Download JEE MCQs for JEE Chemistry: Coordination Compounds
Access targeted multiple-choice questions for Coordination Compounds designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.
Chapter-wise Objective Questions: Coordination Compounds
Access the complete set of multiple-choice questions for Coordination Compounds below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.
Multiple Choice Questions
(a) \( \ce{[Pt(NH3)(Br)(Cl)(Py)]} \) : tetrahedral, EAN = 36
(b) \( \ce{[MnCl4]^{2-}} \) : tetrahedral, zero CFSE
(c) \( \ce{[Fe(CO)2(NO)2]} \) : tetrahedral, EAN = 36
(d) \( \ce{[Co(CO)4]-} \) : tetrahedral, C-O bond longer than in CO
Show Answer & Explanation
Answer: (a) \( \ce{[Pt(NH3)(Br)(Cl)(Py)]} \) : tetrahedral, EAN = 36
Explanation:
1. Pt2+ (d8) with these ligands is square planar, not tetrahedral (and EAN is 84).
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{[Mn(CO)5]} \)
(b) \( \ce{[Fe(CO)5]} \)
(c) \( \ce{[Mn(CO)6]} \)
(d) \( \ce{[Co(CO)4]} \)
Show Answer & Explanation
Answer: (b) \( \ce{[Fe(CO)5]} \)
Explanation:
1. Fe(CO)5: 26 + 10 = 36 (krypton). The others have 35, 37 and 35.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) linkage isomers
(b) coordination isomers
(c) ligand isomers
(d) solvate isomers
Show Answer & Explanation
Answer: (c) ligand isomers
Explanation:
1. The ligands 2-methylpropane-1,3-diamine and butane-1,4-diamine are isomers (both C4H12N2), so the complexes are ligand isomers.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) All of these
(b) The hybridisation of iron is sp3d2 only
(c) Iron and nitrosyl give iron an oxidation state of +1 only
(d) Its magnetic moment is 3.87 BM only
Show Answer & Explanation
Answer: (a) All of these
Explanation:
1. Fe(+1) with NO+ is d7 with three unpaired electrons (3.87 BM), outer orbital sp3d2. All statements hold.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{[PtCl3(C2H4)]-} \)
(b) \( \ce{[CuBr2Cl2]^{2-}} \)
(c) \( \ce{[Co(C2O4)3]^{3-}} \)
(d) \( \ce{[Cr(NH3)2(en)2]^{3+}} \)
Show Answer & Explanation
Answer: (d) \( \ce{[Cr(NH3)2(en)2]^{3+}} \)
Explanation:
1. [Cr(NH3)2(en)2]3+ has trans (1) plus cis (2 enantiomers) = 3. [Co(ox)3]3- has 2, and the other two have 1.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{[Pt(NH3)3Cl]+} \)
(b) \( \ce{[PtCl2(NH3)2]} \)
(c) \( \ce{[PdCl2BrI]} \)
(d) \( \ce{[Pt(NH3)2(Cl)(Br)]} \)
Show Answer & Explanation
Answer: (a) \( \ce{[Pt(NH3)3Cl]+} \)
Explanation:
1. Square planar Ma3b has only one arrangement.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) +2
(b) +4
(c) +6
(d) +3
Show Answer & Explanation
Answer: (d) +3
Explanation:
1. en is neutral, the bridging groups carry -2 (NH) and -1 (OH). For the cation: 2x - 3 = +3, so x = +3.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{[Al(C2O4)3]^{3-}} \) and \( \ce{[BF4]-} \)
(b) \( \ce{[Al(C2O4)3]} \) and \( \ce{[BF4]^{3-}} \)
(c) \( \ce{[Al(C2O4)3]^{3+}} \) and \( \ce{[BF4]^{3+}} \)
(d) \( \ce{[Al(C2O4)3]^{2-}} \) and \( \ce{[BF4]^{2-}} \)
Show Answer & Explanation
Answer: (a) \( \ce{[Al(C2O4)3]^{3-}} \) and \( \ce{[BF4]-} \)
Explanation:
1. Al3+ + 3(-2) = -3 and B3+ + 4(-1) = -1.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) The complex ion is square planar only
(b) All of these
(c) Pt is in +2 oxidation state only
(d) The ethene molecule is perpendicular to the PtCl3 plane only
Show Answer & Explanation
