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JEE • CHEMISTRY • CO-ORDINATION COMPOUNDS

Co-ordination compounds practice · Mixed difficulty

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Showing 61-90 of 326 questions

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61 Medium
61 of 326
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.
62 Medium
62 of 326
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.
63 Easy
63 of 326
Application +4 | -1
Which of the following has least oxidation state of Fe\mathrm{Fe}?
View answer and explanation
Correct option: C
FeSO4(NH4)2SO4  .6H2O\mathrm{FeS}{\mathrm{O}_4}{\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{S}{\mathrm{O}_4}\mathrm{\;}.6{\mathrm{H}_2}\mathrm{O}
Option C (FeSO4(NH4)2SO4  .6H2O\mathrm{FeS}{\mathrm{O}_4}{\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{S}{\mathrm{O}_4}\mathrm{\;}.6{\mathrm{H}_2}\mathrm{O}) is correct because the oxidation state of Fe in this compound is +2, which is the lowest among the given options.
64 Hard
64 of 326
Analyzing +4 | -1
Select the correct statement(s).
View answer and explanation
Correct option: B
The octahedral complex formed by joining the CN−\mathrm{CN}^{-} ligands to Fe+3\mathrm{Fe}^{+3} ion has theoretical value of spin magnetic moment equal to 1.73 B.M.
Option B (The octahedral complex formed by joining the CN−\mathrm{CN}^{-} ligands to Fe+3\mathrm{Fe}^{+3} ion has theoretical value of spin magnetic moment equal to 1.73 B.M.) is correct because the complex is low spin with one unpaired electron.
65 Medium
65 of 326
Remembering +4 | -1
Fe3+\mathrm{Fe}^{3+} cation gives a Prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:
View answer and explanation
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.
66 Medium
66 of 326
Understanding +4 | -1
In the complex K2[Fe(CN)6]{\mathrm{K}_2}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]
View answer and explanation
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.
67 Medium
67 of 326
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.
68 Hard
68 of 326
Analyzing +4 | -1
The equation which is balanced and represents the correct product(s) is
View answer and explanation
Correct option: B
[CoCl(NH3)5]++5H+→CO2++5NH4++Cl−{[CoCl{(N{H_3})_5}]^ + } + 5{H^ + } \to C{O^{2 + }} + 5NH_4^ + + C{l^ - }
Option B is correct because it represents a balanced chemical equation where the ammine complex is destroyed by adding acid.
69 Hard
69 of 326
Analyzing +4 | -1
Assertion : [Co(H2O)6]+3→[Co(H2O)6]+2{\left[ {{\text{Co(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_{\text{6}}}} \right]^{ + 3}} \to {\left[ {{\text{Co(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_{\text{6}}}} \right]^{ + 2}} changes its colour on reduction.

Reason : Crystal field stabilisation energy increases on reduction.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because the color change is due to changes in d-orbital occupancy, not necessarily an increase in CFSE.
70 Easy
70 of 326
Remembering +4 | -1
Carnallite in solution in water shows the properties of
View answer and explanation
Correct option: A
K+,Mg2+,Cl−{\mathrm{K}^ + },\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{C}{\mathrm{l}^ - }
Option A (K+,Mg2+,Cl−{\mathrm{K}^ + },\mathrm{M}{\mathrm{g}^{2 + }},\mathrm{C}{\mathrm{l}^ - }) is correct because carnallite (KCl.MgCl2.6H2O) dissociates into these ions in water.
71 Medium
71 of 326
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.
72 Medium
72 of 326
Analyzing +4 | -1
Which complex compound is most stable?
View answer and explanation
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.
73 Easy
73 of 326
Analyzing +4 | -1
The possible number of isomers for the complex [MCl2Br2]SO4\left[ {\mathrm{MC}{\mathrm{l}_2}\mathrm{B}{\mathrm{r}_2}} \right]\mathrm{S}{\mathrm{O}_4} is:
View answer and explanation
Correct option: D
5
The correct answer is D (5) because the complex [MCl2Br2]SO4 can have five isomers: three geometrical isomers (cis, trans, and another trans) and two ionization isomers.
74 Hard
74 of 326
Analyzing +4 | -1
Assertion : Potassium ferro cyanide is diamagnetic whereas potassium ferricyanide is paramagnetic.

Reason : Crystal field spliting in ferrocyanide ion is greater than that of ferricyanide ion.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because while the assertion is true, the crystal field splitting is actually smaller in ferrocyanide than ferricyanide.
75 Medium
75 of 326
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
View answer and explanation
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.
76 Medium
76 of 326
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.
77 Easy
77 of 326
Remembering +4 | -1
[Co(NH3)6]Cl3\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_3} is called:
View answer and explanation
Correct option: A
Hexaammine cobalt (III) chloride
Option A (Hexaammine cobalt (III) chloride) is correct because it follows the IUPAC nomenclature rules for coordination complexes.
78 Medium
78 of 326
Analyzing +4 | -1
Which among the following is incorrect regarding colour in coordination compounds?
View answer and explanation
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.
79 Easy
79 of 326
Understanding +4 | -1
Geometrical isomerism is found in coordination compounds having coordination number:
View answer and explanation
Correct option: D
6
Option D (6) is correct because geometrical isomerism is commonly found in coordination compounds with coordination number 4 (square planar) and 6 (octahedral).
80 Easy
80 of 326
Understanding +4 | -1
The number of geometrical isomers of [Co(NH3)3(NO2)3]\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}{{\left( {\mathrm{N}{\mathrm{O}_2}} \right)}_3}} \right] are:
View answer and explanation
Correct option: B
2
Option B (2) is correct because the complex can exist as cis and trans isomers.
81 Medium
81 of 326
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.
82 Medium
82 of 326
Remembering +4 | -1
The correct sequence of ligands in the order of decreasing field strength is:
View answer and explanation
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.
83 Medium
83 of 326
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.
84 Easy
84 of 326
Application +4 | -1
The oxidation number of platinum in [Pt(NH3)5Cl]Cl3\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_3} is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the oxidation number of Pt is calculated by considering the charges of the ligands and the overall charge of the complex.
85 Medium
85 of 326
Application +4 | -1
Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?
View answer and explanation
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.
86 Medium
86 of 326
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?
View answer and explanation
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.
87 Hard
87 of 326
Analyzing +4 | -1
Which one of the following species doesn’t have a magnetic moment of 1.73 BM (spin only value)?
View answer and explanation
Correct option: B
CuICuI
Option B (CuICuI) is correct because CuICuI has Cu+Cu^+ which has a d10 configuration and hence no unpaired electrons, resulting in a magnetic moment of 0 BM, not 1.73 BM.
88 Medium
88 of 326
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.
89 Hard
89 of 326
Analyzing +4 | -1
Assertion :Cis−[Fe(en)2Cl2]+{\text{Cis}} - {\left[ {{\text{Fe(en}}{{\text{)}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}} \right]^ + } form racemic mixture.

Reason : Cis−[Fe(en)2Cl2]+{\text{Cis}} - {\left[ {{\text{Fe(en}}{{\text{)}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}} \right]^ + } is square planar complex.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct because the assertion is true, as cis-[Fe(en)2Cl2]+ can form a racemic mixture, but the reason is false since cis-[Fe(en)2Cl2]+ is an octahedral complex, not a square planar complex.
90 Medium
90 of 326
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+}$.
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