CH3COO- Lewis Structure in 6 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

CH3COO- (Acetate ion) lewis structure has 2 Carbon atoms (C) at the center which are surrounded by three Hydrogen atoms (H) and two Oxygen atoms (O). One Oxygen is single bonded and other Oxygen is double bonded with Carbon. There are 3 lone pairs on the single Oxygen atom and 2 lone pairs on double bonded oxygen atom.

If you haven’t understood anything from the above image of CH3COO- lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of CH3COO- ion (acetate ion).

So let’s move to the steps of drawing the lewis structure of CH3COO- ion (acetate ion).

Steps of drawing CH3COO- lewis structure

Step 1: Find the total valence electrons in CH3COO- ion

In order to find the total valence electrons in a CH3COO- ion, first of all you should know the valence electrons present in carbon atom, hydrogen atom as well as oxygen atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of carbon, hydrogen as well as oxygen using a periodic table.

Total valence electrons in CH3COO- ion

→ Valence electrons given by carbon atom:

Carbon is group 14 element on the periodic table. ^[1] Hence the valence electrons present in carbon is 4.

You can see the 4 valence electrons present in the carbon atom as shown in the above image.

→ Valence electrons given by hydrogen atom:

Hydrogen is group 1 element on the periodic table. ^[2] Hence the valence electron present in hydrogen is 1.

You can see that only 1 valence electron is present in the hydrogen atom as shown in the above image.

→ Valence electrons given by oxygen atom:

Oxygen is group 16 element on the periodic table. ^[3] Hence the valence electrons present in oxygen is 6.

You can see the 6 valence electrons present in the oxygen atom as shown in the above image.

Hence, 

Total valence electrons in CH3COO- ion = valence electrons given by 2 carbon atoms + valence electrons given by 3 hydrogen atoms + valence electrons given by 2 oxygen atoms + 1 more electron is added due to 1 negative charge = 4(2) + 1(3) + 6(2) + 1 = 24.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

(Remember: If hydrogen is present in the given molecule, then always put hydrogen outside.)

Now here the given ion is CH3COO- ion and it contains carbon atoms (C), hydrogen atoms (H) and oxygen atoms (O).

So as per the rule we have to keep hydrogen outside.

Now, you can see the electronegativity values of carbon atom (C) and oxygen atom (O) in the above periodic table.

If we compare the electronegativity values of carbon (C) and oxygen (O) then the carbon atom is less electronegative.

So here the carbon atoms (C) are the center atom and the oxygen atoms (O) are the outside atom.

Step 3: Connect each atoms by putting an electron pair between them

Now in the CH3COO molecule, you have to put the electron pairs between the carbon (C) atom, oxygen (O) atom and hydrogen (H) atoms.

This indicates that these atoms are chemically bonded with each other in a CH3COO molecule.

Step 4: Make the outer atoms stable

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of CH3COO molecule, you can see that the outer atoms are hydrogen atoms and oxygen atom.

These hydrogen atoms and oxygen atom are forming a duplet and octet respectively and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the CH3COO- ion.

The CH3COO- ion has a total 24 valence electrons and all these valence electrons are used in the above sketch.

Hence there are no remaining electron pairs to be kept on the central atoms. 

So now let’s proceed to the next step.

Step 5: Check the octet on the central atom. If it does not have octet, then shift the lone pair to form a double bond or triple bond.

In this step, you have to check whether the central carbon atoms (C) are stable or not.

In order to check the stability of the central carbon (C) atoms, we have to check whether they are forming an octet or not.

Unfortunately, one of the carbon atoms is not forming an octet here. It has only 6 electrons and it is unstable.

Now to make this carbon atom stable, you have to shift the electron pair from the outer oxygen atom so that the carbon atom can have 8 electrons (i.e octet).

After shifting this electron pair, the carbon atom will get 2 more electrons and thus its total electrons will become 8.

You can see from the above picture that the carbon atom is forming an octet as it has 8 electrons.

Now let’s proceed to the final step to check whether the lewis structure of CH3COO is stable or not.

Step 6: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of above lewis structure.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on carbon (C) atoms, hydrogen (H) atoms as well as oxygen (O) atoms present in the CH3COO molecule.

For calculating the formal charge, you have to use the following formula;

Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons

You can see the number of bonding electrons and nonbonding electrons for each atom of CH3COO molecule in the image given below.

For Carbon (C) atom:
Valence electrons = 4 (because carbon is in group 14)
Bonding electrons = 8
Nonbonding electrons = 0

For Hydrogen (H) atom:
Valence electron = 1 (because hydrogen is in group 1)
Bonding electrons = 2
Nonbonding electrons = 0

For single bonded Oxygen (O) atom:
Valence electrons = 6 (because oxygen is in group 16)
Bonding electrons = 2
Nonbonding electrons = 6

For double bonded Oxygen (O) atom:
Valence electrons = 6 (because oxygen is in group 16)
Bonding electrons = 4
Nonbonding electrons = 4

Formal charge	=	Valence electrons	–	(Bonding electrons)/2	–	Nonbonding electrons
C	=	4	–	8/2	–	0	=	0
H	=	1	–	2/2	–	0	=	0
O (single bonded)	=	6	–	2/2	–	6	=	-1
O (double bonded)	=	6	–	4/2	–	4	=	0

From the above calculations of formal charge, you can see that the single bonded oxygen (O) atom has -1 charge and the other atoms have 0 charges.

So let’s keep these charges on the respective atoms in the CH3COO molecule.

This overall -1 charge on the CH3COO molecule is represented in the image given below.

In the above lewis dot structure of CH3COO- ion, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of CH3COO- ion.

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

Acetone (C3H6O) Lewis Structure	POCl3 Lewis Structure
HNO2 Lewis Structure	HCO3- Lewis Structure
C3H8 (Propane) Lewis Structure	CH3CN Lewis Structure