Question

Which of the following ion is not tetrahedral in shape:
A.$B{F_4}^ -$
B.$N{H_4}^ +$
C.$S{F_4}$
D.$C{F_4}$

Answer 1

The three dimensional arrangement of atoms that constitute a molecule is known as its molecular geometry. It includes the shape of the molecule as well as bond lengths, bond angles and other geometrical parameters that determines the position of each atom.

Complete step by step answer:
Generally we use a formula for a molecule to calculate its hybridization and from hybridization we calculate the shape of the molecule.
The formula used is:
$H = \dfrac{1}{2}\left[ {V + M - C + A} \right]$
Where,
$H$ is the number of orbitals involved in hybridization.
$V$ is the number valence electrons of a central atom.
$C$ is in charge of cation.
$A$ is in charge of anion.
$M$ is the number of monovalent atoms linked to the central atom.
For $B{F_4}^ -$
$H = \dfrac{1}{2}\left[ {3 + 4 - 0 + 1} \right] = 4$
Therefore its hybridization is $s{p^3}$ and its shape will be tetrahedral.
For $N{H_4}^ +$
$H = \dfrac{1}{2}\left[ {5 + 4 - 1 + 0} \right] = 4$
Therefore its hybridization is $s{p^3}$ and its shape will be tetrahedral.
For $S{F_4}$
$H = \dfrac{1}{2}\left[ {6 + 4 - 0 + 0} \right] = 5$
Therefore its hybridization is $s{p^3}d$ and its shape will be see-saw.
For $C{F_4}$
$H = \dfrac{1}{2}\left[ {4 + 4 - 0 + 0} \right] = 4$
Therefore its hybridization is $s{p^3}$ and its shape will be tetrahedral.
Hence in all these calculations we found that the hybridization of $S{F_4}$ is $s{p^3}d$, molecules with this hybridization does not give tetrahedral shape so the only molecule/ion that has shape other than tetrahedral is $S{F_4}$.

Hence, option (C) is correct.

Note: Some ions/molecules have lone pairs with them which change their shapes from the expected one, same is the case of $S{F_4}$, it has one lone pair which contributes in hybridization and changes its shape to see-saw.