Coordination Compounds MCQ Quiz in मल्याळम - Objective Question with Answer for Coordination Compounds - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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നേടുക Coordination Compounds ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Coordination Compounds MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Coordination Compounds MCQ Objective Questions

Top Coordination Compounds MCQ Objective Questions

Coordination Compounds Question 1:

Comprehension:

Alfred Werner, a Swiss chemist, significantly contributed to the understanding of coordination compounds through his pioneering theory. He distinguished between primary and secondary valences of metal ions and prepared numerous coordination compounds, which he analyzed for their physical and chemical properties. Notably, Werner explored the behavior of cobalt(III) chloride complexes with ammonia, observing variations in the precipitation of chloride ions as AgCl upon addition of excess silver nitrate. This discrepancy in chloride precipitation across different compounds—ranging from full precipitation to partial retention in solution—provided critical evidence supporting his ideas. Werner concluded that up to six groups (chloride ions or ammonia molecules) could bind to cobalt, forming a stable complex. This led to his proposal that compounds consist of a central metal ion surrounded by other molecules or ions, which he defined through the concept of secondary valence. His work laid foundational principles for the structural composition of coordination compounds.

Which of the following compounds prepared by Werner had all chloride ions precipitated as AgCl when treated with excess silver nitrate?

  1. CoCl3.4NH3 (Green)
  2. CoCl3.5NH3 (Purple)
  3. CoCl3.6NH3 (Yellow)
  4. CoCl3.4NH3 (Violet)

Answer (Detailed Solution Below)

Option 3 : CoCl3.6NH3 (Yellow)

Coordination Compounds Question 1 Detailed Solution

The correct answer is CoCl3.6NH3 (Yellow)

Explanation:-

The compoundCoCl3.6NH3 (Yellow) led to the precipitation of all three moles of chloride ions as AgCl when treated with excess silver nitrate, indicating that all chloride ions were ionically bound and not directly coordinated to the cobalt ion. In contrast, other compounds retained some of the chloride ions in solution, implying their direct coordination to cobalt and hence not precipitating as AgCl.

Conclusion:-

So, CoCl3.6NH3 (Yellow) ompounds prepared by Werner had all chloride ions precipitated as AgCl when treated with excess silver nitrate.

Coordination Compounds Question 2:

Catalyst used in Haber-Bosch process for making NH3 is __________.

  1. MnO2
  2. FeSO4
  3. Promoted Fe
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : Promoted Fe

Coordination Compounds Question 2 Detailed Solution

The correct answer is Promoted Fe

Concept:-

  • Heterogeneous catalysis: Heterogeneous catalysis is a type of catalysis in which the catalyst is in a different phase from the reactants. In the Haber-Bosch process, the promoted iron catalyst is in a solid phase, while the hydrogen and nitrogen reactants are in a gas phase.
  • Active site: An active site is a specific site on the surface of a catalyst where the reaction takes place.

Explanation:-

The Haber-Bosch process is a chemical process that is used to produce ammonia from hydrogen and nitrogen gases. The overall reaction for the Haber-Bosch process is:

N2 + 3H2 → 2NH3

  • The catalyst used in the Haber-Bosch process is typically a promoted iron catalyst. Promoted iron catalysts are made by adding small amounts of other metals, such as potassium and aluminum, to iron. These promoters help to improve the activity and selectivity of the catalyst.
  • The catalyst in the Haber-Bosch process plays a crucial role in facilitating the reaction between hydrogen and nitrogen gases to form ammonia. The catalyst provides a surface on which the hydrogen and nitrogen molecules can adsorb and react with each other. The catalyst also helps to lower the activation energy of the reaction, making it more likely to occur.

The mechanism of the Haber-Bosch process is complex and involves a series of steps. However, the overall process can be summarized as follows:

  • Hydrogen and nitrogen molecules adsorb onto the surface of the catalyst.
  • The adsorbed hydrogen and nitrogen molecules react with each other to form ammonia molecules.
  • The ammonia molecules desorb from the surface of the catalyst.

Advantages of promoted iron catalyst: Promoted iron catalysts are typically used in the Haber-Bosch process because they are highly active and selective for the reaction between hydrogen and nitrogen gases to form ammonia. Promoted iron catalysts are also relatively stable and have a long lifespan.

Conclusion:-

So, Catalyst used in Haber-Bosch process for making NH3 is Promoted Fe

Coordination Compounds Question 3:

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 3 Detailed Solution

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.

