Health insurance primer

WEBVTT
Kind: captions
Language: en

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- [Narrator] What we're
gonna do in this video

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is try to break down the terminology

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and a little bit of the
math of health insurance.

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So the first question
that you might wonder is,

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how much does an insurance plan cost?

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In many cases, you might have an employer

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who pays for all or
part of your insurance,

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but in many cases, you have
to get your own insurance.

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And so I'm going to
focus on the second case.

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What I did is, I went to my
state's insurance exchange,

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and I just looked up an indicative
plan for a family of four

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that makes a combined $100,000 a year.

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And this is just one of
the plans that I found.

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I got rid of the company's name.

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And what it has here
is the monthly premium.

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So whenever people talk
about premiums in insurance,

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they're talking about the
amount that you have to pay

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to get the insurance.

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So this saying that

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you to have to pay $629.51 a month

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in order to get this insurance.

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You might also hear things

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in terms of annual premiums

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and you might guess how
to go from a monthly

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to an annual premium.

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If you multiply this number, $629.51,

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times 12, you're going
to get an annual premium

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of $7,554.12.

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So this is how much
someone would have to pay,

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likely you, in order
to get this insurance.

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And what this is telling me on my exchange

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is that this is actually after

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a pretty big subsidy that I'm getting

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from the federal and state government.

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And most of this is happening
from the federal government,

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but we can talk about
that in future videos.

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The big picture here is,

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if you wanna know how much a policy costs,

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that's the premium.

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This is how much per month,
and this is how much per year.

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Now, the next thing that
you are very likely to see

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when getting a policy is a
term called a deductible.

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And one way to think
about a deductible is,

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this is how much you will have to pay

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out of your own pocket

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until the insurance
really starts to kick in.

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So here, we're looking
at actually two columns,

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in network and out of network.

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In network are doctors
and healthcare providers

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that the insurance company
already has put in their network,

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they already have some type
of an arrangement with,

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while out of network, they
don't have an arrangement

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with those parties.

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So for this insurer, if you go in network,

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you would have to spend

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up to $6,300 on an individual

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before the insurance
really starts to kick in.

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So if you had to go and
get some type of a surgery

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that costs $6,000,

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even if you've been paying the premium,

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you would have to pay
$6,000 on top of the premium

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in order to get that surgery.

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If that surgery, on the other
hand, for just one person,

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let's say that surgery costs $10,300.

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Well, in that situation,

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you would have to pay the first 6,300,

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so minus 6,300,

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which means that the insurance company

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would pay the remainder.

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It would pay the remaining $4,000.

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And in that year, if you
were to get other surgeries

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for that individual, let's say it's you,

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it would then be covered.

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The insurance company
would pay above and beyond,

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anything above and beyond the $6,300.

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Now, you also have a deductible
for the entire family.

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So they're not just totalling

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what each individual is spending.

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They're also totalling what
the whole family is spending.

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So let's say I have four
people in my family,

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and over the course of the year,

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I've had to, for each of them,

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I have spent 3,150.

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I'm just using this as an example.

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So each of them have not gotten

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to their individual deductible yet.

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But in total, my family of four,

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3,150 times four, is $12,600.

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After this point, even
though no individual

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has maxed out their deductible yet,

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now the insurance will also kick in.

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Now when they talk
about a drug deductible,

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this is a separate
deductible for the drugs.

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So the one that I looked at before,

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that could be general procedures

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or certain types of care you might need,

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while this right over here is specific

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to what you might need for medications.

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But it's the same idea.

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If it's for any one individual,

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the first $500 on medication in that year,

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they're going to have
to pay out of pocket,

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but then after that,
the insurance kicks in.

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And then once again, $1,000
for the entire family.

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Now, out-of-pocket maximum,
this says, no matter what,

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no individual on an individual basis

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should you have to put
out more than $7,800

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for an individual, no matter what happens.

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And no more than $15,600
for an entire family.

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So this is real money
here, but it's saying that

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you won't have to pay more than that

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out of your actual pocket.

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And then here, the maximum
cost per prescription,

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this is the most that
you will ever have to pay

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to get a prescription.

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So even if there's some
type of exotic drug

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that costs $1,000 per prescription,

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the most that you would have to pay

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is $500 per prescription for that.

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Now that we've dissected
deductibles a good bit,

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another term that you are
likely to see is copay.

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Copay, one way to think about it is,

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something that you pay every
time you go into that service.

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So, for example, a primary care visit.

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This says the first three visits

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are a $65 copay before the deductible.

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So this is saying, even
if you haven't used up

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your $6,300 individual deductible,

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your primary care visits are covered,

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but every visit, you have to pay $65.

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Now, the reason why they call it copay

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is that you and the
insurance company are paying.

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That visit might cost
the insurance company

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another 150, 200, $250,

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but you're paying $65 on every visit.

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So the insurance company,
the insurance is kicking in

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even before the deductible
has been reached.

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Now specialists visits.

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If you go to specialist doctors

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like a cardiologist or a
rheumatologist or a dermatologist,

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these copays only kick
in after the deductible.

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So what that means is,

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before you reach your deductible ceiling,

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you're going to pay whatever
it costs out of pocket.

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So if that specialist visit costs $500,

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you're going to pay $500.

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But once you've reached your deductible,

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then every visit after that

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is going to cost $95 is
what you would contribute,

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and then the insurance
company would pay the rest.

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Now, the last term I'm going
to introduce into this video

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is that of coinsurance.

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So what this is telling us,

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right here they're telling
us how we are covered

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for various drugs.

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So, for example, if it's a generic drug,

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these are things that are off patent,

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then you pay $18 after you
have reached your deductible.

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So no matter what the drug costs,

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after you've reached your deductible,

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every prescription, you pay $18.

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Now these tier two drugs,

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and I'm not gonna go
into detail about this,

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but these are drugs that
tend to still be on patent,

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and they tend to be more expensive.

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What this coinsurance is telling us is,

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you're going to pay
the drugs out of pocket

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until you reach your deductible
for this insurance plan,

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and then after you've
reached the deductible,

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you're going to pay 40% of the cost.

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So whenever someone
talks about coinsurance,

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that means that you are splitting the cost

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with the insurance company at that point,

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and it's usually
specified as a percentage.

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So the 40% is what you
would pay for that drug.

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So let's say it's $100 a month.

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You would pay 40% of that, so $40,

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and you'd only get the 60% subsidy

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from the insurance company

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after your drug deductible
has been reached.

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So I will leave you there.

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I've already introduced
a lot of ideas to you,

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but hopefully that will
already start to help you

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dissect how insurance plans work.


Mapping shapes

WEBVTT
Kind: captions
Language: en

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- [Instructor] We're told that triangles,

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let's see, we have triangle PQR

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and triangle ABC are congruent.

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The side length of each square
on the grid is one unit.

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So each of these is one unit.

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Which of the following sequences

00:00:16.000 --> 00:00:21.000
of transformations maps
triangle PQR onto triangle ABC?

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So we have four different
sequences of transformations.

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And so why don't you pause
this video and figure out

00:00:30.490 --> 00:00:34.720
which of these actually
does map triangle PQR,

00:00:34.720 --> 00:00:38.150
so this is PQR, onto ABC.

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And it could be more than one of these.

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So pause this video and have a go at that.

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All right, now let's do this together.

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So let's first think about sequence A.

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And I will do sequence
A in this purple color.

00:00:50.600 --> 00:00:53.120
So remember, we're
starting with triangle PQR.

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So first it says a rotation
90 degrees about the point R.

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So let's do that and then we'll
do the rest of this sequence

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So if we rotate this 90 degrees,

00:01:05.780 --> 00:01:07.680
so one way to think about it is

00:01:07.680 --> 00:01:12.100
a line like that is then
going to be like that.

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So we're gonna go like that.

00:01:14.358 --> 00:01:16.960
And so R is going to stay where it is.

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You're rotating about
it, but P is now going

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to be right over here.

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One way to think about
it is to go from R to P,

00:01:26.330 --> 00:01:29.270
we went down one and three to the right.

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Now when you do the
rotation, you're going to go

00:01:32.250 --> 00:01:34.980
to the right one and then up three.

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So P is going to be there
and you could see that.

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That's the rotation.

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So that side will look like this.

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So that is P and then Q is
going to go right over here.

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Once again also do a 90
degree rotation about R.

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And so after you do the
90 degree rotation PQR

00:02:00.490 --> 00:02:02.400
is going to look like this.

00:02:02.400 --> 00:02:04.350
So that is Q.

00:02:04.350 --> 00:02:05.460
So we've done that first part.

00:02:05.460 --> 00:02:10.460
Then a translation six units
to the left and seven units up.

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So each of these points
are gonna go six units

00:02:14.890 --> 00:02:16.620
to the left and seven up.

00:02:16.620 --> 00:02:20.670
So if we take point P,
six to the left one, two,

00:02:20.670 --> 00:02:22.540
three, four, five, six.

00:02:22.540 --> 00:02:27.540
Seven units up one, two,
three, four, five, six, seven.

00:02:27.880 --> 00:02:29.640
It'll put it right over there.

00:02:29.640 --> 00:02:31.520
So that is point P.

00:02:31.520 --> 00:02:34.600
If we take point R, we
take six units to the left.

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One, two, three, four, five, six.

00:02:38.310 --> 00:02:43.310
Seven up, one, two, three,
four, five, six, seven

00:02:43.540 --> 00:02:47.470
it gets us right over
there and then point Q.

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If we go six units to the left one, two,

00:02:50.670 --> 00:02:52.880
three, four, five, six.

00:02:52.880 --> 00:02:57.880
Seven up is one, two, three,
four, five, six, seven

00:02:58.170 --> 00:03:00.660
puts us right over there.

00:03:00.660 --> 00:03:02.430
So this looks like it worked.

00:03:02.430 --> 00:03:04.790
Sequence A is good.

00:03:04.790 --> 00:03:08.917
It maps PQR on to ABC.

00:03:10.430 --> 00:03:13.830
This last one isn't an R
this is a Q right over here.

00:03:13.830 --> 00:03:16.840
So that worked, sequence A.

00:03:16.840 --> 00:03:20.370
Now let's work on sequence B
on this in different color.

00:03:20.370 --> 00:03:24.870
A translation eight units
to the left and three up.

00:03:24.870 --> 00:03:26.080
So let's do that first.

00:03:26.080 --> 00:03:29.810
So if we take point Q, eight
to the left and three up.

00:03:29.810 --> 00:03:34.740
One, two, three, four,
five, six, seven, eight.

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Three up, one, two, three.

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So this will be my red Q for now

00:03:41.010 --> 00:03:43.630
and now if I do this point R.

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One, two, three, four,
five, six, seven, eight.

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Let me make sure I do that right.

00:03:50.875 --> 00:03:53.460
One, two, three, four,
five, six, seven, eight.

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Three up, one, two, three.

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So my new R is going to be there

00:03:58.750 --> 00:04:03.500
and then last but not least
point P, eight to the left,

00:04:03.500 --> 00:04:07.420
One, two, three, four,
five, six, seven, eight.

00:04:07.420 --> 00:04:09.960
Three up, one, two, three.

00:04:09.960 --> 00:04:11.160
Goes right there.

00:04:11.160 --> 00:04:16.160
So just that translation
will get us to this point.

00:04:19.270 --> 00:04:20.700
It'll get us to that point.

00:04:20.700 --> 00:04:22.290
So we're clearly not done mapping yet,

00:04:22.290 --> 00:04:26.110
but there's more
transformation to be done.

00:04:26.110 --> 00:04:27.650
So it looks something like that.

00:04:27.650 --> 00:04:29.330
It says then a reflection

00:04:29.330 --> 00:04:33.600
over the horizontal line through point A.

00:04:33.600 --> 00:04:36.630
So point A is right over here.

00:04:36.630 --> 00:04:40.100
The horizontal line is right like that.

00:04:40.100 --> 00:04:42.320
So if I were to reflect,
point A wouldn't change.

00:04:42.320 --> 00:04:45.510
Point R right now is three
below that horizontal line.

00:04:45.510 --> 00:04:48.120
Point R will then be three
above that horizontal line,

00:04:48.120 --> 00:04:52.370
so point R will then go right over there.

00:04:52.370 --> 00:04:55.430
Just from that, I can
see that this sequence

00:04:55.430 --> 00:04:57.410
of transformations is not going to work.

00:04:57.410 --> 00:04:59.550
It's putting R in the wrong place.

00:04:59.550 --> 00:05:03.240
So I'm going to rule out sequence B.

00:05:03.240 --> 00:05:07.760
Sequence C, let me do
that with another color.

00:05:07.760 --> 00:05:10.530
I don't know I will do it
with this orange color.

