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1.

Why are low spin tetrahedral complexes are not formed?

Answer»

Because for tetrahedral complexes the crystal field stabilization energy is lower than pairing energy.

2.

Which is an example for homoleptic complexes(a) [Co(NH3Cl2]+(b) [CoNH3)6]3+(c) [Cr(NH3)(H2O)3]Cl3(d) [CoCl2(en)2]

Answer»

(b) [CoNH3)6]3+

3.

EDTA is a dentate ligand (a) uni dentate (b) bidentate (c) Tridentate (d) hexadentate

Answer»

(d) hexadentate

4.

EDTA combines with cations to form :(a) chelates (b) polymers (c) clathrates (d) non-stoichiometric compounds

Answer»

Option : (a) chelates

5.

The complex which violates EAN rule is :(a) Fe(CO)5 (b) [Fe(CN)6]3-(c) Ni(CO)4(d) [Zn(NH3)4]Cl2

Answer»

Option : (b) [Fe(CN)6]3-

6.

The compounds [Co(SO4)(NH3)5]Br and [Co(SO4)(NH3)5]Cl represent(i) linkage isomerism(ii) ionisation isomerism(iii) coordination isomerism(iv) no isomerism

Answer»

(iv) no isomerism

7.

The oxidation state of cobalt ion in the complex [Co(NH3)5 Br]SO4 is …………a. + 2 b. + 3 c. + 1d. + 4

Answer»

(b) + 3

The oxidation state of cobalt ion in the complex [Co(NH3)5 Br]SO4 is + 3.

8.

EDTA is a ligand of the type : (a) bidentate (b) tridentate (c) tetradentate (d) hexadentate

Answer»

Option : (d) hexadentate

9.

 A cationic complex has two isomers A & B. Each has one Co3+, five NH3 , one Br- and one SO4 2-. A gives a white precipitate with BaCl2 solution while B gives a yellow precipitate with AgNO3 solution. (a) What are the possible structures of the complexes A and B? (b) Will the two complexes have same colour?

Answer»

(a) A= [Co(NH3)5Br]SOB= [Co(NH3)5SO4]Br

(b)No

10.

The cationic complex among the following is :(a) K3[Fe(CN)6] (b) Ni(CO)4 (c) K2HgI4(d) [CO(NH3)6]Cl2

Answer»

Option : (d) [CO(NH3)6]Cl2

11.

What are cationic, anionic and neutral complexes? Give one example of each.

Answer»

(1) Cationic sphere complexes : 

A positively charged coordination sphere or a coordination compound having a positively charged coordination sphere is called cationic sphere complex.

For example :

[Zn(NH3)4]2+ and [Co(NH3)5Cl] SO4 are cationic complexes. The latter has coordination sphere [Co(NH3)5Cl]2+, the anion SO42+ makes it electrically neutral.

(2) Anionic sphere complexes : 

A negatively charged coordination sphere or a coordination compound having negatively charged coordination sphere is called anionic sphere complex. 

For example, 

[Ni(CN)4]2+ and K3 [Fe(CN)6] have anionic coordination sphere; [Fe(CN)6]3- and three K+ ions make the latter electrically neutral.

(3) Neutral sphere complexes : 

A neutral coordination complex does not possess cationic or anionic sphere.

[Pt(NH3)2Cl2] or [Ni(CO)4] are neither cation nor anion but are neutral sphere complexes.

12.

Is the complex [CoF6] cationic or anionic if the oxidation state of cobalt ion is +3?

Answer»

In the complex, Co carries + 3 charge while 6F- carry – 6 charge. Hence the net charge on the complex is – 3. 

Therefore it is an anionic complex.

13.

Why a solution of [Ni(H2O)5]2+ is green while a solution of [Ni(CN)4]2- is colourless? (At. No. of Ni = 28)

Answer»

Unpaired electrons in [Ni(H2O6]2+/d-d transition.

Detailed Answer:

[Ni(CN)4]2- has no unpaired electron in its d-subshell therefore d-d transition is not possible whereas [Ni(H2O)6]2+ has unpaired electron in its d-subshell resulting in d-d transition imparting colour.

14.

[Fe(CN)6]4- and [Fe(H2O)6]2- are of different colours in dilute solutions. Why?