Answer: (b) All of these
Explanation:
1. All three: square planar around Pt(II), and the C=C axis perpendicular to the PtCl3 plane.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) Linkage
(b) Ionisation
(c) Geometrical
(d) Optical
Show Answer & Explanation
Answer: (a) Linkage
Explanation:
1. There is no ambidentate ligand, so linkage isomerism is not possible. Ionisation (Br- with a ligand), geometrical and optical isomerism are all possible.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) 8
(b) 12
(c) 14
(d) 16
Show Answer & Explanation
Answer: (d) 16
Explanation:
1. Four Cu-N coordinate sigma bonds plus 12 N-H sigma bonds = 16.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{[Cr(H2O)6]Cl3} \)
(b) \( \ce{[Cr(H2O)3Cl3]} \)
(c) \( \ce{[Cr(H2O)4Cl2]Cl} \)
(d) \( \ce{[Cr(H2O)5Cl]Cl2} \)
Show Answer & Explanation
Answer: (a) \( \ce{[Cr(H2O)6]Cl3} \)
Explanation:
1. Only [Cr(H2O)6]Cl3 has all three chloride ions outside the coordination sphere.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{[Pd(PPh3)2(NCS)2]} \) and \( \ce{[Pd(PPh3)2(SCN)2]} \)
(b) \( \ce{[Co(NH3)5NO3]SO4} \) and \( \ce{[Co(NH3)5SO4]NO3} \)
(c) \( \ce{[PtCl2(NH3)4]Br2} \) and \( \ce{[PtBr2(NH3)4]Cl2} \)
(d) \( \ce{[Cu(NH3)4][PtCl4]} \) and \( \ce{[Pt(NH3)4][CuCl4]} \)
Show Answer & Explanation
Answer: (a) \( \ce{[Pd(PPh3)2(NCS)2]} \) and \( \ce{[Pd(PPh3)2(SCN)2]} \)
Explanation:
1. NCS- and SCN- are the two binding modes of the thiocyanate ambidentate ligand.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{[Cr(CO)6]} \)
(b) \( \ce{[V(CO)6]-} \)
(c) \( \ce{[Mn(CO)6]+} \)
(d) \( \ce{[Ti(CO)6]^{2-}} \)
Show Answer & Explanation
Answer: (c) \( \ce{[Mn(CO)6]+} \)
Explanation:
1. A positive charge on the metal reduces back-donation into the CO pi* orbital, so the C-O bond stays closest to a triple bond.
Teacher's Note:
1. More electron-rich metal means more back bonding and a weaker C-O bond.
(a) \( \ce{[Co(CN)6]^{3-}} \)
(b) \( \ce{[Ag(CN)2]-} \)
(c) \( \ce{[Cu(CN)4]^{3-}} \)
(d) \( \ce{[Co(CN)6]^{4-}} \)
Show Answer & Explanation
Answer: (d) \( \ce{[Co(CN)6]^{4-}} \)
Explanation:
1. Co(II) in the low spin d7 complex readily loses its eg electron to form Co(III), a very strong reducing agent.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{K3[Co(NO2)6]} \)
(b) \( \ce{Co(NO2)3.2KNO2} \)
(c) \( \ce{K[Co(NO2)4]} \)
(d) \( \ce{Co(NO2)3.3KNO2} \)
Show Answer & Explanation
Answer: (a) \( \ce{K3[Co(NO2)6]} \)
Explanation:
1. K3[Co(NO2)6] gives 3 K+ and one complex ion: 4 particles with total charge 3 + 3 = 6.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{[Co(en)2SO4]} \)
(b) \( \ce{[Co(en)3]SO4} \)
(c) \( \ce{[Co(en)3]2(SO4)3} \)
(d) \( \ce{[Co(en)3](SO4)3} \)
Show Answer & Explanation
Answer: (c) \( \ce{[Co(en)3]2(SO4)3} \)
Explanation:
1. The cation is [Co(en)3]3+ and sulphate is 2-, so the ratio is 2 : 3.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{[Zn(NH3)6]^{2+}} \)
(b) \( \ce{[Ni(NH3)6]^{2+}} \)
(c) \( \ce{[Cr(NH3)6]^{3+}} \)
(d) \( \ce{[Co(NH3)6]^{3+}} \)
Show Answer & Explanation
Answer: (d) \( \ce{[Co(NH3)6]^{3+}} \)
Explanation:
1. Co3+ (d6) with strong field NH3 is low spin d2sp3 and diamagnetic. The Zn complex is outer orbital and the Ni and Cr complexes are paramagnetic.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) [Sc(H2O)6]3+ and [Ti(H2O)6]3+ are both colourless
