Lewis structure of IF5

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Lewis structure of IF5
Lewis structure of IF5 | Image: Root Memory

The Lewis structure of IF5 contains five single bonds, with iodine in the center, and five fluorines on either side. There are three lone pairs on each fluorine atom, and one lone pair on the iodine atom.

Steps

By using the following steps, you can easily draw the Lewis structure of IF5:

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

Now IF5 has one iodine atom and five fluorine atoms.

So the total number of valence electrons = valence electrons of iodine atom + (valence electrons of fluorine atom × 5)

Therefore, the total number of valence electrons = 7 + 35 = 42

  • 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 fluorine as follows:

Electronegativity value of iodine = 2.66
Electronegativity value of fluorine = 3.98

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

So now, put iodine in the center and fluorines on either side. And draw the rough skeleton structure for the Lewis structure of IF5 something like this:

Skeleton structure for Lewis structure of IF5 | Image: Root Memory

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since iodine is surrounded by five fluorines, use ten electrons to show five chemical bonds as follows:

Five chemical bonds shown between atoms | Image: Root Memory

#3 Mark lone pairs

As calculated earlier, we have a total of 42 valence electrons. And in the above structure, we have already used ten valence electrons. Hence, thirty-two valence electrons are remaining.

Two valence electrons represent one lone pair. So thirty-two valence electrons = sixteen lone pairs.

Note that iodine is a period 5 element, so it can keep more than 8 electrons in its last shell. And fluorine is a period 2 element, so it can not 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 fluorines, so each fluorine will get three lone pairs. And the central atom (iodine) will get one lone pair.

So the Lewis structure of IF5 looks something like this:

Lone pairs marked on Lewis structure of IF5 | Image: Root Memory

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

#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 = 2
Bonding electrons = 10

Formal charge = 7 – 2 – ½ (10) = 0

  • For each fluorine 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 IF5 looks something like this:

Formal charges are calculated, and got the stable Lewis structure of IF5 | Image: Root Memory

In the above structure, you can see that the formal charges of both (iodine and fluorine) are zero. Therefore, this is the stable Lewis structure of IF5.

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

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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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