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Why does \[PC{l_5}\] in the solid state exist as cation and anion, i.e. in the form of \[PC{l_4}^ + \] and\[PC{l_6}^{ - 1}\]?

Answer
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Hint: Phosphorus pentachloride exists in ionic form in the solid state. It exists as \[PC{l_4}^ + \] in action from and \[PC{l_6}^{ - 1}\] as anionic form. This is because of the stability reasons. In solid form the stability of these ions is more than that of \[PC{l_5}\]. Also the lattice structure of these ions are more stable than that of \[PC{l_5}\].

Complete Answer:
In solid state, \[PC{l_5}\] exists in ionic form. The cationic form is \[PC{l_4}^ + \] and the anionic form is \[PC{l_6}^{ - 1}\]. This is because of the stability reasons. The formation of these ions can be represented by the following reaction as:
\[PC{l_5}{\text{ }} + {\text{ }}PC{l_5}{\text{ }} + {\text{ }}energy{\text{ }} \to {\text{ }}PC{l_4}^ + {\text{ }} + {\text{ }}PC{l_6}^ - \]
Thus two moles of \[PC{l_5}\] react with some amount of energy absorbed and we get \[PC{l_4}^ + \] and\[PC{l_6}^{ - 1}\]. The stability of any compound depends on its lattice energy and its lattice structure. If a compound has unstable structure or it has high energy then it is considered to be unstable. The lattice energy is the energy which is required to break the lattice structure of the compound. Thus the ions have higher lattice energy and thus we require more energy to break its structure, therefore it is stable in nature. \[PC{l_4}^ + \] has tetrahedral geometry and \[PC{l_6}^{ - 1}\] has octahedral geometry. While \[PC{l_5}\] has trigonal bipyramidal geometry. Thus ionic form of \[PC{l_5}\] has more stable geometry so it exists in ionic form in solid state.

Note:
The geometry of \[PC{l_5}\] , \[PC{l_4}^ + \] and \[PC{l_6}^{ - 1}\] can be find out with the help of VSEPR model. We can find the hybridization of each compound and thus find its geometry. Tetrahedral and octahedral structures are more stable structures than the trigonal bipyramidal structure of a compound. The stability of structure also depends on the angle strain between the bonds.