Skip to main content
Back to CO-ORDINATION COMPOUNDS

Your Progress

0 / 326 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • CO-ORDINATION COMPOUNDS

Co-ordination compounds practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 271-300 of 326 questions

Preparing practice and saved progress…

271 Medium
271 of 326
Analyzing +4 | -1
The incorrect statement among the following is
View answer and explanation
Correct option: D
Red colour of ruby is due to the presence of Co3+C{o^{3 + }}
Option D is the incorrect statement because the red color of ruby is due to the presence of Cr3+ ions, not Co3+ ions, in Al2O3.
272 Easy
272 of 326
Remembering +4 | -1
Chlorophyll is a coordination compound having central atom of:
View answer and explanation
Correct option: B
MgMg
Option B (MgMg) is correct because Chlorophyll is a coordination compound with Magnesium (MgMg) as its central atom.
273 Medium
273 of 326
Analyzing +4 | -1
Complex in which t2g6t_{2 g}^6 electronic configuration is NOT observed
View answer and explanation
Correct option: B
K3[Fe(CN)6]{K}_3\left[{Fe}({CN})_6\right]
Option B (K3[Fe(CN)6]{K}_3\left[{Fe}({CN})_6\right]) is correct because Fe3+{Fe}^{3+} has a d5d^5 configuration, resulting in t2g5t_{2g}^5 in a strong field, not t2g6t_{2g}^6.
274 Medium
274 of 326
Analyzing +4 | -1
Which of the following does not show optical isomerism?
View answer and explanation
Correct option: B
[CO( en )2Cl2]Cl\left[{CO}(\text { en })_2 {Cl}_2\right] {Cl}
The correct answer is A because [Cr(NH3)6][CO(CN)6] does not show optical isomerism due to the absence of chiral centers and planes of symmetry.
275 Medium
275 of 326
Application +4 | -1
How many electrons are present in 3d orbital of tetrahedral K2[NiCl4]{K}_2\left[{NiCl}_4\right] complex ?
View answer and explanation
Correct option: B
8 electrons
Option B (8 electrons) is correct because in the tetrahedral complex K2NiCl4{K}_2{NiCl}_4, Ni is in the +2 oxidation state with a d8 electronic configuration.
276 Medium
276 of 326
Understanding +4 | -1
The compound [CoCl3(C5H5 N)2]Br\left[\mathrm{CoCl}_3\left(\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}\right)_2\right] \mathrm{Br} will show the chemical test for which of the following ions?
View answer and explanation
Correct option: A
Br−\mathrm{Br}^{-}
Option A (Br−\mathrm{Br}^{-}) is correct because only the bromide ion (Br−\mathrm{Br}^{-}) is outside the coordination sphere and will thus give a positive chemical test.
277 Medium
277 of 326
Understanding +4 | -1
Low oxidation states are found in transition elements when a complex compound has :
View answer and explanation
Correct option: C
ligands capable of π\pi acceptor character as well as σ\sigma donor character
Option C is correct. Ligands with both π-acceptor and σ-donor character stabilize low oxidation states by synergic bonding.
278 Medium
278 of 326
Understanding +4 | -1
For the square planar complex [M(a)(b)(c)(d)][M\mathrm{ }(a)(b)(c)(d)] (where , M=M\mathrm{ } = central metal and a,b,c,a,\mathrm{ }b,\mathrm{ }c, and dd are monodentate ligands), the number of possible geometrical isomers are
View answer and explanation
Correct option: C
3
Option C (3) is correct because for a square planar complex with four different monodentate ligands, there are 3 possible geometrical isomers.
279 Medium
279 of 326
Remembering +4 | -1
Prussian blue is obtained by mixing together aqueous solution of Fe3+{Fe}^{3+} salt with
View answer and explanation
Correct option: B
Ferrocyanide
Option B (Ferrocyanide) is correct because Prussian blue, ferric ferrocyanide, is formed by the reaction of Fe3+{Fe}^{3+} ions with ferrocyanide ions.
280 Medium
280 of 326
Understanding +4 | -1
Which type of Isomerism in isomers [Co(NH3)5\left[{Co}\left({NH}_3\right)_5\right. (SO4)]Br\left.\left({SO}_4\right)\right] {Br} and [Co(NH3)5Br]SO4\left[{Co}\left({NH}_3\right)_5 {Br}\right] {SO}_4 ?
