Lewis structure of IBr

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Lewis structure of IBr
Lewis structure of IBr

The Lewis structure of IBr contains a single bond between the iodine atom and bromine atom, and both the atoms have three lone pairs.

Steps

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

#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… both iodine and bromine are the group 17 elements. Hence, both iodine and bromine have seven valence electrons.

Now IBr has one iodine atom and one bromine atom.

So the total number of valence electrons = valence electrons of bromine atom + valence electrons of bromine atom

Therefore, the total number of valence electrons = 7 + 7 = 14

  • 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 iodine and bromine as follows:

Electronegativity value of iodine = 2.66
Electronegativity value of bromine = 2.96

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

So now, put iodine and bromine next to each other. And draw the rough skeleton structure for the Lewis structure of IBr something like this:

Skeleton structure for Lewis structure of IBr

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond as follows:

One chemical bond shown between atoms

#3 Mark lone pairs

As calculated earlier, we have a total of 14 valence electrons. And in the above structure, we have already used two valence electrons. Hence, twelve valence electrons are remaining.

Two valence electrons represent one lone pair. So twelve valence electrons = six lone pairs.

Note that iodine is a period 5 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 atom is bromine, so bromine will get three lone pairs. And the central atom (iodine) will also get three lone pairs.

So the Lewis structure of IBr looks something like this:

Lone pairs marked on Lewis structure of IBr

In the above structure, you can see that the octet is completed on the central atom (iodine), and also on the outside atom. 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.

#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 iodine atom

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

Formal charge = 7 – 6 – ½ (2) = 0

  • For 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 IBr looks something like this:

Formal charges are calculated, and got the stable Lewis structure of IBr

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

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

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External links

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.

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