Skip to main content
Back to CO-ORDINATION COMPOUNDS

Your Progress

0 / 200 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • CO-ORDINATION COMPOUNDS

Co-ordination compounds practice · Medium

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 91-120 of 200 questions

Preparing practice and saved progress…

91 Medium
91 of 200
Analyzing +4 | -1
According to CFTCFT, both high-spin and low-spin octahedral complexes are not possible for the configuration.
View answer and explanation
Correct option: A
d8d^8
The correct answer is d8 because for d8 configuration, low spin octahedral complex is not possible.
92 Medium
92 of 200
Analyzing +4 | -1
Which of the following are false about [Cu(NH3)4]SO4\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right] \mathrm{SO}_4 ?

I. It is paramagnetic with one unpaired electron in the dd-subshell.

II. Its aqueous solution cannot conduct electricity.

III. It gives white BaCl2\mathrm{BaCl}_2 solution.

IV. It is a square planar complex.

Choose the correct option.
View answer and explanation
Correct option: A
I and II
Option A (I and II) is correct because [Cu(NH3)4]SO4 is paramagnetic and its aqueous solution conducts electricity.
93 Medium
93 of 200
Application +4 | -1
Which of the following do not Obeys Sidwick's rule of EAN
View answer and explanation
Correct option: A
[V(CO)6]\left[{V}({CO})_6\right]
Option A ([V(CO)6]) is correct because it does not obey Sidgwick's EAN rule, as its EAN is 35, not the expected 36 for Krypton.
94 Medium
94 of 200
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 in Zeise's salt, the coordination number of Pt is 4, and ethene is indeed a monodentate ligand.
95 Medium
95 of 200
Analyzing +4 | -1
Which of the following will not show geometrical isomerism?
View answer and explanation
Correct option: A
[Co(ox)3]3−\left[\mathrm{Co}(o x)_3\right]^{3-}
Option A ([Co(ox)3]3−\left[\mathrm{Co}(o x)_3\right]^{3-}) is correct because M(aa)3M(a a)_3 type complexes do not exhibit geometrical isomerism.
96 Medium
96 of 200
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 + }}
The correct answer is D ([Co(H2O)6]2+) because H2O is a weak field ligand, resulting in only 3 unpaired electrons for Co2+, leading to minimum paramagnetic behavior.
97 Medium
97 of 200
Application +4 | -1
The coordination number and oxidation state of central atom for [TiF9]6−\left[{TiF}_9\right]^{6-} are respectively
View answer and explanation
Correct option: B
9,+39,+3
The correct answer is B (9, +3) because the coordination number is 9 (number of ligands) and the oxidation state of Ti is +3.
98 Medium
98 of 200
Analyzing +4 | -1
Which of the following complex does not show geometrical isomerism?
View answer and explanation
Correct option: C
[Cr(en)3]3+\left[\mathrm{Cr}(\mathrm{en})_3\right]^{3+}
The correct answer is C ([Cr(en)_3]3+) because this complex does not exhibit geometrical isomerism due to its symmetrical structure.
99 Medium
99 of 200
Understanding +4 | -1
Mercuric chloride is soluble in KI{KI} solution due to:
View answer and explanation
Correct option: A
The formation of complex ion
Option A is correct because mercuric chloride (HgCl2{HgCl}_2) dissolves in KI solution due to the formation of a complex ion, HgI42−{HgI}_4^{2-}.
100 Medium
100 of 200
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}) does not show optical isomerism because it does not contain any chiral center and exists as geometrical isomers (fac and mer).
101 Medium
101 of 200
Understanding +4 | -1
Which of the following is a square planar complex?
View answer and explanation
Correct option: B
[Cu(NH3)4]2+\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}
Option B ([Cu(NH3)4]2+\left[\mathrm{Cu}(\mathrm{NH}_3)_4\right]^{2+}) is correct because [Cu(NH3)4]2+\left[\mathrm{Cu}(\mathrm{NH}_3)_4\right]^{2+} has a d8d^8 configuration and strong field ligands, leading to a square planar geometry.
102 Medium
102 of 200
Understanding +4 | -1
When an aqueous solution of copper (II) sulphate is saturated with ammonia, the blue compound crystallises on evaporation. The formula of this blue compound is:
View answer and explanation
Correct option: A
[Cu(NH3)4]SO4⋅H2O\left[{Cu}\left({NH}_3\right)_4\right] {SO}_4 \cdot {H}_2 {O} (square planar)
Option A ([Cu(NH3)4]SO4⋅H2O\left[{Cu}\left({NH}_3\right)_4\right] {SO}_4 \cdot {H}_2 {O} (square planar)) is correct because copper(II) ions react with ammonia to form a tetraamminecopper(II) complex, which is square planar and includes a water molecule in the crystal structure.
103 Medium
103 of 200
Application +4 | -1
The CFSE for octahedral [CoCl6]4+{\left[ {CoC{l_6}} \right]^{4 + }} is 18,000 cm−1c{m^{ - 1}}. The CFSE for tetrahedral [COCl4]2−{\left[ {COC{l_4}} \right]^{2 - }} will be
View answer and explanation
Correct option: C
8,000  cm−18,000\,\,c{m^{ - 1}}
Option C (8,000  cm−18,000\,\,c{m^{ - 1}}) is correct because the CFSE for tetrahedral complexes is approximately 4/9 of the octahedral CFSE.
104 Medium
104 of 200
Understanding +4 | -1
Consider the following statements: According to the Werner's theory.
(1) Ligands are connected to the central metal ions by ionic bonds.
(2) Secondary valencies have directional properties
(3) Secondary valencies are non-ionisable
The correct statements are:
View answer and explanation
Correct option: B
2 and 3 are correct
Option B is correct because, according to Werner's theory, ligands are connected to the central metal ions by coordinate bonds, not ionic bonds, and secondary valencies (coordination number) are non-ionizable and have directional properties.
105 Medium
105 of 200
Analyzing +4 | -1
Assertion :- Ethanol cannot show position isomerism.

