(b) Why does O₃^- not exist?

Question

Accepted Answer

hey everyone, we're asked to consider the caddy on and the an ion of our beryllium ion and to sketch its molecular orbital diagrams and determine their bond orders. And were asked what these ions exist in the gas phase first. Let's go ahead and start off with our caddy on. We want to calculate the total number of valence electrons we have and we know that brilliant is in our group to a so we're going to multiply two times two since we have two of William. This will get us to four valence electrons. Now since we have that plus one charge, this means we're going to have to subtract one valence electron. This will get us to a total of three failings electrons. Now let's go ahead and draw out our molecular orbital diagram. Starting off with our sigma s we then have our sigma? S anti bonding. Next we have our pipi and next we have our sigma p And again the reason why is because our atomic number is less than eight. And so our signal P is higher in energy than our pipi. Next we have our pi p anti bonding and lastly we have our sigma p anti bonding. Now let's go ahead and fill up our diagram by using Hunt's rule Since we have three valence electrons We're going to input one and and next we have one in our sigma s anti bonding and this gives us a total of three valence electrons to calculate our bond order. We're going to take half times the number of electrons in our bonding molecular orbital in this case it's going to be too and we're going to subtract it by one, which is the number of electrons in our anti bonding molecular orbital. This will get us to a bond order of half. And since we know this, this means that this would exist in our gas phase. Now, let's go ahead and look at our an ion. We're going to do the same steps so we know that beryllium is in our group to a and since we have to we're going to multiply two times two to get four. And since we had that -1 charge, We're going to have to add one valence electron. This will get us to a total of five. Now drawing this out, we're going to draw out the same diagram as we did previously. So we have our sigma s R, sigma s anti bonding, our pipi, followed by our sigma p followed by our pi p anti bonding, followed by our sigma p anti bonding. And we want to fill up our diagram by using hands rule again, we have 1234. And lastly five to calculate our bond order. Again, we're going to take half and we're going to multiply this by the number of electrons in our bonding molecular orbital in this case it's going to be three and we're going to subtract it by two, which is the number of electrons in our anti bonding molecular orbital. This will get us to a bond order of one half, which means it also exists. So these are going to be our final answers for this question. Now, I hope that made sense and let us know if you have any questions.

(b) Why does O₃^- not exist?

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(b) Why does O₃^- not exist?