Question

In certain polar solvents $PC{{l}_{5}}$ undergoes an ionization reaction in which $C{{l}^{-}}$ ion leaves one $PC{{l}_{5}}$molecule and attaches itself to another.
$2PC{{l}_{5}}\rightleftharpoons PCl_{4}^{+}+PCl_{6}^{-}$
What are the changes in the geometrical shape that occur in this ionization?
(a)- Trigonal bipyramidal to tetrahedral and octahedral
(b)- Trigonal bipyramidal to pentagonal planar and square planar
(c)- Pentagonal planar to tetrahedral and octahedral
(d)- None of the above

Answer 1

The shape or structure of the compound can be predicted by calculating the hybridization of the compound. The hybridization can be calculated with the number of valence electrons of the central atom, the number of monovalent atoms/ groups surrounding the central atom, charge on the cation, and charge on the anion.

Complete step by step answer:
To predict the shape of the molecule by:
If the number of hybrid orbitals is equal to the number of surrounding groups, it has a regular geometry and the shape is the same as that predicted by hybridization. If the number of surrounding groups is less than the hybrid orbitals, the difference gives the number of lone pairs present. So, the molecule will have irregular geometry and the shape is predicted by leaving the hybrid orbitals containing the lone pairs.
For calculating the number of hybrid orbital or hybridization of the central atom we can use the formula:
$X=\dfrac{1}{2}\left[ \begin{aligned} & \\{\text{no}\text{. of valence electrons of central atom }\\!\\!\\}\\!\\!\text{ + }\\!\\!\\{\\!\\!\text{ no}\text{. of monovalent atoms }\\!\\!\\}\\!\\!\text{ } \\\ & \text{ - }\\!\\!\\{\\!\\!\text{ charge on cation }\\!\\!\\}\\!\\!\text{ + }\\!\\!\\{\\!\\!\text{ charge on the anion }\\!\\!\\}\\!\\!\text{ } \\\ \end{aligned} \right]$
$X=\dfrac{1}{2}\left[ VE+MA-c+a \right]$
Given the molecule is $PC{{l}_{5}}$ , the central atom is phosphorus, and it has a monovalent atom, chlorine.
The central atom has 5 valence electrons. There are 5 monovalent atoms in $PC{{l}_{5}}$ and it is a neutral molecule.
So, the hybridization will be,
$X=\dfrac{1}{2}\left[ 5+5-0+0 \right]=\dfrac{10}{2}=5$
The value of X is 5, therefore, the hybridization is $s{{p}^{3}}d$
Since the number of surrounding atoms is equal to hybrid orbitals, therefore, there are no lone pairs in $PC{{l}_{5}}$. So, $s{{p}^{3}}d$hybridization has a Trigonal bipyramidal.
Given the molecule is $PCl_{4}^{+}$, the central atom is phosphorus, and it has a monovalent atom, chlorine.
The central atom has 5 valence electrons. There are 4 monovalent atoms in $PCl_{4}^{+}$ and it has a +1 cationic charge
So, the hybridization will be,
$X=\dfrac{1}{2}\left[ 5+4-1+0 \right]=\dfrac{8}{2}=4$
The value of X is 4, therefore, the hybridization is $s{{p}^{3}}$
Since the number of surrounding atoms is equal to hybrid orbitals, therefore, there are no lone pairs in$PCl_{4}^{+}$. So, $s{{p}^{3}}$ hybridization has a Tetrahedral shape.
Given the molecule is $PCl_{6}^{-}$, the central atom is phosphorus, and it has a monovalent atom, chlorine.
The central atom has 5 valence electrons. There are 6 monovalent atoms in $PCl_{6}^{-}$ and it has -1 anionic charge.
So, the hybridization will be,
$X=\dfrac{1}{2}\left[ 5+6-0+1 \right]=\dfrac{12}{2}=6$
The value of X is 6, therefore, the hybridization is $s{{p}^{3}}{{d}^{2}}$
Since the number of surrounding atoms is equal to hybrid orbitals, therefore, there are no lone pairs in$PCl_{6}^{-}$. So, $s{{p}^{3}}{{d}^{2}}$hybridization has an octahedral shape.

Therefore, the correct answer is an option (a)- Trigonal bipyramidal to tetrahedral and octahedral.

Note: For predicting the shape of the molecule, the only number of atoms or groups is considered. Even, if they are a monovalent or divalent atom or molecule.