AsI3 Lewis structure

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AsI₃ (arsenic triiodide) has one arsenic atom and three iodine atoms.

In AsI₃ Lewis structure, there are three single bonds around the arsenic atom, with three iodine atoms attached to it. Each iodine atom has three lone pairs, and the arsenic atom has one lone pair.

Steps

To properly draw the AsI₃ 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, arsenic lies in group 15, and iodine lies in group 17.

Hence, arsenic has five valence electrons and iodine has seven valence electrons.

Since AsI₃ has one arsenic atom and three iodine atoms, so…

Valence electrons of one arsenic atom = 5 × 1 = 5
Valence electrons of three iodine atoms = 7 × 3 = 21

And the total valence electrons = 5 + 21 = 26

Learn how to find: Arsenic 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 arsenic is less electronegative than iodine, assume that the central atom is arsenic.

Therefore, place arsenic 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 13 electron pairs. And three As — I bonds are already marked. So we have to only mark the remaining ten electron pairs as lone pairs on the sketch.

Also remember that arsenic is a period 4 element, so it can keep more than 8 electrons in its last shell. And iodine is a period 5 element, so it can keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are iodines.

So for each iodine, there are three lone pairs, and for arsenic, 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 arsenic atom, formal charge = 5 – 2 – ½ (6) = 0

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

Here, both arsenic 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 (arsenic) 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 AsI₃.

Next: SbF₆^– Lewis structure

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External links

• https://oneclass.com/homework-help/chemistry/6909957-lewis-dot-structure-of-arsenic.en.html
• https://www.chegg.com/homework-help/two-iodides-arsenic-z-33-asi3-asi5-lewis-diagram-conforms-oc-chapter-12-problem-59mcp-solution-9781133114710-exc
• https://www.answers.com/chemistry/How_do_you_draw_a_Lewis_dot_structure_of_AsI3