The calculated spin-only magnetic moments of K₃[Fe(OH)₆]$K_{3}[Fe(OH)_{6}]$ and K₄[Fe(OH)₆]$K_{4}[Fe(OH)_{6}]$ respectively are: (1) 4.90 and 4.90 B.M. (2) 5.92 and 4.90 B.M. (3) 3.87 and 4.90 B.M. (4) 4.90 and 5.92 B.M.

Core Logic

Evaluate the ligand field strength from the spectrochemical series. Halide ligands are weak field ligands, with the order of their strength being I^- < Br^- < Cl^- < F^-$I^- < Br^- < Cl^- < F^-$.

Step 1: Statement I Evaluation

Since Cl^-$Cl^-$ is a stronger ligand than Br^-$Br^-$, the crystal field splitting energy (Δₜ$\Delta_t$) for [CoCl₄]²⁻$[CoCl_4]^{2-}$ is greater than that of [CoBr₄]²⁻$[CoBr_4]^{2-}$. Energy absorbed (E$E$) is directly proportional to Δₜ$\Delta_t$. Therefore, [CoBr₄]²⁻$[CoBr_4]^{2-}$ will absorb lower energy than [CoCl₄]²⁻$[CoCl_4]^{2-}$. Statement I is True.

Step 2: Statement II Evaluation

Comparing [CoI₄]²⁻$[CoI_4]^{2-}$ and [CoCl₄]²⁻$[CoCl_4]^{2-}$: I^-$I^-$ is a weaker ligand than Cl^-$Cl^-$. Therefore, the energy separation (Δₜ$\Delta_t$) in [CoI₄]²⁻$[CoI_4]^{2-}$ will be less than in [CoCl₄]²⁻$[CoCl_4]^{2-}$. Statement II states it is more, which is False.

Pattern Recognition

Spectrochemical series memorization shortcut for halides: I Brought Some Cloth (I- < Br- < S2- < Cl-).

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Core Logic

Analyze the structural and electronic properties of the [Ni(dmg)₂]$[Ni(dmg)_2]$ complex.

Step 1: Evaluating Each Statement

(A) It is a rosy red precipitate. (True) (B) It forms a precipitate in a basic medium (like ammonium hydroxide), indicating it is insoluble in water at pH=9$pH=9$. (False) (C) The Ni²⁺$Ni^{2+}$ ion is 3d⁸$3d^8$. DMG is a strong field ligand in a square planar geometry, causing electron pairing (dsp²$dsp^2$ hybridization). Number of unpaired electrons = 0. (False) (D) Square planar geometry ensures N-Ni-N bond angles are close to 90^°$90^\circ$. (True) (E) The complex contains two 5-membered rings and two 6-membered rings formed by hydrogen bonding. (False)

Step 2: Identifying Incorrect Statements

Statements B, C, and E are incorrect.

Pattern Recognition

Ni-DMG$Ni-DMG$ complex = Rosy red ppt, Square Planar (dsp²$dsp^2$), Diamagnetic (n=0$n=0$), Hydrogen bonded (extra stability).

Chapter Mix

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

Related Formula

(Note: Pairing energy is neglected with respect to Δ₀$\Delta_0$ based on assumption)

Core Logic

Since Δ₀ gg P$\Delta_0 \gg P$ (Strong field logic / low spin complexes), electrons will pair up in the lower energy t2g$t_{2g}$ orbitals before occupying the higher energy eg$e_g$ orbitals. Identify the oxidation state and d$d$-electron count for the central metal in both complexes.

Step 1: Cobalt Complex Analysis

Complex: [Co(ox)₃]³⁻$[Co(\text{ox})_{3}]^{3-}$ Cobalt oxidation state = +3$+3$. Electronic configuration of Co³⁺$Co^{3+}$: [Ar] 3d⁶$[Ar] 3d^6$. Under Δ₀ gg P$\Delta_0 \gg P$, the d⁶$d^6$ configuration is t2g2,2,2 eg0,0$t_{2g}^{2,2,2} e_{g}^{0,0}$ (i.e., t2g⁶ eg⁰$t_{2g}^6 e_g^0$). CFSECo³⁺ = 6 × (-0.4 Δ₀) = -2.4 Δ₀$\text{CFSE}_{Co^{3+}} = 6 \times (-0.4 \Delta_0) = -2.4 \Delta_0$

Step 2: Chromium Complex Analysis

Complex: [Cr(ox)₃]³⁻$[Cr(\text{ox})_{3}]^{3-}$ Chromium oxidation state = +3$+3$. Electronic configuration of Cr³⁺$Cr^{3+}$: [Ar] 3d³$[Ar] 3d^3$. Configuration is t2g1,1,1 eg0,0$t_{2g}^{1,1,1} e_{g}^{0,0}$ (i.e., t2g³ eg⁰$t_{2g}^3 e_g^0$). CFSECr³⁺ = 3 × (-0.4 Δ₀) = -1.2 Δ₀$\text{CFSE}_{Cr^{3+}} = 3 \times (-0.4 \Delta_0) = -1.2 \Delta_0$

Step 3: Calculating n

Ratio n = CFSE of Co³⁺CFSE of Cr³⁺$n = \frac{\text{CFSE of } Co^{3+}}{\text{CFSE of } Cr^{3+}}$ n = (|-2.4 Δ₀|)/(|-1.2 Δ₀|) = 2$n = \frac{|-2.4 \Delta_0|}{|-1.2 \Delta_0|} = 2$

