Coordination Compounds MCQ Quiz - Objective Question with Answer for Coordination Compounds - Download Free PDF

Last updated on Jun 8, 2025

Latest Coordination Compounds MCQ Objective Questions

Coordination Compounds Question 1:

The correct order of the complexes [Co(NH3)5 (H2 O)]3+ (A), [Co(NH3)6 ]3+ (B), [Co(CN)6 ]3– (C) and [CoCl(NH3)5 ]2+ (D) in terms wavelength of light absorbed is : 

  1. D > A > B > C
  2. C > B > D > A
  3. D > C > B > A 
  4. C > B > A > D

Answer (Detailed Solution Below)

Option 1 : D > A > B > C

Coordination Compounds Question 1 Detailed Solution

CONCEPT:

Crystal Field Stabilization Energy (CFSE) and Wavelength of Light Absorbed

E = hν = hC / λ

  • In coordination complexes, the wavelength of light absorbed is inversely related to the energy of the d-d transition. This relationship is expressed as:
  • Where:
    • E is the energy of the absorbed light,
    • h is Planck’s constant,
    • ν is the frequency of light,
    • C is the speed of light,
    • λ is the wavelength of the absorbed light.
  • The Crystal Field Stabilization Energy (CFSE) is related to the ligand field strength. The stronger the ligand field, the higher the CFSE, and the lower the wavelength of light absorbed (since energy is higher).
  • The order of ligand field strength for various ligands is:
    • CN- > NH3 > H2O > Cl-

EXPLANATION:

  • The order of CFSE is therefore: C > B > A > D. Since higher CFSE corresponds to shorter wavelengths of absorbed light, the order of wavelength is the reverse: D > A > B > C.
  • In these complexes, all Co are in the +3 oxidation state, and the ligand field strength determines the CFSE, which in turn affects the wavelength of light absorbed.
  • The order of ligand field strength is: CN- > NH3 > H2O > Cl-
  • Based on this, the CFSE order is: C > B > A > D, and the wavelength order (since E ∝ 1/λ) is: D > A > B > C.

Therefore, the correct answer is: D > A > B > C.

Coordination Compounds Question 2:

Match List I with List II.

  List - I
(Complex)
  List - II
(Hybridisation and Geometry)
A. [MnCl4]2− I. sp3, Tetrahedral
B. [Cu(NH3)4]2+ II. dsp2, Square planar
C. [CoF6]3− III. sp3d2, Outer orbital octahedral
D. [Co(NH3)6]3+ IV. d2sp3, Inner orbital octahedral
    V. d2sp3, Outer orbital octahedral

Choose the correct answer from the options given below

  1. A - IV, B - III, C - I, D - II 
  2. A - III, B - I, C - IV, D - II
  3. A - II, B - I, C - IV, D - III
  4. A - I, B - II, C - III, D - IV

Answer (Detailed Solution Below)

Option 4 : A - I, B - II, C - III, D - IV

Coordination Compounds Question 2 Detailed Solution

CONCEPT:

Hybridisation and Geometry of Coordination Compounds

  • In coordination chemistry, the hybridisation of the central metal ion determines the geometry and magnetic properties of the complex.
  • Common hybridisations and corresponding geometries:
    • sp3: Tetrahedral
    • dsp2: Square planar
    • d2sp3: Inner orbital octahedral
    • sp3d2: Outer orbital octahedral
  • Paramagnetic behavior arises due to the presence of unpaired electrons in d-orbitals.

