Question

In the given, Lewis structure of \[{S_2}{O_3}^{2 - }\;\;\], formal charge present on sulphur atoms 1 and 2 respectively are:
A.Zero, $ + 1$
B.$ + 1$ , $ + 1$
C.$ + 2$ , $ + 2$
D.Zero, zeros

Answer 1

Hint: To solve these types of questions we should first draw the Lewis structure. And then after it, we count the number of total bonds and then divide it by the number of elements in which these bonds are made with the central atom
 The formula used:
 Formal charge is expressed \[ = {\text{V}} - {\text{L}} - \frac{1}{2} \times {\text{S}}\]
Where V = total no. of valence electron in a free atom, L = lone pair electrons and S = bonding electrons

Complete step by step answer:
First drawing lewis structure of \[{S_2}{O_3}^{2 - }\;\;\]

For the sulphur atom given in the Lewis dot structure:
For Sulphur atom (1) we can calculate the formal charge as:
\[{\text{FC}} = 6 - 4 - \dfrac{1}{2} \times 4{\text{ = }}0\]
For Sulphur atom (2) we can formal charge as:
\[{\text{FC}} = 6 - 0 - \dfrac{1}{2} \times 12 = 0\]

Hence, the correct option is option D.

Note:
Sulphur is present in the same group as oxygen. It has a vacant d-orbital and can thus show variable oxidation states. Sulphur can be present at an oxidation state of $ + 6$ after losing the 6 electrons in its valence shell like in $S{O_3}$. It can also be present as $ - 2$ after gaining 2 new electrons to complete its octet like in sulphide ions like ${H_2}S$ .
In most of the compounds, the oxidation number of oxygen is $ - 2$ . There are two exceptions here.
Peroxides: Each oxygen atom exhibits an oxidation number of $ - 1$ . Example, \[N{a_2}{O_2}\]
Superoxide- Every oxygen atom is allocated an oxidation number of \[ - \dfrac{{{\text{ }}1}}{2}\] . Example, \[K{O_2}\]
Oxygen is bonded to fluorine- Example, dioxygen difluoride where the oxygen atom is allocated an oxidation number of $ + 1$ .