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JEE • CHEMISTRY • CO-ORDINATION COMPOUNDS

Co-ordination compounds practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 200 questions
31 Medium
31 of 200
Application +4 | -1
In all the following complexes, the coordination number of iron is six. In which of them is the oxidation state of iron the lowest?
View answer and explanation
Correct option: B
Fe(CO)6\mathrm{Fe}(\mathrm{CO})_6
Option B (Fe(CO)6\mathrm{Fe}(\mathrm{CO})_6) is correct because the oxidation state of iron in this complex is 0, which is the lowest among the given options.
32 Medium
32 of 200
Analyzing +4 | -1
Which of the following complex is optically inactive?
View answer and explanation
Correct option: A
[RhCl(CO)(PPh3)(NH3)]\left[{RhCl}({CO})\left({PPh}_3\right)\left({NH}_3\right)\right]
Option A is correct because [RhCl(CO)(PPh3)(NH3)]\left[{RhCl}({CO})({PPh}_3)({NH}_3)\right] is optically inactive due to the presence of a plane of symmetry.
33 Medium
33 of 200
Remembering +4 | -1
The donar sites of [EDTA]−4\mathrm{[EDTA}]^{-4} are:
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Correct option: C
Two N\mathrm{N} atoms and four O\mathrm{O} atoms
Option C (Two N atoms and four O atoms) is correct because EDTA has two nitrogen and four oxygen atoms that can donate electron pairs to form coordinate bonds.
34 Medium
34 of 200
Understanding +4 | -1
Which of the following complex is expected to be the most stable?
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Correct option: A
[Co(en)3]3+\left[{Co}({en})_3\right]^{3+}
Option A ([Co(en)3]3+\left[{Co}({en})_3\right]^{3+}) is the correct answer due to the chelation effect, which enhances the stability of the complex.
35 Medium
35 of 200
Remembering +4 | -1
The shape of Wilkinson’s catalyst is
View answer and explanation
Correct option: B
Square planar
Option B (Square planar) is correct because Wilkinson's catalyst, [(Ph3P)3RhCl][(Ph_3P)_3RhCl], has a square planar geometry.
36 Medium
36 of 200
Understanding +4 | -1
In [Cr(C2O4)3]3−{\left[ {\mathrm{Cr}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]^{3 - }}, the isomerism shown is:
View answer and explanation
Correct option: B
Optical
Option B is correct because the complex [Cr(C2O4)3]3−{\left[ {\mathrm{Cr}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]^{3 - }} is a tris-chelate complex of the type [M(AA)3]\left[ {\mathrm{M}{{\left( {\mathrm{AA}} \right)}_3}} \right] which exhibits optical isomerism.
37 Medium
37 of 200
Analyzing +4 | -1
Given below are two statements:
Statement I: [Ni(CN)4]2−{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }} is square planar and diamagnetic complex, with dsp2ds{p^2} hybridization for NiNi but [Ni(CO)4]\left[ {Ni{{\left( {CO} \right)}_4}} \right] is tetrahedral, paramagnetic and with sp3s{p^3}-hybridization for NiNi.
Statement II: [NiCl4]2−{\left[ {NiC{l_4}} \right]^{2 - }} and [Ni(CO)4]\left[ {Ni{{\left( {CO} \right)}_4}} \right] both have same dd-electron configuration, have same geometry and are paramagnetic.
In light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: B
Both Statement I and Statement II are false
Option B is correct. Statement I is false because Ni(CO)4 is diamagnetic, and Statement II is false because [NiCl4]2- is tetrahedral and paramagnetic, while Ni(CO)4 is tetrahedral and diamagnetic.
38 Medium
38 of 200
Analyzing +4 | -1
Assertion: [CoF6]3−\left[\mathrm{CoF}_6\right]^{3-} is high spin complex.

Reason: F−\mathrm{F}^{-} is strong field ligand.
View answer and explanation
Correct option: B
Assertion is true but Reason is false
Option B (Assertion is true but Reason is false) is correct because [CoF6]3- is indeed a high spin complex, but F- is a weak field ligand, not a strong field ligand.
39 Medium
39 of 200
Analyzing +4 | -1
Given below are two statements.

Statement-I: In CuSO4.5H2O\mathrm{CuSO}_4 .5 \mathrm{H}_2 \mathrm{O}, Cu−OCu-O bonds are present.

Statement-II: In CuSO4.5H2O\mathrm{CuSO}_4 .5 \mathrm{H}_2 \mathrm{O}, ligands coordinating with Cu(II)\mathrm{Cu}(\mathrm{II}) ion are O\mathrm{O}-and S\mathrm{S}-based ligands.

