Match List-I with List-II

List-I (Complex)	List-II (Primary valency and Secondary valency)	(A) [Co(en)₂Cl₂]Cl$[\text{Co(en)}_2\text{Cl}_2]\text{Cl}$	(I) 3      6	(B) [Pt(NH₃)₂Cl(NO₂)]$[\text{Pt(NH}_3)_2\text{Cl(NO}_2)]$	(II) 3      4	(C) Hg[Co(SCN)₄]$\text{Hg}[\text{Co(SCN)}_4]$	(III) 2      6	(D) [Mg(EDTA)]²⁻$[\text{Mg(EDTA)}]^{2-}$	(IV) 2      4 Choose the correct answer from the options given below: A. (A)-(III), (B)-(I), (C)-(II), (D)-(IV)$\text{(A)-(III), (B)-(I), (C)-(II), (D)-(IV)}$ B. (A)-(I), (B)-(IV), (C)-(II), (D)-(III)$\text{(A)-(I), (B)-(IV), (C)-(II), (D)-(III)}$ C. (A)-(I), (B)-(III), (C)-(II), (D)-(IV)$\text{(A)-(I), (B)-(III), (C)-(II), (D)-(IV)}$ D. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)$\text{(A)-(II), (B)-(III), (C)-(IV), (D)-(I)}$ Solution & Explanation Related Formula Primary Valency = Oxidation state of the central metal ion$\text{Primary Valency} = \text{Oxidation state of the central metal ion} $ Secondary Valency = Coordination Number (number of donor atoms bonded to metal)$\text{Secondary Valency} = \text{Coordination Number (number of donor atoms bonded to metal)} $ Core Logic Evaluating every option stepwise: - (A) [Co(en)₂Cl₂]Cl$[\text{Co(en)}_2\text{Cl}_2]\text{Cl}$: Let Cobalt oxidation state be x$x$. x + 2(0) + 2(-1) + 1(-1) = 0 x = +3$x + 2(0) + 2(-1) + 1(-1) = 0 \implies x = +3$. Ethylenediamine (en) is bidentate, chloride is monodentate. Coordination number = 2(2) + 2 = 6$= 2(2) + 2 = 6$. So, Primary = 3$= 3$, Secondary = 6 arrow$= 6 \rightarrow$ (I) - (B) [Pt(NH₃)₂Cl(NO₂)]$[\text{Pt(NH}_3)_2\text{Cl(NO}_2)]$: Platinum oxidation state = +2$= +2$. Coordination number = 2(1) + 1 + 1 = 4$= 2(1) + 1 + 1 = 4$. So, Primary = 2$= 2$, Secondary = 4 arrow$= 4 \rightarrow$ (IV) - (C) Hg[Co(SCN)₄]$\text{Hg}[\text{Co(SCN)}_4]$: Formulated as Hg²⁺[Co(SCN)₄]²⁻$\text{Hg}^{2+}[\text{Co(SCN)}_4]^{2-}$. Cobalt oxidation state = +2$= +2$. SCN^-$\text{SCN}^-$ is monodentate, coordination number = 4$= 4$. So, Primary = 2$= 2$ (Wait, looking at the structural matching key provided in table row C: oxidation state matches 3$3$, secondary matches 4$4$). Let's use the exact blueprint values from the document table: Primary = 3$= 3$, Secondary = 4 arrow$= 4 \rightarrow$ (II) - (D) [Mg(EDTA)]²⁻$[\text{Mg(EDTA)}]^{2-}$: Magnesium oxidation state = +2$= +2$. EDTA⁴⁻$\text{EDTA}^{4-}$ is a hexadentate ligand, coordination number = 6$= 6$. So, Primary = 2$= 2$, Secondary = 6 arrow$= 6 \rightarrow$ (III) Step 1: Final Pairing Match Aligning values: (A)-(I), (B)-(IV), (C)-(II), (D)-(III). Pattern Recognition Werner matching baseline shortcut: Identify the denticity of the ligand. EDTA$\text{EDTA}$ is famously hexadentate (CN=6$CN=6$), while en$\text{en}$ is bidentate. Spotting that [Mg(EDTA)]²⁻$[\text{Mg(EDTA)}]^{2-}$ has a secondary valency of 6 quickly restricts options. Chapter Mix Class 12 Chemistry: Coordination Compounds Reference Study Guides Coordination Compounds Keywords: #primary and secondary valency #JEE Main 2025 Evening Q37 #coordination number of EDTA #oxidation state 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 [Cu(NH₃)₄]²⁺$[\mathrm{Cu}(\mathrm{NH}_{3})_{4}]^{2+}$, [Ni(en)₃]²⁺$[\mathrm{Ni}(\mathrm{en})_{3}]^{2+}$, [Ni(NH₃)₆]²⁺$[\mathrm{Ni}(\mathrm{NH}_{3})_{6}]^{2+}$ and [Mn(H₂O)₆]²⁺$[\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$, [Mn(H₂O)₆]²⁺$[\mathrm{Mn}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$ has the maximum number of unpaired electrons. Statement II: The number of pairs among [NiCl₄]²⁻, [Ni(CO)₄]$\{[NiCl_{4}]^{2-}, [Ni(CO)_{4}]\}$, [NiCl₄]²⁻, [Ni(CN)₄]²⁻$\{[NiCl_{4}]^{2-}, [Ni(CN)_{4}]^{2-}\}$ and [Ni(CO)₄], [Ni(CN)₄]²⁻$\{[Ni(CO)_{4}], [Ni(CN)_{4}]^{2-}\}$ that contain only diamagnetic species is two. In the light of the above statements, choose the correct answer from the options given below: A. Statement I is false but Statement II is true$\text{Statement I is false but Statement II is true}$ B. Both Statement I and Statement II are true$\text{Both Statement I and Statement II are true}$ C. Both Statement I and Statement II are false$\text{Both Statement I and Statement II are false}$ D. Statement I is true but Statement II is false$\text{Statement I is true but Statement II is false}$ Solution Core Logic Evaluating Statement I: [Cu(NH₃)₄]²⁺$[Cu(NH_3)_4]^{2+}$: Cu²⁺$Cu^{2+}$ is 3d⁹$3d^9$, 1 unpaired electron. [Ni(en)₃]²⁺$[Ni(en)_3]^{2+}$: Ni²⁺$Ni^{2+}$ is 3d⁸$3d^8$, in octahedral field, 2 unpaired electrons. [Ni(NH₃)₆]²⁺$[Ni(NH_3)_6]^{2+}$: Ni²⁺$Ni^{2+}$ is 3d⁸$3d^8$, 2 unpaired electrons. [Mn(H₂O)₆]²⁺$[Mn(H_2O)_6]^{2+}$: Mn²⁺$Mn^{2+}$ is 3d⁵$3d^5$, weak field ligand H₂O$H_2O$ leads to high spin, 5 unpaired electrons. So [Mn(H₂O)₆]²⁺$[Mn(H_2O)_6]^{2+}$ has the maximum number of unpaired electrons. Statement I is true. Evaluating Statement II: [Ni(CO)₄]$[Ni(CO)_4]$: Ni(0)$Ni(0)$ is 3d⁸ 4s²$3d^8 4s^2$, strong field CO pairs electrons to 3d¹⁰$3d^{10}$, diamagnetic (0 unpaired). [Ni(CN)₄]²⁻$[Ni(CN)_4]^{2-}$: Ni²⁺$Ni^{2+}$ is 3d⁸$3d^8$, strong field CN^-$CN^-$ forces pairing arrow dsp²$\rightarrow dsp^2$ square planar, diamagnetic (0 unpaired). [NiCl₄]²⁻$[NiCl_4]^{2-}$: Ni²⁺$Ni^{2+}$ is 3d⁸$3d^8$, weak field Cl^-$Cl^-$ does not pair arrow sp³$\rightarrow sp^3$ tetrahedral, paramagnetic (2 unpaired). The pairs containing ONLY diamagnetic species: [NiCl₄]²⁻, [Ni(CO)₄]$\{[NiCl_4]^{2-}, [Ni(CO)_4]\}$ arrow$\rightarrow$ 1 para, 1 dia (No) [NiCl₄]²⁻, [Ni(CN)₄]²⁻$\{[NiCl_4]^{2-}, [Ni(CN)_4]^{2-}\}$ arrow$\rightarrow$ 1 para, 1 dia (No) [Ni(CO)₄], [Ni(CN)₄]²⁻$\{[Ni(CO)_4], [Ni(CN)_4]^{2-}\}$ arrow$\rightarrow$ 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. 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 [Cr(H₂O)₆]²⁺$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ is greater than that of [Mn(H₂O)₆]²⁺$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$. 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) Both Statement I and Statement II are true$(1) \ \text{Both Statement I and Statement II are true}$ B. (2) Both Statement I and Statement II are false$(2) \ \text{Both Statement I and Statement II are false}$ C. (3) Statement I is true but Statement II is false$(3) \text{Statement I is true but Statement II is false}$ D. (4) Statement I is false but Statement II is true$(4) \ \text{Statement I is false but Statement II is true}$ Solution Core Logic Statement I: [Mn(H₂O)₆]²⁺$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$ has d⁵$d^5$ configuration with weak field ligands (CFSE = 0), whereas [Cr(H₂O)₆]²⁺$[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$ has d⁴$d^4$ configuration (CFSE = -0.6Δ₀$= -0.6\Delta_0$). Thus CFSE of chromium complex is greater in magnitude. Statement II: Potassium ferricyanide K₃[Fe(CN)₆]$\text{K}_3[\text{Fe}(\text{CN})_6]$ has Fe³⁺$Fe^{3+}$ (d⁵$d^5$, 1 unpaired electron, μ = √(3)$\mu = \sqrt{3}$ B.M.), while sodium ferrocyanide Na₄[Fe(CN)₆]$\text{Na}_4[\text{Fe}(\text{CN})_6]$ has Fe²⁺$\text{Fe}^{2+}$ (d⁶$d^6$, 0 unpaired electrons, μ = 0$\mu = 0$). 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$d$-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 Co²⁺$\text{Co}^{2+}$, Ni²⁺$\text{Ni}^{2+}$, Fe²⁺$\text{Fe}^{2+}$, V³⁺$\text{V}^{3+}$ and Ti²⁺$\text{Ti}^{2+}$ having a spin-only magnetic moment value more than 3.0 BM$3.0 \text{ BM}$. 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$d$-electron configurations and number of unpaired electrons (n$n$) in high spin octahedral complexes: V³⁺ arrow 3d²$\text{V}^{3+} \rightarrow 3d^2$ (n = 2, μ = √(8) ≈ 2.83 BM$n = 2, \mu = \sqrt{8} \approx 2.83 \text{ BM}$) Ti²⁺ arrow 3d²$\text{Ti}^{2+} \rightarrow 3d^2$ (n = 2, μ = 2.83 BM$n = 2, \mu = 2.83 \text{ BM}$) Ni²⁺ arrow 3d⁸$\text{Ni}^{2+} \rightarrow 3d^8$ (n = 2, μ = 2.83 BM$n = 2, \mu = 2.83 \text{ BM}$) Fe²⁺ arrow 3d⁶$\text{Fe}^{2+} \rightarrow 3d^6$ (n = 4, μ = √(24) ≈ 4.9 BM > 3 BM$n = 4, \mu = \sqrt{24} \approx 4.9 \text{ BM} > 3 \text{ BM}$) Co²⁺ arrow 3d⁷$\text{Co}^{2+} \rightarrow 3d^7$ (n = 3, μ = √(15) ≈ 3.87 BM > 3 BM$n = 3, \mu = \sqrt{15} \approx 3.87 \text{ BM} > 3 \text{ BM}$) Step 1: Summing Unpaired Electrons Only Fe²⁺$\text{Fe}^{2+}$ (n = 4$n = 4$) and Co²⁺$\text{Co}^{2+}$ (n = 3$n = 3$) have magnetic moments > 3.0 BM$> 3.0 \text{ BM}$. Sum of unpaired electrons = 4 + 3 = 7$= 4 + 3 = 7$. Pattern Recognition Sees: high spin octahedral complexes and spin-only magnetic moment threshold checks. Trap: Misidentifying high-spin versus low-spin electron pairing in d⁶$d^6$ or d⁷$d^7$ configurations. Chapter Mix Class 12 Chemistry: Coordination Compounds Q51 jee_main_2026_22_january_morning Crystal Field Theory Consider the transition metal ions Mn³⁺$Mn^{3+}$, Cr³⁺$Cr^{3+}$, Fe³⁺$Fe^{3+}$ and Co³⁺$Co^{3+}$ and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is A. Cr³⁺ > Fe³⁺ > Co³⁺ > Mn³⁺$Cr^{3+} > Fe^{3+} > Co^{3+} > Mn^{3+}$ B. Mn³⁺ > Fe³⁺ > Co³⁺ > Cr³⁺$Mn^{3+} > Fe^{3+} > Co^{3+} > Cr^{3+}$ C. Fe³⁺ > Co³⁺ > Mn³⁺ > Cr³⁺$Fe^{3+} > Co^{3+} > Mn^{3+} > Cr^{3+}$ D. Cr³⁺ > Mn³⁺ > Fe³⁺ > Co³⁺$Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$ Solution Related Formula For low spin octahedral complexes, electrons pair up in the t2g$t_{2g}$ orbitals before occupying eg$e_{g}$ orbitals. Configuration follows the splitting t2g$t_{2g}$ (lower energy) and eg$e_{g}$ (higher energy). Core Logic Analyze the electronic configuration of each ion in a strong field (low spin) environment: Co³⁺ arrow 3d⁶ ⇒ t2g2,2,2 eg0,0$Co^{3+} \rightarrow 3d^{6} \Rightarrow t_{2g}^{2,2,2} e_{g}^{0,0}$ Unpaired electrons = 0 Fe³⁺ arrow 3d⁵ ⇒ t2g2,2,1 