9.4 Sigma and Pi Bonds Flashcards
All covalent bonds are either sigma or pi bonds.
* All single bonds = sigma bonds
* pi bonds only show up in double/tripple bonds
When you’ve got a double or tripple bond, the first bond is always going to be a sigma. Its the other bonds that are going to be pi bonds
Double bonds = 1 sigma and 1 pi bond
Tripple bond = 1 sigma and 2 pi bonds
Sigma overlap, resulting in sigma bonding, is the end to end overlap of any kind of orbitals. It an be s orbitals, it can be p orbitals, it can be hyrbidized orbitals etc…
* H-H shown below represents a sigma bond (2 s orbitals overlapping)
* H-F orbital (shown below) shows the overlap between a s orbital and a p orbital
* Remember, all single bonds are sigma bonds - anything passed that first bond is a pi bond
Pi overlap can only involve p orbitals
* the greek letter Pi is analgous to our letter P in the alphabet. These only involve p orbitlas
* However, it has to be side to side overlap of p orbitals, not end to end (we showed end to end in the F-F, and H-F)
* This is shown on the left side of the picture below. Note it overlaps in 2 places, however, this is just considered a single pi bond because it only deals w/ 2 different p orbitals overlapping side to side
* You only get these when you have double and tripple bonds
* They might ask a question like “what orbitals are overlapping to make a pi bond - these answer is always p orbitals - this is an easy question. Sigma orbitals are the ones that often vary
With sigma bonds, if an atom is hybridized it will be using one of its hybrids.
* lets look at to the left in the molecule below. We learned that if its tetrahedral (which that first carbon is) it will be utilizing and sp^3 hybrdizied orbital). That means its got an sp^3 hybridized orbital pointing towards every single one of the atoms its single bonded to.
So for that sigma bond between that first carbon and hydrogen below its simply made of that hybrized sp^3 orbital and the 1s orbital.
The 2nd carbon to the left only has 2 electron domains. Meaning its utilizing sp hyrbiziation to bond to the first carbon and the same to bond to the third carbon
* so the sigma bond between carbon 1 and carbon 2 results from the overlap of a sp^3 orbital and an sp orbital
* NOTE: the sp, sp2, and sp3 orbital all look similar (one small end one fat end)
The below is a planar molecule. I turned it sideways so we can see the orbitals better
So, its tetrahedral, both carbons create sp^2 orbitals 120 degrees apart, which link into hydrogens 1s orbitals
to remember, were in the perpindicatular plane below. So those pi bonds (in green) really flank latteraly, not above
* and this is a double bond
* overlaps in 2 places but is a single pi bond
* remember, this is made from that 2p orbital carbons extra valence electron was in that wasnt used
* pi bonds = always p orbitals
sp3
So this is the same idea as the one above. It still has 4 electron domains, so its tetrahedral, meaning they need to be at 109.5 degrees apart to where the hydrogen is. If you used a plain p orbital it would be at the wrong angle
sp^3
This has 5 electron domains
sp^3d
This has 2 electron domains (remember, double bonds count as 1 each)
note, you always hyrbdiize the lower energy ones first
* meaning we have hyrbidized sp orbitals. We also have valence electrons in our 2p orbitals. Remember, these are involved in our pi bonds (make up those double bonds, 1 applies to each double bond so we have the main bonding sigma, and then the pi bond)
answer = sp
sp^3d^2
he didnt really explain d orbitals, but also said they werent that important for these
note: borons in the middle, which only has 3 valence electrons
So we dont have any unused p electrons
sp^2
I used all 3 of the electrons in the sp^2 hyrbidization because its trigonal planar and i need 3 for that.
None of them
1) all the F’s are pulling but they’re pulling 4 times 90 degrees apart from eachother, meaning theres no net magnitude.
* also the lone pairs don’t pull and they need to be oon the ends for this one so they’re 180 degrees apart instead of 90 degrees apart (because they repell the most)
* Also, typically if it has an unbonded lone pair its polar, however, in the case of square planar this doesnt happen (shown below)
2) All the H’s are pulling at 109.5 degrees apart at the same magnitude.
* also, if it has no lone pairs and is surrounded by the same atom it will never be polar
3) This is our second rule break.
* typially if it has lone pairs it will be polar, however in a linear geometric orientation this is not true.
* This is because those electrons are equatoral meaning the two F atoms are 180 degrees apart w/ the same magnitude and opposite direction, meaning they cancel eachother out.
*
3 only
because its tetrahedral so the Cl’s are = in magnitude but not opposite in direction so they don’t cancel out
* tetrahedral ones are a bitch
2/3
2) The Cl’s have the same magnitude, however the hydrogen offsets the whole thing, meaning it will be polar
* same thing for 3
Number 1 is offset by the lone pair, meaning the hydrogens arent offset = polar
2) also polar, the lone pairs dont pull and the hydrogens are pulling at 109.5 degrees (not 180 like indicated above, thats the trick, need to look at actual moelcular geometry, not lewis structures)
3) is also polar because those F’s are pulling at 109.5 degrees. Not the 180 indicated above.
so the answer is that all are polar
So I dont even have to draw all these out ti figure it out
Boron makes 3 bonds, meaning its orientation = trigonal planar. Since it only has 3 cl and no lone pairs it must be non polar
CF4
* Carbon makes 4 bonds, meaning there won’t be any lone pairs
* surrounded by the same kind of atom w/ no lone pairs = non polar
NF3
* Nitrogen likes to make 3 pairs, but it will have a lone pair
* That lone pair is int he trigonal pyramidal orientation, meaning all those F’s are pulling at 109.5 degrees, and do not offset eachother = polar
I did all of the above w/o drawing it, but ill draw it for simplification
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