In the last lecture we have seen the electronic transitions and the vibrational structure of these electronic transitions.

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1 Title: Term vales of the electronic states of the molecle Pae-1 In the beinnin of this modle, we have learnt the formation of the molecle from atoms. We have also learnt the moleclar orbital and the electronic states of the molecles. In the last lectre we have seen the electronic transitions and the vibrational strctre of these electronic transitions. The nomenclatre of these electronic states needs to be nderstood and the transition selection rles need to be evalated so that the electronic transitions between the two electronic states of the molecle cold be explained. There are several ways to determine the term vales of the electronic states. However, in this lectre we will focs only one procedre.

2 Pae- In lectre-6, when we solved the Schrödiner eqation for the hydroen molecle ion to nderstand the stability of the electronic states, we et a nmber of different potential enery crves. We refer to these states as the electronic states of the molecle. The electronic state correspondin to the lowest enery is known as rond electronic state and the others are known as excited electronic states. In analoy to the case for atomic electronic states, we will identify these states sin moleclar term symbols. The method for determinin these moleclar term symbols is somewhat more complicated than that for atomic term symbols. The manifold of the electronic states can be obtained, as for atoms, by the sccessive brinin toether of the parts. This bildin p of the atom can be done only in one way. Bt for the molecle there are three different ways. Process-I: The molecle may be bilt p by brinin the atoms toether. This means that the moleclar states will be the reslt of the iven states of separated atoms. Ths, when we carry ot for all possible combinations of separated atomic states, we obtain the complete manifold of the states of the molecle. Process-II: Instead of developin the moleclar states from the separated atoms, we can start from the so called nited atom and split them hypothetically. Process-III: Since we have already learnt the electronic arranements of molecles in lectre-7, we can start with the electronic confiration of the orbital of the molecle and then determine the moleclar term vales as we have done it in case of atoms. In this lectre we will determine the term vales of the electronic states followin this procedre.

3 Pae-3 The terms tell s the vales of certain anlar momenta. In the case of molecle, we have to define the anlar momenta in terms of the orbital anlar momentm L and the spin anlar momentm S of the atoms. As we know that in the atom L and S are the ood qantm nmbers when we determine the terms becase the motion of the electrons in an atom takes place in a spherically symmetrical field of nclear force. In case of spin orbit coplin, L and S cople to ive total anlar momentm J (= L S). In case of linear molecle, the symmetry of the field in which the electrons move is redced. There is only axial symmetry abot the internclear axis (the cylindrical symmetry) created by the stron electric field of the nclei. This destroys the relationship between J, L and S. Not only that even the L ceases to be the ood qantm nmber. As a conseqence, only the component of L alon the inter-nclear axis is a constant of motion or ood qantm nmber. In an electric field, nlike the manetic field, reversin the directions of motion of electrons does not chane the enery of the system. Which means that the eneries of M L = 1 and M L = -1 will be deenerate. Therefore, it is convenient to classify the electronic states of diatomic molecles accordin to the vales of M L not L. Ths we define the projection of L alon the inter-nclear axis as Λ as shown in fire So, Λ= M L..(34.1) L Λ Fire-34.1

4 Pae-4 Accordin to eqation 34.1, for a iven vale of L, the qantm nmber Λ can take the vales Λ= 0,1,,..., L = λi i..(34.) Where λ is the projection of l of individal electrons alon internclear axis. For example, λ = 0 for a electron, λ = ±1 for a π electron and so on. Dependin on the vales of Λ, the electronic state is defined. For Λ = 0, it is state For Λ = 1, it is state For Λ =, it is state For Λ = 3, it is Φ state Note here that the deenerate π moleclar orbital is defined as π and π -, where and sbscripts represent the clockwise and anticlockwise rotation. An example of the two electrons is shown in fire-34.. For two electrons havin l 1 = 0, l = Λ= 0 Σ state Λ= ± state Λ= ± 1 Π state Fire-34.

5 Pae-5 A moleclar term symbol labels the moleclar states and specifies the total spin and orbital anlar momentm of the molecle, alon with varios other symmetries. The term symbol is written as S 1 ( / ) Λ ( / ) In the followin we will nderstand the meanin of these notations. 1. As defined, Λ is the qantm nmber for the total orbital anlar momentm L of the electrons abot the inter-nclear axis.. As defined, S is the total spin anlar momentm qantm nmber, formed from the individal electron. Spin qantm nmber of sinle electron s = 1/, and m s = ±1/. For one npaired electron, S = s = 1/ (a doblet state with S 1 = ). For two npaired electrons, the possible vales are S = 1 and 0. For S = 1, ives triplet (S 1 = 3) and for S = 0 ives sinlet (S 1 = 1) states. 3. Symmetric molecles are havin a centre of symmetry. The erade () or nerade () sbscripts apply only to these molecles and labels the symmetry of the electronic wavefnction with respect to inversion throh this centre. For the molecle, this can be thoht of as simple mltiplication of fnctions of even or odd symmetry of the individal electrons. In the followin table provides the resltant symmetry of the molecle from two electrons. Individal electrons erade ( ) erade(), even even Gerade () nerade (), even odd nerade () nerade (), odd odd Moleclar erade, nerade, erade, 4. The / sperscript applies only to state (Λ = 0), and labels the symmetry of the electronic wavefnction with respect to reflection in a plane containin the nclei.

6 Pae-6 Now we will carry ot some examples of diatomic molecles to find ot the rond state from its electronic confiration. Closed shell confiration: For any closed shell moleclar electronic confiration, for example s, there is no net orbital and spin anlar momentm. So, Λ = 0 and S = 0. The term will be always, 1 Σ state. One npaired electron: Let s take the case of hydroen moleclar ion H The electronic confiration 1 (fire-34.3). M L = 0 Λ= 0; Σstate (a) (b) S = 1 Moleclar orbital Fire-34.3 Bond order = (no. of electrons in bondin orbital- no. of electrons in anti-bondin orbital) / = (1-0) / = ½ > 0. Moleclar electronic state is stable. Moleclar term for the rond state is Σ He 1 (fire-34.4) M L = 0 Λ= 0; Σstate S = 0 1 = 0 Bond order = ½ > 0. Stable rond state will form. Parity = = Moleclar term for the rond state is Σ (a) Moleclar orbital Fire-34.4 (b)

7 Pae-7 Two electrons system: Hydroen molecle H. Moleclar electronic confiration is (fire-34.5). Bond order = 1; molecle is stable in the rond electronic confiration. Λ= 0; S = 1 or 0 ; Bt de to Pali exclsion principle, S = 0 for closed shell. Parity = = Moleclar term for the rond state (a) Moleclar orbital Fire-34.5 (b) is 1 Σ Let s draw the Λ-M S table as we have done in case of atoms. Here since Λ= λ we i i / / will write as ( λ, λ ) where and represents m S = 1/ and m S = -1/, respectively. 1 For electrons λ = 0, so the Λ-M S table for confiration is M S Λ 0 0 (0, 0 - ) Now, we can write down the wavefnctions. The spatial wavefnction is φ = (1) () spatial Antisymmetric The spin wavefnction is χ = [ α(1) β() β(1) α() ] spin The total moleclar wavefnction is [ ] Ψ( Σ ) = (1) () α(1) β() β(1) α()

8 Pae-8 For He molecle, the electronic confiration Bond order = 0; No bond forms, rond state is not stable. For Li molecle, the electronic confiration Bond order = 1; Grond state is stable. s Λ= 0; S = 1 or 0 ; bt de to Pali exclsion principle, S = 0 So, the moleclar term for the rond state is 1 Σ For O molecle, Moleclar electronic confiration is iven in fire-34.6 s s p pπ pπ and Bond order = (6-4)/ = 1, molecle is stable. 4 π π 4 p π π 4 p s (a) Moleclar orbital Fire-34.6 s (b)

9 Pae-9 So the Λ-M S table for the electronic confiration is iven below. For π electrons, λ 1 = 1 and -1; similarly λ = 1 and -1. For spin, m s = 1/ and m s = -1/. M S Λ (1, 1 - ) (1 -, -1 - ) (1, -1 - ),(1 -, -1 ) (1, -1 ) (-1, -1 - ) ---- Now, we will look at the wavefnctions. The spatial wavefnction for Λ = φ = π (1) π () spatial Since this is symmetric, we have to mltiply with the antisymmetric of spin wavefnction to make the total wavefnction antisymmetric. The antisymmetric spin wavefnction is Antisymmetric 1 χspin = α β β α [ (1) () (1) () ] And the total wavefnction is for (Λ = and term is 1 ), 1 1 [ ] Ψ( ) = π (1) π () α(1) β() β(1) α() For Λ = -, the spatial wavefnction is φ = π (1) π () and this will ive rise to same term 1. So 1 is dobly deenerate. spatial

