Transition Elements and Coordinate Geometry Previous Year Questions With Solutions

Transition Elements and Coordinate Geometry Previous Year Questions with Solutions

1.0Introduction

Transition metals do not typically bond in the same manner as main group elements. Instead, they primarily form coordinate covalent bonds, a type of Lewis acid-base interaction in which both electrons in the bond are donated by a Lewis base (donor) to a Lewis acid (acceptor). In coordination complexes, the central metal ion (or atom)—often a transition or inner transition metal—acts as the Lewis acid, while the ligands (atoms, molecules, or ions) act as Lewis bases by donating electron pairs. These ligands must contain at least one lone pair of electrons capable of forming a coordinate bond with the metal ion.

2.0Key Concepts

Bonding in Coordination Complexes

Covalent vs. Ionic vs. Coordinate Covalent Bonds

Covalent bonds involve the sharing of electrons.
Ionic bonds involve the transfer of electrons.
Coordinate covalent bonds involve the donation of electron pairs from a ligand to a metal center.

For example, in scandium’s octahedral complex [Sc(H₂O)₆]³⁺, the six water molecules donate lone pairs to form coordinate bonds with the Sc³⁺ ion.

Coordination Sphere and Coordination Number

The coordination sphere consists of the central metal ion and its attached ligands, enclosed in brackets in chemical formulas.
The coordination number refers to the number of donor atoms directly bonded to the metal center.
- In [Ag(NH₃)₂]⁺, silver has a coordination number of 2.
- In [CuCl₄]²⁻, copper(II) has a coordination number of 4.
- In [Co(H₂O)₆]²⁺, cobalt(II) has a coordination number of 6.

Ligands: Monodentate, Bidentate, and Polydentate

Monodentate ligands (e.g., NH₃, Cl⁻, H₂O) donate one lone pair.
Bidentate ligands (e.g., ethylenediamine, en) donate two lone pairs.
Polydentate ligands (e.g., EDTA) donate multiple lone pairs.

Chelation and Chelating Agents

A chelate is a coordination complex in which a polydentate ligand binds through multiple donor atoms.
For example, [Co(en)₃]³⁺ contains three bidentate ethylenediamine (en) ligands, giving cobalt a coordination number of 6.
In heme four nitrogen atoms coordinate to an iron (Fe) center, forming a chelating ligand critical in hemoglobin.

3.0Geometries of Coordination Complexes

Transition metal complexes exhibit octahedral (CN = 6), tetrahedral (CN = 4), or square planar (CN = 4, d⁸ metals) geometries, dictated by their coordination number (CN).

Common Geometries

Linear (CN = 2) → [Ag(NH₃)₂]⁺
Trigonal planar (CN = 3) → [Cu(CN)₃]²⁻
Tetrahedral (CN = 4, d⁰/d¹⁰) → [Ni(CO)₄]
Square planar (CN = 4,d⁸) → [Pt(NH₃)₂Cl₂]
Octahedral (CN = 6) → [Co(H₂O)₆]²⁺

4.0Isomerism

Geometric isomers: Cis-[Co(NH₃)₄Cl₂]⁺ (violet, polar) vs. Trans-[Co(NH₃)₄Cl₂]⁺ (green, nonpolar).

Ligand arrangement affects color, solubility, and reactivity.

Optical Isomers & Coordination Complexes

Optical isomers (enantiomers) are nonsuperimposable mirror images, like hands. While they share physical properties, they differ in polarized light interaction and reactions with chiral molecules.

In coordination chemistry, complexes like [M(en)₃]ⁿ⁺ (M = Fe³⁺, Co²⁺) exhibit optical isomerism, interacting selectively with chiral structures like DNA. [Co(en)₂Cl₂]⁺ shows both geometric and optical isomerism, with the cis form being chiral.

Other Isomerism in Complexes

Linkage Isomerism: Ligands bind via different atoms (e.g., SCN⁻ via S or N).
Ionization Isomerism: Exchange of ligands between inner and outer spheres (e.g., [CoCl₆]Br ↔ [CoCl₅Br]Cl).

5.0Important Previous Year Questions

Q.1 Addition of excess aqueous ammonia to a pink coloured aqueous solution of MCl2. 6H2O (X) and NH4Cl gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1 : 3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y.

Among the following options, which statements is(are) correct ?

