Lewis structure

[Rachel Maddiee]

Is this the correct explanation and Lewis structure for ch2O?

 

[hypervalent_iodine]

I assume that this is referring to a Lewis structure you have not shown here? Could you possibly post a picture of it so I can understand if your answer is correct?  I would note that you have said in your answer that the C had a lone pair and 3 bonds, yet only 7 electrons around it - this would not be true based on your description (2 electrons per bond and 2 per lone pair = 8). I would also be careful of your use of the term, "need." Yes, the oxygen and carbon atoms do require that many in this instance so as to minimise formal charge, but don't forget that they can have more or less than those numbers (except carbon, which cannot have more). Consider carbanions, for example, or oxoniums.  

The hand-drawn Lewis structure given underneath the answer is fine. 

 

 

[Rachel Maddiee]

Here is the question. So what changes do I need to make in my description?

[hypervalent_iodine]

Oh how I hate when institutions teach students to draw bonds as dots. 

17 minutes ago, Rachel Maddiee said:

So what changes do I need to make in my description?

What do you think? Most of your description was fine, but your parts about how the octet rule is not satisfied needs to be re-looked at. Count the electrons (dots) around the C and O. What do you know about formal charge and Lewis structures? What charge would the C and O have in that picture? Is that good or bad? 

[MigL]

I'm actually trained in Physics and know very little Chemistry, so take this with a grain of salt.

Look at the periodic table.
Hydrogen has one electron in its outer orbital, the 1S orbital, which needs two to be filled.
Carbon has four electrons in its outer orbital, the 2S orbital, which needs eight to be filled.
Oxygen has six electrons in its outer orbital, also 2S, and also needs eight.

To achieve the lowest energy configuration, all the outer valence orbitals want to be filled.
And the only way for this to happen is the Carbon shares two electrons with the two Hydrogens, and shares four electrons with the Oxygen, IOW, a double bond.

Now I don't really know how to draw Lewis diagrams ( my last Chem. class was Gr. 13 in 1976/77 ) but I would think, if you doubled the dots between the C::O ( like so ), and got rid of the dots under the C and O, all conditions would be satisfied.

However, could be I have no clue, so I defer to Hyper's expertise.

[Rachel Maddiee]

Note that these are valence bonds, not ionic bonds,
so the valence electrons are "shared" between the bonded atoms
rather than being gained by one atom and lost from another.
So would it be an incomplete octet?

oxygen has four non-bonding lone pairs remaining in its outer shell, so it will have two lone pairs.

 

[hypervalent_iodine]

51 minutes ago, MigL said:

Now I don't really know how to draw Lewis diagrams ( my last Chem. class was Gr. 13 in 1976/77 ) but I would think, if you doubled the dots between the C::O ( like so ), and got rid of the dots under the C and O, all conditions would be satisfied.

Spot on. This is more or less what Rachel drew in the picture in the OP, except using the (much more correct) bond notation of lines instead of dots. Many high schools will teach students to draw bonds using dots and x's when teaching Lewis structures, but it is an awful habit and can lead to confusion later on. We will usually deduct marks from students for using this notation when they get to first year (they are told not to). In any case, I don't believe knowing how to draw the correct configuration is the issue here.

You have hinted at the real problem with Rachel's description in this sentence:

51 minutes ago, MigL said:

To achieve the lowest energy configuration, all the outer valence orbitals want to be filled.

And the only way for this to happen is the Carbon shares two electrons with the two Hydrogens, and shares four electrons with the Oxygen, IOW, a double bond.

The way students are generally shown to think about this is by adding up the valence electrons, and matching that up with the Lewis structure. So I don't completely give the answer away, we can consider formic acid (HCO₂H) as an example:

(Image mine, drawn in ChemDraw)

It becomes slightly more complicated when you get to hypervalent compounds (such as (bis(trifluoroacetoxy)iodo)benzene, which is where my username comes from), but only in the sense that you have to consider the possibility of 10 or 12 valence electrons. 

8 minutes ago, Rachel Maddiee said:

Note that these are valence bonds, not ionic bonds,
so the valence electrons are "shared" between the bonded atoms
rather than being gained by one atom and lost from another.
So would it be an incomplete octet?

oxygen has four non-bonding lone pairs remaining in its outer shell, so it will have two lone pairs.

