# Lewis Structures: Shapes and Hybridization & Molecular Orbital Diagram

by heykyou
Tags: diagram, hybridization, lewis, molecular, orbital, shapes, structures
 P: 8 I don't think the problems I'm going to post are very difficult, but I'm totally clueless regarding chemistry, so they're pretty difficult for me ^^;; I also have a lot of questions to ask, so I hope you don't mind my sticking them all into one topic. I'm looking forward to your help. You don't have to answer ALL my questions, since I seem to explain them confusingly...but I'd really appreciate explanations. Thanks in advance! First off is about the Lewis Structures. You have to know the electrons needed and those available and the bonds to be formed, right? But I don't understand the significance of the number of bonds, since it doesn't seem to be followed. (At least this was how we were taught to do LS.) Anyway, while drawing the structure, how would you know which shape to follow? (I've been trying to listen to class as much as possible, but I still can't get it.) And we're also required to determine the hybridization, which is another thing I can't understand. And how do you solve something with a charge? One problem is below: 1. The problem statement, all variables and given/known data Draw the structures and determine the shapes and hybridizations of the following: a. ClO3- 2. Relevant equations no. of electrons needed: 4x8=32 no. of electrons available: Cl=7, O=6x3, total=25, +1 (because of the charge?), total=26 32-26=6 electrons, 3 bonds 3. The attempt at a solution Shape: (I have no idea...) Hybridization: Cl: (Ne) 3s2 3p5 ground state: ↓↑3s ↓↑3p ↓↑3p ↓3p hybrid state: ↓↑3sp3 ↓↑3sp3 ↓↑3sp3 ↓3sp3 Is that correct? If the hybridization is correct (hopefully), I have another question about it. For example, the hybridization for BrF5. Br: (Ar) 3d10 4s2 4p5 ground state: ↓↑4s ↓↑4p ↓↑4p ↓4p 4d 4d 4d 4d 4d Is the hybrid state: ↓↑4sp3d2 ↓4sp3d2 ↓4sp3d2 ↓4sp3d2 ↓4sp3d2 ↓4sp3d2 ...or: ↓4sp3d3 ↓4sp3d3 ↓4sp3d3 ↓4sp3d3 ↓4sp3d3 ↓4sp3d3 ↓4sp3d3 I see both being used, and I don't know how to determine which one to use. Sometimes the ns part is just left as it is, working only with the np and nd, but then sometimes all of them are hybridized or something... I hope you got what I'm blabbering about ^^;; I'd really appreciate help about SHAPES (the seesaw, etc.) because I can't understand them... Well, another question: how do you draw the molecular orbital diagram of something that has a charge? For example, O22-. I only know how to do those without charges, but would it be the same with or without the charge? Another about formal charges: For example for SO4-2. Does the structure need to be drawn first before determining the FC of S? And does S have only one formal charge? Thanks very much in advance
 PF Patron Sci Advisor Emeritus P: 11,137 I haven't read past the first problem, so let's start there. Good attempt on the structure - it's close but you need to keep the following in mind (eveything up to calculating the number of bonds is correct, except for a typo at "O=6X8"): 1. Look at the structure you've drawn and, count (i) the number of electrons provided by each atom (eg: you have 7 electrons from each O atom), and (ii) the total number of electrons around each atom, including shared electrons electrons (eg: around each O atom, there's 8 electrons). The point of this counting excercise is to make sure that the first number matches the valency of the atom (6, for O), and the second number comes out to be 8 (making each atom have a complete octet). But it's not always possible to have a structure where both these criteria are met*. However, whenever there isn't the correct number of valence electrons from a given atom, then the excess or deficit makes up the net charge on the atom. In this case, since O has a valency of 6 and you've got 7 electrons provided by each O atom, that's an excess of 1 electron for each O. That gives each O atom a net charge of 1- (and you need to show this in the diagram by writing "-" or "1-" next to each O atom). Thus the three O atoms have a total charge of 3-. What you need to do, is the same counting procedure for the Cl atom (to see if it has an octet and/or a net charge). And when you've counted the number of electrons provided by the Cl atom, make sure that the total number of electrons (those from the Cl atom + those from the 3 O atoms) provided comes out to be 26. If it doesn't, you've got too many or too few electrons and you need to add or remove some, to make the number right. Finally, make sure that the sum of all the net charges (just as a doublecheck) equals the net chrge on the ion (for ClO3-, the total charge is -1). * NOTE : It may not always be possible to ensure neutral (chargeless) octets for each atom. In that case, the following order tells you the which is the lesser of the alternative evils (topmost being the least evil, or most prefered, and bottommost being the least prefered): I: charged octets (like the O atoms above) II: neutral non-octets III: charged non-octets
 P: 8 Thanks for replying! Whoops...I corrected the typo ^^; Another try at the structure: (I know this isn't correct yet...) There should be 26 electrons, but in my structure there are only 25...and I don't know where to add the other one...to Cl? I read somewhere that those below C, O, N, F and Ne can have more than an octet or something... So if I add an electron to Cl, do I put the net charge there? Ack...something just hit me. So is this more correct... Thanks again
PF Patron
Emeritus
P: 11,137

## Lewis Structures: Shapes and Hybridization & Molecular Orbital Diagram

Good attempt on that last one! You're very close now, so let me point out 2 things to note here:

1. The Cl atom is sharing in 14 electrons - that's a good bit more than the 8 that it would ideally like to have, but clearly can't, in this case. Anything that you can do to reduce this number, without having any of the O atoms lose their octets (or making any excessive net charges) might be useful.

