Match List-I with List-II.
List-I (Complex)List-II (Hybridisation of central metal ion)
(A) [CoF_6]^3-(I) d^2sp^3
(B) [NiCl_4]^2-(II) sp^3
(C) [Co(NH_3)_6]^3+(III) sp^3d^2
(D) [Ni(CN)_4]^2-(IV) dsp^2
Choose the correct answer from the options given below :

Solution & Explanation

### Related Formula Coordination Number 6 corresponds to either d^2sp^3 or sp^3d^2 configuration templates. Coordination Number 4 corresponds to either sp^3 or dsp^2 configuration templates. ### Core Logic Analyzing metal orbital dynamics under varying ligand fields: - **(A) [CoF_6]^3-**: Co^3+ (3d^6) with a weak field ligand (F^-) rightarrow no pairing occurs rightarrow utilizes outer orbitals rightarrow sp^3d^2. - **(B) [NiCl_4]^2-**: Ni^2+ (3d^8) with a weak field ligand (Cl^-) rightarrow no pairing occurs rightarrow tetrahedral profile rightarrow sp^3. - **(C) [Co(NH_3)_6]^3+**: Co^3+ (3d^6) with a strong field ligand (NH_3) rightarrow electrons pair up rightarrow inner orbital configuration rightarrow d^2sp^3. - **(D) [Ni(CN)_4]^2-**: Ni^2+ (3d^8) with a strong field ligand (CN^-) rightarrow forced pairing opens a 3d slot rightarrow square planar geometry rightarrow dsp^2. ### Step 1: Final Pairing Match The completed matching configuration aligns cleanly with: (A)-(III), (B)-(II), (C)-(I), (D)-(IV). ### Pattern Recognition Isolate coordination frameworks quickly: - Nickel(II) with weak field ligands (Cl^-) yields sp^3, while with strong field ligands (CN^-) it yields dsp^2. - Cobalt(III) with weak field ligands (F^-) yields sp^3d^2, while with strong field ligands (NH_3) it yields d^2sp^3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Reference Study Guides

Coordination Compounds

More Coordination Compounds Previous-Year Questions — Page 6

Q40 jee_main_2025_07_april_evening Magnetic Properties and Crystal Field Theory
The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively: [textFe(CN)6]^3-, [textFeF6]^3-, [textCoF6]^3-, [textMn(CN)6]^3-
  • A. 1, 5, 4, 2
  • B. 1, 5, 5, 2
  • C. 1, 1, 4, 2
  • D. 1, 4, 4, 2

Solution

### Related Formula textStrong Field Ligand (SFL) ightarrow textCauses electron pairing in t2g text orbitals textWeak Field Ligand (WFL) ightarrow textHigh-spin state (Follows Hund's rule directly across CFT split) ### Core Logic Analyzing each coordination sphere step-by-step under Crystal Field Theory (CFT): - [textFe(CN)_6]^3-: textFe^3+ (3d^5). textCN^- is a Strong Field Ligand (SFL) implies pairing happens. Configuration is t2g^5 e_g^0 (paired as t2g^2,2,1). Unpaired electrons = 1. [cite: 958, 959] - [textFeF6]^3-: textFe^3+ (3d^5). textF^- is a Weak Field Ligand (WFL) implies no pairing. Configuration is t2g^3 e_g^2. Unpaired electrons = 5. - [textCoF_6]^3-: textCo^3+ (3d^6). textF^- is a Weak Field Ligand (WFL) implies no pairing. Configuration is t2g^4 e_g^2 (paired down to t2g^2,1,1 e_g^1,1). Unpaired electrons = 4. - [textMn(CN)6]^3-: textMn^3+ (3d^4). textCN^- is a Strong Field Ligand (SFL) implies pairing happens. Configuration is t2g^4 e_g^0 (arranged as t2g^2,1,1). Unpaired electrons = 2. ### Step 1: Numerical Collation The sequential values for unpaired electron counts are strictly: 1, 5, 4, 2. ### Pattern Recognition Ligand field shortcut: textCN^- is a strong field ligand that forces pairing, minimizing the spin state. textF^- is a weak field ligand that retains maximum spin values. Tracking textFe^3+ under strong field (3d^5 ightarrow 1) versus weak field (3d^5 ightarrow 5) instantly clarifies the solution sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q49 jee_main_2025_07_april_evening Magnetic Properties and Crystal Field Theory
The number of paramagnetic metal complex species among [textCo(textNH_3)_6]^3+, [textCo(textC_2textO_4)_3]^3-, [textMnCl_6]^3-, [textMn(textCN)_6]^3-, [textCoF_6]^3-, [textFe(textCN)_6]^3- and [textFeF_6]^3- with same number of unpaired electrons is dots.
Numerical Answer. Answer: 1.5 to 2.5