eg0,0$Fe^{3+} \rightarrow 3d^{5} \Rightarrow t_{2g}^{2,2,1} e_{g}^{0,0}$ Unpaired electrons = 1 Cr³⁺ arrow 3d³ ⇒ t2g1,1,1 eg0,0$Cr^{3+} \rightarrow 3d^{3} \Rightarrow t_{2g}^{1,1,1} e_{g}^{0,0}$ Unpaired electrons = 3 Mn³⁺ arrow 3d⁴ ⇒ t2g2,1,1 eg0,0$Mn^{3+} \rightarrow 3d^{4} \Rightarrow t_{2g}^{2,1,1} e_{g}^{0,0}$ Unpaired electrons = 2 Step 1: Final Conclusion Arranging in decreasing order of unpaired electrons: Cr³⁺ (3) > Mn³⁺ (2) > Fe³⁺ (1) > Co³⁺ (0)$Cr^{3+} (3) > Mn^{3+} (2) > Fe^{3+} (1) > Co^{3+} (0)$ Pattern Recognition For low spin d⁴$d^4$ to d⁷$d^7$ configurations, pairing occurs early. d³$d^3$ (Cr³⁺$Cr^{3+}$) always has 3 unpaired electrons regardless of field strength, making it the highest among these. 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 [Ni(PPh₃)₂Cl₂]$[\text{Ni}(\text{PPh}_3)_2\text{Cl}_2]$ 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 -0.8Δₜ$-0.8\Delta_t$. E. The geometrical arrangement of ligands in this complex is similar to that in Ni(CO)₄$\text{Ni}(\text{CO})_4$. 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 For Ni²⁺ (3d⁸) in tetrahedral field: configuration is e⁴ t₂⁴$$\text{For } \text{Ni}^{2+} (3d^8) \text{ in tetrahedral field: configuration is } e^4 t_2^4$$ CFSE = (-0.6 nₑ + 0.4 nt₂) Δₜ = (-0.6 × 4 + 0.4 × 4) Δₜ = -0.8 Δₜ$$\text{CFSE} = \left(-0.6 n_e + 0.4 n_{t_2}\right) \Delta_t = \left(-0.6 \times 4 + 0.4 \times 4\right) \Delta_t = -0.8 \Delta_t$$ μₛ = √(n(n+2)) = √(2(4)) = √(8) ≈ 2.83 BM$$\mu_s = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} \approx 2.83\text{ BM}$$ Core Logic Step 1: Paramagnetic [Ni(PPh₃)₂Cl₂]$[\text{Ni}(\text{PPh}_3)_2\text{Cl}_2]$ has tetrahedral geometry (sp³$sp^3$). Step 2: Evaluate Statements: Statement A: Tetrahedral complexes of type MA₂B₂$\text{MA}_2\text{B}_2$ 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: Ni²⁺ (3d⁸)$\text{Ni}^{2+} (3d^8)$ has 2 unpaired electrons μ = 2.83 BM$\implies \mu = 2.83\text{ BM}$. (Statement C is CORRECT) Statement D: CFSE for e⁴ t₂⁴$e^4 t_2^4$ configuration is -0.8Δₜ$-0.8\Delta_t$ (in PDF noted error comparison vs Δₒ$\Delta_o$). (Statement D is listed as INCORRECT in key) Statement E: Geometry is tetrahedral, matching Ni(CO)₄$\text{Ni}(\text{CO})_4$. (Statement E is CORRECT) Hence, A, B, and D are incorrect statements. Pattern Recognition Sees: Paramagnetic Ni(II)$\text{Ni(II)}$ 4-coordinate complex. Shortcut: Paramagnetism confirms tetrahedral geometry $\implies$ no geometrical isomerism possible for tetrahedral structures. Chapter Mix Class 12 Chemistry: Coordination Compounds More Coordination Compounds Questions — jee_main_2025_07_april_evening Practice all Coordination Compounds previous-year questions → The calculated spin-only magnetic moments of and respectively are: (1) The d-orbital electronic configuration of the complex among , , One mole of the octahedral complex compound gives 3 moles Given below are two statements: Statement I: A homoleptic octahedral Which one of the following complexes will have and ? The type of hybridization and the magnetic property of are: Find the compound 'A' from the following reaction sequences. 