2. Between Cl and O, which element has a higher electronegativity? So, if a single negative charge had to live somewhere, which atom would have a stronger pull on it?

You'll see that both 1 and 2 can be effected through a single modification to your last diagram:

Do you understand what modification I've proposed? Does it make sense to you? Can you now draw the correct structure based on this?
 P: 8 If the charge is moved, do I have to move the one electron from Cl too? Sorry for asking this question, but I'm not sure about it ^^; Can an atom have an odd number of electrons--one isn't paired? And sorry again, but can you explain about the electronegativity? The higher the electronegativity, the more likely an atom is to accept molecules, right? But exactly how I know when it's more likely to accept molecules? (I've been asking my teacher, but my other classmates have been bombarding her with lots of questions already...) Also, I asked her about my tries in solving this problem. She said we didn't even need to draw a Lewis Structure, but just a structure. (The instructions were: "Draw the structures and determine the shapes and hybridizations of the following.") So what's the difference between a LS and just a structure? When drawing the non-Lewis structure, do I still have to compute for the electrons needed, available, etc.? (But I still appreciate the help in LS because I can understand it better now, and am still looking forward to any help regarding that :) )
 PF Patron Sci Advisor Emeritus P: 11,137 Actually, what I was suggesting was that it should rather be: Notice that the process of transfering one bonding electron from Cl to O simply converts the double bond into a single bond. See how this helps with both 1 and 2 above. Electronegativity is the desire to suck out the negative charge from a neighboring atom. The atom with the greater electronegativity wins. When drawing Lewis Structures, and it's possible to come up with many alternatives, always pick the one where the more electronegative atom has more negative charge on it. As for drawing "just structures" as opposed to Lewis Strctures, I have no idea what that means. You certainly can't determine anything about the shape or hybridization, unless you draw the complete Lewis Structure. Let me point out that if this is the first LS you are drawing, you've started off with a pretty difficult example. Ideally, you want to get lots of practice with simpler molecules and ions, before moving on to harder ones. PS: In both my diagrams, I've represented the charge as "-1". The correct (conventional) notation is actually, "1-", as you've done. I'm too lazy to correct the picture now.
 P: 8 My answer was really funny -_-;; Kinda stupid to not figure out what I was supposed to do... But... Wow, everything falls into place! I can't believe that's so hard...and that's just one problem. Well, I'll try solving the other LS on my own, and ask you again when something comes up. Thanks for explaining this topic so well! So, there can never be a single electron (doesn't belong in a pair) sticking to the atom? About the "just structures", I tried to look in the Internet too, but I don't think there are other ways to draw structures except for LS... I guess I'll ask my teacher what she meant when she said that... Umm, can I ask about the hybridization? (Don't mind my first post about it; I already understood that part ^^; Actually I asked my teacher earlier about most of the problems I posted here, but I still can't understand hybridization so...) When doing the hybridization on something with a charge, does the charge affect the electrons on the hybridized atom? What I think: In the example ClO3-, the hybridization of Cl won't be affected since the charge is on an O. But my question is, what if the charge is on the central atom? (Or does the charge NEVER land on the central atom, since it's most likely the least electronegative?)
 PF Patron Sci Advisor Emeritus P: 11,137 Charge does not affect the hybridization. As for the "just structures", don't bother asking your teacher - they don't exist. The only correct way to do this is by drawing complete Lewis Structures, and anyone who suggests otherwise, sadly, is only trying to avoid doing the work (of teaching it properly) themselves. I'll look at the rest of the question sometime over the weekend, if no one else has gotten to it before.
 P: 8 Thanks so much for the help! But I have another question... What about BH4-? My answer is: I'm not sure which of the two is more electronegative, but putting the negative sign on B seems more correct, since B has 3 valence electrons (and in the diagram has 4). And so, the hybridization would be: ground state: ↑↓2s ↑↓2p 2p 2p hybrid state: ↓ ↓ ↓ ↓ sp3 Meaning the charge affected the hybridization? Because if I don't add an extra electron to the hybridization, I'd only end up with sp2, which isn't correct, right? Thanks again and sorry for the trouble ^^;
PF Patron
Emeritus
P: 11,137
 Quote by heykyou Thanks so much for the help! But I have another question... What about BH4-? My answer is:
This is correct.

 I'm not sure which of the two is more electronegative, but putting the negative sign on B seems more correct, since B has 3 valence electrons (and in the diagram has 4).
Actually, H is a wee bit more electronegative than B, but in this case, that doesn't matter, because there is no way of moving that -1 charge over to the H atom (since the B atom is bonded to 4 H atoms, and hence needs a minimum of 4 bonds; this puts a minimum of 4 electrons on the B atom giving it at least a -1 net charge; so there's no way to make the B neutral). What you've drawn is the only correct way to draw this.

 And so, the hybridization would be: ground state: ↑↓2s ↑↓2p 2p 2p hybrid state: ↓ ↓ ↓ ↓ sp3
This is correct too!

 Meaning the charge affected the hybridization? Because if I don't add an extra electron to the hybridization, I'd only end up with sp2, which isn't correct, right?
Well, in a sense this is true. But remember that you can't simply take away the -1 charge and make a molecule like BH4. If you could, it might still have the same hybridization (after all you do need 4 hybrid bonds to bond to the 4 H atoms). What you can have, however, is the molecule BH3, which you will find is sp2 hybrid.

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