EXPLANATION:

  • A - I: [MnCl4]2−
    • Manganese has a +2 oxidation state (d5 configuration).
    • Cl is a weak ligand → no pairing occurs.
    • Hybridisation: sp3 → Geometry: Tetrahedral
    • qImage6820d39397cd103818d73661
  • B - II: [Cu(NH3)4]2+
    • Copper is in +2 state (d9 configuration).
    • Strong field NH3 ligands → pairing occurs → dsp2
    • Geometry: Square planar
    • qImage6820d39497cd103818d73663
  • C - III: [CoF6]3−
    • Co3+ is d6.
    • F is a weak field ligand → no pairing.
    • Uses 4s, 4p, and 4d → sp3d2 hybridisation (outer orbital)
    • Geometry: Octahedral
    • qImage6820d39497cd103818d73665
  • D - IV: [Co(NH3)6]3+
    • Co3+ is d6.
    • NH3 is a strong field ligand → pairing occurs.
    • Uses inner 3d orbitals → d2sp3 hybridisation (inner orbital)
    • Geometry: Octahedral
    • qImage6820d39597cd103818d73666

Therefore, the correct answer is: Option 4) A - I, B - II, C - III, D - IV

Coordination Compounds Question 3:

Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?

  1. [Co(NH₃)₃Cl₃]
  2. [Co(NH₃)₄Cl₂]Cl
  3. [Co(NH₃)₆]Cl₃
  4. [Co(NH₃)₅Cl]Cl

Answer (Detailed Solution Below)

Option 1 : [Co(NH₃)₃Cl₃]

Coordination Compounds Question 3 Detailed Solution

CONCEPT:

Conductance of Ionic Compounds in Solution

  • The conductance of a compound in solution depends on the number of ions produced upon dissociation.
  • More the number of ions, higher the conductance.
  • Complex compounds can dissociate into different numbers of ions depending on their structure and coordination sphere.

EXPLANATION:

  • [Co(NH₃)₃Cl₃]: This is a neutral complex where all three chlorides are coordinated to cobalt. It does not dissociate into ions. Hence, conductance is minimum.
  • [Co(NH₃)₄Cl₂]Cl: This is a 1:2 electrolyte. It dissociates as:

    [Co(NH₃)₄Cl₂]Cl → [Co(NH₃)₄Cl2]+ + Cl-

    Producing 2 ions in solution.
  • [Co(NH₃)₆]Cl₃: This is a 1:3 electrolyte. It dissociates as:

    [Co(NH₃)₆]Cl₃ → [Co(NH₃)₆]3+ + 3Cl-

    Producing 4 ions in solution.
  • [Co(NH₃)₅Cl]Cl: This is a 1:2 electrolyte. It dissociates as:

    [Co(NH₃)₅Cl]Cl → [Co(NH₃)₅Cl]+ + Cl-

    Producing 2 ions in solution.

So, [Co(NH₃)₃Cl₃] produces no ions in solution and will have the minimum conductance.

Therefore, the compound with minimum conductance in solution is [Co(NH₃)₃Cl₃].

Coordination Compounds Question 4:

The correct order of the wavelength of light absorbed by the following complexes is,
A. [Co(NH₃)₆]³⁺
B. [Co(CN)₆]³⁻
C. [Cu(H₂O)₄]²⁺
D. [Ti(H₂O)₆]³⁺
Choose the correct answer from the options below:

  1. B < D < A < C
  2. B < A < D < C
  3. C < D < A < B
  4. C < A < D < B

Answer (Detailed Solution Below)

Option 2 : B < A < D < C

Coordination Compounds Question 4 Detailed Solution

CONCEPT:

Wavelength of Light Absorbed by Complexes

  • The wavelength of light absorbed by a complex depends on the energy gap (Δo, crystal field splitting) between the d-orbitals in the central metal ion.
  • This energy gap is influenced by the nature of the ligands and their position in the spectrochemical series.
  • Strong field ligands (e.g., CN⁻) cause a larger splitting of d-orbitals (higher Δo), leading to the absorption of light with shorter wavelengths (higher energy).
  • Weak field ligands (e.g., H₂O) cause smaller splitting of d-orbitals (lower Δo), leading to the absorption of light with longer wavelengths (lower energy).