 

 

These are covalent bonds, no one is suggesting that they are ionic. Look at the structure again and count the number of electrons around the carbon and then the oxygen. How many do you see?

[Rachel Maddiee]

I’ve been having trouble with the description for this. So carbon has 8 electrons and oxygen has 6?

[hypervalent_iodine]

10 minutes ago, Rachel Maddiee said:

I’ve been having trouble with the description for this. So carbon has 8 electrons and oxygen has 6?

No. I think the problem you are having is that when you count the two electrons between the carbon and oxygen in the question, you are only counting them once for one of the atoms and then treating it as though that atom has sole ownership over them, which is not the case. The question you have to ask yourself when considering the octet rule is how many electrons is each atom surrounded by individually. You are not asked to consider how many electrons the atom possesses in a formal sense (i.e. both electrons from a lone pair and one from each bond), just how many electrons are around it total.

Look at each of them separately, count two electrons for each bond that atom has and two electrons for each lone pair. For carbon, yes, that makes 8 (2 from the lone pair and 6 from each bond). Similarly, for both hydrogens you should count 2 from the bond to the carbon. However, that also makes 8 for oxygen (6 from lone pairs and 2 from the bond to the carbon). Note that this double counting is only when you are looking to see if the octet rule has been satisfied. If you are then counting the total number of valence electrons, you obviously only count the electrons once. Does it help you to visualise it if the structure from your question is redrawn like this?:

 

 

[Rachel Maddiee]

so Since the oxygen and carbon end up with full shells, do they hold to the rule?

[hypervalent_iodine]

Just now, Rachel Maddiee said:

so Since the oxygen and carbon end up with full shells, do they hold to the rule?

Yes, they do. However, the Lewis structure is still incorrect (for other reasons). Have you counted the total number of valence electrons in the Lewis structure? How does that compare to the total number of valence electrons available? 

[Rachel Maddiee]

I know that Carbon has 4 valence electrons and oxygen has 6..

I’m not sure if I’m counting the 6 from the lone pair? 

[hypervalent_iodine]

1 minute ago, Rachel Maddiee said:

I know that Carbon has 4 valence electrons and oxygen has 6..

I’m not sure if I’m counting the 6 from the lone pair? 

You need to do two calculations here, one where you calculate the total number of valence electrons available, and other where you count all of the valence electrons in the Lewis structure (i.e. all of the ones in bonds and lone pairs). If the Lewis structure is correct, the numbers must be the same. If we look at my previous example in my post of formic acid, HCO₂H. Each hydrogen has 1 valence electron, the carbon has 4 and the oxygens 6. Thus, there is a (1x2)+(6x2)+(4x1) valence electrons, which comes to a total of 18 valence electrons available. If we then look at the number of electrons present in structure A, we have 6 lone pairs and 4 bonds with 2 electrons in each, which is equal to a total of 20 valence electrons in the Lewis structure. The numbers don't match, so the Lewis structure (A) is wrong. See if you can apply that to your molecule. 

[Rachel Maddiee]

how do I determine which rule it does not follow?

[hypervalent_iodine]

7 minutes ago, Rachel Maddiee said:

how do I determine which rule it does not follow?

I don’t totally understand the question. You have to check both things - the total number of electrons and whether or not the octet is satisfied. They’re related but separate things in the sense that one being correct doesn’t mean the other will be. 

[Rachel Maddiee]

and counting the valence electrons will help me determine which rule is incorrect?

I’m trying but I don’t understand how you calculated to find the valence electrons 

Edited January 7, 2020 by Rachel Maddiee

[hypervalent_iodine]

I’m still not totally getting your question. There are two things you have to figure out once you have the basic structure down 1) is the octet rule satisfied; and 2) do you have the correct number of electrons? Counting the electrons around each atom in the Lewis structure will help you determine whether or not the octet rule is satisfied. Tallying up the total number of valence electrons based on molecular formula and comparing that to the total number of valence electrons in the entire Lewis structure will tell you if you have too many or not enough electrons. 
 

How many electrons are in the Lewis structure of your compound? How many should there be? Is the octet rule satisfied?

