Given below are two statements: Statement I: Among [mathrmCu(mathrmNH_3)_4]^2+, [mathrmNi(mathrmen)_3]^2+, [mathrmNi(mathrmNH_3)_6]^2+ and [mathrmMn(mathrmH_2mathrmO)_6]^2+, [mathrmMn(mathrmH_2mathrmO)_6]^2+ has the maximum number of unpaired electrons. Statement II: The number of pairs among \[NiCl_4]^2-, [Ni(CO)_4]\, \[NiCl_4]^2-, [Ni(CN)_4]^2-\ and \[Ni(CO)_4], [Ni(CN)_4]^2-\ that contain only diamagnetic species is two. In the light of the above statements, choose the correct answer from the options given below:

Solution & Explanation

### Core Logic Evaluating Statement I: - [Cu(NH_3)_4]^2+: Cu^2+ is 3d^9, 1 unpaired electron. - [Ni(en)_3]^2+: Ni^2+ is 3d^8, in octahedral field, 2 unpaired electrons. - [Ni(NH_3)_6]^2+: Ni^2+ is 3d^8, 2 unpaired electrons. - [Mn(H_2O)_6]^2+: Mn^2+ is 3d^5, weak field ligand H_2O leads to high spin, 5 unpaired electrons. So [Mn(H_2O)_6]^2+ has the maximum number of unpaired electrons. Statement I is true. Evaluating Statement II: - [Ni(CO)_4]: Ni(0) is 3d^8 4s^2, strong field CO pairs electrons to 3d^10, diamagnetic (0 unpaired). - [Ni(CN)_4]^2-: Ni^2+ is 3d^8, strong field CN^- forces pairing rightarrow dsp^2 square planar, diamagnetic (0 unpaired). - [NiCl_4]^2-: Ni^2+ is 3d^8, weak field Cl^- does not pair rightarrow sp^3 tetrahedral, paramagnetic (2 unpaired). The pairs containing ONLY diamagnetic species: - \[NiCl_4]^2-, [Ni(CO)_4]\ rightarrow 1 para, 1 dia (No) - \[NiCl_4]^2-, [Ni(CN)_4]^2-\ rightarrow 1 para, 1 dia (No) - \[Ni(CO)_4], [Ni(CN)_4]^2-\ rightarrow Both dia (Yes) The number of such pairs is exactly ONE. Statement II says two, so it is false. ### Step 1: Final Conclusion Statement I is true, Statement II is false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

Reference Study Guides

Coordination Compounds

More Coordination Compounds Previous-Year Questions — Page 9

Q62 jee_main_2024_29_jan_morning Applications of Coordination Compounds
Match List I with List II
LIST-I (Substances)LIST-II (Element Present)
A. Ziegler catalystI. Rhodium
B. Blood PigmentII. Cobalt
C. Wilkinson catalystIII. Iron
D. Vitamin B_12IV. Titanium
Choose the correct answer from the options given below:
  • A. textA-II, B-IV, C-I, D-III
  • B. textA-II, B-III, C-IV, D-I
  • C. textA-III, B-II, C-IV, D-I
  • D. textA-IV, B-III, C-I, D-II

Solution

### Core Logic Let us identify the central metal atom/element present in each of the given coordination complexes or biological molecules: 1. **Ziegler-Natta catalyst**: Used for polymerization of alkenes. Its chemical composition involves TiCl_4 and (C_2H_5)_3Al. Hence, the transition metal present is Titanium (IV). 2. **Blood pigment (Haemoglobin)**: The oxygen-carrying metalloprotein in red blood cells. It contains an Iron (III) central atom coordinated to a porphyrin ring. 3. **Wilkinson's catalyst**: Used for the hydrogenation of alkenes. Its formula is [RhCl(PPh_3)_3], meaning it contains Rhodium (I). 4. **Vitamin B_12 (Cyanocobalamin)**: A biologically important coordination compound containing Cobalt (II) at the center of a corrin ring. ### Step 1: Final Mapping Matching the elements: (A) Ziegler catalyst rightarrow (IV) Titanium (B) Blood Pigment rightarrow (III) Iron (C) Wilkinson catalyst rightarrow (I) Rhodium (D) Vitamin B_12 rightarrow (II) Cobalt This strictly maps to sequence A-IV, B-III, C-I, D-II. ### Pattern Recognition Always memorize the central metal for famous catalysts and biomolecules: Chlorophyll (Mg), Haemoglobin (Fe), Vitamin B12 (Co), Wilkinson (Rh), Ziegler-Natta (Ti). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds Class 12 Chemistry: d and f Block Elements
Q75 jee_main_2024_29_jan_morning Metal Carbonyls
In which one of the following metal carbonyls, CO forms a bridge between metal atoms?
  • A. left[mathrmCo_2(mathrmCO)_8right]
  • B. [mathrmMn_2(mathrmCO)_10]
  • C. [mathrmOs_3(mathrmCO)_12]
  • D. [mathrmRu_3(mathrmCO)_12]