EXPLANATION:

  • In the given complexes:
    • [Co(NH₃)₆]³⁺: NH₃ is a moderate field ligand and causes moderate splitting of d-orbitals.
    • [Co(CN)₆]³⁻: CN⁻ is a strong field ligand and causes large splitting of d-orbitals, resulting in absorption of shorter wavelengths.
    • [Cu(H₂O)₄]²⁺: H₂O is a weak field ligand and causes small splitting of d-orbitals, resulting in absorption of longer wavelengths.
    • [Ti(H₂O)₆]³⁺: H₂O is also a weak field ligand, but the metal ion Ti³⁺ has a smaller splitting than Cu²⁺.
  • Comparing the strength of the ligands and the splitting:
    • [Co(CN)₆]³⁻ absorbs the shortest wavelength due to CN⁻ (strong field ligand).
    • [Co(NH₃)₆]³⁺ absorbs a shorter wavelength than [Ti(H₂O)₆]³⁺ due to NH₃ (moderate field ligand).
    • [Cu(H₂O)₄]²⁺ absorbs the longest wavelength due to H₂O (weak field ligand) and Cu²⁺ having a higher splitting than Ti³⁺.

So, the correct order is B < A < D < C.

Coordination Compounds Question 5:

Which of the following are paramagnetic?
A. [NiCl₄]²⁻
B. Ni(CO)₄
C. [Ni(CN)₄]²⁻
D. [Ni(H₂O)₆]²⁺
E. Ni(PPh₃)₄
Choose the correct answer from the options given below:

  1. A and C only
  2. B and E only
  3. A and D only
  4. A, D and E only

Answer (Detailed Solution Below)

Option 3 : A and D only

Coordination Compounds Question 5 Detailed Solution

CONCEPT:

Paramagnetism

  • A substance is considered paramagnetic if it has unpaired electrons in its electronic configuration. The presence of unpaired electrons leads to a net magnetic moment.
  • If all the electrons are paired, the substance is diamagnetic and does not exhibit paramagnetism.
  • The electronic configuration and ligand field strength play a major role in determining whether a compound is paramagnetic or diamagnetic.
  • In the case of transition metal complexes, the ligand's field strength determines whether the d-electrons will pair up (low-spin) or remain unpaired (high-spin).

EXPLANATION:

Complex Electronic Configuration Magnetic Behavior
[NiCl₄]²⁻ Nickel (Ni) in +2 state: 3d84s0 Paramagnetic (due to weak field ligand Cl⁻ and two unpaired electrons)
Ni(CO)₄ Nickel (Ni) in 0 state: 3d84s2 Diamagnetic (due to strong field ligand CO causing pairing of electrons)
[Ni(CN)₄]²⁻ Nickel (Ni) in +2 state: 3d8 Diamagnetic (due to strong field ligand CN⁻ causing pairing of electrons)
[Ni(H₂O)₆]²⁺ Nickel (Ni) in +2 state: 3d8 Paramagnetic (due to weak field ligand H₂O and two unpaired electrons)
Ni(PPh₃)₄ Nickel (Ni) in 0 state: 3d84s2 Diamagnetic (due to strong field ligand PPh₃ causing pairing of electrons)

​So, the correct answer is [NiCl₄]²⁻ and [Ni(H₂O)₆]²⁺ are paramagnetic.

Top Coordination Compounds MCQ Objective Questions

The IUPAC name of [Co(NH3)5 ONO]2+ ion is

  1. Pentaamminenitritocobalt (IV) ion
  2. Pentaamminenitrocobalt (IV) ion
  3. Pentaamminenitrocobalt (III) ion
  4. Pentaamminenitritocobalt (III) ion

Answer (Detailed Solution Below)

Option 4 : Pentaamminenitritocobalt (III) ion

Coordination Compounds Question 6 Detailed Solution

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Concept:

  • IUPAC stands for International Union of Pure and Applied Chemistry.
  • In this, the naming of the compound is followed by a certain rule to avoid confusion all over the world.
  • As per IUPAC rules, additive nomenclature was founded in order to describe the structures of coordination entities or complexes, but this method is readily extended to other molecular entities as well. 
  • Mononuclear complexes are considered to consist of a central atom, often a metal ion, which is bonded to surrounding small molecules or ions, which are referred to as ligands.

