Main content
MCAT
Course: MCAT > Unit 9
Lesson 4: Dot structures- Dot Structures Questions
- Drawing dot structures
- Formal charge and dot structures
- Resonance and dot structures
- VSEPR for 2 electron clouds
- VSEPR for 3 electron clouds
- VSEPR for 4 electron clouds
- VSEPR for 5 electron clouds (part 1)
- VSEPR for 5 electron clouds (part 2)
- VSEPR for 6 electron clouds
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Resonance and dot structures
Drawing resonance structures and resonance hybrid using example of nitrate anion. Created by Jay.
Want to join the conversation?
- At, aren't the three dot structures the same, just rotated? Why are they considered different? 4:00(29 votes)
- I wondered the same thing initially. I think that an important thing to consider is that the diagrams are only the-same-but-rotated if you don't care which oxygen atom is which. If you labeled the oxygen atoms, then it wouldn't be the same. So in real life, if you were somehow able to hold the molecule still and look at just one oxygen atom, the three structures would not be the same (it could have either a single or double bond to the nitrogen).(29 votes)
- while drawing the resonance hybrid structure, why does one bond have 4 dots while the other has 5. shouldn't all of the bonds have six dots(7 votes)
- He is just using dotted lines to represent the partial bonds.
The dots do not represent electrons.
The dotted lines could have 3 or 6 or 10 dots. It wouldn't matter.(25 votes)
- I can't find in either the chemistry or organic chemistry a good explanation for "resonance structure". It is kind of clear from the several drawings of molecular structures where there are several equivalent configurations where the electrons may be found, and where the pi bounds are, but when you call it resonance it should have to do with some kind of oscillatory process, as if the molecule would swing back and forth between two or more states. But then there should be a physically measurable evidence for that. Is this covered anywhere here in more detail?(6 votes)
- i think this page might provide the solution to your query http://www.chem.ucla.edu/harding/tutorials/resonance/draw_res_str.html(8 votes)
- If resonance do not actually exist then why they are only made on paper ?
Why not only resonance hybrid structures are directly made ?(4 votes)- Resonance structures show different possible placements on electrons. In reality, the real molecule is a hybrid of all these resonance structures and we draw them so it can help us understand why molecules behave and react the way they do.(12 votes)
- Which molecule is More stable :
A compound in which resonance occurs OR
a compound in which resonance do not occur..
Please help. Than you.(4 votes)- A molecule with resonance will be more stabile since charges are delocalized.(6 votes)
- what will be the oxidation no of N in NO3^- ?(3 votes)
- The oxygen atoms each have a oxidation state of -2 and there are three such oxygens. Therefore, the total is -6. The entire molecule is only -1. The difference is the oxidation state of nitrogen, which is +5.(7 votes)
- What's happening with the orbitals when electrons are delocalized? If a sigma bond is a head on overlap of lobes but a pi bond is an side overlap, then how are resonance electrons being shared?(4 votes)
- Excellent question! This can be best answered by the idea of molecular orbitals. Just as two atomic orbitals with p-character can overlap to form a π bonding orbital, multiple bonding orbitals can overlap to form systems that cover large parts of the molecule. A good example is benzene: if benzene did just have 3 π bonds with no delocalization, all the electrons would be cramped together, hovering above and below the three sides of the hexagonal ring that have the double bond. But, since every carbon has electrons in an atomic orbital that has p character, in reality, the electrons are hovering in big circular "halos" above and below the ring -- with more room to swish and swirl around in, not all cramped up (like they would be if benzene really just had 3 double bonds), and therefore much more stable.(1 vote)
- At, I get confused. When you de-localize the electrons, you added 4 electrons in a line connecting the first Oxygen to Nitrogen. Why does that Oxygen electron not need 8 total electrons anymore? Also, its seems that you added 14 de-localized electrons...why is it 18 total electrons to complete the octets? And doesn't adding those de-localized electrons between the Oxygens and Nitrogen give Nitrogen more than 8 electrons? 5:20(3 votes)
- When he draws in the delocalized electrons, it's not literally showing that 14 of them are added — those dots represent the idea of delocalized electrons generally, not individual electrons. No electrons are added, but rather some of them are no longer bound to or paired with any one atom. The octets of each atom are still satisfied — you can think about it as if those electrons in the structure not associated with any one atom are spending enough time near each oxygen to keep all of them satisfied.(2 votes)
- Can oxygen ever have a single bond in any compound ? I mean shouldn't it have 2 lone pairs and share the third pair in a double bond ? Even if it does have a single bond , shouldn't it be a coordinate bond as none of oxygen's original six electrons are shared ?
P.S. I dont mean single bond as in the resonancee structures of no3. I mean can oxygen be connected with a single bond in any compound ?(2 votes)- How about any alcohol which will have a carbon bonded to an oxygen bonded to a hydrogen(C-O-H) but otherwise, no, oxygen cannot have only 1 single bond because it needs to follow the octet rule(3 votes)
- Does formal Charge still apply to Resonance Dot Structures? If so, how do you calculate the formal charges in the depicted dot structures for NO3-?(2 votes)
- Yes, each atom in a dot structure has a formal charge.
