Shape of Molecules

[w1ll]

Ok So I've noticed these rare questions come up on many practice tests. There always seems to be at least one of these on the entire exam. EK barely goes in to VSEPR Theory and I've already memorized all the shapes and corresponding sp3 orbitals, and how many bonds info. However, what keeps tripping me up on these questions are whether or not the molecules have a lone pair of electrons. The molecules vary a lot including, N, S, Xe, etc. I know N always has a lone pair of electrons, but S and Xe?! Is there like a list or something I'm missing for tackling these questions?

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[tncekm]

Think of what VSEPR Theory means: Valence Shell Electron Pair Repulsion Theory

Largely, atoms will try to space themselves out in a molecule... electrons will behave the same way and the lone pair electrons will be repulsed by each other due to their charges and repulsed from the electrons in the valence shells of other atoms.

So, if something has 6 constituent, two of which are electron pairs then you're going to get a square planar arrangement with 4 atomic constituents equidistant from each other and from the center atom, and two electron pairs as far away as they could possibly be from each other, on polar opposite sides.

If you take away one of these electron pairs and just assume its now a five constituent molecule, then the electron pair will push the other four further away from it due to its electrostatic interactions with their valence electrons. The net result would be a square pyramidal shape.

Get it? You can pretty much visualize the shapes without memorizing them. If you need more assistance, Wiki VSEPR.

 

[w1ll]

Yeah, I totally get the theory behind it and might need a little bit more practice visualizing the shapes. But like I said before my main problem is, I'm not sure which molecules will have lone pairs of electrons and they give random molecules as the central atom (I'm guessing other atoms' lone pairs wont have much effect on overall shape?). And these lone pairs keep throwing my answers off.

 

[werd]

things like to have full orbitals... that means 8 electrons around the atoms for most cases you'll be dealing with. exceptions include H which likes to have 2, and B which is often happy with 6. once you put your bonds in, distribute the remaining electrons around the atoms until each has a valence of 8... if there is no way to do this your structure is probably wrong. when counting, each molecular bond counts as "2" when you're counting the electrons around an atom, and also counts as "2" towards the total number of valence electrons. i was taught to talley the # of valence electrons going into a molecule (this must be preserved at the end) and then make bonds such that each atom can have 8 electrons around it...

for instance, Si has 4 valence electrons to start with. the compound SiCl4 has 4+(4x7)= 32 valence electrons going in... if each Cl single bonds to the Si, then 8 electrons are in those bonds; 32-8= 24 electrons are left to distribute. at this point the Si has 4 bonds around it, since each bond counts as "2" the Si has 4x2=8 valence electrons and is happy (ie no lone pairs around it). the remaining 24 electrons go around the Cl's, 6 going to each... therefore each chlorine has 6 unbound valence electrons and each gets "2" for its molecular bond = 8 valence electrons also.

but for H2O, there are 6+(2x1)=8 valence electrons. bonding each H to the oxygen takes 4 electrons total. both H's now have 1 bond which counts as "2" for their valence shells and they are happy. the oxygen has 2 bonds which counts as 4 for it's total. you have 4 electrons left to distribute (4 of the 8 are in bonds), and the oxygen needs 4 more to have it's 8... you therefore have two lone pairs around the oxygen.

there is often only 1 way to give each atom its full valence shell while preserving the total number of valence electrons that went in. for instance, CO2... there are 4+(2x6)=16 valence electrons going in. if you try to single bond both oxygens to the carbon you have 16-(2x2)=12 electrons left to distribute... but the C needs 4 more and each O needs 6 more to have their 8... this tallies 16 more needed. this wont work. if you double bond each O2 to the carbon you have 16-(2x4)=8 left to distribute... the C needs 0 more to have 8 valence and each oxygen needs 4 more... this situation works. these are the basics...

 

[BerkReviewTeach]

Yeah, I totally get the theory behind it and might need a little bit more practice visualizing the shapes. But like I said before my main problem is, I'm not sure which molecules will have lone pairs of electrons and they give random molecules as the central atom (I'm guessing other atoms' lone pairs wont have much effect on overall shape?). And these lone pairs keep throwing my answers off.

