Download JEE MCQs for JEE Chemistry: Coordination Compounds
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Chapter-wise Objective Questions: Coordination Compounds
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Multiple Choice Questions
(a) polymerisation
(b) oxidation
(c) reduction
(d) hydrogenation
Show Answer & Explanation
Answer: (a) polymerisation
Explanation:
1. It is a titanium-aluminium catalyst that polymerises alkenes such as ethene and propene.
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) \( \ce{Ni(CO)4} \)
(b) \( \ce{[Cu(en)2]^{2+}} \)
(c) \( \ce{[MnCl4]^{2-}} \)
(d) \( \ce{[Ag(NH3)2]^+} \)
Show Answer & Explanation
Answer: (b) \( \ce{[Cu(en)2]^{2+}} \)
Explanation:
1. Cu(II) (d⁹) with strong-field chelating en forms square planar dsp² complexes.
2. \( \ce{[Ag(NH3)2]^+} \) is linear, \( \ce{[MnCl4]^{2-}} \) and \( \ce{Ni(CO)4} \) are tetrahedral.
Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.
(a) \( \ce{[Fe(CN)6]^{3-}} \)
(b) \( \ce{[Fe(Cl)6]^{3-}} \)
(c) \( \ce{[Fe(OH)3]^{3-}} \)
(d) \( \ce{[Fe(H2O)6]^{3+}} \)
Show Answer & Explanation
Answer: (a) \( \ce{[Fe(CN)6]^{3-}} \)
Explanation:
1. Strong-field cyanide forms the most stable complex with Fe(III) by a large margin.
Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.
(a) sodium nitroferricyanide
(b) sodium pentacyanonitrosylferrate(III)
(c) sodium nitroferrocyanide
(d) sodium pentacyanonitrosylferrate(II)
Show Answer & Explanation
Answer: (d) sodium pentacyanonitrosylferrate(II)
Explanation:
1. NO is nitrosyl and carries +1 here. For \( \ce{[Fe(CN)5NO]^{2-}} \): x - 5 + 1 = -2, so Fe = +2.
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) \( \ce{K3[Fe(CN)6]} \)
(b) \( \ce{K4[Fe(CN)6]} \)
(c) \( \ce{[Cu(NH3)4]SO4} \)
(d) \( \ce{[Cr(H2O)6]Cl3} \)
Show Answer & Explanation
Answer: (b) \( \ce{K4[Fe(CN)6]} \)
Explanation:
1. EANs: \( \ce{K4[Fe(CN)6]} \) 36 (krypton), \( \ce{K3[Fe(CN)6]} \) 35, \( \ce{[Cr(H2O)6]^{3+}} \) 33, \( \ce{[Cu(NH3)4]^{2+}} \) 35.
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) hexaamminecobalt(III) chloride
(b) hexaamminecobalt(II) chloride
(c) triamminecobalt(III) trichloride
(d) none of these
Show Answer & Explanation
Answer: (a) hexaamminecobalt(III) chloride
Explanation:
1. Six ammine ligands, Co = +3 balancing three chlorides.
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) potassium ferrocyanide
(b) dimethylglyoxime (DMG)
(c) diphenylbenzidine
(d) resorcinol
Show Answer & Explanation
Answer: (b) dimethylglyoxime (DMG)
Explanation:
1. Dimethylglyoxime gives a bright red precipitate of nickel dimethylglyoximate in ammoniacal solution.
Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.
(a) \( \ce{[FeF6]^{3-}} \)
(b) \( \ce{[Ni(CN)4]^{2-}} \)
(c) \( \ce{[Cu(NH3)4]^{2+}} \)
(d) \( \ce{[Zn(NH3)4]^{2+}} \)
Show Answer & Explanation
Answer: (a) \( \ce{[FeF6]^{3-}} \)
Explanation:
1. Counting ligands: 6, 4, 4 and 4.
2. Only \( \ce{[FeF6]^{3-}} \) has six ligands.
Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.
(a) potassium trioxalatoaluminate(IV)
(b) potassium aluminium(III) trioxalate
(c) potassium trioxalatoaluminate(III)
(d) potassium alumino oxalate
Show Answer & Explanation
Answer: (c) potassium trioxalatoaluminate(III)
Explanation:
1. Oxalate is -2 each: 3 + x - 6 = 0 gives Al = +3.
2. The anionic complex is named trioxalatoaluminate(III).
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) sodium cobalt nitrite
(b) hexanitritocobalt(III) sodium
(c) sodium hexanitritocobaltate(III)
(d) sodium hexanitrocobaltate(III)
Show Answer & Explanation
Answer: (c) sodium hexanitritocobaltate(III)
Explanation:
1. ONO is nitrito-O (bonded through O), unlike nitro (bonded through N).
2. Co = +3, so the name is sodium hexanitritocobaltate(III).
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) lower ionisation potential of the metal ion
(b) lower basicity of the ligand
(c) higher ionic radius of the metal ion
(d) higher charge/size ratio of the metal ion
Show Answer & Explanation
Answer: (d) higher charge/size ratio of the metal ion
Explanation:
1. A small, highly charged metal ion attracts the ligand lone pairs more strongly, so the complex is more stable.
Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.