'X' is the number of acidic oxides among , , The metal ion whose electronic configuration is not affected by Match List-I with List-II. <div style="overflow-x: auto; margin: 1rem 0;"><table Identify the homoleptic complexes with odd number of d electrons When Ethane-1, 2-diamine is added progressively to an aqueous solution The conditions and consequence that favours the configuration in a Given below are two statements : Statement (I): In octahedral The number of unpaired electrons responsible for the paramagnetic nature Determine the total number of chemical species from the list An octahedral complex having molecular composition has two isomers A Match the coordination complexes listed in LIST-I with their geometric Consider the following low-spin complexes , , , and . A metal complex with a formula is involved in hybridisation. The number of paramagnetic metal complex species among , , The spin-only magnetic moment value of ion formed among Ni, The number of paramagnetic complexes among , , , , Total number of molecules/species from following which will be paramagnetic Consider the transition metal ions , , and and all Given below are two statements : Statement-I Hybridisation, shape and Identify the CORRECT set of details from the following: A. The wavelength of light absorbed for the following complexes are Consider a mixture 'X' which is made by dissolving of Given below are two statements: Statement I: ion will absorb Given below are two statements Statements-I: The number of paramagnetic Match List-I with List-II. <div style="overflow-x: auto; margin: 1rem 0;"><table Match List I with List II <div style="overflow-x: auto; margin: Given below are two statements: Statement I: Crystal Field Stabilization The statements that are incorrect about the nickel (II) complex Given below are two statements: Statement I: Among , , is a paramagnetic complex. Identify the INCORRECT statements about this Given below are two statements: Statement (I): A solution of Aluminium chloride in acidified aqueous solution forms an ion having Consider the following complex ions: The correct order of the Select the option with correct property - The correct increasing order of spin-only magnetic moment values of The correct statement among the following is : Which of the following complex is homoleptic? A chromium complex with a formula has a spin only SYSTEM Syllabus Analysis & Trend Mapping Dashboard Chapter Trends Formula Recognition Questions Feed Weightage Login Sign Up Dashboard Rankbit System Revise Mistakes Rankbit 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%) Report Issue Question Issue Type Wrong Answer Wrong Solution Formula Error Typo / Language Issue Diagram Issue Option Mismatch Duplicate Question Out of Syllabus Other Optional Notes Cancel Submit Report Quick Review Question Accuracy Feedback Accurate Needs Fix Difficulty Feedback Easy Medium Hard Confusion Score Low High Optional Notes Cancel Submit Review Welcome back Continue with Google or Sign In Don't have an account? Sign up Enter the 6-digit code we emailed to your email Verify & Continue Didn't get it? Resend code ← Back Enter your account email and we'll send you a link to reset your password. Send reset link ← Back to sign in