Skip to main content
Back to CO-ORDINATION COMPOUNDS

Your Progress

0 / 320 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • 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

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 271-300 of 320 questions

Preparing practice and saved progress…

271 Medium
271 of 320
Analyzing +4 | -1
Consider the following statements
I. Chain and position isomerism are not possible together between two isomers
II. Tautomerism is a chemical phenomenon which is catalysed by acid as well as base
III. Tautomers are always metamers
IV. Tautomers are always functional isomers
Select the correct answer by using the codes given below
View answer and explanation
Correct option: D
I, II and IV are correct
Option D (I, II and IV are correct) is correct because chain and position isomerism are mutually exclusive, tautomerism is acid/base catalyzed, and tautomers are functional isomers.
272 Medium
272 of 320
Remembering +4 | -1
The common name of K[PtCl3(n2C2H4)]{K}\left[{PtCl}_3\left(n^2 {C}_2 {H}_4\right)\right.] is
View answer and explanation
Correct option: B
Zeise's salt
Option B (Zeise's salt) is correct because KPtCl3(n2C2H4){K}{PtCl}_3\left(n^2 {C}_2 {H}_4\right) is the chemical formula for Zeise's salt, a classic example of an organometallic compound.
273 Hard
273 of 320
Analyzing +4 | -1
A certain complex ion has the formula [M(en)2Br2]+\left[M(\mathrm{en})_2 \mathrm{Br}_2\right]^{+}where MM is the metal ion and en == ethylene diamine. How many isomers are possible for this ?
View answer and explanation
Correct option: A
3
Option A (3) is correct because the complex can exist as cis, trans, and mer isomers due to the arrangement of the ligands around the metal center.
274 Hard
274 of 320
Understanding +4 | -1
In an octahedral geometry, the pair of d{d} orbitals involved in d2sp3{d}^2 {sp}^3 hybridization is:
View answer and explanation
Correct option: A
dx2−y2, dz2{d}_{{x}^2-{y}^2}, {~d}_{{z}^2}
The correct answer is A (dx2−y2, dz2{d}_{{x}^2-{y}^2}, {~d}_{{z}^2}). In d2sp3{d}^2 {sp}^3 hybridization, the d{d} orbitals involved are dx2−y2{d}_{{x}^2-{y}^2} and dz2{d}_{{z}^2} because they point directly along the axes.
275 Easy
275 of 320
Remembering +4 | -1
Chromium carbonyl is:
View answer and explanation
Correct option: C
Cr(CO)6\mathrm{Cr}{\left( {\mathrm{CO}} \right)_6}
Option C is correct because chromium carbonyl has the formula Cr(CO)6, where chromium is bonded to six carbonyl ligands.
276 Medium
276 of 320
Application +4 | -1
The EAN of platinum in potassium hexachloroplatinate (IV) is:
View answer and explanation
Correct option: B
86
Option B (86) is correct because the EAN is calculated by adding the number of electrons in the metal ion and the number of electrons donated by the ligands.
277 Hard
277 of 320
Understanding +4 | -1
Fe2(CO)9\mathrm{F}{\mathrm{e}_2}{\left( {\mathrm{CO}} \right)_9} is diamagnetic. Which of the following reasons is correct?
View answer and explanation
Correct option: C
Metal-metal (Fe−Fe\mathrm{Fe} - \mathrm{Fe}) bond in molecule
Option C (Metal-metal (Fe−Fe\mathrm{Fe} - \mathrm{Fe}) bond in molecule) is correct because the presence of an Fe-Fe bond allows for the pairing of electrons between the two iron atoms, resulting in a diamagnetic complex.
278 Medium
278 of 320
Application +4 | -1
Spin only magnetic moment value of [Mn(CN)6]3−{\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{3 - }} ion is
View answer and explanation
Correct option: A
2.83 BM
Option A (2.83 BM) is correct because [Mn(CN)6]3−{\left[ {Mn{{\left( {CN} \right)}_6}} \right]^{3 - }} has a Mn3+Mn^{3+} ion with a d4d^4 configuration. Since CN−CN^- is a strong field ligand, it will pair up the electrons, resulting in 2 unpaired electrons and a magnetic moment of approximately 2.83 BM.
279 Easy
279 of 320
Understanding +4 | -1
[Co(NH3)4Cl2]\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right] possesses:
View answer and explanation
Correct option: D
Octahedral geometry
The correct answer is D (Octahedral geometry) because the complex [Co(NH3)4Cl2]{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}\mathrm{C}{\mathrm{l}_2}} \right]} has a coordination number of 6, leading to an octahedral shape.
280 Easy
280 of 320
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 is correct because Ni(CO)4 is a tetrahedral complex with a d10 configuration, resulting in no d-d transitions and thus is colorless.
281 Medium
281 of 320
Application +4 | -1
Which one of the following has a square planar geometry?
(At.No.Fe=26,Co=27,Ni=28,Pt=78)\left( {At.\mathrm{ }No.\mathrm{ }Fe = 26,\mathrm{ }Co = 27,\mathrm{ }Ni = 28,\mathrm{ }Pt = 78} \right)
View answer and explanation
Correct option: D
[PtCl4]2−{\left[ {\mathrm{PtC}{\mathrm{l}_4}} \right]^{2 - }}
Option D ([PtCl4]2-) is correct because Pt(II) complexes with strong field ligands like Cl- tend to form square planar geometries due to dsp² hybridization.
282 Medium
282 of 320
Understanding +4 | -1
The IUPAC name of [Ni(PPh3)2Cl2]2+{\left[ {\mathrm{Ni}{{\left( {\mathrm{PP}{\mathrm{h}_3}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]^{2 + }} is
View answer and explanation
Correct option: B
Dichloro bis (triphenylphosphine)nickel(II)
Option B (Dichloro bis (triphenylphosphine)nickel(II)) is correct because it correctly names the compound, indicating the dichloro and bis(triphenylphosphine) ligands, and the oxidation state of nickel.
283 Medium
283 of 320
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-colored complex with Nickel(II) ions.
284 Easy
284 of 320
Remembering +4 | -1
Wilkinson’s catalyst, (Ph3P)3RhCl{\left( {\mathrm{P}{\mathrm{h}_3}\mathrm{P}} \right)_3}\mathrm{RhCl} is used for
View answer and explanation
Correct option: C
Hydrogenation of alkenes
Option C (Hydrogenation of alkenes) is correct because Wilkinson's catalyst is primarily used for the hydrogenation of alkenes.
285 Hard
285 of 320
Understanding +4 | -1
Arrange the following cobalt complexes in the order of increasing crystal field stabilization energy (CFSE) value. Complexes
[CoF6]3−(A),   [Co(H2O)6]3+(B),    [Co(NH3)6]3+(C)  and  [Co(en)3]3+(D)\mathop {{{\left[ {Co{F_6}} \right]}^{3 - }}}\limits_{(A)} ,\,\,\,\mathop {{{\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]}^{3 + }}}\limits_{(B)} ,\,\,\,\,\mathop {{{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]}^{3 + }}}\limits_{\left( C \right)} \,\,\mathrm{and}\,\,\mathop {{{\left[ {Co{{\left( {en} \right)}_3}} \right]}^{3 + }}}\limits_{\left( D \right)}
Choose the correct option.
View answer and explanation
Correct option: A
A<B<C<D A < B < C < D\,
Option A (A<B<C<D A < B < C < D\,) is correct because the spectrochemical series dictates the ligand field strength order: F−<H2O<NH3<enF^- < H_2O < NH_3 < en, hence the increasing CFSE order is A<B<C<DA < B < C < D.
286 Hard
286 of 320
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 ([CoCl(NH3)5]++5H+→CO2++5NH4++Cl−{[CoCl{(N{H_3})_5}]^ + } + 5{H^ + } \to C{O^{2 + }} + 5NH_4^ + + C{l^ - }) is correct because it represents a balanced chemical equation with correct products formed when the complex reacts with acid.
287 Hard
287 of 320
Analyzing +4 | -1
Assertion : If in [Co(NH3)6]+3 , NH3{\left[ {{\text{Co(N}}{{\text{H}}_3}{{\text{)}}_{\text{6}}}} \right]^{ + 3}}\,,\,{\text{N}}{{\text{H}}_{\text{3}}} is replaced by H2O{{\text{H}}_{\text{2}}}{\text{O}}, same wave1ength will be absorbed by the complex :

