Adv inorg chem chem 531 exam 2 the molecular orbital mo...

Adv. Inorg. Chem. (CHEM 531) Exam 2 The molecular orbital (MO) diagram of OF is given below. Answer Questions 15-18. 15. How many electrons are in the two degenerate $\pi_{2p}^*$ molecular orbitals (MOs)? A. There are three electrons with one $\pi_{2p}^*$ MO being doubly occupied and the other singly occupied. B. There are two electrons with each $\pi_{2p}^*$ MO being singly occupied. C. There are two electrons that occupy one of the two $\pi_{2p}^*$ MOs. D. There are four electrons that occupy both $\pi_{2p}^*$ MOs. E. There is only one electron that occupies one of the two $\pi_{2p}^*$ MOs. 16. What is the bond order? A. 0.5 B. 1.0 C. 1.5 D. 2.0 E. 2.5 17. The HOMO is ______. A. $\sigma_{2p}^*$ B. $\pi_{2p}^*$ C. $\sigma_{2p}$ D. $\pi_{2p}$ E. $\sigma_{2s}^*$ 18. The LUMO is ______. A. $\sigma_{2p}^*$ B. $\pi_{2p}^*$ C. $\sigma_{2p}$ D. $\pi_{2p}$ E. $\sigma_{2s}^*$

Transcript

00:01 In this question we are going to consider the molecule that is nf now in the first subpart of this question we need to draw a correct molecular orbital energy level diagram for this particular molecule right so here we have to first of all look at the electronic configuration for both of these atoms so for nitrogen the electronic configuration would be 1 s2 2 s2 2p3 because the atomic number for nitrogen is 7 similarly for fluorine having the atomic number 9 the electronic configuration would be 1 as 2 2s2 2p 5 right this is the correct electronic configuration here now the important concept is that in a heterocyclic molecule or or in the heteroatomic molecule right this is a head this is an hetroatomic molecule because in this molecule the two atoms are of different kinds one is nitrogen and the other one is chlorine right so we can say in these kind of molecules the atom that has higher value of electronegativity that atom that has higher value of electronegativity would be having its atomic orbitals and at a comparably lower energy which means let us say that here we have nitrogen and here we have fluorine right so now we are going to just draw the valence shell orbitals that is 2s and 2p for both of them right we can write 2s of nitrogen here and 2p of nitrogen here but since fluorine is having the higher value of electron negativity than nitrogen we can say that the atomic orbitals for fluorine would be shown at a lower energy than that of nitrogen right so we can say 2s of fluorine would be somewhere here and 2p of fluorine would be somewhere here now filling out these orbiters as per their electronic configuration we can say 2 s 2 and 2p3 would be obtained here and then 2s 2 and 2p 5 would be obtained here right so when they just try to combine in form molecule obiters we can say we will get this bonding and this anti -bonding ammo from these from the interaction of the 2s atomic orbital of the two atoms right and similarly we are going to get this kind of 3 bonding and 3 anti -bonding from the interaction of the 2p orbitals of these two atoms here right so this is going to be the structure of or the basic structure of ammo diagram for this nf molecule now filling out the electrons we can say two electrons in both bonding and anti bonding can be filled here and in this particular portion there are a total of eight electrons that can be filled like this right that can be filled like this so this is how we can just draw the correct m o diagram for nf molecule now in the second subpart of this question we are about to find out the bond order for this particular molecule now bond order can be easily obtained by the simple formula number of electrons in the bonding ammo minus number of electrons in the anti -bonding ammo divided by two right this is going to be equal to two and then there are six so total of eight electrons are there in the bonding amos and minus there are total of four electrons in the anti -bonding ammo two are here and two are here and this is going to be divided by two so the overall bond order value comes out as two which is similar to oxygen.

03:00 Now here we can say that we need to find out the correct molecular correct point groups of the molecular orbitals of this molecule, right? so we can just simply find out that if nf is there having a double bond because bond order is two, right? so we can say nf would be somewhere like this and this is going to be a linear molecule, right? this is going to be a linear molecule.

03:21 Now for a linear molecule we can say we can just write its structure like this and we will always have an axis...

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