The easiest thing to do for main group elements is to first identify the column in which the central atom resides. Then, subtract the number of bonds from that number. Upon dividing by two, you'll be left with the number of lone pairs.

What you have done is the following:

The column number describes the nmber of valence electrons for the main group element. For Xe, it's 8.

The number of bonds corresponds to the number of valence electrons shared (assuming that the central atom donates one of the two electrons in the bond). Upon subtracting the bonding electrons from the total number of valence electrons, you are left with the nonbonding electrons.

The nonbonding electrons divided by two tells you how many lone pairs you have.

For XeF4 for instance, you'd have (8-4)/2 = 2 lone pairs. This means that the central Xe has four bonds and two lone pairs, resulting in six electron pairs around the central atom. Placing the lone pairs trans to one another (maximizing the distance between them), you are left with a square planar compound.

For SF4 for instance, you'd have (6-4)/2 = 1 lone pair. This means that the central S has four bonds and one lone pair, resulting in five electron pairs around the central atom. The geometry is trigonal bipyramidal, but the shape is not. You don't count the lone pair in the shape, although you do consider the impact of its repulsion on the position of the bonds. Placing the lone pair axial, you are left with a trigonal pyramidal type structure for the ligands with the central S slightly blow the plane of the base.

Hopefully this helps.

 

[w1ll]

The easiest thing to do for main group elements is to first identify the column in which the central atom resides. Then, subtract the number of bonds from that number. Upon dividing by two, you'll be left with the number of lone pairs.

What you have done is the following:

The column number describes the nmber of valence electrons for the main group element. For Xe, it's 8.

The number of bonds corresponds to the number of valence electrons shared (assuming that the central atom donates one of the two electrons in the bond). Upon subtracting the bonding electrons from the total number of valence electrons, you are left with the nonbonding electrons.

The nonbonding electrons divided by two tells you how many lone pairs you have.

For XeF4 for instance, you'd have (8-4)/2 = 2 lone pairs. This means that the central Xe has four bonds and two lone pairs, resulting in six electron pairs around the central atom. Placing the lone pairs trans to one another (maximizing the distance between them), you are left with a square planar compound.

For SF4 for instance, you'd have (6-4)/2 = 1 lone pair. This means that the central S has four bonds and one lone pair, resulting in five electron pairs around the central atom. The geometry is trigonal bipyramidal, but the shape is not. You don't count the lone pair in the shape, although you do consider the impact of its repulsion on the position of the bonds. Placing the lone pair axial, you are left with a trigonal pyramidal type structure for the ligands with the central S slightly blow the plane of the base.

Hopefully this helps.

Thanks a lot, it does help. In regard to this method, what happens if your subtracted value is odd? For example how do you predict the shape of NO2? I once had a question on a practice asking us to identify the structure of N2H4. I had picked the answer with a triple bond between the N's because thats how the valence electrons would've worked out, but I found out N needs to have 3 bonds and a lone pair. Is this just a rule I need to know for N?

 

[tncekm]

However many electrons an atom is missing in its standard state is how many bonds it can make, usually. E.g. carbon has 4 valence electrons in its standard neutral state, and it wants 8, so it makes 4 bonds. Nitrogen has 5 valence electrons, so its makes 3 bonds, Oxygen has 6 valence electrons, so it makes 2 bonds, and fluorine has 7 valence electrons, so it makes 1 bond. Then for Carbon, if you've got 4 bonds, that's zero lone pair because its sharing each of its 4 electrons to gain 4 more. Nitrogen shares 3 electrons, so makes three bonds and has one lone pair. Oxygen shares 2 electrons to gain 2, so has 2 lone pair, and fluorine shares 1 electron to gain 1 and has 3 lone pair.

Hope that answers your question.

 

[unsung]

Ok So I've noticed these rare questions come up on many practice tests. There always seems to be at least one of these on the entire exam. EK barely goes in to VSEPR Theory and I've already memorized all the shapes and corresponding sp3 orbitals, and how many bonds info. However, what keeps tripping me up on these questions are whether or not the molecules have a lone pair of electrons. The molecules vary a lot including, N, S, Xe, etc. I know N always has a lone pair of electrons, but S and Xe?! Is there like a list or something I'm missing for tackling these questions?

