Chapter 8.3, Problem 46PP

Interpretation Introduction

Interpretation:

The Lewis resonance structure for the ion sulfite ${\left(\text{SO}\right)}_3{}^{2-}$ needs to be explained.

Concept introduction:

Lewis structure is also known as electron − dot structure which is showing the bonding between atoms of a molecule and used for atoms to show their valence electron.

Answer to Problem 46PP

Lewis structure is a representation of the valence electrons of an atom that uses dot around the symbol of the element, where number of dots around of the element equals to the number of valence electrons in the atom.

Total valence electrons (S atom) $=6$

Total valence electrons (O atom) $=6\times 3=18$

Total valenec electrons ${\left(\text{SO}\right)}_3{}^{2-}=6+18+2=26$

Total valence electron pairs

$\begin{array}{l}=\sigma \text{ bond + }\pi \text{ bonds + lone pair at valence shell}\\ {\text{=13e}}^{-}\left(\frac{26}{2}\right)\end{array}$

$\text{O}-\underset{\underset{\text{O}}{\text{I}}}{\text{S}}-\text{O }\left(\text{Rough Sketch}\right)$

In the diagram three $\text{S}-\text{O}$ bonds are the there so only $13-3=10$ valence pairs are remaining.

${}^{-}\text{:}\underset{\text{. }\text{.}}{\stackrel{..}{\text{O}}}-\underset{\underset{\text{:}\underset{.{.}_{-}}{\text{O}}\text{:}}{\text{I}}}{\stackrel{\text{. }{\text{.}}^{+}}{\text{S}}}-\stackrel{..}{\text{O}}{\text{:}}^{-}$

Each oxygen atom will get a -1 change and sulfur atom get a +1 change.

Explanation of Solution

Oxygen atom is holding negative change (greater electro negativity) compared to sulfur atom. But previous structure is not stable as every atom is having charge on it. Then to minimize the charge, lone pair is converted into bond pair. By converting one lone pair of one oxygen atom to bond pair, one S-O bond is formed. After that there will be a double bond between S and O and rest of the two O atoms have single bonds with S.