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Chapter 4.4b Drawing Lewis Structures

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welcome to the second video for chapter 4

section 4 drawing lewis symbols and structures

structures

in this video we'll be focusing on the

step-by-step process to actually draw

lewis structures which depend the

bonding in simple molecules if you've

already got some practice with lewis structures

structures

you will probably be tempted to skip

some of these steps

but i really recommend that you focus on

developing a really solid foundation

with lewis structures

and especially if you've never done

these before or if or if you haven't

done them in a long time this is a

really great review and it's really good

time to focus on the

basics of lewis structures once you get

the hang of it and get your feet under a

little bit

you may be able to start taking some

shortcuts but until you're really really

confident with your lewis structures i

really recommend that you

focus on your foundations and um and

really work through these

this set of of simple steps

i'm going to be showing you these steps

in the context of the formate ion

which is this guy here ch02 minus

just to give you a little bit of context

i think it helps to sort of illustrate

the steps

the first step that you're going to want

to do when you sit down to draw a lewis

structure is calculate the number

of valence electrons in the molecule or ion

ion

essentially we need to know how many

electrons we're working with so that we

can figure out how to distribute them

so we'll do this in uh by looking at the

periodic table so first we'll write down

how many

uh elements or how many atoms of each

type we'll have so we have one carbon

we've got one hydrogen and we've got two

oxygen ions and then plus we just have a

random electron from this

minus charge here so we can go over to

the periodic table

and figure out how many electrons how

many valence electrons each of these

have so i'm using ptable.com because i

think it's a great uh periodic table

uh so if we look here if we look first

at carbon we'll see that we don't

again we don't care about the core

electrons when we're drawing lewis

structures we only care about the

valence electrons

so we're going to be looking for carbon

in the second electron level or

yeah the second level and the second row

when we'll be thinking about

how many electrons it has it's got two

in the s

and it has two in the p so that's going

to be four total

so we'll go ahead and write down this is

going to get us one

molecule times four electrons or one

atom times four electrons is going to be

uh four electrons then we've got hydrogen

hydrogen

we can look that up as well but we also

can remember that hydrogen has one

valence electron it has one total

electron uh so this is going to

contribute one electron to our molecule

and then next we'll go over and look at

oxygen here we've got two of these guys

so whatever number we come up with from

the periodic table we need to multiply

by two

so we'll go and look for oxygen oxygen

is here in the second row

it has two s electrons and it has four

p electrons and so we'll be looking at

six valence electrons for oxygen and

we'll multiply that by 2 and it's going to

get us 12 electrons here

and then we have one random electron

which we just need to add

to our total because we have our anion

this is really important if you have a

cation or an anion don't forget to add

or subtract the amount of electrons that

you've got

if you have a neutral molecule it's

easier you don't have to add or subtract

any electrons because

um you don't have any uh so

here is our total electron count if we

add all of these up we will find that we

have 18 total valence electrons

that we can be working with as we're

drawing the structure

again valence electrons are the

electrons that are involved in bonding

um they're the ones that are going to be shared

shared

by covalent molecules or transferred

during the formation of an ionic compound

compound

we're not going to look at the core

molecule the core of electrons because

core electrons are not involved

in uh bonding all right so now that

we've got our total number of valence electrons

electrons

we can start to figure out what this

thing is going to look like

so we need to draw a skeleton structure

um and what that means is we're going to

just sort of try to arrange our atoms

uh in some kind of rational way

and uh and then we're going to figure

out how how exactly everything is connected

connected

so some sort of hints here the least

electronegative atom should be central

so again we're dealing with this

molecule ch02

minus so we can think about which of

these is going to be the least

electronegative technically it's hydrogen

hydrogen

but hydrogen is not ever going to be a

central atom hydrogen can only ever make

one bond

so we are going to ignore hydrogen um

and so here we're looking at the

the the battle is between carbon or

oxygen to be our central molecule

or our central atom and we're going to

