IF4 Lewis structure

Answer 1

First determine the total number of valence e- for the molecule. I=7, F=(7x4)=28 + the 1 e- due to the ion=36. Now for your drawing to encompass the 36 e- you would sketch it in the following manner. A single bond to each fluorine atom off of the Iodine, this totals to 32 e-, then to amount to the 36 e- total you add 2 lone pairs onto the iodine atom. This works due to Iodine possessing 7 valence e- plus the 1 e-, it was sharing 4 of its e- with F, still leaving 4 needed to complete the structure.

F

F-I-F

F

Answer 2

IF7 has a pentagonal bipyramidal shape. That means 5 I-F bonds on equitorial plane and other two occupy axial positions. you can now draw the structure according, the central atom is I and it is surrounded by 7 F molecules making a pentagonal bipyramidal shape.

Answer 3

This compound forms a cross-shape, having four fluorines at each end and iodine in the middle. Each fluorine has 6 unpaired electrons and a bond to the iodine atom respectively. The iodine atom, too, has 3 unpaired electrons in addition to its 4 bonds. This is because leftover unpaired electrons go to the central atom, iodine, and iodine just happens to be an element that can have an expanded octet. If the compound was IF4-, then 4 unpaired electrons would be on the iodine atom.
Square planar lol

Answer 4

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Answer 5

F-I-F

first fluorine has 6 v e- and a bond with Iodine.

the fluorine has 6 and a bond with iodine shown above.

and the I in the middle has 6 v e- not bonded (single dots) excluding the bonds with fluorine so total v e- are 22.

Answer 6

The Lewis structure of IF4- molecule features an iodine atom at the center with four fluorine atoms bonded around it. Iodine contributes 7 valence electrons, while each fluorine contributes 7 electrons. The structure shows that iodine has one lone pair and each fluorine atom has three lone pairs.