Explanation:

  • From the above explanation we know that in mononuclear complexes we have a central atom which is metal in our case it is Cobalt which is surrounded by ammonia and ONO molecules then its IUPAC name will be Pentaamminenitritocobalt (III) ion

Confusing point:

  • We can see that Pentaamminenitrocobalt (III) ion and  Pentaamminenitritocobalt (III) ion are both isomers and we can distinguish both as shown below 
Structure
to be
named
F1 Jayesh Madhu 14.09.20 D1 F1 Jayesh Madhu 14.09.20 D2
Centre atom  Cobalt III Cobalt III
Name of ligands

(NH3)5⇒ Ammine 

O-N-O (i.e., NO2) ⇒ nitro 

(NH3)5⇒ Ammine 
O-N=O (i.e.,ONO)⇒ nitrito 
Assemble
name
Pentaamminenitrocobalt (III) ion Pentaamminenitritocobalt (III) ion

This both are 

The coordination numbers of Co and AI in [CoCl(en)2]Cl and K3[Al(C2 O4)3] respectively, are: (en = ethane-1, 2-diamine)

  1. 5 and 3
  2. 3 and 3
  3. 6 and 6
  4. 5 and 6

Answer (Detailed Solution Below)

Option 4 : 5 and 6

Coordination Compounds Question 7 Detailed Solution

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Explanation:

Given,

[CoCl(en)2]Cl

⇒ Cl - → monodentate ligand 

⇒ en bidentate ligand

Co-ordination Number of Co = (2 × 2) + 1 = 5

Now,

K3 [Al(C2O4)3]

⇒ C2O4 2-→ bidendate ligand

Co-ordination Number of Al = 3 × 2 = 6

The correct sequence of ligands in the order of decreasing field strength is :

  1. CO > H2O > F- > S2-
  2. -OH > F- > NH3 > CN-
  3. NCS- > EDTA4- > CN- > CO
  4. S2- > -OH > EDTA4- > CO

Answer (Detailed Solution Below)

Option 1 : CO > H2O > F- > S2-

Coordination Compounds Question 8 Detailed Solution

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Explanation

Ligand Field Strength

  • The field strength of ligands is determined by their ability to split the d-orbitals of a central metal ion in a coordination complex.
  • This is often described using the spectrochemical series, which arranges ligands in order of their field strength.
  • Stronger field ligands cause greater splitting of the d-orbitals, leading to higher energy differences between the d-orbital sets.
  • Common order of field strength for some ligands is:
    • ( \(CO > CN^- > NO_2^- > en > NH_3 > NCS^- > H_2O > EDTA^{4-} > OH^- > F^- > S^{2-} \))

Conclusion

Based on the given options and the general spectrochemical series, the correct sequence in decreasing order of field strength is CO > H2O > F- > S2-

The number of water molecule(s) not coordinated to copper ion directly in CuSO4.5H2O is:

  1. 2
  2. 3
  3. 1
  4. 4

Answer (Detailed Solution Below)

Option 3 : 1

Coordination Compounds Question 9 Detailed Solution

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Concept:

According to Werner’s theory,

  • one of the five molecules of H2O molecules will attach to the SO4 molecules in the form of hydrogen bonding.
  • It will act as both a primary and secondary molecules so the coordination number will be 4.

Explanation:

  • Copper sulphate pentahydrate contains copper (II) in a geometry best described as distorted octahedral.
  • The copper (II) is bound to four water molecules in a square-planar geometry and two oxygen atoms from two sulphate ions (one H2O is H-Bonded here).

This salt dissolves in water to produce the pale-blue [Cu(H2O)6]2+ ion, in which two of the water molecules are less tightly held and have longer bond distances.

