Question

What is the Lewis dot structure of $N{a}_{2}S{O}_4$?

Answer 1

Hint :We know that we can draw electron dot structure easily by calculating the valence electrons on each atom present in the given molecules. While drawing the dot structures here remember that the number of valence electrons in silicon is $$4$$and that for sulphur, beryllium, carbon and oxygen is $$6,2,7,4$$ and $$6$$ respectively. Use this to draw the structures.

Complete Step By Step Answer:
We also know Lewis dot structures as electron dot structures. They are diagrams that show the bonding between atoms or molecules and the lone pairs of electrons existing in the molecule. To draw the structure, we need to follow some steps. They are the total number of electrons represented in the electron dot structure is the sum of the total number of valence electrons present in each atom of the molecule. Electrons other than the valence electrons are not shown in a valence dot structure. After we have determined the number of valence electrons, we can place them in the dot structure following these steps; If we are drawing it for a molecule that consists of two or more atoms; the atoms are connected by single bonds.
Electrons should be placed as lone pairs. One pair of dots for each pair of electrons should be placed. -Lone pairs should be placed on outer atoms until each atom has eight electrons. If there are any extra electrons left, they should be placed on the central atom. When all the lone pairs are placed, if the central atom did not complete its octet, a double bond should be placed. Now we can draw the Lewis dot structure of the given structures
For $$S{O}_4^{2-},$$ a reasonable Lewis structure to accommodate the $$32$$ electrons is$${(O=)}_{2}S{(-O^{-})}_2$$, and clearly this is a derivative of the parent sulfuric acid, $${(O=)}_{2}S{(-OH)}_2.$$Given this Lewis structure, there are$$6$$ valence electrons associated with the doubly bound oxygen (and hence these are neutral),$$6$$ valence electrons with the central sulfur (and hence a neutral sulfur), and $$7$$ valence electrons associated with each singly bound oxygen atoms, and hence these center’s bear the $$2$$ required formal negative charges.
Note that all the oxygen atoms are equivalent, and we could even propose a Lewis structure of $${}^{2+}S{(-O^{-})}_4;{(O=)}_{2}S{(-O^{-})}_2$$is a more conventional representation.

Note :
Remember that to draw electron dot structure, it is important that we remember the structure of compounds if the formula is not given. It is also important to remember the atomic numbers in order to find out the number of valence electrons.