Question

How to draw Lewis structure for ${\text{SO}}_{\text{4}}^{\text{2 - }}$ and ${\text{NO}}_{\text{2}}^{\text{ - }}$.

Answer 1

Hint: For drawing the Lewis structure of any compound or molecule we have to know about the number of valence electrons present in that compound and the valency of each atom present in that compound.

Complete step by step answer:
For drawing the Lewis structure of Sulfate ion (${\text{SO}}_{\text{4}}^{\text{2 - }}$) we have to follow the following steps:
-First we calculate the total number of valence electrons present in the sulfate ion and for this we will add all the valence electrons of each atom and charge present on that molecule.
-Valence electrons on sulfur atom ($\text{S}$) = $6$
Valence electrons on oxygen atom ($\text{O}$) = $6$
Electrons on ${\text{SO}}_{\text{4}}^{\text{2 - }}$molecule due to charge = $2$
Total electrons on ${\text{SO}}_{\text{4}}^{\text{2 - }}$ molecule = $6+\left(4\times 6\right)+2=32$
-Now we construct a basic structure of ${\text{SO}}_{\text{4}}^{\text{2 - }}$ molecule by taking sulfur atom as a central atom because it is less electro-positive as compared to oxygen atom.
        

-In the above figure eight electrons are involved in the formation of four bonds and then we have $24$ electrons left. Now we will distribute these $\text{24}{\text{e}}^{\text{ - }}$ among four oxygen atoms as follow:
      

-As we know that sulfur has six valence electrons but in the above structure only four electrons are involved so for satisfying the valency of sulfur, we will convert two lone pairs of electrons from two different oxygen atoms to double bond as follow:
      

Hence Lewis structure of sulfate ion is shown as below:
   

For drawing the Lewis structure of Nitrite ion (${\text{NO}}_2^{\text{ - }}$) we have to follow the following steps:
-First we calculate the total number of valence electrons present in the nitrite ion and for this we will add all the valence electrons of each atom and charge present on that molecule.
-Valence electrons on nitrogen atom ($\text{N}$) = $5$
Valence electrons on oxygen atom ($\text{O}$) = $6$
Electrons on ${\text{NO}}_2^{\text{ - }}$molecule due to charge = $1$
Total electrons on ${\text{NO}}_2^{\text{ - }}$molecule = $5+\left(2\times 6\right)+1=18$
-Now we construct a basic structure of ${\text{NO}}_2^{\text{ - }}$ molecule by taking nitrogen atom as a central atom because it is less electro-positive as compared to oxygen atom.
      

-In the above figure four electrons are involved in the formation of two bonds and then we have $18$ electrons left. Now we will distribute these $\text{18}{\text{e}}^{\text{ - }}$ among two oxygen atoms and one nitrogen atom as follow:
     

-As we know that nitrogen has five valence electrons but in the above structure only four electrons are involved so for satisfying the valency of nitrogen, we will convert one lone pair of electrons from one oxygen atoms to double bond as follow
      

Hence Lewis structure of nitrite ion is shown as below:

Note: Here some of you may draw the wrong structure if you didn’t count the charge present on a molecule as an electron during calculating the total number of valence electrons of the given compound.