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NEET • CHEMISTRY • CO-ORDINATION COMPOUNDS

Co-ordination compounds practice · Mixed difficulty

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61 Easy
61 of 320
Application +4 | -1
Which compound is zero valent metal complex?
View answer and explanation
Correct option: C
[Ni(CO)4]\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]
Option C ([Ni(CO)4]\left[ {\mathrm{Ni}{{\left( {\mathrm{CO}} \right)}_4}} \right]) is correct because in this complex, the oxidation state of Ni is zero. CO is a neutral ligand, and the overall charge of the complex is zero.
62 Easy
62 of 320
Application +4 | -1
The IUPAC name of [CoCl(NO2)(en)2]Cl\left[ {\mathrm{CoCl}\left( {\mathrm{N}{\mathrm{O}_2}} \right){{\left( {\mathrm{en}} \right)}_2}} \right]\mathrm{Cl} is:
View answer and explanation
Correct option: C
Chloro-bisbis(ethylenediamine)nitrocobalt(III) chloride
The correct answer is C (Chloro-bisbis(ethylenediamine)nitrocobalt(III) chloride) because it follows the IUPAC naming conventions for coordination compounds.
63 Medium
63 of 320
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
The correct answer is D (Fe4[Fe(CN)6]3\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3) because Prussian blue is formed by the reaction of Fe3+\mathrm{Fe}^{3+} with ferrocyanide ions, resulting in this specific complex.
64 Easy
64 of 320
Application +4 | -1
The oxidation state of Cr\mathrm{Cr} in [Cr(NH3)4  Cl2]+ is{\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{\;C}{\mathrm{l}_2}} \right]^ + }\,\mathrm{is}
View answer and explanation
Correct option: D
3
Option D (3) is correct because the oxidation state of Cr in [Cr(NH3)4  Cl2]+{\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{\;C}{\mathrm{l}_2}} \right]^ + } is +3, calculated by considering the neutral NH3 ligands and the -1 charge of each Cl ligand, and the overall +1 charge of the complex.
65 Medium
65 of 320
Analyzing +4 | -1
The d−d -electron configurations of Cr2+,  Mn2+,  Fe2+\mathrm{C}{\mathrm{r}^{2 + }},\mathrm{\;M}{\mathrm{n}^{2 + }},\mathrm{\;F}{\mathrm{e}^{2 + }} and Co2+\mathrm{C}{\mathrm{o}^{2 + }} are d4,d5,d6  and  d7{d^4},{d^5},{d^6}\;and\;{d^7} respectively. Which one of the following will exhibit minimum paramagnetic behaviour?
(At. Nos.Cr=24,  Mn=25,  Fe=26,  Co=27\mathrm{Cr} = 24,\mathrm{\;Mn} = 25,\mathrm{\;Fe} = 26,\mathrm{\;Co} = 27)
View answer and explanation
Correct option: D
[Co(H2O)6]2+{\left[ {\mathrm{Co}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}
Option D ([Co(H2O)6]2+{\left[ {\mathrm{Co}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}) is correct because it has the least number of unpaired electrons after pairing in a weak field ligand environment.
66 Hard
66 of 320
Application +4 | -1
A metal complex having composition Cr(NH3)4Cl2Br\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2 \mathrm{Br} has been isolated in two forms A\mathrm{A} and B.AB . A reacts with AgNO3\mathrm{AgNO}_3 producing a white precipitate which was soluble in dilute ammonia solution. BB reacts with AgNO3\mathrm{AgNO}_3 producing a pale yellow precipitate soluble in concentrated ammonia solution. The formulae of AA and BB are respectively
View answer and explanation
Correct option: D
[Cr(NH3)4BrCl]Cl;[Cr(NH3)4Cl2]Br\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{BrCl}\right] \mathrm{Cl} ;\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Br}
Option D ([Cr(NH3)4BrCl]Cl;[Cr(NH3)4Cl2]Br\left[\mathrm{Cr}(\mathrm{NH}_3)_4 \mathrm{BrCl}\right] \mathrm{Cl} ;\left[\mathrm{Cr}(\mathrm{NH}_3)_4 \mathrm{Cl}_2\right] \mathrm{Br}) is correct because it correctly identifies which isomers will precipitate AgCl (soluble in dilute ammonia) and AgBr (soluble in concentrated ammonia).
