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 Multiple Choice QuestionsShort Answer Type

171. Draw all the isomers (geometrical and optical) of
[Co(NH3)Cl(en)2 ]2+ 
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172. Draw all the isomers (geometrical and optical) of
[Co(NH3)2Cl2(en)]+
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173. Write all the geometrical isomers of[Pt(NH3)(Br)Cl(py)] and how many of these will exhibit optical isomers? 
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174. Aqueous copper sulphate solution (blue in colour) gives:
(i) a green precipitate with aqueous potassium fluoride and
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
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175. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution?  
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176.

Draw figure to show splitting of d orbitals in an octahedral crystal field.

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177.

What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.

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178. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory. 
 [Fe(CN)6]4–  
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179. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory. 
[FeF6]3–


Bonding in [FeF6]3–
The oxidation state of Fe is +3 and its coordination number is 6. The complex has octahedral geometry and experimental study shows that it is paramagnetic.

The bonding in this entity is explained on the basis of overlap of sp3d2 hybrid orbitals of Fe3+ ion and six lone pair orbitals of cyanide ligands. Five 3d electrons are unpaired which make it paramagnetic. The outer electronic configuration of Fe3+ ion is 3d5 which is highly stable and no pairing of electrons takes place in presence of weak field of F ions. Here, 4d–orbitals of Fe3+ (which are empty) are involved.



The entity is strongly paramagnetic due to five unpaired electrons and is an outer orbital complex.

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180. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory. 
 [CoF6]3–
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