Lewis Structures & Ionic Compounds

[immike1234]

Although this isn't a problem for the MCATs, I thought that my question made the most sense to ask in this subforum.

For my General Chemistry 1 class, my professor asked for one of the questions, which of the following Lewis Structures is incorrect.

One of the answers was NaOH (with its "lewis structure" drawn out), which we know is incorrect because it doesn't have one.

However, after reviewing the concept of ionic compounds not having bonds between the cation and anion, I stumbled across the example BeF2.

However, how is it that BeF2 has a linear structure of F-Be-F? Wouldn't it make more sense that it would be written as [F-] + [Be2+] [F-] since it is an ionic compound?

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[matlabsux2012]

I was doing some Kaplan practice problems and I had to make 2 valid lewis structures with NO formal charge. The molecule can only contain C, N, H &O. 1 of each atom is the limit.

I submitted... N-C=0
........................l
.......................H

I was told this is not valid but using the Formal Charge = # Valence electrons - (# of bonds + # lone electrons) formula for Nitrogen I get FC = 5- (4+1) = 0.

Is there some rule I am missing or is this an invalid Lewis structure as far as formal charge goes. Thanks in advance.😱

 

[Jepstein30]

I was doing some Kaplan practice problems and I had to make 2 valid lewis structures with NO formal charge. The molecule can only contain C, N, H &O. 1 of each atom is the limit.

I submitted... N-C=0
........................l
.......................H

I was told this is not valid but using the Formal Charge = # Valence electrons - (# of bonds + # lone electrons) formula for Nitrogen I get FC = 5- (4+1) = 0.

Is there some rule I am missing or is this an invalid Lewis structure as far as formal charge goes. Thanks in advance.😱

Little confused about that structure..

Are you not showing all hydrogens because there you have carbon with only 3 bonds (formal charge of -1) and N with only two bonds (formal charge of -1).

HN=C=O

N~C-OH

where ~ is a triple bond. Those are the only two structures with no formal charges.

You can honestly forget that rule and just keep in mind the # of bonds each atom makes. Carbon wants 4, nitrogen 3, oxygen 2. If you get that many bonds and no extra lone electron pairs (oxygen should have two and nitrogen one).. you have no formal charge.

Oh, and it's # of bonds, not total e- in a bond.. so it should be:

FC for N = 5 - (2 + 4) = -1

 

[matlabsux2012]

Sorry is is N-CH=O

N has 1 bond to Carbon and 4 lone electrons FC = 5-(4+1) =0
C has 1 bond to N, 1 bond to H and 2 bonds to O
H has 1 bond to C
O has 2 bonds to C

Why would this be invalid as far as formal charge?

 

[Jepstein30]

Sorry is is N-CH=O

N has 1 bond to Carbon and 4 lone electrons FC = 5-(4+1) =0
C has 1 bond to N, 1 bond to H and 2 bonds to O
H has 1 bond to C
O has 2 bonds to C

Why would this be invalid as far as formal charge?

N doesn't even fulfill the octet rule..

When you're making structures, you have to count the total number of valence electrons.

Here: 5 (N) + 4 (C) + 6 (O) + 1 (H). So you have a total of 16 electrons to place.

Now, look at how many electrons you placed. You have 3 bonds = 6 electrons. And then how many lone pairs? I presume Oxygen has two that you didn't mention and then two more on Nitrogen. So that's 8 electrons there. = 14 total electrons.

You're missing two electrons that both would go on Nitrogen to fulfill its octet.

Therefore, nitrogen would have a formal charge of 5-(1+6) = -2.

Again, just make it so that nitrogen has 3 bonds, carbon 4 and oxygen 2 and you will always have a complete octet (fill in the lone pairs) and no formal charges.

I showed the two answers to the problem above.

 

[Sports Junky]

Sorry is is N-CH=O

N has 1 bond to Carbon and 4 lone electrons FC = 5-(4+1) =0
C has 1 bond to N, 1 bond to H and 2 bonds to O
H has 1 bond to C
O has 2 bonds to C

Why would this be invalid as far as formal charge?

The Nitrogen in the structure you have shown would have to have 3 pairs of lone electrons in order to satisfy it's octet. Therefore the formal charge on Nitrogen would be -2.