00:05:10.530 --> 00:05:15.530
A reflection over the vertical
point through point Q,

00:05:15.870 --> 00:05:19.310
sorry, a reflection over the
vertical line through point Q.

00:05:19.310 --> 00:05:20.220
So let me do that.

00:05:20.220 --> 00:05:23.840
So the vertical line through
point Q looks like this.

00:05:23.840 --> 00:05:25.800
You can draw that vertical line.

00:05:25.800 --> 00:05:28.460
So if you reflect it Q is going to be,

00:05:28.460 --> 00:05:29.930
it's going to stay in place.

00:05:29.930 --> 00:05:31.950
R is one to the right of that,

00:05:31.950 --> 00:05:33.850
so now it's going to
be one to the left once

00:05:33.850 --> 00:05:34.970
you do the reflection.

00:05:34.970 --> 00:05:38.030
And point P is four to the right

00:05:38.030 --> 00:05:39.160
so that's gonna be for the left.

00:05:39.160 --> 00:05:41.050
One, two, three, four.

00:05:41.050 --> 00:05:44.090
So P is going to be there
after the reflection.

00:05:44.090 --> 00:05:47.290
And so it's going to
look something like this

00:05:47.290 --> 00:05:49.220
after that first transformation,

00:05:49.220 --> 00:05:51.600
unless it's getting a little bit messy.

00:05:51.600 --> 00:05:54.300
But this is what you probably
have to go through as well.

00:05:54.300 --> 00:05:56.240
So I'll go through it with you.

00:05:56.240 --> 00:05:58.270
All right, so we did that
first part the reflection,

00:05:58.270 --> 00:06:02.070
then a translation for to
the left and seven units up.

00:06:02.070 --> 00:06:04.280
So four to the left and seven up.

00:06:04.280 --> 00:06:05.930
So let me try that.

00:06:05.930 --> 00:06:07.010
So four to the left,

00:06:07.010 --> 00:06:09.470
one, two, three, four, seven up.

00:06:09.470 --> 00:06:14.170
One, two, three, four, five, six, seven.

00:06:14.170 --> 00:06:17.170
So it's putting Q right over here.

00:06:17.170 --> 00:06:19.870
I'm already suspicious of
it because sequence A worked

00:06:19.870 --> 00:06:21.700
where we put P right over there.

00:06:21.700 --> 00:06:24.400
So I'm already suspicious of
this but let's keep trying.

00:06:24.400 --> 00:06:27.070
So four to the left and seven up.

00:06:27.070 --> 00:06:28.900
One, two, three, four.

00:06:28.900 --> 00:06:33.630
Seven up One, two, three,
four, five, six, seven.

00:06:33.630 --> 00:06:37.630
So R is going to the same
place that sequence A put it

00:06:37.630 --> 00:06:41.950
and then point P one, two, three, four.

00:06:41.950 --> 00:06:45.780
One, two, three, four, five, six, seven.

00:06:45.780 --> 00:06:47.323
Actually it worked.

00:06:49.007 --> 00:06:53.540
And it works because this is
actually an isosceles triangle.

00:06:53.540 --> 00:06:55.240
So this one actually worked out.

00:06:55.240 --> 00:06:59.830
We were able to map a PQR
onto ABC with sequence C.

00:06:59.830 --> 00:07:03.260
So I like I like this one as well

00:07:03.260 --> 00:07:06.740
and then last but not
least, let's try sequence D.

00:07:06.740 --> 00:07:09.290
I'll do that in black
so that we can see it.

00:07:09.290 --> 00:07:11.260
So first we do a translation eight units

00:07:11.260 --> 00:07:12.853
to the left and three up.

00:07:13.715 --> 00:07:15.610
So we start here.

00:07:15.610 --> 00:07:19.380
One, two, three, four,
five, six, seven, eight,

00:07:19.380 --> 00:07:21.674
three up, one, two, three.

00:07:21.674 --> 00:07:25.580
So I'll put my black Q right over there.

00:07:25.580 --> 00:07:29.960
So eight to the left
One, two, three, four,

00:07:29.960 --> 00:07:31.860
five, six, seven, eight.

00:07:31.860 --> 00:07:34.390
Three up, one, two, three.

00:07:34.390 --> 00:07:37.140
I'll put my black R right over there.

00:07:37.140 --> 00:07:37.973
That's actually exactly

00:07:37.973 --> 00:07:40.090
what we did in sequence B the first time.

00:07:40.090 --> 00:07:42.980
So P is going to show up right over there.

00:07:42.980 --> 00:07:45.680
So after that translation sequence...

00:07:45.680 --> 00:07:49.630
First translation in sequence
D it gets us right over there

00:07:49.630 --> 00:07:54.630
Then it says a rotation negative
270 degrees about point A.

00:07:56.370 --> 00:07:58.800
So this is point A right over here

00:07:58.800 --> 00:08:02.260
and negative 270 degrees.

00:08:02.260 --> 00:08:05.020
It's negative, so it's
going to go clockwise

00:08:05.020 --> 00:08:06.970
and I'll see 180 degrees.

00:08:06.970 --> 00:08:09.810
Let's say if we were to take
this line right over here

00:08:09.810 --> 00:08:11.910
if we were to go 180 degrees,

00:08:11.910 --> 00:08:16.850
it would go to this line like that

00:08:16.850 --> 00:08:19.003
and then if we go to another 90 degrees,

00:08:20.074 --> 00:08:24.670
it actually does look like
it would map on to that.

00:08:24.670 --> 00:08:27.500
So this is actually looking pretty good.

00:08:27.500 --> 00:08:29.930
If you were to this line right over here

00:08:29.930 --> 00:08:32.680
well then if you go negative 270 degrees,

00:08:32.680 --> 00:08:37.200
we'll map on to this right over here

00:08:37.200 --> 00:08:40.260
and then that point R will
kind of go along for the ride.

00:08:40.260 --> 00:08:41.939
It is one way to think about it

00:08:41.939 --> 00:08:43.130
and so it'll go right over there as well.

00:08:43.130 --> 00:08:45.840
So I'm actually liking sequence D as well.

00:08:45.840 --> 00:08:49.193
So all of these work
except for sequence B.


The Court in Action

WEBVTT
Kind: captions
Language: en

00:00:00.280 --> 00:00:02.910
- [Instructor] Of the three
branches of the U.S. Government,

00:00:02.910 --> 00:00:04.730
the Judicial branch is the one

00:00:04.730 --> 00:00:08.390
that is least bound by public opinion.

00:00:08.390 --> 00:00:11.800
Supreme Court Justices aren't
elected, they're appointed,

00:00:11.800 --> 00:00:15.280
and they serve for life or
until they decide to retire.

00:00:15.280 --> 00:00:18.300
Usually a Justices serve
on the court for many years

00:00:18.300 --> 00:00:20.590
after the end of the term of the President

00:00:20.590 --> 00:00:21.560
who appointed them.

00:00:21.560 --> 00:00:24.060
One Justice, William O. Douglas,

00:00:24.060 --> 00:00:28.690
was appointed by Franklin
Delano Roosevelt in 1939

00:00:28.690 --> 00:00:33.560
and didn't retire until 1975
when Gerald Ford was President.

00:00:33.560 --> 00:00:35.460
Can you imagine how much changed

00:00:35.460 --> 00:00:37.470
during his tenure on the court?

00:00:37.470 --> 00:00:39.240
The framers designed things this way

00:00:39.240 --> 00:00:42.900
so that the Supreme Court
Justices can be independent

00:00:42.900 --> 00:00:46.670
consulting the Constitution
and precedent to decide

00:00:46.670 --> 00:00:50.020
whether laws an executive
actions are permissible

00:00:50.020 --> 00:00:52.060
without having to worry about keeping

00:00:52.060 --> 00:00:54.240
the public happy to get reelected,

00:00:54.240 --> 00:00:56.850
but this also means that it's not unusual

00:00:56.850 --> 00:01:00.430
that the Supreme Court makes
decisions that the public,

00:01:00.430 --> 00:01:04.280
the President or members
of Congress don't like.

00:01:04.280 --> 00:01:08.820
Brown versus Board of Education
in 1954 is a good example.

00:01:08.820 --> 00:01:12.190
White southerners resisted
the ruling refusing

00:01:12.190 --> 00:01:15.000
to integrate segregated public schools.

00:01:15.000 --> 00:01:18.530
Although it wasn't popular
with the majority population

00:01:18.530 --> 00:01:20.860
of the states that it most affected,

00:01:20.860 --> 00:01:24.680
the ruling protecting minority
rights in Brown versus Board

00:01:24.680 --> 00:01:28.650
was made possible by the
independence of the Judiciary.

00:01:28.650 --> 00:01:30.250
In that case, the Supreme Court

00:01:30.250 --> 00:01:33.080
had the backing of
Presidents Dwight Eisenhower

00:01:33.080 --> 00:01:35.700
and John F. Kennedy
who sent Federal troops

00:01:35.700 --> 00:01:38.490
into the south to enforce the ruling.

00:01:38.490 --> 00:01:40.450
Even though the court is independent,

00:01:40.450 --> 00:01:42.610
it's still important that it maintains

00:01:42.610 --> 00:01:45.530
it's legitimacy in the eyes of the public,

00:01:45.530 --> 00:01:48.760
so that citizens believe
that the court is impartial

00:01:48.760 --> 00:01:52.580
and that it's possible to
challenge the other branches

00:01:52.580 --> 00:01:54.510
through the actions of the court.

00:01:54.510 --> 00:01:56.610
To maintain it's legitimacy,

00:01:56.610 --> 00:01:58.670
Justices on the court are careful

00:01:58.670 --> 00:02:02.980
not to give the appearance of
favoring one political party

00:02:02.980 --> 00:02:05.240
over another by ruling too often

00:02:05.240 --> 00:02:08.510
for or against one parties policies.

00:02:08.510 --> 00:02:10.900
But what happens when the President

00:02:10.900 --> 00:02:14.660
or Congress doesn't support
the Supreme Court's rulings?

00:02:14.660 --> 00:02:18.430
What measures can they take to
affect the court's decisions?

00:02:18.430 --> 00:02:19.450
There are three main ways

00:02:19.450 --> 00:02:21.340
that we'll talk about in this video.

00:02:21.340 --> 00:02:23.480
Future appointments to the court,

00:02:23.480 --> 00:02:25.630
changing the court's jurisdiction

00:02:25.630 --> 00:02:29.170
and refusing to implement
the court's decisions.

00:02:29.170 --> 00:02:32.010
One of the ways that Presidents
can influence the court

00:02:32.010 --> 00:02:34.280
is by appointing Justices.

00:02:34.280 --> 00:02:37.300
Presidents tend to nominate
Supreme Court Justices

00:02:37.300 --> 00:02:40.060
who share their interpretation
of the Constitution,

00:02:40.060 --> 00:02:41.700
so their appointments change

00:02:41.700 --> 00:02:44.320
the ideological composition of the court.

00:02:44.320 --> 00:02:47.980
For example, President Richard
Nixon appointed four Justices

00:02:47.980 --> 00:02:49.790
to the Supreme Court that altered

00:02:49.790 --> 00:02:51.680
the court's ideological leaning.

00:02:51.680 --> 00:02:54.510
He ushered in a transition
from a liberal court

00:02:54.510 --> 00:02:58.900
under Chief Justice Earl Warren
to a more conservative court

00:02:58.900 --> 00:03:01.460
under Chief Justice Warren Burger.

00:03:01.460 --> 00:03:03.240
Congress can also influence the court

00:03:03.240 --> 00:03:05.530
by changing the number of Justices.

00:03:05.530 --> 00:03:06.890
But today, we accept that there are

00:03:06.890 --> 00:03:09.000
nine Justices on the Supreme Court,

00:03:09.000 --> 00:03:12.120
but Article III of the Constitution
doesn't actually specify

00:03:12.120 --> 00:03:13.930
how many Justices are on the court.

00:03:13.930 --> 00:03:17.880
Over time, there have been as
few as six and as many as 10.

00:03:17.880 --> 00:03:19.850
There have been nine since Congress passed

00:03:19.850 --> 00:03:22.330
the Judiciary Act of 1869,

00:03:22.330 --> 00:03:26.150
but Congress had actually
eliminated seats in the 1860's.

00:03:26.150 --> 00:03:29.420
Then in the 1930's,
Franklin Roosevelt proposed

00:03:29.420 --> 00:03:31.500
adding more court seats
to the Supreme Court

00:03:31.500 --> 00:03:34.470
so that it's ideological
balance would move in favor

00:03:34.470 --> 00:03:36.490
of his new deal legislation.

