Formal Charge

[fataliz]

Hi all. as above, the question asks for the formal charges of the 4 atoms. this is the model answer (D)

Atom 1: 5 valence, 8 bonding 0 lone pair. Hence charge = 5-(8/2) =1
Atom 2: 7 valence, 4 bonding 4 lone pair. Hence charge = 7-(4/2)-4 =1
Atom 3: 6 valence, 2 bonding 6 lone pair. Hence charge = 6-(2/2)-6 =-1
Atom 4: 6 valence, 4 bonding 4 lone pair. Hence charge = 6-(4/2)-4=0

This is highly confusing. The number of valence and electrons involved in bonding are easily derived, but how are the lone pair electrons derived when they are not given in the question? usually the questions i see asking for formal charge will include the lone pairs in the questions as well. My query is if i am suppose to deduce this based on this question?

Furthermore, how did this numbers for lone electrons come about??? N has 5 valence electrons, so after contributing 4 for the covalent bonds, shouldn't it have 1 lone electron left? how did it end up as 0 as in the question?

Take also Fluorine. If it has 7 valence, after contributing 2 for the double bond, shouldn't it have 5 lone pair electrons left? how did the model answer assume it has 4?

your help is much appreciated!

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[Cawolf]

From your questions, it seems like you lack the fundamentals of some basic chemistry concepts.

I could explain why your thoughts are incorrect, but I think you would find more value in reviewing the chapter on Lewis structures in any chemistry text.

It would suit you well to be be at a point of understanding where you know N with 4 bonds is +1 and a halogen with 2 bonds is +1.

Non bonding electrons are not shown because it assumed you know where they would be in accordance with hybrid orbital theory and the octet rule.

 

[fataliz]

hi Cawolf. thanks for the reply. i have tried reading from a few sources, but it's always the same issue. i can't understand where the missing or extra non bonding electrons go/come from.

i can use logic to deduce the charge, but mathematically it doesn't make sense to me. what am i missing?

 

[Cawolf]

I will try to explain my thought on one of the atoms.

N has 5 valence electrons, so after contributing 4 for the covalent bonds, shouldn't it have 1 lone electron left? how did it end up as 0 as in the question?

Neutral N has 5 valence electrons, so we look to see how many electrons it is sharing or has to determine the charge. Firstly, it has two single bonds and one double bond - so it is SP2 hybridized. All orbitals are used for bonds so there are no lone pairs. You see it has 4 electrons from it's bonds, but it needs 5 to be neutral, so it is +1.

Formal Charge = Neutral Valence Electrons - # of bonds - # lone pair electrons

=5 - 4 - 0 = +1

The concept of it having 5 electrons and contributing 4 is unfounded, that is not how to approach this problem. Furthermore, an atom with a lone electron is a radical.

 

[edgerock24]

@fataliz , @Cawolf
Am I crazy, or do none of those answer choices look completely correct to me?
First, the lone pairs are not drawn on any of the atoms, which makes sense.

Atom 1: Nitrogen. Has four total bonds, and one lone pair [that isn't drawn]. Needs 5 to be neutral, but, has 6. So it has a -1 charge.
Atom 2: Flourine. Needs 7. Has three lone pairs and one double bond. This makes 8, giving it a -1 charge.
Atom 3: Top Oxygen. Needs 6. Has three lone pairs and one single bond. This makes 7, giving it a -1 charge.
Atom 4: final Oxygen. Needs 6. Has two single bonds and two lone pairs. 0 charge.


Really tired, but the answers here seem incorrect. Am I off my rocker?

 

[Labrat07]

You're off the rock @edgerock24 .

Atom 1: Nitrogen has only 4 bonds and no lone pair. You misunderstood the atomic number with that n. We only consider that 4 bonds. Most can only have 4 bonds except few exceptions.
Atom 2: You have the same error as above. This only have 2 bonds and 4 unpair electron. 7 -4 -2 = +1.

I suggest you review this subject again.

 

[krukshanks]

@edgerock24

Nitrogen has no lone pair with four bonds. The four bonds already have 4*2 = 8 electrons and that's an octet (Nitrogen can never go beyond an octet as it has no d orbitals). Same logic for Fluorine.

 

[edgerock24]

Thanks guys!