choose carbon because it is less electronegative

electronegative

than oxygen so we'll go ahead and stick

carbon in the middle of what's going to

be our skeleton

structure and then we will connect

everybody else to this carbon

using single bonds to start off with so

i'll just draw my hydrogen and then i

will draw an oxygen and then i will draw another

another

oxygen all right and at this point we're

going to need to do some electron math

so i'm going to just work on my electron tally

tally

so i started out with 18 total valence electrons

electrons

and now i need to subtract out how many

i have used up

when i was making my skeleton structure

so each single

bond is two electrons i've got three

single bonds so i've got two

four six electrons used up

in my skeleton structure that means i

have 12 electrons left

to distribute

all right so the next thing that i'm

going to do is distribute the remaining electron

electron

uh around um around the outer

uh the non-central atoms to complete the

octets so

uh what i like to do is figure out if i

have how many electrons i've got left i

like to figure out how many lone pairs

that translates to because it's easier

for me to count in ones

rather than counting in twos um but you

figure out your favorite method

so i've got 12 electrons that contrib

that that means i've got six

lone pairs which are the uh again those

little dots that aren't involved

in uh in bonding so i'm going to

start by distributing those around the

non-central or the terminal atoms

and again that's because i put the more

electronegative atoms at the ends of the

bond rather than the central so i'm

going to start by giving the electrons

to the more electronegative atom

as that's where they're likely to wind

up anyway and then if i have leftovers

then i will give them to any central atoms

atoms

again hydrogen here hydrogen actually

does not

get an octet um it only ever has

two electrons max because it doesn't

have a p orbital

at all it only has the 1s orbital it

only has s orbitals available and so it

can't actually

uh make more than one bond all right so

we're going to start off with these six

lone pairs and we're going to give them

to the oxygens

so one two three

four five six i am out of electrons so i

have to stop

so uh and then i should i'm just gonna

double check my electron count so i have

one two

three four five six seven eight nine

uh pairs of electrons total which is 18

which is what i started with so life is good

good

all right so now i have done step three

which is distribute the electrons around

the terminal ions

or sorry terminal atoms to fulfill the octets

octets

and then step four is to distribute any remaining

remaining

electrons as lone pairs around the

central atom here i'm done

i'm out of electrons i do not have any

more to distribute so

i'm going to just skip this step and

then the last thing that i'm going to do

is step five so i'm going to check the octets

octets

i'm going to redistribute any lone pairs

as double or triple bonds to

the central atom if i need so again i'm

going to check the octets

i will do this in green so i'm looking

at my oxygens

check right two four six eight oxygen

two four six

eight check hydrogen two

check carbon two four six so i

am missing some electrons here i do not

have a full octet

and i'm going to deal with this by

redistributing some lone pairs from my

terminal atoms

to uh as as double or triple bonds until

i have

all my f fulfilled so i'm going to go

ahead and take one of these

and switch it down to a bond

so let me go ahead and just make that

change i'm going to erase that lone pair

and then i'm going to draw it in as a

double bond

it doesn't matter which one i take it

from uh you would generally start with

the less electronegative

central or terminal atom if you have a

choice but we don't hear they're both oxygens

oxygens

so let's go ahead and check my octets

again i didn't change this guy i didn't

change the hydrogen

two four six eight oxygen is still good and

and

carbon two four six eight carbon is now fulfilled

fulfilled

so um this is my structure with one exception

exception

uh this is an anion so i'm going to just

get rid of some of my extra notations

and the last thing that i need to do

here is put this whole thing in brackets

and add a negative sign this is really important

important

um if you don't put the brackets and you

don't put the negative then your

electron count is off by one

because um this this negative tells you

that you have an extra electron in here from

from

a different molecule or uh ion

and you can't form the structure this

way without this negative charge it's

really important to always put your ions

in brackets all right so that is the

structure of the formate anion

i'm going to move on and do some

examples but now is a really good time

to pause the video and try these on your

own so i'll just give you a second to

look at the examples that i'm going to do

i'm going to work on ammonium i'm going

to work on ethylene and then i'm going

to work on

cyanide so take a minute and try to work

