Animated Solution for Chemistry - Coordination Compounds: A list of species having the formula ${\text{XZ}}_4$ is given below : ${\text{XeF}}_4$, ${\text{SF}}_4$, ${\text{SiF}}_4$, ${\text{BF}}_4^{-}$, ${\text{BrF}}_4^{-}$, $[\text{Cu}({\text{NH}}_3{)}_4{]}^{2+}$, $[{\text{FeCl}}_4{]}^{2-}$, $[{\text{CoCl}}_4{]}^{2-}$ and $[{\text{PtCl}}_4{]}^{2-}$. Defining shape on the basis of the location of X and Z atoms, the total number of species having a square planar shape is

Imagine you are handed a mixed bag of chemical species, some from the p-block and some from the d-block, and you are asked to act as the sorting hat. Your mission? To identify exactly how many of these species possess a square planar shape.

This is not just a simple memory test; it is a beautiful intersection of two major chemical theories: the VSEPR theory for main group elements and Crystal Field Theory (CFT) for transition metal complexes. Let's break down this list of nine species systematically.

VSEPR Theory in Action

For p-block elements, the shape is dictated by the repulsion between electron pairs in the valence shell, beautifully summarized by the VSEPR theory. The key metric here is the Steric Number (SN), which is the sum of bond pairs (BP) and lone pairs (LP).

Let's look at ${\text{XeF}}_4$ and ${\text{BrF}}_4^{-}$. Xenon is a noble gas with 8 valence electrons. It forms 4 single bonds with fluorine, utilizing 4 electrons. The remaining 4 electrons form 2 lone pairs. Thus, $SN=4+2=6$. An electron geometry of 6 is octahedral, but with two lone pairs positioned opposite to each other to minimize repulsion, the resulting molecular shape is perfectly square planar.

Similarly, Bromine in ${\text{BrF}}_4^{-}$ has 7 valence electrons, plus 1 extra electron from the negative charge, totaling 8. Just like Xenon, it forms 4 bonds and has 2 lone pairs, resulting in a square planar shape.

What about the others? In ${\text{SF}}_4$, Sulfur has 6 valence electrons, leading to 4 bond pairs and 1 lone pair ($SN=5$). This gives a see-saw shape. Silicon in ${\text{SiF}}_4$ and Boron in ${\text{BF}}_4^{-}$ have 4 and 3(+1) valence electrons respectively, leading to 4 bond pairs and 0 lone pairs ($SN=4$). Both of these are tetrahedral.

Coordination Chemistry

Now we transition to the d-block complexes. For coordination complexes with a coordination number of 4, the geometry hinges on the hybridization of the central metal ion. If the metal undergoes $s{p}^3$ hybridization, the shape is tetrahedral. If it undergoes $ds{p}^2$ hybridization, the shape is square planar.

Let's examine $[{\text{FeCl}}_4{]}^{2-}$ and $[{\text{CoCl}}_4{]}^{2-}$. Iron(II) is a $3d^6$ system and Cobalt(II) is a $3d^7$ system. In both cases, the chloride ion (${\text{Cl}}^{-}$) acts as a weak field ligand. It does not have the strength to force the pairing of electrons against Hund's rule. Consequently, the inner 3d orbitals remain occupied by unpaired electrons, forcing the metal to use its outer 4s and 4p orbitals for bonding. This results in $s{p}^3$ hybridization and a tetrahedral shape.

Here is where the magic happens. Look at $[{\text{PtCl}}_4{]}^{2-}$. You might think, "Chloride is a weak field ligand, so it must be tetrahedral!" But there is a massive catch. Platinum is a 5d transition metal. For metals in the 4d and 5d series, the effective nuclear charge is so high that all ligands, regardless of their nature, act as strong field ligands. This forces electron pairing, freeing up an inner d-orbital. The hybridization becomes $ds{p}^2$, making the complex square planar.

Finally, we have the classic JEE favorite: $[\text{Cu}({\text{NH}}_3{)}_4{]}^{2+}$. Copper(II) is a $3d^9$ system. Ammonia is a strong field ligand. To achieve a more stable configuration, one unpaired electron from the 3d orbital is excited to a higher 4p orbital. This vacates an inner d-orbital, allowing for $ds{p}^2$ hybridization. This phenomenon, often explained by Jahn-Teller distortion, results in a square planar shape.

After carefully analyzing all nine species, we have successfully identified the square planar ones: 1. ${\text{XeF}}_4$ 2. ${\text{BrF}}_4^{-}$ 3. $[\text{Cu}({\text{NH}}_3{)}_4{]}^{2+}$ 4. $[{\text{PtCl}}_4{]}^{2-}$

The total count is exactly 4. This problem is a masterclass in applying fundamental rules while staying hyper-vigilant for the beautiful exceptions that make chemistry so fascinating.