(A) The hybridization of the central metal ion in Y is d2sp3

(B) Z is tetrahedral complex

(C) Addition of silver nitrate to Y gives only two equivalents of silver chloride

(D) When X and Z are in equilibrium at 0°C, the colour of the solution is pink

Ans. (A, B, D)

Solution

(A) Hybridisation of (Y) is d²sp³as NH₃ is strong field ligand

(B) [CoCl₄]₂– have sp³ hybridisation as Cl^– is weak field ligand

So, correct answer is (A ,B & D)

Q 2 The complex(es), which can exhibit the type of isomerism shown by [Pt(NH₃)₂Br₂], is(are)

[en = H₂NCH₂CH₂NH₂]

(A) [Pt(en)(SCN)₂]

(B) [Zn(NH₃)2Cl₂]

(C) [Pt(NH₃)₂Cl₄]

(D) [Cr(en)₂(H₂O)(SO₄)]⁺

Ans. (C, D)

Solution

Hybridisation : dsp² , geometry : square planar

(A) [Pt(en)(SCN)₂] : square planar, cis–trans not possible

(B) [Zn(NH₃)2Cl₂] : tetrahedral, cis–trans not possible

(D) [Cr(en)₂(H₂O)(SO₄)]⁺

Q.3 Which of the following is correct order of ligand field strength?

(1) $CO < e n < N H_{3} < C_{2} O_{4}^{2 -} < S^{2 -}$

(2) $S^{2 -} < C_{2} O_{4}^{2 -} < N H_{3} < e n < CO$

(3) $N H_{3} < e n < CO < S^{2 -} < C_{2} O_{4}^{2 -}$

(4) $S^{2 -} < N H_{3} < e n < CO < C_{2} O_{4}^{2 -}$

Ans. (2)

Solution: The increasing order of field strength of ligands (according to spectrochemical series)

Q.4. Given below are two statements : one is labelled as “Assertion A” and the other is labelled as “Reason R”.

Assertion (A) : In the complex Ni(CO)₄ and Fe(CO)₅, the metals have zero oxidation state.

Reason (R) : Low oxidation states are found when a complex has ligands capable of p-donor character in addition to the s-bonding.

In the light of the above statements, choose the most appropriate answer from the option given below.

(1) A is correct but R is not correct

(2) A is not correct but R is correct

(3) Both A and R are correct but R is NOT the correct explanation of A

(4) Both A and R are correct and R is the correct explanation of A.

Ans. (1) A is correct but R is not correct

The low oxidation state of metals can stabilized by synergic bonding so the ligand has to be p-acceptor.

Q.4. Transition metal complex with highest value of crystal field splitting (D₀) will be

(1) $[C r (H_{2} O)_{6}]^{3 +}$ (2) $[M o (H_{2} O)_{6}]^{3 +}$

(3) $[F e (H_{2} O)_{6}]^{3 +}$ (4) $[O s (H_{2} O)_{6}]^{3 +}$

Ans. (4)

Solution : CFSE of octahedral complexes with water is greater for 5d series metal centre ion as compared to 3d and 4d series metal centre.

Q.5. Fe³⁺ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of:

[Fe(H₂O)6]₂[Fe(CN)₆]
Fe₂[Fe(CN)₆]₂
Fe₃[Fe(OH)₂ (CN)₄]₂
Fe₄[Fe(CN)₆]₃

Ans. (4)

Solution : $4 F e^{3 +} + 3 [F e (CN)_{6}]^{- 4} \to 4 F e_{4} [F e (CN)_{6}]_{3}$

Prussian Blue

Q.6. The total number of relatively more stable isomer(s) possible for octahedral complex [Cu(en)₂(SCN)₂] will be ___________.

Ans. (3)

Solution : [Cu(en)₂(SCN)₂]

Q.7. Given below are two statements :

Statement I The identification of Ni²⁺ is carried out by dimethyl glyoxime in the presence of NH₄OH.

Statement II :The dimethyl glyoxime is a bidentate neutral ligand.

In the light of the above statements, choose the correct answer from the options given below: (1) Statement I is false but Statement II is true.

(2) Both Statement I and Statement II are false.

(3) Statement I is true but Statement II is false.

(4) Both Statement I and Statement II are true.

Ans. (3)

Solutions: Neutral dimethyl glyoxime does not act as ligand.

When Ni²⁺ reacts with dimethyl glyoxime in presence of NH4OH, it produce dimethyl glyoximate then it form rosy red ppt.

Q.8 The type of hybridisation and magnetic property of the complex [MnCl₆]^3–, respectively, are :

(1) sp³d² and diamagnetic

(2) d²sp³ and diamagnetic

(3) d²sp³ and paramagnetic

(4) sp³d² and paramagnetic

Ans. (4) sp³d² and paramagnetic

Solution: Paramagnetic and having 4 unpaired electrons.

Q.9 The number of geometrical isomers possible in triamminetrinitrocobalt (III) is X and in trioxalatochromate (III) is Y. Then the value of X + Y is _______.

Ans. (2) X + Y = 2 + 0 = 2.0

Solution:

Triamminetrinitrocobalt(III) → [Co(NO₂)₃(NH₃)₃]

trioxalatochromate(III) ion → [Cr(C₂O₄)₃]^3– [Co(NO₂)₃(NH₃)₃]

Q.10 Choose the correct tests with respective observations.

(A) CuSO₄ (acidified with acetic acid) + K₄[Fe(CN)₆] →Chocolate brown precipitate.

