The iodine bromide molecule, IBr, is an interhalogen compound. Assume that the molecular orbitals of IBr are analogous to the homonuclear diatomic molecule F2. (c) One of the valence MOs of IBr is sketched here. Why are the atomic orbital contributions to this MO different in size?

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hey everyone in this example, we need to determine the valence atomic orbital's of iodine and chlorine involved in the construction of the molecular orbital's for the iodine chlorine molecule, An inter halogen compound assuming that it has molecular orbital similar to the home, a nuclear di atomic molecules of F2 which is flooring as a diatonic molecule. So our first step is to draw out the molecular orbital diagram for F two. So we want to go ahead and recall the total valence electrons first in F2 and we would recognize that flooring on our periodic table is in group Seven a across period two. And so we would recall that the group number corresponds to the number of valence electrons. So we would have for the two atoms of Florin multiply this by these there seven valence electrons. And so that would give us two times seven which would give us 14 electrons total for its molecular orbital diagram. And as we said, flooring is across the second period of the periodic table. So we would begin our molecular orbital diagram for flooring or for F two rather at the bonding sigma two s molecular orbital where we're going to place our first two electrons here. We're then going to go ahead and move up an energy to the anti bonding sigma to s molecular orbital will replace our next two electrons. So now we have four filled in. We need to fill in 10 more electrons. So we're going to move up in energy to go to the sigma two p bonding molecular orbital where we will fill in a total of two more electrons Moving up in energy. We then have our PIN two p Bonding molecular orbital where we have to orbital's here filled in with a total of four electrons pairing are electrons up last to honor the poly exclusion principle. And now we just have four more electrons to fill in because we filled in a total of 10. So moving up in energy, we would go to the pie anti bonding to p molecular orbital where we have to orbital's where we're going to fill in our last four electrons. So we would have 123 and four. And so now we have all 14 of our valence electrons filled in. And just to label these two orbital's in the anti bonding pi two P and bonding pi two P molecular orbital's. This would be P. Why and then pC. And we would have the same labels here, P Y and P. Z. So our next step is to now draw out the electron configurations for our I dine and chlorine atoms given in the prompt. And this is due to the fact that if iodine chlorine has similar molecular orbital's to our homo nuclear di atomic molecules F two, then the valence orbital's of iodine and chlorine will be used to form the molecular orbital's of iodine chlorine chlorine as a molecule. So we're going to draw out the or the electron configurations of first iodine here and we would use the shorthand configuration. So we're going to find the noble gas that comes before iodine, which is across period five on our periodic tables. And that would be the noble gas krypton. So we're going to place that in brackets because krypton comes before I dine on the periodic table. And so it's going to start off our configuration. We would recognize that iodine is in R. P block of our periodic table, meaning we have to pass through the D block where our transition metal section are transition metal section of our periodic table is located in the middle. And we would recall that at the fifth period where I dine is located on our periodic table. The D block begins at the fourth energy level. And because we need to pass through the D block, we would have four D. With all 10 electrons filled in to continue our configuration. We also would recognize that again because I dine is on the fifth period of the periodic table, that we need to pass through the S block, which begins at the fifth energy level here where we would fill in its two electrons. And then, as we stated, iodine is in the P block of our periodic table. And so we would be at the fifth energy level because iodine is in the fifth period and we would count for a total of five units in R P block to land on our atom of iodine corresponding to the five electrons that we would fill in our p block which can hold a maximum of six electrons. So now this completes our configuration for iodine and now we want to go ahead and look at the configuration of chlorine again. We would see chlorine on the periodic table across period three and group seven A. We want to find the noble gas that comes before period three which would correspond to our noble gas being neon to start off our configuration. So once we get to period three we need to pass through our S sub level which begins at the third energy level here. So we would have three S two filled in With two electrons. And then as we stated, we would go into the P block to land on our atom of chlorine. So we would begin at the third energy level of R. P block. So we would have three P. And we would count for a total of five units to land on our atom of chlorine, meaning we would fill in five electrons for our configuration of chlorine. So as we stated, we have 14 electrons total making up our molecular orbital diagram for F two molecule here. And so we want to ensure that our valence orbital's for iodine and chlorine. Give us a total of 14. So we would have at the fifth energy level which is our valence energy level here. A total of five electrons from the p orbital and then two more from the s orbital here. So five plus two is seven electrons in the valence or sorry, in the valence orbital from our configuration of iodine. And then here in the valence orbital for chlorine, we would be at the third energy level. We would account five electrons from the p orbital and two from the S sub level. And so we would have again, five plus two giving us seven electrons in the valence orbital. And we would recall that seven plus seven as a sum is 14. And so it would make sense that these valence electrons from aydin and chlorine would be used to form the molecular orbital's of iodine chlorine as a molecule Since it has similar molecular orbital's to our di atomic molecules F two. And so for our answer, we can say therefore our valence atomic orbital's for iodine first would be our five S orbital As well as five PX, five P y And five p. z. To construct the molecular orbital's for the iodine chlorine molecule. And then for chlorine we would use the three S valence orbital since it was the outermost energy level orbital. And I'm just going to scroll down for more space here Where we would go ahead and fill in our three PX three py and three pz molecular orbital's to form our molecular orbital's for the iodine chlorine molecule. And this part here are valence atomic orbital's from iodine and chlorine that we've listed out here are our final answers 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.

The iodine bromide molecule, IBr, is an interhalogen compound. Assume that the molecular orbitals of IBr are analogous to the homonuclear diatomic molecule F2. (a) Which valence atomic orbitals of I and of Br are used to construct the MOs of IBr?

Verified Solution

Key Concepts

Molecular Orbitals (MOs)

Atomic Orbitals of Iodine and Bromine

Interhalogen Compounds