Consider the following low-spin complexes K₃[Co(NO₂)₆]$K_{3}[Co(NO_{2})_{6}]$, K₄[Fe(CN)₆]$K_{4}[Fe(CN)_{6}]$, K₃[Fe(CN)₆]$K_{3}[Fe(CN)_{6}]$, Cu₂[Fe(CN)₆]$Cu_{2}[Fe(CN)_{6}]$ and Zn₂[Fe(CN)₆]$Zn_{2}[Fe(CN)_{6}]$. The sum of the spin-only magnetic moment values of complexes having yellow colour is ________ B.M. (answer is nearest integer)

Core Logic

The complex molecular composition is Co · 5NH₃ · Cl · SO₄$\mathrm{Co \cdot 5NH_3 \cdot Cl \cdot SO_4}$. Let's formulate the formulas for the two isomers:

• Isomer A: Gives a white precipitate of AgCl$\mathrm{AgCl}$ when reacted with AgNO₃$\mathrm{AgNO_3}$. This means free chloride ions (Cl^-$\mathrm{Cl}^-$) are present in the outer ionization sphere:

• Isomer B: Gives a white precipitate of BaSO₄$\mathrm{BaSO_4}$ when reacted with BaCl₂$\mathrm{BaCl_2}$. This means free sulphate ions (SO₄²⁻$\mathrm{SO}_4^{2-}$) are present in the outer ionization sphere:

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:

• AgNO₃$\mathrm{AgNO_3}$ PPT arrow$\rightarrow$ free halide ion in outer sphere.
• BaCl₂$\mathrm{BaCl_2}$ PPT arrow$\rightarrow$ free sulphate ion in outer sphere.
• Outer-inner ion exchanges are always called Ionisation isomerism.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Core Logic

Let's systematically analyze the coordination, ligand strength, hybridization, and magnetic behavior of each complex:

• [FeF₆]³⁻$[\mathrm{FeF}_{6}]^{3-}$:
• Fe³⁺$\mathrm{Fe}^{3+}$ (3d⁵$3\mathrm{d}^5$). F^-$\mathrm{F}^-$ is a weak-field ligand (WFL). No pairing occurs.
• Outer-orbital complex: sp³d²$\mathrm{sp}^3\mathrm{d}^2$.
• Paramagnetic (5$5$ unpaired electrons).
• [Fe(CN)₆]³⁻$[\mathrm{Fe(CN)}_{6}]^{3-}$:
• Fe³⁺$\mathrm{Fe}^{3+}$ (3d⁵$3\mathrm{d}^5$). CN^-$\mathrm{CN}^-$ is a strong-field ligand (SFL). Pairing occurs.
• Config: t2g⁵ eg⁰$\mathrm{t}_{2\mathrm{g}}^5\ \mathrm{e}_{\mathrm{g}}^0$ (one unpaired electron remains $\implies$ Paramagnetic).
• Inner-orbital complex: d²sp³$\mathrm{d}^2\mathrm{sp}^3$.
• [Mn(CN)₆]³⁻$[\mathrm{Mn(CN)}_{6}]^{3-}$:
• Mn³⁺$\mathrm{Mn}^{3+}$ (3d⁴$3\mathrm{d}^4$). CN^-$\mathrm{CN}^-$ is an SFL. Pairing occurs.
• Config: t2g⁴ eg⁰$\mathrm{t}_{2\mathrm{g}}^4\ \mathrm{e}_{\mathrm{g}}^0$ (two unpaired electrons remain $\implies$ Paramagnetic).
• Inner-orbital complex: d²sp³$\mathrm{d}^2\mathrm{sp}^3$.
• [Co(C₂O₄)₃]³⁻$[\mathrm{Co(C}_2\mathrm{O}_4)_3]^{3-}$:
• Co³⁺$\mathrm{Co}^{3+}$ (3d⁶$3\mathrm{d}^6$). Oxalate is a chelating SFL here. Full pairing occurs.
• Config: t2g⁶ eg⁰$\mathrm{t}_{2\mathrm{g}}^6\ \mathrm{e}_{\mathrm{g}}^0$ (zero unpaired electrons $\implies$ Diamagnetic).
• Inner-orbital complex: d²sp³$\mathrm{d}^2\mathrm{sp}^3$.
• [MnCl₆]³⁻$[\mathrm{MnCl}_{6}]^{3-}$:
• Mn³⁺$\mathrm{Mn}^{3+}$ (3d⁴$3\mathrm{d}^4$). Cl^-$\mathrm{Cl}^-$ is a WFL. No pairing occurs.
• Outer-orbital complex: sp³d²$\mathrm{sp}^3\mathrm{d}^2$.
• Paramagnetic (4$4$ unpaired electrons).
• [CoF₆]³⁻$[\mathrm{CoF}_{6}]^{3-}$:
• Co³⁺$\mathrm{Co}^{3+}$ (3d⁶$3\mathrm{d}^6$). F^-$\mathrm{F}^-$ is a WFL. No pairing occurs.
• Outer-orbital complex: sp³d²$\mathrm{sp}^3\mathrm{d}^2$.
• Paramagnetic (4$4$ unpaired electrons).

