[Daisie]

Reviewing for my AP Chem final, I came upon an online question that gave me pause.

How many of the following molecules can have their structures drawn within a plane?
CH4
H2O
CO2
BF3

The correct answer is 3, but I don't have any more information. My guess is that all but H2O can be drawn within a plane because they have no lone pairs. Is that right?

Thanks so much.

[sjb]

Reviewing for my AP Chem final, I came upon an online question that gave me pause.

How many of the following molecules can have their structures drawn within a plane?
CH4
H2O
CO2
BF3

The correct answer is 3, but I don't have any more information. My guess is that all but H2O can be drawn within a plane because they have no lone pairs. Is that right?

Thanks so much.

 No, this is not correct. I don't know what level AP Chem final is, but do you know the shape of the molecules? It may be useful to know some basic tenets of geometry, too.

[Borek]

It is not about lone pairs (these can stick out), it is about localization of atoms.

[Natalia]

Hello!

I believe you are wrong... All choices but CH₄ are planar, in fact.

This problem can be solved more or less by the VSEPR theory, which consists of some basic rules:
1) The shape of a molecule is determined by the number of electron densities around the central atom. (By electron density we mean any of the following: lone pair, single electron, single bond, double bond, triple bond etc.)
2) Electron densities will arrange themselves in space around the central atom with the greatest angle possible (because they repel each other!). So every time you want to figure out a structure, you want to imagine how the molecule should be so that electron pairs (etc.) are as far from each other as possible.
3) The biggest repulsion occurs from non bonding electron pairs. A little weaker is the repulsion of bonding ones and the repulsion of single electrons is the weakest (of the three). So, when we have e.g. a tetrahedral electron arrangement, with all of the pairs bonded, all angles of the molecule will be equal (to ~109.5 deg.). But if two pairs are bonding and two are not, the repulsion is stronger from the side of the non-bonding ones, so, molecule's angles will be unequal.

If you already knew about that and I misunderstood your question, I'm sorry for posting it again.
Back to your question now.

I think you considered CH₄ to have a "cross-like" planar structure. But CH4 is tetrahedral, because you have to consider an arrangement in space.
According to the VSEPR theory...
In CH₄ there are 4 pairs, all bonding, which, due to repulsion, need to be arranged with the greatest angle possible - so the electron arrangement, as well as its structure, is tetrahedral, obviously not something that can be drawn within one plane.

H₂O has four electron pairs around it, two of which are bonding with hydrogen atoms, and the other two being lone. The electron arrangement is tetrahedral (so, not planar) but - be careful now- the shape of the actual molecule is "V" (planar) because it consists of just three atoms. Three atoms can never form something that is NOT planar, think about it!

The remaining two, CO2 and BF3 are indeed planar, but not because they have no lone pairs, but due to the VSEPR theory rules...
http://en.wikipedia.org/wiki/VSEPR_theory Check out this article and see the table with the shapes according to the number of bonding and non-bonding electron pairs.
Now a quiz... Can you try applying these rules for BF₃ and CO₂ ?

If there's something you don't understand or I haven't explained correctly, tell me!

[Twigg]

Remember the O in H2O has four electron groups and two lone pairs, so VESPR requires a bent geometry.
CH4 cannot be planar because there are four electron groups, zero lone pairs, and the farthest distance you can get between electron groups is a tetrahedron, not in a plane.
CO2 is linear: two bonding groups and no lone pairs.
BF3 is the only tricky one here. I don't know what the empirical evidence suggests, but VESPR says (and the AP test agrees) that it's trigonal planar.

@Borek I don't think an AP test would be so cruel to pull a trick molecular orbital model here. That would be plain cruel. There's no way to know if you even got the right answer in that case (unless you've got one of those modeling computers in your pocket). 

[Natalia]

BF3 is the only tricky one here. I don't know what the empirical evidence suggests, but VESPR says (and the AP test agrees) that it's trigonal planar.

BF3 is not tricky either. Its configuration is 1s2 2s2 2p1 (or simply (2.3) and it only has 3 valence electrons. the farthest angle is 120 deg. and it has to be planar. Try it with a set of toy-magnets to see

Hope I helped,
Nat