SeO42- Lewis structure

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SeO₄^2- (selenate) has one selenium atom and four oxygen atoms.

In SeO₄^2- Lewis structure, there are two double bonds and two single bonds around the selenium atom, with four oxygen atoms attached to it. The oxygen atom with double bonds has two lone pairs, and the oxygen atom with single bonds has three lone pairs.

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

Steps

To properly draw the SeO₄^2- Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary
#4 Minimize formal charges by converting lone pairs of the atoms
#5 Repeat step 4 if necessary, until all charges are minimized

Let’s break down each step in more detail.

#1 Draw a rough sketch of the structure

• First, determine the total number of valence electrons

In the periodic table, both selenium and oxygen lie in group 16.

Hence, both selenium and oxygen have six valence electrons.

Since SeO₄^2- has one selenium atom and four oxygen atoms, so…

Valence electrons of one selenium atom = 6 × 1 = 6
Valence electrons of four oxygen atoms = 6 × 4 = 24

Now the SeO₄^2- has a negative (-2) charge, so we have to add two more electrons.

So the total valence electrons = 6 + 24 + 2 = 32

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

• Second, find the total electron pairs

We have a total of 32 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 = 32 ÷ 2 = 16

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since selenium is less electronegative than oxygen, assume that the central atom is selenium.

Therefore, place selenium in the center and oxygens on either side.

• And finally, draw the rough sketch

#2 Next, indicate lone pairs on the atoms

Here, we have a total of 16 electron pairs. And four Se — O bonds are already marked. So we have to only mark the remaining twelve electron pairs as lone pairs on the sketch.

Also remember that selenium is a period 4 element, so it can keep more than 8 electrons in its last shell. And oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are oxygens.

So for each oxygen, there are three lone pairs, and for selenium, there is zero lone pair because all twelve electron pairs are over.

Mark the lone pairs on the sketch as follows:

#3 Indicate formal charges on the atoms, if necessary

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

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

For selenium atom, formal charge = 6 – 0 – ½ (8) = +2

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

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

The above structure is not a stable Lewis structure because both selenium and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

#4 Minimize formal charges by converting lone pairs of the atoms

Convert a lone pair of the oxygen atom to make a new Se — O bond with the selenium atom as follows:

#5 Repeat step 4 (minimize charges again)

Since there are charges on selenium and oxygen atoms, again convert a lone pair of the oxygen atom to make a new Se — O bond with the selenium atom as follows:

In the above structure, you can see that the central atom (selenium) forms an octet. And the outside atoms (oxygens) also form an octet. Hence, the octet rule is satisfied.

Now there is still a negative (-1) charge on the two oxygen atoms.

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.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the most stable Lewis structure of SeO₄^2-.

And since the SeO₄^2- has a negative (-2) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: BrCl₄^– Lewis structure

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External links

• https://homework.study.com/explanation/draw-the-lewis-structure-for-seo42.html
• https://www.chegg.com/homework-help/questions-and-answers/choose-best-lewis-structure-seo42–q1725231
• https://www.numerade.com/ask/question/part-a-draw-a-lewis-diagram-for-seo42-in-which-all-atoms-have-a-formal-charge-of-zero-part-b-draw-a-lewis-structure-for-seo42-in-which-the-octet-rule-is-satisfied-on-all-atoms-and-show-all-n-96122/
• https://geometryofmolecules.com/so42-lewis-structure-hybridization-bond-angle-and-molecular-geometry/
• https://oneclass.com/homework-help/chemistry/7032595-seo4-2-lewis-structure.en.html
• https://www.answers.com/chemistry/What_is_the_shape_of_SeO42