Ionization of ${\rm{K}}\left[ {{\rm{Ag}}{{\left( {{\rm{CN}}} \right)}_2}} \right]$ will give:

${{\rm{K}}^ + }$ and ${\left[ {{\rm{Ag}}{{\left( {{\rm{CN}}} \right)}_2}} \right]^ - }$ ion

${{\rm{K}}^ + },{\rm{\;A}}{{\rm{g}}^ + },{\rm{\;C}}{{\rm{N}}^ - }$

In the complexes ${\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right]^{3 + }},{\rm{\;}}{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]^{3 - }},{\rm{\;}}{\left[ {{\rm{Fe}}{{\left( {{{\rm{C}}_2}{{\rm{O}}_4}} \right)}_3}} \right]^{3 - }}{\rm{\;and}}\,{\rm{\;}}{\left[ {{\rm{FeC}}{{\rm{l}}_6}} \right]^{3 - }},$ more stability is shown by:

${\left[ {{\rm{Fe}}{{\left( {{{\rm{C}}_2}{{\rm{O}}_4}} \right)}_3}} \right]^{3 - }}$

${\left[ {{\rm{Fe}}{{\left( {{{\rm{H}}_2}{\rm{O}}} \right)}_6}} \right]^{3 + }}$

${\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]^{3 - }}$

${\left[ {{\rm{FeC}}{{\rm{l}}_6}} \right]^{3 - }}$

$\left[\mathrm{MnCl}_6\right]^{3-}$ - Paramagnetic

$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ - Diamagnetic

$\left[\mathrm{NiCl}_4\right]^{2-}$ - Paramagnetic

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JEE • CHEMISTRY • CO-ORDINATION COMPOUNDS

Co-ordination compounds practice · Mixed difficulty

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Showing 1-30 of 326 questions

1 Medium

1 of 326

Analyzing +4 | -1

Which has maximum paramagnetic character?

View answer and explanation

Correct option: D

{\left[ {\mathrm{Mn}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}

Option D (

{\left[ {\mathrm{Mn}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{2 + }}

) is correct because it has the maximum number of unpaired electrons (5).

2 Easy

2 of 326

Understanding +4 | -1

Ionization of

\mathrm{K}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]

will give:

View answer and explanation

Correct option: A

{\mathrm{K}^ + }

and

{\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - }

ion

Option A (

{\mathrm{K}^ + }

and

{\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - }

ion) is correct because the complex salt dissociates into its constituent ions in solution.

3 Easy

3 of 326

Understanding +4 | -1

The cation that does not form an ammine complex with excess of ammonia is:

View answer and explanation

Correct option: A

\mathrm{A}{\mathrm{l}^{3 + }}

Option A (

\mathrm{A}{\mathrm{l}^{3 + }}

) is correct because

Al^{3+}

does not form stable ammine complexes in aqueous solution due to its high charge density and strong polarizing power, which favors hydrolysis over ammine complex formation.

4 Medium

4 of 326

Understanding +4 | -1

The complexes

\left[ {\mathrm{PtC}{\mathrm{l}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{B}{\mathrm{r}_2}\mathrm{\;and\;}\left[ {\mathrm{PtB}{\mathrm{r}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{C}{\mathrm{l}_2}\mathrm{\;}

are example for isomerism

View answer and explanation

Correct option: C

Ionization

Option C (Ionization) is correct because the complexes differ in the counter ions (Br⁻ and Cl⁻) that can exchange with ligands inside the coordination sphere.

5 Medium

5 of 326

Understanding +4 | -1

The correct name of

\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]\left[\mathrm{PtCl}_4\right]

is:

View answer and explanation

Correct option: A

Tetraammine dichloro platinum (IV) tetrachloro platinate (II)

Option A (Tetraammine dichloro platinum (IV) tetrachloro platinate (II)) is correct because it follows the IUPAC naming conventions for coordination compounds, where the cation is named before the anion, ligands are named alphabetically, and oxidation states are indicated in Roman numerals.

