Animated Solution for Chemistry - Coordination Compounds: The geometries of the ammonia complexes of ${\text{Ni}}^{2+}$, ${\text{Pt}}^{2+}$ and ${\text{Zn}}^{2+}$, respectively , are :

Welcome to the fascinating world of coordination chemistry! Today, we are embarking on a journey to decode the three-dimensional architectures of three distinct metal complexes.

Imagine you are an architect, but instead of bricks and mortar, your building blocks are metal ions and ammonia molecules. Our task is to determine the geometries of the ammonia complexes formed by ${\text{Ni}}^{2+}$, ${\text{Pt}}^{2+}$, and ${\text{Zn}}^{2+}$.

This isn't just about memorizing shapes; it is about understanding the profound rules of Crystal Field Theory and hybridization that govern the microscopic world. Let's break them down one by one.

The Octahedral Giant

Our first candidate is the Nickel ion, ${\text{Ni}}^{2+}$. When placed in an aqueous solution of ammonia, it surrounds itself with six ammonia ligands, forming the hexaammine complex, $[\text{Ni}({\text{NH}}_3{)}_6{]}^{2+}$.

To understand its geometry, we must first look at its electronic passport. Nickel is a $3d$ transition metal, and in its $+2$ oxidation state, it possesses a $[\text{Ar}]3d^8$ configuration.

Now, ammonia is generally considered a moderate to strong field ligand. However, we have a spatial constraint here. With eight electrons in the $3d$ subshell, the orbitals are quite crowded. Even if we tried to pair up the unpaired electrons, we would only free up a single $d$-orbital.

But wait! To form an inner-orbital octahedral complex ($d^{2}s{p}^3$), we strictly need two empty inner $d$-orbitals. Since that is mathematically impossible for a $3d^8$ system, Nickel has no choice but to look outward.

It utilizes its empty outer $4d$ orbitals, resulting in $s{p}^3{d}^2$ hybridization. This specific hybridization perfectly corresponds to a symmetrical Octahedral geometry.

The Power of Heavy Metals

Next, we shift our focus to Platinum, ${\text{Pt}}^{2+}$. It forms a tetraammine complex, $[\text{Pt}({\text{NH}}_3{)}_4{]}^{2+}$, meaning it has a coordination number of four.

Platinum is a heavy hitter. It resides deep down in the periodic table, belonging to the $5d$ transition series. Its electronic configuration is $[\text{Xe}]4f^{14}5d^8$.

Here is where the magic happens—a concept that is an absolute favorite in JEE and NEET exams! As we move down a group to the $4d$ and $5d$ series, the $d$-orbitals become significantly larger and more diffuse. This allows ligands to approach much closer, creating an incredibly strong electrostatic interaction.

Because of this intense interaction, the crystal field splitting energy (${\Delta }_o$) becomes exceptionally high. It is so high, in fact, that all ligands act as strong field ligands when bonded to $4d$ or $5d$ metals!

This massive energy gap forces the unpaired electrons in the $5d$ orbitals to pair up against their natural repulsion. This pairing clears out exactly one inner $d$-orbital. Platinum eagerly grabs this empty orbital, leading to $ds{p}^2$ hybridization.

And as the laws of quantum mechanics dictate, $ds{p}^2$ hybridization always manifests as a flat, elegant Square Planar geometry.

When the Inn is Full

Finally, let's examine Zinc, ${\text{Zn}}^{2+}$. Like Platinum, it forms a complex with four ammonia ligands: $[\text{Zn}({\text{NH}}_3{)}_4{]}^{2+}$.

However, Zinc is unique. It sits at the very end of the $3d$ series. In its $+2$ oxidation state, its electronic configuration is $[\text{Ar}]3d^{10}$.

Take a moment to visualize that. The $3d$ subshell is completely, 100% full. There is absolutely no room at the inn! No matter how strong the ligand is, you cannot pair up electrons that are already paired, and you cannot create an empty inner $d$-orbital out of thin air.

Therefore, inner orbital hybridization is completely off the table for Zinc. It is forced to rely entirely on its outer, empty orbitals—specifically, one $4s$ and three $4p$ orbitals.

This results in $s{p}^3$ hybridization. In the realm of VSEPR theory and coordination chemistry, $s{p}^3$ hybridization invariably leads to a Tetrahedral geometry.

We have successfully decoded the architectural blueprints of all three complexes!

The Nickel complex, constrained by its $3d^8$ configuration, expands outward to form an Octahedral shape. The Platinum complex, wielding the immense splitting power of a $5d$ metal, forces electron pairing to achieve a Square Planar shape. The Zinc complex, with its completely filled $d$-orbitals, has no choice but to adopt a Tetrahedral shape.

Matching our rigorous derivations with the given options, we find that the sequence "octahedral, square planar, and tetrahedral" aligns perfectly with Option (A).

Always remember these fundamental principles: the nature of the ligand matters, but the intrinsic properties of the metal—its series ($3d$ vs $5d$) and its available orbitals—are the true master architects of molecular geometry!