(b) Only [Co(NH3)4Br2]Cl shows isomerism
(c) None of these
(d) [Co(NH3)4Br2]Cl shows ionisation and geometrical isomers, and [Pd(NO2)2(NH3)2] is square planar showing geometrical and linkage isomers
Show Answer & Explanation
Answer: (d) [Co(NH3)4Br2]Cl shows ionisation and geometrical isomers, and [Pd(NO2)2(NH3)2] is square planar showing geometrical and linkage isomers
Explanation:
1. Ti3+ is d1 and hence coloured, so the first statement is false. The first listed statement holds true for both complexes.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{CoCl3.4NH3} \) > \( \ce{CoCl3.5NH3} \) > \( \ce{CoCl3.6NH3} \)
(b) \( \ce{CoCl3.5NH3} \) > \( \ce{CoCl3.6NH3} \) > \( \ce{CoCl3.4NH3} \)
(c) \( \ce{CoCl3.6NH3} \) > \( \ce{CoCl3.5NH3} \) > \( \ce{CoCl3.4NH3} \)
(d) \( \ce{CoCl3.6NH3} \) > \( \ce{CoCl3.4NH3} \) > \( \ce{CoCl3.5NH3} \)
Show Answer & Explanation
Answer: (c) \( \ce{CoCl3.6NH3} \) > \( \ce{CoCl3.5NH3} \) > \( \ce{CoCl3.4NH3} \)
Explanation:
1. They ionise to 4, 3 and 2 ions respectively: [Co(NH3)6]Cl3, [Co(NH3)5Cl]Cl2 and [Co(NH3)4Cl2]Cl.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{Ni^{2+}} \) : t2g6 eg2, CFSE = -1.6 Delta0
(b) \( \ce{Mn^{2+}} \) : t2g3 eg2, CFSE = 0
(c) \( \ce{V^{2+}} \) : t2g3 eg0, CFSE = -1.2 Delta0
(d) \( \ce{Ti^{2+}} \) : t2g2 eg0, CFSE = -0.8 Delta0
Show Answer & Explanation
Answer: (a) \( \ce{Ni^{2+}} \) : t2g6 eg2, CFSE = -1.6 Delta0
Explanation:
1. Ni2+ (d8): 6(-0.4) + 2(0.6) = -1.2 Delta0, not -1.6.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) amount of AgCl precipitated only
(b) conductivity measurement only
(c) heating with conc. H2SO4 only
(d) all of these methods
Show Answer & Explanation
Answer: (d) all of these methods
Explanation:
1. They give different amounts of AgCl, different conductivities and lose different amounts of water with conc. H2SO4.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) oxidation of Co
(b) reduction of Co only
(c) dimerisation only
(d) both reduction of Co and dimerisation
Show Answer & Explanation
Answer: (d) both reduction of Co and dimerisation
Explanation:
1. Co(CO)4 has an odd electron. It can gain an electron to give [Co(CO)4]- (36) or dimerise to Co2(CO)8 with a Co-Co bond.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{Co^{2+}} \)
(b) \( \ce{Fe^{2+}} \)
(c) \( \ce{Mn^{2+}} \)
(d) \( \ce{Fe^{3+}} \)
Show Answer & Explanation
Answer: (b) \( \ce{Fe^{2+}} \)
Explanation:
1. Fe2+ (d6) has four unpaired electrons (4.9 BM) in a high spin complex and zero in a low spin complex.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) \( \ce{[Fe(C2O4)3]^{3-}} \)
(b) \( \ce{[Fe(CN)6]^{3-}} \)
(c) \( \ce{[Fe(H2O)6]^{3+}} \)
(d) \( \ce{[Fe(CN)6]^{4-}} \)
Show Answer & Explanation
Answer: (b) \( \ce{[Fe(CN)6]^{3-}} \)
Explanation:
1. Fe3+ is a harder, higher charged ion than Fe2+ and forms the most stable cyano complex.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
(a) 2, 2
(b) 4, 2
(c) 3, 2
(d) 2, 3
Show Answer & Explanation
Answer: (a) 2, 2
Explanation:
1. Ma2b4 octahedral: cis and trans. Square planar Ma2b2: cis and trans.
Teacher's Note:
1. Work out the oxidation state, coordination number and geometry step by step.
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Multiple Choice Questions (MCQs) for JEE Chemistry Coordination Compounds
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