00:03:36.490 --> 00:03:39.710
So, although, the number
of Justices on the court

00:03:39.710 --> 00:03:43.210
hasn't actually changed in 150 years,

00:03:43.210 --> 00:03:47.320
Congress does retain the
power to adjust that number.

00:03:47.320 --> 00:03:50.410
Another way that Congress can
influence the Supreme Court

00:03:50.410 --> 00:03:52.730
is by changing it's jurisdiction

00:03:52.730 --> 00:03:54.400
or the field of authority

00:03:54.400 --> 00:03:57.550
that it has to hear and decide cases.

00:03:57.550 --> 00:03:59.290
There are two places in the Constitution

00:03:59.290 --> 00:04:01.550
that say that Congress can do this.

00:04:01.550 --> 00:04:04.090
Article I, Section 8, which says

00:04:04.090 --> 00:04:08.220
that Congress has the power to
constitute tribunals inferior

00:04:08.220 --> 00:04:09.640
to the Supreme Court,

00:04:09.640 --> 00:04:12.040
and Article III, Section 1, which says

00:04:12.040 --> 00:04:16.020
that judicial power shall be
vested in the Supreme Court,

00:04:16.020 --> 00:04:18.760
and such inferior courts as the congress

00:04:18.760 --> 00:04:21.620
may from time to time
ordain and establish.

00:04:21.620 --> 00:04:24.080
So Congress may add new Federal courts

00:04:24.080 --> 00:04:26.210
by splitting up an existing District

00:04:26.210 --> 00:04:29.610
or subtract them by combining Districts.

00:04:29.610 --> 00:04:32.170
Congress may also pass legislation

00:04:32.170 --> 00:04:34.330
stripping the Supreme Court's jurisdiction

00:04:34.330 --> 00:04:37.810
to hear certain types of cases on appeal.

00:04:37.810 --> 00:04:40.410
Some recent examples
of Congress attempting

00:04:40.410 --> 00:04:43.660
to strip the Supreme Court
of jurisdiction concern

00:04:43.660 --> 00:04:47.600
the detainees at Guantanamo
Bay detention camp in Cuba.

00:04:47.600 --> 00:04:48.810
In the mid 2000's,

00:04:48.810 --> 00:04:52.050
the detainee's petitioned
the Supreme Court to review

00:04:52.050 --> 00:04:54.060
whether their detention was legal,

00:04:54.060 --> 00:04:57.520
and Congress passed two
laws stripping detainees

00:04:57.520 --> 00:05:00.220
of the right to petition
courts for review.

00:05:00.220 --> 00:05:03.270
The Supreme Court overturned
both of those laws.

00:05:03.270 --> 00:05:06.350
The last way that the other
branches can push back

00:05:06.350 --> 00:05:10.230
against a court decision is
by refusing to implement it

00:05:10.230 --> 00:05:13.560
or by doing as little as
possible to implement it.

00:05:13.560 --> 00:05:17.060
Remember, the court itself
has no enforcement power.

00:05:17.060 --> 00:05:19.750
It has to rely on the executive branch

00:05:19.750 --> 00:05:21.610
to enforce it's rulings.

00:05:21.610 --> 00:05:24.130
So what happens when
the President disagrees

00:05:24.130 --> 00:05:25.320
with the ruling?

00:05:25.320 --> 00:05:27.750
In most cases, Presidents enforce

00:05:27.750 --> 00:05:30.510
the Supreme Court's rulings
whether they like them or not

00:05:30.510 --> 00:05:33.370
seeing it as part of their
duty to uphold the government.

00:05:33.370 --> 00:05:34.700
But in rare cases,

00:05:34.700 --> 00:05:37.960
a President might choose not
to enforce a court ruling,

00:05:37.960 --> 00:05:40.710
especially if their
party controls Congress.

00:05:40.710 --> 00:05:43.060
An early example of this was in 1808

00:05:43.060 --> 00:05:46.910
when the Supreme Court ruled
Thomas Jefferson's Embargo Act,

00:05:46.910 --> 00:05:49.230
the one with the famous,
Oh Grab Me cartoon,

00:05:49.230 --> 00:05:50.800
was unconstitutional.

00:05:50.800 --> 00:05:53.100
Jefferson instructed customs agents

00:05:53.100 --> 00:05:56.410
to keep on enforcing the
Embargo and they did.

00:05:56.410 --> 00:05:59.280
So the important takeaway
here is that even though

00:05:59.280 --> 00:06:02.320
the Supreme Court determines
the law of the land,

00:06:02.320 --> 00:06:05.640
the system of checks and
balances still applies to it.

00:06:05.640 --> 00:06:08.190
Congress and the President
have several ways

00:06:08.190 --> 00:06:11.353
to influence the court's
power, present and future.


Help Jason Give Back to Khan Academy

WEBVTT
Kind: captions
Language: en

00:00:01.450 --> 00:00:04.080
- My name is Jason Spyres,
and at the age of 19,

00:00:04.080 --> 00:00:06.500
I made a stupid decision to sell cannabis

00:00:06.500 --> 00:00:08.420
and ended up in prison.

00:00:08.420 --> 00:00:11.560
Fortunately, my mother sent
me Khan Academy transcripts

00:00:11.560 --> 00:00:13.800
to start improving my education.

00:00:13.800 --> 00:00:16.240
And while I was doing
that, other inmates noticed

00:00:16.240 --> 00:00:19.530
and would come up to me and
ask if I could help them.

00:00:19.530 --> 00:00:21.117
When I'd ask them, "Why
don't you try to get

00:00:21.117 --> 00:00:22.560
"in the GED classes?"

00:00:22.560 --> 00:00:25.120
Some of the waiting lists was too long.

00:00:25.120 --> 00:00:27.960
Others, I realize, what it is is,

00:00:27.960 --> 00:00:30.990
it's hard to admit you
don't know something,

00:00:30.990 --> 00:00:33.420
and it's embarrassing when
you're a full-grown man

00:00:33.420 --> 00:00:34.600
to say that you don't know

00:00:34.600 --> 00:00:37.500
that there's three different
ways to spell the word there.

00:00:38.570 --> 00:00:39.573
So I'd help them.

00:00:40.470 --> 00:00:43.970
And when I got out, I used
Khan Academy's SAT program

00:00:43.970 --> 00:00:45.270
to prepare myself,

00:00:45.270 --> 00:00:47.623
and I ended up getting a 1440 on the SATs.

00:00:48.870 --> 00:00:51.150
And after that, I applied and was accepted

00:00:51.150 --> 00:00:53.423
into Stanford University.

00:00:55.710 --> 00:00:57.160
I'm making this video today

00:00:57.160 --> 00:01:00.300
'cause I wanna tell you that
when you support Khan Academy,

00:01:00.300 --> 00:01:02.550
you're not only supporting people like me

00:01:02.550 --> 00:01:05.330
and helping me get to
places like Stanford.

00:01:05.330 --> 00:01:08.180
You're supporting a
mission that strips away

00:01:08.180 --> 00:01:11.270
the embarrassment factor
of having to admit

00:01:11.270 --> 00:01:12.500
you don't know something

00:01:12.500 --> 00:01:14.950
because you wanna learn something.

00:01:14.950 --> 00:01:17.620
The best teacher can only help the student

00:01:17.620 --> 00:01:20.140
if they'll walk into the classroom,

00:01:20.140 --> 00:01:21.630
and what Khan Academy does

00:01:21.630 --> 00:01:25.000
is they put a classroom
in everybody's home

00:01:25.000 --> 00:01:27.800
on their time, on their schedule

00:01:27.800 --> 00:01:30.890
without any embarrassment.

00:01:30.890 --> 00:01:32.820
So I hope you will continue to support

00:01:32.820 --> 00:01:34.820
Khan Academy's mission,

00:01:34.820 --> 00:01:37.200
and I hope to continue
to support the mission

00:01:37.200 --> 00:01:39.400
of increasing a world-class education

00:01:39.400 --> 00:01:41.463
for everybody everywhere.


Formal charge

WEBVTT
Kind: captions
Language: en

00:00:00.150 --> 00:00:01.290
- [Instructor] In this video, we're going

00:00:01.290 --> 00:00:06.050
to introduce ourselves to
the idea of formal charge,

00:00:06.050 --> 00:00:09.220
and as we will see, it
is a tool that we can use

00:00:09.220 --> 00:00:12.890
as chemists to analyze molecules.

00:00:12.890 --> 00:00:15.780
It is not the charge on
the molecule as a whole,

00:00:15.780 --> 00:00:17.640
it's actually a number
that we can calculate

00:00:17.640 --> 00:00:21.060
for each of the individual
atoms in a molecule,

00:00:21.060 --> 00:00:22.610
and as we'll see in future videos,

00:00:22.610 --> 00:00:25.670
it'll help us think about
which resonance structures,

00:00:25.670 --> 00:00:29.480
which configurations of a
molecule will contribute most

00:00:29.480 --> 00:00:31.520
to a resonance hybrid.

00:00:31.520 --> 00:00:33.610
So before going too deep into that,

00:00:33.610 --> 00:00:36.740
let's just give ourselves a
definition for formal charge,

00:00:36.740 --> 00:00:38.660
and then as practice,
we're going to calculate

00:00:38.660 --> 00:00:41.090
the formal charge on the various atoms

00:00:41.090 --> 00:00:45.500
in each of these resonance
structures for nitrous acid.

00:00:45.500 --> 00:00:47.630
These are both legitimate Lewis diagrams.

00:00:47.630 --> 00:00:49.900
They're both legitimate
resonance structures

00:00:49.900 --> 00:00:52.040
for nitrous acid, but we'll think about

00:00:52.040 --> 00:00:54.630
which one contributes more
to the resonance hybrid

00:00:54.630 --> 00:00:57.220
based on formal charge.

00:00:57.220 --> 00:00:59.500
So the definition of formal charge,

00:00:59.500 --> 00:01:02.430
and we're going to do this
for each atom in our molecule,

00:01:02.430 --> 00:01:06.250
for each atom, we're going
to calculate the number

00:01:06.250 --> 00:01:10.470
of valence electrons in free,

00:01:10.470 --> 00:01:13.000
in free neutral,

00:01:13.000 --> 00:01:14.423
neutral atom,

00:01:15.290 --> 00:01:16.780
atom.

00:01:16.780 --> 00:01:21.610
From that, we are going
to subtract the number

00:01:21.610 --> 00:01:25.890
of valence electrons allocated,

00:01:25.890 --> 00:01:28.993
allocated to bonded,

00:01:30.050 --> 00:01:31.690
bonded atom.

00:01:31.690 --> 00:01:33.170
And so you're next question is,

00:01:33.170 --> 00:01:36.230
what does is mean to be allocated?

00:01:36.230 --> 00:01:39.430
Well, I will break up this
definition a little bit.

00:01:39.430 --> 00:01:41.130
So if we want to think
about the valence electrons

00:01:41.130 --> 00:01:44.480
that are allocated to a bonded atom,

00:01:44.480 --> 00:01:49.470
these are going to be the
number of lone pair electrons,

00:01:49.470 --> 00:01:53.543
number of lone pair electrons

00:01:56.290 --> 00:02:01.290
plus one half of the
number of shared electrons.

00:02:03.780 --> 00:02:05.240
So lets try and make sense of this

00:02:05.240 --> 00:02:07.940
by applying this
definition of formal charge

00:02:07.940 --> 00:02:10.680
to the constituents of nitrous acid.

00:02:10.680 --> 00:02:14.290
So let's start with
this hydrogen over here.

00:02:14.290 --> 00:02:16.330
So what's the number of valence electrons

00:02:16.330 --> 00:02:18.720
in a free, neutral atom of hydrogen?

00:02:18.720 --> 00:02:20.080
Well we've seen this multiple times,

00:02:20.080 --> 00:02:22.540
you could look at this on the
periodic table of elements,

00:02:22.540 --> 00:02:25.380
free neutral hydrogen
has one valence electron.

00:02:25.380 --> 00:02:27.640
Now how many valence electrons

00:02:27.640 --> 00:02:30.400
are allocated to the bonded atom?

00:02:30.400 --> 00:02:31.680
Well one way to think about it is,

00:02:31.680 --> 00:02:36.390
draw a circle around that
atom in the molecule,

00:02:36.390 --> 00:02:38.530
and you want to capture
all of the lone pairs,

00:02:38.530 --> 00:02:40.000
and you want to capture,
you can think of it

00:02:40.000 --> 00:02:42.990
as half the bond, you
could say for each bond,

00:02:42.990 --> 00:02:44.670
it's going to be one electron

00:02:44.670 --> 00:02:46.510
'cause it's half of the shared electrons,

00:02:46.510 --> 00:02:49.100
each bond is two shared electrons,

00:02:49.100 --> 00:02:50.430
but you're gonna say half of those,

00:02:50.430 --> 00:02:52.800
and then you have no lone pairs over here,

00:02:52.800 --> 00:02:56.000
so the number of valence electrons
allocated to bonded atom,

00:02:56.000 --> 00:02:58.420
in the case of hydrogen here, is one,

00:02:58.420 --> 00:03:01.170
and so we are dealing with a formal charge

00:03:01.170 --> 00:03:03.910
of zero for this hydrogen.