Answer»

The colour of a particular coordination compound depends on the magnitude of crystal field splitting energy ∆. This CFSE in turn depends on the nature of the ligand. In case of [Fe (CN)6)4- and [Fe(H2O)6]2- the colour differs because there is a difference in the CFSE. Now, CN- is a strong field ligand having a higher CFSE value as compared to the CFSE value of water. This means that absorption of energy for the extra d- d transition also differed. Hence, the transmitted colour also different.

15.

A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2- is colourless. Explain.

Answer»

In [Ni(H2O)6]2+ is a weak field ligand. Therefore there are unpaired electrons in Ni2+. In this complex the cl electrons from the lower energy level can be excited to the higher energy level is the possiblity of d- d transition is present. Hence [Ni(H2O)6]2+ is coloured.

In [Ni(CN)4]2- CN is a strong field ligand. Therefore, d- d transition is not possible due to absence of unpaired electron (CN initiates pairing of d elcctrons ). Hence it is colourless.

16.

Why is [NiCl4]2- paramagnetic while [Ni(CN)4]2- is diamagnetic? (Atomic number of Ni = 28)

Answer»

In [NiCl4]2-, due to the presence of Cla weak field ligand no pairing occurs whereas in [Ni(CN)4]2-, CN- is a strong field ligand and pairing takes place/diagrammatic represenlation. 

17.

What is crystal field splitting?

Answer»

The splitting of five degenerate d-orbitals of the transition metal ion into different sets of orbitals (to2g and eg) having different energies in the presence of ligands in the complex is called crystal field splitting.

18.

Explain the factors affecting Crystal Field Splitting parameter (Δ0).

Answer»

Crystal Field Splitting parameter (Δ0) depends on. 

(a) Strength of the ligands and 

(b) Oxidation state of the metal.

(a) Strength of the ligands : Since strong field ligands like CN-, en, etc. approach closer to the central metal ion, it results in a large crystal field splitting and hence Δ0has higher values.

(b) Oxidation state of the metal : A metal ion with the higher positive charge draws the ligands closer to it which results in large separation of t2g and eg set of orbitals. The complexes involving metal ions with low oxidation state have low values of Δ0

For example,

[Fe(NH3)6]3+ has higher Δ0 than [Fe(NH3)6]2+. H

19.

Arrange following complex ions in increasing order of crystal field splitting energy (ΔO) :[Cr(Cl)6]3–, [Cr(CN)6]3–, [Cr(NH3)6]3+.

Answer»

Crystal field splitting energy increases in the order [Cr(Cl)6]3– < [Cr(NH3)6]3+ < [Cr(CN)6]3-

20.

What is crystal field splitting energy? How does the magnitude of ∆o decide the actual configuration of d orbitals in a coordination entity?

Answer»

The degenerate d-orbitals (in a spherical field environment) split two level i.e. eg and t2g in the presence of ligands. The splitting of the degenerate orbitals in the presence of ligands is called crystal field splitting and the energy difference between the two levels (e and t2g) is called the crystal field splitting energy. It is denoted by ∆o. After the orbitals have split, the filling of the electrons takes place. After 1 electron (each) has filled in the three t2g orbitals, the filling of the electrons takes place in 2 ways.

It can enter the orbital (giving) rise to t3g eg, like electronic configuration on the pairing of the electrons can take place in the t2g orbitals (giving rise to t42g eg0 like electronic configuration). If the ∆o value of a ligand is less than the pairing energy, then the electrons enter the eg orbital. On the other hand, if the ∆o value of a ligand is more than the pairing energy, then the electrons enter the t2g orbitals.

21.

Explain why [Fe(H2O)6]3+ has magnetic moment value of 5.92 BM whereas [Fe(CN)6]3– has a value of only 1.74 BM.

Answer»

[Fe(CN)6]3– involves d2sp3 hybridisation with one unpaired electron and [Fe(H2O)6]3+ involves sp3d2 hybridisation with five unpaired electrons. This difference is due to the presence of strong ligand CN– and weak ligand H2O in these complexes.

22.

On the basis of crystal field theory write the electronic configuration for d4 ion if Δ0 &gt; p.

Answer»

Electronic configuration is t42g or by diagram.

23.

When are optical isomers called chiral?

