Lewis structure of SBr2 - Root Memory (2024)

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Lewis structure of SBr2 - Root Memory (1)

The Lewis structure of SBr2 contains two single bonds, with sulfur in the center, and two bromines on either side. There are three lone pairs on each bromine atom, and two lone pairs on the sulfur atom.

Contents

Steps

By using the following steps, you can easily draw the Lewis structure of SBr2.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)

Let’s one by one discuss each step in detail.

#1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

  • Let’s calculate the total number of valence electrons

We know that… sulfur is a group 16 element and bromine is a group 17 element. Hence, sulfur has six valence electrons and bromine has seven valence electrons.

Now SBr2 has one sulfur atom and two bromine atoms.

So the total number of valence electrons = valence electrons of sulfur atom + (valence electrons of bromine atom × 2)

Therefore, the total number of valence electrons = 6 + 14 = 20

  • Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for sulfur and bromine as follows:

Electronegativity value of sulfur = 2.58
Electronegativity value of bromine = 2.96

Obviously, sulfur is less electronegative than bromine. Hence, assume that sulfur is the central atom.

So now, put sulfur in the center and bromines on either side. And draw the rough skeleton structure for the Lewis structure of SBr2 something like this:

Lewis structure of SBr2 - Root Memory (2)

Also read: How to draw Lewis structure of NBr3 (4 steps)

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since sulfur is surrounded by two bromines, use four electrons to show two chemical bonds as follows:

Lewis structure of SBr2 - Root Memory (3)

Also read: How to draw Lewis structure of HOCl (4 steps)

#3 Mark lone pairs

As calculated earlier, we have a total of 20 valence electrons. And in the above structure, we have already used four valence electrons. Hence, sixteen valence electrons are remaining.

Two valence electrons represent one lone pair. So sixteen valence electrons = eight lone pairs.

Note that sulfur is a period 3 element, so it can keep more than 8 electrons in its last shell. And bromine is a period 4 element, so it can keep more than 8 electrons in its last shell.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are bromines, so each bromine will get three lone pairs. And the central atom (sulfur) will get two lone pairs.

So the Lewis structure of SBr2 looks something like this:

Lewis structure of SBr2 - Root Memory (4)

In the above structure, you can see that the octet is completed on the central atom (sulfur), and also on the outside atoms. Therefore, the octet rule is satisfied.

After completing the octet, one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

Also read: How to draw Lewis structure of SeF4 (4 steps)

#4 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

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

Collect the data from the above structure and then, write it down below as follows:

  • For sulfur atom

Valence electrons = 6
Nonbonding electrons = 4
Bonding electrons = 4

Formal charge = 6 – 4 – ½ (4) = 0

  • For each bromine atom

Valence electrons = 7
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of SBr2 looks something like this:

Lewis structure of SBr2 - Root Memory (5)

In the above structure, you can see that the formal charges of both (sulfur and bromine) are zero. Therefore, this is the stable Lewis structure of SBr2.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

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Related

  • Lewis structure of NBr3
  • Lewis structure of HOCl
  • Lewis structure of SeF4
  • Lewis structure of NH2Cl
  • Lewis structure of H3PO4

External links

Lewis structure of SBr2 - Root Memory (6)

Deep

Rootmemory.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.

Lewis structure of SBr2 - Root Memory (2024)
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