Coordination Chemistry

01

PYQ 2017

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chemistry ID: iit-jam-

The number of unpaired electron(s) in K NiF is ...........

02

PYQ 2017

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chemistry ID: iit-jam-

Jahn-Teller distortion is observed in octahedral complexes with d-electron configuration of

1

d high spin

2

d low spin

3

d high spin

4

d low spin

03

PYQ 2017

medium

chemistry ID: iit-jam-

The coordination number of Al in crystalline AlCl and liquid AlCl, respectively, is

1

4 and 4

2

6 and 6

3

6 and 4

4

3 and 6

04

PYQ 2017

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chemistry ID: iit-jam-

Nitrosyl ligand binds to d-metal atoms in linear and bent fashion and behaves, respectively, as

1

NO and NO

2

NO and NO

3

NO and NO

4

NO and NO

05

PYQ 2017

medium

chemistry ID: iit-jam-

The correct order of wavelength of absorption () of the Cr-complexes is (en = ethylenediamine)

1

[CrF] >[Cr(H O)] >[Cr(en)] >[Cr(CN)]

2

[Cr(H O)] >[CrF] >[Cr(en)] >[Cr(CN)]

3

[Cr(CN)] >[Cr(en)] >[Cr(H O)] >[CrF]

4

[Cr(en)] >[Cr(CN)] >[Cr(H O)] >[CrF]

06

PYQ 2021

medium

chemistry ID: iit-jam-

$According to the crystal field theory, d-d transition observed in [Ti(H O)] is$

1

$Laporte forbidden and spin forbidden$

2

$Laporte allowed and spin forbidden$

3

$Laporte allowed and spin allowed$

4

$Laporte forbidden and spin allowed$

Official Solution

Correct Option: (4)

$Step 1: Electronic configuration. In [Ti(H O)], Ti = 3d configuration. Hence, there is only one possible transition: . Step 2: Selection rules. (a) Laporte rule: Transitions within the same parity (g \to g or u \to u) are forbidden. As this is an octahedral complex (center of symmetry present), the - transition is Laporte forbidden. (b) Spin selection rule: Transitions that do not change spin multiplicity are spin allowed. Here, only one unpaired electron is present (no spin change occurs). Hence, spin allowed. Step 3: Conclusion. The transition is Laporte forbidden but spin allowed .$

07

PYQ 2021

medium

chemistry ID: iit-jam-

$Reaction of [Ni(CN)] with metallic potassium in liquid ammonia at -33 °C yields complex E. The geometry and magnetic behavior of E, respectively, are$

1

$Square planar and diamagnetic$

2

$Tetrahedral and diamagnetic$

3

$Octahedral and paramagnetic$

4

$Square pyramidal and paramagnetic$

Official Solution

Correct Option: (2)

$Step 1: Reaction type. Reduction of [Ni(CN)] with metallic potassium produces [Ni(CN)], where Ni is in the +0 oxidation state. Step 2: Electronic configuration. Ni : 3d configuration \to fully filled d-orbitals, no unpaired electrons \to diamagnetic. Step 3: Geometry. In the zero oxidation state, Ni forms tetrahedral complexes with CN ligands. Therefore, geometry = tetrahedral and magnetic nature = diamagnetic. Step 4: Conclusion. The correct answer is (B) Tetrahedral and diamagnetic .$

08

PYQ 2021

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chemistry ID: iit-jam-

$Spin-only magnetic moments (in BM) of [NiCl (PPh)] and [Mn(NCS)], respectively, are$

1

$0.00 and 5.92$

2

$2.83 and 1.89$

3

$0.00 and 1.89$

4

$2.83 and 5.92$

Official Solution

Correct Option: (4)

$Step 1: For [NiCl (PPh)]. Ni is in +2 oxidation state \to 3d configuration. Cl and PPh are weak field ligands \to complex is paramagnetic . Unpaired electrons = 2. Spin-only magnetic moment, Step 2: For [Mn(NCS)] . Mn is in +2 oxidation state \to 3d configuration. All ligands are weak field (high-spin complex). Unpaired electrons = 5. Step 3: Conclusion. Hence, magnetic moments are 2.83 BM and 5.92 BM, respectively.$

09

PYQ 2021

medium

chemistry ID: iit-jam-

$If the crystal field splitting energy of [Co(NH)] is 5900 cm, then the magnitude of its crystal field stabilization energy, in kJ mol (rounded off to one decimal place), is ________.$

Official Solution

Correct Option: (1)

$Step 1: Recall CFSE for octahedral (high-spin) complex. For high-spin configuration (in octahedral field): Step 2: Convert to energy. However, the complex [Co(NH)] is low-spin (), so Step 3: Conclusion. CFSE = 141.5 kJ mol .$

10

PYQ 2021

medium

chemistry ID: iit-jam-

$The total number of optically active isomers of dichloridobis(glycinato)cobalt(III) ion is ________.$

Official Solution

Correct Option: (1)

$Step 1: Identify coordination geometry. The complex is [Co(gly) Cl], where gly = glycinate (bidentate ligand). Co(III) forms an octahedral complex with two bidentate (gly) and two monodentate (Cl) ligands. Step 2: Types of isomerism. This complex can exhibit: - Geometrical isomerism (cis and trans forms). - The cis form is optically active due to lack of symmetry, while the trans form is optically inactive (has a plane of symmetry). Step 3: Count optical isomers. The cis form has two non-superimposable mirror images (enantiomers). Step 4: Conclusion. Therefore, the number of optically active isomers is 2 .$

