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.
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
10 answers
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!!!
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!!!
The answer was "e"...I had to include the 120-degrees as well.
This is a trigonal bipyramid (now I tell you) so it has 120, 89, and 180.
It has a seesaw shape, so it is therefore 90,120,and 180 degrees
Jordan is correct, it has a seesaw shape.
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It's definitely see-saw shape, just like Jordan said. That means that the answer is e).
90,120,and 180
and its a seesaw shape