6 Medium

6 of 326

Understanding +4 | -1

Which of the following pairs of isomers and types of isomerism are correctly matched?

(I)

\left[{Co}\left({NH}_3\right)_5\left({NO}_2\right)\right] {Cl}_2

and

\left[{Co}\left({NH}_3\right)_5({ONO})\right] {Cl}_2 \ldots \ldots \ldots

Linkage

(II)

\left[{Cu}\left({NH}_3\right)_4\right]\left[{PtCl}_4\right] \,\, \& \,\, \left[{Pt}\left({NH}_3\right)_4\right]\left[{CuCl}_4\right] \ldots \ldots \ldots

Coordinations

(III)

\left[{Pt}\left({NH}_3\right)_4 {Cl}_2\right] {Br}_2 \, \, \& \, \, \left[{Pt}\left({NH}_3\right)_4 {Br}_2\right]{Cl}_2 \ldots \ldots \ldots

Ionization

Select the correct answer using the codes given below:

View answer and explanation

Correct option: D

I, II, III

Option D (I, II, III) is correct because all three pairs correctly identify the type of isomerism present: linkage, coordination, and ionization, respectively.

7 Medium

7 of 326

Application +4 | -1

Which of the following is a pair of diamagnetic complex?

View answer and explanation

Correct option: A

\left[{Co}({ox})_3\right]^{3-},\left[{Fe}({CN})_6\right]^{3-}

Option A (

\left[{Co}({ox})_3\right]^{3-},\left[{Fe}({CN})_6\right]^{3-}

) is correct because both complexes are diamagnetic, meaning they have no unpaired electrons.

8 Easy

8 of 326

Application +4 | -1

EAN of Fe in

{\left[ {Fe{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{3 - }}

View answer and explanation

Correct option: C

Option C (35) is correct because the Effective Atomic Number (EAN) is calculated using the atomic number, oxidation number, and coordination number of the metal.

9 Easy

9 of 326

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

\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_3}

Option C (

\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Cl}} \right]\mathrm{C}{\mathrm{l}_3}

) is correct because it produces four ions upon ionization, with three chloride ions outside the coordination sphere.

10 Hard

10 of 326

Analyzing +4 | -1

Identify the homoleptic complexes with odd number of d electrons in the central metal.

(A)

\left[\mathrm{FeO}_4\right]^{2-}

(B)

\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}

(C)

\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{2-}

(D)

\left[\mathrm{CoCl}_4\right]^{2-}

(E)

\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{F}_3\right]

Choose the correct answer from the options given below:

View answer and explanation

Correct option: A

(B) and (D) only

Option A ((B) and (D) only) is correct because only

\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}

and

\left[\mathrm{CoCl}_4\right]^{2-}

are homoleptic complexes with an odd number of d electrons in the central metal ion.

11 Medium

11 of 326

Application +4 | -1

The IUPAC name of the complex

[Ag{({H_2}O)_2}][Ag{(CN)_2}]

View answer and explanation

Correct option: A

diaquasilver(I) dicyanidoargentate(I)

The correct answer is A (diaquasilver(I) dicyanidoargentate(I)) because it correctly names the complex with the correct oxidation state and ligands.

12 Medium

12 of 326

Analyzing +4 | -1

Assertion (A) :

{CoCl}_3 \cdot 4{NH}_3

gives two ions when ionised.

Reason (R) : Complex is

\left[{Co}\left({NH}_3\right)_4\right] {Cl}_3

View answer and explanation

Correct option: C

\mathbf{A}

is true but

\mathbf{R}

is false.

Option C (

\mathbf{A}

is true but

\mathbf{R}

is false.) is correct because the complex gives three ions, not two, and the formula is incorrect.

13 Easy

13 of 326

Understanding +4 | -1

Which is not a

\pi -

bonded complex?