00:03:03.910 --> 00:03:05.580
Now what about this oxygen here?

00:03:05.580 --> 00:03:07.300
Well we do the same exercise,

00:03:07.300 --> 00:03:11.070
I like to draw a little
bit of a circle around it.

00:03:11.070 --> 00:03:12.970
And so the number of valence electrons

00:03:12.970 --> 00:03:15.750
in a free, neutral oxygen
we've seen multiple times,

00:03:15.750 --> 00:03:18.320
that is six, and then from that,

00:03:18.320 --> 00:03:21.000
we're going to subtract the
number of valence electrons

00:03:21.000 --> 00:03:22.880
allocated to the bonded atom.

00:03:22.880 --> 00:03:26.350
So the bonded atom has
two lone pair electrons,

00:03:26.350 --> 00:03:28.520
and then it gets half
of the shared electrons,

00:03:28.520 --> 00:03:30.810
so half of the shared electrons
would be one from this bond,

00:03:30.810 --> 00:03:33.030
one from that bond,
and one from that bond.

00:03:33.030 --> 00:03:35.940
So you add them all together,
two, three, four, five.

00:03:35.940 --> 00:03:40.940
So six minus five is
equal to positive one,

00:03:41.150 --> 00:03:44.000
and so the formal charge
on this oxygen atom,

00:03:44.000 --> 00:03:48.590
in this configuration of
nitrous acid is positive one.

00:03:48.590 --> 00:03:50.010
Now what about the nitrogen?

00:03:50.010 --> 00:03:52.710
Well we'll do a similar exercise there.

00:03:52.710 --> 00:03:56.260
A free neutral nitrogen
has five valence electrons,

00:03:56.260 --> 00:03:57.450
we've seen that multiple times,

00:03:57.450 --> 00:03:59.540
you can look at that from the
periodic table of elements,

00:03:59.540 --> 00:04:02.200
and then from that,
we're going to subtract

00:04:02.200 --> 00:04:03.470
the number of valence electrons

00:04:03.470 --> 00:04:06.340
allocated to the bonded to nitrogen,

00:04:06.340 --> 00:04:09.930
well we see one, two, three,

00:04:09.930 --> 00:04:14.030
and then two more lone pair
electrons, so that is five,

00:04:14.030 --> 00:04:16.870
and so you have zero formal charge there.

00:04:16.870 --> 00:04:19.460
And then let's look at this last oxygen.

00:04:19.460 --> 00:04:22.550
So this last oxygen, a free neutral oxygen

00:04:22.550 --> 00:04:24.600
has six valence electrons,

00:04:24.600 --> 00:04:27.140
from that, we're going
to subtract the number

00:04:27.140 --> 00:04:29.610
of valence electrons
allocated to the bonded atom,

00:04:29.610 --> 00:04:33.440
so two, four, six lone pair electrons,

00:04:33.440 --> 00:04:36.650
plus half of this bond, so that's seven

00:04:36.650 --> 00:04:38.330
allocated valence electrons,

00:04:38.330 --> 00:04:43.170
six minus seven equals negative one.

00:04:43.170 --> 00:04:46.926
So this oxygen has a formal
charge of negative one,

00:04:46.926 --> 00:04:48.940
and I really want to remind you,

00:04:48.940 --> 00:04:51.800
we're not talking about the
charge of the entire molecule,

00:04:51.800 --> 00:04:54.730
formal charge is really
a mathematical tool

00:04:54.730 --> 00:04:58.210
we use to analyze this configuration,

00:04:58.210 --> 00:05:01.200
but one way you can kind
of conceptualize it is,

00:05:01.200 --> 00:05:04.460
in this configuration,
this oxygen on average

00:05:04.460 --> 00:05:06.900
has one more electron hanging around it,

00:05:06.900 --> 00:05:09.110
one more valence electron
hanging around it

00:05:09.110 --> 00:05:11.010
than a free neutral oxygen would.

00:05:11.010 --> 00:05:15.060
This oxygen has one less valence
electron hanging around it

00:05:15.060 --> 00:05:18.300
than a neutral free oxygen would.

00:05:18.300 --> 00:05:20.860
Now let's look at this
configuration down here,

00:05:20.860 --> 00:05:23.080
well this hydrogen is
identical to this hydrogen,

00:05:23.080 --> 00:05:24.910
it has no lone pair electrons

00:05:24.910 --> 00:05:27.670
and it just has one
covalent bond to an oxygen,

00:05:27.670 --> 00:05:29.550
so we would do the same analysis

00:05:29.550 --> 00:05:32.630
to get that its formal charge is a zero,

00:05:32.630 --> 00:05:35.870
but now let's think about
this oxygen right over here.

00:05:35.870 --> 00:05:39.940
A free neutral oxygen has
six valence electrons,

00:05:39.940 --> 00:05:41.920
the number of valence
electrons allocated to this one

00:05:41.920 --> 00:05:46.500
is two, four, five, and six,

00:05:46.500 --> 00:05:50.220
so six minus six is
zero, no formal charge,

00:05:50.220 --> 00:05:52.150
and we go to this nitrogen.

00:05:52.150 --> 00:05:56.010
Free nitrogen has five valence electrons,

00:05:56.010 --> 00:06:00.600
this nitrogen has two, three, four,

00:06:00.600 --> 00:06:03.690
five valence electrons allocated to it,

00:06:03.690 --> 00:06:06.520
so minus five, it has zero formal charge.

00:06:06.520 --> 00:06:09.800
And then last but not least,
this oxygen right over here.

00:06:09.800 --> 00:06:12.900
A free neutral oxygen has
six valence electrons,

00:06:12.900 --> 00:06:17.900
this one has two, four,
five, six valence electrons

00:06:18.800 --> 00:06:21.240
allocated to the bonded atom,

00:06:21.240 --> 00:06:23.650
and so minus six is equal to zero.

00:06:23.650 --> 00:06:26.410
And so what we see is
this first configuration,

00:06:26.410 --> 00:06:28.090
or you could say this
first resonance structure

00:06:28.090 --> 00:06:30.720
for nitrous acid had some formal charge,

00:06:30.720 --> 00:06:32.210
it had a plus one on this oxygen

00:06:32.210 --> 00:06:33.710
and minus one on this oxygen,

00:06:33.710 --> 00:06:36.330
while the one down here
had no formal charge,

00:06:36.330 --> 00:06:39.250
everything had a formal charge of zero,

00:06:39.250 --> 00:06:41.370
and as we'll see in future videos,

00:06:41.370 --> 00:06:46.060
the closer the individual atom
formal charges are to zero,

00:06:46.060 --> 00:06:48.270
the more likely that that structure,

00:06:48.270 --> 00:06:49.550
that resonance structure,

00:06:49.550 --> 00:06:52.960
will contribute more to
the resonance hybrid,

00:06:52.960 --> 00:06:55.420
but we'll talk about that
more in future videos,

00:06:55.420 --> 00:06:56.380
the whole pint of this one

00:06:56.380 --> 00:06:59.420
is just to get comfortable
calculating formal charge

00:06:59.420 --> 00:07:02.063
for the individual atoms in a molecule.


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WEBVTT
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Language: en

00:00:00.063 --> 00:00:02.730
(upbeat music)


Resonance

WEBVTT
Kind: captions
Language: en

00:00:00.610 --> 00:00:02.790
- [Instructor] Let's see if
we can draw the Lewis diagram

00:00:02.790 --> 00:00:05.530
for a nitrate anion.

00:00:05.530 --> 00:00:10.530
So a nitrate anion has one
nitrogen and three oxygens,

00:00:10.960 --> 00:00:14.030
and it has a negative charge.

00:00:14.030 --> 00:00:15.460
I'll do that in another color.

00:00:15.460 --> 00:00:16.470
It has a negative charge.

00:00:16.470 --> 00:00:18.550
So pause this video and
see if you can draw that,

00:00:18.550 --> 00:00:21.033
the Lewis structure for a nitrate anion.

00:00:21.890 --> 00:00:24.020
All right, well we've
done this many times.

00:00:24.020 --> 00:00:25.470
The first step is to just account

00:00:25.470 --> 00:00:27.230
for the valence electrons.

00:00:27.230 --> 00:00:31.890
Nitrogen has one, two, three,
four, five valence electrons

00:00:31.890 --> 00:00:34.090
in its outer shell, and
in that second shell,

00:00:34.090 --> 00:00:36.990
if it's a neutral, free nitrogen atom.

00:00:36.990 --> 00:00:39.710
So we have five valence electrons there.

00:00:39.710 --> 00:00:43.030
Oxygen has one, two, three, four, five,

00:00:43.030 --> 00:00:44.610
six valence electrons.

00:00:44.610 --> 00:00:45.670
But if you have three oxygens,

00:00:45.670 --> 00:00:47.860
you're going to have six times three.

00:00:47.860 --> 00:00:50.470
And so if you just add
up the valence electrons,

00:00:50.470 --> 00:00:52.900
if these were free, neutral atoms,

00:00:52.900 --> 00:00:57.900
you would get five plus 18,
which is 23 valence electrons.

00:00:59.010 --> 00:01:00.390
Now the next thing we have to keep

00:01:00.390 --> 00:01:02.250
in mind is this is an anion.

00:01:02.250 --> 00:01:05.190
This has a negative one
charge right over here.

00:01:05.190 --> 00:01:08.430
So it's going to have
one more extra electron,

00:01:08.430 --> 00:01:10.660
one more extra valence
electron than you would expect

00:01:10.660 --> 00:01:12.920
if these were just free
atoms that were neutral.

00:01:12.920 --> 00:01:16.600
So let's add one valance electron here.

00:01:16.600 --> 00:01:20.210
So that gets us to 24 valence electrons.

00:01:20.210 --> 00:01:21.640
And then the next step is let's try

00:01:21.640 --> 00:01:23.880
to actually draw this structure.

00:01:23.880 --> 00:01:25.610
And the way we do it is we try

00:01:25.610 --> 00:01:27.460
to pick the least electronegative atom

00:01:27.460 --> 00:01:29.960
that is not hydrogen
to be the central atom.

00:01:29.960 --> 00:01:31.700
And in this case it is nitrogen.

00:01:31.700 --> 00:01:34.270
It's to the left of oxygen
in that second period.

00:01:34.270 --> 00:01:37.930
So let's put nitrogen in the center,

00:01:37.930 --> 00:01:38.763
right over there.

00:01:38.763 --> 00:01:40.544
And around it let's put three oxygens.

00:01:40.544 --> 00:01:45.544
So one, two, three oxygens.

00:01:45.570 --> 00:01:48.890
Let's put a single bond between them.

00:01:48.890 --> 00:01:50.960
And so so far we, and let me do that

00:01:50.960 --> 00:01:53.150
in another color, so we
can account for it better.

00:01:53.150 --> 00:01:54.860
So I'll do them in purple.

00:01:54.860 --> 00:01:56.970
So so far we have accounted for two, four,

00:01:56.970 --> 00:01:58.370
six valence electrons.

00:01:58.370 --> 00:02:01.780
So minus six valence electrons gets us

00:02:01.780 --> 00:02:04.510
to 18 valence electrons.

00:02:04.510 --> 00:02:07.020
The next step is we would try to allocate

00:02:07.020 --> 00:02:11.100
as many of these as possible
to our terminal atoms,

00:02:11.100 --> 00:02:12.240
the oxygens over here.

00:02:12.240 --> 00:02:14.700
Try to get them to a full octet.

00:02:14.700 --> 00:02:15.753
So let's do that.

00:02:15.753 --> 00:02:17.670
This, each of these oxygens,

00:02:17.670 --> 00:02:19.170
they're already participating in one

00:02:19.170 --> 00:02:20.230
of these covalent bonds,

00:02:20.230 --> 00:02:23.090
so they already have two
valence electrons hanging out.

00:02:23.090 --> 00:02:25.940
So let's see if we can
give them each another six,

00:02:25.940 --> 00:02:26.840
to get to eight.

00:02:26.840 --> 00:02:30.583
So two, four, six.

00:02:31.660 --> 00:02:34.976
Two, four, six.

00:02:34.976 --> 00:02:39.520
And then two, four, and six.