  • In CuSO4.5H2O, four H2O molecules are directly coordinated to the central metal ion (Cu+2) while one H2O molecule is hydrogen bonded with SO42-.

09.04.2019 Shift 1 Synergy JEE Mains D88

[Cu(NH3)4]SO4 is a coordinated complex with a coordination number of

  1. 0
  2. 4
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 2 : 4

Coordination Compounds Question 10 Detailed Solution

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Concept-

Crystal field theory:

  • A metal ion in a complex is surrounded by coordinating ligands anions or negative ends.
  • An electric field is produced at the metal ion by the surrounding ligands.
  • This electrostatic approach towards the interaction between metals and ligands is known as crystal field theory.
  • CFT theory considers the ligands as point charges.
  • There is no overlap between the ligand orbitals and metal ion orbitals.
  • Ligands donate lone pair of electrons to the metal atoms and form coordinate complexes.

Explanation:

  • The metal ion has two valencies:
    • The oxidation state or the primary valency.
    • The co-ordination number or the secondary valency.
  • The secondary valencies are satisfied by co-ordinating with primary ligands.
  • The number of ligands satisfying the secondary valency is called the coordination number of the metal.
  • The bond formed by the ligands and the metal ion forms a coordination sphere.
  • The coordination sphere is non-ionizable in solution because of the co-ordinate bond between them.
  • The oxidation number or primary valency of the metal ion is satisfied by the side ions which form the ionizable sphere.
  • In the complex [Cu(NH3)4]SO4[Cu(NH3)4]2+ is the coordination sphere. The number of ligands attached to the metal Cu is four.

F1 Puja.J 29-01-21 Savita D11

Hence, the coordination number of the complex [Cu(NH3)4]SOis four.

Additional Information

The geometry of the complex formed.

  • Placement of the donor atoms at different stereochemical positions will perturb the metal ions to a different extent.
  • This will result in the formation of different stereochemistry or structure.
  • The stereochemistry depends on the co-ordination number.
C.N number Geometry
2 linear
3 trigonal planar
4 square planar, tetrahedral
5 trigonal bipyramidal
6 inner orbital octahedral, outer orbital octahedral
7 pentagonal bipyramidal
 

The number of bridging CO ligand(s) and Co-Co bond Co2(CO)8, respectively are

  1. 2 and 0
  2. 0 and 2
  3. 4 and 0
  4. 2 and 1

Answer (Detailed Solution Below)

Option 4 : 2 and 1

Coordination Compounds Question 11 Detailed Solution

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Concept:

The structure of Co2(CO)8 (a polynuclear metal carbonyl) can be written as:

The total number of bridging CO ligands = 2 and the Co-Co bond =1.

The species that can have a trans-isomer is:

(en=ethane-1, 2-diamine, ox=oxalate)

  1. [Zn(en)Cl2]
  2. [Pt(en)Cl2]
  3. [Cr(en)2(ox)]+
  4. [Pt(en)2Cl2]2+

Answer (Detailed Solution Below)

Option 4 : [Pt(en)2Cl2]2+

Coordination Compounds Question 12 Detailed Solution

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Concept:

Cis-Trans Isomers are isomers that differ in the arrangement of two ligands in square planar and octahedral geometry.

Cis isomers are isomers where the two ligands are 90 degrees apart from one another in relation to the central molecule. This is because Cis isomers have a bond angle of 90o, between two same atoms.

Trans isomers are isomers where the two ligands are on opposite sides in a molecule because trans isomers have a bond angle of 180o, between the two same atoms.

When naming cis or trans isomers, the name begins either with cis or trans, whichever applies, followed by a hyphen and then the name of a molecule.

Cis-trans Isomerism is possible with [Pt(en)2Cl2]2+

Only square planar and octahedral geometries can have cis or trans isomers.

(Dichloro(ethelyenediamine)platinum (II)) shows only optical isomerism. The other complexes do not show stereoisomerism.

