The energy-level diagram in Figure 9.36 shows that the
sideways ...

Question

The energy-level diagram in Figure 9.36 shows that the 
sideways overlap of a pair of p orbitals produces two molecular 
orbitals, one bonding and one antibonding. In ethylene there 
is a pair of electrons in the bonding π orbital between the two 
carbons. Absorption of a photon of the appropriate wavelength 
can result in promotion of one of the bonding electrons 
from the p2p to the p*2p molecular orbital. (c) Is the C¬C bond 
in ethylene stronger or weaker in the excited state than in the 
ground state? Why?

Accepted Answer

Welcome back everyone. We're told to consider the molecular orbital diagram for our nitrogen gas molecule. We're told that there is a pair of electrons in the bonding sigma orbital that through the absorption of a photon of the appropriate wavelength can result in the promotion of one of the bonding electrons. From the sigma two p molecular orbital to the P two p anti bonding molecular orbital. Were asked whether the nitrogen nitrogen bond is stronger or weaker in the excited state than versus our ground state. So we need to draw out molecular orbital Diagrams and calculate our bond orders for our nitrogen molecule in both of these states. So our first step before we do so is to calculate our total valence electrons for our nitrogen molecule. And we would recall that nitrogen is located in group five a on the periodic table corresponding to five valence electrons. And because we have the subscript of two, we would say that we have two multiplied by the five valence electrons of these nitrogen atoms. And that would give us a total number of valence electrons being 10 valence total. And so that is how much we need to include in our molecular orbital diagram configuration. And so we should recall that because nitrogen is also located across period two, we're going to go all the way up to the two p bonding molecular orbital because we also recognize that nitrogen is in the P block on our periodic table. And so we would begin with the lowest orbital energy here and that would begin at the sigma two s bonding molecular orbital, which we recall only has one orbital. And following Huntsville, we are going to fill this orbital up with our first two electrons of opposite spins. This is our lowest energy orbital for our ground state nitrogen. Moving on up in energy, we would proceed to the sigma anti bonding to s molecular orbital, which also is just one orbital here and filling that in. We have opposite spin electrons. So we've filled in a total of four of our 10 electrons, meaning we have six more electrons to fill in our molecular orbital diagram. And so moving on up further in energy, we have our pi two p bonding molecular orbital where we have a total of two orbital's here and following Hunt's rule, we're going to fill them each in first individually and then pairing them up last. So that leaves us with two last electrons to fill in. So moving on up in energy, we have our sigma two p bonding molecular orbital which consists of just one orbital and we would fill in our last two electrons of opposite spins here. And this gives us our total of 10 electrons filled in our ground state molecular orbital diagram for nitrogen. And as we stated, we want to recall how to calculate bond order and we would recall that that is taking one half and multiplying it by our number of electrons in our bonding molecular orbital which is then subtracted from our number of electrons in our anti bonding molecular orbital. So for our ground state bond order of nitrogen we would say that we have a bond order equal to one half, multiplied by our electrons in our bonding molecular orbital we recall are the orbital's without the asterix. So that would be 2468. So we would have eight electrons subtracted From our electrons and our anti bonding molecular orbital. So the molecular orbital with the aspect where we only have two electrons here. So that would be 8 -2 electrons multiplied by one half. And that is going to give us a bond order equal to a value of three. And now we want to compare this to our excited state molecular orbital diagram and bond order for nitrogen. So starting off with our lower energy orbital, we begin with the sigma two s bonding molecular orbital where we have our first two electrons again moving on up an energy, we have our sigma anti bonding to s molecular orbital with one orbital filled in with the next two electrons. We then have our pi two p bonding molecular orbital with two orbital's filled in with our next four electrons. So we fill them in individually first before pairing them up and then after that we have our sigma two p bonding molecular orbital with just one orbital and this is where things get interesting here because we have according to the prompt one of our bonding electrons that is excited from the sigma two p bonding molecular orbital to the pi two P anti bonding molecular orbital. And so we'll draw out our pi two P anti bonding molecular orbital which we should recall has to orbital's. And because we would have originally placed two electrons here but we now have an excited electron up to the sigma. Or sorry, the pi two p anti bonding molecular orbital. That means we're going to fill in just one electron here. And then we are exciting the next electron up in energy to the anti bonding pi two P molecular orbital. And this is now going to give us a different bond order where we would say our bond order is equal to one half, multiplied by our electrons in our bonding molecular orbital. For the excited state of nitrogen where we have 246 and then seven. So we have seven electrons in our bonding molecular orbital subtracted from our electrons in our anti bonding molecular orbital, the ones with the asterix, we would count two and then three. So that would be seven minus. We have three anti bonding molecular orbital electrons and this is all multiplied by one half, which is going to give us a bond order equal to a value of two. And because we have this bond order equal to two, we would say that therefore the bond order of nitrogen in its excited state is lower. And so thus the nitrogen nitrogen bond is weaker in the excited state. And so for our final answer, we say that the nitrogen nitrogen bond is going to be weaker. So this completes this example. As our final answer. 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.

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