Skip to main content
Back to CO-ORDINATION COMPOUNDS

Your Progress

0 / 37 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • CO-ORDINATION COMPOUNDS

Co-ordination compounds practice · Hard

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 1-30 of 37 questions

Preparing practice and saved progress…

1 Hard
1 of 37
Analyzing +4 | -1
Identify the homoleptic complexes with odd number of d electrons in the central metal.

(A) [FeO4]2−\left[\mathrm{FeO}_4\right]^{2-}

(B) [Fe(CN)6]3−\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}

(C) [Fe(CN)5NO]2−\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{2-}

(D) [CoCl4]2−\left[\mathrm{CoCl}_4\right]^{2-}

(E) [Co(H2O)3F3]\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 [Fe(CN)6]3−\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} and [CoCl4]2−\left[\mathrm{CoCl}_4\right]^{2-} are homoleptic complexes with an odd number of d electrons in the central metal ion.
2 Hard
2 of 37
Application +4 | -1
Low spin complex of d6d^6-cation in an octahedral field will have the following energy:
View answer and explanation
Correct option: B
−125Δ0+3P\frac{-12}{5} \Delta_0+3 {P}
Option B (−125Δ0+3P\frac{{-12}}{{5}} \Delta_0+3 {P}) is correct because it represents the crystal field stabilization energy for a low spin d6d^6 complex with pairing energy.
3 Hard
3 of 37
Analyzing +4 | -1
The total number of possible isomers for the complex compound [Cu(NH3)4][PtCl4]\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\left[ {\mathrm{PtC}{\mathrm{l}_4}} \right]
View answer and explanation
Correct option: A
6
The correct answer is A (6) because the complex compound [Cu(NH3)4][PtCl4] can form six different isomers through ligand exchange between the two coordination complexes.
4 Hard
4 of 37
Analyzing +4 | -1
Choose the correct order for Δ0\Delta_{0} for the following complexes.

(I) [Co(H2O)6]2+\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\quad (II) [Co(H2O)6]3+\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}

(III) [Fe(H2O)6]3+\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}\quad (IV) [Fe(CN)6]3−\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}
View answer and explanation
Correct option: B
I << III << II << IV
Option B (I << III << II << IV) is correct because it correctly orders the crystal field splitting energy based on charge, metal, and ligand strength.
5 Hard
5 of 37
Analyzing +4 | -1
Which of the following will have maximum stabilization due to crystal field?
View answer and explanation
Correct option: C
[Co(CN)6]−3{[Co{(CN)_6}]^{ - 3}}
Option C ([Co(CN)6]−3{[Co{(CN)_6}]^{ - 3}}) is correct because CN−CN^- is a strong field ligand, leading to a large crystal field splitting and maximum stabilization for the d6d^6 configuration.
6 Hard
6 of 37
Analyzing +4 | -1
Assertion :- Neopentyl bromide and 2-bromo-2-methyl butane are enantiomers.

Reason :- Both have same molecular formula as well as structure formula.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because neopentyl bromide and 2-bromo-2-methyl butane are not enantiomers, and they do not have the same structural formula.
7 Hard
7 of 37
Analyzing +4 | -1
The correct order for the wavelengths of absorption in the visible region for the following is