Answer»

When the mirror images of optical isomers of the complex are non-superimposable they are said to be chiral. 

For example, 

[Co(en)2(NH3)2]3+.

24.

Explain : (1) Plane polarised light (2) Optical activity.

Answer»

(1) Plane polarised light : A monochromatic light having vibrations only in one plane is called a plane polarised light. This light is obtained by passing monochromatic light through NICOL prism.

(2) Optical activity : A phenomenon of rotating a plane of a plane polarised light by an optically active substance is called optical activity. This substance is said to be optically active.

25.

Define : (1) Optical isomerism (2) Optical isomers.

Answer»

(1) Optical isomerism : The phenomenon of isomerism in which different coordination compounds having same molecular formula have different optical activity is called optical isomerism.

(2) Optical isomers : Different coordination compounds having same molecular formula but different optical activity are called optical isomers.

26.

What type of isomerism is shown by the complex [Co(NH3)5 (SC]2+?

Answer» Linkage isomerism
27.

Write IUPAC name of the following Complex  [Cr(NH3)3Cl3]

Answer»

Triamminetrichloridochromium (III)

28.

What are ligands? What are their types? Give one example of each type.

Answer»

Ligands : The neutral molecules or negatively charged anions (or rarely positive ions) which are bonded by coordinate bonds to the central metal atom or metal ion in a coordination compound are called ligands or donor groups. For example in [Cu(CN)4]2-, four CN- ions are ligands coordinated to central metal ion Cu2+. Ligands can be classified on the basis of number of electron donor atoms in the ligand i.e. denticity.

(1) Monodentate or unidentate ligand : A ligand molecule or an ion which has only one donor atom with a lone pair of electrons forming only one coordinate bond with metal atom or ion in the complex is called monodentate or unidentate ligand. 

For example NH3, Cl-, OH-, H2O, etc.

(2) Polydentate or multidentate ligand : A ligand molecule or an ion which has two or more donor atoms with the lone pairs of electrons forming two or more coordinate bonds with the central metal atom or ion in the complex is called polydentate or multidentate ligand. For example, ethylene diamine, H2N – (CH2)2 – NH2. According to the number of donor atoms they are classified as follows :

  • Bidentate ligand : This ligand has two donor atoms in the molecule or ion. For example, ethylenediamine, H2N – (CH2)2 – NH2.
  • Tridentate ligand : This ligand molecule has three donor atoms or three sites of attachment.

E.g. Diethelene triamine, H2N – CH2 – CH2 – NH – CH2 – CH2 – NH2. This has three N donor atoms.

  • Tetradentate (or quadridentate) ligand : This ligand molecule has four donor atoms. 

Eg. Triethylene tetraamine which has four N donor atoms.

  • Hexdentate ligand : This ligand molecule has six donor atoms. E.g. Ethylenediamine tetracetato.

(3) Ambidentate ligand : A ligand molecule or an ion which has two or more donor atoms, however in the formation of a complex, only one donor atom is attached to the metal atom or an ion is called ambidentate ligand. For example, NO-2 which has two donor atoms N and O forming a coordinate bond, M ← ONO (nitrito) or M ← NO2 (nitro). 

(4) Bridging ligand : A monodentate ligand having more than one lone pairs of electrons, hence can attach to two or more metal atoms or ions and hence acts as a bridge between different metal atoms is called bridging ligand. For example : OH-, F-, SO4-2, etc.

29.

Define the term Ligands.

Answer»

The neutral or negative ions bound to the central metal or ion in the coordination entity. These donate a pair/s of electrons to the central metal atom /ion. 

30.

When CuSO4 is mixed with an excess of NH3 , a deep blue coloured solution is obtained. a) Write the formula of the compound formed. b) What is the IUPAC name of the compound? c) What do you understand by the term coordination number and ligand in a coordination compound? d) Give the oxidation number and coordination number of the central metal atom of the deep blue coloured compound.

Answer»

a) [Cu (NH3)4]SO4 

b) Tetraammine copper (ll) sulphate 

c) It is the number of ligand atoms to which the metal is directly bonded. Ligands are neutral molecules or ions bounded to the central atom/ion in the coordination entity. 

d) Oxidation number of Cu in [Cu(NH3)4 ]SO4 is +2 Coordination number of Cu in [Cu(NH3)4 ]SO4 is 4.

