XeI2 Lewis structure

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XeI₂ has one xenon atom and two iodine atoms.

In XeI₂ Lewis structure, there are two single bonds around the xenon atom, with two iodine atoms attached to it, and each atom has three lone pairs.

Steps

To properly draw the XeI₂ 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

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, xenon lies in group 18, and iodine lies in group 17.

Hence, xenon has eight valence electrons and iodine has seven valence electrons.

Since XeI₂ has one xenon atom and two iodine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of two iodine atoms = 7 × 2 = 14

And the total valence electrons = 8 + 14 = 22

Learn how to find: Chlorine valence electrons

• Second, find the total electron pairs

We have a total of 22 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 = 22 ÷ 2 = 11

• Third, determine the central atom

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

Since xenon is less electronegative than iodine, assume that the central atom is xenon.

Therefore, place xenon in the center and iodines on either side.

• And finally, draw the rough sketch

#2 Next, indicate lone pairs on the atoms

Here, we have a total of 11 electron pairs. And two Xe — I bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.

Also remember that both (xenon and iodine) are the period 5 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 iodines.

So for each atom, there are three lone pairs.

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 xenon atom, formal charge = 8 – 6 – ½ (4) = 0

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

Here, both xenon and iodine atoms do not have charges, so no need to mark the charges.

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

Therefore, this structure is the stable Lewis structure of XeI₂.

Next: IOF₅ Lewis structure

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External links

• https://www.chegg.com/homework-help/questions-and-answers/draw-best-lewis-structure-xei2-q1658571
• https://www.bartleby.com/questions-and-answers/choose-the-best-lewis-structure-for-xei2.-a-b-c-d-e/abcdfc15-ee9d-4545-a0aa-0516e3107a48
• https://www.numerade.com/ask/question/draw-a-correct-lewis-structure-for-xei2-xe-in-middle-surrounded-by-is-that-puts-a-0-formal-charge-on-all-atoms-how-many-lone-pairs-unshared-pairs-are-on-the-central-atom-2-a-03876-gram-sampl-28624/
• https://geometryofmolecules.com/xef2-lewis-structure-polarity-hybridization-and-shape/