Reason : It is a high spin species.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because replacing NH3 with H2O will change the ligand field strength and hence the wavelength absorbed, and [Co(NH3)6]+3 is a low spin complex, not high spin.
288 Easy
288 of 320
Application +4 | -1
A complex of platinum, ammonia and chlorine produces four ions per molecule in the solution. The structure consistent with the observation is:
View answer and explanation
Correct option: C
[Pt(NH3)5Cl]Cl3\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_3}
Option C ([Pt(NH3)5Cl]Cl3) is correct because it dissociates into one complex ion [Pt(NH3)5Cl]3+ and three chloride ions, resulting in a total of four ions.
289 Medium
289 of 320
Application +4 | -1
The magnetic moment (in B.M.) of complex [Co(H2O)6]3+\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_{\mathrm{6}}\right]^{3+} is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because Co3+Co^{3+} in a weak field ligand like water undergoes pairing and becomes diamagnetic, resulting in zero unpaired electrons and thus zero magnetic moment.
290 Medium
290 of 320
Application +4 | -1
The oxidation state of Mo\mathrm{Mo} in its oxo-complex species [Mo2O4(C2H4)2(H2O)2]2− is{\left[ {\mathrm{M}{\mathrm{o}_2}{\mathrm{O}_4}{{\left( {{\mathrm{C}_2}{\mathrm{H}_4}} \right)}_2}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_2}} \right]^{2 - }}\,\mathrm{is}
View answer and explanation
Correct option: B
3
The correct answer is 3 because the oxidation state of Mo in [Mo2O4(C2H4)2(H2O)2]2−{\left[ {\mathrm{M}{\mathrm{o}_2}{\mathrm{O}_4}{{\left( {{\mathrm{C}_2}{\mathrm{H}_4}} \right)}_2}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_2}} \right]^{2 - }} is +3.
291 Easy
291 of 320
Remembering +4 | -1
In Cr(NH3)4Cl2]Cl\mathrm{Cr}{\left( {\mathrm{N}{\mathrm{H}_3}} \right)_4}\mathrm{C}{\mathrm{l}_2}]\mathrm{Cl} the ligands are:
View answer and explanation
Correct option: C
Both NH3  and  Cl−\mathrm{N}{\mathrm{H}_3}\mathrm{\;and\;C}{\mathrm{l}^ - }
Option B (Chelates) is correct because complexes formed with bidentate or polydentate ligands are called chelates.
292 Easy
292 of 320
Understanding +4 | -1
The coordination number of CrCr in [Cr(NH3)3(H2O)3]Cl3 is:\left[ {\mathrm{Cr}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_3}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_3}} \right]\mathrm{C}{\mathrm{l}_3}\,\mathrm{is}:
View answer and explanation
Correct option: C
6
Option C (6) is correct because the coordination number is the number of ligands directly attached to the central metal ion, which in this case is 3 NH3 and 3 H2O, totaling 6.
293 Easy
293 of 320
Analyzing +4 | -1
The complex ion which has no ‘d’-electron in the central metal atom is:
View answer and explanation
Correct option: A
[MnO4]−{\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - }
Option A ([MnO4]−{\left[ {\mathrm{Mn}{\mathrm{O}_4}} \right]^ - }) is correct because Mn in MnO4- has a +7 oxidation state, resulting in a d0 configuration, meaning no d-electrons.
294 Medium
294 of 320
Analyzing +4 | -1
Which of the following 0.1 M complex compound solutions will have the minimum electrical conductivity?
View answer and explanation
Correct option: D
Trichloro triammine platinum (IV) chloride
Option D (Trichloro triammine platinum (IV) chloride) is correct because it has the fewest ions in solution, leading to the lowest electrical conductivity.
295 Medium
295 of 320
Understanding +4 | -1
When ammonia is added to green aqueous solution of nickel(II) sulphate, the colour of the solution changes to blue violet. This is caused by:
View answer and explanation
Correct option: B
Ammonia molecules replacing water molecules surrounding nickel
The correct answer is B. The blue-violet color arises because ammonia molecules replace water molecules in the coordination sphere of nickel, leading to a different complex with a different color.
296 Easy
296 of 320
Understanding +4 | -1
Which of the following statements is correct?
View answer and explanation
Correct option: C
In K3[Fe(CN)6,  {\mathrm{K}_3}[\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6},\; the ligand has satisfied only the secondary valency of ferric ion
The correct answer is C. In K3[Fe(CN)6]{\mathrm{K}_3}[\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}], the ligand (CN-) satisfies only the secondary valency (coordination number) of the ferric ion (Fe3+).
297 Hard
297 of 320
Application +4 | -1
Which of the following compound is not colored?
View answer and explanation
Correct option: B
Na2Cd⋅Cl4{Na}_2 {Cd} \cdot {Cl}_4
Option B (Na2Cd⋅Cl4{Na}_2 {Cd} \cdot {Cl}_4) is correct because compounds with filled d-orbitals are generally colorless.
298 Hard
298 of 320
Application +4 | -1
The magnetic moment is measured in Bohr Magneton (BM).