Solution

### Related Formula textParamagnetic species: Complexes with unpaired electron count (n) > 0 ### Core Logic Let's perform electron tracking across every entry using CFT parameters: 1. [textCo(textNH_3)_6]^3+: textCo^3+ (3d^6), textNH_3 is SFL implies t2g^6 e_g^0, unpaired electrons = 0 (Diamagnetic). 2. [textCo(textC_2textO_4)_3]^3-: textCo^3+ (3d^6), Oxalate acts as SFL here implies t_2g^6 e_g^0, unpaired electrons = 0 (Diamagnetic). 3. [textMnCl_6]^3-: textMn^3+ (3d^4), textCl^- is WFL implies t_2g^3 e_g^1, unpaired electrons = 4. 4. [textMn(textCN)_6]^3-: textMn^3+ (3d^4), textCN^- is SFL implies t_2g^4 e_g^0, unpaired electrons = 2. 5. [textCoF_6]^3-: textCo^3+ (3d^6), textF^- is WFL implies t_2g^4 e_g^2, unpaired electrons = 4. 6. [textFe(textCN)_6]^3-: textFe^3+ (3d^5), textCN^- is SFL implies t_2g^5 e_g^0, unpaired electrons = 1. 7. [textFeF_6]^3-: textFe^3+ (3d^5), textF^- is WFL implies t_2g^3 e_g^2, unpaired electrons = 5. ### Step 1: Finding Common Electronic Counts Reviewing unpaired counts among paramagnetic entities: - n=1: 1 complex ([textFe(textCN)_6]^3-) - n=2: 1 complex ([textMn(textCN)_6]^3-) - n=4: 2 complexes ([textMnCl_6]^3- and [textCoF_6]^3-) - n=5: 1 complex ([textFeF_6]^3-) The highest matching sub-group frequency has a count of 2. ### Pattern Recognition CFT Shortcut tracking: For 3d^4 weak field and 3d^6 weak field systems, the unpaired counts identically match (n=4). Spotting that textMn^3+text/WFL and textCo^3+text/WFL both leave 4 electrons unpaired immediately provides the pair answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q33 jee_main_2025_24_jan_evening Qualitative Analysis of Cations
Find the compound 'A' from the following reaction sequences. mathrmA xrightarrowtextaqua-regia mathrmB xrightarrowtext(1) mathrmKNO2 | mathrmNH4mathrmOH, text (2) mathrmAcOH textyellow ppt
  • A. \text{ZnS}
  • B. \text{CoS}
  • C. \text{MnS}
  • D. \text{NiS}

Solution

### Core Logic This pathway corresponds to the standard confirmatory test for cobalt (mathrmCo^2+) ions in qualitative inorganic analysis: 1. mathrmCoS dissolves in aqua regia to yield cobalt chloride (mathrmCoCl_2): mathrmCoS + textaqua regia ightarrow mathrmCoCl_2 2. Treating this solution with potassium nitrite (mathrmKNO_2) in the presence of acetic acid (mathrmAcOH) oxidizes mathrmCo^2+ to mathrmCo^3+, precipitating potassium cobaltinitrite as a characteristic yellow solid: mathrmCoCl_2 + 7mathrmKNO_2 + 2mathrmCH_3mathrmCOOH ightarrow mathrmK_3[mathrmCo(mathrmNO_2)_6]downarrow (textyellow) + 2mathrmNaCl + mathrmNO + 2mathrmCH_3mathrmCOOK + mathrmH_2mathrmO ### Pattern Recognition A yellow precipitate formed specifically upon adding mathrmKNO_2 and acetic acid is a definitive signature of potassium cobaltinitrite, mathrmK_3[mathrmCo(mathrmNO_2)_6]. This confirms the starting sulfide was mathrmCoS. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: Qualitative Analysis
Q37 jee_main_2025_24_jan_evening Spectrochemical Series and Colour
When Ethane-1, 2-diamine is added progressively to an aqueous solution of Nickel (II) chloride, the sequence of colour change observed will be:
  • A. \text{Pale Blue } ightarrow \text{ Blue } ightarrow \text{ Green } ightarrow \text{ Violet}
  • B. \text{Pale Blue } ightarrow \text{ Blue } ightarrow \text{ Violet } ightarrow \text{ Green}
  • C. \text{Green } ightarrow \text{ Pale Blue } ightarrow \text{ Blue } ightarrow \text{ Violet}
  • D. \text{Violet } ightarrow \text{ Blue } ightarrow \text{ Pale Blue } ightarrow \text{ Green}