Coordination Compounds Question 4:

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 4 Detailed Solution

[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)

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[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)

Coordination Compounds Question 5:

Consider the following complexes

(A) [CoCl(NH3)5]2+

(B) [Co(CN)6]3-

(C) [Co(NH3)5(H2O)]3+

(D) [Cu(H2O)4]2+

The correct order of A, B, C and D in terms of wavenumber of light absorbed is : 

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

Answer (Detailed Solution Below)

Option 2 : D < A < C < B 

Coordination Compounds Question 5 Detailed Solution

CONCEPT:

 

Determining Wavenumber of Light Absorbed by Complexes

  • The wavenumber of light absorbed by a complex is influenced by the ligand field strength, which causes a splitting of the d-orbitals.
  • Stronger field ligands cause a larger splitting (higher ∆oct), thus absorbing higher energy (higher wavenumber) light.
  • The ligand field strength order from weak to strong is as follows: H2O < NH3 < Cl < CN.
  • As ligand field increases, light of more energy is absorbed

    Energy ∝ wave number\((\bar{v})\)

Explanation:-

  • (A) [CoCl(NH3)5]2+:
    • Contains Cl, a weaker field ligand compared to NH3.
  • (B) [Co(CN)6]3−:
    • Contains CN, a very strong field ligand.
  • (C) [Co(NH3)5(H2O)]3+:
    • Contains NH3 and H2O, with NH3 being a stronger field ligand than H2O.
  • (D) [Cu(H2O)4]2+:
    • Contains H2O, the weakest field ligand of those listed.

The correct order of the complexes in terms of increasing wavenumber of light absorbed (weakest ligand field to strongest ligand field) 
The correct answer is D < A < C < B

Coordination Compounds Question 6:

Comprehension:

According to the valence bond theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. It is usually possible to predict the geometry of a complex from the knowledge of its magnetic behaviour on the basis of the valence bond theory. Consider the formation of [Co(NH3)5Cl] Cl2 and answer the following question:

The primary valence of Co in above coordination entity is

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

Answer (Detailed Solution Below)

Option 3 : 3

Coordination Compounds Question 6 Detailed Solution

[Co(NH3)5Cl]Cl: x + 5(0) + 3(-1) = 0

⇒ x = +3 ⇒ Co3+

Primary valence = oxidation state of central metal = + 3

Coordination Compounds Question 7:

The correct IUPAC nomenclature of [NiCl2(PPh3)2] is 

  1. Bischloridobis(triphnylphosphine)nickel(II)
  2. Dichloridobis(triphenylphosphine)nickel(II)
  3. Dichloridodi(triphenylphosphine)nickel(II)
  4. Bischloridodi(triphenylphosphine)nickel(II)

Answer (Detailed Solution Below)

Option 2 : Dichloridobis(triphenylphosphine)nickel(II)

Coordination Compounds Question 7 Detailed Solution

Explanation-

The standard rules that must be followed in the nomenclature of coordination compounds are described below.

  1. The ligands are always written before the central metal ion in the naming of complex coordination complexes.
  2. When the coordination center is bound to more than one ligand, the names of the ligands are written in an alphabetical order which is not affected by the numerical prefixes that must be applied to the ligands.
  3. When there are many monodentate ligands present in the coordination compound, the prefixes that give insight into the number of ligands are of the type: di-, tri-, tetra-, and so on.
  4. When there are many polydentate ligands attached to the central metal ion, the prefixes are of the form bis-, tris-, and so on.
  5. The names of the anions present in a coordination compound must end with the letter ‘o’, which generally replaces the letter ‘e’. Therefore, the sulfate anion must be written as ‘sulfato’ and the chloride anion must be written as ‘chlorido’.
  6. The following neutral ligands are assigned specific names in coordination compounds: NH3 (ammine), H2O (aqua or aquo), CO (carbonyl), and NO (nitrosyl).
  7. After the ligands are named, the name of the central metal atom is written. If the complex has an anionic charge associated with it, the suffix ‘-ate’ is applied.
  8. When writing the name of the central metallic atom in an anionic complex, priority is given to the Latin name of the metal if it exists (with the exception of mercury).
  9. The oxidation state of the central metal atom/ion must be specified with the help of roman numerals that are enclosed in a set of parentheses.
  10. If the coordination compound is accompanied by a counter ion, the cationic entity must be written before the anionic entity.

The oxidation state of an atom can be defined as the hypothetical charge that would be held by that atom if all of its bonds to other atoms were completely ionic in nature.

Given data and Analysis-

Given compound is [NiCl2(PPh3)2

The oxidation state of Ni-

Let the oxidation state of Ni is x. Final Charge in the compound = 0

Charge of Cl = -1, Charge of PPh3 = 0

So, x - 2 + (0 × 2) = 0, x = +2

So according to rules nomenclature will be

As two chloride anion is present so Dichlorido, then two triphenylphosphines are present so bis will be used.

So the final name will be Dichloridobis(triphenylphosphine)nickel(II).

Coordination Compounds Question 8:

Which of the following exhibits Linkage isomerism?

  1. [Cr(H2O)6]Cl3
  2. [Co(NH3)5(SCN)]
  3. [Cr(NH3)4Cl2]Cl
  4. [Cr(NH3)3Cl3]

Answer (Detailed Solution Below)

Option 2 : [Co(NH3)5(SCN)]

Coordination Compounds Question 8 Detailed Solution

Concept:

  • In the area of coordination, chemistry isomerism arises from varying chemical linkages and complexity in stereochemistry.