FC = Valence - Non-bonded - (Bonded/2) = 5 - 6 - (2/1) = -2

As a general rule, Nitrogen likes to make 3 bonds in order to satisfy it's octet with a formal charge of 0. You can remember this using the HONC rule (1,2,3,4). Hydrogen likes to make 1 bond, Oxygen 2, Nitrogen 3, and Carbon 4.

 

[matlabsux2012]

Maybe you did not understand my structure

N has 1 bond to C, 1 lone pair, and 2 lone electrons
C has 1 bond to N, 2 bonds to O, and 1 bond to H (8 electrons around it)
H has 1 bond to C (2 electrons around it)
O has 2 bonds to C and 2 lone pairs (sorry, cant draw lone pairs with just normal keyboard)

So I have 4 bonds (8 electrons), 2 lone pairs on 0 (4e), 1 lone pair on N (2e), and 2 other un-bonded electrons on N (2e).
Total # electrons is 8+4+2+2 = 16 electrons.

So the reason this fails outright is because N doesn't satisfy the octet Rule and thus, is an invalid structure?

 

[Jepstein30]

Maybe you did not understand my structure

N has 1 bond to C, 1 lone pair, and 2 lone electrons
C has 1 bond to N, 2 bonds to O, and 1 bond to H (8 electrons around it)
H has 1 bond to C (2 electrons around it)
O has 2 bonds to C and 2 lone pairs (sorry, cant draw lone pairs with just normal keyboard)

So I have 4 bonds (8 electrons), 2 lone pairs on 0 (4e), 1 lone pair on N (2e), and 2 other un-bonded electrons on N (2e).
Total # electrons is 8+4+2+2 = 16 electrons.

So the reason this fails outright is because N doesn't satisfy the octet Rule and thus, is an invalid structure?

huh

lone pair = 2 unbonded electrons

there's no such thing as "other unbonded electrons".. they are either in a bond or are considered a lone pair.

Nitrogen in your example has 3 lone pairs and 1 bond.

 

[Sports Junky]

Maybe you did not understand my structure

N has 1 bond to C, 1 lone pair, and 2 lone electrons
C has 1 bond to N, 2 bonds to O, and 1 bond to H (8 electrons around it)
H has 1 bond to C (2 electrons around it)
O has 2 bonds to C and 2 lone pairs (sorry, cant draw lone pairs with just normal keyboard)

So I have 4 bonds (8 electrons), 2 lone pairs on 0 (4e), 1 lone pair on N (2e), and 2 other un-bonded electrons on N (2e).
Total # electrons is 8+4+2+2 = 16 electrons.

So the reason this fails outright is because N doesn't satisfy the octet Rule and thus, is an invalid structure?

Yes, that is exactly the reason why; all of the elements must have a full octet.

Edit: I did not see the 'unbonded electrons'... what exactly do you mean by that?

 

[Jepstein30]

Yes, that is exactly the reason why; all of the elements must have a full octet.

His nitrogen DOES have a full octet.. it just has 3 lone electron pairs and one bond, giving it a formal charge of -2

 

[matlabsux2012]

Nitrogen only has 5 electrons around it. Normally it has 1 non bonded pair and 3 lone electrons used to form bonds. If N only forms 1 bond (to C), it still has the lone pair and the other 2 non-bonded electrons. THAT is what I mean by 2 other non-bonded electrons. I guess then N can be considered to have 2 lone pairs and 1 bond (5 electrons around it).

So, even if the FC equations shows the FC = 5- (4+1) =0, it fails before that because the N only has 5 electrons around it. Is that right.

 

[Jepstein30]

Nitrogen only has 5 electrons around it. Normally it has 1 non bonded pair and 3 lone electrons used to form bonds. If N only forms 1 bond (to C), it still has the lone pair and the other 2 non-bonded electrons. THAT is what I mean by 2 other non-bonded electrons. I guess then N can be considered to have 2 lone pairs and 1 bond (5 electrons around it).

So, even if the FC equations shows the FC = 5- (4+1) =0, it fails before that because the N only has 5 electrons around it. Is that right.

Those 2 non-bonded electrons ARE additional lone electron pairs. N would have, in your example, 3 lone pairs and one bond.