67 Medium
67 of 320
Application +4 | -1
The oxidation number of Fe\mathrm{Fe} in [Fe(CN)6]4−,Co[\mathrm{Fe}(\mathrm{CN})_6]^{4-}, \mathrm{Co} in [Co(NH3)3(NO2)3]\left[\mathrm{Co}(\mathrm{NH}_3)_3\left(\mathrm{NO}_2\right)_3\right] and Ni\mathrm{Ni} in [Ni(CO)4][\mathrm{Ni}(\mathrm{CO})_4] are respectively
View answer and explanation
Correct option: A
+2,+3,0+2,+3,0
Option A (+2, +3, 0) is correct because it accurately calculates the oxidation states of Fe, Co, and Ni in the given complexes.
68 Easy
68 of 320
Analyzing +4 | -1
Assertion(A) Ni(CO)4{Ni}({CO})_4 is diamagnetic in nature.

Reason(R) Ni-atom undergoes sp3s p^3 hybridization.
View answer and explanation
Correct option: B
If both (A) and (R) are correct but (R) is not correct explanation for (A).
Option B (If both (A) and (R) are correct but (R) is not correct explanation for (A)) is correct because both the assertion and reason are true, but the reason does not explain why Ni(CO)4 is diamagnetic.
69 Easy
69 of 320
Understanding +4 | -1
[Co(NH3)5SO4]Br    and  [Co(NH3)5  Br]SO4\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{S}{\mathrm{O}_4}} \right]\mathrm{Br\;}\,\,\mathrm{and}\,\,\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{\;Br}} \right]\mathrm{S}{\mathrm{O}_4} are a pair of ….. Isomers.
View answer and explanation
Correct option: A
Ionisation
Option A (Ionisation) is correct because the given compounds exchange ions in solution, leading to different counter-ions.
70 Easy
70 of 320
Understanding +4 | -1
Each metal possesses:
View answer and explanation
Correct option: D
All of the above
The correct answer is D (All of the above) because metals in coordination complexes possess primary valencies satisfied by anions, secondary valencies satisfied by donor molecules, and a coordination number.
71 Medium
71 of 320
Understanding +4 | -1
Name the metal MM which is extracted on the basis of following reactions,
4M+8CN−+2H2O+O2→4[M(CN)2]−+4OH−4M + 8\mathrm{C}{\mathrm{N}^ - } + 2{\mathrm{H}_2}\mathrm{O} + {\mathrm{O}_2} \to 4{\left[ {M{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - } + 4\mathrm{O}{\mathrm{H}^ - }
2[M(CN)2]−+Zn→[Zn(CN)4]2−+2M2{\left[ {M{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - } + \mathrm{Zn} \to {\left[ {\mathrm{Zn}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }} + 2M:
View answer and explanation
Correct option: B
Silver
The correct answer is B (Silver). The given reactions are characteristic of the cyanide process used for the extraction of silver from its ore.
72 Easy
72 of 320
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 is KCl.MgCl2.6H2OKCl.MgCl_2.6H_2O, which dissociates into K+K^+, Mg2+Mg^{2+}, and Cl−Cl^- ions in water.
73 Easy
73 of 320
Application +4 | -1
The oxidation number of Fe\mathrm{Fe} in K4[Fe(CN)6] is{\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]\,\mathrm{is}
View answer and explanation
Correct option: C
2
The correct answer is 2 because the oxidation state of Fe in K4[Fe(CN)6]{\mathrm{K}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right] is +2.
