Question

Indicate the molecular geometry, electronic geometry and polarity of the following molecules or ions?
(A) $Br{F_5}$
(B) $S{O_4}^{2 - }$

Answer 1

Based upon the valence electrons that were involved as bonding pair and lone pair electrons, the electron geometry and electron geometry will be determined. Molecular geometry can be written by considering total bonding pair and lone pair electrons. electron geometry can be written based on the bonding pair of electrons only.

Complete answer:
Given molecules or ions are $Br{F_5}$ , and $S{O_4}^{2 - }$
In $Br{F_5}$ , bromine is the central metal atom, consisting of $7$ valence electrons. out of which $5$ electrons were shared with five fluorine atoms leading to the $s{p^3}{d^2}$ hybridization.
The molecular geometry of $Br{F_5}$ is octahedral, due to the presence of one lone pair the electron geometry is square pyramidal.
There are five $Br - F$ bonds, out of which four $Br - F$ are exactly opposite to each other. But ${5^{th}}$ , $Br - F$ bonds do not have opposite bonds. Thus, $Br{F_5}$ is a polar molecule.
In $S{O_4}^{2 - }$ , sulphur is the central metal atom consisting of $6$ valence electrons. But two electrons were gained by the sulphur atom. Thus, the total valence electrons were $8$ . These $8$ electrons involved in the bond formation with four oxygen atoms leads to the hybridization of $s{p^3}$ . As there were no lone pairs of electrons the electron geometry and molecular geometry is tetrahedral only. In $S{O_4}^{2 - }$ there are opposite bonds that lead to the non-polar molecule.

Note:
In any molecule, the atom with low electronegativity only occupies the central position. In $S{O_4}^{2 - }$ , sulphur is the atom with low electronegativity and is a central atom, while calculating the hybridization the electrons gained which were indicated in negative charge should also be considered and these were the electrons added to a central atom only.