View answer and explanation

Correct option: D

Tetraethyl lead

Option D (Tetraethyl lead) is correct because it is an organometallic compound that does not involve

\pi

-bonding.

14 Medium

14 of 326

Analyzing +4 | -1

S-I:

\left[{Ni}({CN})_4\right]^{2-}

is a diamagnetic complex.

S-II: Compound is low spin complex.

View answer and explanation

Correct option: A

Statement-I and Statement-II are true

Option A (Statement-I and Statement-II are true) is correct because

\left[{Ni}({CN})_4\right]^{2-}

is indeed a diamagnetic complex due to the strong field ligand

CN^-

, which causes pairing of electrons, making it a low spin complex.

15 Medium

15 of 326

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

\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]

Option D (

\left[ {\mathrm{Pt}{{\left( {\mathrm{en}} \right)}_2}\mathrm{C}{\mathrm{l}_2}} \right]

) is correct because it can exhibit both geometrical (cis/trans) and optical isomerism (for the cis isomer).

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16 Easy

16 of 326

Understanding +4 | -1

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{Br}} \right]\mathrm{S}{\mathrm{O}_4}\mathrm{\;}

and

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{S}{\mathrm{O}_4}} \right]

\mathrm{Br}

are examples of which type of isomerism?

View answer and explanation

Correct option: D

Ionisation

Option D (Ionisation) is correct because the isomers differ in the counter-ion that is directly bonded to the metal center.

17 Easy

17 of 326

Understanding +4 | -1

Change in composition of coordination sphere yields which type of isomer?

View answer and explanation

Correct option: B

Ionization

Option B is correct because ionization isomers arise when the composition of the coordination sphere changes, resulting in different ions being present in the solution.

18 Easy

18 of 326

Remembering +4 | -1

The coordination number of a metal in coordination compound is

View answer and explanation

Correct option: C

Same as secondary valency

Option C (Same as secondary valency) is correct because the coordination number is defined as the number of ligands directly attached to the central metal ion, which corresponds to the secondary valency.

19 Medium

19 of 326

Remembering +4 | -1

{K}_3\left[{Al}\left({C}_2 {O}_4\right)_3\right]

is called

View answer and explanation

Correct option: B

Potassium trioxalatoaluminate(III)

Option B (Potassium trioxalatoaluminate(III)) is correct because it follows the IUPAC naming conventions for coordination compounds.

20 Medium

20 of 326

Understanding +4 | -1

The number of ions formed when a molecule of

{\mathrm{K}_4}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}

dissociates is:

View answer and explanation

Correct option: B

Option B (5) is correct because the complex dissociates into 4 potassium ions and one complex ion, resulting in a total of 5 ions.

21 Medium

21 of 326

Understanding +4 | -1

Ammonia forms the complex ion

{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }}

with copper ions in alkaline solution but not in acidic solution. What is the reason for it?

View answer and explanation

Correct option: D

In acidic solutions, protons coordinate with ammonia molecules forming

\mathrm{NH}_4^ +

ions and

\mathrm{N}{\mathrm{H}_3}

molecules are not available

Option D is correct. In acidic solutions, ammonia molecules are protonated to form ammonium ions (

\mathrm{NH}_4^ +

), which cannot act as ligands.

22 Medium

22 of 326

Analyzing +4 | -1

Statement I: Cadmium sulphide is yellow coloured

Statement II: The d-d transition does not take place in this compound

View answer and explanation

Correct option: B

Statement I and statement II are correct but statement II is not the correct explanation of statement I

Option B is correct because both statements are correct, but the absence of d-d transition is not the reason for the yellow color of cadmium sulfide.

23 Medium

23 of 326

Understanding +4 | -1

Which of the following compounds does not dissolve in conc.

{\mathrm{H}_\mathrm{2}}\mathrm{S}{\mathrm{O}_\mathrm{4}}

even on warning?