these out on your own

and then come back and i will show you

all right so we're going to start out

with ammonium

uh so ammonium we're going to just do

all these really systematically we're

going to walk through the first

the steps so the first step is always

going to be calculate your valence electrons

electrons

so if i look on my periodic table i will

find that nitrogen has five valence electrons

electrons

and i've got one nitrogen uh hydrogen has

has

four uh sorry there's four hydrogens

but each of them has one valence

electron so i'll multiply that number by four

four

so this is the number of valence

electrons and then i multiply by four

and then i have a plus charge so this

means that this whole structure has lost

one electron so i actually need to

subtract one electron to get my final

electron count and so when i do this

math very carefully i will find that my

structure is going to have eight electrons

electrons

total all right so step two is to start drawing

drawing

my skeleton and i will do that by

putting the

uh any central atom that can be central

in the middle uh or the least electronegative

electronegative

atom that can be central in the middle

uh hydrogen again cannot be central

because it can only ever make one bond

so i'm stuck here with nitrogen

so i'll put my nitrogen in the middle

and then i will connect everybody else

to the nitrogen

all right so that is step two i need to

do some

electron mass at this point and keep

track of how many electrons i've used up

um so i will go ahead and do this in red

so i have used up two four

six eight electrons i need to subtract

eight electrons

and well i'm out of electrons um

i'm just going to go ahead and walk

through the uh just talk you through the

last three steps just to

for completeness but i'm out of

electrons i can't really do anything here

here

um so the next step is to distribute any

remaining electrons on

external the terminal ions are terminal

atoms i don't have any electrons and

also hydrogen can't take any more

anyway step four is to add any leftover

electrons to the central atom i'm still

out of electrons and then step five

is to check the octets and rearrange as

necessary so here i will check the octets

octets

um hydrogen doesn't get on octet it gets

a duet so i will

check all these guys off because they

each have two electrons

as they are sharing two electrons with

that central nitrogen

and then nitrogen two four six eight

it's got an octet

life is good so my last step uh

anytime i have an ion my last step is to

put that guy in

brackets and apply the appropriate charge

charge

so i will do that so that is the

structure for the ammonium

cation all right and then we will move

on to ethylene

so again first step is to count the

number of valence electrons here i have

carbon carbon has four valence electrons

i've got two carbons so i'll multiply

that by two

and then i have four hydrogens each

hydrogen has one

valence electron and i will multiply

that by four so when i do

my careful addition i will find out that this

this

structure has 12 electrons in it

all right so step two is to sketch this out

out

uh using a a skeleton structure with the

least electronegative atom in the center

here i've got

some hydrogens and i've got some carbons

hydrogen can never be central so i'm

going to go with carbon

uh i have two carbons here which is kind

of interesting and so

i'm going to go ahead and either i can

pick one of them to be central or

i can make a guess that there's probably

some symmetry here and go ahead and

attach those guys and they're gonna sort of

share the the duty of being the central atom

atom

um when you when you work on this if you

started out with a different

arrangement you would figure out pretty

quickly that you have to put the two

carbons in the middle

um so i'll go ahead and put the

because that's our skeleton structure

and then i'm going to do

my math my bookkeeping so i will just

count up how many elections i've used

i've used one two three

four five pairs which is 10 electrons

so i have two electrons left which is

one lone pair

so um step three is to

assign any remaining electrons to my

terminal or external atoms

uh hydrogens can't take any so i can

just give them to a carbon so i'll just

go ahead and pick one of my carbons to give

give

that lone pair to um uh

because i that's my central so that's

technically step four is adding the

uh remaining electrons to the central

atoms and then step five is to check my octets

octets

so i'm going to go ahead and check my

octets hydrogen

these are duets they're all good they've

got one bond they are satisfied

carbon two four six eight this carbon is good

good

carbon two one uh two four six

nope this guy does not have a full octet

so anytime i don't have a full octet i'm

going to go ahead and

grab some of these grab a lone pair from

somebody that does have an octet

um usually it's a terminal atom this guy

it is technically a central atom

but it's the only guy that's got a lone

pair that might be able to share it

so you're looking for somebody that does

