This O3 Lewis Diagram looks weird -

why does atom with 8 electrons positive,

why is atom with 4 electrons neutral,

why is atom with 6 electrons negative

In CO3 -2 Why there is a -2 charge ?

What is happening in HNO3 ?

N is positive that means it is 5 electrons short of Nobel config.

Why there is a negative charge on bottom oxygen ?

Are these Lewis Diagram possible or covalent and electrovalent bonds ?

[NO2]-

is there 1 electron extra on oxygen ?

[NO2]-

Edited August 23Aug 23 by HbWhi5F

57 minutes ago, HbWhi5F said:

This O3 Lewis Diagram looks weird -

why does atom with 8 electrons positive,

why is atom with 4 electrons neutral,

why is atom with 6 electrons negative

I will answer this, even though you did not respond to my question abour homework in the maths thread.

Please tell us a bit more about what you are studying (not personal details) as it helps enormously in answering.

OK so firstly the Oxygen molecule O₂

_{What you need to understand to start with is that the bonding atoms have two types of electrons.}

_{Those electrons involved in the bond with another atom or bonding electrons.}

_{these are represented in lewis by lines, not dots.}

_and

_{Those which are not involved in the bonding or non bonding electrons.}

These are represented in lewis by dots not lines

_{So the oxygen molecule is symmetrical.}

_{Neither of the oxygen atoms are different, both have 6 electrons}

_{represented in lewis by four non bonding dots and two lines ( a double bond)}

The bonding is purely covalent since no + or - signs appear against any atom.

Now look at the ozone molecule.

Firstly what you call the atom with four electrons actually has six

Four non bonding dots and two double bond lines

this is just like each atom in the O₂ molecule.

Next the atom with the + sign has 5 electrons

two non bonding dots two double bond electrons and one single bond electron.

That is it has lost one electron so is positive.

Finally the last oxygen atom has gained that electron and has a negative sign by it

So it has six non bonding dots and one single bond electron, making seven in all.

The bond between the last two oxygen atoms is therefore partly (50%) ionic and partly covalent.

So in Lewis covalent bonds are shown with a line and Ionic bonds with a + and - to show complete electron transfer.

Following this reasoning can you work through your other examples for yourself and come back with any further questions ?

Edited August 23Aug 23 by studiot

Update: How is unequal sharing of electrons is possible ?

I'll explain O₃ then leave CO₂^2– and HNO₃ for you to explain.

A neutral oxygen atom has 6 electrons in its valence shell. If it gains two electrons, completing the octet, it will have a charge of –2. Each of the three oxygen atoms of ozone has a complete octet. Now consider that each of the oxygen atoms has donated electrons to the oxygen atoms to which it is bonded. Each donated electron changes the charge by +1. The oxygen atom on the left has donated 2 electrons to the central oxygen atom, and therefore its charge is neutral. The oxygen atom on the right has donated 1 electron to the central oxygen atom, and therefore its charge is –1. The central oxygen atom has donated 3 electrons to the two bonded oxygen atoms (2 to the left oxygen atom and 1 to the right oxygen atom), and therefore its charge is +1.

5 minutes ago, HbWhi5F said:

Update: How is unequal sharing of electrons is possible ?

This representation is missing a negative charge on the carbon atom, and a positive charge on the oxygen atom.

9 minutes ago, KJW said:

I'll explain O₃ then leave CO₂^2– and HNO₃ for you to explain.

A neutral oxygen atom has 6 electrons in its valence shell. If it gains two electrons, completing the octet, it will have a charge of –2. Each of the three oxygen atoms of ozone has a complete octet. Now consider that each of the oxygen atoms has donated electrons to the oxygen atoms to which it is bonded. Each donated electron changes the charge by +1. The oxygen atom on the left has donated 2 electrons to the central oxygen atom, and therefore its charge is neutral. The oxygen atom on the right has donated 1 electron to the central oxygen atom, and therefore its charge is –1. The central oxygen atom has donated 3 electrons to the two bonded oxygen atoms (2 to the left oxygen atom and 1 to the right oxygen atom), and therefore its charge is +1.

A donated electron = an ionic bond

Which is polar (has a negative end and a positve end)

A shared electron = a covalent bond

A shared electron = a covalent bond where the atom looses (donates) a part only share in one of its electrons but also gains a part shart in an electron from another atom.
Thus remains electrically neutral.

8 minutes ago, studiot said:

A donated electron = an ionic bond

Which is polar (has a negative end and a positve end)

A shared electron = a covalent bond

A shared electron = a covalent bond where the atom looses (donates) a part only share in one of its electrons but also gains a part shart in an electron from another atom.
Thus remains electrically neutral.