Thus, only [Fe(CN)₆]³⁻$[\mathrm{Fe(CN)}_{6}]^{3-}$ and [Mn(CN)₆]³⁻$[\mathrm{Mn(CN)}_{6}]^{3-}$ are both paramagnetic and involve d²sp³$\mathrm{d}^2\mathrm{sp}^3$ hybridization.

Pattern Recognition

VBT shortcut:

• Strong-field ligand complexes with d⁴--d⁶$d^4\text{--}d^6$ central ions form inner-orbital d²sp³$\mathrm{d}^2\mathrm{sp}^3$ complexes.
• Of those, check the number of electrons: d⁶$d^6$ is completely paired (diamagnetic), but d⁵$d^5$ ([Fe(CN)₆]³⁻$[Fe(CN)_6]^{3-}$) and d⁴$d^4$ ([Mn(CN)₆]³⁻$[Mn(CN)_6]^{3-}$) both leave unpaired electrons in the t2g$t_{2g}$ 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 [Ma₆]$[Ma_6]$. 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(NH₃)₂Cl₂]$[Pt(NH_3)_2Cl_2]$. 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.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Core Logic

Let us systematically determine the hybridization configuration of each species:

• [Co(NH₃)₆]³⁺$[\text{Co(NH}_3)_6]^{3+}$: Co³⁺$\text{Co}^{3+}$ has a 3d⁶$3d^6$ configuration. Ammonia (NH₃$\text{NH}_3$) 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 d²sp³$d^2sp^3$ hybridization.
• SF₆$\text{SF}_6$: Central sulfur has 6 valence electrons, forming 6 single bonds. Steric number = 6$= 6$, resulting in a regular outer octahedral sp³d²$sp^3d^2$ hybridization.
• [CrF₆]³⁻$[\text{CrF}_6]^{3-}$: Cr³⁺$\text{Cr}^{3+}$ has a 3d³$3d^3$ configuration. The t2g$t_{2g}$ subshell holds 3 unpaired electrons, leaving the two eg$e_g$ orbitals empty regardless of ligand field strength. This results in a d²sp³$d^2sp^3$ hybridization.
• [CoF₆]³⁻$[\text{CoF}_6]^{3-}$: Co³⁺$\text{Co}^{3+}$ has a 3d⁶$3d^6$ 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 sp³d²$sp^3d^2$ configuration.
• [Mn(CN)₆]³⁻$[\text{Mn(CN)}_6]^{3-}$: Mn³⁺$\text{Mn}^{3+}$ has a 3d⁴$3d^4$ configuration. Cyanide (CN^-$\text{CN}^-$) is a Strong Field Ligand (SFL), inducing pairing to leave two 3d$3d$ slots vacant, giving a d²sp³$d^2sp^3$ hybridization.
• [MnCl₆]³⁻$[\text{MnCl}_6]^{3-}$: Mn³⁺$\text{Mn}^{3+}$ has a 3d⁴$3d^4$ configuration. Chloride (Cl^-$\text{Cl}^-$) is a Weak Field Ligand (WFL) and cannot cause pairing. It utilizes the outer 4d$4d$ shell, resulting in an outer orbital sp³d²$sp^3d^2$ hybridization.

