Animated Solution for Chemistry - Coordination Compounds: The number of geometrical isomers found in the metal complexes $[PtC{l}_2(N{H}_3{)}_2]$, $[Ni(CO{)}_4]$, $[Ru(H_{2}O{)}_{3}C{l}_3]$ and $[CoC{l}_2(N{H}_3{)}_4{]}^{+}$ respectively, are

Welcome to a thrilling journey through the spatial arrangements of coordination compounds! Today, we are tasked with finding the number of geometrical isomers for four distinct metal complexes. Geometrical isomerism arises when ligands can be arranged in different relative positions around the central metal atom. Let's break down each complex and visualize its geometry.

Our first candidate is $[PtC{l}_2(N{H}_3{)}_2]$. Platinum in the $+2$ oxidation state is famous for forming square planar complexes. This specific complex falls under the $M{A}_2{B}_2$ category, where 'M' is the metal and 'A' and 'B' are two different types of ligands.

Imagine a square table with the Platinum atom sitting right in the center. The four ligands occupy the four corners. Now, how can we arrange two Chlorine atoms and two Ammonia molecules?

If we place the two Chlorine atoms on adjacent corners (at a $90^{\circ }$ angle to each other), we get the cis-isomer. Conversely, if we place them on opposite corners (at a $180^{\circ }$ angle), we get the trans-isomer.

Therefore, $[PtC{l}_2(N{H}_3{)}_2]$ exhibits exactly 2 geometrical isomers.

Next up is $[Ni(CO{)}_4]$, nickel tetracarbonyl. Nickel is in a $0$ oxidation state here, and the strong carbonyl ligands force it into a tetrahedral geometry. This is an $M{A}_4$ type complex.

Here is the catch: in a perfect tetrahedron, every corner is adjacent to every other corner. There is no "opposite" position like in a square plane. Furthermore, all four ligands are identical! No matter how you try to swap the carbonyl groups, the resulting molecule is completely superimposable on the original.

Because all relative positions are equivalent, $[Ni(CO{)}_4]$ shows 0 geometrical isomers.

Moving to the third complex, $[Ru(H_{2}O{)}_{3}C{l}_3]$. With a coordination number of 6, this complex adopts an octahedral geometry. It is a classic $M{A}_3{B}_3$ type complex, a favorite among examiners!

For an $M{A}_3{B}_3$ complex, we look for two very special spatial arrangements. If the three identical ligands (say, the three Chlorines) occupy the corners of one triangular face of the octahedron, they form the facial or fac-isomer.

If, instead, the three identical ligands are arranged around the meridian (the equator) of the octahedron, they form the meridional or mer-isomer.

Thus, $[Ru(H_{2}O{)}_{3}C{l}_3]$ gives us exactly 2 geometrical isomers.

Finally, let's examine $[CoC{l}_2(N{H}_3{)}_4{]}^{+}$. Like the Ruthenium complex, it has a coordination number of 6 and is octahedral. However, its formula fits the $M{A}_4{B}_2$ pattern.

To find its isomers, we focus on the two identical 'B' ligands—the Chlorines. Just like in the square planar case, we can place these two Chlorines adjacent to each other at a $90^{\circ }$ angle to form the cis-isomer. If we place them at opposite poles of the octahedron at a $180^{\circ }$ angle, we get the trans-isomer.

This gives us 2 geometrical isomers for the Cobalt complex.

Summarizing our spatial detective work, the number of geometrical isomers are: - $[PtC{l}_2(N{H}_3{)}_2]\to 2$ - $[Ni(CO{)}_4]\to 0$ - $[Ru(H_{2}O{)}_{3}C{l}_3]\to 2$ - $[CoC{l}_2(N{H}_3{)}_4{]}^{+}\to 2$

The correct sequence is 2, 0, 2, 2, which perfectly matches option (c). Understanding the 3D geometry of these complexes is the ultimate key to mastering isomerism!