The types of isomerism are:

  • Ionization isomerism:
    • Ionization isomers are produced by an interchange of positions of the coordinating anions insides the complex sphere and anions outside the complex sphere.
    • Two ionization isomers yield two different ions in solution for example, [Co(Br)(NH3)3]SO4 and [Co(SO4)(NH3)3]Br.
  • Hydrate isomer:
    • ​This isomerism arises due to the different dispositions of aqua molecules inside and outside the first sphere of attraction.
    • Examples are [Cr(Cl)(H2O)5]Cl2 blue-green and [Cr(Cl)2(H2O)5]Cl.H2O which is violet.
  • Coordination isomerism:
    • ​Such isomeric forms are exhibited when both the cation and the anion of salt are complexes and there occurs a redistribution of the ligands between the two coordination zones.
  • For example [Cr(NH3)]6[Co(Ox)3] and [Co(NH3)]6[Cr(Ox)3].
  • Linkage isomerism:
    • If two ligands are isomeric to each other then their complexes are also isomeric in nature.
    • Linkage isomerism some coordinating ligands possess two different donor atoms.
    • Linkage isomerism arises out of two modes of attachment of the ligand to the central metal ion via two different donor atoms.
    • Linkage isomers are generally identified by their distinguishing vibrational spectra and electronic spectra.
  • Stereoisomerism:
    • This is a form of isomerism in which two complexes of the identical first spheres of co-ordination differ in the relative positions of the coordinating groups.
    • Such isomerism can be divided into geometrical and optical isomerism.
    • Geometrical isomerism: This is cis-trans isomerism.
    • Optical isomerism is shown by the isomers which are non-superimposable mirror images of each other.

Explanation:

  • In the complex [Co(NH3)5(SCN)], thiocyanate SCN is an ambidentate ligand, which has two different donor sites sulphur and nitrogen but can coordinate to the metal atom using only one site at a time.
  • Thus it can link via two ways to the metal atom and shows linkage isomerism.
  • The two complexes are:

F1 Puja J 25.3.21 Pallavi D4

  • The complex [Cr(H2O)6]Cl3 shows hydrate isomerism.
  • Examples of isomers are [Cr(Cl)(H2O)5]Cl2 blue-green and [Cr(Cl)2(H2O)5]Cl.H2O which is violet.

Hence, [Co(NH3)5(SCN)] shows linkage isomerism.

Coordination Compounds Question 9:

The spin only magnetic moment of Hexacyanidomanganate(II) ion is BM.

  1. 5.90
  2. 1.73
  3. 4.90
  4. 3.87

Answer (Detailed Solution Below)

Option 2 : 1.73

Coordination Compounds Question 9 Detailed Solution

CONCEPT:

Spin-Only Magnetic Moment

  • The spin-only magnetic moment ( μ  ) of a complex is determined by the number of unpaired electrons in the d-orbitals of the central metal ion.
  • The spin-only magnetic moment is given by the formula:

    \(μ_{so} = \sqrt{n(n+2)} BM\)

  • where n is the number of unpaired electrons.
  • The unit of the magnetic moment is Bohr Magneton (BM).

EXPLANATION:

  • Hexacyanidomanganate(II) ion is [Mn(CN)6]{4-}.
  • Manganese in the +2 oxidation state (Mn2+) has an electron configuration of [Ar] 3d5.
  • Since cyanide (CN-) is a strong field ligand, it causes pairing of electrons.
  • This results in the low-spin configuration:

    \(\text{Mn}^2+ : t_{2g}^6 e_g^0\)

  • However, with d5 low-spin configuration, there is one unpaired electron.
  • Using the spin-only formula:

    \(μ_{so} = \sqrt{1(1+2)} BM \\ μ_{so} = \sqrt{3} BM \\ μ_{so} \approx 1.73 BM\)

Therefore, the spin-only magnetic moment of the Hexacyanidomanganate(II) ion is 1.73 BM.

Coordination Compounds Question 10:

Consider the following reaction 

MnO2 + KOH + O2 A + H2O.

Product ‘A’ in neutral or acidic medium disproportionate to give products ‘B’ and ‘C’ along with water. The sum of spin-only magnetic moment values of B and C is __________ BM. (nearest integer)

(Given atomic number of Mn is 25)

Answer (Detailed Solution Below) 4

Coordination Compounds Question 10 Detailed Solution

MnO2 + KOH + O2 K2MnO4 + H2O

(A)

K2MnO4 \(\xrightarrow{\text{Neutral/acidic solution}}\) KMnO4 + MnO2

Mn+4 :- [Ar]3d3

n = 3, \(\mu = \sqrt{3(3+2)} \) = 3.87 B.M.

Nearest integer is (4)

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