In the nitrate ion, the N atom is +1 and the two single-bonded O atoms are each -1.(3 votes)
Video transcript
Now that we know how
to draw dot structures, let's apply our rules
to the nitrate anion. And we're going to see that we
can draw a few different dot structures for this anion. And we're going to
call those resonance structures of each other. But first, we need to
calculate the total number of valence electrons. And so nitrogen is in Group
5 in the period table, therefore, five
valence electrons. Oxygen is in Group 6,
therefore, six valence electrons for each oxygen. I have three of them. So 6 times 3 is 18 valence
electrons, plus the 5 from the nitrogen gives me 23. And I have a negative charge. This is an anion here. So we have to add
one electron to that. So 23 plus 1 gives us a
total of 24 valence electrons that we need to represent
in our dot structure. So we know that nitrogen is
going to go in the center, because oxygen is
more electronegative. So nitrogen goes in the center. Nitrogen is bonded
to three oxygens. So I can go ahead and put
them in there like that. And let's see. How many valence electrons
have we represented so far? 2, 4, and 6. Therefore, 24 minus 6 gives us
18 valence electrons left over. We're going to put those
leftover valence electrons on our terminal atoms,
which are our oxygens. And oxygen's going to
follow the octet role. Currently, each oxygen has two
valence electrons around it, the ones in magenta. So if each oxygen
has two, each oxygen needs six more to
complete the octet. And so I go ahead and put
six more valence electrons on each one of my oxygens. Now each oxygen is surrounded
by eight electrons. So the oxygens are happy. We added a total of six valence
electrons to three oxygens. So 6 times 3 is 18. So we've used up
all of the electrons that we need to represent. And so this dot
structure, so far, it has all of our
valence electrons here. Oxygen has an octet. So oxygen is happy. But nitrogen does
not have an octet. If you look at the
electrons in magenta, there are only six electrons
around the nitrogen. And so the nitrogen
wants to get to an octet. And there are a couple
of different ways that we could give
nitrogen an octet. For example, we could take
a lone pair of electrons from this top oxygen here
and move them into here to share those electrons
between that top oxygen and that nitrogen. So let's go ahead and draw
that resulting dot structure. So we would have
our nitrogen now with a double bond
to our top oxygen. Our top oxygen had three
lone pairs of electrons. But now it has only two,
because electrons in green moved in to form a double bond. This nitrogen is
bonded to an oxygen on the bottom left and an
oxygen on the bottom right here. So this is a valid
dot structure. We followed our steps. And we'll go ahead and
put this in brackets and put a negative charge
outside of our brackets like that. So that's one possible
dot structure. But we didn't have to take
a lone pair of electrons from the top oxygen. We could've taken a lone pair
of electrons from the oxygen on the bottom left here. So if those electrons
in blue moved in here, we could have drawn
another dot structure which would have been equally valid. We could have shown this oxygen
on the bottom left now bonded to this nitrogen, and it used
to have three lone pairs. Now it has only two. And now this top oxygen
is still a single bond with three lone pairs around it. And this bottom
right oxygen is still a single bond with three
lone pairs around it. So this is a valid
dot structure as well. So let's go ahead
and put our brackets with a negative charge. And then, of course,
we could have taken a lone pair of
electrons from the oxygen on the bottom right. So I could have moved these
in here to form a double bond. And so now, we would have
our nitrogen double bonded to an oxygen on
the bottom right. The oxygen on the
bottom right now has only two lone
pairs of electrons. The oxygen at the top, single
bond with three lone pairs. And then the same situation for
this oxygen on the bottom left. And so this is, once again,
another possible dot structure. And so these are
considered to be resonance structures
of each other. And the way to
represent that would be this double-headed
resonance arrow here. And I think when students
first see resonance structures, the name implies
that, in this case, the ion is resonating back
and forth between these three different possible, equally
valid dot structures. And that's not quite
what's going on here. Each of these dot
structures is an attempt to represent the
structure of the ion. But they're really not the
best way of doing that. You need to think about
combining these three dot structures in a resonance
hybrid of each other. And so let's go ahead and draw
just a simple representation of a way of thinking
about a resonance hybrid. So if I combined all
three of my dot structures here into one picture,
I had a double bond to one oxygen in each of my
three resonance structures here. And so the top oxygen had a
double bond in one of them, the bottom left in the middle
one, and then the bottom right in the third one. So, in reality, if we take a
hybrid of all those things, we could think about the
electrons being delocalized or spread out among all
three of our oxygens. And so instead of giving
our top nitrogen-oxygen, instead of making
that a double bond, we can just show some electrons
being delocalized in that area, so stronger than a single
bond, but not as strong as a double bond. And we could do the same
thing between this nitrogen and this oxygen. So the electrons are
delocalized a little bit here. It's not a double bond. It's not a single bond. And the same idea for this
nitrogen-oxygen in here. And one way we know that the
ion looks more like this hybrid is because of bond length. When the ion is measured in
terms of the bond length, all the nitrogen and oxygen
bonds are the same length. And of course, if
we thought about one of these resonance
structures as being the true picture
of the ion-- let's say this one, for
example-- that wouldn't be the case for this ion,
because this double bond here, we know that would
be shorter than one of these single
nitrogen-oxygen bonds. And so it's actually
more of a hybrid with the electrons
delocalized throughout. And that's the idea of
resonance structures here.