# SeOBr2 lewis structure

SeOBr2 (selenium oxybromide) has one selenium atom, one oxygen atom, and two bromine atoms.

In SeOBr2 lewis structure, there is one double bond and two single bonds around the selenium atom, with one oxygen atom and two bromine atoms attached to it. Each bromine atom has three lone pairs, the oxygen atom has two lone pairs, and the selenium atom has one lone pair.

Contents

## Steps

To properly draw the SeOBr2 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, and bromine lies in group 17.

Hence, both selenium and oxygen have six valence electrons, and bromine has seven valence electrons.

Since SeOBr2 has one selenium atom, one oxygen atom, and two bromine atoms, so…

Valence electrons of one selenium atom = 6 × 1 = 6
Valence electrons of one oxygen atom = 6 × 1 = 6
Valence electrons of two bromine atoms = 7 × 2 = 14

And the total valence electrons = 6 + 6 + 14 = 26

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

• Second, find the total electron pairs

We have a total of 26 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 = 26 ÷ 2 = 13

• Third, determine the central atom

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

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

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

• And finally, draw the rough sketch

### #2 Next, indicate lone pairs on the atoms

Here, we have a total of 13 electron pairs. And three bonds are already marked. So we have to only mark the remaining ten electron pairs as lone pairs on the sketch.

Also remember that oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell. And both (selenium and bromine) are the period 4 elements, so they can keep more than 8 electrons in their last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are oxygen and bromines.

So for oxygen and each bromine, there are three lone pairs, and for selenium, there is one lone pair.

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 – 2 – ½ (6) = +1

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

For each bromine atom, formal charge = 7 – 6 – ½ (2) = 0

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: