At high temperatures, sulfur vapor is predominantly in the
form ...

Question

At high temperatures, sulfur vapor is predominantly in the 
form of S21g2 molecules. 
(d) When two electrons are added to S2, the disulfide ion 
S22- is formed. Is the bond length in S22- likely to be 
shorter or longer than the bond length in S2? Explain.

Accepted Answer

Hello. Everyone in this video we're trying to do is compare our 02 molecule with the 022 minus or the peroxide. An ion. Or trying to compare specifically is going to be the bond length if it's longer shorter or the same exact length. So how can do this by drawing out the molecular orbital diagram which is also known as the M. O diagram. So the bond length is inversely related to the bond order. As one increases, the other one will decrease or vice versa. Let's go ahead and draw the mo diagram for each. I have to start off with our neutral one first which is going to be our 02. So going from bottom to top will have increasing energy here. Alright, so the central is going to be our M. O. Like an orbital. So that's going to be of our 02. And off to the right and left. We'll have our auction. That's our atomic orbital's. So because we're on the upper right hand corner of our periodic table we're dealing with are two S and two P orbital's. So we can say first we'll have our two S interactions. So then for the molecule will have our sigma and sigma star. Alright. So of course these do have interactions and that's why we represent this with a dotted line. All right, this is it for the two s orbital's we'll go ahead and move on to two P. So same exact thing. We'll have three lines this time for R. Two P. C. For the left side we'll have 12 and three again that's two P. Then we'll have our bonding and anti bonding. So on the bottom we'll have the bonding and then on top we'll have our anti bonding start from the top. We'll have sigma star pi star going down, we have pi and sigma and again these do have interactions and which represent this with a dotted line. Alrighty. But putting in the valence electrons. So we know that auction in the group 68. That means we have six valence electrons per auction adam. So let's go ahead and fill out our atomic orbital's first. So right or left doesn't matter. So just put 12345 and six. Same thing for the right side. We have 12345 and six. Now for the molecular orbital we have two or two atoms of auctions so we'll have a total of 12 fans electrons. So we'll have 123456789, 10, 11 and 12. Alright, so that's the M. O diagram for our 02. Let's now go ahead and calculate our bond order. So the bond order equals have multiplied by all the equation is electrons in our bonding. So let's go ahead and count that. That's just anything without the star. And our medical bills here in central outside so we'll have 1234567 and eight. So let's put that in our equation first. Then subtracted from our anti bonding. So anti bonding will have 123 and four electrons. So -4. So 8 -4 is equal to four times half. Or divided by two, that's equal to two. So for R. O. to hear a bond orders equal to two. Let's go ahead and do the exact same thing. But now we're doing it for the peroxide and against going from bottom to top of increasing energy. This time we're dealing with 02 to minus the same thing for the bottom here from Lockport orbital's and then we'll have the same thing for atomic rules. So again we're dealing with R two S and then we have our anti bonding. So sigma star and bonding just sigma will represent these interactions with a bonded or dotted line. Alright now we're done with the two s. Let's move on to p so 1232 P another 12 and three. So we'll have to pee then we'll have our anti bonding on top and bonding in the bottom. Of course that's just sigma star, pi star pi and sigma. And of course we have our interactions represented with a dollar line. All right now filling in the atomic orbital electrons. First of course we just have six. So 12345 and six. Same thing for the right side. We have 12345 and six. Now for the middle. This time we have a ion. So normally we'll have 12 total valence electrons because of the two minus, it means we have added two extra electrons. So 12 plus two is equal to 14. So we'll go ahead and add 14 electrons here. So we have 123456789, 10, 11, 12, 13 and 14. Alright, let's go ahead and count for our bond order then then we have half and then we go ahead and count the number of electrons in our bonding. So that's 1234567 and eight. Let's go ahead and count for our anti bonding electrons. Now that's 12345 and six. So putting that into the equation eight minus six is equal to two to tempt halve or divide by two is equal to one. The bond order of our peroxide and ion is equal to one. Again, the question is asking is the bond length of R to minus shorter, longer, same length as our 02 are neutral molecule. The answer that we see since we already calculated for our bond order two verses one, the peroxide and I on 02 to minus does have a lower bond order, meaning it has a longer bond length, let's go ahead and put that down. So we can say that 02 to minus has a lower bond order and therefore we can conclude that we have a longer bond length. Alright, so we'll go ahead and highlight this, and this right here is going to be my final answer for this problem.

At high temperatures, sulfur vapor is predominantly in the form of S21g2 molecules. (d) When two electrons are added to S2, the disulfide ion S22- is formed. Is the bond length in S22- likely to be shorter or longer than the bond length in S2? Explain.

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