Stereoisomerism, is also called as spatial isomerism, is a form of isomerism in which molecules have the same molecular formula and sequence of bonded atoms (constitution), but differ in the three-dimensional orientations of their atoms in space. This contrasts with structural isomers which shares the same molecular formula, but the bond connections or their order differs.

10.04.2019 Shift 1 Synergy JEE Mains D15

Wilkinson catalyst is

  1. [(Et3P)3 RhCl]
  2. [(Et3P)3 IrCl]  (Et=C2H5)
  3. [(Ph3P)3 RhCl]
  4. [(Ph3P)3 IrCl]

Answer (Detailed Solution Below)

Option 3 : [(Ph3P)3 RhCl]

Coordination Compounds Question 13 Detailed Solution

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Concept:

Wilkinson's catalyst is an σ –bonded organometallic compound [(Ph3P)3 RhCl]. It is commercially used for hydrogenation of alkenes and vegetable oils (unsaturated).

IUPAC name is Chloridotris (triphenylphosphene) rhodium (I).

How many ions are produced from the complex [Co(NH3)6] Cl2 in solution?

  1. 9
  2. 8
  3. 3
  4. 2

Answer (Detailed Solution Below)

Option 3 : 3

Coordination Compounds Question 14 Detailed Solution

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Concept:

Co-ordination complexes and CFT:

  • A metal ion in a complex is surrounded by coordinating ligands anions or negative ends.
  • An electric field is produced at the metal ion by the surrounding ligands.
  • This electrostatic approach towards the interaction between metals and ligands is known as crystal field theory.
  • CFT theory considers the ligands as point charges.
  • There is no overlap between the ligand orbitals and metal ion orbitals.
  • Ligands donate lone pair of electrons to the metal atoms and form coordinate complexes.

Explanation:

  • The metal ion has two valencies:
    • The oxidation state or the primary valency.
    • The coordination number or the secondary valency.
  • The secondary valencies are satisfied by coordinating with primary ligands.
  • The bond formed by the ligands and the metal ion forms a coordination sphere.
  • The coordination sphere is non-ionizable in solution because of the co-ordinate bond between them.
  • The oxidation number or primary valency of the metal ion is satisfied by the side ions which form the ionizable sphere.

F2 Madhuri Engineering 03.01.2022 D1 V2

  • The compound will thus dissociate as:

\(\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_2} \to {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{ + 2}} + 2Cl^-\)

Thus, the total number of ions is = 3.

Hence, the number of ions that are produced from the complex [Co(NH3)6] Cl2 in solution is 3.

Number of complexes from the following with even number of unpaired "d" electrons is____.

[V(H2O)6]3+, [Cr(H2O)6]2+, [Fe(H2O)6]3+[Ni(H2O)6]3+, [Cu(H2O)6]2+

[Given atomic numbers : V = 23, Cr = 24, Fe = 26, Ni = 28, Cu = 29]

  1. 2
  2. 4
  3. 5
  4. 1

Answer (Detailed Solution Below)

Option 1 : 2

Coordination Compounds Question 15 Detailed Solution

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[V(H2O)6]3+ → d2sp3

23V :- [Ar]3d34s2

V+3 :- [Ar]3d2 , n = 2 (even number of unpaired e)

[Cr(H2O)6]2+ → sp3d2

24Cr :- [Ar]3d54s1

Cr+2 :- [Ar]3d4 , n = 4 (even number of unpaired e)

qImage6697dc8a69f9d98675591c83

[Fe(H2O)6]3+ sp3d2

Fe3+ :- [Ar]3d54s0

n = 5 (odd number of unpaired e)

[Ni(H2O)6]3+ sp3d2

Ni :- [Ar]3d84s2

Ni+3 :- [Ar]3d7 , n = 3 (odd number of unpaired e)

[Cu(H2O)6]2+ sp3d2

Cu :- [Ar]3d94s0

n = 1 (odd number of unpaired e)

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