Average Bond Order Example 1

Jules Bruno
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here, It says, What is the average bond order of the sulfur oxygen bonds within the sulfide ion. So within the sulfide ion, we have one sulfur that is basically double bonded toe one oxygen and single bonded together to the steps we take here. It given on Lee the molecular formula, then draw one of the residents structures. If multiple resident structures are given, choose only one. So let's just choose this first one. We're gonna count the total number of bonds between the surrounding elements involved, and then we're gonna divide the number of bonds by the total number of those surrounding elements. So how many total bonds do we have between sulfur and oxygen? 123 four. So we have four total bonds, and we're gonna divided by the total number of surrounding elements Who are the surrounding elements? Oxygen so divided by three surrounding elements. Okay, so that's gonna be 4/3 or 1.33 So the average bond order off a sulfur toe oxygen bond within the sulfide ion is 1.33 What does that mean? Well, each oxygen, for sure it's single, bonded to the sulfur, but that pesky pie bond because of resonance. It doesn't fully exist with this oxygen or this oxygen or this oxygen. The possibility is it could be with either of them. So they're going to share that one bond so that one bond is cut into three. So each one has ah, third possession of that pie bond. So that's why it comes out toe 1.33 But just remember, just count the total number of bonds and divided by the number of surrounding elements involved, you'll always be able to calculate your average bond order.