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 8

Q67 jee_main_2024_01_february_morning Colour in Coordination Compounds
Given below are two statements: Statement (I): A solution of [Ni(H_2O)_6]^2+ is green in colour. Statement (II): A solution of [Ni(CN)_4]^2- is colourless. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. textBoth Statement I and Statement II are incorrect
  • B. textBoth Statement I and Statement II are correct
  • C. textStatement I is incorrect but Statement II is correct
  • D. textStatement I is correct but Statement II is incorrect

Solution

### Core Logic [Ni(H_2O)_6]^2+: Water is a weak field ligand. Ni^2+ is a 3d^8 system. In an octahedral weak field, it has 2 unpaired electrons (t_2g^6 e_g^2). Due to the presence of unpaired electrons, d-d transition is possible, making the solution green in colour. [Ni(CN)_4]^2-: CN^- is a strong field ligand. The complex is square planar (dsp^2 hybridization). All 8 electrons are paired in the lower energy d-orbitals. Because there are no unpaired electrons (diamagnetic), d-d transition does not fall in the visible region, and it is colourless. ### Step 1: Evaluate Statements Statement I is correct (Green due to unpaired electrons). Statement II is correct (Colourless as it is diamagnetic). ### Pattern Recognition Strong field ligands (like CN^-) with d^8 metal ions (Ni^2+, Pd^2+, Pt^2+) generally force pairing, creating square planar, diamagnetic, and often colourless complexes unless charge transfer occurs. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q71 jee_main_2024_01_february_morning Definitions of Some Important Terms Pertaining to Coordination Compounds
Which of the following complex is homoleptic?
  • A. [Ni(CN)_4]^2-
  • B. [Ni(NH_3)_2Cl_2]
  • C. [Fe(NH_3)_4Cl_2]^+
  • D. [Co(NH_3)_4Cl_2]^+

Solution

### Core Logic A homoleptic complex is one in which the central metal atom/ion is bound to only one kind of donor group (ligand). A heteroleptic complex is one in which the central metal atom/ion is bound to more than one kind of donor group. ### Step 1: Analyze Options (1) [Ni(CN)_4]^2-: Only one type of ligand (CN^-). Homoleptic. (2) [Ni(NH_3)_2Cl_2]: Two types of ligands (NH_3 and Cl^-). Heteroleptic. (3) [Fe(NH_3)_4Cl_2]^+: Two types of ligands. Heteroleptic. (4) [Co(NH_3)_4Cl_2]^+: Two types of ligands. Heteroleptic. ### Pattern Recognition Homo = same, leptic = ligands. Look for the formula bracket containing only one symbol type after the central metal. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q66 jee_main_2024_29_january_evening IUPAC Nomenclature of Coordination Complexes
The correct IUPAC name of mathrmK_2mathrmMnO_4 is
  • A. Potassium tetraoxopermanganate (VI)
  • B. Potassium tetraoxidomanganate (VI)
  • C. Dipotassium tetraoxidomanganate (VII)
  • D. Potassium tetraoxidomanganese (VI)

Solution

### Related Formula textOxidation State of Mn Evaluation: 2(+1) + x + 4(-2) = 0 ### Core Logic Solving for x: 2 + x - 8 = 0 implies x = +6 Since the complex is anionic, the metal name ends with the suffix '-ate', making it 'manganate(VI)'. The ligands are oxygen atoms, designated systematically as 'tetraoxido' or 'tetraoxo' according to newer recommendations. ### Step 1: Assembly Combining parts systematically yields the correct IUPAC string: Potassium tetraoxidomanganate(VI). ### Pattern Recognition Anionic metal centers must include the trailing '-ate' modifier followed immediately by their absolute Roman oxidation state indicators. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q82 jee_main_2024_29_january_evening Brown Ring Complex Oxidation State
The oxidation number of iron in the compound formed during brown ring test for mathrmNO_3^- ion is ________.
Numerical Answer. Answer: 1 to 1

Solution

### Related Formula textFormula of Brown Ring Complex: [mathrmFe(mathrmH_2mathrmO)_5(mathrmNO)]^2+ ### Core Logic In this specific coordination complex, charge transfer occurs where nitric oxide transfers an electron to the iron center. As a result, textNO exists as a positive textNO^+ ligand, and iron drops to an unusual oxidation state: x + 5(0) + 1(+1) = +2 x + 1 = 2 implies x = +1 ### Step 1: Final Value Assignment Solving this charge balance confirms that the oxidation number of iron in the brown ring complex is +1. ### Pattern Recognition The brown ring test features a rare +1 oxidation state for iron because textNO coordinates as the positive nitrosonium cation (textNO^+). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q68 jee_main_2024_27_jan_morning Magnetic Properties of Coordination Complexes
Consider the following complex ions: P=[FeF_6]^3- Q=[V(H_2O)_6]^2+ R=[Fe(H_2O)_6]^2+ The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :
  • A. R < Q < P
  • B. R < P < Q
  • C. Q < R < P
  • D. Q < P < R

Solution

### Step 1: Evaluation of P=[FeF_6]^3- textFe^3+ rightarrow 3textd^5 Since textF^- is a weak field ligand, no pairing occurs. Number of unpaired electrons (n) = 5. mu = sqrt5(5+2) = sqrt35text BM ### Step 2: Evaluation of Q=[V(H_2O)_6]^2+ textV^2+ rightarrow 3textd^3 Number of unpaired electrons (n) = 3. mu = sqrt3(3+2) = sqrt15text BM ### Step 3: Evaluation of R=[Fe(H_2O)_6]^2+ textFe^2+ rightarrow 3textd^6 Since textH_2textO is a weak field ligand, configuration is textt_2textg^4 texte_textg^2. Number of unpaired electrons (n) = 4. mu = sqrt4(4+2) = sqrt24text BM ### Step 4: Comparison Comparing values: mu(Q) < mu(R) < mu(P) implies Q < R < P ### Pattern Recognition Count unpaired electrons strictly accounting for weak field vs strong field rules. Order scales monotonically with n. ### Chapter Mix Class 12 Chemistry: Coordination Compounds

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