Video: Identifying the Bond Strength of Carbon Oxides According to Lewis Theory of Bonding

Which of the following oxides and oxyanions of carbon has the greatest C–O bond strength according to Lewis bond theory? [A] CO₂ [B] CO [C] CO₃²⁻ [D] CO₄⁴⁻ (theoretical)


Video Transcript

Which of the following oxides and oxyanions of carbon has the greatest C–O bond strength according to Lewis bond theory? A) CO₂, B) CO, C) CO₃²⁻, or D) CO₄⁴⁻, theoretical.

Let’s start by getting some clues about how to tackle this from the question itself. The question tells us that we’re going to need to use Lewis bond theory. When you think about Lewis bond theory, you might think of Lewis structures and Lewis dot-cross diagrams. So this could be helpful in helping answer the question. We’re also asked to compare the CO bond strengths in each of these oxides and oxyanions. So let’s think, how could we do that? One of the easiest ways to compare bond strengths is to compare their bond orders. A higher order bond is stronger than a lower order bond. For example, a single bond is not as strong as a double bond. So let’s look at each of the potential answers in turn and try drawing a dot-cross diagram for each.

We’ll begin with carbon dioxide, CO₂. Let’s begin by considering how many electrons are in the valence shell of both carbon and oxygen. We know that carbon has four electrons in its outer shell, while oxygen has six. And for the purpose of these diagrams, we’ll draw carbon’s electrons with a cross and oxygens with a dot. Let’s begin by drawing in one bond between each carbon and each oxygen. This has used up one electron from each oxygen and two of the electrons in carbon. Let’s now try adding in another bond between the central carbon and each of the outer oxygens. Doing this has now used up all of the carbon’s valence electrons. And it has also given a full outer shell of eight.

We simply now need to deal with the two oxygens. Each oxygen has four electrons left over. So we can arrange these in two lone pairs each. We can now count up all the electrons. And we can see that each has had all of its valence electrons accounted for. And also each now has a full octet. So here we see that we have a carbon oxygen double bond. Let’s see how that compares to carbon monoxide. Let’s begin with carbon monoxide simply by drawing in a single bond. This leaves us with three electrons on the carbon and five on the oxygen, which we need to arrange. Let’s try drawing in a second bond. Now this is where it gets a little tricky. Currently, we need four more electrons to fill our carbon’s outer shell. Two of its initial electrons have already been used. And two are left over. If we arrange, it’s left over two electrons as a lone pair. We still need two more electrons in carbon’s outer shell for it to fill the octet role.

Luckily, our oxygen atom still has four remaining electrons. We can arrange two of these electrons as a dative covalent bond and the last two as a lone pair. So carbon monoxide has a triple bond between the C and the O. This has a higher bond order than carbon dioxide. And therefore, it will be a stronger bond. Let’s see how this compares to CO₃²⁻. Remember that the charge on here, the two minus charge, means that we have two more electrons to arrange. Once again, let’s begin by drawing one bond between each of the oxygens to the carbon. If we focus on our carbon, we’ve used up three of its initial electrons, leaving us one more to arrange. It also requires two more electrons to fulfill the octet role. This seems ideal for drawing a second bond between the carbon and one of the oxygens. Once we’ve drawn in this second bond, we can see that our carbon now has a full outer shell. And all four of its initial electrons have been accounted for.

Let’s now look at the oxygen, which is doubly bonded to the carbon. It has four more of its initial electrons left over to arrange. So this can be done as two lone pairs. This leaves it with a full outer shell. And we can move on to the last two oxygens. Each of these oxygens has five electrons left over. We can see that if we arrange these, we’re left with an outer shell that still needs one more electron to fill the octet role. Don’t forget that this is actually a two minus anion. So we have two more electrons that we need to place. If we add one to each of the oxygen atoms, we’re left with everything having a full outer shell. So in this instance, we have two carbon-oxygen single bonds and one carbon-oxygen double bond.

Finally, let’s move on to CO₄⁴⁻. Once again, let’s begin by placing a single bond between the carbon and each of the oxygens. We can see from simply doing this that the carbon’s outer shell is already full. And all of its initial electrons are accounted for. This leaves us with a similar dilemma to that of the last anion, where we have oxygen atoms that still have five of their initial electrons left to arrange. Again, don’t forget that this is actually an anion, this time a four minus anion. So we still have four electrons, which we can arrange. If we arrange them like this, we can see that all of our atoms now have a full outer shell.

So we’ve now drawn dot and cross diagrams and Lewis structures for each of our four potential answers. We now need to decide which will have the greatest carbon-oxygen bond strength. And as we saw before, this means finding whichever answer has the highest CO bond order. Answer A has double bonds. B has a triple bond. C has both single and double bonds. And D only has single bonds.

This means that the answer with the highest bond order and therefore the greatest bond strength is B, carbon monoxide with a triple bond.

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