Coordination Compounds Online Mock Test: Practice SetBy admin / January 12, 2025 Uncategorized Rate this post 0% 0 Report a question What’s wrong with this question? You cannot submit an empty report. Please add some details. Coordination Compounds Online Mock Test: Practice Set 1 / 40 Match the complexes from List I with their corresponding type of isomerism from List II: A. A-II, B-III, C-IV, D-I B. A-I, B-IV, C-III, D-II C. A-II, B-IV, C-III, D-I D. A-II, B-III, C-IV, D-I A → linkage, B → ionization, C → coordination, D → solvate. 2 / 40 The hybridization involved in the complex [Ni(CN)₄]²⁻ is: A. dsp² B. sp³ C. d²sp² D. d²sp³ Explanation: Strong field CN⁻ → square planar → dsp² hybridization. 3 / 40 The expected number of unpaired electrons for the complex ion [Cr(NH₃)₆]³⁺ are: A. 2 B. 3 C. 4 D. 5 Explanation: Cr³⁺ → d³ configuration → 3 unpaired electrons. 4 / 40 The IUPAC name of the complex [Cr(en)₃]Cl₃ is: A. Triethylenediaminechromium (III) chloride B. Tris(ethylenediammine)chromium (II) chloride C. Tris(ethylenediamine)chromium (III) chloride D. Tris(ethylenediammine)chromium (II) chloride Explanation:en is a neutral bidentate ligand → “tris(ethylenediamine)”.Oxidation state of Cr = +3 → chromium (III). 5 / 40 The isomerism displayed by K₃[Co(NO₂)₆] and K₃[Co(ONO)₆] is: A. Linkage B. Co-ordination C. Ionization D. Geometrical Explanation: NO₂⁻ is ambidentate ligand → nitro and nitrito forms. 6 / 40 What is the correct electronic configuration of the central atom in K₄[Fe(CN)₆] based on crystal field theory? A. e⁴ t₂g² B. t₂g⁴ eg² C. t₂g⁶ eg⁰ D. e³ t₂g³ Fe²⁺ (d⁶) with strong field ligand CN⁻ forms low-spin complex. 7 / 40 Match List-I with List-II: A. (III), (II), (I), (IV) B. (II), (III), (IV), (I) C. (III), (II), (I), (IV) D. (IV), (III), (II), (I) Explanation:A → Linkage isomerismB → Coordination isomerismC → Ionization isomerismD → Solvate isomerism 8 / 40 The complex that is used as an anticancer agent is: A. mer-[Co(NH₃)₃Cl₃] B. cis-[PtCl₂(NH₃)₂] C. cis-K₂[PtCl₂Br₂] D. Na₂CoCl₄ Explanation: Cisplatin is an important anticancer drug. 9 / 40 The electronic configuration that represents the maximum magnetic moment is: A. d³ B. d² C. d⁸ D. d⁶ (high spin) Explanation: Maximum unpaired electrons = 4 → highest magnetic moment. 10 / 40 IUPAC name of [Pt(NH₃)₃(Br)(NO₂)Cl]Cl is: A. Triamminebromidochloridonitroplatinum(IV) chloride B. Triamminebromonitrochloroplatinum(IV) chloride C. Triamminechlorobromonitroplatinum(IV) chloride D. Triamminenitrochlorobromoplatinum(IV) chloride Explanation: Ligands arranged alphabetically: ammine, bromido, chlorido, nitro. 11 / 40 Given below are two statements:Statement I: [Co(NH₃)₆]³⁺ is a homoleptic complex, whereas [Co(NH₃)₄Cl₂]⁺ is a heteroleptic complex.Statement II: Complex [Co(NH₃)₆]³⁺ has only one kind of ligand but [Co(NH₃)₄Cl₂]⁺ has more than one kind of ligands. A. Both Statement I and Statement II are False B. Statement I is True but Statement II is False C. Statement I is False but Statement II is True D. Both Statement I and Statement II are True Explanation: Homoleptic → one type ligand; heteroleptic → more than one type. 12 / 40 The species that cannot act as an ambidentate ligand is: A. CN⁻ B. NO₂⁻ C. SCN⁻ D. NH₃ Explanation: NH₃ has only one donor atom (N). 