Download JEE MCQs for JEE Chemistry: Coordination Compounds
Review structured MCQ sets for JEE Chemistry Coordination Compounds. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Chapter-wise Objective Questions: Coordination Compounds
Navigate directly to the 50 objective questions for Coordination Compounds using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.
Multiple Choice Questions
(a) [CrCl3(H2O)3]·3H2O
(b) [CrCl2(H2O)4]Cl·2H2O
(c) [CrCl(H2O)5]Cl2·H2O
(d) [Cr(H2O)6]Cl3
Show Answer & Explanation
Answer: (d) [Cr(H2O)6]Cl3
Explanation:
1. All three chloride ions precipitate, so all are ionisable and outside the coordination sphere.
2. Hence the complex is [Cr(H2O)6]Cl3.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Cr(H2O)4Cl2]+
(b) [Pt(NH3)3Cl]
(c) [Co(NH3)6]3+
(d) [Co(CN)5(NC)]3-
Show Answer & Explanation
Answer: (a) [Cr(H2O)4Cl2]+
Explanation:
1. An octahedral Ma4b2 complex has cis and trans forms.
2. Ma6 and Ma5b cannot show geometrical isomerism, and a 3-coordinate Pt(II) species is not a normal square planar complex.
3. Answer: (a)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Cu(NH3)4]2+
(b) [Cu(H2O)4]2+
(c) [Cu(en)2]2+
(d) [Cu(CN)4]2-
Show Answer & Explanation
Answer: (c) [Cu(en)2]2+
Explanation:
1. Ethane-1,2-diamine (en) is a bidentate chelating ligand.
2. Chelate rings give extra stability (chelate effect), so [Cu(en)2]2+ is the most stable.
3. Answer: (c)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) Linkage isomerism
(b) Ionisation isomerism
(c) Coordination isomerism
(d) No isomerism
Show Answer & Explanation
Answer: (d) No isomerism
Explanation:
1. Isomers must have the same molecular formula.
2. These two compounds have different counter ions (Br and Cl), so their formulae differ and they are not isomers.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Co(en)3]3+
(b) trans-[Co(en)2Cl2]+
(c) cis-[Co(en)2Cl2]+
(d) Both [Co(en)3]3+ and cis-[Co(en)2Cl2]+
Show Answer & Explanation
Answer: (d) Both [Co(en)3]3+ and cis-[Co(en)2Cl2]+
Explanation:
1. Optical activity needs the absence of a plane of symmetry.
2. [Co(en)3]3+ and the cis form of [Co(en)2Cl2]+ are non-superimposable on their mirror images, while the trans form has a plane of symmetry.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [CoBrCl(NH3)4]Br
(b) [CoCl(NH3)4]Br2
(c) [CoBr2Cl(NH3)4]
(d) [CoBr2(NH3)4]Cl
Show Answer & Explanation
Answer: (d) [CoBr2(NH3)4]Cl
Explanation:
1. Only Cl- precipitates with Ag+, so Cl is outside the coordination sphere and both Br are inside.
2. This gives [CoBr2(NH3)4]Cl, which has two ions per formula unit.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) sp3
(b) sp2d
(c) sp3d2
(d) dsp2
Show Answer & Explanation
Answer: (d) dsp2
Explanation:
1. [Cu(NH3)4]2+ is square planar.
2. The textbook model promotes one \( 3d \) electron and uses dsp2 hybridisation.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Cr(H2O)6]3+
(b) [Fe(H2O)6]2+
(c) [Cu(H2O)6]2+
(d) [Zn(H2O)6]2+
Show Answer & Explanation
Answer: (b) [Fe(H2O)6]2+
Explanation:
1. Unpaired electrons: Cr3+ d3 = 3, Fe2+ d6 (high spin) = 4, Cu2+ d9 = 1, Zn2+ d10 = 0.
2. Fe2+ has the most, so highest paramagnetism.
3. Answer: (b)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Co(en)3]Cl3 - 3
(b) [Co(en)2Cl2]Cl - 2
(c) [Co(NH3)3Cl3] - 2
(d) [Pt(NH3)2Cl2] - 3
Show Answer & Explanation
Answer: (c) [Co(NH3)3Cl3] - 2
Explanation:
1. [Co(en)3]3+: 2 (a pair of enantiomers). [Co(en)2Cl2]+: 3 (trans \( + a \) cis pair).
2. [Co(NH3)3Cl3]: 2 (fac and mer). [Pt(NH3)2Cl2]: 2 (cis, trans).
