### Drawing the Lewis Structure for ICl4-

Viewing Notes:

• In the Cl4- Lewis structure Iodine (I) is the least electronegative atom and goes in the center of the Lewis structure.
• The ICl4- Lewis structure you'll need to put more than eight valence electrons on the Iodine atom.
• In the Lewis structure for Cl4- there are a total of 36 valence electrons.

Transcript: Let's do the ICl4- Lewis structure. Iodine, 7 valence electrons; Chlorine, 7 as well, we have four Chlorines; and then we have this up here so we're going to add an additional valence electron for a total of 36 valence electrons. Iodine's the least electronegative, we'll put that at the center, and then we'll put four Chlorines around it. We'll form chemical bonds between the I and the Chlorines. So we've used 2, 4, 6, 8 valence electrons. And then we'll complete the octets around the outer atoms. So we've used 2, 4, 6, 8, 10, 12, and 32.

So we've completed the octets on the Chlorines, but we still have four valence electrons left. Since Iodine is in period 5 of the periodic table, it can hold more than eight valence electrons. So we're going to put those remaining four valence electrons on the Iodine. Put a pair here and then a pair over here. So we've used all of the valence electrons up. And if we check our formal charges, you'll find that the Iodine has a formal charge of negative one, and the Chlorines all have formal charges of zero. That makes sense: negative one, we have a negative one up here. So that makes this the best structure for ICl4-. One last thing: we need to put brackets around the structure to show that it is an ion and it has a negative charge.

And that's the Lewis structure for ICL4-. This is Dr. B., and thanks for watching.

 Search our 100+ Lewis Structures   (Opens New Window) See the Big List of Lewis Structures Frequently Tested Lewis Structures Basic CH4,   NH3,   C2H4,   O2,   N2 Intermediate O3,   BBr3,   I3-,   BrF5,   NO Advanced SO3,   H2SO4,   OCN-,   XeO3,   ClO4-