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Coordination Compounds appeared 68 times across 3 years — 7.9% of Chemistry. This question is from Isomerism in Coordination Compounds.

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Questions 19 34 15 68

A metal complex with a formula MC ₄·3NH₃ is involved in sp³ d² hybridisation. It upon reaction with excess of AgNO₃ solution gives 'x' moles of AgCl. Consider 'x' is equal to the number of lone pairs of electron present in central atom of BrF₅ . Then the number of geometrical isomers exhibited by the complex is

Numerical Answer Type:
Enter a numerical value Answer: 1.9 to 2.1 +4 marks

Solution & Explanation

Core Logic
  • Determine the value of x:
  • The central Bromine atom in BrF₅ has 7 valence electrons. It forms 5 single bonds with fluorine, leaving 2 remaining electrons.
  • Therefore, the number of lone pairs on Br in BrF₅ is exactly 1 x = 1.
  • Formulate the coordination sphere formula:
  • Since x = 1, the complex yields 1 mole of AgCl precipitate upon reaction with excess AgNO₃, meaning exactly 1 chloride ion sits outside the coordination sphere as an counter-ion.
  • Rearranging the formula components around an octahedral coordination number of 6 gives the complex configuration:
[M(NH₃)₃Cl₃]Cl
Step 1: Isomer Analysis

Facial and meridional isomers representation for Q48
Facial and meridional isomers representation for Q48

An octahedral complex of the type [Ma₃b₃] exhibits exactly 2 geometrical isomers:

  • Facial (fac) isomer
  • Meridional (mer) isomer
Pattern Recognition

For [Ma₃b₃] octahedral coordination types, don't waste time looking for optical active configurations. It splits cleanly into exactly two classical geometric forms: facial (all three identical ligands adjacent on a face) and meridional (ligands trace a meridian plane).

Chapter Mix

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

Reference Study Guides

More Coordination Compounds Previous-Year Questions — Page 10

Q33 jee_main_2025_24_jan_evening Qualitative Analysis of Cations
Find the compound 'A' from the following reaction sequences. A aqua-regia B (1) KNO2 | NH4OH, (2) AcOH yellow 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 (Co²⁺) ions in qualitative inorganic analysis:

  • CoS dissolves in aqua regia to yield cobalt chloride (CoCl₂):
CoS + aqua regia arrow CoCl₂
  • Treating this solution with potassium nitrite (KNO₂) in the presence of acetic acid (AcOH) oxidizes Co²⁺ to Co³⁺, precipitating potassium cobaltinitrite as a characteristic yellow solid:
CoCl₂ + 7KNO₂ + 2CH₃COOH arrow K₃[Co(NO₂)₆] (yellow) + 2NaCl + NO + 2CH₃COOK + H₂O
Pattern Recognition

A yellow precipitate formed specifically upon adding KNO₂ and acetic acid is a definitive signature of potassium cobaltinitrite, K₃[Co(NO₂)₆]. This confirms the starting sulfide was CoS.

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 } \rightarrow \text{ Blue } \rightarrow \text{ Green } \rightarrow \text{ Violet}
  • B. \text{Pale Blue } \rightarrow \text{ Blue } \rightarrow \text{ Violet } \rightarrow \text{ Green}
  • C. \text{Green } \rightarrow \text{ Pale Blue } \rightarrow \text{ Blue } \rightarrow \text{ Violet}
  • D. \text{Violet } \rightarrow \text{ Blue } \rightarrow \text{ Pale Blue } \rightarrow \text{ Green}

Solution

Core Logic

An aqueous nickel (II) chloride solution contains the green hexaquarickel(II) complex, [Ni(H₂O)₆]²⁺. Ethane-1,2-diamine ('en') is a bidentate ligand that binds more strongly than water, shifting the crystal field splitting parameter (Δₒ) to higher energies as it replaces water molecules:

  • Initial state:
[Ni(H₂O)₆]²⁺ (Green)
  • Adding 1 equivalent of 'en' forms a mono-en complex:
[Ni(H₂O)₆]²⁺ + en arrow [Ni(H₂O)₄(en)]²⁺ (Pale Blue) + 2H₂O
  • Adding a 2nd equivalent forms a bis-en complex:
[Ni(H₂O)₄(en)]²⁺ + en arrow [Ni(H₂O)₂(en)₂]²⁺ (Blue / Purple) + 2H₂O
  • Adding a 3rd equivalent forms the tris-en complex:
[Ni(H₂O)₂(en)₂]²⁺ + en arrow [Ni(en)₃]²⁺ (Violet) + 2H₂O

This progressive ligand replacement shifts the absorption spectrum, changing the solution's visible color from Green arrow Pale Blue arrow Blue arrow Violet.