74 Medium
74 of 320
Application +4 | -1
Show the coordination number of the metal ion, its oxidation number, the number of electrons in d−d -orbitals and the number of unpaired electrons d−d -orbitals respectively in complex [Co(H2O)4SO3]Cl.\left[ {\mathrm{Co}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}\mathrm{S}{\mathrm{O}_3}} \right]\mathrm{Cl}.
View answer and explanation
Correct option: C
5, 3, 6, 4
Option A is correct. The coordination number is 5, the oxidation number of Co is +3, the number of d-electrons is 6, and the number of unpaired electrons is 4.
75 Easy
75 of 320
Analyzing +4 | -1
[Fe(en)2(H2O)2]2++\left[\mathrm{Fe}({en})_2\left({H}_2 {O}\right)_2\right]^{2+}+ en →\rightarrow complex (X)({X}). The correct statement about the complex (X)(X) is-
View answer and explanation
Correct option: D
(1) and (2) both
Option D ((1) and (2) both) is correct because after the reaction, the complex becomes [Fe(en)3]2+\left[\mathrm{Fe}({en})_3\right]^{2+}, which is a low spin complex and diamagnetic due to the strong field ligand en.
76 Hard
76 of 320
Application +4 | -1
What is the ratio of uncomplexed to complexed Zn2+\mathrm{Zn}^{2+} ion [Zn(NH3)4]2+\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_4\right]^{2+} is 3×1093 \times 10^9 ?
View answer and explanation
Correct option: C
3.3×10−143.3 \times 10^{-14}
Option C (3.3×10−143.3 \times 10^{-14}) is correct because it represents the reciprocal of the formation constant divided by the concentration of the complex.
77 Medium
77 of 320
Understanding +4 | -1
Among the following pair of complexes, in which case Δ0{\Delta _0} value is higher for the second one
View answer and explanation
Correct option: C
[Co(H2O)6]+3  and  [Rh(H2O)6]+3{\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{ + 3}}\,\,and\,\,{\left[ {Rh{{\left( {{H_2}O} \right)}_6}} \right]^{ + 3}}
Option C ([Co(H2O)6]+3 and [Rh(H2O)6]+3{\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{ + 3}}\,and\,{\left[ {Rh{{\left( {{H_2}O} \right)}_6}} \right]^{ + 3}}) is correct because as we go down the group, the value of Δ0{\Delta _0} increases.
78 Medium
78 of 320
Analyzing +4 | -1
Which of the following does not show optical isomerism?
View answer and explanation
Correct option: C
[Co(NH3)3Cl3]0{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{C}{\mathrm{l}_3}} \right]^0}
Option C ([Co(NH3)3Cl3]0{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}\mathrm{C}{\mathrm{l}_3}} \right]^0}) is correct because this complex can exist as fac and mer isomers, but neither are chiral and thus do not show optical isomerism.
79 Medium
79 of 320
Analyzing +4 | -1
Assertion: In zieses salt coordination no. of PtPt is five.
Reason: Ethene is monodentate ligand.
View answer and explanation
Correct option: D
Assertion is false, Reason is true
Option D (Assertion is false, Reason is true) is correct because the coordination number of Pt in Zeise's salt is 3+1=4, and the reason correctly states that ethene is a monodentate ligand.
80 Medium
80 of 320
Remembering +4 | -1
In the lassigne’s lest for Nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of
View answer and explanation
Correct option: B
Fe4[Fe(CN)6]3F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}
Option B (Fe4[Fe(CN)6]3F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}) is correct because the Prussian blue color in Lassaigne's test is due to the formation of ferric ferrocyanide.
81 Medium
81 of 320
Remembering +4 | -1
The ligands in anti-cancer drug ciscis-platin are:
View answer and explanation
Correct option: A
NH3,Cl{NH}_3, {Cl}
Option A is correct. The ligands in cis-platin are ammonia (NH3NH_3) and chloride (ClCl).
82 Medium
82 of 320
Application +4 | -1
Which one of the following has lowest value of paramagnetic behaviour?
