Animated Solution for Chemistry - Coordination Compounds: The complex that can show fac- and mer-isomers is :

Coordination chemistry isn't just about writing formulas; it's about visualizing molecules in three-dimensional space. When a central metal atom is surrounded by ligands, the specific spatial arrangement of these ligands can give rise to entirely different molecules called isomers.

Today, we are diving into a very specific and fascinating type of geometrical isomerism: Facial (fac) and Meridional (mer) isomerism. This is a classic concept that frequently appears in competitive exams like JEE and NEET.

Before we even look at the options, we need to establish the golden rule. Facial and meridional isomerism is a property exclusive to octahedral complexes (coordination number 6) that possess a very specific general formula: $M{a}_3{b}_3$.

In this formula, '$M$' represents the central metal atom, '$a$' represents one type of monodentate ligand, and '$b$' represents another type of monodentate ligand. The key is that there must be exactly three of each type. If you can identify this pattern, you have already solved the problem!

Let's systematically break down the given options to see which one fits our golden rule:

Option (a): $[Pt(N{H}_3{)}_{2}C{l}_2]$ This complex has a coordination number of 4. Platinum(II) complexes are typically square planar. Its general formula is $M{a}_2{b}_2$. While it can show cis and trans isomerism, it cannot show fac/mer isomerism because it lacks the octahedral geometry.

Option (c): $[Co(N{H}_3{)}_{4}C{l}_2{]}^{+}$ This is an octahedral complex, but its general formula is $M{a}_4{b}_2$. Complexes of this type can exhibit cis and trans isomerism depending on whether the two identical '$b$' ligands are adjacent ($90^{\circ }$) or opposite ($180^{\circ }$) to each other. It does not fit the $M{a}_3{b}_3$ requirement.

Option (d): $[CoC{l}_2(en{)}_2]$ Here, 'en' stands for ethylenediamine, which is a bidentate ligand. The general formula is $M(AA{)}_2{b}_2$. This complex is famous for showing both geometrical (cis/trans) and optical isomerism (the cis form is chiral). However, it still doesn't match our target formula.

Option (b): $[Co(N{H}_3{)}_3(N{O}_2{)}_3]$ Bingo! We have a central Cobalt atom surrounded by three ammine ($N{H}_3$) ligands and three nitro ($N{O}_2$) ligands. This is a perfect match for the $M{a}_3{b}_3$ general formula. Therefore, this is the complex that will exhibit fac and mer isomerism.

Imagine an octahedron as two square pyramids joined at their bases. It has eight triangular faces.

In the fac-isomer, the three identical ligands (let's say the three $N{H}_3$ groups) are positioned at the corners of one of these triangular faces. Because they occupy a single face, they form a tight cluster. In this arrangement, any two identical ligands are adjacent to each other, meaning the bond angle between them is exactly $90^{\circ }$.

Now, let's rearrange those three $N{H}_3$ ligands. Instead of clustering them on a face, let's spread them out along the equator, or the meridian, of the octahedron.

In the mer-isomer, the three identical ligands lie in the same plane that bisects the octahedron. Because they are spread out along this arc, two of the identical ligands will be situated directly opposite each other, creating a $180^{\circ }$ bond angle, while the third sits at a $90^{\circ }$ angle to the other two.

Mastering coordination chemistry requires a blend of pattern recognition and spatial visualization. Whenever you are asked about fac and mer isomers, immediately scan the options for the $M{a}_3{b}_3$ signature. Once you spot it, you can confidently lock in your answer and move on to the next challenge!