11

PYQ 2021

medium

chemistry ID: iit-jam-

$The complex(es) that show(s) Jahn-Teller distortion is (are)$

1

$[Co(CN) (H O)]$

2

$[NiF]$

3

$[Mn(CNMe)]$

4

$[Co(en)]$

Official Solution

Correct Option: (1)

$Step 1: Recall Jahn-Teller theorem. Jahn-Teller distortion occurs in coordination complexes with unevenly occupied degenerate orbitals (especially or levels). Step 2: Check electronic configurations. (A) Co(III): d low-spin \to (no degeneracy) \to no distortion. (B) Ni(II): d \to in octahedral field \to, unevenly filled orbitals \to Jahn-Teller active . (C) Mn(II): d high-spin \to symmetric half-filled \to no distortion. (D) Co(III): d low-spin \to \to no distortion. Step 3: Conclusion. The only complex showing Jahn-Teller distortion is [NiF] (Option B)}.$

12

PYQ 2021

medium

chemistry ID: iit-jam-

$The CORRECT statement(s) about sodium nitroprusside is (are)$

1

$It is a paramagnetic complex$

2

$Nitroprusside ion is formed in the brown ring test for nitrates$

3

$It is used for the detection of S in aqueous solution$

4

$It contains nitrosyl ligand as NO$

Official Solution

Correct Option: (3)

$Step 1: Structure and oxidation states. Sodium nitroprusside = Na [Fe(CN) (NO)]. Here Fe is in +2 oxidation state and NO acts as a nitrosyl cation (NO). Step 2: Properties. - The complex is paramagnetic due to one unpaired electron. - It is used for qualitative detection of S ions (formation of purple color due to [Fe(CN) (NOS)] complex). - The nitrosyl group is linear, indicating NO character. Step 3: Conclusion. Correct statements are (A), (C), and (D).$

13

PYQ 2025

medium

chemistry ID: iit-jam-

$A yellow compound X is produced after the reaction of with excess of at -33°C. The oxidation state of Ni in the compound X is ..........$

Official Solution

Correct Option: (1)

$The given compound is, where the oxidation state of Ni is +2. On reacting with excess, a yellow compound is formed. This typically suggests a reduction of Ni(II) to Ni(0) since yellow complexes are often associated with nickel in the zero oxidation state. Thus, the oxidation state of Ni in compound X is 0. Final Answer:$

14

PYQ 2025

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chemistry ID: iit-jam-

$The correct statement(s) about the octahedral Mn-complex with spin only magnetic moment of approximately 6.0 is(are)$

1

$Possible oxidation state of Mn in the complex is +4$

2

$Possible oxidation state of Mn in the complex is +2$

3

$Ligands associated with the complex is strong field ligand$

4

$Ligands associated with the complex is weak field ligand$

Official Solution

Correct Option: (2)

$Step 1: Understanding spin-only magnetic moment. The spin-only magnetic moment is calculated using the formula: where is the number of unpaired electrons. Given that the magnetic moment is approximately 6.0, solving for gives approximately 4 unpaired electrons, suggesting a configuration. Step 2: Oxidation state of Mn. - Mn(II): The configuration corresponds to Mn(II) with 4 unpaired electrons in a weak field octahedral complex. - Mn(IV): Would have a configuration, which would not give 4 unpaired electrons. Step 3: Conclusion. The possible oxidation state of Mn in this complex is +2, with weak field ligands. Final Answer:$

15

PYQ 2025

medium

chemistry ID: iit-jam-

$In alkaline medium, which of the following metal ion(s) form(s) red precipitate/coloration with dimethylglyoxime?$

1

$Ni(II)$

2

$Bi(III)$

3

$Zn(II)$

4

$Fe(II)$

Official Solution

Correct Option: (1)

$Step 1: Dimethylglyoxime and metal ion complexes. Dimethylglyoxime is commonly used as a reagent to form colored complexes, particularly with metal ions such as Ni(II). The complex formed with Ni(II) is typically red. Step 2: Analyzing each metal ion. - Ni(II): Forms a red precipitate with dimethylglyoxime, indicating a positive test for nickel. - Bi(III): Does not form a red precipitate with dimethylglyoxime. - Zn(II): Forms a white precipitate with dimethylglyoxime, not a red one. - Fe(II): Forms a light green precipitate with dimethylglyoxime, not a red one. Step 3: Conclusion. Only Ni(II) forms a red precipitate with dimethylglyoxime. Final Answer:$

16

PYQ 2026

medium

chemistry ID: iit-jam-

$Which of the following has tetrahedral coordination?$

1

$CdI$

2

$CsCl$

3

$ZnS (sphalerite)$

4

$ZnS (wurtzite)$

Official Solution

Correct Option: (3)

$Step 1: Understand tetrahedral coordination. In tetrahedral coordination, each cation is surrounded by 4 anions arranged at the corners of a tetrahedron. Coordination number = 4. Step 2: Analyze each compound. (A) CdI \to Layered structure \to Octahedral coordination (Coordination number = 6). (B) CsCl \to Cubic structure \to Coordination number = 8 (body-centered cubic type). (C) ZnS (sphalerite) \to Zinc blende structure \to Each Zn surrounded by 4 S \to Tetrahedral coordination. (D) ZnS (wurtzite) \to Hexagonal structure \to Also tetrahedral coordination (Coordination number = 4). Step 3: Conclusion. Both sphalerite and wurtzite forms of ZnS have tetrahedral coordination.$

High-Yield Trend

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16 MCQs

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Coordination Chemistry

High-Yield Trend

Chapter Questions 16 MCQs

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Chapter Questions
16 MCQs