Lewis Structure of SeO4 2- (With 5 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of SeO4 2- in just 5 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of SeO4 2- ion.

So, if you are ready to go with these 5 simple steps, then let’s dive right into it!

Lewis structure of SeO4 2- ion contains two double bonds and two single bonds between the Selenium (Se) atom and Oxygen (O) atoms. The Selenium atom (Se) 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 SeO4^2- ion).

5 Steps to Draw the Lewis Structure of SeO4 2- ion

Step #1: Calculate the total number of valence electrons

Here, the given ion is SeO4^2-. In order to draw the lewis structure of SeO4^2- ion, first of all you have to find the total number of valence electrons present in the SeO4^2- 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 SeO4 2- ion

• For Selenium: 

Selenium is a group 16 element on the periodic table.

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

• For Oxygen:

Oxygen is a group 16 element on the periodic table.

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

Hence in a SeO42- ion, 

Valence electrons given by Selenium (Se) atom = 6
Valence electrons given by each Oxygen (O) atom = 6
Electron due to -2 charge, 2 more electron are added
So, total number of Valence electrons in SeO4^2- 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). ^[1]

Here in the SeO4^2- ion, if we compare the selenium atom (Se) and oxygen atom (O), then selenium is less electronegative than oxygen.

So, selenium 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 SeO4, put the two electrons (i.e electron pair) between each selenium atom and oxygen atom to represent a chemical bond between them.

These pairs of electrons present between the Selenium (Se) and Oxygen (O) atoms form a chemical bond, which bonds the selenium and oxygen atoms with each other in a SeO4^2- 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 SeO4^2- 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 SeO4^2- 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 SeO4^2- 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 selenium), 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 SeO4^2- ion.

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

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

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

Formal charge	=	Valence electrons	–	Nonbonding electrons	–	(Bonding electrons)/2
Se	=	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 SeO4^2- ion.

As we know that selenium 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 selenium atom becomes “zero”, and this gives us the most stable lewis structure of SeO4^2-.

There are two -ve charges left on the oxygen atom, which indicates the 2- formal charge on the SeO4^2- ion.

Each electron pair (:) in the lewis dot structure of SeO4^2- ion represents the single bond ( | ). So the above lewis dot structure of SeO4^2- ion can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of BrCl4-
Lewis Structure of SeO
Lewis Structure of ICN
Lewis Structure of P2H4
Lewis Structure of SI6 

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