OH- Lewis structure

The information on this page is ✔ fact-checked.

OH^– (hydroxide) has one oxygen atom and one hydrogen atom.

In the OH^– Lewis structure, there is a single bond between the oxygen and hydrogen atom, and on the oxygen atom, there are three lone pairs.

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

Steps

Here’s how you can easily draw the OH^– Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, oxygen lies in group 16, and hydrogen lies in group 1.

Hence, oxygen has six valence electrons and hydrogen has one valence electron.

Since OH^– has one oxygen atom and one hydrogen atom, so…

Valence electrons of one oxygen atom = 6 × 1 = 6
Valence electrons of one hydrogen atom = 1 × 1 = 1

Now the OH^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 6 + 1 + 1 = 8

Learn how to find: Oxygen valence electrons and Hydrogen valence electrons

• Second, find the total electron pairs

We have a total of 8 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 = 8 ÷ 2 = 4

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Hence, here we have to assume that the central atom is oxygen.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have a total of 4 electron pairs. And one O — H bond is already marked. So we have to only mark the remaining three electron pairs as lone pairs on the sketch.

Also remember that oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell. And hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atom is hydrogen. But no need to mark on hydrogen, because hydrogen already has two electrons.

So for oxygen, there are three lone pairs.

Mark the lone pair on the sketch as follows:

#3 If needed, mention formal charges on the atoms

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

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

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

For hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

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

In the above structure, you can see that the central atom (oxygen) forms an octet. And the outside atom (hydrogen) also forms a duet. Hence, the octet rule and duet rule are satisfied.

Now there is still a negative (-1) charge on the 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 OH^–.

And since the OH^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: N₂H₂ Lewis structure

📝 Your feedback matters. Visit our contact page.

External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/OHMinus-Lewis-Structure.html
• https://www.chemistryscl.com/general/OH–lewis-structure/index.php
• https://lambdageeks.com/oh-lewis-structure/
• https://socratic.org/questions/how-do-you-write-the-electron-dot-diagram-for-the-hydroxide-ion-oh