# Logic Question; Molecular Shapes

1. Oct 20, 2013

### Qube

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

Number 17:

http://i.minus.com/jbqSc694thPeBA.jpg [Broken]

2. Relevant equations

I'm not even sure where to start.

3. The attempt at a solution

I know that tetrahedral shapes are non-polar if all of the attached elements are the same (e.g. CH4 or CCl4).

I know all the answer choices are factually correct. I'm just unsure of how any of them would justify that all the others are tetrahedral.

Well I know that the answer is 3, but why? This is the best explanation I've been able to come up with; that if CH3CL were square planar and had two lone pairs, one on top, and one below, the tetrahedron, the resultant net dipoles (circled in the second diagram I drew below) would cancel out. The first diagram I drew is just of the regular tetrahedron CH3Cl. We know that CH3Cl has to be polar but if it were square planar it wouldn't be polar anymore and this would justify the fact that all the carbon compounds in the answer choices are tetrahedral.

http://i1.minus.com/jBchuhvMGOTFh_e.jpg [Broken]

Last edited by a moderator: May 6, 2017
2. Oct 20, 2013

### Staff: Mentor

How do you know that the answer is 3?

3. Oct 20, 2013

### Qube

I have the answer I just dint have the method or justification.

4. Oct 20, 2013

### Staff: Mentor

I think that the answer is wrong and (2) is correct.

5. Oct 20, 2013

### Qube

Wait why

6. Oct 20, 2013

### Staff: Mentor

There are two parts to each answer: "only one compound is known" and polar/nonpolar. Of all the molecules listed, there is only one that is singular only in a tetrahedral configuration, and for which the polarity is only unambiguous in a tetrahedral configuration.

7. Dec 3, 2013

### Qube

I stand corrected. I was making the question way too complicated with the discussion of lone pairs in a square planar arrangement. For this question these lone pairs can be ignored as they always cancel each other out.

I drew this diagram. Basically, this is a proof through contradiction question. We are given a series of statements, all factually correct. We just have to find which statement is incongruent with the proposition that the carbon molecule is square planar. We can also group the molecules into pairs since it makes no difference in terms of molecular geometry whether we're talking about CF4 or CH4 and CH2Cl2 or CH2Br2, since the both molecules in the former pair are tetrahedral and are non-polar and both molecules in the latter pair are tetrahedral and polar.

The resulting properties, regardless of the attachments, are all the same; in this question we are only discussing the resultant properties - molecular geometry and molecular polarity.

To this end I group CH4 and CCl4 together. Both are non-polar as their answer choices state, and are unambiguously non-polar when tetrahedral AND when square planar. When the attachments are all identical on a tetrahedral molecule, the dipoles all cancel out. When square planar, the dipoles similarly cancel out as the attachments are all directly across from each other (imagine tugging on a square quilt at each of its four corners ... you get nowhere).

I also group CH3Cl and CHCl3 together. These two molecules are polar, and are unambiguously so when tetrahedral. No matter what isomer, they are polar. Same with square planar. One corner of the quilt is going to be tugged more strongly than the other corners are. There is polarity no matter how you arrange the attachments.

Finally, we come to the lone (no chemistry pun) one out. As commonly seen on standardized testing, the odd answer choice out is usually the correct answer choice. As a tetrahedral molecule, CH2Cl2 is polar. All tetrahedral molecules are polar unless all four attachments are identical.

But as a square planar molecule, there is room for ambiguity. If identical attachments were placed diagonally from each other, the molecule would be non-polar (see picture). This contradicts the assertion in the answer choice that only one CH2Cl2 molecule is known and it is actually polar. Therefore choice 2 is correct, as DrClaude stated and explained earlier. Therefore, with respect to the original question, choice 2 provides the best evidence that the carbon compounds are tetrahedral and not square planar as choice 2 provides that if the carbon compounds were square planar, that would lead to ambiguities regarding the polarity of any CX2Y2 molecule - the molecular geometries would imply the presence of polar and non-polar isomers, while empirical data would show that CX2Y2 is only polar.

https://scontent-b-dfw.xx.fbcdn.net/hphotos-prn2/v/1462198_10201195942842374_1739555558_n.jpg?oh=61a9e18fcbaca46041a0cd97cdae6a51&oe=529F97B9

Last edited: Dec 3, 2013