[butterwings]

1. The problem statement, all variables and given/known data


For the molecule [SIZE="5"]ICl3[/SIZE]

I) How many lone pairs of electrons are on the central atom?

II) How many other atoms is the central atom bonded to ?

III) What is the molecular geometry of the molecule?

IV) What is the electron-pair geometry?


a) triangular planar        b) tetrahedral

c) triangular pyramidal     d) octahedral

e) triangular bipyramidal   f) linear

g) angular                  h) square planar

i) see-saw                  j) T shaped

k) square pyramid


2. Relevant equations

No formula needed.

3. The attempt at a solution
I want to know if I am correct: or anyone will agree that
Correct Answers?
I. 2
II. 5
III. e
IV. j
Thank you!

[Schrödinger]

You are right about the lone pairs, molecular geometry and electron pair geometry.
But number of atoms...re-check the structure.

[cth]

3. The attempt at a solution
I want to know if I am correct: or anyone will agree that
[SIZE="5"][COLOR="Red"]Correct Answers?
I. 2
II. 5
III. e
IV. j
[/COLOR][/SIZE]
Thank you!

I) I agree with you. There are 2 lone pairs.

II) 5? I don't think so. It asks about "other atoms" bonded to the central one. So, I think you shouldn't count the two lone pairs. There are 3 Cl atoms bonded to the central I atom.

III) If you consider the geometry of the three Cl atoms and the two lone pairs (5 "objects" around the iodine), then it is triangular bipyramidal. The Cl atoms are on the three equatorial positions of the triangular bipyramidal geometry, where they are as far apart from one another (angle Cl-I-Cl of 120 between them). If you place one above or below the three equatorial positions, then you have some Cl-I-Cl angles of 90 degree. Try to minimise the repulsion between Cl atoms by putting them as far apart as you can. The two lone pairs are above and below the plane formed by the three chlorine atoms.

So, the four atoms are in the same plan. The iodine is central and the three chlorine atoms form a triangle geometry around it.

IV) From the previous question, you see that the lone pairs are above and below the plane of the molecule. The are linear in respect with the iodine.

[Schrödinger]

@cth:

I think your arguments for III and IV are wrong.

The magnitude of repulsions is as follows:

LP-LP > LP-BP > BP-BP

LP=Lone pair
BP = Bond Pair

So, the original answers for III and IV were right.

[butterwings]

Thank you guys but i tried,

I. 2
II. 3
III. e
IV. f

But it's incorrect. I also tried

I. 2
II. 3
III. e
IV. j

 which is incorrect as well. Please help? Am I missing some point? :[

[Schrödinger]

What do we mean when we are talking about molecular geometry?

[cth]

Thank you guys but i tried,

I. 2
II. 3
III. e
IV. f

But it's incorrect. I also tried 23ej, which is incorrect as well. Please help? Am I missing some point? :[

Could you try III. a? Everything else the same.
ICl3 is planar with a triangular geometry.

By the way, ICl3 can actually dimerise to form I₂Cl₆ which is planar http://en.wikipedia.org/wiki/ICl3. The two lone pairs on each iodine are above and below the plane.

[cth]

@cth:

I think your arguments for III and IV are wrong.

The magnitude of repulsions is as follows:

LP-LP > LP-BP > BP-BP

LP=Lone pair
BP = Bond Pair

So, the original answers for III and IV were right.

Yes, I agree about LP-LP > LP-BP > BP-BP. (LP: lone pair ; BP: bounding pair)

But, in this case you deal with Cl, not hydrogen or carbon atoms. Chlorine is bigger and larger. So, overall, you should take into account the Cl-Cl repulsion, and not only the BP-BP repulsion.

[Schrödinger]

According to your argument, the triiodide ion I₃^- mustn't be linear. Iodine is way bigger than Chlorine. It must be bent, like a water molecule.

But it is linear.

So, i think the rule is applicable to all atoms alike, without any exceptions.

[cth]

According to your argument, the triiodide ion I₃^- mustn't be linear. Iodine is way bigger than Chlorine. It must be bent, like a water molecule.

But it is linear.

So, i think the rule is applicable to all atoms alike, without any exceptions.

For I₃^-, I think my argument still holds.

In I₃^-, the central I atom has three lone pairs and two I atoms bonded to it. So, it has as well 5 "objects" around it: trigonal bipyramidal.
The iodine atoms are very large, so you should minimise the repulsion between them. The best way to do it is to put the three lone pair into the plane and the two I atoms above and below the plane. That way, the angle I-I-I is 180 degrees. It is indeed linear.

That's how I see it.

[butterwings]

Thank you guys but i tried,

I. 2
II. 3
III. e
IV. f

But it's incorrect. I also tried 23ej, which is incorrect as well. Please help? Am I missing some point? :[

Could you try III. a? Everything else the same.
ICl3 is planar with a triangular geometry.

By the way, ICl3 can actually dimerise to form I₂Cl₆ which is planar http://en.wikipedia.org/wiki/ICl3. The two lone pairs on each iodine are above and below the plane.

this answer 23af is incorrect too...

[cth]

Sorry that we couldn't help any better. 23af is what I would have tried

Once you get the answer, I would be curious to know about it.

[metrostars40]

hey im taking the same quiz for my class and i got the correct answer just figured id help out, the correct answer is "23je"

[butterwings]

thank you so much guys! yah xD i input 23je n i got it correct!!! thank you to each and everyone of you!!! luv u guys! cth ,metrostars40,Schrödinger 

[cth]

Nice one 

For III), Schrödinger was right. The LP-LP > LP-BP > BP-BP applies here. Well done.

For IV), it is obvious. It asks for "elctron-pair geometry", both lone pairs and bonding pairs. I misunderstood it and answered for lone pairs only. I feel ashamed.