6.4 quantum numbers and atomic orbitals / 6.5 electron config

Question 1

wave functions = different shapes of orbitals

Question 2

Any orbital in the universe can hold a max of 2 electrons
* you can find orbitals empty, you can find them half ocupied with just one electron, or you can find them full being ocupied w/ 2 electrons

Question 3

Where these shapes come from:
* mathmatical 3D equations that we call wave functions
* so the pictures below, think of as 3D not 2D

P orbitals are dumbell shaped (3D)

D orbitals are either going to be 4 leaf clover shaped (3D) or torodial shaped (3D)

Theres a single S orbital (meaning it can only hold 2 electrons)

There are 3 P orbitals

And there are 5 D orbitals
* 4 out of the 5 look like a 4 leaf clover - I didnt draw them all below but they’re shown on the next slide. Its the filled in color thats each peice of the 4 leaf clover and when you combine them all you get the clover - so whats different about them is how they’re oriented on that 4 leaf clover in space
* dxy is in the xy plane
* dxz is in the xz plane
* dyz is in the yz plane
* dx^2-y^2 is also in the xy plane, where dxy is inbetween the x and y axis this one is right on the axis
* this is shown much more clearly below, make sure to look at the albeled axis’s because the change
* and then theres the 1 torodial
* dz^2 is the lobes are on the big lobes are on the z axis and the circular portion is in the xy plane
* remember, these are all individual orbitals and you can hold a max of 2 electrons in each orbital (for instance dxy is 1 orbital that can hold 2 electrons)

There are 7 F orbitals

Answer 3

Question 4

better look at d orbitals

Question 5

as you go up the orbitals, like from 1s –> 2s they orbital itself gets bigger, but it gets a node in it (hole in the middle where you dont find electrons)

This getting bigger actually means they’re higher energy (because remember, the further away from the nucleus the more portnaial energy it has in it) and were thinking of these like orbitals where the electron could be anywhere in them, but sicne its bigger than means its like further away = more energy

Reason further orbitals = more E:
* The electron is farther from the nucleus on average –> less attracted –> less tightly bound
* Since its less tightly bound, it has more potential energy (its less negative)
* The total energy (rememberL total = potential + kinetic) becomes less negative which = higher energy
* the 2s is wider than 1s orbital meaning the electron is further form the nucleus

what bhor thought was the electrons circled the nucleus in 2D circles, whats really true that its 3D orbitals that electrons actually live in.

Question 6

So this is a super important concept from the lession prior

We find that electrons in the closest orbital have the least energy or the most negative energy. As the deviate to the more peripheral orbitals they get less negative energy (but not 0) <— 0 energy would be more than these electrons have

In atomic physics, we define zero energy as a state where:
* The electron is completely free - its infiently far from the nucleus and is no longer feeling any attraction to the nucleus - this is why when we divide by infinity in the equation from last lession it yields 0 energy, which is more than when its in any of the oribtals
* so anything thats less than infinity (i.e., bound to the nucleus) has negative energy because you’d have to add energy to get it to 0 (think a photon hitting it and sending it out of its orbital and away from the nucleus)

So the key is that you’d have to add e to get it out of the orbitals, and you’d need more E to detach it when its in the closer orbitals, making the electrons in those orbitals have more negative E (takes more E to detach them)

so its getting to 0 E

Question 7

Quantum Numbers. What are their names and what are they

Answer 7

n = Principal
l = Azimuthal
m sub l = magnetic
m sub s = spin

Question 8

these n numbers are the same as the ones from last chapter. Telling you basically which orbit they’re in (w/ 1 being the closest)

Lowest shell = shell 1… infinity
* and just like w/ bhors orbits it goes all the way to infinity
* However, w/ normal atoms we don’t really see anything passed the 7th shell, however, we can verify that the different shells do exist

sub shell can be s,p,d,f
* describes where an electron is found in an atom
* So if i said l = 0, i should know that that electron is in an s orbital somewhere
* 0 = s
* 1 = p
* 2 = d
* 3 = f

For instance when l = 1, its in the p orbital, meaning its in 1 of the 3 different p orbitals (doesnt specify which one)

Question 9

Polyexclusion principles: excludes electrons from having the exact same 4 quantum #’s
* no electron in an atom can be identical –> this is like their individual finger print
* No 2 electrons can live in the same orbital and have the same spin
* Think of these #’s like the electrons address. Each one is specific to each electron

Question 10

Lowest energy state = 1s (because thats the one thats closest to the nucleus, meaning its most attracted to the nucelus, meaning it would take the most energy to detach it from the atom entirely and send it out into space)

Each orbital can hold 2 electrons w/ different spin direction (represented w/ up or down arrows)

based on where the electron is found (its actual orbital), we can assign different values to n, l, m sub l, and m sub s

Question 11

Name the quantum #’s of the picture below

Answer 11

Question 12

Name the quantum #’s of the picture below

Answer 12

notice, as we go up were increasing in energy (getting further from the nucleus = needs less E to detach = we have electrons w/ a less negative energy = more energy)
* basically saying 1s has less energy than 2s, and 2s has less energy than 2d

As shown below p is higher energy than s, d is higher energy than p, and f is higher energy than d.
* the only exception is atoms w/ 1 electron (hydrogen), where all the subshells have an = amount of E. meaning s = p = d = f energy levels

Question 13

name the energy levels of the subshells from least E to most

Answer 13

s –> p –> d –> f

obvisouly have to be at the same levels

Question 14

the reason n takes on values from 1 –> infinity is the orbital its in, and they technically go up to infiinity, w/ infinity being the most E at 0 (because it takes no energy to detach it)

Question 15

remember L goes from 0 –> (n-1)

Question 16

remember, m sub l does from a value of -l to +l

Question 17

Question 18

Answer 18

Question 19

Answer 19

Question 20

Answer 20

Question 21

Answer 21

3

Question 22

Answer 22

Question 23

Every box represents an orbital, and every orbital can hold at most 2 electrons w/ different spins

Question 24

Aufbau Principle - Electrons fill orbitals in order of increasing energy
* meaning they fill the lowest energy orbitals first (like 1s)
* Meaning the orbitals that would take the most energy to detach them (so most negtaive E or closest to the nucleus, or least potential E orbitals are filled first - also called the most stable orbitals = least E orbitals)

Doesnt matter if you do spin up or spin down first, as long as its filling the lowest E orbital first
* however, they need to have opposite spins so they dont have the same 4 quantum #’s (polyexclusion principle)

Question 25

What does degenerate mean?

Answer 25

Equal in energy

Question 26

So the subshells are are different energies (s, p, d, f) even when they're in the same shell (meaning they have the same principle #) * So 3s is at a slightly different energy than 3p * and s has the lowest energy followed by p --> d --> f, meaning s is the most stable **However they all have the same energy equivilence if its a 1 electron system, most notably hydrogen** * meaning that in hydrogen 3s = 3p = 3d * All the subshells have the same energy level * However, if they're in different shells they have different energy states still, like 5s is not equal to 1s in a 1 electron system (think hydrogen) 2s and 2p would be **degenerate**, meaning they're equal in energy. Because they're in the same shell, but have different subshells * however, this would not be true in a multi electron system (so think any other element) - this is depected below * so this means in a normal system it would fill from lowest energy to highest energy 4s --> 4p --> 4d --> 4f --> 5s etc...

Question 27

if you're filling the 3p orbitals, they would be considered degenerate orbitals (because they're all the same shell and subshell) **it turns out when you're filling degenerate orbitals, every square gets 1 electron before we do any pairing (meaning all the spin ups fill first or all the spin downs fill first, its not that we fill 1 w/ a spin up and a spin down and move on to the next** * so you wouldnt want to pair 2 untill the rest of the degenerate orbitals are filled because you'll create "pairing energy". Because those 2 electrons repell eachother and it would cost energy putting them in the same orbital * We also have to fill them in ordr of all spin up or all spin down for the entire subshell group (in the case below its 2p) - when the spins allign it cost energy. Every degenerate orbital gets 1 electron w/ the same spin before we start pairing They will fill as dipicted below

Question 28

The orbitals dont fill in the order you would expect (for instance 4s fills before 3d, which fucking sucks) this is one way to remember, it, you can also remember it w/ the periodic table (better way)

Question 29

So the atomic # refers to the # of protons in the nucleus, and if they're elements on the periodic table, those protons are balanced w/ the same # of electrons. So that atomic #, corresponds to the # of electrons for neutral atoms. So the two electrons that are in the s orbitals (1-6), correspond to the first two groups of the periodic table * these are the s block for p, the max they can hold is 6 electrons and that corresponds to the 6 elements on the nonmetal side (groups 13-18, not counting He) * this group of elements is refered to as the p block * Notice theres no such thing as the 1p's the d's correspond to the transition metals right in the middle of the table * remember the d's can hold 10 electrons, which makes since because this is groups 3-12 (10 total) * notice, these are always 1 row behind where you think they would be. For instance it goes 4s --> 3d --> 4p. and this has to do w/ the odd filling order shown above * So you fill the 4s before you fill the 3d, then fill the 4p lanthanides and actindes at the bottom, have 14 spaces to fit the 7f orbitals * 2 rows behind where you would think they would be

Question 30

Hydrogen is classfied as 1s^1 * that 1 means theres only 1 electron in the 1s orbital Helium is classfied as 1s^2 * that 2 means theres 2 electrons in that 1s orbital w/ opposite spins Li is classfied as 1s^2, 2s^1 (because its still in the s block) Be = 1s^2, 2s^2 B = 1s^2, 2s^2, 2p^1 * notice it falls in the p block * So this has an unpaired electron in the 2p subshell. This subshell can hold 6 electrons and only has 1 in it. To find unpaired electrons you only need to look at the 2p subshell, not any of the other ones. C = 1s^2, 2s^2, 2p^2 * **a common question is how many unpaired electrons does carbon have in the ground state** * So we need to look at the 2p^2 part. * We can see that there are 2 unpaired electrons because they dont fill in in pairs carbon ended w/ 2p^2 at the end of its configeration (because its in the p block. You can actually just go straight down and see that Si would end w/ 3p^2, and Ge would end w/ 4p^2. etc... * so as long as they're in the same block you can follow this rule (because theres 6 total in each p orbital, and 6 in the row)

Answer 30

Question 31

Standard Electron configuration for sodium * how many unpaired? * Write it in noble gas configuration

Answer 31

Standard electron configuration 1s^2, 2s^2, 2p^6, 3s^1 1 unpaired electron (looking at 3s^1, the s orbital holds 2 electrons and it only has 1 full) notice we can also follow our rule to predict 3s^1 * We know hydrogen is 1s^1, straight down to Li is 2s^1, which means Na must be 3s^1 * Makes since because these are the s block and they only hold 2 electrons * This is a great way to eliminate wrong answers on a multiple choice test Noble gas config: [Ne] 3s^1 * notice Ne matches the first part of our electron configuration above (1s^2, 2s^2, 2p^6) --> so it basically just saves us the time of writing it out like this. * We can also predict that final number in the element were looking for then just write the noble gas to save us loads of time

Question 32

Write the standard electron configuration of Ti * write the noble gas config of Ti * How many unpaired electrons

Answer 32

Standard = 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, **3**d^2 * notice we fill the 4s orbital before the 3d orbital. Have to memorize this or look at the 3 above Sc to know thats the third shell still Noble gas config: [Ar] 4s^2, 3d^2 Unpaired electrons = 2. Theres 2 electrons in that d orbital that dont have a match (remember this orbital can hold 10 total, but it fills them one at a time, not pairing them up to start) NOTICE: noble gases have completely full orbitals, which is why they're so stable

Question 33

5 exceptions What you think Cu would be: [Ar] 4s^2, 3d^9 * this is wrong. * copper steals and electron from the s and puts it in the d orbital * [Ar] 4s^1, 3d^10 <--- ground state electron config * **an atom is more stable if it completely fills a subshell, or halfway fills up a subshell** * So in this case it did both Ag/Au follow the same rules as Cu, but just 1 row further down * so they're s1 and d10, instead of s2 and d9 Cr/Mo do something similar * Cr: [Ar] 4s^2, 3d^4 is what we think it would be, however its 4s^1, 3d^5. Gets them both half full which lowers the energy and puts it in its ground state

Question 34

Mg = [Ne]3s^2 * this would have a +2 charge * So what is Mg^2+ electron configuration going to look like? (this is an example of a cation) * We know its missing 2 electrons so they cant be in their shells * **[Ne] is all it would look like. It would have the same electron configuration as the nobel gas (which makes since because nobles gases are very stable, and fill their electron shell entirely). Thats why it forms the +2 charge (so it gets back to the energy state of that stable nobel gas fulling filling its shell)** * NOTE: when it loses electrons it loses the highest E ones. which are the ones in 3s^2 for mg * We would call this isoelectronic as neon, meaning it has the same electron configuration (but doesnt have the same proton configuration) S = [Ne]3s^2, 3p^4 * sulfur typically carries a -2 charge to be isoelectronic w/ argon (meaning it has the same # of electrons as Ar, but not the same number of protons) * S^-2 = [Ar] can also be written as [Ne]3s^2, 3p^4 * This means we added to electrons, so added them to that 3p^4 portion to fully fill the p subshell and make it 3p^6 Now lets look at reansition metals Fe = [Ar] 4s^2, 3d^6 Fe^2+ = [Ar] * so our rule is that we take away the last ones that were filled in, which in this case would be the ones in 3d. However we dont follow this rule for Fe * the 4s and 3d subshells are very close in E and when they're empty it turns putting a secone electron in the 4s would be lower E than putting it in a 3d orbital. Thats why we filling the 4s first. However, once you actaully put electrons in them they experience rupulsions w/ the other electrons and react differently. So now they switch when removing electrons. Now the 3d is lower than the 4s. So when you remove the higehst E electrons you need to remove the 4s ones before the 3d ones * Fe^2+ = [Ar]3d^6 * rule for this **when you go to remove electrons from a cation, always remove from the highest shell number first** - in this case the 4s's were in shell # 4 and the 3d's were in shell number 3. So remove from 4 first * But again were adding to the 4s first here Fe3+ = [Ar] 3d^5 * follows the same rule as above

Question 35

In atoms we break up the electrons * outer most shell = valence electrons - these are what are involved in chemical rxns. * Inner electrons = the core electrons - anything closer to the nucelus than the valence electrons Valence electrons per group * Group 1 = 1 * group 2 = 2 * group 13 = 3 * Group 14 = 4 * group 15 = 5 * group 16 = 6 * group 17 = 7 * group 18 = 8 Transition metals dont have a set # of valence To determine these w/ electron config w can basically just look at whats past the noble gas * Mg = [Ne] 3s^2, meaning it has 2 valence electrons (has 2 filled in the s orbital passed the nobel gas) * S = [Ne] 3s^2, 3P^4 = 6 valence electrons (2 in the S orbitals, 4 in the P orbitals) = 6 total. (meaning it really wants to grab onto 2, to have a full shell and match the nobel Ar full valence shell confirmation) <-- which is why it carries a -2 charge Cl: [Ne] 3s^2, 3p^5 * = 7 valence electrons Br: [Ar] 4s^2, 3d^10, 4p^5 * = 17 valence electrons * however it just has 7 valence electrons (its in the halogens group) * So this means we don't count those 10 d electrons **this rule only applys if the d orbital is entirely full w/ 10 electrons** Fe: [Ar] 4s^2, 3d^6 * **so were still filling the D's so these are still considered the outermost shell so we count it** * = 8 valence electrons

Question 36

Fe[Ar] = 4s^1, 3d^6, 6p^1 * this is the excited state * This means that this atom got w/ just the right E photon to promote the electron to transition up and put 1 of the 4s electrons into the 6p shell * So this would be the excited state, not the ground state * **The rules we've learned applied specifically to the ground state** * So this would just be one of many possibilites for the excited state of an iron atom - so this would not be the only possible exited states

Question 37

Answer 37

Aufbau Principle

Question 38

Answer 38

1s^2, 2s^2, 2p^4

Question 39

Answer 39

i litteraly just found out that it was 3d1 as the last one and that was the only 1 that had it * remember, the first period of the D's starts w/ a 3 not a 4 (because the 4's fill before the 3d)

Question 40

Answer 40

Remember this is one of our exceptions * the rule is that its more stable when an orbit is completely full or half full originally you would think it would be: [Kr] 5s^2, 4d^9 * however, that d wants to be full so it lets the s be half full The answer is [Kr] 5s^1, 4d^10

Question 41

Answer 41

so that means it gained 3 electrons to fill its valence shell and match its noble has stable orbit * so just plain [Ar] would be an answer, however, this wasnt an answer to I just mapped the last configuration of Ar which was 3p^6 and there was only 1 choice w/ this (1s^2, 2s^2, 2p^6, 3s^2, 3p^6)

Question 42

Answer 42

So that means it lost 2 electrons So answer = [Ar] because its losing electrons to match the nearest noble gas confirmation (because it has a full electron shell)

Question 43

good problem

Answer 43

[Ar]3d^2 For transition metals cations we must remember to remove the 4s electrons before the 3d as they are in a higher shell, making the ground state config whats above This is specific to transition metal cations * because you're removing electrons to get it to that 2+ charge state So for Ti you would writre it out like this: * [Ar]4s^2, 3d^2 * However 2 are remove * So for transition metal cations you start by removing the ones w/ the highest number (because these are technically the higher E electrons when removing, not filling] * So the answer is [Ar] 3d^2 <--- because we needed to remove 2 electrons to get that +2 charge

Question 44

how many valence electrons does na have?

Answer 44

Go by the number at the top. Its 1 * makes sense why it gets a +1 charge. It tries to lose that valence electron to match the noble gas, and get its octet

Question 45

How many valence electrons does Se have?

Answer 45

6 [Ar] 4s^2, 3d^10, 4p^4 only count the outer shell electrons because those are the ones that are valence electrons remember, the rule is to skip the d orbitals when they're full and count the rest * 4s^2 = 2 valence electrons * 3d^10 = none becuase its full * 4p^4 = 4 * 6 total

Question 46

Answer 46

So we count the d orbital because because its not full [Ar] 4s^2, 3d^3 thats 5

Question 47

Answer 47

Si

Question 48

Answer 48

meaning one of the normal config is knocked up to a higher state normal config = 1s^2, 2s^2, 2p^3 answer = 1s^2, 2s^2, 2p^2, 4s^1 * the third one on the 2p subshell was knocked up into a higher E state

Question 49

Answer 49

c

Question 50

Answer 50

all you have to do is look at the last bit of the confirmation because everything before that is paired 1s^2, 2s^2, 3p^3 * so p has 6 total spots and fills the first 3 first w/o pairing * Meaning it has 3 unpaired (because all 3 are filled in w/ the same spin first)

Question 51

Answer 51

Something that is paramagnetic has unpaired electrons p is the only one that has unpaired electrons

Question 52

Answer 52

2 and 3

Question 53

Answer 53

Question 54

max number of electrons that can ocupy the third shell so this is talking about anything that start w/ d * 3s, 3p, and 3d * 3s = 2, 3p = 6, 3d = 10 = **18** * can also count this off on the periodic table