View answer and explanation
Correct option: D
[Co(CN)6]3−{\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}
Option D ([Co(CN)6]3−{\left[ {\mathrm{Co}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}) is correct because Co3+Co^{3+} in a strong field ligand environment like CN−CN^- results in a low-spin complex with no unpaired electrons, leading to the lowest paramagnetic behavior.
83 Easy
83 of 320
Analyzing +4 | -1
The pair of the compounds in which both the metals are in the higher possible oxidation state is
View answer and explanation
Correct option: A
CrO2Cl2,MnO4−\mathrm{CrO}_2 \mathrm{Cl}_2, \mathrm{MnO}_4^{-}
Option A (CrO2Cl2,MnO4−\mathrm{CrO}_2 \mathrm{Cl}_2, \mathrm{MnO}_4^{-}) is correct because Cr and Mn exhibit their highest possible oxidation states (+6 and +7, respectively) in these compounds.
84 Medium
84 of 320
Remembering +4 | -1
π\pi acid ligand among the following is
View answer and explanation
Correct option: D
CO{CO}
Option D (CO{CO}) is correct because CO is a π-acid ligand, meaning it can accept electron density from the metal through back-bonding into its π* antibonding orbitals.
85 Easy
85 of 320
Remembering +4 | -1
Estimation of calcium and magnesium is done by
View answer and explanation
Correct option: A
EDTA
Option A (EDTA) is correct because EDTA (ethylenediaminetetraacetic acid) is used in complexometric titrations to estimate the concentration of calcium and magnesium ions.
86 Hard
86 of 320
Analyzing +4 | -1
The total number of possible structural isomers of the compound [CuII(NH3)4][PtIICl4]\left[ {\mathrm{C}{\mathrm{u}^{\mathrm{II}}}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\left[ {\mathrm{P}{\mathrm{t}^{\mathrm{II}}}\mathrm{C}{\mathrm{l}_4}} \right] are:
View answer and explanation
Correct option: C
4
Option C (4) is correct because there are four possible linkage isomers for the given coordination compound.
87 Easy
87 of 320
Application +4 | -1
The IUPAC name of compound K3[Fe(CN)5NO]{\mathrm{K}_3}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_5}\mathrm{NO}} \right] is
View answer and explanation
Correct option: D
Potassium pentacyanonitrosyl ferrate (II)
Option D (Potassium pentacyanonitrosyl ferrate (II)) is correct because it correctly names the compound, indicating the potassium cation, the pentacyano and nitrosyl ligands, the ferrate anion, and the oxidation state of iron.
88 Hard
88 of 320
Analyzing +4 | -1
Among the following pairs of complexes, in which case the Δ0\Delta_0 value is higher for the first one?
View answer and explanation
Correct option: D
[Rh(H2O)6]3+\left[\mathrm{Rh}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} and [Co(H2O)6]3+\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}
Option D is correct because, for the same ligand, the Δ0\Delta_0 value increases down the group due to increased metal-ligand orbital overlap.
89 Easy
89 of 320
Remembering +4 | -1
[EDTA]4− is  a{\left[ {\mathrm{EDTA}} \right]^{4 - }}{\mkern 1mu} \mathrm{is}{\mkern 1mu} {\mkern 1mu} \mathrm{a}
View answer and explanation
Correct option: D
Hexadentate ligand
Option D (Hexadentate ligand) is correct because EDTA has six donor atoms that can bind to a central metal ion.
90 Hard
90 of 320
Analyzing +4 | -1
Which one of the following has largest number of isomers?
(R == alkyl group, en == ethylenediamine)
View answer and explanation
Correct option: D
[CO(en)2Cl2]+{\left[ {\mathrm{CO}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]^ + }
Option D is correct because [CO(en)2Cl2]+{\left[ {\mathrm{CO}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]^ + } exhibits both geometrical (cis-trans) and optical isomerism, leading to a larger number of isomers.
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