I. [Ni(NO2)6]4−\left[\mathrm{Ni}\left(\mathrm{NO}_2\right)_6\right]^{4-} II. [Ni(NH3)6]2+\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+} III. [Ni(H2O)6]2+\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}
View answer and explanation
Correct option: C
III >> II >> I
Option C (III > II > I) is correct because the wavelengths of absorption are inversely proportional to the crystal field splitting energy (CFSE), which depends on the ligand field strength (H2O < NH3 < NO2-).
8 Hard
8 of 37
Analyzing +4 | -1
For an octahedral complex, which of the following dd - electron configuration will give maximum CFSE?
View answer and explanation
Correct option: C
Low spin d5d^5
Option D (MgO\mathrm{MgO}) is correct because MgO has the highest lattice energy due to the higher charges of Mg2+ and O2- compared to the other options.
9 Hard
9 of 37
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.
10 Hard
10 of 37
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.
11 Hard
11 of 37
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.
12 Hard
12 of 37
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.
13 Hard
13 of 37
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.
14 Hard
14 of 37
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.
15 Hard
15 of 37
Understanding +4 | -1
The mismatched combinations are
A. Chlorophyll - Co\mathrm{Co}
B. Water hardness - EDTA
C. Photography −[Ag(CN)2]−-\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^-
D. Wilkinson catalyst −[(Ph3P)3RhCl]-\left[\left(\mathrm{Ph}_{3} \mathrm{P}\right)_{3} \mathrm{RhCl}\right]
E. Chelating ligand - D-Penicillamine
Choose the correct answer from the options given below.
View answer and explanation
Correct option: B
A and C only
Option B (A and C only) is correct because Chlorophyll contains Mg, not Co, and Photography involves AgBr, not [Ag(CN)₂]⁻.

Explore Other Chapters in CHEMISTRY

View All →
P-BLOCK ELEMENTS 736 MCQs Equilibrium 696 MCQs STRUCTURE OF ATOM 634 MCQs THERMODYNAMICS 536 MCQs CHEMICAL BONDING & MOLECULAR STRUCTURE 514 MCQs SOLUTIONS 480 MCQs
16 Hard
16 of 37
Analyzing +4 | -1
In which of the following compound, C – O bond length is maximum
View answer and explanation
Correct option: A
[Fe(CO)5]−{\left[ {Fe{{\left( {CO} \right)}_5}} \right]^ - }
Option A ([Fe(CO)5]−{\left[ {Fe{{\left( {CO} \right)}_5}} \right]^ - }) is correct because the negative charge on the complex increases back-bonding, weakening the C-O bond and thus increasing its length.
17 Hard
17 of 37
Analyzing +4 | -1
Assertion :- cis-3-bromo propenoic acid is more stable than its trans form.

Reason :- Dipole moment of cis form is greater than trans form.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because both the Assertion and the Reason are false. The trans form is more stable due to less steric hindrance, and the dipole moment of the cis form is greater.
18 Hard
18 of 37
Remembering +4 | -1
The increasing order of crystal field splitting strength of the given ligands is
View answer and explanation
Correct option: C
Cl−<F−<H2O<NH3<CN−<CO{Cl}^{-} < {F}^{-} < {H}_2 {O} < {NH}_3 < {CN}^{-} < {CO}
The correct answer is C (Cl−<F−<H2O<NH3<CN−<CO{Cl}^{-} < {F}^{-} < {H}_2 {O} < {NH}_3 < {CN}^{-} < {CO}) because it represents the correct increasing order of crystal field splitting strength based on the spectrochemical series.
19 Hard
19 of 37
Application +4 | -1
The oxidation state of platinum in Na[PtBrCl(NO2)(NH3)]{Na}\left[{PtBrCl}\left({NO}_2\right)\left({NH}_3\right)\right] is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the oxidation state of platinum is determined by balancing the charges of the complex ion and the counter ion.
20 Hard
20 of 37
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 ([Rh(H2O)6]3+[\mathrm{Rh}(\mathrm{H}_2 \mathrm{O})_6]^{3+} and [Co(H2O)6]3+[\mathrm{Co}(\mathrm{H}_2 \mathrm{O})_6]^{3+}) is correct because for the same ligand and oxidation state, the complex with the metal from the later period (Rh) will have a higher Δ₀ value due to increased interaction with the ligands.
21 Hard
21 of 37
Analyzing +4 | -1
Assertion : AgI\mathrm{AgI} is coloured while AgF\mathrm{AgF} is colourless.