Pattern Recognition

Replacing weak-field ligands (like H₂O) with stronger bidentate chelating ligands (like 'en') increases crystal field splitting. For Ni²⁺, this ligand substitution always follows the specific chromatic progression: Green arrow Pale Blue arrow Blue arrow Violet.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q38 jee_main_2025_24_jan_evening Crystal Field Theory
The conditions and consequence that favours the t2g³ eg¹ 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 d⁴ transition metal ion configuration:

  • Weak Field Ligand (WFL):
  • The crystal field splitting energy is smaller than the pairing energy (Δₒ < P). Consequently, electrons prefer to occupy the higher-energy eg orbitals rather than pair up in the lower-energy t2g orbitals. This leads to a high spin complex with the configuration:

t2g³ eg¹
  • Strong Field Ligand (SFL):
  • The splitting energy is larger than the pairing energy (Δₒ > P). Electrons pair up in the t2g orbitals before occupying the eg subshell, resulting in a low spin complex with the configuration:

t2g⁴ eg⁰
Pattern Recognition

An electron occupying an eg orbital before the t2g 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.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q28 jee_main_2025_24_jan_morning Werner's Theory of Coordination Compounds
One mole of the octahedral complex compound Co(NH₃)₅Cl₃ gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with excess of AgNO₃ solution to yield two moles of AgCl(s). The structure of the complex is:
  • A. [Co(NH₃)₅Cl]Cl₂
  • B. [Co(NH₃)₄Cl].Cl₂.NH₃
  • C. [Co(NH₃)₄Cl₂]Cl.NH₃
  • D. [Co(NH₃)₃Cl₃].2NH₃

Solution

Related Formula
Moles of AgCl precipitated = Moles of ionizable Cl⁻ ions outside the coordination sphere
Core Logic

Since 1 mole of the complex yields 2 moles of AgCl(s), there must be exactly 2 chloride ions outside the coordination sphere to undergo precipitation:

[Co(NH₃)₅Cl]Cl₂ arrow [Co(NH₃)₅Cl]²⁺(aq) + 2Cl⁻(aq)

This dissociation produces a total of 3 moles of ions per mole of the complex, perfectly consistent with the problem constraints.

Pattern Recognition

Number of precipitated AgCl moles directly equates to the count of counter-anions located outside the square brackets.

Chapter Mix

Class 12 Chemistry: Coordination Compounds

Q36 jee_main_2025_28_jan_evening Valence Bond Theory and Hybridization
Match List-I with List-II.
List-I (Complex)List-II (Hybridisation of central metal ion)
(A) [CoF₆]³⁻(I) d²sp³
(B) [NiCl₄]²⁻(II) sp³
(C) [Co(NH₃)₆]³⁺(III) sp³d²
(D) [Ni(CN)₄]²⁻(IV) dsp²
Choose the correct answer from the options given below :
  • A. (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
  • B. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  • C. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  • D. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Solution

Related Formula

Coordination Number 6 corresponds to either d²sp³ or sp³d² configuration templates. Coordination Number 4 corresponds to either sp³ or dsp² configuration templates.

Core Logic

Analyzing metal orbital dynamics under varying ligand fields:

  • (A) [CoF₆]³⁻: Co³⁺ (3d⁶) with a weak field ligand (F^-) arrow no pairing occurs arrow utilizes outer orbitals arrow sp³d².
  • (B) [NiCl₄]²⁻: Ni²⁺ (3d⁸) with a weak field ligand (Cl^-) arrow no pairing occurs arrow tetrahedral profile arrow sp³.
  • (C) [Co(NH₃)₆]³⁺: Co³⁺ (3d⁶) with a strong field ligand (NH₃) arrow electrons pair up arrow inner orbital configuration arrow d²sp³.
  • (D) [Ni(CN)₄]²⁻: Ni²⁺ (3d⁸) with a strong field ligand (CN^-) arrow forced pairing opens a 3d slot arrow square planar geometry arrow dsp².
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 sp³, while with strong field ligands (CN^-) it yields dsp².
  • Cobalt(III) with weak field ligands (F^-) yields sp³d², while with strong field ligands (NH₃) it yields d²sp³.
Chapter Mix

Class 12 Chemistry: Coordination Compounds

More Coordination Compounds Questions — jee_main_2025_04_april_evening

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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%)