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**Hint :**To check for a compound to what shape it is going to have, you have to first determine the hybridisation of the orbitals so as the compound can be formed. Also, this is true only for d block elements. These d block elements exist in the middle of the table.

**Complete Step By Step Answer:**

In $ PtC{l_4}^{2 - } $ , we will first find out the oxidation state of platinum, as in this case chlorine is the higher electronegative element which will possess $ - 1 $ charge and the total charge on the ion is $ - 2 $ -, let the oxidation state of platinum be $ x $ , therefore, it can be written as

$ x + \{ 4 \times ( - 1)\} = - 2 \\

\Rightarrow x - 4 = - 2 \\

\Rightarrow x = + 2 \\ $

$ x $ being the oxidation state of platinum.

Now we will write electronic configuration of $ P{t^{2 + }} $ , normally platinum has 78 electrons but here it has only 76

$ P{t^{2 + }} = [Xe]5{d^8}6{s^0} $

Now as the highly electronegative chlorine ion with extra electron attack the platinum attack the platinum ion, it distorts and hybridises the orbitals and therefore the d orbital has four pairs of electrons and the four chlorine atoms assume the $ ds{p^2} $ hybridisation.

Now comparing the hybridisation with its general structure of its molecules or ions. As the above-mentioned ion has $ ds{p^2} $ hybridisation therefore the complex should be tetrahedral but here there is one anomaly that is the size of platinum is too big that it forms bonds with ligands. There is strong repulsion between the electron of platinum and ligand resulting in strong crystal field splitting. It stabilizes the square planar structure more.

**The structure of $ PtC{l_4}^{2 - } $ is square planar.**

**Note :**

Ligand is any ion or molecule having a pair of electrons such that it can be attached to the metal atom by coordinate bonding. Examples of such are water, ammonia, etc.

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