Question

The structure of ${\mathrm{PCl}}_5$ in the solid-state is

• A. Square Planar ${\left[{\mathrm{PCl}}_4\right]}^{+}$and Octahedral ${\left[{\mathrm{PCl}}_6\right]}^{-}$
• B. Tetrahedral ${\left[{\mathrm{PCl}}_4\right]}^{+}$and Octahedral ${\left[{\mathrm{PCl}}_6\right]}^{-}$
• C. Trigonal bipyramidal
• D. Square pyramidal

Answer 1

The correct option is B

Tetrahedral ${\left[{\mathrm{PCl}}_4\right]}^{+}$and Octahedral ${\left[{\mathrm{PCl}}_6\right]}^{-}$

The explanation for the correct option:

B: Tetrahedral ${\left[{\mathrm{PCl}}_4\right]}^{+}$and Octahedral ${\left[{\mathrm{PCl}}_6\right]}^{-}$

Existence of ${\mathrm{PCl}}_5$:

• Ionic bonding promotes the crystalline structure of PCl₅, which likes to exist in the solid form as oppositely charged ions like ${\left[{\mathrm{PCl}}_4\right]}^{+}\mathrm{and}{\left[{\mathrm{PCl}}_6\right]}^{–}$ .
• Since it takes on the tetrahedral and octahedral shapes in the ionic form, the shape is more stable in this state.
• In solid state ${\mathrm{PCl}}_5$ exist in lone-pair i.e. ${\left[{\mathrm{PCl}}_4\right]}^{+}\mathrm{and}{\left[{\mathrm{PCl}}_6\right]}^{–}$.
• Where, ${\left[{\mathrm{PCl}}_4\right]}^{+}\mathrm{is}{\mathrm{sp}}^3\mathrm{tetrahedral}$ and ${\left[{\mathrm{PCl}}_6\right]}^{–}\mathrm{is}{\mathrm{sp}}^3{\mathrm{d}}^2\mathrm{octahedral}$.

The explanation for the incorrect options:

A: Square Planar and ${\left[{\mathrm{PCl}}_4\right]}^{+}$ and Octahedral ${\left[{\mathrm{PCl}}_6\right]}^{-}$

• ${\left[{\mathrm{PCl}}_4\right]}^{+}$ exists in the shape of tetrahedral, i.e. why it can't be shown in the shape of the square planer.

C: Trigonal bipyramidal

• The third period's elements are made up of d orbitals as well as s and p orbitals. The $3\mathrm{d}$ orbitals' energy is comparable to that of the $3\mathrm{s}$ and $3\mathrm{p}$ orbitals.

D. Square pyramidal

• The term "square pyramid" refers to a three-dimensional geometric form with a square base and four triangular faces or sides that meet at a single point (known as the vertex).

Therefore, option B Tetrahedral ${\left[{\mathrm{PCl}}_4\right]}^{+}$and Octahedral ${\left[{\mathrm{PCl}}_6\right]}^{-}$ is the correct option.