00:02:39.520 --> 00:02:40.650
And so just like that

00:02:40.650 --> 00:02:43.550
we have allocated 18 valence electrons,

00:02:43.550 --> 00:02:45.120
six, 12, 18.

00:02:45.120 --> 00:02:48.390
So minus 18 valence electrons.

00:02:48.390 --> 00:02:52.830
And we are now left with no
further valence electrons

00:02:52.830 --> 00:02:54.080
to allocate.

00:02:54.080 --> 00:02:56.220
But let's see how our atoms are doing.

00:02:56.220 --> 00:02:58.250
We know the oxygens have a full octet,

00:02:58.250 --> 00:03:00.680
but the nitrogen only has two, four,

00:03:00.680 --> 00:03:04.290
six valance electrons hanging around.

00:03:04.290 --> 00:03:06.580
It would be great if there
was a Lewis structure

00:03:06.580 --> 00:03:09.010
where we could have
eight valence electrons

00:03:09.010 --> 00:03:10.470
for that nitrogen.

00:03:10.470 --> 00:03:13.110
Well one way to do that is to take one

00:03:13.110 --> 00:03:15.650
of these lone pairs
from one of the oxygens

00:03:15.650 --> 00:03:19.120
and turn that into another covalent bond.

00:03:19.120 --> 00:03:20.278
So let's do that.

00:03:20.278 --> 00:03:23.893
So let me just erase this
pair right over here,

00:03:24.760 --> 00:03:26.030
and I'm just going to turn that

00:03:26.030 --> 00:03:28.460
into another covalent bond.

00:03:28.460 --> 00:03:30.814
And this is looking pretty good.

00:03:30.814 --> 00:03:32.800
We have eight valence
electrons around each

00:03:32.800 --> 00:03:34.170
of the oxygens.

00:03:34.170 --> 00:03:35.840
And now we have eight for the nitrogen,

00:03:35.840 --> 00:03:37.180
two, four, six, eight.

00:03:37.180 --> 00:03:38.780
And we have to remind ourselves

00:03:38.780 --> 00:03:40.720
that this is an anion.

00:03:40.720 --> 00:03:42.580
It has a negative one charge.

00:03:42.580 --> 00:03:44.020
So to finish the Lewis diagram

00:03:44.020 --> 00:03:45.880
we would just put that
negative charge there.

00:03:45.880 --> 00:03:48.200
And this is all well and good,

00:03:48.200 --> 00:03:51.590
but if this was the only
way that nitrate existed

00:03:51.590 --> 00:03:55.040
when we observed nitrate
anions in the world,

00:03:55.040 --> 00:03:58.060
we would expect to see one shorter bond

00:03:58.060 --> 00:04:01.010
and two longer bonds,
and we would expect one

00:04:01.010 --> 00:04:02.400
of the bonds to have a different energy

00:04:02.400 --> 00:04:03.890
than the other two.

00:04:03.890 --> 00:04:06.200
But in the real world we don't see that.

00:04:06.200 --> 00:04:10.480
We see that all of the bonds
actually have the same length,

00:04:10.480 --> 00:04:12.920
and they actually have the same energy.

00:04:12.920 --> 00:04:15.370
And so an interesting
question is why is that?

00:04:15.370 --> 00:04:17.610
And one thing that you
might appreciate is,

00:04:17.610 --> 00:04:19.230
when I took that lone pair

00:04:19.230 --> 00:04:20.460
to create this covalent bond,

00:04:20.460 --> 00:04:22.160
I could have done it with that top oxygen.

00:04:22.160 --> 00:04:24.400
I could have done it with
this bottom-left oxygen.

00:04:24.400 --> 00:04:26.740
Or I could have done it with
that bottom-right oxygen.

00:04:26.740 --> 00:04:30.490
And so there's actually
three valid Lewis structures

00:04:30.490 --> 00:04:31.470
that we could have had.

00:04:31.470 --> 00:04:33.353
Not only could we have
had this Lewis structure,

00:04:33.353 --> 00:04:35.360
we could have had this one,

00:04:35.360 --> 00:04:38.405
and I'll draw it all in
yellow to save us some time,

00:04:38.405 --> 00:04:40.220
where you have this nitrogen.

00:04:40.220 --> 00:04:42.660
It has a single bond with that top oxygen.

00:04:42.660 --> 00:04:44.793
And so that top oxygen
still has six electrons

00:04:44.793 --> 00:04:46.289
in lone pairs.

00:04:46.289 --> 00:04:48.130
And maybe it forms a double bond

00:04:48.130 --> 00:04:50.010
with the bottom-left oxygen.

00:04:50.010 --> 00:04:52.970
So this bottom-left oxygen
only has two lone pairs.

00:04:52.970 --> 00:04:55.155
One of them would have gone
to form the double bond.

00:04:55.155 --> 00:04:58.010
And then this oxygen would look the same.

00:04:58.010 --> 00:05:02.826
So what I am drawing here is
another valid Lewis structure.

00:05:02.826 --> 00:05:04.819
Or the double bond might have formed

00:05:04.819 --> 00:05:06.500
with this bottom-right oxygen,

00:05:06.500 --> 00:05:09.740
so let me draw that.

00:05:09.740 --> 00:05:12.560
So another valid Lewis
structure could look like this.

00:05:12.560 --> 00:05:17.560
So nitrogen bonded to that
oxygen has three lone pairs.

00:05:18.160 --> 00:05:22.450
This oxygen also has three lone pairs.

00:05:22.450 --> 00:05:24.260
And now this one has the double bond

00:05:24.260 --> 00:05:27.200
and only has two lone pairs.

00:05:27.200 --> 00:05:28.690
And whenever we see a situation

00:05:28.690 --> 00:05:31.633
where we have three
valid Lewis structures,

00:05:31.633 --> 00:05:34.690
we call this resonance.

00:05:34.690 --> 00:05:36.253
Resonance.

00:05:37.220 --> 00:05:38.350
Resonance.

00:05:38.350 --> 00:05:39.610
And we'll put an arrow,

00:05:39.610 --> 00:05:42.000
these two-way arrows
between these structures.

00:05:42.000 --> 00:05:43.780
And when you hear the word resonance,

00:05:43.780 --> 00:05:45.460
it sometimes conjures up this image

00:05:45.460 --> 00:05:47.790
that you're bouncing
back, you're resonating

00:05:47.790 --> 00:05:49.050
between these structures.

00:05:49.050 --> 00:05:50.760
But that's actually not right.

00:05:50.760 --> 00:05:52.690
What the right way to think about it is,

00:05:52.690 --> 00:05:55.540
these different ways of
visualizing the nitrate,

00:05:55.540 --> 00:05:58.160
these contribute to a resonance hybrid,

00:05:58.160 --> 00:05:59.670
which is really the true way

00:05:59.670 --> 00:06:01.040
that the nitrate exists.

00:06:01.040 --> 00:06:03.330
And so, if we wanted to
draw a resonance hybrid,

00:06:03.330 --> 00:06:05.230
it would look like this.

00:06:05.230 --> 00:06:07.140
You have the nitrogen in the center.

00:06:07.140 --> 00:06:11.670
You have your oxygens, one, two, three.

00:06:11.670 --> 00:06:15.430
I can draw our first
covalent bond like that.

00:06:15.430 --> 00:06:16.430
And then you would show the bond

00:06:16.430 --> 00:06:17.420
between nitrogen and each

00:06:17.420 --> 00:06:21.220
of these oxygens are a
hybrid between someplace

00:06:21.220 --> 00:06:23.680
between a single bond and a double bond.

00:06:23.680 --> 00:06:26.510
And so instead of just one of
them having the double bond

00:06:26.510 --> 00:06:28.270
and the other two having single bonds,

00:06:28.270 --> 00:06:30.120
they're all somewhere in between.

00:06:30.120 --> 00:06:32.148
So maybe you draw a dotted line,

00:06:32.148 --> 00:06:37.148
something like that, to
show what the reality is,

00:06:37.460 --> 00:06:39.590
is that you actually have three bonds

00:06:39.590 --> 00:06:41.720
that are someplace in between a single

00:06:41.720 --> 00:06:44.159
and a double bond, because the electrons

00:06:44.159 --> 00:06:46.880
in this molecule are
delocalized throughout.

00:06:46.880 --> 00:06:48.550
And of course you wanna make sure,

00:06:48.550 --> 00:06:50.950
you always wanna make
sure that people recognize

00:06:50.950 --> 00:06:53.320
that this is a anion.

00:06:53.320 --> 00:06:55.110
So this is the idea of resonance.

00:06:55.110 --> 00:06:57.650
You have multiple valid Lewis structures.

00:06:57.650 --> 00:07:01.380
They all contribute to a resonance hybrid,

00:07:01.380 --> 00:07:03.070
which is actually what we observe.

00:07:03.070 --> 00:07:04.093
We're not just bouncing between

00:07:04.093 --> 00:07:05.980
these different structures.

00:07:05.980 --> 00:07:09.870
The actual observation will
be a hybrid of the three.

00:07:09.870 --> 00:07:11.030
Now what we just drew here,

00:07:11.030 --> 00:07:12.804
these three are all equivalent.

00:07:12.804 --> 00:07:14.580
But in certain cases, we'll see this

00:07:14.580 --> 00:07:17.500
in future videos, you don't
have equivalent structures,

00:07:17.500 --> 00:07:19.620
and some of them might contribute more

00:07:19.620 --> 00:07:21.230
to the resonance hybrid than others.

00:07:21.230 --> 00:07:23.083
But we'll see that in future videos.


Exceptions to the octet rule

WEBVTT
Kind: captions
Language: en

00:00:00.270 --> 00:00:02.040
- [Instructor] In this video,
we're gonna start talking

00:00:02.040 --> 00:00:04.250
about exceptions to the octet rule,

00:00:04.250 --> 00:00:05.850
which we've talked about
in many other videos.

00:00:05.850 --> 00:00:07.070
The octet rule is this notion

00:00:07.070 --> 00:00:09.490
that atoms tend to react in ways

00:00:09.490 --> 00:00:12.100
that they're able to
have a full outer shell.

00:00:12.100 --> 00:00:14.550
They're able to have
eight valence electrons.

00:00:14.550 --> 00:00:16.610
Now, we've already talked
about some exceptions,

00:00:16.610 --> 00:00:17.930
things like hydrogen.

00:00:17.930 --> 00:00:19.730
Its outer shell is that first shell

00:00:19.730 --> 00:00:22.460
which gets full with two electrons.

00:00:22.460 --> 00:00:25.090
So it's trying to get to that duet rule.

00:00:25.090 --> 00:00:27.090
But as we'll see, there
are other exceptions.

00:00:27.090 --> 00:00:29.420
Boron and aluminum, for example,

00:00:29.420 --> 00:00:31.960
they can form stable
molecules where the boron

00:00:31.960 --> 00:00:36.630
or the aluminum only have six
valence electrons, not eight.

00:00:36.630 --> 00:00:38.720
And there are exceptions
in the other direction.

00:00:38.720 --> 00:00:41.630
As you get to the third period and beyond,

00:00:41.630 --> 00:00:44.650
we'll actually see atoms
that can maintain more

00:00:44.650 --> 00:00:46.330
than eight valence electrons.

00:00:46.330 --> 00:00:47.650
And we're actually going to see an example

00:00:47.650 --> 00:00:50.070
of that with xenon.

00:00:50.070 --> 00:00:51.870
So let's just go into a few examples.

00:00:51.870 --> 00:00:53.740
Given what I've told you,

00:00:53.740 --> 00:00:57.620
see if you can come up
with the Lewis diagram

00:00:57.620 --> 00:01:00.070
for aluminum hydride.

00:01:00.070 --> 00:01:03.850
So aluminum hydride has one aluminum

00:01:03.850 --> 00:01:05.470
and three hydrogens.

00:01:05.470 --> 00:01:08.600
See if you can draw the
Lewis diagram for that.

00:01:08.600 --> 00:01:10.220
All right, now let's do this together.

00:01:10.220 --> 00:01:12.070
So the first thing you wanna do is account

00:01:12.070 --> 00:01:14.174
for all of the valence electrons.

00:01:14.174 --> 00:01:16.080
Aluminum's outer shell is the third shell,

00:01:16.080 --> 00:01:17.180
so the third period here,

00:01:17.180 --> 00:01:20.740
and it has one, two,
three valence electrons.

00:01:20.740 --> 00:01:23.500
And then we have three hydrogens,

00:01:23.500 --> 00:01:26.890
and each hydrogen has
one valence electron.

00:01:26.890 --> 00:01:29.510
And so you add all of this up together,

00:01:29.510 --> 00:01:34.380
three plus three is equal
to six valence electrons

00:01:34.380 --> 00:01:36.760
in aluminum hydride.

00:01:36.760 --> 00:01:39.400
Now, the next step after that is

00:01:39.400 --> 00:01:43.350
to try to draw the structure
with some covalent bonds.

00:01:43.350 --> 00:01:45.590
We don't wanna make
hydrogen our central atom.

00:01:45.590 --> 00:01:48.240
That would be very atypical.

00:01:48.240 --> 00:01:51.030
And so let's put aluminum in the center.

00:01:51.030 --> 00:01:53.300
And then we're gonna have three hydrogens.

00:01:53.300 --> 00:01:57.060
So one, two, and three.

00:01:57.060 --> 00:02:00.220
And then let's put some
covalent bonds in here.

00:02:00.220 --> 00:02:03.550
And so let's see, how
many valence electrons

00:02:03.550 --> 00:02:04.810
have we now accounted for?

00:02:04.810 --> 00:02:06.530
This is two in this covalent bond.

00:02:06.530 --> 00:02:07.900
Another two gets us to four.

00:02:07.900 --> 00:02:09.570
Another two gets us to six.

00:02:09.570 --> 00:02:14.120
So we have just accounted for
all six valence electrons.

00:02:14.120 --> 00:02:16.470
So we have no more valence
electrons to play with.

00:02:16.470 --> 00:02:19.190
Let's think about how the
various atoms are doing.

00:02:19.190 --> 00:02:22.170
So the hydrogens are all
meeting their duet rule.

00:02:22.170 --> 00:02:25.720
These two electrons in this
bond are hanging around hydrogen

00:02:25.720 --> 00:02:27.230
and around the aluminum.

00:02:27.230 --> 00:02:30.130
But from hydrogen's point
of view, it has a full duet,

00:02:30.130 --> 00:02:32.410
that hydrogen as well and
that hydrogen as well.

00:02:32.410 --> 00:02:34.410
But notice the aluminum over here,

00:02:34.410 --> 00:02:37.120
it has two, four, six electrons,

00:02:37.120 --> 00:02:41.140
valence electrons around it,
and so it's not a full octet.

00:02:41.140 --> 00:02:43.470
But aluminum hydride is actually something

00:02:43.470 --> 00:02:45.860
that has been observed.

00:02:45.860 --> 00:02:47.690
Let's think about another example.

00:02:47.690 --> 00:02:51.793
Let's think about xenon pentafluoride.

00:02:53.520 --> 00:02:57.870
Xenon pentafluoride cations,

00:02:57.870 --> 00:03:00.400
a positively charged ion here.

00:03:00.400 --> 00:03:01.260
Pause this video and see

00:03:01.260 --> 00:03:03.853
if you can draw the
Lewis diagram for this.

00:03:05.010 --> 00:03:06.080
All right, now let's do this together.

00:03:06.080 --> 00:03:07.170
If any of this seems unfamiliar,

00:03:07.170 --> 00:03:09.360
I encourage you to watch the video

00:03:09.360 --> 00:03:11.920
on introduction to drawing Lewis diagrams.

00:03:11.920 --> 00:03:13.830
But what we'd wanna do is first think

00:03:13.830 --> 00:03:16.360
about our valence electrons.

00:03:16.360 --> 00:03:19.890
So xenon right over here,
it's actually a noble gas.

00:03:19.890 --> 00:03:23.600
It already has a full
octet in its outer shell,

00:03:23.600 --> 00:03:26.120
so it has eight valence electrons.

00:03:26.120 --> 00:03:28.620
So xenon has eight valence electrons.

00:03:28.620 --> 00:03:31.640
And then fluorine, we've
seen this multiple times,

00:03:31.640 --> 00:03:34.160
has one, two, three, four, five, six,

00:03:34.160 --> 00:03:37.510
seven valence electrons,
but there's five of them.

00:03:37.510 --> 00:03:39.400
So five times seven.

00:03:39.400 --> 00:03:41.370
I'm gonna be drawing a
lot of electrons in this.

00:03:41.370 --> 00:03:44.440
So this gives us a total of eight plus 35,

00:03:44.440 --> 00:03:48.380
which is 43 valence electrons.

00:03:48.380 --> 00:03:50.120
But we have to be careful.

00:03:50.120 --> 00:03:51.350
This is a cation.

00:03:51.350 --> 00:03:53.640
It is a positively charged molecule.

00:03:53.640 --> 00:03:55.520
It has a positive one charge.

00:03:55.520 --> 00:03:58.350
So we have to take one
electron away because of that.

00:03:58.350 --> 00:04:03.350
So let's take away one valence
electron to get that cation.

00:04:03.380 --> 00:04:06.060
And so we are left with 42.

00:04:06.060 --> 00:04:08.700
42 valence electrons.

00:04:08.700 --> 00:04:11.180
So the next step is to
try to draw its structure

00:04:11.180 --> 00:04:13.800
with some basic single covalent bonds.

00:04:13.800 --> 00:04:16.470
And xenon would be our
preferred central atom

00:04:16.470 --> 00:04:19.060
because fluorine is more electronegative.

00:04:19.060 --> 00:04:21.510
It's actually the most
electronegative element.

00:04:21.510 --> 00:04:23.483
So let's put xenon in the middle,

00:04:24.410 --> 00:04:26.890
and then let's put some
fluorines around it,

00:04:26.890 --> 00:04:28.790
five of them to be specific.

00:04:28.790 --> 00:04:31.183
So one, two,

00:04:32.130 --> 00:04:34.860
three, four.

00:04:34.860 --> 00:04:36.440
I'm having trouble writing an F.

00:04:36.440 --> 00:04:39.900
Four and then five fluorines.

00:04:39.900 --> 00:04:42.510
And now let me make five covalent bonds.

00:04:42.510 --> 00:04:45.930
One, two, three, four, five.

00:04:45.930 --> 00:04:48.410
So just like that, I have
accounted for 10 valence electrons

00:04:48.410 --> 00:04:49.620
because you have two valence electrons

00:04:49.620 --> 00:04:50.850
in each of these covalent bonds,

00:04:50.850 --> 00:04:52.780
two, four, six, eight, 10.

00:04:52.780 --> 00:04:55.990
So let me subtract 10 valence electrons.

00:04:55.990 --> 00:05:00.020
And then we are left with
32 valence electrons.

00:05:00.020 --> 00:05:02.880
Now, the next step is to
try to allocate some more

00:05:02.880 --> 00:05:05.170
of these valence electrons
to the terminal atom

00:05:05.170 --> 00:05:06.780
so that they get to a full octet.

00:05:06.780 --> 00:05:08.340
So let me do that to the fluorines.

00:05:08.340 --> 00:05:10.510
Each of these fluorines
already are participating

00:05:10.510 --> 00:05:13.100
in a covalent bond, so they already have

00:05:13.100 --> 00:05:15.400
two valence electrons
hanging out with them,

00:05:15.400 --> 00:05:17.270
so let's give 'em each six more.

00:05:17.270 --> 00:05:21.080
So let's give that fluorine six,

00:05:21.080 --> 00:05:23.780
and that fluorine gets six,

00:05:23.780 --> 00:05:28.570
and that fluorine gets
six valence electrons,

00:05:28.570 --> 00:05:31.550
and that fluorine gets
six valence electrons,

00:05:31.550 --> 00:05:32.890
and then last but not least,

00:05:32.890 --> 00:05:35.610
this fluorine gets six valence electrons.

00:05:35.610 --> 00:05:38.750
So I have just given away
six valence electrons

00:05:38.750 --> 00:05:41.110
to each of five fluorine atoms.

00:05:41.110 --> 00:05:45.560
So that is 30 valence electrons
that I have just allocated.

00:05:45.560 --> 00:05:48.470
And then what does that leave me with?

00:05:48.470 --> 00:05:51.760
That leaves me with two valence electrons

00:05:51.760 --> 00:05:53.370
that have gone unallocated.

00:05:53.370 --> 00:05:56.560
And the only place to now
put them is on the xenon.

00:05:56.560 --> 00:05:58.440
And as I said, things that are lower down

00:05:58.440 --> 00:05:59.950
in that periodic table of elements,

00:05:59.950 --> 00:06:03.770
especially as we get
below the third period,

00:06:03.770 --> 00:06:06.440
these can defy the octet rule.

00:06:06.440 --> 00:06:08.820
Xenon already has 10 valence electrons,

00:06:08.820 --> 00:06:13.460
and I'm about to allocate it
two more to it just like that.

00:06:13.460 --> 00:06:15.360
So you allocate those two more.

00:06:15.360 --> 00:06:18.260
And then we have allocated
all of our valence electrons.

00:06:18.260 --> 00:06:22.730
And I wanna make sure I
remind myself and everyone

00:06:22.730 --> 00:06:24.700
that this is a cation.

00:06:24.700 --> 00:06:28.030
So I have to put that plus
charge just like this,

00:06:28.030 --> 00:06:31.050
but this is something
that has been observed

00:06:31.050 --> 00:06:34.990
where you can actually have
a central atom like this

00:06:34.990 --> 00:06:38.610
that goes beyond an octet
number of valence electrons.

00:06:38.610 --> 00:06:41.290
In this case, it has two, four, six,

00:06:41.290 --> 00:06:45.470
eight, 10, 12 valence electrons.

00:06:45.470 --> 00:06:48.560
Now, an interesting question
is how do these atoms

00:06:48.560 --> 00:06:51.690
that are in the third
period or beyond handle more

00:06:51.690 --> 00:06:53.570
than eight valence electrons?

00:06:53.570 --> 00:06:56.300
And it is a matter of debate,

00:06:56.300 --> 00:06:59.150
but some chemists believe
that it's possible

00:06:59.150 --> 00:07:01.560
because they're able to
place their electrons

00:07:01.560 --> 00:07:04.210
in their empty valence d-orbitals.

00:07:04.210 --> 00:07:05.890
But once again, this is controversial

00:07:05.890 --> 00:07:08.113
in the chemistry community.


Worked example: Lewis diagram of xenon difluoride (XeF₂)

WEBVTT
Kind: captions
Language: en

00:00:00.310 --> 00:00:01.530
- [Instructor] Let's do one more example

00:00:01.530 --> 00:00:03.460
of constructing a Lewis diagram

00:00:03.460 --> 00:00:05.680
that might be a little bit interesting.

00:00:05.680 --> 00:00:08.130
So let's say we wanted to
construct the Lewis structure

00:00:08.130 --> 00:00:11.963
or Lewis diagram for xenon difluoride.

00:00:13.589 --> 00:00:16.193
So pause this video and have a go of that.

00:00:17.070 --> 00:00:20.090
All right, now let's work
through this together.

00:00:20.090 --> 00:00:20.940
So first step,

00:00:20.940 --> 00:00:23.920
we just have to account
for the valence electrons.

00:00:23.920 --> 00:00:25.640
Xenon right over here.

00:00:25.640 --> 00:00:27.700
It is a noble gas.

00:00:27.700 --> 00:00:30.280
It has eight valence electrons.

00:00:30.280 --> 00:00:34.620
One, two, three, four,
five, six, seven, eight

00:00:34.620 --> 00:00:36.700
in that fifth shell.

00:00:36.700 --> 00:00:38.220
It's in the fifth period.

00:00:38.220 --> 00:00:40.930
So it has eight valence electrons.

00:00:40.930 --> 00:00:42.740
And then fluorine,

00:00:42.740 --> 00:00:44.850
we have looked at fluorine multiple times,

00:00:44.850 --> 00:00:47.210
we know that it has
seven valence electrons.

00:00:47.210 --> 00:00:50.830
One, two, three, four, five, six, seven

00:00:50.830 --> 00:00:52.760
in that second shell.

00:00:52.760 --> 00:00:54.450
And we have two of these fluorines.

00:00:54.450 --> 00:00:56.390
So two times seven.

00:00:56.390 --> 00:00:59.210
And then this gives us a total of

00:00:59.210 --> 00:01:02.430
eight plus 14 valence electrons

00:01:02.430 --> 00:01:07.430
which gets us to 22
valence electrons in total.

00:01:07.510 --> 00:01:09.900
Now the next step, and we've
done this multiple times,

00:01:09.900 --> 00:01:11.160
in multiple videos now,

00:01:11.160 --> 00:01:14.730
is we would try to draw the structure

00:01:14.730 --> 00:01:16.840
with some single covalent bonds

00:01:16.840 --> 00:01:20.250
and we would put xenon as our central atom

00:01:20.250 --> 00:01:23.560
because it is less
electronegative than fluorine.

00:01:23.560 --> 00:01:26.040
So let's put a xenon there.

00:01:26.040 --> 00:01:29.620
And let's put two
fluorines on either side.

00:01:29.620 --> 00:01:31.170
So fluorine there

00:01:31.170 --> 00:01:32.850
and a fluorine there.

00:01:32.850 --> 00:01:36.820
And let's set up some
single covalent bonds.

00:01:36.820 --> 00:01:39.870
And so how many of our valence electrons

00:01:39.870 --> 00:01:41.400
have we now accounted for?

00:01:41.400 --> 00:01:44.610
Well two in that bond and
then two in that bond.

00:01:44.610 --> 00:01:45.870
So we've accounted for four.

00:01:45.870 --> 00:01:48.670
So minus four valence electrons.

00:01:48.670 --> 00:01:52.970
We now have a total of
18 valence electrons.

00:01:52.970 --> 00:01:55.520
Now the next step is
we wanna allocate them

00:01:55.520 --> 00:02:00.520
to our terminal atoms and try
to get them to a full octet.

00:02:00.650 --> 00:02:05.170
Each of these fluorines already
have two valence electrons

00:02:05.170 --> 00:02:06.130
that they are sharing.

00:02:06.130 --> 00:02:08.380
So we need to give each of them six more.

00:02:08.380 --> 00:02:12.260
So two, four, six.

00:02:12.260 --> 00:02:15.660
Two, four, six.

00:02:15.660 --> 00:02:18.940
So I've just allocated 12
more valence electrons.

00:02:18.940 --> 00:02:21.970
So minus 12 valence electrons

00:02:21.970 --> 00:02:24.910
means that we still have
six valence electrons

00:02:24.910 --> 00:02:25.860
left to allocate.

00:02:25.860 --> 00:02:27.340
And there's only one place

00:02:27.340 --> 00:02:31.100
where we can allocate those
left over six valence electrons

00:02:31.100 --> 00:02:32.240
and that's at the central atom.

00:02:32.240 --> 00:02:33.500
At the xenon.

00:02:33.500 --> 00:02:34.333
So let's do that.

00:02:34.333 --> 00:02:39.320
So, two, four, and six.

00:02:39.320 --> 00:02:40.350
And there you have it.

00:02:40.350 --> 00:02:43.310
We have the Lewis diagram,
the Lewis structure

00:02:43.310 --> 00:02:45.340
for xenon difluoride.

00:02:45.340 --> 00:02:47.020
Now what's interesting here is

00:02:47.020 --> 00:02:50.300
our fluorines they have an
octet of valence electrons.

00:02:50.300 --> 00:02:52.120
But what's going on with xenon?

00:02:52.120 --> 00:02:56.180
Xenon has two, four, six, eight,

00:02:56.180 --> 00:02:59.370
10 valence electrons hanging around.

00:02:59.370 --> 00:03:00.840
So this is one of those examples

00:03:00.840 --> 00:03:02.880
of an exception to the octet rule

00:03:02.880 --> 00:03:05.600
where we go beyond eight valence electrons

00:03:05.600 --> 00:03:07.860
which is possible for elements

00:03:07.860 --> 00:03:10.173
in the third or higher period.


Worked example: Lewis diagram of the cyanide ion (CN⁻)

WEBVTT
Kind: captions
Language: en

00:00:00.210 --> 00:00:01.470
- [Instructor] In this
video we're gonna try to get

00:00:01.470 --> 00:00:04.210
more practice constructing Lewis diagrams.

00:00:04.210 --> 00:00:08.750
And we're gonna try to do
that for a cyanide anion.

00:00:08.750 --> 00:00:09.660
So this is interesting.

00:00:09.660 --> 00:00:11.280
This is the first time we're constructing

00:00:11.280 --> 00:00:13.960
a Lewis diagram for an ion.

00:00:13.960 --> 00:00:17.550
So pause this video and see
if you can have a go at that.

00:00:17.550 --> 00:00:20.780
All right, now let's do this together.

00:00:20.780 --> 00:00:23.030
So we've already seen in many videos,

00:00:23.030 --> 00:00:25.990
the first step is to essentially count

00:00:25.990 --> 00:00:29.210
the total valence electrons
that we're dealing with.

00:00:29.210 --> 00:00:30.610
And the reason why we
do that is to make sure

00:00:30.610 --> 00:00:32.940
that we're allocating all
the valence electrons.

00:00:32.940 --> 00:00:33.773
To help us there

00:00:33.773 --> 00:00:35.790
we can look at a periodic
table of elements.

00:00:35.790 --> 00:00:37.440
You might already know that carbon has

00:00:37.440 --> 00:00:40.100
one, two, three, four valence electrons

00:00:40.100 --> 00:00:42.670
in that second shell,
it's in the second period,

00:00:42.670 --> 00:00:45.010
so you have four valence
electrons from carbon.

00:00:45.010 --> 00:00:48.691
Nitrogen has one, two, three,
four, five valence electrons

00:00:48.691 --> 00:00:51.780
in its second shell, it's
in that second period.

00:00:51.780 --> 00:00:55.730
And so the valence electrons
from a neutral carbon

00:00:55.730 --> 00:00:57.730
and a neutral nitrogen-free atom

00:00:57.730 --> 00:01:01.310
would be a total of
nine valence electrons.

00:01:01.310 --> 00:01:02.730
But we are not done yet.

00:01:02.730 --> 00:01:04.910
Because this is not a neutral molecule.

00:01:04.910 --> 00:01:06.370
We have a negative charge here.

00:01:06.370 --> 00:01:08.630
It is an anion, it has
a negative one charge.

00:01:08.630 --> 00:01:10.460
And so because of that negative one

00:01:10.460 --> 00:01:13.630
we can think about it having
an extra valence electron.

00:01:13.630 --> 00:01:16.200
So let's add a valence electron here.

00:01:16.200 --> 00:01:17.180
Why do we do it?

00:01:17.180 --> 00:01:19.056
Because of this negative charge.

00:01:19.056 --> 00:01:24.056
So we're dealing with a total
of 10 valence electrons.

00:01:24.100 --> 00:01:28.190
Now, the next step is to
try to draw single bonds.

00:01:28.190 --> 00:01:33.190
Try single bonds, and
identify a central atom.

00:01:33.860 --> 00:01:35.770
Now, we only have two atoms here,

00:01:35.770 --> 00:01:39.300
so really neither feels central,

00:01:39.300 --> 00:01:41.440
so let me just put a carbon

00:01:41.440 --> 00:01:44.090
and a nitrogen next to each other here.

00:01:44.090 --> 00:01:47.830
And then let me draw one single bond.

00:01:47.830 --> 00:01:49.530
So by drawing that one single bond

00:01:49.530 --> 00:01:52.544
I have now accounted for
two valence electrons.

00:01:52.544 --> 00:01:56.440
So now I am left with
eight valence electrons,

00:01:56.440 --> 00:01:58.130
and so that's the next step,

00:01:58.130 --> 00:02:00.330
allocate remaining valence electrons,

00:02:00.330 --> 00:02:03.690
allocate valence electrons.

00:02:03.690 --> 00:02:05.950
So let me start with the
more electronegative.

00:02:05.950 --> 00:02:07.340
Let's try to get nitrogen to eight.

00:02:07.340 --> 00:02:08.680
It already has two.

00:02:08.680 --> 00:02:11.070
So let's give it three more lone pairs.

00:02:11.070 --> 00:02:16.070
So we have two, four, six, eight.

00:02:16.230 --> 00:02:19.350
So I have just used up six

00:02:19.350 --> 00:02:23.160
of these remaining valence
electrons, six, so minus six,

00:02:23.160 --> 00:02:25.860
so I have two left to allocate.

00:02:25.860 --> 00:02:30.390
So let me give carbon two
valence electrons, like that.

00:02:30.390 --> 00:02:33.658
And there I have used up all of my,

00:02:33.658 --> 00:02:36.640
all of my valence electrons.

00:02:36.640 --> 00:02:38.830
Now let's see how happy everyone is.

00:02:38.830 --> 00:02:41.790
Nitrogen has eight valence
electrons hanging around,

00:02:41.790 --> 00:02:44.070
two, four, six, eight.

00:02:44.070 --> 00:02:47.550
But carbon only has four, two and four.

00:02:47.550 --> 00:02:49.200
So this is where we think about

00:02:49.200 --> 00:02:53.500
whether we would want to
have some extra bonds,

00:02:53.500 --> 00:02:57.470
extra bonds, or higher-order bonds.

00:02:57.470 --> 00:03:00.227
So how can we give carbon
more valence electrons?

00:03:00.227 --> 00:03:01.710
Well, what we could do is

00:03:01.710 --> 00:03:03.800
we could take some of these
lone pairs around nitrogen

00:03:03.800 --> 00:03:06.310
and then use them, turn
this single covalent bond

00:03:06.310 --> 00:03:08.270
into a higher-order bond.

00:03:08.270 --> 00:03:11.190
So let's see, if we were to take these two

00:03:11.190 --> 00:03:14.320
and turn it into another covalent bond,

00:03:14.320 --> 00:03:15.810
what is going to happen?

00:03:15.810 --> 00:03:17.560
Let me erase all of these,

00:03:17.560 --> 00:03:22.560
and then I'll just draw
another covalent bond.

00:03:25.430 --> 00:03:28.360
So nitrogen still has eight
electrons hanging around.

00:03:28.360 --> 00:03:29.820
Carbon now has six.

00:03:29.820 --> 00:03:32.550
So maybe we can do that again.

00:03:32.550 --> 00:03:36.150
So let me erase these two characters.

00:03:36.150 --> 00:03:38.670
Let me erase these two characters

00:03:38.670 --> 00:03:41.380
and make another covalent
bond out of them.

00:03:41.380 --> 00:03:46.110
So let me make a covalent
bond out of them.

00:03:46.110 --> 00:03:48.050
And so now what's going on?

00:03:48.050 --> 00:03:50.930
Carbon has two, four, six, eight

00:03:50.930 --> 00:03:52.380
valence electrons hanging around.

00:03:52.380 --> 00:03:55.230
Nitrogen has two, four, six, eight

00:03:55.230 --> 00:03:57.370
valence electrons hanging around.

00:03:57.370 --> 00:03:59.125
So this is looking pretty good.

00:03:59.125 --> 00:04:01.410
But are we done yet?

00:04:01.410 --> 00:04:02.670
The simple answer is no.

00:04:02.670 --> 00:04:05.810
We still haven't represented
this negative charge

00:04:05.810 --> 00:04:07.490
in our Lewis diagram.

00:04:07.490 --> 00:04:09.130
The way that we would do that is say hey,

00:04:09.130 --> 00:04:11.850
this entire molecule, you
put brackets around it,

00:04:11.850 --> 00:04:13.840
has a negative charge.

00:04:13.840 --> 00:04:15.100
And now we're done.

00:04:15.100 --> 00:04:17.640
We've allocated all of
our valence electrons,

00:04:17.640 --> 00:04:20.380
we have our octet rule on all of our atoms

00:04:20.380 --> 00:04:22.260
that are not hydrogen,
there's no hydrogen here,

00:04:22.260 --> 00:04:25.360
and we're showing that
this indeed is an anion,

00:04:25.360 --> 00:04:27.053
and now we are done.


Drawing Lewis diagrams

WEBVTT
Kind: captions
Language: en

00:00:00.540 --> 00:00:01.830
- [Sal] In this video we're going

00:00:01.830 --> 00:00:04.530
to think about constructing
Lewis diagrams,

00:00:04.530 --> 00:00:06.000
which you've probably seen before.

00:00:06.000 --> 00:00:07.840
They're nice ways of visualizing

00:00:07.840 --> 00:00:10.980
how the atoms in a molecule
are bonded to each other

00:00:10.980 --> 00:00:14.580
and what other lone pairs
of valence electrons

00:00:14.580 --> 00:00:16.780
various atoms might have.

00:00:16.780 --> 00:00:18.820
And so let's just start with an example,

00:00:18.820 --> 00:00:20.570
then we'll come up with some rules

00:00:20.570 --> 00:00:23.100
for trying to draw these Lewis diagrams.

00:00:23.100 --> 00:00:25.300
So the first example that we will look at

00:00:25.300 --> 00:00:30.300
is silicon tetrafluoride,
and tetrafluoride

00:00:30.580 --> 00:00:32.860
is just a fancy way of
saying four fluorines,

00:00:32.860 --> 00:00:33.903
so tetrafluoride.

00:00:35.810 --> 00:00:37.297
Now the first step is to say,

00:00:37.297 --> 00:00:40.400
"Well, what are the electrons
that are of interest to us?"

00:00:40.400 --> 00:00:41.873
And if we're talking about the electrons

00:00:41.873 --> 00:00:43.830
that are likely to react, we're talking

00:00:43.830 --> 00:00:47.790
about the valence electrons,
so V.E. for short,

00:00:47.790 --> 00:00:49.240
valence electrons.