have an octet and a lone pair to share

with the atom that doesn't have a lone pair

pair

or sorry it doesn't have a full octet so

we'll go ahead and just transfer this over

over

so i'm just going to clean up my

structure at this point i will clean up

my little notations here

um and we will move

that lone pair to

being a double bond all right so now

we'll check again the hydrogens are still

still

all fulfilled carbon two four

six eight and then the other carbon two

four six eight so this is my structure

and it turns out this is in fact the

structure of ethylene

um just real quick i wanna show you what

happens if you draw

this um out a different way so let's say

you put a carbon

and then you put a carbon and then you

put three hydrogens around one side

and one hydrogen around the other side

um when

you try well let's go ahead and just do this

this

so our step two would be to assign any

remaining lone pairs so let me just

count up one two three four five i still

have the same number of

electrons left over i've got one lone

pair left over so i'm going to go ahead

and give it to

my well most external atom that doesn't have

have

a full octet and i'll give it to carbon

so what happens is

the next uh step that right to move this

guy around to fulfill an octet

i can't because this one already this

two four six it doesn't this is this

doesn't have an octet

two four six eight this already has an

octet the only option here is to move

this guy and make a double bond with

this carbon

um and when you do that we have a big

problem here

which is this if we look here this

carbon still doesn't have a phyllo octet

two four six this carbon is still only

has six electrons and this carbon

two four six eight ten this has ten

electrons around it we're

we're breaking the octet this is a this

is not a good structure you can't do this

this

um so anytime this happens where you

have somebody that doesn't have an octet

and you can't

do anything without breaking somebody

else's octet um with too many electrons

what that means is that one of these

hydrogens is in the wrong spot

um so that's a good clue to you need to

start moving

you need to redraw your skeleton

structure with a different arrangement

and uh and try again

all right uh so we're going to go

look at uh cyanide here for our last

example uh here before we talk about our

expanded octets or exceptions to the octet

octet

so cyanide we're going to start off with

the exact same

process that we have for the last ones

so we need to figure out how many

valence electrons

cyanide has we've got one carbon atom

carbon has four electrons

and we've got one nitrogen nitrogen has

five electrons

and then we have a negative charge uh

which means we have one

additional electron so when we add this

all up we wind up with 10

electrons total in our cyanide

anion all right the next step is to draw

the skeleton structure

there isn't really a central atom when

you only have two atoms so we'll just go

ahead and connect those guys

whoops helps if i draw the correct atom

we'll just go ahead and correct our two

atoms with a single bond so that's step two

two

then step three is to distribute any

remaining lone pairs

so i need to do my bookkeeping and

figure out how many remaining lone pairs

i've got

all right so i used up two electrons i

will subtract two

i wind up with eight electrons which

means that's four

lone pairs so i've got four pairs of

electrons to assign

so i'm going to go ahead and assign them

here i don't have a central atom so i

don't really have a terminal atom that i should

should

choose first but i am going to choose

nitrogen first because it is more electronegative

electronegative

so i'm going to give i'm going to give

the nitrogen the extra electrons before

i give them to the less electronegative atom

atom

all right so i have four lone pairs to

work with and i need to fill my octets

so nitrogen already has two four

six eight there's my octet and then i

have one more lone pair to give and that

goes to carbon

because is the other one all right step

four is to distribute any remaining

electrons i have already done that

and then step five is to check my octets

so here i'm gonna look

nitrogen i filled up my octet as i was

doing it two four six

eight carbon two four so i've got a

problem uh all right so i need to start rearranging

rearranging

my lone pairs into double bonds so i'll

start with one

move this guy down again i'm already

always going to start with a lone pair from

from

the guy that's already got electrons

that already has a full octet and try to

share them with the atom that doesn't

have a full octet so let me just go

ahead and erase

that and redraw it and then i'm going to

check my

octet again nitrogen still has my still

has an octet carbon

2 4 6 still doesn't have an octet so i'm

going to take another lone pair off

the nitrogen and share that again so i

will erase this

just for clarity and draw that out

now i'm going to check this again 2 4 6

8 nitrogen still has an octet

2 4 6 8. now carbon has an octet

um so my last step as always is to

put some brackets around this and redraw

it uh sorry and add the at the charge

i can if i want to redraw this a little

bit more neatly

with um and this is just a thing that

we'll talk about when we get to the geometry

geometry

section but we tend to put uh we tend to

try to distribute the

lone pairs sort of evenly around

our atoms it doesn't matter um that's

just sort of

because it looks nicer and it's a little

bit more realistic that the electrons

are further away from the other

electrons as we'll talk about

in the geometry section but this is

all right so now we're going to move on

to talking about exceptions to the octet

rule and how that changes

or doesn't change our our process of drawing

drawing

lewis structures so we talked before

about the fact that some molecules are

going to have odd electrons

they're going to have an odd number of

valence electrons which means that just

by definition not all atoms will have an octet

octet

because it's not possible um so what's

going to happen is we're going to have

one unpaired electron

and we need to give that to the less

electronegative atom

that is more likely to have an unfilled octet

octet

so we'll just do this exactly the same

way we start off with

figuring out how many valence electrons

we have so nitrogen has five valence electrons

electrons

oxygen has six so we wind up with 11

electrons all right then our step two is

to draw our skeleton structure

we don't have a central atom so we'll

just go ahead and connect our

two atoms with a single bond and then we

will do our electron bookkeeping and

subtract two electrons

so we are left with nine electrons

all right so our next step is to assign

these lone pairs which

here is where it gets a little weird

right we've got basically four

lone pairs plus one electron so we need

to assign these to

our molecule and we'll start with the

more electronegative atoms and we'll

move to the less electronegative atoms

so we've got four lone pairs to

distribute we need to fill octets oxygen

already has two

so we'll give it six more

so that's three of our lone pairs so

we'll give nitrogen our last lone pair

and we will also give it the unpaired electron

electron

so uh that's all the electrons we've got

to work with we are done

so then step five is to check our

valence uh our octets oxygen two four six

six

eight has our octet nitrogen two four

five doesn't have an octet and in fact

it's pretty far off from an octet

um and so with these ones it's not so

much meeting the octet

but trying to get to seven um we can't

get to eight but we can get to seven

so let's go ahead and reassign one of

these lone pairs

to be a double bond so when we do that

we will find uh now we'll double check

again we've got two four six

eight still hasn't still has an octet

and then two four

six seven so nitrogen doesn't

technically have an octet here but it is

close as

as we can get uh given that it it is

stuck with that odd electron

um and this is not this is uh it's not a charged

charged

particle it's not an ion or anything so

we don't put it in brackets we just

leave it alone

all right um next we will talk about atoms

atoms

that are electron deficient so uh for

example bf3

where uh somebody in here might not have

a filled octet uh this can happen with

groups two

and thirteen um

all right so we'll start off exactly the

same way we're gonna start off by

uh calculating the number of valence

electrons boron has

three valence electrons and fluorine

each one has seven

and then we need to multiply that by

three because we have uh

three fluorines and so when we finish

off our

math we will find that this is going to

have 24

electrons all right our step two

is to draw a skeleton structure boron

goes in the middle because it is less

electronegative than fluorine

and then we will surround it with the three

three

fluorines all right and then we will do

our electron bookkeeping

two four six subtract six electrons so

we've got eighteen

electrons left over which means six

uh nope nope it does not mean six it

all right so now we need to go ahead and

assign them

uh so step three is to assign them

assign them to our external

um external atoms so we'll go ahead and

do that

uh so each of these fluorines has two so

far so it can

have it needs three more lone pairs

we'll go ahead and do that so

one two three

one two three all right so that is

actually all of our electrons so

2 4 6 8 10 12 14 16 18

20 22 24. so that's it we're done this

is all of our electrons we don't have

anybody left over to give to the boron

so normally our uh our step four we

don't have anybody left step five we

would we would be checking our

our octets and we would see that boron

doesn't have an octet it's only got six

electrons around it

and we would be choosing one of these

guys to give to make a double bond

um and it turns out that when you're

dealing with a group

uh 2 or 13 element

you're just not going to do that you're

just going to leave leave that lone pair

alone and let that boron

stay with six electrons um this is what

