Resonance Structures - Online Tutor, Practice Problems & Exam Prep

Resonance structures are multiple valid Lewis dot structures for polyatomic ions with pi bonds, illustrating different possible arrangements of electrons. These structures are connected by double-sided arrows, signifying their equivalency in depicting the molecule's structure. However, not all resonance structures are actually equivalent; some contribute more to the actual structure of the molecule than others, which becomes more apparent in advanced chemistry studies. The true representation of the molecule is the resonance hybrid, an average of all significant resonance structures. It is depicted by placing dotted lines where pi bonds may exist, as seen in the example of the nitrite ion. This concept is crucial for understanding the delocalization of electrons within certain molecules, which impacts their chemical behavior and stability.

Resonance Structures

Now with resident structures, we're going to say a set of two or more valid Lewis dot structures for polyatomic ions possessing at least one π bond. Now we're going to say in a resonant structure we have the movement of only electrons from either a π bond or a lone pair here. This depicts two different ways that we can show our nitride ion. Here the π bond could be either on the left oxygen or it could be on the right oxygen. Both are valid depictions of the nitride ion.

Now here to show resonance structures we utilize double sided arrows. They're used to show that the resonance structures are equal or equivalent with each other. At this level of chemistry, we're going to say that they're equivalent. But once you get to higher levels of chemistry, such as organic chemistry, you're later learn that not all resonance structures are equivalent to one another.

Now we're going to say the real structure is represented by the composite of the resonance structures, and it's called the resonance hybrid. Now the resonance hybrid is the composite of all major resonance structures. When we say composite, we really mean an average of the two resonant structures. To draw the resonance hybrid, we place a dotted line anywhere a π bond has been. So there was a π bond on the left oxygen, so we do a dotted line there. There was a double bond on the right oxygen, so we show a dotted line there. So this would represent the resonance hybrid of our nitrite ion.

Resonance Structures Example 1

Here it says determine the remaining resonance structures possible for the carbonate ion, which is CO3-2. So here we have our carbonate ion, we have our pie, our double bond on the oxygen on the top. Realize here that the oxygens that are single bonded have formal charges equal to -1, so they're depicted that way and we'd have to remember to carry over those formal charges of -1 for any other oxygens that are single bonded.

So remember in resonance we're just moving around the electrons from the Pi bonds. So way another resonance structure I could show is what if it's not the top oxygen that's double bonded, but it's the one on the left? Then we notice here that the double bonded oxygen has 2 lone pairs. So here's 2 lone pairs, and the single bonded oxygens have 3 lone pairs. Remember the single bonded oxygens have formal charges of -1 and here we put it in brackets because it has a charge.

But remember they're connected with double bonds. But who says that the oxygen on the left side is double bonded? What if it's the oxygen on the right side that's double bonded? The ones that are single bonded remember, have 3 lone pairs. The one that's double bonded has two. Remember, the single bonded oxygens each have a formal charge of -1. Put it in brackets with a charge on the outside.

And remember, these double sided arrows are what connects these resonance structures to one another. So just remember we're just showing different places the Python could have been within each of these structures. Each of them represents A resonance structure for the carbonate ion.

Draw all possible resonance structures for the chlorate ion, ClO₃^–?

Here it says to show all possible resonance structures for the Chloride ion. Alright. So, Chloride ion, chlorine is in group 7a, so it has 7 valence electrons. Oxygen is in group 6a, so it has 6 valence electrons. Minus 1 means we've gained 1 electron from an outside source. So here, that comes out to be 26 total valence electrons. So chlorine goes in the center, we're going to have it connected to our 3 oxygens. Remember, we have to follow the octet rule. And here, though, this would not be the best depiction of this structure because if we calculated the formal charges—remember, formal charges are group number minus bonds the element is making plus non-bonding electrons. If you calculated the formal charges of the oxygens, each one would be minus 1. And then if you calculated for the chlorine, the chlorine would be 7 for group number minus 3 bonds plus 2 non-bonding electrons; it'd be plus 2. That would be a plus 2 formal charge. Remember, we always want the formal charge to be negative 1, 0, or plus 1. So, to prevent this error, what we would do is we would make 2, keep this lone pair there, but we'd make 2 pi bonds. So when we do 2 pi bonds now, that would make that chlorine have a formal charge of 0 because then it would become group 7a, we see it making 5 bonds, it has 2 electrons not bonded, so that's 0. So this would be the correct depiction of the structure. Now, how do we show its other resonance structures? We're saying here that only one of the oxygens is not double bonded. Okay. But who says it has to be that one on the right? What if it is the one on the left that's singly bonded? So that's another resonance structure. But who says it's the one on the left? What about the one on the top? What if it's the single bond? So these 3 would represent the resonance structures for the chloride ion. All we're showing are all the possibilities that exist. Who tells us that the oxygen on the right is singly bonded? It could be any of the oxygens that are singly bonded. So you have to show all those possibilities. Each one represents a resonance structure.

