# SeO32- Lewis Structure

SeO32- (selenite) has one selenium atom and three oxygen atoms. In the lewis structure of SeO32-, there is one double bond and two single bonds around the selenium atom, with three oxygen atoms attached to it. The oxygen atom with a double bond has two lone pairs, the two oxygen atoms with single bonds have three lone pairs, and the selenium atom has one lone pair.

Also, there is a negative (-1) charge on the two oxygen atoms with single bonds.

## Steps

Here’s how you can draw the SeO32- lewis structure step by step.

Step #1: draw sketch
Step #2: mark lone pairs
Step #3: mark charges
Step #4: minimize charges
Step #5: minimize charges again (if there are)

Let’s break down each step in detail.

### #1 Draw Sketch

• First, determine the total number of valence electrons

In the periodic table, both selenium and oxygen lie in group 16.

Hence, both selenium and oxygen have six valence electrons.

Since SeO32- has one selenium atom and three oxygen atoms, so…

Valence electrons of one selenium atom = 6 × 1 = 6
Valence electrons of three oxygen atoms = 6 × 3 = 18

Now the SeO32- has a negative (-2) charge, so we have to add two more electrons.

So the total valence electrons = 6 + 18 + 2 = 26

• Second, find the total electron pairs

We have a total of 26 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 26 ÷ 2 = 13

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since selenium is less electronegative than oxygen, assume that the central atom is selenium.

Therefore, place selenium in the center and oxygens on either side.

• And finally, draw the rough sketch

### #2 Mark Lone Pairs

Here, we have a total of 13 electron pairs. And three Se — O bonds are already marked. So we have to only mark the remaining ten electron pairs as lone pairs on the sketch.

Also remember that selenium is a period 4 element, so it can keep more than 8 electrons in its last shell. And oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are oxygens.

So for each oxygen, there are three lone pairs, and for selenium, there is one lone pair.

Mark the lone pairs on the sketch as follows:

### #3 Mark Charges

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For selenium atom, formal charge = 6 – 2 – ½ (6) = +1

For each oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

Here, both selenium and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable lewis structure because both selenium and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Minimize Charges

Convert a lone pair of the oxygen atom to make a new Se — O bond with the selenium atom as follows:

In the above structure, you can see that the central atom (selenium) forms an octet. Hence, the octet rule is satisfied.

Now there is a negative (-1) charge on the two oxygen atoms.

This is okay, because the structure with a negative charge on the most electronegative atom is the best lewis structure. And in this case, the most electronegative element is oxygen.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the most stable lewis structure of SeO32-.

And since the SeO32- has a negative (-2) charge, mention that charge on the lewis structure by drawing brackets as follows: