9.5 - molecular orbital theroy Flashcards
Molecular orbital theroy
* this theory gives the best description of reality - but it sucks
* this is our best theory of bonding
remember, all our atomic orbitals are wave functions (math equations taht are solutions to the sheroters equation)
* These wave functions are 3D math equations themselves
* When you overlap these wave functions you can combine them in different ways.
Looking at the SIN function
* From 0 –> 180 it takes on positive values
* From 180 –> 360 it takes on negative values
* Right in the middle, it takes on a value of 0
* for a wave function we refer to that 0 spot and a node - this is just a place where the function has a value of 0
* So for 1s orbitals theres no node, but for pretty much all the other orbitals theres a node somewhere, like p orbitals always have a node at the nucleus
* so theres a 0% chance of finding the electron at that node, because the wave function doesnt encompass that space
If we have 2 SIN functions where the positive parts are perfectly alligned and the negative parts are perfectly alligned - we refer that to being “In phase”
* they add together essentially and you get constructive overlap and the function is amplifed (shown below)
You can also have the exactly out of phase where the + and - align and they add together to = nothing
* they comepletely cancel when they’re exactly out of phase and have the same magnitude
* this is refered to as destructive interference
This is important when we start adding wave functions because the atomic orbitals are 3D mathmatetical equations in some cases have positive values for the function and in other cases they’ll take on negative values for the functions
Now looking at the 1s orbitals below
* they can take on either positive values or negative values
* for the 1s orbital it could be either one. It could be all positive or all negative
So looking below when 2 positive, when they over lap the region they overlap will become amplified, much like the SIN function from before.
* remember, these wave functions describe where an electron might be found
So below the most amplified portion is the point right between the two nuclei, where the overlap was occuring
* so that means that area is the most proable place to find an elevtron
* This is a good thing because its between the two positively charged nucli and its attracted to both of them so thats the place it wants to be most
* What that really means is thats where the electrons are going to be lowest in energy (have the most negative energy, meaning it will take the most positive energy to remove them)
* When they have the most attraction to the things they’re attracted to they have the lowest energy.
These two oribinal atomic orbitals (the 1s orbitals) have now combined to form 1 gaint molecular orbital (because its around a molecule)
* It turns out this molecular orbital is lower in energy than either of the 2 atomic orbitals
* This is because the electrons are now attracted to 2 positive nucli instead of 1
* and the most proably place they’ll be found is exactly between these two nuceli, which lowers the energy even more (more attracted to something positive = more E needed to remove them = they have more negative E = lower E)
Another rule says the larger the orbital, the lower its energy. And just looking at the combined orbital below you can see its bigger than either of the 1s orbitals
So you can see this is end to end overlap, which is sigma overlap (remember, side to side is Pi overlap), its the name Sigma 1s for the molecular orbital below
So now lets talk about if one of those 1s orbitals has the negative version of the wave function (the atomic orbital is negative)
Instead of constructive overlap and amplification, were going to get destructive overlap and it will cancel out this region right down the middle
So the picture I drew is still considered 1 molecular orbital by combining 2 atomic orbitals. It just has a massive node down the middle, subdividing it into 2 peices. But the key is that its still 1 molecular orbital
* on this node out wave function has a value of 0, meaning no electrons can possibly be in it
* and remember, thats like the actual best place an electron would want to be - perfectly in the middle of the 2 positive nucli
* so if letting the electrons be where they want to be lowers the energy (showed in the card before), than not letting the electrons be where they want to be raises the energy
* Again this makes since, now they arent as stuck in there because they arent between the 2 positive nucli, meaning it would be easier to detach them (would take less energy to detach them), meaning they’re less negative, meaning they’re in a higher energy state.
Also we had destructive overlap, meaning the total volume of the molecular orbital is smaller in this one, and smaller orbitals = higher energy
So this molecular orbital is higher in energy
called sigma 1s*
NOTE: these molecular orbital we made can also hold 2 electrons, just like the atomic orbitals above.
* any orbital you’ll come across will hold a max of 2 electrons –> whether its the atomic orbitals, or the molecule orbitals.
having the sigma 1s orbital thats lower in E than the original atomic orbitals, means that putting electrons here, is likely to hold these atoms together into a molecule, rather than them being seperate and apart
* theres an advantage to being together [its lower E] so thats what they do
However, the sigma 1s* is higher in E than the original 1s orbitals, so putting electrons in this orbital is worse than having the seperate atoms.
* so the electrons will just stay seperate (because it keeps them at a lower E) than come together when they would be assuming a 1s* orbital.
We call this 1s* orbital an antimolecular bonding orbital
When you get the overlap between atomic orbitals, it always makes a lower E bonding moleculr orbital, from constructive overlap and a higher energy anti bonding molecular orbital from destructive overlap. Both happen at the same time.
* so the total # of orbitals you have will always be unchanged. You start w/ 2 1s orbital that came together those 1s orbitals ceased to exist once we formed a molecule. It created 2 molecular orbitals (shown below) a Sigma 1s and a Sigma 1s*
In this class were only going to look at the same kinds of orbitals overlapping. However you still could have things like an s overlapping w/ a P
Combination of P orbitals
* Remember, for P orbitals theres a node right at the nucleus, so you will never find an electron at the nucleus (the function goes to 0)
* This makes since, because on one side of that P orbital it will take on positive values, while on the other side it takes on negative values
* So when you pass through that node you’re going from positive values –> negative and theres nothing at 0 (or from negative –> positive)
To represent positives and negatives we atually just shade one side
* we don’t actaully define what the shaded side means, we just need opposites
* again, remember, this is not assosicated w/ charge but the mathmatical sign of the function
In the picture below we’ve drawn it in such a way that the lobes of the orbital have the same sign (they’re over lapping)
* This is a sigma bond because they’re end to end (not the pi bond, pi bonds are P orbitals, but are side to side typically creating double and tripple bonds)
So where they’re overlapping they have the same sign, either both positive or both negative (we didnt define what the shading meant)
* This will lead to constructive overlap (will amplify what it is)
* Construcive overlap leads to a lower E molecular orbital that is a bonding molecular orbital
You can see in the molecular orbital below that the overlap region is amplified (which is why its bigger in the molecular orbital)
* you can also see the change in sign once you pass the node
* once again those nodes are present at the nucli
* When you get constructive overlap you dont create any new nodes
* you will get a region of amplification right inbetween the atoms
* called sigma 2p because we dont have a 1p
Remember, those electrons want to be between the two nuclei, and we’ve amplified their proability of being there, which is why this is considered lower E. (more stable)
Shown below is destructive
The region where they overlap is going to have opposite signs and will cancel each other out
* This will create a new molecular orbital that is anti bonding and higher moleculear E
* So the way you draw this is the same as the bonding one except you earase that peice down the middle (where the electrons would ideally be)
So you can see just the region they overlap is a node (not the region immediatley before that because thats not overlapping).
* So they completely 100% cancel each other out because they’re the same magnitude and opposite signs, leaving a node where electrons cannot be found
* This becomes a higher energy molecular orbital, because those electrons are not bound as tightly to the nuclei (arent found directly between them where they would have the most attractive forces), meaning it would take less energy to detach them. So this system is has less negative E = higher Esystem (because it takes less E to detach them) = less stable.
NOTE: w/ an antibonding orbital destructive overlap will always lead to the creation of a new node. Right between the two nuclei
This is called O2p*
So the above was showing the orientation of p orbitals coming together end to end to make a sigma bond.
However, what about a different orientation of the p orbital that comes together side to side
* These would be those pi bonds (always p orbitals)
So in the picture below those 2 p orbitals are “in phase”
* They have constructive overlap in 2 regions, amplifying it (the shaded part is overlapping and the white part is overlapping)
In the hybridized orbital there is a node. However, we didnt create that, it was there to being w/ (drawn in red on the overlap portion of the original non hyrbidized one)
* theres always a node at the nucli for p orbitals
* in the hyrbizied orbital theres amplification on either side of the nucli (because it was constructive overlap)
This is called the pi2p molecular orbital
* just like any other orbital it can hold 2 electrons (one spin up one spin down)
This is now showing the destructive overlap between two p orbitals and are forming a pi bond (are side to side instead of head on)
So that overlap region will cancel itself out, meaning electrons cannot possibly be there
* this will create a second node (not we already had one node where the nucli are. However, this is creating a second place where electrons cant possibly be)
So now we’ve create a second perpendicular node right down the middle, this is the space where the overlap was, so it cancels out)
This is the pi2p* orbital
We know when 2 hydrogen atoms come together their S orbitals are going to overlap to create a lower energy bonding orbital that we call sigma 1s
* sigma because its face to face (single bond)
* Its lower energy again because they wouldnt bond if it didnt put them in a lower energy state (meaning those electrons and bound closer to the protons in the nucli - remember, here its between the 2 nucli which is a lower energy state [harder to detach] than just near 1 nucli’s protons)
So below its showing those 2 hydrogens coming together and the two bonding orbitals that are created (1 being higher energy, and the other being lower energy)
So we show each of the 1s electrons in 2 seperate boxes before the hydrogens have bound (thats what the lateral boxes are) - they’re still in their atomic orbitals here
* The molecular orbitals are post bonding
remember, the lowest energy spots are filled first, so the signma 1s box is filled (and the sigma 1s* box remains empty)
So now you can understand why hydrogen exists as a diatomic. The after bonding picture is actaully lower in energy than the before bonding picture. So it will want to bond to other hydrogen atoms to lower its energy and put it in a more stable state
* Better said, the electrons when paired in the sigma 1s box are lower in energy than the single atomic orbitals that a single electron of hydrogen would exist in before bonding.
* again, lower energy states are spontaneous, so these hydrgoens bonding would happen automatically.
Now were going to show why He is not a diatomic (while H was as shown above)
So each of these He’s has 2 electrons in the 1s orbital (one spin up, one spin down)
So now we’ve got 4 total electrons to fill into our orbitals
* remember the two lateral don’t actually exist one the molecular orbital is made (so those dont fill)
* So now 2 electrons will be put into that sigma 1s orbital, but that means the other 2 have to go into the sigma 1s* orbital (which is higher E) - again remember, those 1s orbitals simply don’t exist anymore (these are atomic orbitals, once is comes together these are deleted and 2 molecular orbitals are synthesized)
Putting electrons in a bonding molecular orbital is equivilant to a bond
* however, putting those 2 electrons in the anti bonding molecular orbital cancels it out (its an anti bond)
* So the 2 bonding electrons are canceled out by the 2 anti bonding electrons - so its kind of like it has no bonds - which is why it doesnt exist
Bond order: helps us put a mathmatical calculation on this
* from the molecular orbitals take your # of bonding electrons - anti bonding and divide the entire thing by 2.
Above in the hydrogen example we got (2-0)/2 = 1
* a 1 is equivilent to a single bond
With the He we got 2-2/2 = 0
* so we have 0 bonds for He - meaning theres no bond there
* This means that that diatomic He doesnt exist
He2^+1
* The question is if a diatomic of helium w/ a + charge exists (we already showed that He2 didnt exist)
So this means its missing an electron
* so instead of having 4 electrons, it would only have 3
So looking at the atomic orbitals below, just earase 1 of the electrons out (again this is before they combined). This means 1 of those 2 He electrons lost an electron which is why it has a _ charge.
* So instead of having 4 electrons total it only has 3.
* Since lower E orbitals ALWAYS fill first when combined it fills the sigma 1s orbital first with 2 electrons, then fills the Sigma 1s* with 1 electron (instead of 2 like before
We find this bond order to be 0.5, which is > than 0. Its the equivilent to like half a bond.
* because its >0 He2^1+ actually has a chance of existing in nature
Homonuclear = 2 identical atoms
So the diagram for O2, F2, and Ne2 would like the below
* You’re going to get the lower energy 1s orbitals overlapping to get the lower E bonding sigma 1s and the higher E sigma 1s* anti bonding
* You’re going to get the 2s orbitals overlapping to create the sigma 2s bonding and sigma 2s anti bonding orbitals
* Then the P orbitals are going to overlap. The P sub Z’s overlap to give the lower energy sigma 2p (bonding) and the higher energy sigma 2p antibonding orbital.
* P sub x and P sub y overlap side to side to make a lower energy pi 2p molecular orbital (bonding), and pi 2p anti bonding molecular orbital (higher E)
Were going to start by mapping O2
* each oxygen looks like: 1s^2, 2s^2, 2p^4
So Remember, it fills from the lowest energy firt. So those 1s electrons are entirely full, so both those sigma 1s bonding and sigma 1s* antibonding fill
* note a lot of times we only draw out the valence and not the entire structure (like shown below) because the core ones are always going to do the same thing
The valence electrons here all are in the 2 orbital (so its the highest number in the configuration = where the valence electrons are. In this example its the electrons in 2s^2, 2p^4)
So the 2s fills entirely
* makes since, the 2s orbital is entirely full for BOTH oxygens
* Remember, were mapping 2 oxygen atoms here, not just one
However, the 2p’s for both oxygens are not entirely full
* Theres 4 electrons in each of the 2p orbitals (for each oxygen)
* so we have a total of 8 electrons to fill into the molecular orbitals.