Atoms in covalent bonds get a full octet, not just the valence number. N needs 8 electrons so if you're only using one bond, you have 6 electrons around it (3 pairs).. all 3 pairs need to be included in the formal charge calculation.

Again, much simpler if you just remember that nitrogen makes 3 bonds and has one lone pair.

If it has more than one lone pair, it will have a negative FC.
If it doesn't have any lone pairs (i.e. 4 bonds), it will have a positive FC.

 

[matlabsux2012]

Those 2 non-bonded electrons ARE additional lone electron pairs. N would have, in your example, 3 lone pairs and one bond.

Atoms in covalent bonds get a full octet, not just the valence number. N needs 8 electrons so if you're only using one bond, you have 6 electrons around it (3 pairs).. all 3 pairs need to be included in the formal charge calculation.

Again, much simpler if you just remember that nitrogen makes 3 bonds and has one lone pair.

If it has more than one lone pair, it will have a negative FC.
If it doesn't have any lone pairs (i.e. 4 bonds), it will have a positive FC.

Where would these other lone pairs come from exactly? As we said, the molecule only has 16 electrons. I know ideally N has 3 bonds & 1 lone pair. Here I am saying that in forming only 1 bond, it still results in a formal charge of 0. You cannot just stick electrons anywhere. In my structure:

N has 2 lone pairs & 1 bond to Carbon Only.
C has 4 bonds total. 2 to O, 1 to H, ! to N
H has 1 bond total, to C
O has 2 bonds (2 to C) and 2 lone pairs.

There cannot b3 3 lone pairs on N. Where would they come from?

 

[Jepstein30]

Where would these other lone pairs come from exactly? As we said, the molecule only has 16 electrons. I know ideally N has 3 bonds & 1 lone pair. Here I am saying that in forming only 1 bond, it still results in a formal charge of 0. You cannot just stick electrons anywhere. In my structure:

N has 2 lone pairs & 1 bond to Carbon Only.
C has 4 bonds total. 2 to O, 1 to H, ! to N
H has 1 bond total, to C
O has 2 bonds (2 to C) and 2 lone pairs.

There cannot b3 3 lone pairs on N. Where would they come from?

That's the point...

Your structure is impossible.. N would need to have 3 lone pairs to complete its octet and if it did, it would have a formal charge of -2 and would need another atom involved somewhere.

 

[Sports Junky]

Where would these other lone pairs come from exactly? As we said, the molecule only has 16 electrons. I know ideally N has 3 bonds & 1 lone pair. Here I am saying that in forming only 1 bond, it still results in a formal charge of 0. You cannot just stick electrons anywhere. In my structure:

N has 2 lone pairs & 1 bond to Carbon Only.
C has 4 bonds total. 2 to O, 1 to H, ! to N
H has 1 bond total, to C
O has 2 bonds (2 to C) and 2 lone pairs.

There cannot b3 3 lone pairs on N. Where would they come from?

If Nitrogen only has 6 electrons in your proposed structure, than it is invalid. Quite simply, the structure is not valid without satisfying the octet rule.

 

[matlabsux2012]

Ok, then you misunderstood my original Q. All I wanted to know was why this structure failed despite satisfying (technically) the formal charge rule.

Long story short, it is because N fails to complete its octet that the structure is invalid. Thanks everyone for the assist.

 

[Jepstein30]

Ok, then you misunderstood my original Q. All I wanted to know was why this structure failed despite satisfying (technically) the formal charge rule.

Long story short, it is because N fails to complete its octet that the structure is invalid. Thanks everyone for the assist.

Well, okay.

It fails because if it were a structure, it would be negatively charged overall (-2) and have a formal charge of -2 on the Nitrogen.

 

[matlabsux2012]

Well, okay.

It fails because if it were a structure, it would be negatively charged overall (-2) and have a formal charge of -2 on the Nitrogen.

Your assumption of having 3 lone pairs on N is what allows you to say N has a FC of -2.

MY hypothetical structure technically had a FC of 0, but was impossible because N cannot have only 5 electrons around it (1 bond and 2 lone pairs).

Thanks again!

 

[shaboobly]

the way i count formal charge is to count all the bonds (1 e- per bond) and the lone pairs.. it's easier then trying to remember yet another formula and is technically the same thing.

 

[SN2ed]

Thread closed. These types of questions belong in the Study Q&A forum.