Look at the periodic table. I've memorized a section of it inadvertently just through using it so much (i.e. H through to F), but if they throw something like S have me, I'll still look just to double check it's in the same group as oxygen, and therefore has the same number of valence electrons.

The point is not to memorize a bunch of facts about which atoms have how many lone pairs. Just look at how many valence electrons it should have, based on the column it's in, and always draw out your Lewis structures based on that.

 

[unsung]

Thanks a lot, it does help. In regard to this method, what happens if your subtracted value is odd? For example how do you predict the shape of NO2? I once had a question on a practice asking us to identify the structure of N2H4. I had picked the answer with a triple bond between the N's because thats how the valence electrons would've worked out, but I found out N needs to have 3 bonds and a lone pair. Is this just a rule I need to know for N?

For N2H4, first think to yourself, how many valence electrons does N have? I always think back carbon has 4, nitrogen is to the right of it, so it's got 5 (and oxygen has 6, fluorine has 7, and so on).

We know H can only really form one bond with anything.
So, basically we're just going to have N-N bonding and N-H bonding (no H-H bonding). Okay, that was a no-brainer, but that's just the thought process you want to go through.

Okay, so next Q is- how many Hs are bonded to each N?

Consider H3N-NH. Is this plausible? The N with 3 Hs attached is going to carry a + charge. An easy way to figure that out really quickly is to see how many sigma bonds (i.e. single bonds) it has, then count 1 electron from each bond. So, that N has 4 electrons. It usually likes to have 5 electrons to be neutral (here's where the periodic table comes in, if you're ever unsure). Take 5-4=1, so we see N has a +1 charge.

The other N with only 1 H attached is dreadfully unhappy. We know N2H4 is a neutral compound, but one of the Ns as we just discussed is +1, so we know the other N is carrying a -1. But to figure it out from figuring out the formal charge, this N has one single bond to H, one single bond to N, and importantly, two lone pairs. So to figure out the charge on it, we count one electron from each single bond, and we count lone pairs as 2 electrons (which is exactly what a lone pair is). Therefore, we get 6 electrons" belonging to that nitrogen, and we know nitrogen likes to have 5, so that means it's got an overall charge of 5-6=-1.

This is a pretty sh~tty resonance structure. It's got separation of charge, it's unstable, it's not going to be a major resonance contributor.

Skipping forwards a bunch of steps. Consider H2N-NH2. This is hydrazine! It makes sense structurally as each nitrogen "owns" 5 electrons (each nitrogen has 3 single bonds and 1 lone pair, so we count 1 electron from each single bond, and we count the 2 electrons of the lone pair). This is exactly what nitrogen wants, so it's happy and neutral and stable.

Basically, you just draw out possibilities, then try them out. But first of all, make sure they are *valid* resonance structures!

For ex, the two nitrogens can't possibly be triple bonded together, and still have a molecular formula of N2H4. This would require each N to have 5 bonds: 3 as part of a triple bond to the other N, and 2 bonded to H. Remember the Octet Rule: we want each atom to have 8 electrons total for all its bonds. (Exceptions are big atoms like sulfur that can do the expanded octet thing).

Anyway, I just realized that was a way too long explanation and really this is all just about knowing how to draw Lewis dot structures, understanding formal charge, etc. If you can draw correct structures, then it's just a matter of deciding which of the *valid* structures is most stable. But if you haven't gotten to the point where you're drawing *valid* structures even, then it's really not a matter of memorizing "facts" like "nitrogen has 3 bonds" or whatever. Make sure you can eliminate structures that are unfeasible.

 

[w1ll]

Thanks a lot for the long but useful help =)

BTW can any of you tell me why Al is a p block element? And how you knew/figured this out?

NVM this info wasn't in my EK books but it was in my TPR. I got my TPR book recently and now I wished I used both earlier =/ (April 19th exam)

I get 10's and 9's in PS and BS and 11-9 in VR but never broke 30 on AAMC's wth!