View answer and explanation
Correct option: B
Ionization
Option B (Ionization) is correct because the isomers differ in the ions they produce in solution, with one containing SO42−SO_4^{2-} outside the coordination sphere and the other containing Br−Br^- outside.
281 Easy
281 of 326
Understanding +4 | -1
The IUPAC name of the compound [CuCl2(CH3NH2)2[\mathrm{CuC}{\mathrm{l}_2}{\left( {\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}} \right)_2} is
View answer and explanation
Correct option: B
Dichloro bis (methyl amine) copper(II)
Option B is correct because it follows IUPAC naming conventions: dichloro (two chloride ligands), bis (two methylamine ligands), copper(II) (copper with a +2 oxidation state).
282 Easy
282 of 326
Application +4 | -1
What is the EAN of nickel in [Ni(CN)4]2−?{\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}?
View answer and explanation
Correct option: C
34
Option C (34) is correct because the Effective Atomic Number (EAN) is calculated by subtracting the oxidation state of the metal from its atomic number and adding twice the coordination number.
283 Easy
283 of 326
Remembering +4 | -1
According to postulates of Werner’s theory for coordination compounds, which of the following is true?
View answer and explanation
Correct option: A
Primary valencies are ionizable
Option A (Primary valencies are ionizable) is correct because, according to Werner's theory, primary valencies are ionizable and correspond to the oxidation state of the metal.
284 Medium
284 of 326
Understanding +4 | -1
An example of a sigma bonded organometallic compound is
View answer and explanation
Correct option: C
Grignard’s reagent
Option C (Grignard’s reagent) is correct because Grignard reagents (RMgX) have a direct sigma bond between the carbon atom of the alkyl group (R) and the magnesium atom.
285 Medium
285 of 326
Understanding +4 | -1
Which one group is trivalent in nature?
View answer and explanation
Correct option: A
Benzo
Option A (nucleophilic substitution reaction) is correct because ammonia acts as a nucleophile, substituting the halogen atom in the alkyl or benzyl halide.
286 Easy
286 of 326
Remembering +4 | -1
Which of the following is colorless?
View answer and explanation
Correct option: A
Ni(CO)4\mathrm{Ni}(\mathrm{CO})_4
Option A (Ni(CO)4\mathrm{Ni(CO)_4}) is correct because nickel tetracarbonyl is a colorless liquid.
287 Hard
287 of 326
Understanding +4 | -1
K4[Fe(CN)6]{K}_4\left[{Fe}({CN})_6\right] reacts with FeCl3{FeCl}_3 to form:
View answer and explanation
Correct option: B
KFe[Fe(CN)6]{KFe}\left[{Fe}({CN})_6\right]
Option B (KFeFe(CN)6{KFe}{Fe}({CN})_6) is correct because K4Fe(CN)6{K}_4{Fe}({CN})_6 reacts with FeCl3{FeCl}_3 to form Prussian blue, which then reacts with more K4Fe(CN)6{K}_4{Fe}({CN})_6 to form KFeFe(CN)6{KFe}{Fe}({CN})_6.
288 Medium
288 of 326
Analyzing +4 | -1
[Ni(CO)4]\left[{Ni}({CO})_4\right] and [Ni(CN)4]2−\left[{Ni}({CN})_4\right]^{2-} show similar
View answer and explanation
Correct option: C
Magnetic behaviour
Option C is correct because both complexes exhibit diamagnetic behavior due to the absence of unpaired electrons after pairing.
289 Medium
289 of 326
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 carbon monoxide (CO{CO}) is a π\pi-acid ligand, meaning it can accept electron density from the metal center through back-bonding.
290 Medium
290 of 326
Application +4 | -1
The magnetic moment of K3[Fe(CN)6]{\mathrm{K}_3}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right] is found to be 1.7 BM. How many unpaired electron (s) is/are present per molecule?
View answer and explanation
Correct option: A
1
Option A (1) is correct because the magnetic moment of 1.7 BM corresponds to approximately one unpaired electron in the complex.
291 Medium
291 of 326
Analyzing +4 | -1
Assertion :- 4-Methylhex-2-ene shows stereoisomerism.

