2]$. While they are indeed heteroleptic complexes, they are coordination coordinates of **Platinum (Pt$Pt$)**, not Palladium (Pd$Pd$). **Statement II is false.**

### Pattern Recognition Always read element symbols with immense focus in coordination chemistry. Changing a single letter from Pt to Pd creates a false assertion trap designed to test parsing alertness rather than chemical difficulty. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### Core Logic Let us systematically determine the hybridization configuration of each species: 1. **[textCo(NH_3)_6]^3+$[\text{Co(NH}_3)_6]^{3+}$**: textCo^3+$\text{Co}^{3+}$ has a 3d^6$3d^6$ configuration. Ammonia (textNH_3$\text{NH}_3$) acts as a Strong Field Ligand (SFL), forcing the pairing of 3d$3d$ electrons. This leaves two internal 3d$3d$ orbitals vacant, leading to an **inner orbital** d^2sp^3$d^2sp^3$ hybridization. 2. **textSF_6$\text{SF}_6$**: Central sulfur has 6 valence electrons, forming 6 single bonds. Steric number = 6$= 6$, resulting in a regular outer octahedral **sp^3d^2$sp^3d^2$** hybridization. 3. **[textCrF_6]^3-$[\text{CrF}_6]^{3-}$**: textCr^3+$\text{Cr}^{3+}$ has a 3d^3$3d^3$ configuration. The t_2g$t_{2g}$ subshell holds 3 unpaired electrons, leaving the two e_g$e_g$ orbitals empty regardless of ligand field strength. This results in a d^2sp^3$d^2sp^3$ hybridization. 4. **[textCoF_6]^3-$[\text{CoF}_6]^{3-}$**: textCo^3+$\text{Co}^{3+}$ has a 3d^6$3d^6$ configuration. Fluoride (F^-$F^-$) is a Weak Field Ligand (WFL) and cannot induce spin pairing. Thus, the complex utilizes outer shell 4d$4d$ orbitals, yielding an **outer orbital** **sp^3d^2$sp^3d^2$** configuration. 5. **[textMn(CN)_6]^3-$[\text{Mn(CN)}_6]^{3-}$**: textMn^3+$\text{Mn}^{3+}$ has a 3d^4$3d^4$ configuration. Cyanide (textCN^-$\text{CN}^-$) is a Strong Field Ligand (SFL), inducing pairing to leave two 3d$3d$ slots vacant, giving a d^2sp^3$d^2sp^3$ hybridization. 6. **[textMnCl_6]^3-$[\text{MnCl}_6]^{3-}$**: textMn^3+$\text{Mn}^{3+}$ has a 3d^4$3d^4$ configuration. Chloride (textCl^-$\text{Cl}^-$) is a Weak Field Ligand (WFL) and cannot cause pairing. It utilizes the outer 4d$4d$ shell, resulting in an **outer orbital** **sp^3d^2$sp^3d^2$** hybridization. Counting the outer-orbital sp^3d^2$sp^3d^2$ species: textSF_6$\text{SF}_6$, [textCoF_6]^3-$[\text{CoF}_6]^{3-}$, and [textMnCl_6]^3-$[\text{MnCl}_6]^{3-}$. Total count = 3$= 3$. ### Pattern Recognition Outer orbital complexes (sp^3d^2$sp^3d^2$) require weak field ligands (like F^-, Cl^-$F^-, Cl^-$) paired with metal configurations where internal d$d$-orbitals cannot be cleared by pairing (d^4, d^5, d^6$d^4, d^5, d^6$). SF_6$SF_6$ is a primary group molecule that always uses outer-shell d-orbitals. This brings our total to 3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Chemical Bonding and Molecular Structure