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6.05b Determine the formal charge on atoms in H2CO

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化学

肯塔基大学

4.7（663 個評分） | 67K 名學生已註冊

This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

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從本節課中

Week 6

This unit looks more in depth at molecular compounds to see how they are bonded together by looking at Lewis structures. These structures provide information about the types of bonds (single, double, or triple) as well as the connectivity of atoms. By knowing the Lewis structure, we can also predict the three-dimensional geometry of an individual molecule.

6.01 Lewis symbols 4:36

6.02 Covalent bonds 5:19

6.03 Electronegativity 11:21

6.04 Lewis Structures Part 1 13:11

6.04 Lewis Structures Part 2 18:49

6.04a Draw the Lewis structure for H2O 1:13

6.04b Draw the Lewis structure for H2CO 2:00

6.04c Draw the Lewis structure for CCL4 1:21

6.04d Draw the Lewis structure for NH3 1:16

6.05 Resonance & Formal Charge 16:27

6.05a Determine the formal charg on atoms in NH4+ 2:24

6.05b Determine the formal charge on atoms in H2CO 3:17

6.06 Exceptions to the Octet Rule 8:48

6.06a Draw the Lewis structure for I3- 1:16

6.06b Draw the Lewis structure for ClF4 1:41

6.06c Draw the Lewis structure for XeF4 1:23

6.07 VSEPR Part 1 11:01

6.07 VSEPR Part 217:07

6.07a Determine the electron pair and molecular geometries for I3- 1:42

6.07b Determine the electron pair and molecular geometries for NH3 1:22

6.07c Determine the electron pair and molecular geometries for ICl4- 1:57

6.07d Determine the electron pair and molecular geometries for PF5 1:52

6.07e Determine the electron pair and molecular geometries for XeF2 2:16

教學方

Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer

腳本

Determine the formal charge on atoms in H2CO.

So the first thing I need to do is draw my Lewis structure,

because formal charge depends on knowing how atoms are connected to one another.

So I'm going to first count the number of valence electrons I have in the molecule.

For hydrogen I have two hydrogen atoms, each with one valence, plus four for

the carbon, plus six for the oxygen.

And so I have a total of 12 electrons in my structure.

I see that carbon is going to be a central atom.

Hydrogen always has to be terminal atoms, and oxygen will

be a terminal atom because carbon is less electro-negative than oxygen.

So I've filled out my skeletal structure, I've used up six of my electrons I have

six electrons remaining, so I'm going to fill in the octet on my terminal atoms so

it's really only the oxygen I have to worry about the hydrogen only wants two

electrons so it's perfectly fine there.

Now I see that, while my oxygen has an octet, my carbon does not and so

I need to move electrons from that oxygen atom into

forming a double bond here between the carbon and the oxygen.

So I'm going to redraw this structure, so I have an oxygen now with only two lone

pairs of electrons, with a double bond to carbon and a single bond to hydrogen.

Now I've used up my 12 electrons.

And I have everybody with an octet, well for hydrogen it has a duet.

So now that gives me the Lewis structure for H2CO.

And I can use that now to find the formal charges.

So now I need to look at each atom to determine the formal charge on each of

these atoms.

So I'm going to start with my hydrogen atom, and

what I look at is the number of electrons assigned to hydrogen in an isolated atom,

which would just be one electron.

I subtract the number of non bonding electrons assigned to it in

the Lewis Structure which is zero in this case, and half of the bonding electrons.

So it has two bonding electrons around it I take half of that, and

I see that I get a formal charge of zero for my hydrogen.

So we’re going to put a little zero there to record that.

I notice that it's going to be the same for both hydrogens, because they both have

the same type of structure in this particular Lewis structure.

Now, I'm going to look at my carbon atom.

It would have four electrons in the isolated atom.

It has no non bonding electrons.

And it has eight electrons around it.

So I take half of eight and

that's going to give me a formal charge of zero as well.

Now we can look at the oxygen atom.

[SOUND] And we see that in oxygen,

it has six electrons assigned to it in an isolated atom.

We subtract off the number of non binding electrons assigned to it in the structure

which in this case is four, cause we have two pairs of non binding electrons,

minus half of our bonding electrons, so

we have four bonding, and I also get a formal charge of zero.

And this is actually as good as it gets to have formal charges all of zero.

Remember we always want to get those values as close to zero as possible, and

we can't get any closer than this.

And that just indicates is that we have a very stable structure here.

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