Given below are two statements: Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism. Statement II: cis- and trans-platin are heteroleptic complexes of Pd. In the light of the above statements, choose the correct answer from the options given below:

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