13 / 40 The correct statement among the following regarding Fe(CO)₅ is: A. Tetranuclear B. Mononuclear C. Trinuclear D. Dinuclear Explanation: Fe(CO)₅ contains one Fe atom. 14 / 40 The correct IUPAC name for [CrF₂(en)₂]Cl is: A. Chlorodifluoridoethylenediaminechromium(III) chloride B. Bis-(ethylenediamine)difluoridochromium(III) chloride C. Difluorobis-(ethylenediamine)chromium(III) chloride D. Chlorodifluoridobis(ethylenediamine)chromium(III) Explanation: Ligands arranged alphabetically → difluorido + bis(ethylenediamine); Cr is +3. 15 / 40 The type of isomerism shown by the complex [CoCl₂(en)₂]⁺ is: A. Geometrical isomerism B. Coordination isomerism C. Ionization isomerism D. Linkage isomerism Explanation: Octahedral complex with bidentate ligand (en) shows cis-trans isomerism. 16 / 40 An ion, among the following, that has a magnetic moment of 2.84 BM is:(At. no. Ni = 28, Ti = 22, Cr = 24, Co = 27) A. Ni²⁺ B. Ti³⁺ C. Cr²⁺ D. Co²⁺ Explanation: Ni²⁺ has 2 unpaired electrons → μ ≈ √(n(n+2)) = √8 ≈ 2.84 BM. 17 / 40 The correct order of the stoichiometries of AgCl formed when AgNO₃ in excess is treated with the complexes CoCl₃·6NH₃, CoCl₃·5NH₃, CoCl₃·4NH₃ respectively is: A. 3AgCl, 1AgCl, 2AgCl B. 3AgCl, 2AgCl, 1AgCl C. 2AgCl, 3AgCl, 1AgCl D. 1AgCl, 3AgCl, 2AgCl Explanation: Number of ionizable Cl⁻ decreases with increasing coordinated NH₃. 18 / 40 The complex that can exhibit linkage isomerism is: A. [Co(NH₃)₅H₂O]Cl₃ B. [Co(NH₃)₅(NO₂)]Cl₂ C. Co(NH₃)₅NO₃ D. [Co(NH₃)₅Cl]SO₄ Explanation: NO₂⁻ is ambidentate ligand → linkage isomerism. 19 / 40 The low spin complex among the following is: A. [Fe(CN)₆]³⁻ B. [Co(NO₂)₆]³⁻ C. [Mn(CN)₆]³⁻ D. All of the above Explanation: CN⁻ and NO₂⁻ are strong field ligands → low spin. 20 / 40 [Cr(NH₃)₆]³⁺ is paramagnetic, while [Ni(CN)₄]²⁻ is diamagnetic because: A. Electrons in the 3d orbitals remain unpaired in [Ni(CN)₄]²⁻ B. Electrons in the 3d orbitals remain unpaired in [Cr(NH₃)₆]³⁺ C. Electrons in the 3p orbitals remain unpaired in [Cr(NH₃)₆]³⁺ D. Electrons in the 3p orbitals remain unpaired in [Ni(CN)₄]²⁻ Explanation: Cr³⁺ is d³ with unpaired electrons; [Ni(CN)₄]²⁻ is low-spin diamagnetic. 21 / 40 The correct electronic configuration of the central atom in K₄[Fe(CN)₆] based on crystal field theory is: A. e⁴t₂g² B. t₂g⁴eg² C. t₂g⁶eg⁰ D. e³t₂g³ Explanation: Fe²⁺ with strong field CN⁻ gives low-spin octahedral complex. 22 / 40 An ion that has magnetic moment 2.83 BM is:(Atomic Number: Ti = 22, Cr = 24, Mn = 25, Ni = 28) A. Ti³⁺ B. Ni²⁺ C. Cr³⁺ D. Mn²⁺ Explanation: Ni²⁺ has 2 unpaired electrons → μ = √8 ≈ 2.83 BM. 23 / 40 Iron carbonyl, Fe(CO)₅, is classified as: A. Tetranuclear B. Mononuclear C. Trinuclear D. Dinuclear Explanation: Only one Fe atom is present. 24 / 40 Optical isomerism is exhibited by among the following is/are: A. [Cr(Ox)₃]³⁺ B. [PtCl₂(en)₂]²⁺ C. [Cr(NH₃)₂Cl₂en]⁺ D. All of the above Explanation: All complexes can show chirality. 25 / 40 Ethylenediaminetetraacetate (EDTA) ion is: A. Bidentate ligand with two "N" donor atoms B. Tridentate ligand with three "N" donor atoms C. Hexadentate ligand with four "O" and two "N" donor atoms D. Unidentate ligand Explanation: EDTA binds through 6 donor atoms → hexadentate. 26 / 40 The correct statement, among the following, regarding [Mn(CN)₆]³⁻ is: A. It is sp³d² hybridised and octahedral in shape B. It is sp³d² hybridised and tetrahedral in shape C. It is d²sp³ hybridised and octahedral in shape D. It is dsp² hybridised and square planar in shape Explanation: CN⁻ is strong field ligand → low spin inner orbital complex. 