you find experimentally you do find that

this uh this atom or that sorry this

this molecule does act like a boron with

three single bonds around it um we can

tell with bond length then we can also

look at reactivity

this is a super reactive molecule it it

does act like the boron

is electron deficient

all right so our last kind of

exceptions to the octet rule are

hypervalent molecules

where the central atom can exceed the octet

octet

so as i mentioned before only elements

in the third and higher period

uh can have this happen and it's because

they have mtd orbitals

so phosphorus in the third energy level

doesn't have any electrons in the 3d uh orbitals

orbitals

because they they don't fill up until

the fourth energy level but they do

exist and so

um basically they can because there are

those empty d orbitals

um elements in the third row and below

can exceed the octet

all right so let's go ahead and talk

about how you would draw

some of this so if we look at

phosphorous and phosphorus pentachloride

five chlorines

it's exactly the same process we start

off with figuring out how many electrons

um there are how many valence electrons

we've got

so phosphorus lives just below nitrogen

in the periodic table it's going to have

five valence electrons and then we have

chlorines which each have seven valence

electrons and there are five of them

and when we figure this out we find out

we have 40 electrons

in our structure then we will go ahead

and draw out a skeleton structure and

we'll put the least electronegative atom

in the center and then everybody else

attached to it by a single bond

here phosphorus is less electronegative

than chlorine and we'll go ahead and

draw this

and yeah we're going to go ahead and put

five chlorines around

our phosphorus um

so here's our five chlorines around our

phosphorus um it feels weird but it it

this is what actually uh happens in uh

in reality

all right so now i need to do my

electron bookkeeping two four six

eight ten subtract 10 electrons

which is 15 lone pairs all right and

then i'm going to start assigning them

to my

terminal external more electronegative

atoms first i'll go ahead and do this in green

green

so i'm just going to keep track of how

many electrons i use up

one two three four

five six

7 8 9

10 11 12

13 14 15. all right that's it that's as

many electrons as i've got

um i should go ahead and double check my

octets um and i i can do that so

look at the chlorines each of these

chlorines has eight electrons around it

uh they are all fulfilled uh phosphorus

technically has 10 electrons around it so

it is exceeding its octet um it is

fulfilled i don't need to rearrange

anybody to

try to get that phosphorus more electron

all right so then we're going to look at

one more example and this is

uh this is um xenon

tetrafluoride xenon is a noble gas so it

seems very strange

xenon actually can make uh stable

compounds under certain conditions and

this is one of them

so we're gonna exactly the same way we

would we're just gonna treat this

um like like any other compound we're

going to go ahead and figure out how

many valence electrons this structure

should have

um xenon has eight valence electrons and

the four fluorines each have seven so

we'll do

seven times four um and so when we

finish off our electron math we're going

to find

that this structure has 36 electrons

all right then we're going to put the uh

least electronegative atom in the center

which turns out to be

xenon and then we will arrange the

fluorines around

it attached by single bonds

and then step three after i do my electron

electron

bookkeeping so subtract one two three four

four

four lone pairs or eight electrons so

subtract eight electrons

we wind up with 28 electrons left over

which is 14 millimeters

all right so now we're going to start

assigning these lone pairs to my

terminal or external

atoms so 14 lone pairs so let's go

1 2 3 4

5 6 7

8 9 10

11 12. all right so i have used up 12

lone pairs

i have two lone pairs left

all right um step four is to assign lone

pairs to

my central atom so even though xenon

technically has

eight electrons here i have two more

electrons i need to deal with

and i'm going to put them around my

xenon so

this is what it looks like fluorine we

can check our octet so fluorine

atoms each have eight electrons and the xenon

xenon

atom has 12. um so and

yeah again this is this is what we

actually see we can we can

we can look at some properties and

understand that this is is in fact

likely the correct structure for this

molecule um it's important to note we

can extend the octet right we've

extended it

uh phosphorus here has 10 uh 10

electrons around it instead of eight

here xenon's got 12. we don't generally

see anything above 12 when we're looking

at main group elements so

um six lone pairs or six six

regions of density is pretty much the

max for

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YouTube Transcript:Chapter 4.4b Drawing Lewis Structures

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Chapter 4.4b Drawing Lewis Structures