Question

What are the expected bond angles in ICl4+ (or) ICl_4^+ ??

I think that I have the stucture drawn correcty: I is the central atom with one lone pair above it. I then have the Cl's with 6dots around them & then bonded, singly, to the "I". Two "Cl's" go straight out to the right & left of the "I" & the other two bend under. Is this correct? I know that the lone pair, that I figure is above the "I" will push down the "Cl" atoms.
Basically I think that ICl_4+ is see-saw shaped.

*****Would the correct answer, of the following, be "d" 90-degrees only???

a) 180 degrees
b) 120 degrees
c) 109.5 degrees
d) 90 degrees
e) 90, 120, and 180 degrees
f) 90 and 180 degrees

Answer 1

I tried for "d" 90-degrees only & it stated the following:

Also consider the angle between the equatorial atoms.

would this then be: "f" 90 & 180 degrees?? I don't see where 120-degrees plays in this structure & I am obviously missing another to add to the 90degrees, I think.

Answer 2

Ok...I really don't have much of a clue now. "f" was also incorrect.

So...does this problem need me to add 120-degrees in there then as well:

"e"- 90, 120, & 180...or am I seriously still way off!!!

Answer 3

The answer was "e"...I had to include the 120-degrees as well.

Answer 4

This is a trigonal bipyramid (now I tell you) so it has 120, 89, and 180.

Answer 5

It has a seesaw shape, so it is therefore 90,120,and 180 degrees

Answer 6

Jordan is correct, it has a seesaw shape.

Answer 7

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Answer 8

It's definitely see-saw shape, just like Jordan said. That means that the answer is e).

Answer 9

90,120,and 180

Answer 10

and its a seesaw shape