AsO33- Lewis structure

The information on this page is ✔ fact-checked.

AsO33- Lewis Structure
AsO33- Lewis structure

AsO33- has one arsenic atom and three oxygen atoms.

In AsO33- Lewis structure, there are three single bonds around the arsenic atom, with three oxygen atoms attached to it. Each oxygen atom has three lone pairs, and the arsenic atom has one lone pair.

Also, there is a negative (-1) charge on each oxygen atom.

Steps

Use these steps to correctly draw the AsO33- Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

#1 First draw a rough sketch

  • First, determine the total number of valence electrons
Periodic table

In the periodic table, arsenic lies in group 15, and oxygen lies in group 16.

Hence, arsenic has five valence electrons and oxygen has six valence electrons.

Since AsO33- has one arsenic atom and three oxygen atoms, so…

Valence electrons of one arsenic atom = 5 × 1 = 5
Valence electrons of three oxygen atoms = 6 × 3 = 18

Now the AsO33- has a negative (-3) charge, so we have to add three more electrons.

So the total valence electrons = 5 + 18 + 3 = 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 arsenic is less electronegative than oxygen, assume that the central atom is arsenic.

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

  • And finally, draw the rough sketch
AsO33- Lewis Structure (Step 1)
Rough sketch of AsO33- Lewis structure

#2 Mark lone pairs on the atoms

Here, we have a total of 13 electron pairs. And three As — 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 arsenic 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 arsenic, there is one lone pair.

Mark the lone pairs on the sketch as follows:

AsO33- Lewis Structure (Step 2)
Lone pairs marked on AsO33- Lewis structure

#3 Calculate and mark formal charges on the atoms, if required

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

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

For arsenic atom, formal charge = 5 – 2 – ½ (6) = 0

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

Here, each oxygen atom has a charge, so mark it on the sketch as follows:

AsO33- Lewis Structure (Step 3)
Formal charges marked, and got the most stable Lewis structure of AsO33-

In the above structure, you can see that the central atom (arsenic) forms an octet. And the outside atoms (oxygens) also form an octet. Hence, the octet rule is satisfied.

Now there is still a negative (-1) charge on each oxygen atom.

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.

Therefore, this structure is the most stable Lewis structure of AsO33-.

And since the AsO33- has a negative (-3) charge, mention that charge on the Lewis structure by drawing brackets as follows:

AsO33- Lewis Structure (Final)
AsO33- Lewis structure showing a negative (-3) charge

Next: C4H6 Lewis structure

External links

Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.

Leave a Comment