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JEE • CHEMISTRY • CO-ORDINATION COMPOUNDS

Co-ordination compounds practice · Mixed difficulty

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121 Medium
121 of 326
Understanding +4 | -1
Assertion: The ligands nitro and nitrito are called ambidentate ligands.

Reason: These ligands give linkage isomers.
View answer and explanation
Correct option: A
Assertion is true, Reason is true and Reason is correct explanation for Assertion
The correct answer is A (Assertion is true, Reason is true and Reason is correct explanation for Assertion) because ambidentate ligands can bind through different atoms, leading to linkage isomers.
122 Medium
122 of 326
Analyzing +4 | -1
Among the following metal carbonyls, C−O\mathrm{C} - \mathrm{O} bond order is lowest in
View answer and explanation
Correct option: B
[Fe(CO)5]\left[ {\mathrm{Fe}{{\left( {\mathrm{CO}} \right)}_5}} \right]
Option B ([Fe(CO)5]) is correct because it has the highest number of electrons available for back-bonding, leading to the lowest C-O bond order.
123 Medium
123 of 326
Analyzing +4 | -1
Which of the following statements is not correct?
View answer and explanation
Correct option: D
The complexes [NiCl4]2−  and   [Ni(CN)4]2−{\left[ {\mathrm{NiC}{\mathrm{l}_4}} \right]^{2 - }}\mathrm{\;and}\,\mathrm{\;}{\left[ {\mathrm{Ni}{{\left( {\mathrm{CN}} \right)}_4}} \right]^{2 - }} differ in primary valencies of nickel.
Option C (Both statements are correct) is incorrect. Both statements are incorrect because the heat of neutralization of HF(aq) with NaOH(aq) is *more* than 13.7 kcal due to the high hydration energy of F-, and some heat is *released* during the hydration of F-.
124 Hard
124 of 326
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.
125 Medium
125 of 326
Understanding +4 | -1
The IUPAC name of K3[Ir⁡(C2O4)3]\mathrm{K}_3\left[\operatorname{Ir}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right] is
View answer and explanation
Correct option: B
potassium trioxalatoiridate (III)
Option D (potassium tris (oxalato) iridate (III)) is correct because it follows IUPAC nomenclature rules for coordination compounds, naming the cation first, then the ligand alphabetically with appropriate prefixes, and ending with the metal name with its oxidation state in Roman numerals and \"-ate\" suffix because the complex is an anion.
126 Medium
126 of 326
Understanding +4 | -1
Assertion: (Ph3P)3RhCl\left({Ph}_3 P\right)_3 {RhCl} is Wilkinson's catalyst.
Reason: (Ph3P)3RhCl\left({Ph}_3 {P}\right)_3 {RhCl} can show optical isomerism but does not show geometrical isomerism.
View answer and explanation
Correct option: C
Assertion is true, Reason is false
Option C (Assertion is true, Reason is false) is correct because Wilkinson's catalyst is indeed (Ph3P)3RhCl(Ph_3P)_3RhCl, but it does not show optical isomerism or geometrical isomerism.
127 Medium
127 of 326
Application +4 | -1
Spin only magnetic moment of which complex is highest?
View answer and explanation
Correct option: C
[FeF6]3−\left[\mathrm{FeF}_{{6}}\right]^{3-}
The correct answer is C ([FeF_{{6}}]3-) because it has the highest number of unpaired electrons, resulting in the highest spin-only magnetic moment.
128 Easy
128 of 326
Application +4 | -1
Which among the following doesn't form a white precipitae with AgNO3{AgNO}_3 solution?
View answer and explanation
Correct option: D
[CO(NH3)3Cl3]\left[{CO}\left({NH}_3\right)_3 {Cl}_3\right]
Option D ([CO(NH3)3Cl3]) is correct because it does not have any chloride ions outside the coordination sphere to react with AgNO3 to form a white precipitate of AgCl.
129 Easy
129 of 326
Understanding +4 | -1
The IUPAC name of the complex ion formed when gold dissolves in aqua-regia is
View answer and explanation
Correct option: A
Tetrachloridoaurate (iii)
Option A is correct because when gold dissolves in aqua regia, it forms the complex ion [AuCl4]−[AuCl_4]^−, which is named tetrachloridoaurate(III).
130 Medium
130 of 326
Analyzing +4 | -1
Which of the following complex has highest C−O\mathrm{C}-\mathrm{O} bond energy?
View answer and explanation
Correct option: B
[Co(CO)6]3+\left[\mathrm{Co}(\mathrm{CO})_6\right]^{3+}
Option B is correct because a higher positive charge on the metal center reduces back-bonding, leading to a stronger C-O bond.
131 Medium
131 of 326
Understanding +4 | -1
Which of the following ligand is tridentate type?
View answer and explanation
Correct option: C
Diethylamine (dien)
Option C (Diethylamine (dien)) is correct because diethylamine (dien) acts as a tridentate ligand, coordinating through three nitrogen atoms.
132 Medium
132 of 326
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
The correct answer is D (I, II and IV are correct) because chain and position isomerism cannot occur together, tautomerism is acid/base catalyzed, and tautomers are functional isomers.
133 Medium
133 of 326
Understanding +4 | -1
Coordination number of Co\mathrm{Co} in the complex [Co(en)3]3∗\left[\mathrm{Co}(\mathrm{en})_3\right]^{3 *} is
View answer and explanation
Correct option: C
66
Option C (6) is correct because ethylenediamine (en) is a bidentate ligand, and three en ligands coordinate to the cobalt ion, resulting in a coordination number of 6.
134 Easy
134 of 326
Understanding +4 | -1
Number of ions for the complex [Co(CH3)4Cl]Cl2\left[{Co}\left({CH}_3\right)_4 {Cl}\right] {Cl}_2 shown by the conductivity measurement, is
View answer and explanation
Correct option: C
3
The correct answer is C. The complex dissociates into three ions: one complex ion and two chloride ions.
135 Medium
135 of 326
Analyzing +4 | -1
Which of the following complexes will have four different isomers?
View answer and explanation
Correct option: B
[Co(en)(NH3)2Cl2]Cl\left[ {Co(en){{(N{H_3})}_2}C{l_2}} \right]Cl
Option B ([Co(en)(NH3)2Cl2]Cl) is correct because this complex can exhibit both geometrical (cis/trans) and optical isomerism, resulting in a total of four isomers.
136 Medium
136 of 326
Understanding +4 | -1
Which of the following is not correctly matched?