Counting the outer-orbital sp³d²$sp^3d^2$ species: SF₆$\text{SF}_6$, [CoF₆]³⁻$[\text{CoF}_6]^{3-}$, and [MnCl₆]³⁻$[\text{MnCl}_6]^{3-}$. Total count = 3$= 3$.

Pattern Recognition

Outer orbital complexes (sp³d²$sp^3d^2$) require weak field ligands (like F^-, Cl^-$F^-, Cl^-$) paired with metal configurations where internal d$d$-orbitals cannot be cleared by pairing (d⁴, d⁵, d⁶$d^4, d^5, d^6$). SF₆$SF_6$ is a primary group molecule that always uses outer-shell d-orbitals. This brings our total to 3.

Chapter Mix

Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

Let us apply Valence Bond Theory (VBT) and crystal field rules to evaluate each coordination complex:

• A. [Ni(CO)₄]$[\text{Ni(CO)}_4]$: Nickel is in the 0$0$ oxidation state (3d⁸ 4s²$3d^8 4s^2$). Carbon monoxide (CO$\text{CO}$) is a strong field ligand, forcing the 4s$4s$ electrons into the 3d$3d$ shell to produce a fully paired 3d¹⁰$3d^{10}$ configuration. The vacant 4s$4s$ and three 4p$4p$ orbitals hybridize into an sp³$sp^3$ tetrahedral geometry. All spins are paired, so μ = 0 BM$\mu = 0 \text{ BM}$. Thus, A arrow III$\text{A} \rightarrow \text{III}$.

  

• B. [Ni(CN)₄]²⁻$[\text{Ni(CN)}_4]^{2-}$: Nickel is in the +2$+2$ state (3d⁸$3d^8$). Cyanide (CN^-$\text{CN}^-$) is a strong field ligand, forcing the pairing of the two unpaired 3d$3d$ electrons. This leaves one internal 3d$3d$ orbital vacant, leading to dsp²$dsp^2$ square planar hybridization with zero unpaired electrons (μ = 0 BM$\mu = 0 \text{ BM}$). Thus, B arrow II$\text{B} \rightarrow \text{II}$.

  

• C. [NiCl₄]²⁻$[\text{NiCl}_4]^{2-}$: Nickel is in the +2$+2$ state (3d⁸$3d^8$). Chloride (Cl^-$\text{Cl}^-$) is a weak field ligand, leaving the two 3d$3d$ electrons unpaired (n = 2$n = 2$). The system adopts sp³$sp^3$ tetrahedral hybridization with a spin-only moment of μ = √(2(2+2)) = √(8) ≈ 2.8 BM$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.8 \text{ BM}$. Thus, C arrow I$\text{C} \rightarrow \text{I}$.

  

• D. [MnBr₄]²⁻$[\text{MnBr}_4]^{2-}$: Manganese is in the +2$+2$ state (3d⁵$3d^5$). Bromide (Br^-$\text{Br}^-$) is a weak field ligand, preserving five unpaired parallel spins (n = 5$n = 5$). The geometry is sp³$sp^3$ tetrahedral with a maximum spin-only moment of μ = √(5(5+2)) = √(35) ≈ 5.9 BM$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \text{ BM}$. Thus, D arrow IV$\text{D} \rightarrow \text{IV}$.

  

Step 1: Alignment Summary

Consolidating our results:

This matches Option (3).

Pattern Recognition

Nickel complexes provide classic benchmarks: Nickel zero tetracarbonyl is always tetrahedral diamagnetic. Nickel +2$+2$ tetracyanide is square planar diamagnetic due to strong ligand field pairing. Spotting these properties cuts down the problem solving time significantly.

Chapter Mix

Class 12 Chemistry: Coordination Compounds