3. Answer: (c)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) Tetrahedral and paramagnetic
(b) Tetrahedral and diamagnetic
(c) Square planar and diamagnetic
(d) Square planar and paramagnetic
Show Answer & Explanation
Answer: (d) Square planar and paramagnetic
Explanation:
1. The complex is [Cu(NH3)4]2+, square planar.
2. Cu2+ is d9 with one unpaired electron, so it is paramagnetic.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Co(NH3)4Cl2]+
(b) [Ni(en)(NH3)4]2+
(c) [Ni(C2O4)(en)2]0
(d) [Cr(SCN)2(NH3)4]+
Show Answer & Explanation
Answer: (d) [Cr(SCN)2(NH3)4]+
Explanation:
1. [Co(NH3)4Cl2]+ has 2 isomers, [Ni(en)(NH3)4]2+ has 1, [Ni(C2O4)(en)2] has 2 (optical).
2. In [Cr(SCN)2(NH3)4]+ cis/trans combined with \( S/N \) linkage options give 6 isomers.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Cr(H2O)6]3+
(b) trans-[CrCl2(ox)2]3-
(c) cis-[CrCl2(NH3)4]+
(d) cis-[CrCl2(ox)2]3-
Show Answer & Explanation
Answer: (d) cis-[CrCl2(ox)2]3-
Explanation:
1. Only the cis form of an M(AA)2b2 complex lacks a plane of symmetry.
2. cis-[CrCl2(NH3)4]+ is Ma4b2, which has planes of symmetry.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Fe(H2O)6]3+
(b) [Fe(C2O4)3]3-
(c) [Fe(NH3)6]3+
(d) [FeCl4]-
Show Answer & Explanation
Answer: (b) [Fe(C2O4)3]3-
Explanation:
1. Oxalate is a bidentate chelating ligand, forming three five-membered rings.
2. The chelate effect makes this complex the most stable.
3. Answer: (b)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) cis-Platin
(b) Zeise's salt
(c) Tollen's reagent
(d) Sodium nitroprusside
Show Answer & Explanation
Answer: (b) Zeise's salt
Explanation:
1. An organometallic compound has at least one metal-carbon bond to an organic group.
2. Zeise's salt K[PtCl3(C2H4)] has a Pt-ethene bond.
3. Answer: (b)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) d5 (octahedral, Δo > P)
(b) d8 (tetrahedral)
(c) d6 (high spin)
(d) d9 (octahedral)
Show Answer & Explanation
Answer: (c) d6 (high spin)
Explanation:
1. Unpaired electrons: low spin d5 = 1, d8 = 2, high spin d6 = 4, d9 = 1.
2. d6 high spin has the most.
3. Answer: (c)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) +2
(b) +3
(c) +1
(d) 0
Show Answer & Explanation
Answer: (c) +1
Explanation:
1. The brown ring is [Fe(H2O)5(NO)]2+.
2. NO is present as NO+, so \( x + 1 = +2 \) and \( x = +1 \).
3. Answer: (c)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) Both square planar
(b) Tetrahedral and square planar
(c) Both tetrahedral
(d) Square planar and tetrahedral
Show Answer & Explanation
Answer: (c) Both tetrahedral
Explanation:
1. Ni(0) in Ni(CO)4 is sp3, tetrahedral.
2. [Ni(PPh3)2Cl2] is a tetrahedral Ni(II) complex with weak field Cl- and bulky PPh3.
3. Answer: (c)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Ag(CN)(NH3)]
(b) Na2[Cd(NO2)4]
(c) [PtCl4I2]2-
(d) [PtCl(NH3)3][Au(CN)4]
Show Answer & Explanation
Answer: (c) [PtCl4I2]2-
Explanation:
1. [PtCl4I2]2- is octahedral of type Ma4b2, so cis and trans forms exist.
2. A linear complex and a tetrahedral complex of type M(NO2)4 cannot show it.
3. Answer: (c)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Cr(NH3)6][Co(CN)6]
(b) [Co(en)2Cl2]Cl
(c) [Pt(NH3)4][PtCl6]
(d) [Co(NH3)4Cl2]NO2
Show Answer & Explanation
Answer: (b) [Co(en)2Cl2]Cl
Explanation:
1. The cis form of [Co(en)2Cl2]+ has no plane of symmetry and exists as a pair of enantiomers.
2. The other options have symmetric structures.
3. Answer: (b)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) CuCl
(b) K3[Cu(CN)4]
(c) CuF2
(d) [Cu(CH3CN)4]BF4
Show Answer & Explanation
Answer: (c) CuF2
Explanation:
1. Colour from \( d-d \) transitions needs a partly filled \( d \) subshell.
2. Cu+ (d10) in the other three has none, while Cu2+ (d9) in CuF2 does.
3. Answer: (c)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) Geometry
(b) Magnetic nature
(c) Hybridisation
(d) Oxidation state
Show Answer & Explanation
Answer: (d) Oxidation state
Explanation:
1. [NiCl4]2- is tetrahedral, paramagnetic (sp3), but [Ni(CN)4]2- is square planar, diamagnetic (dsp2).