Yes, but I'm actually talking about charge bookkeeping.

I think you need a quick tutorial on the Lewis representation of bonds.

Other than the first row of the Periodic table, which has a maximum of 2 electrons in its 1S orbital, all others contain up to 8 ( 2S, 2P, ... ), because the central Transition Metals can be disregarded ( their outer electrons are not valence, but rather conduction ).
Atoms, whether free or bonded to others are 'happiest' when their lowest orbitals are full with 8 electrons, so sometimes they steal, or sometimes borrow, electrons from their bonded neighbor.
Two dots, inline between neighboring atoms is equivalent to a line ( .. > - ).
Four dots, where two are inline is equivalent to two lines ( :: > = ).
And this can go to three lines or ( possibly ? ) to four.

Sometimes this electron 'book-keeping does not take the protons in the nucleus into account, so when one of the pair 'steals' electrons from the other, you end up with 'excess' charge on opposite ends, and this ionic bond is polar ( as Studiot and KJW are discussing above ).
A simple example would be water ( H₂O ) where the electronegative Oxygen ( 6 protons ) now has 8 electrons due to uneven sharing, making its side negative, while leaving the Hydrogen sides with an excess positive charge.

You know, your best resource for learning IS your teacher ( at least until you have enough foundational knowledge, discipline and confidence for self-learning ).
If something is not explained to your satisfaction, keep asking questions; they owe you this knowledge.

Edited August 23Aug 23 by MigL

1 hour ago, KJW said:

Yes, but I'm actually talking about charge bookkeeping.

And I wasn't since the resultant species are all electrrically neutral (unlike say the ammonium ion).

Charge bookkeeping leads to seriously difficulties with later more advanced stuff.

15 minutes ago, MigL said:

I think you need a quick tutorial on the Lewis representation of bonds.

Agreed except that an in-depth tutorial would be more beneficial.

This is partly why I asked what is being studied as there are many other subjects that consider chemical bonding besides Chemistry itself.

1 hour ago, MigL said:

or ( possibly ? ) to four.

Quadruple bonds are possible, although they are quite rare, found in some transition metal complexes. According to Wikipedia, the first chemical compound containing a quadruple bond to be synthesised was chromium(II) acetate, Cr₂(μ-O₂CCH₃)₄(H₂O)₂, which contains a Cr–Cr quadruple bond.

It's worth noting that delta bonds are also possible. Similar to the way pi bonds can be formed from overlapping p orbitals, delta bonds can be formed from overlapping d orbitals.

In the case of chromium(II) acetate, the quadruple bond is considered to arise from the overlap of four d orbitals on each metal with the same orbitals on the other metal: the d_z² orbitals overlap to give a sigma bonding component, the d_xz and d_yz orbitals overlap to give two pi bonding components, and the d_xy orbitals give a delta bond.

Edited August 23Aug 23 by KJW

1 hour ago, studiot said:

Agreed except that an in-depth tutorial would be more beneficial.

I'm not sure the forum discussion process is suitable for that.
I know you often try, but I think the best we can do is 'steer them' in the right direction.

17 minutes ago, KJW said:

Quadruple bonds are possible

Thanks.
I wasn't sure; my last Chemistry course was a long time ago ( Gr. 13 in 76/77 ).

Even quintuple bonds exist. These involve all five d orbitals on each metal, participating in one sigma, two pi, and two delta bonds.

3 hours ago, KJW said:

Even quintuple bonds exist. These involve all five d orbitals on each metal, participating in one sigma, two pi, and two delta bonds.

How strong are these higher order bonds? Can they beat the triple bond in carbon monoxide?

I think @HbWhi5F is just starting to study bonding in a pretty conventional way.

This is to distinguish and identify two types of bonds viz ionic and covalent bonds and understand the difference between them and also understand the equation

Number of bonds + Number of unbonded outer electrons = total number of outer electrons in each atom.

More advanced schemes like sigma Pi delta bonds, metallic bonding, hydrogen bonbding and so on come later as do molecular orbitals, resonance and other stuff.

Here I don't think our OP is yet clear about ionic and covalent bonding.

Often the questions

Why do atoms bond at all ?

are never asked

Similarly for

why do chemical reactions happen ?

So HbWhi5F, have you done any quantum theory at all ? Have you heard of 'orbitals' ?

Help us to help you.

Edited August 23Aug 23 by studiot

1 hour ago, sethoflagos said:

How strong are these higher order bonds? Can they beat the triple bond in carbon monoxide?

