# How do you draw the Lewis structure for SeO_2. What is the electron geometry around the central atom?

Sep 3, 2016

#### Explanation:

Here are the steps I follow when drawing a Lewis structure.

1. Decide which is the central atom in the structure.

That will normally be the least electronegative atom ($\text{Se}$).

2. Draw a skeleton structure in which the other atoms are single-bonded to the central atom:

$\text{O-Se-O}$

3. Draw a trial structure by putting electron pairs around every atom until each gets an octet.

In this editor, I will have to write it as

:stackrel(. .)("O")-stackrel(. .)("Se")-stackrel(. .)("O"":
$\textcolor{w h i t e}{l l} \stackrel{. . \textcolor{w h i t e}{m m} . . \textcolor{w h i t e}{m m l l} . .}{\textcolor{w h i t e}{m m m m m l}}$

4. Count the valence electrons in your trial structure (20).

5. Now count the valence electrons you actually have available.

$\text{1 Se + 2 O = 1×6 + 2×6 = 18}$.

The trial structure has two extra electrons.

6. Draw a new trial structure, this time inserting one double bond for each extra pair of electrons:

$\text{O=Se-O}$ and $\text{O-Se=O}$

7. As before, add valence electrons to give each atom an octet:

:stackrel(. .)("O")="Se"-stackrel(. .)("O"":
$\textcolor{w h i t e}{l l} \stackrel{. . \textcolor{w h i t e}{m m} . . \textcolor{w h i t e}{m m l l} . .}{\textcolor{w h i t e}{m m m m m l}}$

and

:stackrel(. .)("O")-"Se"=stackrel(. .)("O"":
$\textcolor{w h i t e}{l l} \stackrel{. . \textcolor{w h i t e}{m m} . . \textcolor{w h i t e}{m m l l} . .}{\textcolor{w h i t e}{m m m m m l}}$

8. Calculate the formal charge on each atom.

:stackrel(. .)("O")=stackrel(+1)("Se")-stackrel(. .)("O"):^"-1"
$\textcolor{w h i t e}{l l} \stackrel{. . \textcolor{w h i t e}{m m} . . \textcolor{w h i t e}{m m l l} . .}{\textcolor{w h i t e}{m m m m m l}}$

and

$\text{^"-1":stackrel(. .)("O")-stackrel(+1)("Se")=stackrel(. .)("O} :$
$\textcolor{w h i t e}{m l l} \stackrel{. . \textcolor{w h i t e}{m m} . . \textcolor{w h i t e}{m m l l} . .}{\textcolor{w h i t e}{m m m m m l}}$

9. We see that some of the atoms have formal charges.

The “best” Lewis structure is one in which has the fewest formal charges.

We can generate a structure with zero formal charges if we move a lone pair from the single-bonded $\text{O}$ to make a double bond to the $\text{S}$.

This gives us a third possibility:

$: \stackrel{. .}{\text{O")=stackrel(. .)("Se")=stackrel(. .)("O}} :$

We now have a structure in which $\text{S}$ has ten valence electrons.

However, that’s OK, because $\text{S}$ can “expand” its octet.

We have three different structures, differing ONLY in the locations of the electrons.

We say that these are resonance structures of ${\text{SeO}}_{2}$.

The actual structure of ${\text{SeO}}_{2}$ is a resonance hybrid of all three structures.

In all three structures, there are three electron domains about the $\text{Se}$ atom: the lone pair and the bonds on either side.

The electron geometry around $\text{Se}$ is trigonal planar.