2. Both contain Ni in the \( +2 \) state.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) -0.6 Δo
(b) -1.8 Δo
(c) -1.6 Δo + P
(d) -1.2 Δo
Show Answer & Explanation
Answer: (a) -0.6 Δo
Explanation:
1. High spin d4 is t2g3eg1.
2. CFSE = \( 3(-0.4) + 1(+0.6) = -0.6 \) Δo.
3. Answer: (a)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Fe(CN)6]4-
(b) [Fe(CN)6]3-
(c) [Fe(H2O)6]2+
(d) [Fe(H2O)6]3+
Show Answer & Explanation
Answer: (b) [Fe(CN)6]3-
Explanation:
1. CN- gives a larger splitting than H2O.
2. Between the two cyanide complexes, Fe3+ (higher charge, smaller radius) has the larger Δo.
3. Answer: (b)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) A-R, B-P, C-R, D-R
(b) A-Q, B-P, C-P, D-P
(c) A-P, B-R, C-P, D-P
(d) A-Q, B-Q, C-P, D-P
Show Answer & Explanation
Answer: (b) A-Q, B-P, C-P, D-P
Explanation:
1. [Co(en)2Cl2]+ has 3 (trans + cis pair).
2. [Co(en)3]3+, square planar [Pt(NH3)2Cl2] and octahedral [Pt(NH3)3Cl3]+ (fac, mer) each have 2.
3. Answer: (b)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [CrCl2(CN)2(NH3)2]3+
(b) [CrCl2(CN)2(NH3)2]3-
(c) [CrCl2(CN)2(NH3)2]
(d) [CrCl2(CN)2(NH3)2]-
Show Answer & Explanation
Answer: (d) [CrCl2(CN)2(NH3)2]-
Explanation:
1. Charge = \( +3 + 2(-1) + 2(-1) + 0 = -1 \).
2. So the complex is the anion [CrCl2(CN)2(NH3)2]-, consistent with the suffix -ate.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) (η6-Benzene)tricarbonylchromium(0)
(b) Tricarbonyl(η6-benzene)chromate(0)
(c) Tricarbonyl(η6-benzene)chromium(0)
(d) (η6-Benzene)tricarbonylchromate(0)
Show Answer & Explanation
Answer: (a) (η6-Benzene)tricarbonylchromium(0)
Explanation:
1. Ligands are named alphabetically: benzene before carbonyl (prefix tri- is ignored).
2. The complex is neutral, so the metal is named chromium(0).
3. Answer: (a)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) Fe(η5-C5H5)2
(b) Cr(η6-C6H6)2
(c) K[Pt(η2-C2H4)Cl3]
(d) K[Pt(η2-C2H4)2Cl2]
Show Answer & Explanation
Answer: (c) K[Pt(η2-C2H4)Cl3]
Explanation:
1. Zeise's salt is potassium trichlorido(η2-ethene)platinate(II).
2. The other two named formulas are ferrocene and dibenzenechromium.
3. Answer: (c)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) [Ca(EDTA)]2-
(b) CaC2O4
(c) [Ca(C2O4)2]2-
(d) CaSO4·2H2O
Show Answer & Explanation
Answer: (a) [Ca(EDTA)]2-
Explanation:
1. EDTA is hexadentate, so it gives coordination number 6.
2. Two oxalate ligands would give only 4.
3. Answer: (a)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) d2sp3, octahedral, 4.9 BM
(b) sp3d2, octahedral, 4.9 BM
(c) dsp2, square planar, 4.9 BM
(d) sp3, tetrahedral, 4.9 BM
Show Answer & Explanation
Answer: (b) sp3d2, octahedral, 4.9 BM
Explanation:
1. Co is \( +3 \) (d6) and carbonate is a weak field ligand, so the complex is high spin.
2. Four unpaired electrons give \( \sqrt{4 x 6} = 4.9 \) BM, with sp3d2 octahedral geometry.
3. Answer: (b)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) With d2sp3 hybridisation [FeCl(CN)4(O2)]4- is diamagnetic
(b) [NiCl4]2- is more stable than [Ni(dmg)2]
(c) [V(CO)6] is not very stable and easily reduces to [V(CO)6]-
(d) Ligands such as CO, CN- and NO+ are pi-donors
Show Answer & Explanation
Answer: (c) [V(CO)6] is not very stable and easily reduces to [V(CO)6]-
Explanation:
1. [V(CO)6] has 17 valence electrons, so it readily gains one electron to reach the 18-electron anion.
2. The chelate [Ni(dmg)2] is more stable than [NiCl4]2-, and CO, CN-, NO+ are pi-acceptors.
3. Answer: (c)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
(a) 5
(b) 3
(c) 6
(d) None of these
Show Answer & Explanation
Answer: (d) None of these
Explanation:
1. The anion is [PtY6]2-, since \( 4 - 6 = -2 \).
2. So 2 H+ are needed and \( x = 2 \), which is not listed.
3. Answer: (d)
Teacher's Note:
1. Work through each option against the rule or structure involved before choosing. Remember the exceptions your textbook lists.
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FAQs
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