Carbon monoxide, CO, is isoelectronic to N2. (d) Would you expect the p2p MOs of CO to have equal atomic orbital contributions from the C and O atoms? If not, which atom would have the greater contribution?

Video transcript

welcome back everyone in this example, we need to determine which of the below statement is a true statement given that oxygen and nitrogen flooring are Isil electronics. So we first should recognize that electronic means that we have the same configuration for these two molecules and because they have the same configuration, they're going to have the same number of electrons. Now looking at our answer choices, we can see that they're discussing our atomic orbital's of each of our atoms and their contribution to the bonding pi two P molecular orbital. So we want to recall that atomic orbital's of more electro negative elements have lower energy than atomic orbital's of the less electro negative elements. We also should recall that when Adams of two elements combined, the lower energy atomic orbital or we'll say orbital's make a greater contribution two, the bonding molecular orbital's so we'll underline lower energy and greater contribution. So with these two facts we've recalled, we want to compare our Adams nitrogen versus flooring and we would recognize that based on our electro negativity trend on our periodic table, which we recall increases as we go from the left to the top right of our periodic table. Flooring is the most electro negative element on the periodic table. And due to that we would say that therefore florins atomic orbital has less energy and thus a greater contribution to the pi two P bonding molecular orbital and we'll write out bonding molecular orbital. And because we would recognize that based on what we've outlined, nitrogen would therefore be less electro negative and therefore have more energy and a lesser contribution to our bonding molecular orbital. We would rule out Choice A. As well as Choice C and E. And that would just leave us with Choice B, which states that the atomic orbital of nitrogen has a greater or has a smaller contribution to the pi two P bonding molecular orbital's of nitrogen flooring than the atomic orbital of flooring. And so B would be our final answer to complete this example. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.