Question

Which of the following structure is similar to $S{O_3}^{2 - }?$
A. ${F_2}SeO$
B. ${F_2}Se{O_2}$
C. $S{O_4}^{2 - }$
D. $S{O_2}$

Answer 1

Hint: We can determine the structure of molecule by predicting the shape of the molecule. The shape of the molecule can be obtained by identifying the number of bond pairs and lone pairs present in the molecule.

Complete step by step answer: We know that the shape of sulfite ion is trigonal pyramidal structure. The arrangement of electron pairs is tetrahedral and the number of electron pairs is four.
We know that the arrangement of electron pairs in $S{O_2}$ is trigonal planar and the number of electron pairs is three. From the arrangement of electron pairs, the shape of the molecule would be v-shaped (or) bent.
We know that the electron pair arrangement in $S{O_4}^{2 - }$ is tetrahedral and the number of electron pairs is three. The shape of the molecule $S{O_4}^{2 - }$ would be tetrahedral.
We know that in $F{e_2}SeO,$ the central metal atom is selenium, and the hybridization around the central atom is $s{p^3}.$ For a molecule, that has $s{p^3}$ hybridization, the electron pair geometry could be tetrahedral and the shape of the molecule is trigonal pyramidal.
From the above predicted, the structure of $F{e_2}S{O_4}$ is similar to the structure of $S{O_3}^{2 - }.$
Hence, option (A) is correct.

Note: Lewis structure helps to determine the molecules geometry. Since, it helps us to identify the outermost electrons. We could predict the geometry of a compound by,
- Drawing its Lewis structure
- Counting the number of electron pairs
- Arranging the electron pairs to reduce repulsion
- Arranging the atoms to reduce the lone pair-lone pair repulsion
- Naming the molecular geometries from the positions of the atom