Spin only magnetic moment of Fe\mathrm{Fe} in [Fe(H2O)6]3+\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} and [Fe(CN)6]3−\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} complexes respectively is:
View answer and explanation
Correct option: C
5.92 B.M. and 1.732 B.M.
Option C (5.92 B.M. and 1.732 B.M.) is correct because it accurately calculates the spin-only magnetic moment based on the number of unpaired electrons in each complex.
299 Medium
299 of 320
Analyzing +4 | -1
Which of the following complexes will show geometrical as well as optical isomerism? (en ==ethylene diamine)
View answer and explanation
Correct option: D
[Pt(en)2Cl2]\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]
Option D ([Pt(en)2Cl2]\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]) is correct because it can exist as cis and trans isomers, and the cis isomer is chiral, exhibiting optical isomerism.
300 Easy
300 of 320
Analyzing +4 | -1
Which statement is not correct in the case of [Co(NH3)6]3+{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }} complex?
View answer and explanation
Correct option: B
It involves d2sp2{d^2}s{p^2} - hybridization
Option B (It involves d2sp2{d^2}s{p^2} - hybridization) is not correct because [Co(NH3)6]3+{\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]^{3 + }} involves d2sp3d^2sp^3 hybridization.
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 320 questions

Your practice

0 / 320 answered · 320 remaining · 0% accuracy

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

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

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.