The structure of $$\text{PCl}_5$$ in the solid state is:

We recall first that $$\text{PCl}_5$$ shows different structures in different physical states. In the gaseous or liquid phase it behaves as a discrete, covalent molecule and adopts the $$\text{trigonal\ bipyramidal}$$ geometry predicted from VSEPR theory for five electron pairs around phosphorus. However, when we cool the substance sufficiently to obtain the solid state, experimental X-ray diffraction studies reveal that individual $$\text{PCl}_5$$ molecules do not survive intact. Instead, each molecule ionises inside the crystal lattice.

The ionisation process can be written as

$$\text{PCl}_5 \longrightarrow [\text{PCl}_4]^+ + [\text{PCl}_6]^-. $$

Now we identify the geometries of the two ions produced.

1. For $$[\text{PCl}_4]^+$$ we have a central phosphorus atom surrounded by four chloride ions. According to VSEPR theory, “four regions of electron density” around a central atom lead to the $$\text{tetrahedral}$$ arrangement because a tetrahedron keeps the four bonding pairs as far apart as possible, minimising repulsion. Therefore $$[\text{PCl}_4]^+$$ is tetrahedral.

2. For $$[\text{PCl}_6]^-$$ there are six chloride ions bonded to the central phosphorus atom. VSEPR tells us that “six regions of electron density” produce an $$\text{octahedral}$$ shape, again because an octahedron best minimises repulsions among six bonding pairs. Hence $$[\text{PCl}_6]^-$$ is octahedral.

Combining these observations we see that the solid contains discrete $$[\text{PCl}_4]^+$$ tetrahedral cations and $$[\text{PCl}_6]^-$$ octahedral anions packed together in an ionic lattice. This description matches exactly the statement given in Option A.

Hence, the correct answer is Option A.