(a) -4
(b) -2
(c) +4
(d) +2
Show Answer & Explanation
Answer: (d) +2
Explanation:
1. The ion carries a charge of +2 since ammonia is neutral and copper is Cu(II).
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) 54
(b) 23
(c) 18
(d) 35
Show Answer & Explanation
Answer: (d) 35
Explanation:
1. Fe(III) has 23 electrons; six CN- add 12: 35.
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) \( \ce{[CdBr4]^{2-}} \)
(b) \( \ce{[CdCl4]^{2-}} \)
(c) \( \ce{[Cd(CN)4]^{2-}} \)
(d) \( \ce{[CdI4]^{2-}} \)
Show Answer & Explanation
Answer: (c) \( \ce{[Cd(CN)4]^{2-}} \)
Explanation:
1. Cyanide is a much stronger ligand than the halides, giving by far the most stable cadmium complex.
Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.
Reason (R): These ligands give linkage isomers.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false
Show Answer & Explanation
Answer: (a) Both A and R are true and R is the correct explanation of A
Explanation:
1. Each can bind through either of two donor atoms (N or O).
2. This produces linkage isomerism, so R explains A.
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) \( sp^3d^2 \) hybridised
(b) \( sp^3d \) hybridised
(c) \( dsp^2 \) hybridised
(d) \( sp^3 \) hybridised
Show Answer & Explanation
Answer: (c) \( dsp^2 \) hybridised
Explanation:
1. Square planar geometry corresponds to dsp² hybridisation.
Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.
(a) optical
(b) linkage
(c) geometrical
(d) ionisation
Show Answer & Explanation
Answer: (b) linkage
Explanation:
1. The ambidentate nitro/nitrito ligand gives linkage isomers.
Teacher's Note:
1. Identify the type of isomerism from what changes: the ligands' arrangement, their donor atom, or which ions are inside the coordination sphere.
2. For magnetism, find the metal's oxidation state and d-electron count, then decide high or low spin from the ligand field.
(a) CO is a gas
(b) \( \ce{Ni(CO)4} \) has zero oxidation number for Ni
(c) Ni is a metal
(d) \( \ce{Ni(CO)4} \) has oxidation number +4 for Ni
Show Answer & Explanation
Answer: (d) \( \ce{Ni(CO)4} \) has oxidation number +4 for Ni
Explanation:
1. CO is neutral, so in the neutral molecule \( \ce{Ni(CO)4} \) nickel is in the zero oxidation state.
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) \( \ce{Ti(OCOCH3)4} \)
(b) \( \ce{Ti(C2H5)4} \)
(c) \( \ce{Ti(OC6H5)4} \)
(d) \( \ce{Ti(OC2H5)4} \)
Show Answer & Explanation
Answer: (b) \( \ce{Ti(C2H5)4} \)
Explanation:
1. Only \( \ce{Ti(C2H5)4} \) has a direct metal-carbon bond; the others are bonded through oxygen.
Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.
(a) ethyl lithium
(b) lithium carbide
(c) lithium acetate
(d) lithium ethoxide
Show Answer & Explanation
Answer: (a) ethyl lithium
Explanation:
1. Ethyllithium has a direct Li-C bond.
Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.
(a) 6
(b) 4
(c) 2
(d) -4
Show Answer & Explanation
Answer: (b) 4
Explanation:
1. Cuprammonium sulphate is \( \ce{[Cu(NH3)4]SO4} \).
2. Four ammonia ligands are bonded to Cu.
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) \( \ce{F^-} \)
(b) \( \ce{CN^-} \)
(c) \( \ce{NH3} \)
(d) all
Show Answer & Explanation
Answer: (d) all
Explanation:
1. All three have lone pairs to donate, so all act as Lewis bases.
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
(a) \( \ce{[Zn(H2O)6]^{2+}} \)
(b) \( \ce{[Cu(H2O)6]^{2+}} \)
(c) \( \ce{[Cr(H2O)6]^{3+}} \)
(d) \( \ce{[Fe(H2O)6]^{2+}} \)
Show Answer & Explanation
Answer: (d) \( \ce{[Fe(H2O)6]^{2+}} \)
Explanation:
1. Unpaired electrons: Cr(III) d³ = 3, Fe(II) d⁶ (high spin, water is weak field) = 4, Cu(II) d⁹ = 1, Zn(II) d¹⁰ = 0.
Teacher's Note:
1. Identify the type of isomerism from what changes: the ligands' arrangement, their donor atom, or which ions are inside the coordination sphere.
2. For magnetism, find the metal's oxidation state and d-electron count, then decide high or low spin from the ligand field.
(a) \( dsp^2 \)
(b) \( sp^3d^2 \)
(c) \( d^2sp^3 \)
(d) \( dsp^3 \)
Show Answer & Explanation
Answer: (c) \( d^2sp^3 \)
Explanation:
1. CN- is a strong-field ligand; Fe(III) pairs its d electrons (t₂g⁵) and uses two inner 3d orbitals: d²sp³.
Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.
(a) basic in nature
(b) some acidic and others basic
(c) acidic in nature
(d) neither acidic nor basic
Show Answer & Explanation
Answer: (a) basic in nature
Explanation:
1. Ligands donate electron pairs, which is Lewis basic behaviour.
Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.
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Coordination Compounds Objective Questions & Solutions for JEE Chemistry
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FAQs
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