31.

Say TRUE or FALSE.[Co(NO3)(NH3)5]SO4 and [Co(NO3)(NH3)4 (SO4)] (NH3) are ionisation isomers.

Answer»

FALSE.

[Co(NO3)(NH3)5]SO4 and [Co(NO3)(NH3)4 (SO4)] (NH3) are ionisation isomers.i.e., False.

32.

Oxidation number of cobalt in K[COCI4] is : (a) +1 (b) -1 (c) +3 (d) -3

Answer»

Option : (b) -1

33.

Explain the oxidation state of a metal in a complex.

Answer»
  • The oxidation state of a metal atom or ion in the complex is the apparent charge carried by it in the complex. 
  • It depends upon the atomic number and electronic configuration of the metal atom or ion. 
  • The coordination number, the formula and geometry of a complex depend upon the oxidation state of the metal atom or ion. 
34.

Calculate the oxidation state of a metal in the following complexes :(a) [Fe(NH3)6](NO3)3(b) Ni(CO)5.

Answer»

(a) [Fe(NH3)6](NO3)3 ⇌ [Fe(NH3)6]3+ + 3NO3-

NH3 is a neutral ligand, and the charge number of complex ion is + 3. 

If the oxidation state of Fe is x then,

+ 3 = x + 6(0) 

∴ x = + 3 

∴ The oxidation state of Fe is +3.

(b) Ni(CO)5 is a neutral complex and CO is a neutral ligand. 

If the oxidation state of Ni is x, then 

zero = x + 5 x (zero) 

∴ x = zero. 

The oxidation state of Ni is zero.

35.

What is meant by Coordination Polyhedron?

Answer»

The spatial arrangement of the ligand atoms which are directly attached to the central metal atom/ion. They are commonly Octahedral, Square-planar or Tetrahedral. 

36.

What is Oxidation number?

Answer»

The charge that the central atom would carry if all the ligands are removed along with their pairs of electrons shared with the central atom. It is represented in parenthesis. 

37.

CuSO4.5H2O is blue in colour while CuSO4 is colourless. Why?

Answer»

In CuSO4.5H2O, water acts as ligand as a result it causes crystal field splitting. Hence d—d transition is possible in CuSO4.5H2O and shows colour. In the anhydrous CuSO4 due to the absence of water (ligand), crystal field splitting is not possible and hence no colour.

38.

Match the complex ions given in Column I with the colours given in Column II and assign the correct code :Column I (Complex ion)Column II (Colour)A. [Co(NH3)6]3+1. VioletB. [Ti(H2O)6]3+2. GreenC. [Ni(H2O)6]2+3. Pale blueD. (Ni (H2O)4 (en)]2+ (aq)4. Yellowish orange5. BlueCode :(i) A (1) B (2) C (4) D (5)(ii) A (4) B (3) C (2) D (1)(iii) A (3) B (2) C (4) D (1)(iv) A (4) B (1) C (2) D (3)

Answer»

(ii) A (4) B (3) C (2) D (1)

39.

Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code :Column I (Complex ion)Column II (Hybridisation, number of unpaired electrons)A. [Cr(H2O)6]3+1. dsp2, 1B. [Co(CN)4]2–2. sp3d2, 5C. [Ni(NH3)6]2+3. d2sp3, 3D. [MnF6]4–4. sp3, 45. sp3d2, 2Code :(i) A (3) B (1) C (5) D (2)(ii) A (4) B (3) C (2) D (1)(iii) A (3) B (2) C (4) D (1)(iv) A (4) B (1) C (2) D (3

Answer»

(ii) A (4) B (3) C (2) D (1)

40.

Match the compounds given in Column I with the oxidation state of cobalt present in it (given in Column II) and assign the correct code.Column I (Compound)Column II (Oxidation state of Co)A. [Co(NCS)(NH3)5](SO3)1. + 4B. [Co(NH3)4Cl2]SO42. 0C. Na4[Co(S2O3)3]3. + 1D. [Co2(CO)8]4. + 25. + 3Code :(i) A (1) B (2) C (4) D (5)(ii) A (4) B (3) C (2) D (1)(iii) A (5) B (1) C (4) D (2)(iv) A (4) B (1) C (2) D (3)

Answer»

(i) A (1) B (2) C (4) D (5)

41.

