CBSE Class 12 Chemistry Coordination Compunds Questions Answers

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COORDINATION COMPOUNDS

1 mark questions

1. Explain coordination entity with example.

Ans: it constitute a central metal atom or ions bonded to a fixed number of molecules or ions ( ligands) .eg. [Co(NH3)3Cl3].

2. What do you understand by coordination compounds?

Ans: coordination compounds are the compounds which contains complex ions. These compounds contain a central metal atom or cation which is attached with a fixed number of anions or molecules called ligands through coordinate bonds. eg. [Co(NH3)3Cl3]

3. What is coordination number?

Ans: the coordination number of a metal ion in a complex may be defined as the total number of ligand donor atoms to which the metal ion is directly bonded. Eg. In the complex ion [Co(NH3)6]3+ has 6 coordination number.

4. Name the different types of isomerisms in coordination compounds.

Ans: structural isomerism and stereoisomerism.

5. Draw the structure of xenon difluoride.

Ans: structure :trigonalbipyramidal

Shape: linear

6. What is spectrochemical series ?

Ans: the series in which ligands are arranged in the order of increasing field strength is called spectrochemical series. The order is : 

I-<Br-<SCN-<Cl-<S2-<F-<OH-<C2O42-<H2O<NCS-<EDTA4-<NH3<en<CN-<CO

7. What do you understand by denticity of a ligand ?

Ans: the number of coordinating groups present in ligand is called denticity of ligand. Eg.Bidentate ligand ethane-1,2-diamine has 2 donor nitrogen atoms which can link to central metal atom.

8. Why is CO a stronger ligand than Cl-?

Ans: because CO has π bonds.

9. Why are low spin tetrahedral complexes not formed ?

Ans : because for tetrahedral complexes, the crystal field stabilisation energy is lower than pairing energy.

10. Square planar complexes with coordination number 4 exhibit geometrical isomerism whereas tetrahedral complexes do not. Why?

Ans: tetrahedral complexes do not show geometrical isomerism because the relative positions of the ligands attached to the central metal atom are same with respect to each other.

11. What are crystal fields ?

Ans: the ligands has around them negatively charged field because of which they are called crystal fields.

12. What is meant by chelate effect ? Give an example .

Ans: when a didentate or polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal atom, a 5 or 6 membered ring is formed , the effect is called chelate effect. Eg. [PtCl2(en)]

13. What do you understand by ambidentate ligand ?

Ans: a ligand which contains two donor atoms but only one of them forms a coordinate bond at a time with central metal atom or ion is called an ambidentate ligand. Eg.nitrito-N and nitrito-O.

14. What is the difference between homoleptic and heteroleptic complexes ?

Ans: in homoleptic complexes the central metal atom is bound to only one kind of donor groups whereas in heteroleptic complexes the central metal atom is bound to more than one type of donor atoms.

15. Give one limitation for crystal field theory.

Ans: i) as the ligands are considered as point charges, the anionic ligands should exert greater splitting effect. However the anionic ligands are found at the low end of the spectrochemical series.

ii) it does not take into account the covalent character of metal ligand bond. ( any one )

16. How many ions are produced from the complex: [Co(NH3)6]Cl2

Ans: 3 ions

17. The oxidation number of cobalt in K[Co(CO)4]

Ans: -1

18. Which compound is used to estimate the hardness of water volumetrically ?

Ans: EDTA

19. Magnetic moment of [MnCl4]2- is 5.92B.M explain with reason.

Ans: the magnetic moment of 5.9 B.M. corresponds to the presence of 5 unpaired electrons in the d- orbitals of Mn2+ ion. As a result the hybridisation involved is sp3 rather than dsp2. Thus tetrahedral structure of [MnCl4]2- complex will show 5.92 B.M magnetic moment value.

20. How many donor atoms are present in EDTA ligand ?

Ans: 6

2 marks questions

1. Give the electronic configuration of the following complexes on the basis of crystal field splitting theory.

i) [CoF6]3-

ii) [Fe(CN)6]4-

Ans: i) Co3+ (d6) t2g4eg2

iii) Fe2+ d6t2g6eg

2. Explain the following with examples:

i) Linkage isomerism

ii) Outer orbital complex

Ans: i) this type of isomerism arises due to the presence of ambidentate ligand in a coordination compound. Eg. [Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2 

iii) When ns, np and nd orbitals are involved in hybridisation , outer orbital complex is formed. Eg. [CoF6]2- in which cobalt is sp3d2 hybridised.

3. i)Low spin octahedral complexes of nickel are not found . Explain why ?

ii)theπ complexes are known for transition elements only.explain.

Ans: i) nickel in its atomic or ionic state cannot afford 2 vacant 3d orbitals and hence d2sp3 hybridisation is not possible.

ii) transition metals have vacant d orbitals in their atoms or ions into which the electron pairs can be donated by ligands containing πelectrons.eg. benzene, ethylene etc. thus dπ-pπ bonding is possible.

4. How would you account for the following:

i) [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.

ii) [Ni(CO)4] possess tetrahedral geometry whereas [Ni(CN)4]2- is square planar. 

Ans: i) due to the presence of 1 electron in 3d subshell in [Ti(H2O)6]3+ complex d-d transition takes place by the absorption of visible light. Hence the complex appears coloured. On the other hand, [Sc(H2O)6]3+ does not possess any unpaired electron .Hence d-d transition is not possible (which is responsible for colour) in this complex is not possible, therefore it is colourless.

ii) Ni in [Ni(CO)4] is sp3 hybridised. Hence it is tetrahedral. Whereas for [Ni(CN)4]2- is dsp2 hybridised hence it has square planar geometry.

5. State reasons for each of the following:

i) All the P—Cl bonds in PCl5 molecule are not equivalent.

ii) S has greater tendency for catenation than O.

Ans: i) in P Cl5 the 2 axial bonds are longer than 3 equatorial bonds. This is due to the fact that the axial bond pairs suffers more repulsion as compared to equatorial bond pairs.

ii) The property of catenation depends upon the bond strength of the element.

As S—S bond is much stronger (213kJ / mole) than O—O bond (138 kJ/mole), S has greater tendency for catenation than O. 

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