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Ch.9 - Molecular Geometry and Bonding Theories

Chapter 9, Problem 67b

Predict the molecular geometry of each of the following molecules: (b) H O C O C O O H

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Hello everyone. So in this video we're given this structure right over here and we're trying to see what the molecular geometry around each labeled atom is. So usually we only have one central atom but this molecule has two so well separated into this here and this. Alright, so for a our central atom is going to be our carbon. So let's go ahead and count how many groups are attached to it. So we have 12 and three groups. So because carbon has three groups bonded to it, this means that the molecular geometry is going to be tribunal player geometry. So say A is tribunal. Let me go ahead and rewrite that a little bit better. So we have trig no plainer geometry. Okay then next is B and B are essential atom is labeled as our nitrogen. So we have 123 bonds are three groups attached to our nitrogen and then we have one lone pair because nitrogen has three groups and one lone pair. The molecular geometry for that is going to be tricky. No parameter. Alright. And those two are going to be my final answer for this question again for our carbon. That's tribunal planner and nitrogen is tribunal parameter. Alright, thank you all so much for watching