Reason :- Ethanol cannot show isomerism.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct because while ethanol cannot show position isomerism, it *can* show isomerism (constitutional isomerism with ethers).
106 Medium
106 of 200
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 produces the fewest number of ions in solution, resulting in the lowest electrical conductivity.
107 Medium
107 of 200
Understanding +4 | -1
Given below are two statements:

Statement (I): A solution of [Ni(H2O)6]2+\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} is green in colour.

Statement (II): A solution of [Ni(CN)4]2−\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} is colourless.

In the light of the above statements, choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: B
Both Statement I and Statement II are correct
Option B (Both Statement I and Statement II are correct) is correct because [Ni(H2O)6]2+\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} is green due to d-d transitions, and [Ni(CN)4]2−\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} is colorless due to being diamagnetic.
108 Medium
108 of 200
Understanding +4 | -1
Which of the following has the highest molar conductivity in solution?
View answer and explanation
Correct option: A
[Pt(NH3)6]Cl4\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_4}
Option A ([Pt(NH3)6]Cl4\left[ {\mathrm{Pt}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_6}} \right]\mathrm{C}{\mathrm{l}_4}) is correct because it dissociates into the largest number of ions in solution, leading to the highest molar conductivity.
109 Medium
109 of 200
Analyzing +4 | -1
Assertion: [Fe(CN)6]3−\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} is more paramagnetic than [FeF6]3−\left[\mathrm{FeF}_6\right]^{3-}.

Reason: [FeF6]3−[\mathrm{FeF} 6]^{3-} has more number of unpaired electrons.
View answer and explanation
Correct option: D
Assertion is wrong, but reason is correct statement.
Option D is correct because the assertion is wrong, and the reason is correct. [FeF6]3- has more unpaired electrons than [Fe(CN)6]3-.
110 Medium
110 of 200
Analyzing +4 | -1
Which is high spin complex?
View answer and explanation
Correct option: D
All are correct
The correct answer is D (All are correct) because all the given complexes have weak field ligands, resulting in high spin complexes.
111 Medium
111 of 200
Understanding +4 | -1
The crystal field splitting energy for Cr+3\mathrm{Cr}^{+3} ion in an octahedral field increases for the ligands I−,H2O,NH3\mathrm{I}^{-}, \mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3, CN−\mathrm{CN}^{-}and the order is such that
View answer and explanation
Correct option: A
I−<H2O<NH3<CN−\mathrm{I}^{-} < \mathrm{H}_2 \mathrm{O} < \mathrm{NH}_3 < \mathrm{CN}^{-}
Option A (I−<H2O<NH3<CN−\mathrm{I}^{-} < \mathrm{H}_2 \mathrm{O} < \mathrm{NH}_3 < \mathrm{CN}^{-}) is correct because it represents the correct order of ligands in the spectrochemical series, which determines the crystal field splitting energy.
112 Medium
112 of 200
Analyzing +4 | -1
A s- block element (M) reacts with oxygen to form an oxide of the formula MO2.M{O_2}. The oxide is pale yellow in colour and paramagnetic. The element(M) is
View answer and explanation
Correct option: D
K
Option D (K) is correct because potassium reacts with oxygen to form potassium superoxide (KO2KO_2), which is paramagnetic due to the presence of an unpaired electron.
113 Medium
113 of 200
Analyzing +4 | -1
Of the following complex ions, which is diamagnetic in nature?
View answer and explanation
Correct option: C
[Ni(CN)4]2−{\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}
Option C ([Ni(CN)4]2−{\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }}) is correct because it has a d8 configuration with strong field ligands causing pairing of electrons, resulting in a diamagnetic complex.
114 Medium
114 of 200
Analyzing +4 | -1
Among the complex ions given below which is/are outer - orbital's complex