Solution

### Core Logic An aqueous nickel (II) chloride solution contains the green hexaquarickel(II) complex, [mathrmNi(mathrmH_2mathrmO)_6]^2+. Ethane-1,2-diamine ('en') is a bidentate ligand that binds more strongly than water, shifting the crystal field splitting parameter (Delta_o) to higher energies as it replaces water molecules: 1. Initial state: [mathrmNi(mathrmH_2mathrmO)_6]^2+text (Green) 2. Adding 1 equivalent of 'en' forms a mono-en complex: [mathrmNi(mathrmH_2mathrmO)_6]^2+ + mathrmen ightarrow [mathrmNi(mathrmH_2mathrmO)_4(mathrmen)]^2+text (Pale Blue) + 2mathrmH_2mathrmO 3. Adding a 2nd equivalent forms a bis-en complex: [mathrmNi(mathrmH_2mathrmO)_4(mathrmen)]^2+ + mathrmen ightarrow [mathrmNi(mathrmH_2mathrmO)_2(mathrmen)_2]^2+text (Blue / Purple) + 2mathrmH_2mathrmO 4. Adding a 3rd equivalent forms the tris-en complex: [mathrmNi(mathrmH_2mathrmO)_2(mathrmen)_2]^2+ + mathrmen ightarrow [mathrmNi(mathrmen)_3]^2+text (Violet) + 2mathrmH_2mathrmO This progressive ligand replacement shifts the absorption spectrum, changing the solution's visible color from Green ightarrow Pale Blue ightarrow Blue ightarrow Violet. ### Pattern Recognition Replacing weak-field ligands (like mathrmH_2mathrmO) with stronger bidentate chelating ligands (like 'en') increases crystal field splitting. For mathrmNi^2+, this ligand substitution always follows the specific chromatic progression: Green ightarrow Pale Blue ightarrow Blue ightarrow Violet. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q38 jee_main_2025_24_jan_evening Crystal Field Theory
The conditions and consequence that favours the t_2g^3 e_g^1 configuration in a metal complex are:
  • A. \text{weak field ligand, high spin complex}
  • B. \text{strong field ligand, high spin complex}
  • C. \text{strong field ligand, low spin complex}
  • D. \text{weak field ligand, low spin complex}

Solution

### Core Logic Consider an octahedral coordination environment for a d^4 transition metal ion configuration: * Weak Field Ligand (WFL): The crystal field splitting energy is smaller than the pairing energy (Delta_o < P). Consequently, electrons prefer to occupy the higher-energy e_g orbitals rather than pair up in the lower-energy t_2g orbitals. This leads to a high spin complex with the configuration: t_2g^3 e_g^1 * Strong Field Ligand (SFL): The splitting energy is larger than the pairing energy (Delta_o > P). Electrons pair up in the t_2g orbitals before occupying the e_g subshell, resulting in a low spin complex with the configuration: t_2g^4 e_g^0 ### Pattern Recognition An electron occupying an e_g orbital before the t_2g orbitals are fully paired requires a weak-field ligand. This configuration maximizes the number of unpaired electrons, which is the defining characteristic of a high-spin complex. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

More Coordination Compounds Questions — jee_main_2025_28_jan_evening

Practice all Coordination Compounds previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)