Match the complex species given in Column I with the possible isomerism given in Column II and assign the correct code :Column I (Complex species)Column II (Isomerism)A. [Co(NH3)4Cl2]+1. opticalB. cis-[Co(en)2Cl]+2. ionisationC. [Co(NH3)5(NO2)]Cl23. coordinationD. [Co(NH3)6][Cr(CN)6]4. geometrical5. linkageCode :(i) A (1) B (2) C (4) D (5)(ii) A (4) B (3) C (2) D (1)(iii) A (4) B (1) C (5) D (3)(iv) A (4) B (1) C (2) D (3)

Answer»

(iv) A (4) B (1) C (2) D (3)

42.

Match the coordination compounds given in Column I with the central metal atoms given in Column II and assign the correct code :Column I (Coordination Compound)Column II (Central metal atom)A. Chlorophyll1. rhodiumB. Blood pigment2. cobaltC. Wilkinson catalyst3. calciumD. Vitamin B124. iron5. magnesiumCode :(i) A (5) B (4) C (1) D (2)(ii) A (3) B (4) C (5) D (1)(iii) A (4) B (3) C (2) D (1)(iv) A (3) B (4) C (1) D (2)

Answer»

(i) A (5) B (4) C (1) D (2)

43.

Explain the bonding in coordination compounds in terms of Werner’s postulates.

Answer»

Werner’s postulates explain the bonding in coordination compounds as follows: 

(i) A metal exhibits two types of valencies ‘ namely, primary and secondary valencies. Primary valencies are satisfied by negative ions while secondary valences are satisfied by both negative and neutral ions. (In modem terminology the primary valency corresponds to the oxidation number of the metal ion, whereas the secondary valency refers to the coordination number of the metal ion) 

(ii) A metal ion has a definite number of secondary valences around the central atom. Also, these valences project in a specific direction in the space assigned to the definite geometry of the coordination compound. 

(iii) Primary valences are usually ionisable, while secondary valences are non ionisable.

44.

Coordination compounds are those compounds which retain their identity even in solution and it is essential for all living matter.1. Name a coordination compound containing Magnesium, which is essential for plants.2. When we coordinate EDTA with any metal, we get a ring structure. What is this process called?3. Explain.

Answer»

1. Chlorophyll. 

2. Chelation. 

3. When a poly dentate ligand is coordinated to the metal, a ring structure is obtained, called chelate complex. Such complexes are more stable than similar complexes containing unidentate ligands.

45.

What is the importance of the following coordination compounds is different fields?1. EDTA 2. Gold cyanide [AU(CN)2] 3. cis-platin 4. [Ag(S2O3)]3-

Answer»

1. EDTA → Estimation of hardness of water

2. AU(CN)2 → Metallurgy

3. cis-platin → Cancer therapy

4. [Ag(S2O3)]3- → Photography

46.

The number of unpaired electrons in a low spin octahedral complex ion of d1 is :(a) 0 (b) 1 (c) 2 (d) 3

Answer»

Option : (b) 1

47.

The number of unpaired electrons in a high spin octahedral complex ion of d7 is :(a) 0 (b) 1 (c) 2 (d) 3

Answer»

Option : (d) 3

48.

On the basis of CFT predict the number of unpaired electrons in [CrF6]3⊕.a. 1 b. 2 c. 3 d. 4

Answer»

Correct answer is

(c) 3

49.

IUPAC name of [Pt (NH3)2 Cl(NO2)] is :(i) Platinum diaminechloronitrite(ii) Chloronitrito-N-ammineplatinum (II)(iii) Diamminechloridonitrito-N-platinum (II)(iv) Diamminechloronitrito-N-platinate (II)

Answer»

(iii) Diamminechloridonitrito-N-platinum (II)

50.

When a coordination compound CrCl3.6H2O is mixed with AgNO3 solution 3 moles of AgCl are precipitated per mole of the compound. Write:(i) Structural formula of the complex(ii) IUPAC name of the complex(iii) Magnetic and spin behavior of the complex

Answer»

(i) [Cr(H2O)6]Cl3

(ii) Hexaaquachromium (III) chloride

(iii) Paramagnetic and high spin