Question

In solid state $PC{{l}_{5}}$ is a:
A-Covalent solid
B-Octahedral structure
C-Ionic solid with ${{\left[ PC{{l}_{6}} \right]}^{+}}$ octahedral and ${{\left[ PC{{l}_{4}} \right]}^{-}}$ tetrahedral
D-Ionic solid with ${{\left[ PC{{l}_{4}} \right]}^{+}}$ tetrahedral and ${{\left[ PC{{l}_{6}} \right]}^{-}}$ octahedral

Answer 1

The hybridisation is $s{{p}^{3}}d$ and in gaseous and liquid states has trigonal bipyramidal geometry. It has three $P-Cl~$equatorial bonds, and two $P-Cl~$axial bonds. $PC{{l}_{5}}$ finds use as a chlorinating reagent. It is a solid that is colourless, waterproof and resilient to moisture, while commercial samples can be yellowish and polluted with hydrogen chloride.

Complete answer:
The phosphorus chloride structures are generally in line with the principle of VSEPR. $PC{{l}_{5}}$ 's composition depends on its context. Gaseous and molten is a neutral molecule with bipyramidal trigonal geometry and symmetry.

The hypervalent character of this species ( and of $\text{ }PCl_{6}^{-}$ ) the presence of nonbonding molecular orbitals ( molecular orbital theory ) or of resonance (valence bond theory) .
In solid condition: $PC{{l}_{5}}$ Tries to live as opposite charged ions such as $PCl_{4}^{+}\text{ }$ and $\text{ }PCl_{6}^{-}$ As ionic bonding strengthens the structure of crystalline. Likewise $PCl_{4}^{+}$ Is tetrahedral, or $\text{ }PCl_{6}^{-}$ That's octahedral. These structures fit well into each other, giving the solid structure more stability.

That's why option D is right.

Additional information:
The attractive forces in an ionic crystal are much greater than the dipole-dipole or dispersion of covalent substances in crystals. Lattice energy is usually the biggest factor in determining the stability of an ionic solid. The extra energy gained from the lattice energy more than compensates for the energy required to transfer a chloride ion from one $PC{{l}_{5}}$ molecule to another.

Note: $PC{{l}_{5}}$ undergoes self ionization in polar solvent solutions. Dilute solutions dissociate according to the equilibrium below:
$PC{{l}_{5}}\,\,\rightleftharpoons PCl_{4}^{+}+C{{l}^{-}}$
A second equilibrium becomes more prevalent at higher concentrations:
$2PC{{l}_{5}}\,\,\rightleftharpoons PCl_{4}^{+}+Cl_{6}^{-}$
The Cation $PCl_{4}^{+}$ and the $PCl_{6}^{-}$ anion Are both tetrahedral and octahedral. $PC{{l}_{5}}$ was thought to form a dimeric structure in solution at one time,${{P}_{2}}C{{l}_{10}}$.