* we fill from lowest E first
* Its degenerate meaning, everyone gets 1 electron before you double up (from the bottom up)
Bond order = 2
* so this means theres a double bond here in O2
If we’d left out the core electrons enirely we would’ve gotten same same # for bonding order
Paramagnetic = attracted in a magnetic field
Diamagnetic = weak repulsion in a magnetic field
* Electrons have a spin associated w/ them, however, when you pair up electrons their spins are opposite and they cancel entirely.
* If all the electrons are paired up, then there spins cancel, and you dont experience an attraction in a magnetic field. You get a weak reupulsion. thats what makes you diamagnetic
* So all electrons paired = diamagnetic = repulsion
According to the lewis structure of O2 it looks like O2 should be diamagnetic. however, it experiences an attraction when palced in a magnetic field (like paramagnetic), lewis doesnt explain this well.
* however molecular orbital theory explains this well.
* Look at the Pi 2p* electrons in the molecular orbital theory boxes (at the top), its showing 2 unpaired electrons
* So molecular orbital theory explains that O2 is actually paramagnetic (attracted in a magnetic field) because it has 2 unpaired electrons (they arent all paired like lewis structures would say they are)
So one of the best things about molecular orbital theory is that it allows us to explain paramagnitism and diamagnitism unlike lewis structures every could
So I’ve mapped out F2.
* each one has 1 more electron than oxygen, however combined they have 2 more electrons than oxygen. So we add 2 electrons to that anti bonding Pi 2p* box, meaning that entire box is now full. Meaning we don’t have any unpaired electrons.
Bond order = 1
* matches the lewis structure
F2 is diamagnetic because all the electrons are paired
Now lets talk about Ne2
* Ne2 does not exist - its a noble gas
* th molecular orbital theroy explains this
So the bond order yields 0, meaning no bonds are made, meaning this diatomic Ne molecule is impossible and does not happen in nature (only things over 0 have the ability to happen)
One other note
* this is 1s^2, 2s^2, 2p^6
* we only look at the ones that start w/ 2 (because this is the most external orbital) to find those valence electrons
* This means we have 8 valence electrons
* you can look at the 2s orbital and see that for both Ne’s its completely full (2 for each atom, and together theres 4 here, which we can see is true because theres 2 electrons in the hybrid sigma 2s and antibonding sigma 2s* orbital)
Moving to the 2p^6 you can see that one of those p orbitals is face to face (the z one) which is the sigma 2p box, and the sigma anti bonding 2p* box
* theres 2 boxes for pi 2p, those represent the other 2 p orbitals that exist as pi bonds (because p has 3 orbitals). And ofc theres also anti bonding to match.
For the below the major change from the last one is that the sigma 2p and the pi 2p orbitals have swapped places
So on the original diagram that 2s and 2p haev energies far apart, and they dont mix at all. However, with the current one they’re much closer together, meaning theres some meaningful mixing of orbitals. called sp mixing.
* so this is true of the elements in period 2 not mapped yet
* Li2, Be2, B2, C2, N2
NOTE: I left the 1s off because the non valence electrons dont change anything, and these are core electrons here.
Im now going to map N2
Nitrogen is shown to have a tripple bond on the lewis structure.
* notice we filled the bonding for the pi 2p, and sigma 2p. Showing a tripple bond.
* bond order of 3 = tripple bond (shown below)
We also have all paired electrons, meaning its diamagnetic (is repelled slightly by the magnetic field)
I should know how to draw the MO diagram for any one of the homonuclear diatomics. Lots of classes dont actually make you memorize hwo to draw these (im going to) but only have to fill them in.
below O2, F2, and Ne2 follow the first one shown below
Li2, Be2, B2, C2, and N2 follow the second one shown below
For this molecul all the oxygens have the formal charge. Thats fine because the more electronegative atom getting the negative formal charge is an okay thing.
answer = 4 electron domains
Be careful on these questions. You cant see how many lone pairs it has based off the formula alone, which is why I missed this
Br = less electronegative
tricky part is that they were asking for the electron goemetry not the molecular geometry. The electron geometry is trigonal bipyramidal while the electron geometry is t-shaped
Trick is that boron doesnt follow the octet rule. It only needs 6 total to pair its 3 unpaired electrons.
* its stable at 6 total (makes since, when an electron is paired its more stable (lower E state)
180 degrees
