[hawkslax7]

I'm having trouble drawing the lewis stucture of IF5. I was wondering if anyone could help.

Thanks.

[agrobert]

For 6 X^- (with 7e^- each) -->

7e^-(6) = 42e^- - 6e^- (5 F with three sets of lone pairs each) =

12e^- - 2e^-(5 I-F bonds sharing electrons) =

2e^- which accounts for lone pair on I

So

7(6) total electrons = 42

42-30 (non-bonding on F) = 12

12-10 (bonding I-F) = 2

2 (lone pair on I)

Now try drawing the Lewis structure if you understand this.

Understand that some halogens and other elements can have expanded octets and higher oxidation states.

http://en.wikipedia.org/wiki/Iodine_pentafluoride

[AWK]

http://en.wikipedia.org/wiki/PF5

[agrobert]

http://en.wikipedia.org/wiki/PF5

This is not the same

[AWK]

That's right, but we do not give here an exact solution, in this case for comparison with you good answer.
http://en.wikipedia.org/wiki/VSEPR
http://www.albion.edu/chemistry/LBL/S07%20Molecular%20Shape.ppt