Lewis Structure Of SO4 2- (With 5 Simple Steps To Draw!)

I’m super excited to teach you the lewis structure of SO4 2- in just 5 simple steps.Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of SO4 2- ion.So, if you are ready to go with these 5 simple steps, then let’s dive right into it!

Lewis structure of SO4 2- ion (Sulfate ion) contains two double bonds and two single bonds between the Sulfur (S) atom and Oxygen (O) atoms. The Sulfur atom (S) is at the center and it is surrounded by 4 Oxygen atoms (O).

Let’s draw and understand this lewis dot structure step by step.

(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of SO42- ion).

5 Steps to Draw the Lewis Structure of SO4 2- ion

Step #1: Calculate the total number of valence electrons

Here, the given ion is SO42-. In order to draw the lewis structure of SO42- ion, first of all you have to find the total number of valence electrons present in the SO42- ion.(Valence electrons are the number of electrons present in the outermost shell of an atom).

So, let’s calculate this first.

Calculation of valence electrons in SO4 2- ion

• For Sulfur:

Sulfur is a group 16 element on the periodic table. [1]

Hence, the valence electrons present in sulfur is 6 (see below image).

• For Oxygen:

Oxygen is also a group 16 element on the periodic table. [2]

Hence, the valence electron present in oxygen is also 6 (see below image).

Hence in a SO42- ion,

Valence electrons given by Sulfur (S) atom = 6 Valence electrons given by each Oxygen (O) atom = 6Electron due to -2 charge, 2 more electron are addedSo, total number of Valence electrons in SO42- ion = 6 + 6(4) + 2 = 32

Step #2: Select the center atom

While selecting the center atom, always put the least electronegative atom at the center.

(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table). [3]

Here in the SO42- ion, if we compare the sulfur atom (S) and oxygen atom (O), then sulfur is less electronegative than oxygen.

So, sulfur should be placed in the center and the remaining 4 oxygen atoms will surround it.

Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of SO4, put the two electrons (i.e electron pair) between each sulfur atom and oxygen atom to represent a chemical bond between them.

These pairs of electrons present between the Sulfur (S) and Oxygen (O) atoms form a chemical bond, which bonds the sulfur and oxygen atoms with each other in a SO42- ion.

Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom

Don’t worry, I’ll explain!

In the Lewis structure of SO42- ion, the outer atoms are oxygen atoms.

So now, you have to complete the octet on these oxygen atoms (because oxygen requires 8 electrons to have a complete outer shell).

Now, you can see in the above image that all the oxygen atoms form an octet.

Also, all the 32 valence electrons of SO42- ion (as calculated in step #1) are used in the above structure. So there are no remaining electron pairs.

Hence there is no change in the above sketch of SO42- ion.

Let’s move to the next step.

Step #5: Check the formal charge

You can see from the above image that the central atom (i.e sulfur), is having 8 electrons. So it fulfills the octet rule.

But, in order to get the most stable lewis structure, we have to check the formal charge on SO42- ion.

For that, you need to remember the formula of formal charge;

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

• For Sulfur: Valence electrons = 6 (as it is in group 16)Nonbonding electrons = 0Bonding electrons = 8
• For Oxygen: Valence electron = 6 (as it is in group 16)Nonbonding electrons = 6Bonding electrons = 2

Formal charge	=	Valence electrons	–	Nonbonding electrons	–	(Bonding electrons)/2
S	=	6	–	0	–	8/2	=	+2
O	=	6	–	6	–	2/2	=	-1

Let’s keep these charges on the atoms in the above lewis structure of SO42- ion.

As we know that sulfur can hold more than 8 electrons, we can further reduce the charges as mentioned below.

By doing this, the formal charges on the two oxygen atoms as well as one central sulfur atom becomes “zero”, and this gives us the most stable lewis structure of SO42-.

There are two -ve charges left on the oxygen atom, which indicates the 2- formal charge on the SO42- ion.

Each electron pair (:) in the lewis dot structure of SO42- ion represents the single bond ( | ). So the above lewis dot structure of SO42- ion can also be represented as shown below.

Related lewis structures for your practice:Lewis Structure of ClO2Lewis Structure of Br2Lewis Structure of BeCl2Lewis Structure of CH3COO-Lewis Structure of Acetone (C3H6O)

Article by;

Jay Rana

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. With a desire to make learning accessible for everyone, he founded Knords Learning, an online learning platform that provides students with easily understandable explanations. Read more about our Editorial process.

Author

Jay Rana

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. With a desire to make learning accessible for everyone, he founded Knords Learning, an online learning platform that provides students with easily understandable explanations.

Read more about our Editorial process.