00:00:49.240 --> 00:00:51.520
So first let's think about how
many total valence electrons

00:00:51.520 --> 00:00:55.020
are involved in silicon tetrafluoride.

00:00:55.020 --> 00:00:56.390
Well, to think about that, we could think

00:00:56.390 --> 00:00:59.130
about how many valence
electrons does silicon have,

00:00:59.130 --> 00:01:00.490
and then how many valence electrons

00:01:00.490 --> 00:01:02.910
does each of the fluorines have

00:01:02.910 --> 00:01:05.490
if they were just free atoms and neutral,

00:01:05.490 --> 00:01:06.960
and then multiply that times four,

00:01:06.960 --> 00:01:08.080
'cause you have four fluorines.

00:01:08.080 --> 00:01:11.140
So let's get out our
periodic table of elements,

00:01:11.140 --> 00:01:14.380
and then you can see here that silicon,

00:01:14.380 --> 00:01:16.690
its outer shell is the third shell,

00:01:16.690 --> 00:01:19.096
and in that third shell it has one, two,

00:01:19.096 --> 00:01:22.620
three, four valence electrons.

00:01:22.620 --> 00:01:26.520
So silicon here has
four valence electrons,

00:01:26.520 --> 00:01:29.270
and then to that, we're going
to add the valence electrons

00:01:29.270 --> 00:01:31.040
from the four fluorines.

00:01:31.040 --> 00:01:34.430
A free, neutral fluorine atom,

00:01:34.430 --> 00:01:36.210
its outer shell is the second shell,

00:01:36.210 --> 00:01:38.680
and in that outer shell, it has one, two,

00:01:38.680 --> 00:01:42.080
three, four, five, six, seven electrons.

00:01:42.080 --> 00:01:45.320
So each of these fluorines
has seven valence electrons,

00:01:45.320 --> 00:01:46.830
but there are four of them.

00:01:46.830 --> 00:01:49.790
So one silicon tetrafluoride molecule

00:01:49.790 --> 00:01:53.840
is gonna have four plus
28 valence electrons.

00:01:53.840 --> 00:01:57.743
So this is going to be a total of 32.

00:01:58.640 --> 00:02:00.630
Now the next step is to think

00:02:00.630 --> 00:02:03.490
about how might these be configured?

00:02:03.490 --> 00:02:05.020
And as a general rule of thumb,

00:02:05.020 --> 00:02:07.370
we'd wanna put the least
electronegative atom

00:02:07.370 --> 00:02:09.720
that is not hydrogen at the center.

00:02:09.720 --> 00:02:11.870
And we've talked about this before,

00:02:11.870 --> 00:02:14.070
but you can even see from the
periodic table of elements,

00:02:14.070 --> 00:02:17.110
fluorine is actually the
most electronegative element,

00:02:17.110 --> 00:02:20.680
and so we would at least try
to put silicon at the center

00:02:20.680 --> 00:02:23.140
and make fluorine a terminal atom,

00:02:23.140 --> 00:02:24.630
something on the outside.

00:02:24.630 --> 00:02:26.070
So let's try to do that.

00:02:26.070 --> 00:02:29.200
So let's put silicon in the center,

00:02:29.200 --> 00:02:31.870
and then we have to put the
four fluorines some place.

00:02:31.870 --> 00:02:34.880
Let's just put one fluorine
there, one fluorine there,

00:02:34.880 --> 00:02:38.780
one fluorine there,
and one fluorine there.

00:02:38.780 --> 00:02:40.820
Now the next step is, well let's just say

00:02:40.820 --> 00:02:43.230
for simplicity that we
just have single bonds

00:02:43.230 --> 00:02:45.530
between the silicon and
each of the fluorines.

00:02:45.530 --> 00:02:46.990
So let's do that.

00:02:46.990 --> 00:02:50.770
So one bond, a bond, a bond, a bond.

00:02:50.770 --> 00:02:52.700
Now each of these covalent bonds,

00:02:52.700 --> 00:02:55.380
each of these lines in our Lewis diagram,

00:02:55.380 --> 00:02:57.190
they represent two electrons.

00:02:57.190 --> 00:03:00.590
So for example, this one right over here

00:03:00.590 --> 00:03:03.590
that I'm doing in yellow,
that represents two electrons

00:03:03.590 --> 00:03:06.460
that are shared by this
fluorine and this silicon.

00:03:06.460 --> 00:03:08.260
This represents another two electrons

00:03:08.260 --> 00:03:10.830
that is shared between this
fluorine and the silicon.

00:03:10.830 --> 00:03:12.235
This is another two
electrons that's shared

00:03:12.235 --> 00:03:14.787
between this fluorine and this silicon.

00:03:14.787 --> 00:03:16.980
And this is another two electrons

00:03:16.980 --> 00:03:19.910
shared between that
fluorine and the silicon.

00:03:19.910 --> 00:03:22.900
So, so far, how many electrons
have we accounted for?

00:03:22.900 --> 00:03:25.240
Well, each of these
represent two electrons,

00:03:25.240 --> 00:03:28.520
so two, four, six, eight electrons.

00:03:28.520 --> 00:03:31.550
So if we subtract eight
from this, we are left

00:03:31.550 --> 00:03:36.550
with 24 electrons to account
for, 24 valence electrons.

00:03:37.830 --> 00:03:40.660
So now, our general
rule of thumb would be,

00:03:40.660 --> 00:03:43.150
try to put those on those terminal atoms

00:03:43.150 --> 00:03:45.130
with the goal of getting
those terminal atoms

00:03:45.130 --> 00:03:47.620
to having eight valence electrons.

00:03:47.620 --> 00:03:49.220
In general we try to get the octet rule

00:03:49.220 --> 00:03:50.720
for any atom except for hydrogen.

00:03:50.720 --> 00:03:53.300
Hydrogen, you just need to get
to two in that outer shell.

00:03:53.300 --> 00:03:54.760
But fluorine, you want to get it to eight.

00:03:54.760 --> 00:03:57.500
It already has two that it can share,

00:03:57.500 --> 00:03:59.970
so it needs six more, so let's add that.

00:03:59.970 --> 00:04:03.460
Two, four, six.

00:04:03.460 --> 00:04:05.140
Let's do that again for this fluorine.

00:04:05.140 --> 00:04:08.860
Two, four, six.

00:04:08.860 --> 00:04:10.430
Do it again for this fluorine.

00:04:10.430 --> 00:04:13.780
Two, four, six.

00:04:13.780 --> 00:04:15.570
And then last but not
least, for this fluorine.

00:04:15.570 --> 00:04:19.450
Two, four, and six.

00:04:19.450 --> 00:04:21.580
Now how many more electrons
are now accounted for?

00:04:21.580 --> 00:04:24.190
Well, six in this fluorine,
six in this fluorine,

00:04:24.190 --> 00:04:26.520
six in this fluorine,
six in this fluorine,

00:04:26.520 --> 00:04:28.860
so six times four, we've now accounted

00:04:28.860 --> 00:04:30.960
for 24 more electrons.

00:04:30.960 --> 00:04:34.360
We've now used up all of
the valence electrons.

00:04:34.360 --> 00:04:36.890
Now that's good, because
we wanted to account

00:04:36.890 --> 00:04:38.270
for all of the valence electrons.

00:04:38.270 --> 00:04:41.020
We wanna represent them
somehow in this Lewis diagram.

00:04:41.020 --> 00:04:44.450
The next thing to check
for is how satisfied

00:04:44.450 --> 00:04:48.670
the various atoms are
relative to to the octet rule.

00:04:48.670 --> 00:04:50.530
We've already seen that the fluorines

00:04:50.530 --> 00:04:51.840
are feeling pretty good.

00:04:51.840 --> 00:04:55.290
They each have six electrons
that are not in a bond,

00:04:55.290 --> 00:04:57.360
and then they're able
to share two electrons

00:04:57.360 --> 00:04:58.806
that are in a bond, so each of them

00:04:58.806 --> 00:05:01.720
can kind of feel like they
have eight outer electrons,

00:05:01.720 --> 00:05:03.940
eight valence electrons
hanging out with them.

00:05:03.940 --> 00:05:07.610
And then the silicon is
able to share in four bonds.

00:05:07.610 --> 00:05:10.150
Each of those bonds have two electrons,

00:05:10.150 --> 00:05:13.710
so the silicon is also feeling
good about the octet rule.

00:05:13.710 --> 00:05:15.730
So I would feel very confident

00:05:15.730 --> 00:05:18.790
in this being the Lewis diagram,

00:05:18.790 --> 00:05:20.330
sometimes called the Lewis structure,

00:05:20.330 --> 00:05:22.340
for silicon tetrafluoride.

00:05:22.340 --> 00:05:24.860
So just to hit the point
home on what we just did,

00:05:24.860 --> 00:05:26.360
I will give you these steps,

00:05:26.360 --> 00:05:28.400
but hopefully you find
them pretty intuitive.

00:05:28.400 --> 00:05:30.144
That's why I didn't wanna
show you from the beginning.

00:05:30.144 --> 00:05:32.410
But as you see, step one was,

00:05:32.410 --> 00:05:34.250
find the total number
of valence electrons.

00:05:34.250 --> 00:05:35.083
We did that.

00:05:35.083 --> 00:05:36.340
That's the four from silicon

00:05:36.340 --> 00:05:38.640
and then the 28 from the fluorines.

00:05:38.640 --> 00:05:41.210
It says add an electron
for every negative charge.

00:05:41.210 --> 00:05:43.290
Subtract an electron for
every positive charge.

00:05:43.290 --> 00:05:44.780
We didn't have to do that in this example

00:05:44.780 --> 00:05:46.650
because it's a neutral molecule.

00:05:46.650 --> 00:05:48.630
Then it says decide the central atom,

00:05:48.630 --> 00:05:51.360
which should be the electronegative
except for hydrogen.

00:05:51.360 --> 00:05:52.510
That's why we picked silicon,

00:05:52.510 --> 00:05:55.100
because fluorine is the
most electronegative atom.

00:05:55.100 --> 00:05:57.290
And then we drew the bonds.

00:05:57.290 --> 00:06:00.610
We saw that the bonds
accounted for eight electrons,

00:06:00.610 --> 00:06:02.220
and we subtracted those electrons

00:06:02.220 --> 00:06:04.660
from the total in step
one, and that's just

00:06:04.660 --> 00:06:06.933
to keep track of the
number of valence electrons

00:06:06.933 --> 00:06:09.340
that we are accounting for.

00:06:09.340 --> 00:06:11.200
And then we had 24 left over.

00:06:11.200 --> 00:06:13.850
And then the next step, it says
assign the valence electrons

00:06:13.850 --> 00:06:15.180
to the terminal atoms.

00:06:15.180 --> 00:06:18.160
That's where we assigned these
extra lone pair electrons

00:06:18.160 --> 00:06:21.370
to the various fluorines,
giving them an extra six each

00:06:21.370 --> 00:06:24.530
so that they were all able
to fulfill the octet rule.

00:06:24.530 --> 00:06:26.820
And then we subtracted
that from the total,

00:06:26.820 --> 00:06:28.090
really just to account, to make sure

00:06:28.090 --> 00:06:29.890
that we're using all of our electrons.

00:06:29.890 --> 00:06:31.610
It says it right here:
subtract the electrons

00:06:31.610 --> 00:06:32.850
from the total in step two.

00:06:32.850 --> 00:06:35.070
And then we saw that all of our electrons

00:06:35.070 --> 00:06:36.230
were accounted for.

00:06:36.230 --> 00:06:38.710
But then in step four,
it says if necessary,

00:06:38.710 --> 00:06:41.350
assign any leftover electrons
to the central atom.

00:06:41.350 --> 00:06:43.230
We didn't have to do that in this example.

00:06:43.230 --> 00:06:45.110
If the central atom has an octet

00:06:45.110 --> 00:06:47.540
or exceeds an octet, you are usually done.

00:06:47.540 --> 00:06:50.400
In this case, it had an
octet, so we felt done.

00:06:50.400 --> 00:06:52.110
And it finally says, if a central atom

00:06:52.110 --> 00:06:55.200
does not have an octet,
create multiple bonds.

00:06:55.200 --> 00:06:56.820
Once again, in this example we were able

00:06:56.820 --> 00:06:59.000
to stay pretty simple
with just single bonds.

00:06:59.000 --> 00:07:00.620
But in future examples, we're going to see

00:07:00.620 --> 00:07:03.930
where we might have to do some
of these more nuanced steps.

00:07:03.930 --> 00:07:05.350
So I will leave you there,

00:07:05.350 --> 00:07:07.200
and I'll see you in the next example.