Question

The pair having similar geometry is:
A. ${\text{PC}}{{\text{l}}_3},\,{\text{NH}}_4^ + $
B. ${\text{BeC}}{{\text{l}}_{\text{2}}}{\text{,}}{{\text{H}}_{\text{2}}}{\text{O}}$
C. ${\text{C}}{{\text{H}}_{\text{4}}}{\text{,}}\,{\text{CC}}{{\text{l}}_{\text{4}}}$
D. ${\text{I}}{{\text{F}}_5},\,\,{\text{P}}{{\text{F}}_5}$

Answer 1

Hint: We can draw the Lewis structure of all the molecules and by using valence shell electron pair repulsion theory the geometry can be determined. On the basis of that we can predict which pairs are having similar geometry.

Complete step by step answer:
Write the Lewis structure as follows:
Write the basic structure. Write the central atom around which writes all atoms of the molecule. The least electronegative atom is the central atom.
Count total valence electrons.
Two electrons are used in the formation of a bond.
Count the total electron used in bond formation.
Subtracts the electrons used in bond formation from the total valence electrons.
Arrange the remaining electrons around each atom to complete the octet.
The valence shell electron pair repulsion theory is as follows:
Electron pair is the number of electron pairs present around the central atom in a molecule.
According to VSEPR the electron pairs present around the central atom repel each other. So, all the pairs get arranged to minimize the repulsion. Based on the number of electron pair the geometry can be determined as follows:

Number of electron pair	Geometry
2	Linear
3	Trigonal planar
4	Tetrahedral
5	Trigonal Bipyramidal
6	Octahedral

Lewis structure of the pair ${\text{PC}}{{\text{l}}_3},\,{\text{NH}}_4^ + $ is as follows:
Total valence electrons in ${\text{PC}}{{\text{l}}_3}$ are as follows:
$ = \,\left( {5 \times 1} \right) + \left( {7 \times 3} \right)$
= 26
Total valence electrons in ${\text{NH}}_4^ + $ are as follows:
$ = \,\left( {5 \times 1} \right) + \left( {1 \times 4} \right) - 1$
= 8

The total electron pair around phosphorus in ${\text{PC}}{{\text{l}}_3}$ is four and around nitrogen in ${\text{NH}}_4^ + $ is also four. So, the geometry of both${\text{PC}}{{\text{l}}_3}$ and ${\text{NH}}_4^ + $ is tetrahedral. So, both have the same geometry. So, option (A) is correct.
Lewis structure of the pair ${\text{BeC}}{{\text{l}}_{\text{2}}}{\text{,}}{{\text{H}}_{\text{2}}}{\text{O}}$ is as follows:
Total valence electrons in ${\text{BeC}}{{\text{l}}_{\text{2}}}$ are as follows:
$ = \,\left( {2 \times 1} \right) + \left( {7 \times 2} \right)$
= 16
Total valence electrons in ${{\text{H}}_{\text{2}}}{\text{O}}$ are as follows:
$ = \,\left( {1 \times 2} \right) + \left( {6 \times 1} \right)$
= 8

The total electron pair around beryllium in ${\text{BeC}}{{\text{l}}_2}$ is two and around oxygen in ${{\text{H}}_{\text{2}}}{\text{O}}$ is four. So, the geometry of ${\text{BeC}}{{\text{l}}_2}$ is linear and ${{\text{H}}_{\text{2}}}{\text{O}}$ is tetrahedral. So, both have a different geometry. So, option (B) is incorrect.
Lewis structure of the pair ${\text{C}}{{\text{H}}_{\text{4}}}{\text{,}}\,{\text{CC}}{{\text{l}}_{\text{4}}}$ is as follows:
Total valence electrons in ${\text{C}}{{\text{H}}_{\text{4}}}$ are as follows:
$ = \,\left( {4 \times 1} \right) + \left( {1 \times 4} \right)$
= 8
Total valence electrons in ${\text{CC}}{{\text{l}}_{\text{4}}}$ are as follows:
$ = \,\left( {4 \times 1} \right) + \left( {7 \times 4} \right)$
= 32

The total electron pair around carbon in both ${\text{C}}{{\text{H}}_{\text{4}}}$ and ${\text{CC}}{{\text{l}}_{\text{4}}}$ is four. So, the geometry of both ${\text{C}}{{\text{H}}_{\text{4}}}$ and ${\text{CC}}{{\text{l}}_{\text{4}}}$ is tetrahedral. So, both the molecules have the same geometry. So, option (C) is correct.
Lewis structure of the pair ${\text{I}}{{\text{F}}_5},\,\,{\text{P}}{{\text{F}}_5}$ is as follows:
Total valence electrons in ${\text{I}}{{\text{F}}_5}$ are as follows:
$ = \,\left( {7 \times 1} \right) + \left( {7 \times 5} \right)$
= 42
Total valence electrons in ${\text{P}}{{\text{F}}_5}$ are as follows:
$ = \,\left( {5 \times 1} \right) + \left( {7 \times 5} \right)$
= 40

The total electron pair around iodine in ${\text{I}}{{\text{F}}_{\text{5}}}$ is six and around phosphorous in ${\text{P}}{{\text{F}}_5}$ is five. So, the geometry of ${\text{I}}{{\text{F}}_{\text{5}}}$ is octahedral and ${\text{P}}{{\text{F}}_5}$ is trigonal bipyramidal. So, both have a different geometry. So, option (D) is incorrect.

Therefore, option (A) ${\text{PC}}{{\text{l}}_3},\,{\text{NH}}_4^ + $ and (C) ${\text{C}}{{\text{H}}_{\text{4}}}{\text{,}}\,{\text{CC}}{{\text{l}}_{\text{4}}}$ are correct.

Note: To determine the total valence electrons of a molecule, sum all the valence electrons of the atoms present in the molecule. Subtract one for every positive charge and add one for every negative charge.