[Rachel Maddiee]

You said that you only count the valence electrons once. How do I do that?

Number of electrons = 6 + 8 = 14 

CH2O There are 4 valence electrons in carbon, 1 each in hydrogen and 6 in oxygen, so there are 12 electrons total.

[hypervalent_iodine]

44 minutes ago, Rachel Maddiee said:

You said that you only count the valence electrons once. How do I do that?

Number of electrons = 6 + 8 = 14 

CH2O There are 4 valence electrons in carbon, 1 each in hydrogen and 6 in oxygen, so there are 12 electrons total.

Sorry, I seem to have caused confusion or you have misunderstood me. When I said you only count the valence electrons once, I meant when you look at the Lewis structure and you are tallying up electrons you only count each electron once. You still have to compare that number to the total number of valence electrons from the molecular formula. 
 

When you are determining whether or not the octet rule has be obeyed, you have to look at each atom separately and look at how many electrons are around it. The oxygen has 6 in lone pairs and 2 from the bond to the C, which makes 8. The carbon has 2 each from the bonds to the hydrogens, 2 from the bond to the oxygen, and 2 from the lone pair, which again makes 8. Each hydrogen has 2 electrons around it from the bond to the carbon. They all check out. 
 

Then you have to count all of the electrons in the bonds and lone pairs and see if they match with the total number of valence electrons as per the molecular formula. You’re half way with this. You now need to count up all of the electrons in bonds and lone pairs in the Lewis structure (these are the valence electrons in the Lewis structure) and check that it matches with the number of valence electrons you have calculated from the molecular formula. Is it the same number?
 

These are two separate actions and you should do both to check that the Lewis structure is correct. 

[Rachel Maddiee]

Would there be 14 total valence electrons?

 

the numbers don’t match 

Edited January 7, 2020 by Rachel Maddiee

[studiot]

12 hours ago, hypervalent_iodine said:

Oh how I hate when institutions teach students to draw bonds as dots. 

As hypervalentiodine says there are many problems associated with Lewis structures.

Originally they were simple dot diagrams, but some authors now connect this to also show bonds as valency lines.

I suggest you stick to the dots.

Because of the many exceptions the dot structure is taught and then students move on.

So I also suggest you do enough to earn the marks (since you must and HI doesn't need to) but remember better methods follow.

Here is a presentation going only as far as you need to answer your question (which it does).

 

This does include Hypervalent's rules as she already set out, but the slightly different explanation may help.

 

 

[Rachel Maddiee]

this is really confusing me. This is what I’m up to 

 

Incorrect structure: 8 bonding electrons and 2 lone pairs Number of electrons = 6 + 8 = 14 total electrons 

Correct structure; CH2O There are 4 valence electrons in carbon, 1 each in hydrogen and 6 in oxygen, so there are 12 electrons total.

In the structure carbon has one lone pair and oxygen has three lone pairs which is incorrect.

[studiot]

57 minutes ago, Rachel Maddiee said:

8 bonding electrons and 2 lone pairs Number of electrons = 6 + 8 = 14 total electrons 

2 lone pairs  = 4 electrons

plus the 8 bonding electrons.

 

So where do you get 6 + 8 from?

[Rachel Maddiee]

sorry, I counted wrong.. Would I add them together? 2+4+ 8

I did 3 bond pairs=6 elections and 4 lone pairs=8 elections.. total 14 elections

Oxygen has one bond pair and 3 lonepairs.

Carbon has thrree bonds and a lone pair.

 

[Rachel Maddiee]

The Lewis structure is not correct. There should be two pairs of electrons (a double bond) between the C and O, and no lone pair on the C. Carbon needs 4 bonds and Oxygen 2. Carbon has 4 valence electrons and oxygen has 6 valence electrons. Carbon shares two electrons with the two Hydrogens, and shares four electrons with the Oxygen, IOW, a double bond.
2(1 for each H) + 4(for C) + 6(for O) = 12 valence electrons 
Carbon has formed three covalent bonds in the structure so it would have 7 electrons around it (not 8 as shown) Likewise the oxygen fails to follow the rule.

This is what I have so far but I’m still working on the rule to finish it.