SeO4 2- Lewis Structure in 5 Steps (With Images)

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

Let me explain the above image in short.

SeO4 2- lewis structure has a Selenium atom (Se) at the center which is surrounded by four Oxygen atoms (O). There are 2 single bonds and 2 double bonds between the Selenium atom (Se) and each Oxygen atom (O). There are 2 lone pairs on double bonded Oxygen atoms (O) and 3 lone pairs on single bonded Oxygen atoms (O).

If you haven’t understood anything from the above image of SeO4 2- lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of SeO4^2- ion.

So let’s move to the steps of drawing the lewis structure of SeO4^2- ion.

Steps of drawing SeO4 2- lewis structure

Step 1: Find the total valence electrons in SeO4 2- ion

In order to find the total valence electrons in SeO4 2- ion, first of all you should know the valence electrons present in selenium 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 selenium as well as oxygen using a periodic table.

Total valence electrons in SeO4 2- ion

→ Valence electrons given by selenium atom:

Selenium is a group 16 element on the periodic table. ^[1] Hence the valence electrons present in selenium is 6.

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

→ Valence electrons given by oxygen atom:

Oxygen is group 16 element on the periodic table. ^[2] 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 SeO4^2- ion = valence electrons given by 1 selenium atom + valence electrons given by 4 oxygen atoms + 2 more electrons are added due to 2 negative charges = 6 + 6(4) + 2 = 32.

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.

Now here the given ion is SeO4 2- ion and it contains selenium atom (Se) and oxygen atoms (O).

You can see the electronegativity values of selenium atom (Se) and oxygen atom (O) in the above periodic table.

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

So here the selenium atom (Se) is the center atom and the oxygen atoms (O) are the outside atoms.

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

Now in the SeO4 molecule, you have to put the electron pairs between the selenium atom (Se) and oxygen atoms (O).

This indicates that the selenium (Se) and oxygen (O) are chemically bonded with each other in a SeO4 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 SeO4 molecule, you can see that the outer atoms are oxygen atoms.

These outer oxygen atoms are forming an octet and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the SeO4^2- ion.

The SeO4^2- ion has a total 32 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 atom. 

So now let’s proceed to the next step.

Step 5: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of SeO4^2- ion.

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 selenium (Se) atom as well as oxygen (O) atoms present in the SeO4 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 SeO4 molecule in the image given below.

For Selenium (Se) atom:
Valence electrons = 6 (because selenium is in group 16)
Bonding electrons = 8
Nonbonding electrons = 0

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

From the above calculations of formal charge, you can see that the selenium (Se) atom has +2 charge while all the oxygen atoms have -1 charge.

So let’s keep these charges on the respective atoms of the SeO4 molecule.

The above image shows that the lewis structure of SeO4 is not stable.

So we have to minimize these charges by shifting the electron pair from the oxygen atom to the selenium atom.

After shifting the electron pair from the oxygen atom to the selenium atom, the charges on selenium and two oxygen atoms become zero. And this is a more stable lewis structure. (see below image).

There are two -ve charges left on the oxygen atoms, which gives -2 formal charge on the SeO4 molecule.

This overall -2 charge on the SeO4 molecule is represented in the image given below.

In the above lewis dot structure of SeO4 2- ion, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of SeO4 2- 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:

About 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. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.

Read more about our Editorial process.