In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: C
Statement-I is correct but Statement-II is incorrect
Option C is correct because in CuSO4.5H2O\mathrm{CuSO}_4 .5 \mathrm{H}_2 \mathrm{O}, only oxygen atoms of water molecules coordinate with Cu(II)\mathrm{Cu}(\mathrm{II}) ion, not sulfur.
40 Medium
40 of 200
Remembering +4 | -1
Fe3+\mathrm{Fe}^{3+} cation gives a Prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
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Correct option: D
Fe4[Fe(CN)6]3\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3
Option D (Fe4[Fe(CN)6]3\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3) is correct because it represents the formula of Prussian blue, the precipitate formed when Fe3+\mathrm{Fe}^{3+} reacts with potassium ferrocyanide.
41 Medium
41 of 200
Understanding +4 | -1
In the complex K2[Fe(CN)6]{\mathrm{K}_2}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]
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Correct option: B
Both Fe{\mathrm{Fe}} atoms are in the same oxidation state
Option B (Both Fe{\mathrm{Fe}} atoms are in the same oxidation state) is correct because the complex should be K4[Fe(CN)6]{\mathrm{K}_4}[{\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}}] and Fe is in +2 oxidation state.
42 Medium
42 of 200
Remembering +4 | -1
Dimethyl glyoxime forms a coloured complex with
View answer and explanation
Correct option: B
NiNi
Option B (NiNi) is correct because dimethyl glyoxime (DMG) forms a characteristic red precipitate complex with nickel(II) ions.
43 Medium
43 of 200
Understanding +4 | -1
The complex cation which has two isomers is:
View answer and explanation
Correct option: C
[Co(NH3)5NO2]2+\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}
Option C ([Co(NH3)5NO2]2+[Co(NH_3)_5NO_2]^{2+}) is correct because it can exhibit linkage isomerism due to the ambidentate ligand NO2−NO_2^-, which can coordinate through either the nitrogen or the oxygen atom.
44 Medium
44 of 200
Analyzing +4 | -1
Which complex compound is most stable?
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Correct option: A
[CoCl2(en)2]NO3\left[\mathrm{CoCl}_2(\text {en})_2\right] \mathrm{NO}_3
Option A ([CoCl2(en)2]NO3) is correct because chelating ligands like ethylenediamine (en) form more stable complexes than monodentate ligands due to the chelate effect.
45 Medium
45 of 200
Application +4 | -1
The magnetic moment of an octahedral Co\mathrm{Co} (II) complex is 4.0 B.M (nearly) The electronic configuration of the complex is
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Correct option: A
t2g5eg2t_{2 g}{ }^5 e_g{ }^2
Option A (1:41:4) is correct because the ratio of diffusion rates is inversely proportional to the square root of the ratio of their molar masses.
46 Medium
46 of 200
Understanding +4 | -1
Type of isomerism which exists between [Pd(C6H5)2(SCN)2]\left[{Pd}\left({C}_6 {H}_5\right)_2({SCN})_2\right] and [Pd(C6H5)2(NCS)2]\left[{Pd}\left({C}_6 {H}_5\right)_2({NCS})_2\right] is :
View answer and explanation
Correct option: A
Linkage isomerism
The correct answer is A (Linkage isomerism) because the SCN ligand can coordinate through either the sulfur or the nitrogen atom.
47 Medium
47 of 200
Analyzing +4 | -1
Which among the following is incorrect regarding colour in coordination compounds?
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Correct option: D
Gradual addition of ethylene diamine to [Ni(H2O)6]2+\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} does not affect its colour observed
Option D is correct because the gradual addition of ethylenediamine to [Ni(H2O)6]2+\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} *does* affect the observed color due to ligand exchange.
48 Medium
48 of 200
Understanding +4 | -1
The IUPAC name of Ni(CO)4{Ni}({CO})_4 is :
View answer and explanation
Correct option: C
Tetracarbonylnickel(0)
Option C (Tetracarbonylnickel(0)) is correct because the carbonyl ligand is neutral, so the oxidation state of nickel is 0.
49 Medium
49 of 200
Remembering +4 | -1
The correct sequence of ligands in the order of decreasing field strength is:
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Correct option: A
CO>H2O>F−>S2−\mathrm{CO} > \mathrm{H}_2 \mathrm{O} > \mathrm{F}^{-} > \mathrm{S}^{2-}
Option A (CO>H2O>F−>S2−\mathrm{CO} > \mathrm{H}_2 \mathrm{O} > \mathrm{F}^{-} > \mathrm{S}^{2-}) is correct because it represents the correct decreasing order of ligand field strength according to the spectrochemical series.
50 Medium
50 of 200
Application +4 | -1
One mode of a complex compound Co(NH3)5Cl3\mathrm{Co}{\left( {\mathrm{N}{\mathrm{H}_3}} \right)_5}\mathrm{C}{\mathrm{l}_3} gives three moles of ions on dissolution in water. One of the same complex reacts with two moles of AgNO3  \mathrm{AgN}{\mathrm{O}_3}\mathrm{\;} solution to yield two moles of AgCl(s)\mathrm{AgCl}\left( s \right). The structure of the complex is