Coordination polyhedronShape(A)[Co(NH3)6]3 + Octahedral(B)[Ni(CO)4]Square planar(C)[PtCl4]2 - Tetrahedral\begin{array}{cccccccccccccccccccc} {}&{{\text{Coordination}}\,{\text{polyhedron}}}&{{\text{Shape}}} \\ {{\text{(A)}}}&{{{{\text{[Co(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}{\text{]}}}^{{\text{3 + }}}}}&{{\text{Octahedral}}} \\ {{\text{(B)}}}&{{\text{[Ni(CO}}{{\text{)}}_{\text{4}}}{\text{]}}}&{{\text{Square}}\,{\text{planar}}} \\ {{\text{(C)}}}&{{{{\text{[PtC}}{{\text{l}}_{\text{4}}}{\text{]}}}^{{\text{2 - }}}}}&{{\text{Tetrahedral}}} \end{array}
View answer and explanation
Correct option: B
B and C
Option B (B and C) is correct because [Ni(CO)4] is tetrahedral and [PtCl4]2- is square planar, so the shapes listed are incorrect.
137 Medium
137 of 326
Analyzing +4 | -1
Amongst the following statements.

I: The structure of [BrF4]+\left[\mathrm{BrF}_4\right]^{+} is regular tetrahedral

II: Bond order of O2\mathrm{O}_2 decreases by the removal of an electron

III: Br3−\mathrm{Br}_3^{-} and Br3+\mathrm{Br}_3^{+} are having same shape and same structure

IV Hybrid orbital of P in PCl5\mathrm{PCl}_5 is sp3dxy\mathrm{sp}^3 \mathrm{d}_{\mathrm{xy}}.

The incorrect statements is/ are
View answer and explanation
Correct option: D
I, II, III, IV
Option B (I, III, IV only) is correct because statements I, III, and IV are all incorrect regarding the structure and properties of the given chemical species.
138 Hard
138 of 326
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.
139 Medium
139 of 326
Analyzing +4 | -1
Which of the following is false?
View answer and explanation
Correct option: B
[FeF6]3−{\left[ {Fe{F_6}} \right]^{3 - }} is an inner orbital complex
The correct answer is B ([FeF6]3−{\left[ {Fe{F_6}} \right]^{3 - }} is an inner orbital complex) because fluoride is a weak field ligand, and therefore [FeF6]3−{\left[ {Fe{F_6}} \right]^{3 - }} is an outer orbital complex, not an inner orbital complex.
140 Medium
140 of 326
Understanding +4 | -1
Assertion :- Organic compounds which do not contain chiral carbon atoms can be optically
active.

Reason :-Presence of chiral carbon is an essential condition for optical activity.
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 optical activity can occur without chiral carbon atoms, and the presence of a chiral carbon is not an essential condition.
141 Medium
141 of 326
Analyzing +4 | -1
The [Fe(CN)6]3−\left[{Fe}({CN})_6\right]^{3-} complex ion
View answer and explanation
Correct option: C
Should be very stable
Option C (Should be very stable) is correct because the high charge density of Fe3+Fe^{3+} and the strong basic character of CN−CN^- contribute to the stability of the complex.
142 Medium
142 of 326
Analyzing +4 | -1
Wrong match is
View answer and explanation
Correct option: D
NH4ClN{H_4}Cl - Both cationic and anionic hydrolysis
Option D (NH4ClN{H_4}Cl - Both cationic and anionic hydrolysis) is the wrong match because NH4ClNH_4Cl only undergoes cationic hydrolysis.
143 Hard
143 of 326
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.
144 Medium
144 of 326
Analyzing +4 | -1
The crystal field stabilization energy (CFSE) is highest for
View answer and explanation
Correct option: C
[Co(NH3)6]3+\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}
Option C ([Co(NH3)6]3+[Co(NH_3)_6]^{3+}) is correct because NH3NH_3 is a strong field ligand, leading to a higher crystal field stabilization energy (CFSE).
145 Medium
145 of 326
Analyzing +4 | -1
Statement I: Under the influence of a strong field ligand, d7d^7-system will have only one unpaired electron either in coordination number six or four

Statement II: t2 g6eg1\mathrm{t}_{2 \mathrm{~g}}{ }^6 \mathrm{e}_{\mathrm{g}}{ }^1 is the electronic configuration for both cases
View answer and explanation
Correct option: C
Statement I is correct but statement II is false
Statement I is correct, but Statement II is false because the electronic configuration for a strong field d7d^7 system is t2g6eg1t_{2g}^6e_g^1 in an octahedral field and e4t23e^4t_2^3 in a tetrahedral field.
146 Hard
146 of 326
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.
147 Hard
147 of 326
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.
148 Medium
148 of 326
Analyzing +4 | -1
The most stable ion is
View answer and explanation
Correct option: C
[Fe(CN)6]3−{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]^{3 - }}
Option C is correct because CN⁻ is a strong field ligand and forms a stable complex with Fe³⁺.
149 Medium
149 of 326
Analyzing +4 | -1
According to valence bond theory which of the following statement is correct about the complexes Ni(CO)4{Ni}({CO})_4 and [Ni(CN)4]2−\left[{Ni}({CN})_4\right]^{2-} if both are diamagnetic in nature
View answer and explanation
Correct option: D
One is tetrahedral and other is square planar
The correct answer is D because Ni(CO)4{Ni}({CO})_4 is tetrahedral due to sp3sp^3 hybridization, while [Ni(CN)4]2−\left[{Ni}({CN})_4\right]^{2-} is square planar due to dsp2dsp^2 hybridization.
150 Medium
150 of 326
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.
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