Reason : Unpaired e−{{\text{e}}^ - } is present in AgI\mathrm{AgI}
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 AgI is colored and AgF is colorless, the reason is due to charge transfer spectra, not unpaired electrons.
22 Hard
22 of 37
Analyzing +4 | -1
Which of the following coordination entities should be expected to absorb light of lowest frequency?
View answer and explanation
Correct option: B
[CrCl6]3−\left[\mathrm{CrCl}_6\right]^{3-}
Option B is correct because [CrCl6]3- has the weakest field ligand (Cl-), leading to the smallest crystal field splitting and absorption of light at the lowest frequency.
23 Hard
23 of 37
Analyzing +4 | -1
Which of the following ions can exhibit optical isomerism-
View answer and explanation
Correct option: C
Cis−Cis- [Co(en)2Cl2]+\left[{Co}({en})_2 {Cl}_2\right]^{+}
Option C (Cis- [Co(en)2Cl2]+) is correct because the cis isomer of [M(aa)2b2] complexes can exhibit optical isomerism, while the trans isomer cannot.
24 Hard
24 of 37
Understanding +4 | -1
[Fe(CN)6]4−\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} and [Fe(H2O)6]2+\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} are of different colours in dilute solutions. This is because of
View answer and explanation
Correct option: A
difference in the nature of ligands
Option A (difference in the nature of ligands) is correct because different ligands cause different crystal field splitting, leading to different absorption of light and thus different colors.
25 Hard
25 of 37
Analyzing +4 | -1
Amongst the following, the most stable complex is
View answer and explanation
Correct option: D
[FeCl6]3−\left[\mathrm{FeCl}_6\right]^{3-}
Option C is correct because [Fe(C2O4)3]3−\left[\mathrm{Fe}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-} forms a chelate due to the bidentate nature of oxalate ions, making it the most stable complex.
26 Hard
26 of 37
Understanding +4 | -1
Assertion : Square planar complex Ma2b2{\text{M}}{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}} has two optical isomers.

Reason : Mirror image of Ma2b2{\text{M}}{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{2}}} is non-super imposable.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because square planar complexes of the type Ma2b2 do not exhibit optical isomerism, and their mirror images are superimposable.
27 Hard
27 of 37
Application +4 | -1
The conductivity measurement of a coordination compound of Cobalt (III) shows that it dissociates into 3 ions in solution. The compound is
View answer and explanation
Correct option: D
Pentaamminechloridocobalt(III) chloride
Option D (Pentaamminechloridocobalt(III) chloride) is correct because it dissociates into three ions in solution: [Co(NH₃)₅Cl]²⁺ and 2Cl⁻.
28 Hard
28 of 37
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 (Na2[CdCl4]) is correct because cadmium has a d10 configuration, meaning no d-d transitions are possible, hence it's colorless.
29 Hard
29 of 37
Analyzing +4 | -1
Assertion : Triethylenediamine is a bidentate mono anion.

Reason : Complex containing propylenediamine ligand shows ligand isomerism.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because triethylenediamine is a neutral bidentate ligand, and complexes with propylenediamine do not show ligand isomerism.
30 Hard
30 of 37
Analyzing +4 | -1
Which of the following statement is false
View answer and explanation
Correct option: A
Of the d4{d^4} species, manganese (III) is strongly reducing while Cr2+C{r^{2 + }} is strongly oxidising.
Option A (Of the d4{d^4} species, manganese (III) is strongly reducing while Cr2+C{r^{2 + }} is strongly oxidising) is false because Mn(III) is oxidizing and Cr(II) is reducing.
Revision notes and key takeaways

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

Key Takeaways

  • ✓ This set contains 37 single-correct-answer MCQs on CO-ORDINATION COMPOUNDS.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
Showing 1-30 of 37 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING CHEMISTRY - CO-ORDINATION COMPOUNDS MCQs (All) 326 MCQs → ENGINEERING CHEMISTRY - CO-ORDINATION COMPOUNDS MCQs (Easy) 89 MCQs → ENGINEERING CHEMISTRY - CO-ORDINATION COMPOUNDS MCQs (Medium) 200 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ P-BLOCK ELEMENTS 736 MCQs → Equilibrium 696 MCQs → STRUCTURE OF ATOM 634 MCQs → THERMODYNAMICS 536 MCQs

Your practice

0 / 37 answered · 37 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:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

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.

Continue with Free Account →