Resonance Structures Example 2

Here we say that the average charge is the charge an element possesses from the overall charge of all its resident structures. Here it says determine the average charge of the oxygen atom within the phosphate ion. So phosphate ion basically has phosphate double bonded to one oxygen but then single bonded to all the others. It gives us 4 resonant structures.

Now, if only given the molecular formula, then draw one of the resonance structures, and if given multiple resonance structures, again just choose one. Here we're going to calculate the formal charge for the elements and add them to determine their overall charge. So we're going to focus on one of them. Let's focus on this one. Remember formal charge equals group number minus the bonds the element is making plus each nonbonding electron calculated or added individually.

So let's say that the double bonded oxygen is oxygen A and all these single bonded ones are oxygen B. So oxygen A is in Group 6A. It's making two bonds and it has four electrons that are not bonding, so its formal charge is 0. Then oxygen B, they are in Group 6A as well. One of them is making one bond and it has six non-bonding electrons, so this one is 0 and these are all -1.

So now we're going to add up all the formal charges together, so 0+-1+-1+-1=-3 for the overall charge. Step 4 says divide the overall charge by the total number of those elements. So we have a -3 overall charge, and we're dealing with four oxygen. So each of those oxygens would have a charge of -34, so we'd say their average charges is -34.

It's as simple as that. Draw the molecule correctly. Then you're going to figure out the total overall charge of all the elements in question and divide by the total number that exists. That gives us each of their individual or average charges.

Determine the average charge of the oxygen atoms within the chlorite ion, ClO₂^–.

• A. +1/2
• B. -1/2
• C. +1
• D. -1

Determine which of the following drawings would be the best structure for the N₂O molecule.

 

• A. a
• B. b
• C. c
• D. All are equally stable

Which of the following phosphate, PO₄^3- Lewis structures is the best, most valid resonance structure?

• A.
• B.
• C.
• D.
• E.

Draw all the resonance structures for the following ionic compound:RbIO₂

Here it says to draw all the resonance structures for the following ionic compound. So, we have RbIO2. It is an ionic compound, so you must break it up into its 2 ions. Rb is in group 1A, so it's +1, and then we have the iodide ion. Iodine is in group 7A, so it has 7 valence electrons. Oxygen is in group 6A, so that's times 2. And then we have a minus charge, which means an outside electron has been added. So that's going to be 7 + 12 + 1, which gives me 20 total valence electrons. Iodine goes in the center. It's going to be single bonded to the oxygens. Remember, the surrounding oxygens have to have their octet rule followed, so we've used 16 of our electrons. We have 4 remaining. Right now, we're putting them on the iodine. But here's the thing. Remember, resonance structures exist when there is at least one pi (π) bond present. Right now, there isn't any π bond. So, that must mean a double bond exists somewhere within this structure. So we just need to show it. So here, I'm going to use one of these lone pairs to make a π bond here. Now when I do that, let's calculate the formal charges. So we're going to say here iodine is in group 7A, we see it making 3 bonds and it has 4 electrons not bonding, so that's 0. Let's say this is oxygen A that's double bonded and oxygen B that's single bonded. So here, it’s in group 6A, we see it making 2 bonds and it has 4 electrons that are not bonding. Oxygen B, we see it’s in group 6A. We have 1 bond and 6 electrons, so this is minus 1. When we add up all the formal charges, it gives me back the charge of the ion, which is minus 1. So, this here is the correct interpretation of the iodide ion. Now, if we show it this way, our other resonance structure would be instead of the oxygen on the left being double bonded, let’s make it the oxygen on the right that's double bonded. So here, we’d still have our 2 lone pairs on the iodine, but now it's going to be the oxygen on the right that's double bonded. And because they have charges, we put them in brackets with the charge on the outside. So these would be the two acceptable resonance structures for the iodide ion.

Here’s what students ask on this topic:

What are resonance structures?

Resonance structures are a way of representing the delocalization of electrons within certain molecules or ions where the chemical connectivity is the same but the electrons are distributed differently around the structure. This occurs in systems that can't be accurately represented by a single Lewis structure. Instead, multiple Lewis structures, called resonance structures, are used to model the molecule's electronic structure more accurately.