Pattern Recognition

d⁶$d^6$ low-spin always gives max CFSE for octahedral (-2.4 Δ₀$\Delta_0$). d³$d^3$ is strictly half of that (-1.2 Δ₀$\Delta_0$). Ratio is always 2.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Core Logic

Evaluate each complex individually using Valence Bond Theory:

(A) [Co(NH₃)₆]³⁺$[Co(NH_{3})_{6}]^{3+}$: Central metal is Co³⁺$Co^{3+}$ (3d⁶$3d^{6}$). NH₃$NH_3$ acts as a strong field ligand (SFL) for Co³⁺$Co^{3+}$, causing pairing. This leads to d²sp³$d^2sp^3$ hybridization, forming an inner orbital complex. (Correct)

(B) [MnCl₆]³⁻$[MnCl_{6}]^{3-}$: Central metal is Mn³⁺$Mn^{3+}$ (3d⁴$3d^{4}$). Cl⁻$Cl^{-}$ is a weak field ligand (WFL), so no pairing occurs. It utilizes outer 4d$4d$ orbitals for hybridization (sp³d²$sp^3d^2$), forming an outer orbital complex. (Correct)

(C) [CoF₆]³⁻$[CoF_{6}]^{3-}$: Central metal is Co³⁺$Co^{3+}$ (3d⁶$3d^{6}$). F⁻$F^{-}$ is a weak field ligand (WFL), causing no pairing. It undergoes sp³d²$sp^3d^2$ hybridization (outer orbital complex). The statement says it is d²sp³$d^2sp^3$ hybridized, which is incorrect. (Incorrect)

(D) [FeF₆]³⁻$[FeF_{6}]^{3-}$: Central metal is Fe³⁺$Fe^{3+}$ (3d⁵$3d^{5}$). F⁻$F^{-}$ is a weak field ligand (WFL), leading to no pairing. It undergoes sp³d²$sp^3d^2$ hybridization, making it an outer orbital complex. (Correct)

(E) [Ni(CN)₄]²⁻$[Ni(CN)_{4}]^{2-}$: Central metal is Ni²⁺$Ni^{2+}$ (3d⁸$3d^{8}$). CN⁻$CN^{-}$ is a strong field ligand (SFL), causing pairing. It undergoes dsp²$dsp^2$ hybridization (inner orbital complex/square planar), not sp³$sp^3$. (Incorrect)

Step 1: Final Conclusion

Only statements A, B, and D are correct.

Pattern Recognition

Spectrochemical series dictates SFL vs WFL. Co³⁺$Co^{3+}$ with NH₃$NH_3$ is a classic exception to memorize: NH₃$NH_3$ behaves as a SFL with Co³⁺$Co^{3+}$ (pairing occurs), whereas it acts as a WFL with many +2$+2$ ions.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Core Logic

Let's determine the electronic configuration, oxidation state, and ligand nature for each complex:

• [Ni(CO)₄]$[Ni(CO)_4]$:

Oxidation state of Ni = 0. Configuration: 3d⁸ 4s²$3d^8 4s^2$. CO is a strong field ligand. The 4s$4s$ electrons are pushed into the 3d$3d$ orbital, making it a 3d¹⁰$3d^{10}$ configuration. It is sp³$sp^3$ hybridized and diamagnetic. Unpaired electrons = 0.

• [NiCl₄]²⁻$[NiCl_4]^{2-}$:

Oxidation state of Ni = +2$+2$. Configuration: 3d⁸$3d^8$. Cl^-$Cl^-$ is a weak field ligand, so no pairing of the d⁸$d^8$ electrons occurs. The configuration is t2g⁶ eg²$t_{2g}^6 e_g^2$. It is sp³$sp^3$ hybridized. Unpaired electrons = 2.

• [PtCl₂(NH₃)₂]$[PtCl_2(NH_3)_2]$:

Oxidation state of Pt = +2$+2$. Configuration: 5d⁸$5d^8$. For 4d and 5d series metals, nearly all ligands act as strong field ligands. This causes pairing, leading to a dsp²$dsp^2$ hybridized square planar geometry. Unpaired electrons = 0.

• [Ni(CN)₄]²⁻$[Ni(CN)_4]^{2-}$:

Oxidation state of Ni = +2$+2$. Configuration: 3d⁸$3d^8$. CN^-$CN^-$ is a strong field ligand, forcing electron pairing. It becomes dsp²$dsp^2$ hybridized. Unpaired electrons = 0.

• [Pt(CN)₄]²⁻$[Pt(CN)_4]^{2-}$:

Oxidation state of Pt = +2$+2$. Configuration: 5d⁸$5d^8$. As established, 5d metals strictly form low spin complexes. CN^-$CN^-$ causes pairing (dsp²$dsp^2$). Unpaired electrons = 0.

Step 1: Total Sum

Summing all unpaired electrons across all listed complexes: 0 + 2 + 0 + 0 + 0 = 2$0 + 2 + 0 + 0 + 0 = 2$.

Pattern Recognition

Metals from 4d and 5d series (like Pd, Pt) ALWAYS form inner-orbital/low-spin complexes regardless of the ligand strength. So, d⁸$d^8$ configurations in Pd²⁺$Pd^{2+}$ and Pt²⁺$Pt^{2+}$ will always pair up to yield zero unpaired electrons.

Chapter Mix

Class 12 Chemistry: Coordination Compounds