Question

Pair of molecules having identical geometry is:
(A) \[B{F_3},\,N{H_3}\]
(B) \[B{F_3},\,Al{F_3}\]
(C) \[Be{F_2},\,{H_2}O\]
(D) \[BC{l_3},\,PC{l_3}\]

Answer 1

Hint:With the help of lewis dot structure, we can determine the presence of lone pair and bond pair, But the shape and geometry of a molecule is determined by the Valence Shell Electron Repulsion Theory or VSEPR theory. In VSEPR theory, the geometry is determined by the repulsion due to the presence of the lone pairs.

Complete step-by-step answer:To find the molecules having the same geometry we have to analyse the geometry and shape of the molecule one by one.
\[B{F_3}\]: In case of Boron Trifluoride, the boron atom is bonded to three fluorine atoms. The electronic configuration of boron is $1{s^2}2{s^2}2{p^1}$. Hence it has $3$ valence electrons in its outermost shell. These electrons are shared mutually to form $3$$B - F$ bonds. The presence of lone pairs of electrons on fluorine will create repulsion among other fluorine atoms. Hence to minimize the repulsion, the $3$$B - F$ bonds will be placed at an angle of ${120^0}$ leading to the formation of a trigonal planar geometry.
\[N{H_3}\]: In case of ammonia, nitrogen atom is bonded to three hydrogen atoms. Nitrogen atoms have $5$ electrons in its valence shell. The presence of a lone pair of electrons will create a repulsion between the $3$$N - H$ bonds and force the molecule to attain a tetrahedral geometry to minimize the repulsion.
\[Al{F_3}\]: In\[Al{F_3}\], Aluminium atoms are bonded to three fluorine atoms. The $3$$Al - F$ bonds are arranged in a trigonal planar geometry to minimize the repulsion with a bond angle of ${120^0}$.
\[{H_2}O\]: In \[{H_2}O\], the oxygen atom is bonded to $2$ hydrogen atoms. The presence of $2$ lone pairs of electrons increases the repulsion between bond pairs and lone pairs that distort the shape of the molecule resulting in the formation of tetrahedral geometry.
\[Be{F_2}\]: In \[Be{F_2}\], the beryllium atom is bonded to two fluorine atoms. They are arranged in a linear geometry with a bond angle of ${180^0}$ to minimize the repulsion.
$BC{l_3}$: In $BC{l_3}$, Boron atom is bonded to three chlorine atoms. The $3$$B - Cl$ bonds are arranged in a trigonal planar geometry to minimize the repulsion with a bond angle of ${120^0}$.
$PC{l_3}$: In $BC{l_3}$, a Phosphorus atom is three chlorine atoms. The $3$$P - Cl$ bonds are arranged in a trigonal pyramidal geometry to minimize the repulsion due to the lone pair on phosphorus with a bond angle of ${109^0}$.
Hence, from the given data we can analyse that the molecules that have the same geometry are \[B{F_3},\,Al{F_3}\].

Therefore, Option (B) is correct.

Note:In VSEPR theory, the geometry is determined by the repulsion between the electron pairs present in the molecule. This repulsion could be in lone pairs-lone pair, Lone pair-Bond pair, Bond pair-Bond pair. The order of the repulsion is: $lp - lp > lp - bp > bp - bp$.