I) [Co(CN)6]4−\left[\mathrm{Co}(\mathrm{CN})_6\right]^{4-}

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

III) [FeF6]3−\left[\mathrm{FeF}_6\right]^{3-}

IV) [CoF6]3−\left[\mathrm{CoF}_6\right]^{3-}
View answer and explanation
Correct option: A
II, III, IV
Option C (If A is correct but R is incorrect) is correct because at very high pressures, the compressibility factor is greater than 1, but the reason provided about neglecting 'b' is incorrect.
115 Medium
115 of 200
Analyzing +4 | -1
Select the correct statement set?

A) Bis-(glycinato) Zinc (II) is optically active

B) [NiCl4]2−\left[{NiCl}_4\right]^{2-} and [PtCl4]2−\left[{PtCl}_4\right]^{2-} differ in shapes

C) [Ni(CN)4]2−\left[{Ni}({CN})_4\right]^{2-} and [Ni(CO)4]\left[{Ni}({CO})_4\right] have same magnetic moment

D) cis-[Co(NH3)2(en)2]\left[{Co}\left({NH}_3\right)_2({en})_2\right] cannot show optical activity
View answer and explanation
Correct option: C
A, B, C
The correct answer is B (Due to formation of CH₃COOH which acts as a weak acid) because the hydrolysis of methyl acetate produces acetic acid, which neutralizes the NaOH, causing the phenolphthalein to become colorless.
116 Medium
116 of 200
Analyzing +4 | -1
Comment on the magnetic behavior of K2[NiF6]\mathrm{K}_2\left[\mathrm{NiF}_6\right] and K4[NiF6]\mathrm{K}_4\left[\mathrm{NiF}_6\right].
View answer and explanation
Correct option: B
Diamagnetic and paramagnetic
Option B is correct because K₂[NiF₆] is diamagnetic due to strong field caused by high charge on Ni, while K₄[NiF₆] is paramagnetic due to weak field ligand F⁻.
117 Medium
117 of 200
Application +4 | -1
Specify the coordination number of the metal in [CoBr2( en )2]+\left[{CoBr}_2(\text { en })_2\right]^{+}
View answer and explanation
Correct option: D
6
Option D (6) is correct because the coordination number is the number of ligands directly attached to the central metal ion, which is 6 in this case.
118 Medium
118 of 200
Analyzing +4 | -1
Maximum C−O\mathrm{C}-\mathrm{O} bond length present in:
View answer and explanation
Correct option: B
[Fe(CO)5]−\left[\mathrm{Fe}(\mathrm{CO})_5\right]^{-}
Option B (Fe(CO)5−\mathrm{Fe}(\mathrm{CO})_5^{-}) is correct because the negative charge on the metal increases back-bonding, which weakens the C-O bond and thus increases the C-O bond length.
119 Medium
119 of 200
Analyzing +4 | -1
S-I: K3[Fe(CN)6]{K}_3\left[{Fe}({CN})_6\right] is a low spin complex.

S-II: Fe3+{Fe}^{3+} ion in this complex undergoes sp3d2s p^3 d^2 hybridization.
View answer and explanation
Correct option: B
Statement-I is true, Statement-II is false
Option B (Statement-I is true, Statement-II is false) is correct because K3[Fe(CN)6]{K}_3\left[{Fe}({CN})_6\right] is indeed a low spin complex, but the hybridization of Fe3+{Fe}^{3+} is d2sp3d^2sp^3.
120 Medium
120 of 200
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.
Revision notes and key takeaways

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

Showing 91-120 of 200 questions

Your practice

0 / 200 answered · 200 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.