View answer and explanation

Correct option: C

Hexane

Option C (Hexane) is correct because alkanes like hexane are generally unreactive and do not dissolve in concentrated sulfuric acid, even upon warming.

24 Easy

24 of 326

Remembering +4 | -1

Which is the strongest field ligand?

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{N}^ - }

Option A (

\mathrm{C}{\mathrm{N}^ - }

) is correct because cyanide (

CN^−

) is a strong field ligand, causing a large splitting of d-orbitals.

25 Medium

25 of 326

Analyzing +4 | -1

In the complexes

{\left[ {\mathrm{Fe}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_6}} \right]^{3 + }},\mathrm{\;}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }},\mathrm{\;}{\left[ {\mathrm{Fe}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]^{3 - }}\mathrm{\;and}\,\mathrm{\;}{\left[ {\mathrm{FeC}{\mathrm{l}_6}} \right]^{3 - }},

more stability is shown by:

View answer and explanation

Correct option: C

{\left[ {\mathrm{Fe}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]^{3 - }}

Option C (

{\left[ {\mathrm{Fe}{{\left( {{\mathrm{C}_2}{\mathrm{O}_4}} \right)}_3}} \right]^{3 - }}

) is correct because

{\mathrm{C}_2}\mathrm{O}_4^{2 - }

is a bidentate ligand, leading to a more stable complex due to chelation.

26 Medium

26 of 326

Understanding +4 | -1

Assertion: Carbon monoxide is a deadly poison.

Reason:

CO

can form strong complexes with haemoglobin.

View answer and explanation

Correct option: A

Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.

Option A (Both assertion and reason are correct statements, and reason is the correct explanation of the assertion) is correct because carbon monoxide is a deadly poison because it binds strongly to hemoglobin, preventing oxygen transport.

27 Easy

27 of 326

Analyzing +4 | -1

Assertion : On cooling a freezing mixture, colour of the mixture turns to pink from deep blue for a reaction.

{Co}\left({H}_2 {O}\right)_6^{2+}{ }_{(a q)}+4 {Cl}_{(a q)}^{-} \rightleftharpoons

{CoCl}_4^{2-}(a q)+6 {H}_2 {O}_{(l)}

.

Reason : Reaction is endothermic so on cooling, the reaction moves to backward direction.

View answer and explanation

Correct option: A

If both assertion and reason are true and the reason is the correct explanation of the assertion.

Option A is correct because both the assertion and reason are true, and the reason correctly explains why cooling shifts the equilibrium towards the pink complex due to Le Chatelier's principle.

28 Medium

28 of 326

Understanding +4 | -1

Which one of the following will not show geometrical isomerism?

View answer and explanation

Correct option: C

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{N}{\mathrm{O}_2}} \right]\mathrm{C}{\mathrm{l}_2}

Option C (

\left[ {\mathrm{Co}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_5}\mathrm{N}{\mathrm{O}_2}} \right]\mathrm{C}{\mathrm{l}_2}

) is correct because complexes of the type

MA_5B

(where all ligands except one are the same) do not exhibit geometrical isomerism.

29 Medium

29 of 326

Analyzing +4 | -1

Wrong match is

View answer and explanation

Correct option: A

\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}

- Diamagnetic

Option A (8) is correct because there are 4 geometrical isomers for the first complex and 4 stereoisomers for the second complex, summing to 8.

30 Hard

30 of 326

Application +4 | -1

Low spin complex of

d^6

-cation in an octahedral field will have the following energy:

View answer and explanation

Correct option: B

\frac{-12}{5} \Delta_0+3 {P}

Option B (

\frac{{-12}}{{5}} \Delta_0+3 {P}

) is correct because it represents the crystal field stabilization energy for a low spin

d^6

complex with pairing energy.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 326 single-correct-answer MCQs on CO-ORDINATION COMPOUNDS.
✓ Try each question before opening its answer. Read the explanation where one is available.
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