Given that the triple bond of carbon monoxide is the strongest known chemical bond, I would say "no". The Wikipedia articles on quadruple bonds and quintuple bonds seem to focus more on bond length than on bond strength, so I don't know if these bonds are particularly strong.

10 hours ago, KJW said:

A neutral oxygen atom has 6 electrons in its valence shell. If it gains two electrons, completing the octet, it will have a charge of –2. Each of the three oxygen atoms of ozone has a complete octet. Now consider that each of the oxygen atoms has donated electrons to the oxygen atoms to which it is bonded. Each donated electron changes the charge by +1. The oxygen atom on the left has donated 2 electrons to the central oxygen atom, and therefore its charge is neutral. The oxygen atom on the right has donated 1 electron to the central oxygen atom, and therefore its charge is –1. The central oxygen atom has donated 3 electrons to the two bonded oxygen atoms (2 to the left oxygen atom and 1 to the right oxygen atom), and therefore its charge is +1.

Looking at Google, I came across the following formula for the formal charge of an atom in a molecule:

Formal charge of atom = (Number of valence electrons in free atom) – (Number of non-bonding electrons) – (Half the number of bonding electrons)

Admittedly, I had not encountered this explicit formula before. Instead, I would work out the formal charge by some ad hoc calculation that would ultimately lead to the same answer. The formula works by starting from an atom which has had all its valence electrons removed, adding the number of non-bonding electrons, then adding the number of bonding electrons that contribute to the formal charge of the atom, which is half the number of bonding electrons, the other half contributing to the formal charge of the other atom. Note that the subtractions in the formula are because electrons are negatively charged.

Although I didn't use an explicit formula to calculate the formal charges of the atoms of ozone, the above description corresponds to the formula:

Formal charge of atom = (Number of valence electrons in free atom) – (Number of electrons) + (Half the number of bonding electrons)

Because (Number of electrons) = (Number of non-bonding electrons) + (Number of bonding electrons), these two seemingly different formulae will give the same answer. However, by adding (Half the number of bonding electrons) instead of subtracting, conceptually I'm removing electrons that do not contribute to the formal charge of the atom.

9 hours ago, KJW said:

Looking at Google, I came across the following formula for the formal charge of an atom in a molecule:

Formal charge of atom = (Number of valence electrons in free atom) – (Number of non-bonding electrons) – (Half the number of bonding electrons)

This approach present difficulties, even at introductory level.

One of the examples in the OP is the carbonate ion, and I have already mentions the ammonium ion.
These are chemical species, not atoms and the OP actually heads the table 'molecule/ion' to acknowledge this.
Molecules, like atoms are electrically neutral, unlike the ion species and the OP avoids allocating formal charge to any atom in the carbonate.
In contrast, atomic charge allocation works in the neutral HNO3 molecule. But what about when it dissociates ?

26 minutes ago, studiot said:

This approach present difficulties, even at introductory level.

One of the examples in the OP is the carbonate ion, and I have already mentions the ammonium ion.
These are chemical species, not atoms and the OP actually heads the table 'molecule/ion' to acknowledge this.
Molecules, like atoms are electrically neutral, unlike the ion species and the OP avoids allocating formal charge to any atom in the carbonate.
In contrast, atomic charge allocation works in the neutral HNO3 molecule. But what about when it dissociates ?

It's not an "approach", it's a skill that chemists, especially organic chemists, need to master.

I don't see the problem with the ammonium ion - the nitrogen atom has a +1 charge. As for the carbonate ion, each resonance structure has its own Lewis structure with a –1 charge on the two singly bonded oxygen atoms. Or one could use a single abbreviated structure that manifests the symmetry with a –²/₃ charge on all three oxygens (averaging the three resonance structures). As for nitric acid, dissociates in what way? Anyway, the products of dissociation have their own Lewis structures requiring formal charge allocation.

1 hour ago, KJW said:

It's not an "approach", it's a skill that chemists, especially organic chemists, need to master.

I don't see the problem with the ammonium ion - the nitrogen atom has a +1 charge. As for the carbonate ion, each resonance structure has its own Lewis structure with a –1 charge on the two singly bonded oxygen atoms. Or one could use a single abbreviated structure that manifests the symmetry with a –²/₃ charge on all three oxygens (averaging the three resonance structures). As for nitric acid, dissociates in what way? Anyway, the products of dissociation have their own Lewis structures requiring formal charge allocation.

Clearly the OP was having a problem or he would not have started this thread.

So I don't see how saying that you don't see a problem is helpful.

There is no one right approach.

Once we know where he is coming from we can help him chart his own path through a complicated subject.

7 hours ago, studiot said:

I don't see how saying that you don't see a problem is helpful.

