Match the coordination complexes listed in LIST-I with their geometric shape and magnetic moment characteristics described in LIST-II:

LIST-I (Complex/Species)	LIST-II (Shape & magnetic moment)
A. [textNi(CO)_4]$[\text{Ni(CO)}_4]$	I. Tetrahedral, 2.8 BM
B. [textNi(CN)_4]^2-$[\text{Ni(CN)}_4]^{2-}$	II. Square planar, 0 BM
C. [textNiCl_4]^2-$[\text{NiCl}_4]^{2-}$	III. Tetrahedral, 0 BM
D. [textMnBr_4]^2-$[\text{MnBr}_4]^{2-}$	IV. Tetrahedral, 5.9 BM

Choose the correct answer from the options given below:

### Related Formula Crystal field splitting values relation for matching configuration choices: Delta_mathrmt = frac49Delta_0$$\Delta_{\mathrm{t}} = \frac{4}{9}\Delta_0$$ ### Core Logic Let's verify both rules based on Crystal Field Theory principles: * **Statement I**: In octahedral configurations, if pairing penalty energy P$P$ exceeds field split magnitude Delta_0$\Delta_0$, electrons prefer moving to upper sub-shells, creating high-spin states. Conversely, if Delta_0 > P$\Delta_0 > P$, forced pairing occurs, creating low-spin complexes. (Statement I is accurate). * **Statement II**: Because tetrahedral configurations separate by an extremely narrow gap magnitude Delta_mathrmt$\Delta_{\mathrm{t}}$ (about half of octahedral field splits), the value almost never exceeds standard pairing energy P$P$. Electrons consistently choose higher sub-levels rather than pairing up, meaning low-spin arrangements are extremely rare. (Statement II is accurate). ### Step 1: Verdict Therefore, both Statement I and Statement II are correct. ### Pattern Recognition Tetrahedral configurations are systematically assumed to be high-spin unless special structural properties dictate otherwise, due to the Delta_mathrmt < P$\Delta_{\mathrm{t}} < P$ constraint. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

### 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