Lewis lab explaination PDF

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lecture noes from professor and TA's...

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If you have an old lab/quiz/exam grade that doesn't match what is posted on d2l, you have until Tuesday to bring it to my attention. Bring that item with you and I'll check to see where the problem is and update it. Grades updated for Exam 7 and the last lab (Periodic Trends in Chem Reactivity) One lab remains - the geometric structure (Lewis Lab/VSEPR/Valence Bond Sketch lab) Do not do the pre or post lab. DO work on the Lewis Structures - you should have all of the AE's, Lewis Structures, VSPER sketchs, and molecular descriptions done byTuesday. I'll cover VB sketches (valence bond sketches) and work with you on a few (including the bonding description), but you won't have to do all of them due to the limited time left in the semester. The Quiz: will be at the end of lecture on Monday so that it can include resonance and formal charge calculations. Which I will go over with you on Monday - but it would help if you had some idea of what they are in advance so you can confirm your knowledge rather than learn it on the fly. So, for the quiz, Be able to calculate AE (available electrons) Be able to draw a Lewis Structure for an ionic compound (just separate the ions and do the Lewis Structure for the monatomic ions or polyatomic ions) Be able to draw a Lewis structure to predict: 1) the correct atomic skeleton

(H-O-H-O

or H-O-O-H ?)

2) the correct arrangement of electrons around that skeleton (i.e. the location of lone pair electrnos, radicals, and bonds (single, double, and triple) 3) resonance (if there is an energetically equivalent choice for an extra bond, then draw all of the equivalelent resonance forms). e.g. ozone O-O=O O=O-O (include the lone pair electrons that I left out) This isn't an issue for CO2 O=C=O OtripleO-O O-OtripleO (where triple is a triple bond) since having the triple bond creates formal charges that destabilize it) 0 0 0 for the all double bonds one and +1 0 -1 and -1 0 +1 for the two other combinations 4) using formal charge calculations to determine if expanded octets are more likely

than strictly following the octet rule FC= the # of electrons in the free atom - # of electrons that an atom controls e.g. H:O:H

where the O has two additional lone pairs on it

to count electrons for FC, the electrons are not double counted as they are when checking the octet rule, instead, the bond is sliced exactly in half so that each atom gets one electron. so H:O one

becomes H.

.O

the H gets one electron from that bond and the O gets

A free atom of H has 1 electron (column 1) and the H in water has one bond, ergo 1 e, so it is neutral or FC= #e free atom- #e controlled = 1-1=0 Oxgyen free atom has 6 valence electrons (column 6) O in water has 1 from each bond plus it completely controlls the 4 electrons in the lone pair So it has control of six and a neutral one has 6, ergo it is zero FC = 6-6 = 0 If you look at your prelab, the first Lewis structure is of KClO4. That is ionic - so separate into two seaprate Lewis Structures: K+ o.k. - done with that (guess what the formal charge is?) and ClO4- that's trickier. 7+4*6+1 = 32 eCl has the greatest bonding capacity (it isn't 1, it's actually 7 - it has d orbitals that enable it to expand its octet) So put Cl in the middle and then a single bond to each O and an octet for each O. Check the octet rule and it seems that everything is hunky dory - but it isn't. Calculate the formal charges: Each oxygen has a FC of -1

and the Cl has a FC of +3 !!!

The most likely Lewis Structure is the one that minimizes the number of charges (i.e. the sum of the absolute values of all of the formal charges). This sum = 1+1+1+1+3 = 7 charges!!! (summing the absolute values, remember?) We can decrease the number of charges by sharing the electrons. That does mean,

however, that Cl will have to expand it's octet by using d orbitals. No problem, it can (it's in the third shell and the third shell has 3 subshells (l=0,1,2 the s the p and the d subshells) So the best/most likely one is three Cl=O and 1 Cl-O This gives the double bonded O's FC's of 0, the Cl an FC of 0, and the oxygen with a single bond a -1 FC so the sum = 1, much better than 7. But, we're not done, since we have a choice as to where to put the single bonds and double bonds. FOcus on the single bond, it could be to either the "top" O, the "left" one, the "bottom" one, or the "right" one. Draw all 4 resonance forms with arrows inbetween them and you have the correct answer.... whew... Now look at the sulfate example in the prelab. You might think sulfate would be S in the middle with a single bond to each O. That does satisfy the octet rule but since S does have the ability to expand its octet, you need to check the formal charges. Not good. The sulfur is +2 and each oxygen is -1 so the stability number (number of charges: sum of the absolute values = +6! To get the most likely form, look at the sulfur - to take it to zero, it needs six electrons. That means add two more bonds (move lone pairs from two oxygens to make two double bonds.) THe stability number drops from +6 down to +2. You can't improve it any more since sulfate has a -2 charge. There are a whole bunch of resonance forms (six, I think) and those also have to be drawn. Carbon monoxide is unusual. :CtripleO: is the Lewis structure. You will not often see carbon with a lone pair. It almost always has 4 bonds (4 single bonds, 1 double + 2 single, 2 double, or 1 triple and 1 single). Carbon has 4 valence e 's in the free atom (column 4). lone pair + 1 from each bond (3 more) FC= -1 !

In CO it has 2 from the

FC=4-5 = -1 ! !

O has 1 from each bond (3) plus 2 from the lone pair (5 total). That's one less than the free atom, hence it is +1 !! FC= 6-5= +1 This is unusual. Normally, the negative FC goes on the more electronegative atom. In the case of carbon monoxide, the only way the fc can be minimized is by going to a double bond and moving the electrons from that bond to the oxygen as a lone pair. THat gives O formal charge of 0 (2 from the bonds and 4 from the 2 lone pairs) so FC=6-6=0

BUT that was at the expense of Carbon which is now deficient of an Octet!! (only six electrons count towards its octet - that's a no-no. Although minimizing formal charge is an important factor, it won't do so at the expense of having LESS than an octet (it will, however, go BEYOND an octet (if possible - i.e. 3rd shell or higher atom in the middle) if that minmimizes the FC's...