27 / 40 Cobalt(III) chloride forms several octahedral complexes with ammonia. A compound among the following that does not give a test for chloride ions with silver nitrate at 25°C is: A. CoCl₃·3NH₃ B. CoCl₃·4NH₃ C. CoCl₃·5NH₃ D. CoCl₃·6NH₃ Explanation: All Cl⁻ are coordinated in [Co(NH₃)₆]Cl₃, so no free Cl⁻ to give AgCl test. 28 / 40 The complex ion among the following that cannot absorb visible light is: A. [Ni(CN)₄]²⁻ B. [Cr(NH₃)₆]³⁺ C. [Fe(H₂O)₆]²⁺ D. [Ni(H₂O)₆]²⁺ Explanation: It is square planar d⁸ diamagnetic complex with large splitting; does not absorb visible light significantly. 29 / 40 The geometry and magnetic properties of [NiCl₄]²⁻, respectively, are: A. Tetrahedral, Paramagnetic B. Tetrahedral, Diamagnetic C. Square planar, Paramagnetic D. Square planar, Diamagnetic Explanation: Weak field ligand Cl⁻ → high spin tetrahedral complex with unpaired electrons. 30 / 40 For a tetrahedral complex [MCl₄]²⁻, the spin-only magnetic moment is 3.83 B.M. The element M is: A. Co B. Cu C. Mn D. Fe Explanation: μ = 3.83 BM → 3 unpaired electrons → d⁶ high spin → Fe²⁺. 31 / 40 The correct splitting diagram of d orbitals in an octahedral crystal field is: 1 2 3 4 32 / 40 Ethylenediaminetetraacetate ion is a/an: A. Hexadentate ligand B. Ambidentate ligand C. Monodentate ligand D. Bidentate ligand Explanation: EDTA coordinates through 6 donor atoms. 33 / 40 A diamagnetic compound among the following is: A. [Co(F₆)]³⁻ B. [Ni(CN)₄]²⁻ C. [NiCl₄]²⁻ D. [Fe(CN)₆]³⁻ Explanation: CN⁻ is strong field → low spin d⁸ → all electrons paired → diamagnetic. 34 / 40 An anticancer agent among the following is: A. mer-[Co(NH₃)₃Cl] B. cis-[PtCl₂(NH₃)₂] C. cis-K₂[PtCl₂Br₂] D. NH₂CoCl₄ Explanation: Cisplatin is used in cancer chemotherapy. 35 / 40 A metal ion present in vitamin B₁₂ is: A. Fe(II) B. Co(III) C. Zn(II) D. Ca(II) Explanation: Vitamin B₁₂ contains cobalt metal ion. 36 / 40 Type of isomerism exhibited by compounds [Cr(H₂O)₆]Cl₃, [Cr(H₂O)₅Cl]Cl₂·H₂O, [Cr(H₂O)₄Cl₂]Cl·2H₂O and the value of coordination number (CN) of central metal ion in all these compounds, respectively, is: A. Geometrical isomerism, CN = 2 B. Optical isomerism, CN = 4 C. Ionisation isomerism, CN = 4 D. Solvate isomerism, CN = 6 Explanation: Difference in water inside/outside coordination sphere → solvate (hydrate) isomerism; CN = 6. 37 / 40 Which of the following is not an ambidentate ligand? A. C₂O₄²⁻ B. SCN⁻ C. NO₂⁻ D. CN⁻ Explanation: Oxalate is bidentate, not ambidentate. 38 / 40 The denticity of EDTA is: A. Monodentate B. Hexadentate C. Bidentate D. Tridentate Explanation: EDTA has 6 donor atoms. 39 / 40 Among the following, the compound that has the maximum coordination number is/are: A. [Co(NH₃)₆]³⁺ B. [Cr(C₂O₄)₃]³⁻ C. [CoCl₃(H₂O)₃] D. All of the above have the same coordination number Explanation: All complexes have coordination number 6. 40 / 40 The correct representation of Potassium tris(oxalato) chromate(III) is: A. K₂[Cr(C₂O₄)₃] B. K₃[Cr(C₂O₄)₃] C. K₃[Cr(C₂O₄)₂] D. K₂[Cr(C₂O₄)₄] Explanation: Cr³⁺ with three oxalato ligands gives complex charge −3. Your score isThe average score is 0% 0% Restart quiz Post Views: 37
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