Solution

### Core Logic Let's examine the structure of each polynuclear metal carbonyl: 1. **[Co_2(CO)_8]**: Solid octacarbonyldicobalt(0) exists in a bridged structure. It has a Co-Co bond and two bridging CO ligands, along with three terminal CO ligands on each cobalt atom. So, it contains bridging CO groups. 2. **[Mn_2(CO)_10]**: Decacarbonyldimanganese(0) has a single Mn-Mn bond and all 10 CO ligands are terminal. It does NOT have any bridging carbonyls. 3. **[Os_3(CO)_12]**: Dodecacarbonyltriosmium(0) forms a triangular cluster of Os atoms with all 12 CO ligands being terminal. 4. **[Ru_3(CO)_12]**: Dodecacarbonyltriruthenium(0) similarly forms a triangular Ru_3 cluster where all 12 CO ligands are terminal. ### Step 1: Conclusion Only [Co_2(CO)_8] features CO forming a bridge between the metal atoms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q74 jee_main_2024_30_january_evening Metal Carbonyls
The coordination geometry around the manganese in decacarbonyldimanganese(0)
  • A. textOctahedral
  • B. textTrigonal bipyramidal
  • C. textSquare pyramidal
  • D. textSquare planar

Solution

### Core Logic Decacarbonyldimanganese(0) is Mn_2(CO)_10. In this binuclear carbonyl, there is one Mn-Mn bond. Each Mn atom is additionally bonded to five carbonyl (CO) ligands. Therefore, the coordination number of each Mn atom is 1 + 5 = 6. The arrangement of these 6 bonds around each Mn atom is octahedral.
Octahedral geometry around Mn diagram for Q74 - JEE Main 2024 Evening
Octahedral geometry around Mn diagram for Q74 - JEE Main 2024 Evening
### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q81 jee_main_2024_30_january_evening Isomerism in Coordination Compounds
Number of complexes which show optical isomerism among the following is cis-[mathrmCr(ox)_2mathrmCl_2]^3-, [mathrmCo(en)_3]^3+ cis-[mathrmPt(en)_2mathrmCl_2]^2+, cis-[mathrmCo(en)_2mathrmCl_2]^+ trans-[mathrmPt(en)_2mathrmCl_2]^2+, trans-[mathrmCr(ox)_2mathrmCl_2]^3-
Numerical Answer. Answer: 4 to 4

Solution

### Core Logic Optical isomerism is shown by complexes that are chiral, meaning they lack both a plane of symmetry (POS) and a center of symmetry (COS). 1. cis-[mathrmCr(ox)_2mathrmCl_2]^3-: Cis configuration with bidentate ligands lacks a POS/COS. rightarrow Shows optical isomerism. 2. [mathrmCo(en)_3]^3+: Tris(bidentate) complex lacks POS/COS. rightarrow Shows optical isomerism (exists as Delta and Lambda enantiomers). 3. cis-[mathrmPt(en)_2mathrmCl_2]^2+: Cis configuration lacks POS/COS. rightarrow Shows optical isomerism. 4. cis-[mathrmCo(en)_2mathrmCl_2]^+: Cis configuration lacks POS/COS. rightarrow Shows optical isomerism. 5. trans-[mathrmPt(en)_2mathrmCl_2]^2+: Trans configuration is highly symmetric and contains a POS (along the square plane). rightarrow Cannot show optical isomerism. 6. trans-[mathrmCr(ox)_2mathrmCl_2]^3-: Trans configuration contains a POS/COS. rightarrow Cannot show optical isomerism. ### Step 1: Final Tally Out of the 6 complexes, 4 can show optical isomerism. The answer is 4. ### Pattern Recognition For octahedral complexes with bidentate ligands: cis isomers generally show optical isomerism, whereas trans isomers have a plane of symmetry and are optically inactive. [M(L-L)_3] types are always chiral. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q67 jee_main_2024_30_jan_morning Coordination Geometry
Aluminium chloride in acidified aqueous solution forms an ion having geometry
  • A. textOctahedral
  • B. textSquare Planar
  • C. textTetrahedral
  • D. textTrigonal bipyramidal

Solution

### Core Logic AlCl_3 dissolves in acidified water to form the hexaquaaluminium(III) complex ion. The reaction can be represented as: AlCl_3 + 6H_2O rightarrow [Al(H_2O)_6]^3+ + 3Cl^- ### Step 1: Analyzing Geometry The central aluminium atom coordinates with 6 water molecules, giving a coordination number of 6. A coordination number of 6 results in an octahedral geometry (sp^3d^2 hybridization). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds Class 11 Chemistry: The p-Block Elements

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