Question

Which one of the following will not show geometrical isomerism?
A. \[\left[ {Cr{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]Cl\]
B. $\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Cl$
C. $\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]C{l_2}$
D. $\left[ {Pt{{\left( {N{H_3}} \right)}_2}C{l_2}} \right]$

Answer 1

Hint: We need to know what is geometrical isomerism and its concept in coordination compounds. In chemistry, isomerism is the phenomenon where molecules have the same molecular formula, i.e., the same number of atoms but their arrangements in space are different and these molecules are known as isomers. There are two primary types of isomerism- Structural Isomerism and Stereoisomerism. In structural isomerism, the atoms, and functional groups of the molecule (isomer) are linked in different ways.

Complete step by step answer:
In organic compounds, Geometric isomerism is also known as cis-trans isomerism. They have different spatial arrangements in three-dimensional space. This can be explained with the help of but-2-ene molecules. In the cis-but-2-ene molecule, the methyl groups are on the same side of the double bond whereas in the trans-but-2-ene molecule, the methyl groups are on opposite sides of the double bond. Similarly, coordination compounds exhibit the same types of geometrical isomerism as organic compounds.
Coordination complexes that differ only in which ligands are adjacent to one another (cis) or directly across from one another (trans) in the coordination sphere of the metal exhibit geometrical isomerism. The basic formula for writing a coordination compound is \[{[M{(Ligand)_x}]^ + }io{n^ - }\] . To study isomerism, for the sake of convenience, we can write the complex as $M{A_x}{B_y}$ where $M$ is the central metal atom, $A$ is one ligand and $B$ is another ligand and $x$ and $y$ are the number of the ligands $A$ and $B$ respectively.
Let us consider the example of a coordination complex \[M{A_2}{B_2}\] . Here either the two ligands \[A\]can be adjacent to one another (cis isomers), in which case the two ligands \[B\] should also be in the cis position by default, or the two ligands $A$ can be diagonal to one another (trans isomers), in which case the two $B$ ligands should also be in trans position. Hence for \[M{A_2}{B_2}\] , there are two possible isomers (cis \[AA,BB\] and trans $AB,AB$ ).
We know the study of the isomerism of the given coordination complexes.
\[\left[ {Cr{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]Cl\]: This complex is of the type \[M{A_4}{B_2}\] . The possible isomers are: \[AA,AA,BB\left( {cis} \right)\] and \[AB,AB,AA\left( {trans} \right)\;\].
$\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]Cl$: This complex is of the type \[M{(AA)_2}{B_2}\] . This is because en is a bidentate ligand. The possible isomers are: \[AA,AA,BB\left( {cis} \right)\] and \[AB,AB,AA\left( {trans} \right)\].
$\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]C{l_2}$: This complex is of the type $M{A_5}B$ . The only possible orientation possible is \[AA,AA,AB\] . Since there is only one B ligand present, there are no further possible combinations hence cannot exhibit cis-trans isomerism.
$\left[ {Pt{{\left( {N{H_3}} \right)}_2}C{l_2}} \right]$: This complex is of the type \[M{A_2}{B_2}\]. The possible isomers are: \[AA,BB\left( {cis} \right)\] and \[AB,AB\left( {trans} \right)\] .

So, the correct answer is Option C.

Note: We must note that some cis-trans isomers can be drawn in more than one type. For the \[M{A_2}{B_2}\] complex, it is possible to draw the cis isomer in four different ways and the trans isomer in two different ways. But all members of each set are chemically equivalent. Also, the most important criteria for coordination complexes to exhibit geometrical isomerism is the presence of two ligands with a minimum of 2 of each type.has been used to produce chlorinated compounds and to produce amyl-naphthalene and isoprene.