The actual molecule is a hybrid of these different structures, meaning it has characteristics of all the resonance forms rather than existing as any one of them. The electrons are not static; they are spread out or resonate across all the bonds, which can lead to greater stability for the molecule. Resonance is an important concept in understanding the behavior of molecules, especially in organic chemistry, where the delocalization of electrons can significantly affect a molecule's reactivity and properties.

How can resonance structures be drawn?

To draw resonance structures for a molecule or ion, you should follow these steps:

1. Identify the compound: Ensure you know the molecular formula and the connectivity of the atoms within the molecule.
2. Locate electrons: Identify all the valence electrons associated with the atoms in the molecule. Pay special attention to pi electrons and lone pairs that can be delocalized.
3. Determine the need for resonance: Look for atoms that can have multiple bonding arrangements without changing the positions of the atoms themselves. Common indicators include conjugated pi systems, atoms with lone pairs adjacent to pi bonds, or atoms that can have expanded octets.
4. Move electrons, not atoms: Draw resonance structures by moving electrons, not atoms. You can shift pi electrons or lone pairs to form new double bonds, and you can convert lone pairs on adjacent atoms into bonds.
5. Use curved arrows: When drawing, use curved arrows to show the movement of electrons from one location to another. The tail of the arrow starts where the electron pair is currently located, and the head points to where the electrons are moving to.
6. Keep the skeleton the same: The connectivity of the atoms should not change between resonance structures, only the placement of electrons.
7. Ensure valid structures: Each resonance structure must be a valid Lewis structure; that is, atoms

Why can't Lewis structures accurately depict molecules that exhibit resonance?

Lewis structures are a way to represent molecules by showing how electrons are distributed among atoms. However, they have limitations when it comes to depicting resonance. Resonance occurs when a molecule has multiple valid Lewis structures that contribute to the actual structure of the molecule. These structures, known as resonance structures, are like snapshots of the various ways electrons can be arranged within the molecule's framework of atomic nuclei.

The reason Lewis structures fall short in accurately depicting molecules with resonance is that they are static representations. They can show the different possible arrangements, but they cannot convey that the actual molecule is a hybrid of these structures. In reality, the electrons are delocalized and not confined to one position or bonding pattern as a Lewis structure might suggest. The molecule exists as a weighted average of the resonance structures, with the electrons spread over the atoms in a way that cannot be fully captured by a single, static Lewis structure. This delocalization often leads to greater stability and is a key feature of molecules like benzene, where the electrons are shared over several atoms.

What is a resonance hybrid?

A resonance hybrid is a concept from chemistry, specifically in the context of molecular structure and bonding. It represents the overall structure of a molecule that cannot be adequately described by a single Lewis structure. In molecules where resonance occurs, there are multiple valid Lewis structures, known as resonance structures, that can depict the possible distributions of electrons in the molecule. These structures differ only in the position of electrons, not the position of atoms.

The resonance hybrid is essentially a blend or an average of these resonance structures, where the actual molecule is thought to have a structure that is intermediate between them. It incorporates the delocalization of electrons across the molecule, which is a more accurate depiction of the electron distribution. The resonance hybrid is depicted by drawing the common structure and using dashed lines to indicate partial bonds or delocalized electrons, showing that the electrons are not localized to a single atom or pair of atoms but are spread out over the entire molecule or a portion of it. This concept helps to explain the stability and properties of certain molecules that cannot be captured by a single static structure.

How can one draw a resonance hybrid?

To draw a resonance hybrid, you must first identify a molecule that has resonance structures. These are molecules that have the same arrangement of atoms but different arrangements of electrons. Here's how to draw a resonance hybrid:

1. Identify Resonance Structures: Determine all the possible resonance structures for the molecule. These are valid Lewis structures that differ only in the position of the electrons, not the atoms.
2. Draw Lewis Structures: Sketch each resonance structure using Lewis dot structures. Make sure to show the movement of electrons with arrow notation.
3. Overlap Structures: Overlay the resonance structures on top of each other. This can be done mentally or by physically drawing one structure on a transparent sheet and placing it over another.
4. Combine Electron Features: In the resonance hybrid, show shared electron features, such as bonds, as solid lines. For electrons that are delocalized (those that move between structures), use dashed lines or a circle to indicate that the electrons are spread out over several atoms.
5. Indicate Partial Charges: If applicable, indicate partial charges on atoms that gain or lose electron density in the different resonance structures.

The final resonance hybrid is a representation that averages the resonance structures, reflecting the delocalization of electrons within the molecule. Remember, a resonance hybrid is more stable than any individual resonance structure and better represents the true electron distribution.