View answer and explanation
Correct option: D
[Co(NH3)5Cl]Cl2\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_2}
Option D ([Co(NH3)5Cl]Cl2) is correct because it gives three ions in solution and two moles of AgCl precipitate with two moles of AgNO3, indicating two ionizable chloride ions.
51 Medium
51 of 200
Application +4 | -1
Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?
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Correct option: C
[Ni(NH3)6]2+{\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}
Option C ([Ni(NH3)6]2+{\left[ {\mathrm{Ni}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{2 + }}) is correct because it forms an outer orbital complex with unpaired electrons, exhibiting paramagnetic behavior.
52 Medium
52 of 200
Application +4 | -1
Atomic numbers of   Cr  and  Fe\;Cr\,\,\mathrm{and}\,\,Fe are respectively 24 and 26. Which of the following is paramagnetic with the spin of the electron?
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Correct option: D
[Cr(NH3)6]3+{\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}
The correct answer is D ([Cr(NH3)6]3+{\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }}) because Cr³⁺ has a 3d³ configuration with 3 unpaired electrons, making it paramagnetic.
53 Medium
53 of 200
Analyzing +4 | -1
Which of the following does not have optical isomer?
View answer and explanation
Correct option: C
[Co(NH3)3Cl]\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{Cl}} \right]
The correct answer is C ([Co(NH3)3Cl3]) because complexes of the type MA3B3 do not exhibit optical isomerism.
54 Medium
54 of 200
Understanding +4 | -1
NO3−{NO}_3^{-}is detected by ring test. Ring formed has formula
View answer and explanation
Correct option: C
[Fe(H2O)5⋅NO]2+\left[{Fe}\left({H}_2 {O}\right)_5 \cdot {NO}\right]^{2+}
Option C is correct because the brown ring formed in the nitrate test is due to the formation of the complex ion [Fe(H2O)5⋅NO]2+\left[{Fe}\left({H}_2 {O}\right)_5 \cdot {NO}\right]^{2+}$.
55 Medium
55 of 200
Application +4 | -1
What is the magnetic moment of [FeF6]3−{\left[ {\mathrm{Fe}{\mathrm{F}_6}} \right]^{3 - }}?
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Correct option: A
5.92
Option A (5.92) is correct because the magnetic moment is calculated using the number of unpaired electrons in the complex ion, which is 5 for [FeF6]3-.
56 Medium
56 of 200
Understanding +4 | -1
In which of the following octahedral complexes of Co{\mathrm{Co}} (at. No. 27), will be magnitude of Δ0{\Delta _0} be the highest?
View answer and explanation
Correct option: A
[Co(CN)6]3−{\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}
Option A ([Co(CN)6]3−{\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}) is correct because cyanide (CN−^-) is a strong field ligand, resulting in a larger crystal field splitting energy (Δ0{\Delta _0}).
57 Medium
57 of 200
Application +4 | -1
The hybridization involved in complex [Ni(CN)4]2−{\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_\mathrm{4}}} \right]^{\mathrm{2 - }}} is: (Atomic number of Ni= 28\mathrm{Ni} = \,28)
View answer and explanation
Correct option: C
dsp2\mathrm{ds}{\mathrm{p}^2}
Option C (dsp2\mathrm{ds}{\mathrm{p}^2}) is correct because the complex is square planar, requiring dsp² hybridization due to the strong field ligand.
58 Medium
58 of 200
Application +4 | -1
Which one of the following complexes is diamagnetic?
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Correct option: B
[Ni(CN)4]2−\left[{Ni}({CN})_4\right]^{2-}
Option B ([Ni(CN)4]2−\left[{Ni}({CN})_4\right]^{2-}) is correct because it is a square planar complex with strong field ligands, resulting in pairing of electrons and thus diamagnetism.
59 Medium
59 of 200
Understanding +4 | -1
Yellow compound of lead chromate gets dissolved on treatment with hot NaOH\mathrm{NaOH} solution. The product of lead formed is a :
View answer and explanation
Correct option: D
Dianionic complex with coordination number four
Option D (Dianionic complex with coordination number four) is correct because lead chromate (PbCrO4PbCrO_4) reacts with hot NaOH to form the tetrahydroxoplumbate(II) ion, which is a dianionic complex, [Pb(OH)4]2−[Pb(OH)_4]^{2-}, with a coordination number of four.
60 Medium
60 of 200
Understanding +4 | -1
The magnetic behavior of Li2O,Na2O2\mathrm{Li}_{2} \mathrm{O}, \mathrm{Na}_{2} \mathrm{O}_{2} and KO2\mathrm{KO}_{2}, respectively, are
View answer and explanation
Correct option: B
Diamagnetic, diamagnetic and paramagnetic
The correct answer is B (Diamagnetic, diamagnetic and paramagnetic) because Li2O and Na2O2 are diamagnetic while KO2 is paramagnetic due to the presence of unpaired electrons in O2-.
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