Chemistry
Coordination Chemistry
319 Questions
Coordination chemistry focuses on coordination compounds, their structures, and their magnetic properties. The questions cover primary and secondary valencies, ligand types, and geometries of complexes. This topic is heavily tested in chemistry competitive exams and requires a good grasp of molecular structures.
primary and secondary valencycomplex geometriesligand field theorymagnetic properties of complexesisomerism in complexes
Coordination Chemistry Questions
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[W(CO)6] and [Cr(NH3)6]3+
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[Co(NH3)6]3+ and [W(CO)6]
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[Fe(CN)6]3- and [Co(NH3)6]3+
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[Co(NH3)6]3+ and [Cr(NH3)6]3+
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[Cr(NH3)6]3+ and [Fe(CN)6]3-
E
Correct answer
Explanation
[Cr(NH3)6]3+: t2g3 eg0; 3 unpaired electrons; paramagnetic
[Fe(CN)6]3-: t2g5 eg0; one unpaired electron; paramagnetic
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Only 1
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Only 2
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Only 3
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Only 1 and 2
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Only 2 and 3
B
Correct answer
Explanation
Correct order of nephelauxetic effect is F− < H2O < NH3 < en < NCS− < Cl− < CN− < Br− < N3− < I−.
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[Zn(NH3)2Cl2]
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[Co(NH3)6]3+
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[Pt(NH3)2Cl2]
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[Ni(CN)4]2-
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All of the above
C
Correct answer
Explanation
[Pt(NH3)2Cl2] is a heteroleptic square planar complex of Ma2b2 type. In a square planar complex of formula [Ma2b2], the two similar ligands may be arranged adjacent to each other in a cis isomer or opposite to each other in a trans isomer.
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1 < 2 < 3 < 4
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1 < 3 < 2 < 4
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2 < 4 < 3 < 1
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4 < 2 < 3 < 1
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4 < 3 < 2 < 1
D
Correct answer
Explanation
The metal ions can be arranged in order of increasing Δ, and this order is largely independent of the identity of the ligand.
Mn2+ < Ni2+ < Co2+ < Fe2+ < V2+ < Fe3+ < Cr3+ < V3+ < Co3+
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[Mn(CN)6]3-, [Fe(CN)6]3- and [Co(C2O4)3]3- are diamagnetic inner orbital complexes that involve d2sp3 hybridisation.
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The ligand field strength increases in the order SCN- < C2O42- < NCS- < en < CO.
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[Mn2(CO)10] has one Mn-Mn bond and [Co2(CO)8] has one Co-Co bond.
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[MnCl6]3-, [FeF6]3- and [CoF6]3- are paramagnetic outer orbital complexes that involve sp3d2 hybridisation.
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M-C bond length in [Fe(CN)6]3- is lower than that in [Mn(CN)6]3-.
A
Correct answer
Explanation
Correct Answer: [Mn(CN)6]3-, [Fe(CN)6]3- and [Co(C2O4)3]3- are diamagnetic inner orbital complexes that involve d2sp3 hybridisation.
A
Correct answer
Explanation
In K4[Fe(CN)6], the primary valency corresponds to the oxidation state of the central metal ion. Since the total charge of the complex is 0 and CN is -1, Fe + 6(-1) = -4, so Fe = +2.
A
Correct answer
Explanation
The complex [Pt(NH3)2Cl2] exhibits geometrical isomerism, having two isomers: the cis-isomer and the trans-isomer.
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[CO (NH3)6] Cl3
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[CO (NH3)3 Cl3]
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[CO (NH3)4 Cl2] Cl
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[CO (NH3)5 Cl] Cl2
B
Correct answer
Explanation
A complex is non-ionisable if there are no counter-ions outside the coordination sphere. In [CO(NH3)3Cl3], all ligands are inside the coordination sphere, so it does not ionize in water.
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[Pt (NH3)2Cl2]
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[CO (en)2 Cl2] Cl
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[Cr (NH3)4Cl2] Cl
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[CO (NH3)5 NO2] Cl2
D
Correct answer
Explanation
Geometrical isomerism requires different spatial arrangements of ligands. [CO(NH3)5NO2]Cl2 is an octahedral complex where all positions are occupied by NH3 or NO2, and it does not allow for cis/trans arrangements that result in distinct isomers.
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Zeise's salt
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Ferrocene
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tetra ethyl lead
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dibenzene chromium
C
Correct answer
Explanation
Pi-bonded organometallic complexes involve pi-electron systems (like alkenes or aromatic rings) donating to the metal. Tetraethyl lead is a sigma-bonded organometallic compound.
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Ni (Co)4
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[NiCl4]2-
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[Ni (CN)6]2-
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[Cu (NH3)4]2+
D
Correct answer
Explanation
[Cu(NH3)4]2+ has a dsp2 hybridization, which corresponds to a square planar geometry.
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Linkage isomerism
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Geometrical isomerism
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Ionization isomerism
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Co-ordination isomerism
C
Correct answer
Explanation
These complexes exhibit ionization isomerism because they produce different ions in solution (Br- vs SO4^2-).
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Acetate
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Oxalate
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Cyanide
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Ammonia
B
Correct answer
Explanation
Oxalate (C2O4^2-) is a bidentate ligand that can bind to a metal ion through two oxygen atoms, forming a ring structure known as a chelate.
B
Correct answer
Explanation
In [Cr(H2O)4Cl2]+, the oxidation state of Cr is calculated as: Cr + 4(0) + 2(-1) = +1, so Cr = +3.