Question

A reaction of Cobalt $ (III) $ Chloride and ethylenediamine in a 1:2 mole ratio generates two isomeric products A (violet colored) B (green colored). A can show optical activity, B is optically inactive. What type of isomers does A and B represent?
(A) Geometrical isomers
(B) Ionization isomers
(C) Coordination isomers
(D) Linkage isomers

Answer 1

Hint :Ethylenediamine is a dentate ligand and $ C{o^{3 + }} $ forms an octahedral complex having coordination number $ 6 $ . The $ 2 $ moles of ethylenediamine can satisfy four coordination numbers so that the remaining two would be satisfied by four Chloride ions.

Complete Step By Step Answer:
The reaction:
 $ CoC{l_3} + 2{C_2}{H_8}{N_2} \to \left[ {CoC{l_2}{{\left( {en} \right)}_2}} \right]Cl $
According to the given ratio, only the above product can be formed and since there are two products, so another product should be an isomer.
There are two possibilities for the two Cl ions – they can either be in cis-form or trans-form. There is no plane of symmetry on cis-form and therefore it is chiral and optically active; and on the other hand, the trans-form will be optically inactive.
Hence, they are Geometrical isomers of each other.
Option (A) is correct.

Note :
Geometrical isomers possess the same molecular and structural formula but differ in the arrangement of groups in space. Geometrical isomerism can only be observed in compounds containing double bonds and is also known as cis-form.