Animated Solution for Chemistry - Coordination Compounds: The one that is not expected to show isomerism is

The concept of isomerism in coordination compounds is a fascinating puzzle of 3D geometry. When we are asked to find a complex that does not show isomerism, we must become molecular detectives, examining the coordination number, the nature of the ligands, and the resulting spatial arrangement of each candidate.

Let's begin by looking at the complexes with a coordination number of 6, which adopt an octahedral geometry.

Option (a) presents us with $[Ni(N{H}_3{)}_4(H_{2}O{)}_2{]}^{2+}$. This is a classic $M{a}_4{b}_2$ type complex. In an octahedron, the two $H_{2}O$ ligands can either be placed adjacent to each other (at a $90^{\circ }$ angle) to form the cis-isomer, or opposite to each other (at a $180^{\circ }$ angle) to form the trans-isomer. Because it can exist in these two distinct spatial arrangements, it definitely exhibits geometrical isomerism.

Option (b) is $[Ni(en{)}_3{]}^{2+}$. Here, we have three ethylenediamine (en) ligands. Since 'en' is a symmetrical bidentate ligand, the complex is of the $M(a-a{)}_3$ type. The three bidentate rings wrap around the central nickel ion like the blades of a propeller. This specific geometry lacks any plane or center of symmetry, making the molecule chiral. Consequently, it exists as two non-superimposable mirror images (the d-form and l-form), meaning it shows optical isomerism.

Before we look at option (c), let's jump to Option (d): $[Pt(N{H}_3{)}_{2}C{l}_2]$. Platinum in the $+2$ oxidation state is notorious for forming square planar complexes, regardless of the ligand strength, due to its high crystal field splitting energy. This complex is of the $M{a}_2{b}_2$ type. In a square plane, the two identical ligands (like the two chloridos) can be placed on the same side to form the cis-isomer (famous as the anti-cancer drug cisplatin) or on opposite sides to form the trans-isomer. Thus, it clearly shows geometrical isomerism.

Finally, we arrive at Option (c): $[Ni(N{H}_3{)}_{2}C{l}_2]$. Nickel is in the $+2$ oxidation state ($3d^8$ configuration). The ligands $N{H}_3$ and $C{l}^{-}$ are not strong enough to force the pairing of the $3d$ electrons in this specific mixed setup, leading to $s{p}^3$ hybridization.

This results in a tetrahedral geometry. The defining feature of a tetrahedron is that all four positions are perfectly symmetrical and adjacent to one another, separated by an angle of $109.5^{\circ }$. Because there is no "opposite" position in a tetrahedron, you cannot create cis or trans arrangements. Furthermore, since the complex is of the $M{a}_2{b}_2$ type, it possesses a plane of symmetry, ruling out any optical isomerism.

Having systematically eliminated the other options, we can confidently conclude that the tetrahedral complex $[Ni(N{H}_3{)}_{2}C{l}_2]$ is the only one that does not exhibit any form of isomerism.