(B) FeCl₃ + K₄[Fe(CN)₆] → Prussian blue precipitate.

(C) ZnCl₂+ K₄[Fe(CN)₆], neutralised with NH₄OH → White or bluish white precipitate.

(D) MgCl₂+ K₄[Fe(CN)₆] → Blue precipitate.

(E) BaCl₂ + K₄[Fe(CN)₆], neutralised with NaOH → White precipitate.

Choose the correct answer from the options given below :

(1) A, D and E only (2) B, D and E only

(3) A, B and C only (4) C, D and E only

Q.11 Given below are two statements :

Statement I : A homolepitc octahedral complex, formed using monodentate ligands, will not show stereoisomerism.

Statement II : cis– and trans– platin are heteroleptic complexes of Pd.

In the light of the above statements, choose the correct answer from the options given below.

(1) Both statement I and Statement II are false.

(2) Statement I is false but Statement II is true.

(3) Both statement I and Statement II are true.

(4) Statement I is true but Statement II is false.

Ans. (4)

Homoleptic complexes of type [Ma₆] (Where a Þ monodentate ligand) cannot show geometrical as well as optical isomerism.

Cis-platin and trans-platin have formula [Pt(NH₃)₂Cl₂] which is a heteroleptic complex of platinum.

Q. 12 Match the LIST-I with LIST-II

LIST-I (Complex/Species)		LIST-II (Shape & magnetic moment)
A.	[Ni(CO)₄]	I.	Tetrahedral, 2.8 BM
B.	[Ni(CN)₄]^2–	II.	Square planar, 0 BM
C.	[NiCl₄]^2–	III.	Tetrahedral, 0 BM
D.	[MnBr₄]^2–	IV	Tetrahedral, 5.9 BM

Choose the correct answer from the options given below :

(1) A-III, B-IV, C-II, D-I

(2) A-I, B-II, C-III, D-IV

(3) A-III, B-II, C-I, D-IV

(4) A-IV, B-I, C-III, D-II

Ans. (3)

Solution

(A) [Ni(CO)₄] , Ni⁰ : [Ar]3d⁸ 4s²

Valence orbitals of Ni⁰ in pre-hybridisation state :

Tetrahedral, Diamagnetic, µ = 0 B.M.

(B) [Ni(CN)₄]^2– , Ni⁺² : [Ar]3d⁸4s⁰

Valence orbitals of Ni⁺² in pre-hybridisation state :

Square planar, Diamagnetic, µ = 0 B.M.

(C) [NiCl₄]^2– , Ni⁺² : [Ar]3d⁸4s⁰

Valence orbitals of Ni⁺² in ground state :

Tetrahedral, paramagnetic, µ = √8 = 2.8 B.M.

(D) [MnBr₄]^2– , Mn⁺² : [Ar]3d⁵

Valence orbitals of Mn⁺² in ground state :

Q.13 Pair of transition metal ions having the same number of unpaired electrons is :

(1) V²⁺, Co²⁺(2) Ti²⁺, Co²⁺(3) Fe³⁺, Cr²⁺ (4) Ti³⁺, Mn²⁺

Ans. (1)

		Configuration	No. of unpaired e–
-1	V3⁺	[Ar]3d³4s⁰	3
-1	Co²⁺	[Ar]3d⁷4s⁰	3
-2	Ti²⁺	[Ar]3d²4s0	2
-2	Co²⁺	[Ar]3d⁷4s⁰	3
-3	Fe³⁺	[Ar]3d⁵4s⁰	5
-3	Cr²⁺	[Ar]3d⁴4s⁰	4
-4	Tⁱ³⁺	[Ar]3d¹4s⁰	1
-4	Mn²⁺	[Ar]3d⁵4s⁰	5

So V²⁺ & Co²⁺ same number of unpaired electron.

Q.14 The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively :

[Fe(CN)₆]^3–, [FeF₆]^3–, [CoF₆]^3–, [Mn(CN)₆]^3–

(1) 1, 5, 4, 2 (2) 1, 5, 5, 2 (3) 1, 1, 4, 2 (4) 1, 4, 4, 2

Ans. (1)

Solution

[Fe(CN)₆]^3–	Fe³⁺	3d⁵	unpaired e– = 1
[FeF₆]^3–	Fe³⁺	3d⁵	unpaired e– = 5
[CoF₆^]3–	Co³⁺	3d⁶	unpaired e– = 4
[Mn(CN)₆^]3–	Mn³⁺	3d⁴	unpaired e– = 2

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Transition Elements and Coordinate Geometry Previous Year Questions with Solutions

1.0Introduction

2.0Key Concepts

Bonding in Coordination Complexes

3.0Geometries of Coordination Complexes

4.0Isomerism

Optical Isomers & Coordination Complexes

Other Isomerism in Complexes

5.0Important Previous Year Questions

Table of Contents

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