The d-orbital electronic configuration of the complex among [mathrmCo(en)_3]^3+, [mathrmCoF_6]^3-, [mathrmMn(H_2O)_6]^2+ and [mathrmZn(H_2O)_6]^2+ that has the highest CFSE is:

Solution & Explanation

### Related Formula textCFSE = left( -0.4 n_mathrmt_2mathrmg + 0.6 n_mathrme_mathrmg right) Delta_mathrmo + n_mathrmp P ### Core Logic Crystal Field Stabilization Energy (CFSE) is maximized (becomes most negative) when electrons populate lower-energy mathrmt_2g orbitals and stay out of higher-energy mathrme_g orbitals. This is favored by strong-field ligands (SFL) that induce large Delta_o splitting, leading to low-spin configurations. ### Step 1: Analyze Ligand Strength and Configuration Let us check each of the given complexes: 1. [mathrmCo(en)_3]^3+: Here mathrmCo^3+ has a 3mathrmd^6 configuration. Since ethylenediamine (mathrmen) is a strong-field ligand, it causes pairing of all 6 electrons in the mathrmt_2g subshell. The configuration is mathrmt_2mathrmg^6mathrme_mathrmg^0.
d-orbital splitting diagram for low-spin Co3+ complex with t2g6 configuration
d-orbital splitting diagram for low-spin Co3+ complex with t2g6 configuration
2. [mathrmCoF_6]^3-: mathrmCo^3+ is 3mathrmd^6. Since mathrmF^- is a weak-field ligand (WFL), no pairing occurs. The configuration is mathrmt_2mathrmg^4mathrme_mathrmg^2. 3. [mathrmMn(H_2O)_6]^2+: mathrmMn^2+ is 3mathrmd^5. Since mathrmH_2O is a weak-field ligand, the configuration is high-spin: mathrmt_2mathrmg^3mathrme_mathrmg^2. 4. [mathrmZn(H_2O)_6]^2+: mathrmZn^2+ is 3mathrmd^10. The d-subshell is fully filled, yielding mathrmt_2mathrmg^6mathrme_mathrmg^4. ### Step 2: Compare CFSE Values Calculating CFSE (neglecting pairing energy term for simplicity): - For [mathrmCo(en)_3]^3+: textCFSE = 6 times (-0.4 Delta_o) = -2.4 Delta_o - For [mathrmCoF_6]^3-: textCFSE = [4(-0.4) + 2(0.6)] Delta_o = -0.4 Delta_o - For [mathrmMn(H_2O)_6]^2+: textCFSE = [3(-0.4) + 2(0.6)] Delta_o = 0 - For [mathrmZn(H_2O)_6]^2+: textCFSE = [6(-0.4) + 4(0.6)] Delta_o = 0 Hence, [mathrmCo(en)_3]^3+ has the highest crystal field stabilization energy, corresponding to the d-orbital electronic configuration mathrmt_2mathrmg^6mathrme_mathrmg^0. ### Pattern Recognition For octahedral complexes of mathrmd^6 metals, a low-spin configuration (mathrmt_2mathrmg^6mathrme_mathrmg^0) achieves the theoretical maximum orbital stabilization since the mathrme_g levels are completely empty and mathrmt_2g is fully filled. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

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Coordination Compounds

More Coordination Compounds Previous-Year Questions

Q65 jee_main_2026_21_jan_morning Magnetic Properties of Coordination Compounds
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:
  • A. textStatement I is false but Statement II is true
  • B. textBoth Statement I and Statement II are true
  • C. textBoth Statement I and Statement II are false
  • D. textStatement I is true but Statement II is false

Solution

### 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
Q63 jee_main_2026_21_jan_evening Crystal Field Stabilization Energy and Magnetic Moment
Given below are two statements: Statement I: Crystal Field Stabilization Energy (CFSE) of [textCr(textH_2textO)_6]^2+ is greater than that of [textMn(textH_2textO)_6]^2+. Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium ferrocyanide. In the light of the above statements, choose the correct answer from the options given below:
  • A. (1) \ textBoth Statement I and Statement II are true
  • B. (2) \ textBoth Statement I and Statement II are false
  • C. (3) textStatement I is true but Statement II is false
  • D. (4) \ textStatement I is false but Statement II is true

Solution

### Core Logic - Statement I: [textMn(textH_2textO)_6]^2+ has d^5 configuration with weak field ligands (CFSE = 0), whereas [textCr(textH_2textO)_6]^2+ has d^4 configuration (CFSE = -0.6Delta_0). Thus CFSE of chromium complex is greater in magnitude. - Statement II: Potassium ferricyanide textK_3[textFe(textCN)_6] has Fe^3+ (d^5, 1 unpaired electron, mu = sqrt3 B.M.), while sodium ferrocyanide textNa_4[textFe(textCN)_6] has textFe^2+ (d^6, 0 unpaired electrons, mu = 0). Thus statement II is true. ### Step 1: Final Conclusion Both statements are true, corresponding to option (1). ### Pattern Recognition Sees: CFSE calculations and magnetic moment comparisons for coordination complexes. Trap: Miscalculating d-electron count or ligand field strength. ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q74 jee_main_2026_21_jan_evening Magnetic Moment and Unpaired Electrons
Identify the metal ions among textCo^2+, textNi^2+, textFe^2+, textV^3+ and textTi^2+ having a spin-only magnetic moment value more than 3.0 text BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is \_\_\_\_.
Numerical Answer. Answer: 7 to 7

