Question

What is the molecular shape for both $N{F_3}$ and $XeC{l_4}$?

Answer 1

Hint: The Valence Shell Electron Pair Repulsion (VSEPR) theory can be used to predict the form of a covalent molecule. VSEPR theory states that valence electron pairs in a molecule can organize themselves around the molecule's central atom(s) in order to minimize the repulsion between their negative charges.

Complete answer:
Since, we know that the molecular shape of any molecule is defined on the basis of Valence Shell Electron Pair Repulsion (VSEPR) theory. So, the shape of $N{F_3}$ and $XeC{l_4}$ is also formed on the basis of this theory.
Molecular Shape of $N{F_3}$ -
The molecule's central atom is nitrogen, which is bound to three fluorine atoms for a total of three group atoms. Nitrogen has a total of $5$ valence electrons, three of which are shared, giving it a valency of 3 and a charge of 0 over the molecule. Now, we will substitute the values in order to get the hybridization of the molecule –
$H = GA + \dfrac{{VE - V - C}}{2}$
Where $H = $Hybridization
            $GA = $Number of groups of atoms attached to the central atom.
           $VE = $ Valence electron present in the central atom.
           $V = $Valency of the central atom
           $C = $ Charge over the molecule
Substituting the values, we get –
$\
  H = 3 + \dfrac{{(5 - 3 - 0)}}{2} \\
  H = 3 + \dfrac{2}{2} \\
  H = 3 + 1 \\
  H = 4 \\
\ $
So, the hybridization of $N{F_3}$ molecule is 4.
Its time to calculate the number of bond pair and lone pair of electrons present in the molecule.
Formula to calculate lone pair is given below –
$L.P. = \dfrac{{VE - M - B}}{2}$
Where $L.P. = $Lone pair
           $M = $ Monovalent atoms attached to the central atom.
           $B = $Number of bivalent atom attached to the central atom.
Substituting the values, we get –
$\
  L.P. = \dfrac{{5 - 3 - 0}}{2} \\
  L.P. = \dfrac{2}{2} \\
  L.P. = 1 \\
\ $
So, there is a total $1$ lone pair of electrons in the molecule. According to VSEPR theory, a molecular shape of the compound having $4$ hybridization and $1$ lone pair is trigonal pyramidal structure.
Hence, the molecular shape of $N{F_3}$ molecule is trigonal pyramidal.
Molecular shape of $XeC{l_4}$ -
There are $4$ bonding pairs around the xenon atom, it means $4$ chlorine atoms are bonding with Xenon (central atom) and there are two lone pairs on Xenon. The central atom Xenon in $XeC{l_4}$ is $s{p^3}{d^2}$hybridized, implying an Octahedral gross geometry with the non-bonding electron pairs trans with respect to each other. When using VSEPR rules to minimize repulsion between lone pairs, bond pairs, and lone pair-bond pairs, we find that if the lone pairs are anti to one another and the fluorine atoms are in Equatorial position, the repulsion is minimal.
Hence, the $XeC{l_4}$ molecule's geometry is square planar since molecular geometry is defined using real atoms.

Note:
Molecular shapes are critical in predicting how one molecule will react with another and determining macroscopic properties like melting and boiling points.