Reason :- It can show Geometrical isomerism.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because 4-Methylhex-2-ene shows stereoisomerism (both geometrical and optical), and the reason is that it can show Geometrical isomerism, but it is not the complete explanation.
292 Medium
292 of 326
Analyzing +4 | -1
S-I: Complex compound [Ni(en)3]Cl2\left[{Ni}(e n)_3\right] {Cl}_2 has lower stability than [Ni(NH3)6]Cl2.\left[{Ni}\left({NH}_3\right)_6\right] {Cl}_2.

S-II: In [Ni(en)3]Cl2\left[{Ni}(e n)_3\right] {Cl}_2, geometry around Ni2+{Ni}^{2+} is octahedral.
View answer and explanation
Correct option: C
Statement-I is false, Statement-II is true
Statement-I is false and Statement-II is true because chelating ligands like ethylenediamine (en) provide more stability, and the geometry around Ni2+ in [Ni(en)3]Cl2 is indeed octahedral.
293 Hard
293 of 326
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}
The correct answer is D ([Cr(NH_3)_4BrCl]Cl ; [Cr(NH_3)_4Cl_2]Br) because it correctly identifies which ions are outside the coordination sphere based on their reaction with AgNO_3.
294 Medium
294 of 326
Application +4 | -1
Arrange the following complexes in increasing order toward the wavelength of light they absorb? Where M{M} is metal ion. [M(NH3)]3+=a,[M(CN)6]3−=b,[M(C2O4)3]3−=c\left[{M}\left({NH}_3\right)\right]^{3+}={a},\left[{M}({CN})_6\right]^{3-}={b},\left[{M}\left({C}_2 {O}_4\right)_3\right]^{3-}={c}, [MF6]3−=d\left[{MF}_6\right]^{3-}={d},
View answer and explanation
Correct option: C
b,a,c,db, a, c, d
Option A (d,c,a,bd, c, a, b) is incorrect. The correct order of increasing wavelength of light absorbed is b,a,c,db, a, c, d based on the spectrochemical series and crystal field splitting energy.
295 Medium
295 of 326
Analyzing +4 | -1
Assertion : Cu2+{Cu}^{2+} and Cd2+{Cd}^{2+} are separated from each other by first adding KCN{KCN} solution and then passing H2 S{H}_2 {~S} gas.

Reason : KCN{KCN} reduces Cu2+{Cu}^{2+} to Cu+{Cu}^{+}and forms a complex with it.
View answer and explanation
Correct option: B
If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
Option B is correct because both the assertion and reason are correct, but the reason doesn't fully explain the separation mechanism.
296 Medium
296 of 326
Analyzing +4 | -1
Which of the following is tetrahedral and paramagnetic complex?
View answer and explanation
Correct option: A
[NiCl4]2−\left[\mathrm{NiCl}_4\right]^{2-}
Option A ([NiCl4]2−\left[\mathrm{NiCl}_4\right]^{2-}) is correct because it is tetrahedral and has unpaired electrons, making it paramagnetic.
297 Medium
297 of 326
Analyzing +4 | -1
Among the following ions, which one has the highest unpaired electrons?
View answer and explanation
Correct option: C
[Fe(H2O)6]2+{\left[ {\mathrm{Fe}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}
The correct answer is C ([Fe(H2O)6]2+{\left[ {\mathrm{Fe}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}) because it has the highest number of unpaired electrons (4) among the given options.
298 Medium
298 of 326
Analyzing +4 | -1
Assertion: [Ni(CN)4]−2\left[{Ni}({CN})_4\right]^{-2} has zero unpaired electron while that of [NiCl4]−2\left[{NiCl}_4\right]^{-2} has two unpaired e−{e}^{-}

Reason: [Ni(CN)4]−2\left[{Ni}({CN})_4\right]^{-2} has strong crystal field while [NiCl4]−2\left[{NiCl}_4\right]^{-2} has weak crystal field.
View answer and explanation
Correct option: A
Assertion is true, Reason is true and Reason is correct explanation for Assertion
Option A (Assertion is true, Reason is true and Reason is correct explanation for Assertion) is correct because the cyanide ligand (CN−CN^−) is a strong field ligand causing pairing of electrons, while chloride (Cl−Cl^−) is a weak field ligand resulting in unpaired electrons.
299 Hard
299 of 326
Analyzing +4 | -1
Each of the compounds Pt(NH3)6Cl4⋅Cr(NH3)6Cl3,Co(NH3)4Cl3{Pt}\left({NH}_3\right)_6 {Cl}_4 \cdot {Cr}\left({NH}_3\right)_6 {Cl}_3, {Co}\left({NH}_3\right)_4 {Cl}_3 and K2Pt.Cl6{K}_2 {Pt} . {Cl}_6 were dissolved in water to make its 0.001 M0.001 \mathrm{~M} solution. The correct order of their increasing conductivity in solution is
View answer and explanation
Correct option: C
Co(NH3)4Cl3<K2Pt.Cl6<Cr(NH3)6Cl3<Pt(NH3)6Cl4Co(NH_3)_4 Cl_3 < K_2Pt.Cl_6 < Cr(NH_3)_6 Cl_3 < Pt(NH_3)_6 Cl_4
Option C is correct because the conductivity increases with the number of ions produced in solution, and the order reflects the increasing number of ions.
300 Medium
300 of 326
Application +4 | -1
Which of the following compounds has same oxidation state of the central metal atom in the cationic and anionic part?
View answer and explanation
Correct option: B
[Pt(py)4][PtCl4]\left[\mathrm{Pt}(\mathrm{py})_4\right]\left[\mathrm{PtCl}_4\right]
Option B ([Pt(py)4][PtCl4]) is correct because the oxidation state of Pt is +2 in both the cationic and anionic parts of the complex.
Revision notes and key takeaways

Review CO-ORDINATION COMPOUNDS in your textbook or class notes, then use this set to check your understanding.

Showing 271-300 of 326 questions

Your practice

0 / 326 answered · 326 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.