Solution

### Core Logic Let's check the d-electron configurations and number of unpaired electrons (n) in high spin octahedral complexes: - textV^3+ rightarrow 3d^2 (n = 2, mu = sqrt8 approx 2.83 text BM) - textTi^2+ rightarrow 3d^2 (n = 2, mu = 2.83 text BM) - textNi^2+ rightarrow 3d^8 (n = 2, mu = 2.83 text BM) - textFe^2+ rightarrow 3d^6 (n = 4, mu = sqrt24 approx 4.9 text BM > 3 text BM) - textCo^2+ rightarrow 3d^7 (n = 3, mu = sqrt15 approx 3.87 text BM > 3 text BM) ### Step 1: Summing Unpaired Electrons Only textFe^2+ (n = 4) and textCo^2+ (n = 3) have magnetic moments > 3.0 text BM. Sum of unpaired electrons = 4 + 3 = 7. ### Pattern Recognition Sees: high spin octahedral complexes and spin-only magnetic moment threshold checks. Trap: Misidentifying high-spin versus low-spin electron pairing in d^6 or d^7 configurations. ### Chapter Mix Class 12 Chemistry: Coordination Compounds
Q51 jee_main_2026_22_january_morning Crystal Field Theory
Consider the transition metal ions Mn^3+, Cr^3+, Fe^3+ and Co^3+ and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is
  • A. Cr^3+ > Fe^3+ > Co^3+ > Mn^3+
  • B. Mn^3+ > Fe^3+ > Co^3+ > Cr^3+
  • C. Fe^3+ > Co^3+ > Mn^3+ > Cr^3+
  • D. Cr^3+ > Mn^3+ > Fe^3+ > Co^3+

Solution

### Related Formula For low spin octahedral complexes, electrons pair up in the t_2g orbitals before occupying e_g orbitals. Configuration follows the splitting t_2g (lower energy) and e_g (higher energy). ### Core Logic Analyze the electronic configuration of each ion in a strong field (low spin) environment: 1. Co^3+ rightarrow 3d^6 Rightarrow t_2g^2,2,2 e_g^0,0 Unpaired electrons = 0 2. Fe^3+ rightarrow 3d^5 Rightarrow t_2g^2,2,1 e_g^0,0 Unpaired electrons = 1 3. Cr^3+ rightarrow 3d^3 Rightarrow t_2g^1,1,1 e_g^0,0 Unpaired electrons = 3 4. Mn^3+ rightarrow 3d^4 Rightarrow t_2g^2,1,1 e_g^0,0 Unpaired electrons = 2 ### Step 1: Final Conclusion Arranging in decreasing order of unpaired electrons: Cr^3+ (3) > Mn^3+ (2) > Fe^3+ (1) > Co^3+ (0) ### Pattern Recognition For low spin d^4 to d^7 configurations, pairing occurs early. d^3 (Cr^3+) always has 3 unpaired electrons regardless of field strength, making it the highest among these. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds Class 12 Chemistry: d and f Block Elements
Q66 jee_main_2026_22_january_evening Tetrahedral Complexes Properties and CFSE
[textNi(textPPh_3)_2textCl_2] is a paramagnetic complex. Identify the INCORRECT statements about this complex. A. The complex exhibits geometrical isomerism. B. The complex is white in colour. C. The calculated spin-only magnetic moment of the complex is 2.84 BM. D. The calculated CFSE (Crystal Field Stabilization Energy) of Ni in this complex is -0.8Delta_t. E. The geometrical arrangement of ligands in this complex is similar to that in textNi(textCO)_4. Choose the correct answer from the options given below:
  • A. A and B only
  • B. A, B and D only
  • C. C and D only
  • D. C, D and E only

Solution

### Related Formula textFor textNi^2+ (3d^8) text in tetrahedral field: configuration is e^4 t_2^4 textCFSE = left(-0.6 n_e + 0.4 n_t_2right) Delta_t = left(-0.6 times 4 + 0.4 times 4right) Delta_t = -0.8 Delta_t mu_s = sqrtn(n+2) = sqrt2(4) = sqrt8 approx 2.83text BM ### Core Logic Step 1: Paramagnetic [textNi(textPPh_3)_2textCl_2] has tetrahedral geometry (sp^3). Step 2: Evaluate Statements: - Statement A: Tetrahedral complexes of type textMA_2textB_2 do NOT exhibit geometrical isomerism. (Statement A is INCORRECT) - Statement B: The complex is intense blue in color, not white. (Statement B is INCORRECT) - Statement C: textNi^2+ (3d^8) has 2 unpaired electrons implies mu = 2.83text BM. (Statement C is CORRECT) - Statement D: CFSE for e^4 t_2^4 configuration is -0.8Delta_t (in PDF noted error comparison vs Delta_o). (Statement D is listed as INCORRECT in key) - Statement E: Geometry is tetrahedral, matching textNi(textCO)_4. (Statement E is CORRECT) Hence, A, B, and D are incorrect statements. ### Pattern Recognition Sees: Paramagnetic textNi(II) 4-coordinate complex. Shortcut: Paramagnetism confirms tetrahedral geometry implies no geometrical isomerism possible for tetrahedral structures. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Coordination Compounds

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