That was addressed to you. I believe I gave a clear answer to the OP, and until the OP comes back with a request for clarification, I see no reason to say anything different.

1. Somethings charge is mentioned sometimes not, either then octet is filled or not, Why ?

2. Covalent bonds are only (most) created to complete the octet ?

3. Completing octet doesn't charge the atoms? Seems like sometimes it does.

CO3 (-1) charges - C -2 O +2 on top and x2 O -2 ?

NF3 should also have charges indicated ? x3 F has -1 each and N has -3.

In HNO3 left O doesn't have charge it should be +2. N has +4. RIght O should be +2. H should be +1

In O2 they both share 2 eletrons to become noble but that also makes each O, -2 charged right ?

@studiot @KJW @MigL

Edited August 25Aug 25 by HbWhi5F

Atoms are not charged; if they were they'd be ions.
A Hydrogen atom has 1 electron ( 2S orbital ) and one proton in the nucleus, making it neutral.
A Nitrogen atom has 7 electrons and 7 protons in the nucleus, making it neutral.
Two of these electrons are in the inner 1S orbital, and 5 electrons are in the outer valence 2S orbital.
The 2S orbital is 'happiest' when it is totally filled with 8 electrons, so when a Nitrogen atom comes across some Hydrogen atoms (given the right circumstances ) it 'steals the electrons of three Hydrogen atoms, forming NH₃; while NH₃ is still a neutral atom, its charge distribution is imbalanced, as the side of the tetrahedron with three Hydrogen atoms is predominantly triple positive ( 3 'bare' protons ), and the side with the Nitrogen atom now has 3 extra electrons, making it predominantly triple negative, and this imbalance of charge leads to Ammonia being a polar ( preferred charge direction ) molecule.

12 hours ago, MigL said:

Two of these electrons are in the inner 1S orbital, and 5 electrons are in the outer valence 2S orbital.
The 2S orbital is 'happiest' when it is totally filled with 8 electrons,

I'm sure you know this is not quite accurate, perhaps you were trying to simplify too much.

It was recognized early on that the electrons are arranged in shells.

An individual shell is a collective name for electrons thought to be at a similar distance from the nucleus.

They were called shells because it was thought that were indeed shells in space like the layers of an onion.

Later is became understood that these shell groupings are actually arranged in terms of energy not spatial separation and actually partially overlap each other.

It was also realised that there are also sub groupings within the shells ie the energies of electrons in a given shell are similar but not necessarily the same.

Originally these shells were also called K, L, M N and so on. (capital letters)

When subsehells were intorduced the letters were replaced by a new system, using numbers instead K, L, M, N etc being replaced by 1, 2, 3, 4 etc.

The subshells were then labelled with lower case letters, s, p, d, f and so on

It is the shells that are being considered for the octet rule.

12 hours ago, MigL said:

while NH₃ is still a neutral atom

Ammonia is a neutral molecule.

Pure substances (matter) can comprise molecules, which can be further divided into pure elements.

And elements which cannot be divided.

An atom is the smallest particle of an 'element' of matter.

A molecule is the smallest particle of a pure substance.

But it can be divided into atoms of consitiuents elements.

4 hours ago, studiot said:

Ammonia is a neutral molecule.

Yeah.
My bad.

And I realize that it's not an accurate description, but every ( layman's ) pictorial representation of 'shells' shows higher energies further from the nucleus, such that total ionization is at infinite separation.
So I thought my explanation appropriate for the 'level' of the discussion.

18 minutes ago, MigL said:

Yeah.
My bad.

And I realize that it's not an accurate description, but every ( layman's ) pictorial representation of 'shells' shows higher energies further from the nucleus, such that total ionization is at infinite separation.
So I thought my explanation appropriate for the 'level' of the discussion.

I think the main part of the message is that a shell is not an orbital.

A shell contains orbitals.

And a 2s orbital can hold exactly 2 electrons (not 8), just like any other s orbital.

But I am not sure if the op understands the word orbital or the s, p, d etc notation evryone is bandying about here.

The question in this thread is at quite a different level from his one about oxides of nitrogen.

Like you I am trying to help, but it is difficult to know where to start, without further information.

My bad again.
In my original post, I should have said

22 hours ago, MigL said:

A Hydrogen atom has 1 electron ( 1S orbital ) and one proton in the nucleus, making it neutral.
...
Two of these electrons are in the inner 1S orbital, and 5 electrons are in the outer valence 2S/2P orbitals.
The 2S/2P orbital shell is 'happiest' when it is totally filled with 8 ( 2+6 ) electrons

My apologies for the confusion.
( speed kills AND adds to confusion )