10 Pae-10 Now for Λ = 0, the spatial wavefnction is 1 φspatial = π π ± π π [ (1) () (1) () ] The symmetric wavefnction is symmetric 1 φspatial = π π π π [ (1) () (1) () ] Becase by exchanin the electrons, wavefnction does not chane. symmetric 1 φspatial (,1) = π () π (1) π () π (1) φ symmetric spatial symmetric (1, ) = φ (,1) spatial [ ] We have to mltiply with the antisymmetric of spin wavefnction to make the total wavefnction antisymmetric. Since Λ = 0, M S = 0, it ives rise to 1 Σ and the total wavefnction is, Ψ( Σ ) = π (1) π () π (1) π () α(1) β() β(1) α() 1 1 [ ][ ] The antisymmetric spatial wavefnction is Antisymmetric 1 φspatial = π π π π [ (1) () (1) () ] Antisymmetric 1 φspatial (,1) = π () π (1) π () π (1) φ Antisymmetric spatial [ ] Antisymmetric (1, ) = φ (,1) spatial Since this is antisymmetric, we have to mltiply with the symmetric of spin wavefnction to make the total wavefnction antisymmetric. The symmetric spin wavefnctions are Symmetric χspin = α(1) α() = β(1) β() 1 = [ α(1) β() β(1) α() ]

11 Pae-11 Since Since Λ = 0, M S = 1, it ives rise to 3 Σ and the total wavefnction is, 3 1 [ ] Ψ( Σ ) = π (1) π () π (1) π () α(1) α() 1 = [ π (1) π () π (1) π () ] β(1) β() 1 = [ π (1) π () π (1) π () ][ α(1) β() β(1) α() ] So the rnd state electronic confiration of oxyen ives three terms namely, 3 Σ, 1 and 1 Σ. For enery orderin, we will se the Hnd s rle. Hnd s rles are applied to determine the enery orderin of terms arisin from the rond state electron confiration of a molecle: 1. The term with the hihest spin mltiplicity, S 1, is lowest in enery. This stems from the electron spin correlation: electrons with parallel spins have a tendency to spend more time frther apart, on averae, than those with paired spins.. For terms of the same mltiplicity, the term with the larest orbital anlar momentm, iven by Λ, is lowest in enery. Accordin to Hnd s rle, the terms arisin from the rond state of O lie in the order 3 Σ < 1, < 1 Σ. For Be molecle, Atomic electronic confiaration : s Moleclar electronic confiration: s s Bond order = 0: no bond formation: molecle is not formed with Be.

12 Pae-1 The electric dipole selection rles for the electronic transitions for linear molecles can be determined from the followin 1. Total anlar momentm selection rle: The total anlar momentm chane i.e Λ = 0, ± 1. That means Σ Σ, Π Π, will be allowed becase Λ = 0. Similarly, Σ Π, Π, Π Σ, Π will be allowed becase Λ = ±1.. Since we are takin electric dipole transitions, the spin shold be conserved. So the spin selection rle is S = 0. Please note that this is tre in the absence of spin orbit mixin i.e. spin orbit coplin. 3. The selection rles for the plane of reflection symmetry, Σ Σ and Σ Σ 4. Since the electric dipole transition operator chanes parity by one in the wavefnction, the transition selection rles for the parity are and.

13 Pae-13 Total anlar momentm of the electrons: Spin-orbit coplin The total electronic anlar momentm abot the inter-nclear axis, denoted by Ω, is determined by addin Λ and M S, as we have done it in case of atom. So the total qantm nmber of the resltant anlar momentm abot inter-nclear axis is Ω= Λ M s For example, if the electronic term of a molecle is Π, then S = 1/ and Λ = ±1. Then the projection of spin alon the inter-nclear axis M S = ±1/. So the total anlar momentm is 1 3 Ω= 1 = 1 1 Ω= 1 = 1 1 Ω= 1 = 1 3 Ω= 1 = Ths, the term Π splits into two mltiplets i.e. spin-orbit levels Ω=± 3 and Ω=± 1. Both these levels are deenerate. The symbol for these mltiplets are Π 3 and Π 1 Pae-14

14 Recap In this lectre we have learnt that becase of the stron inter-nclear force, the individal qantm nmbers of the electrons in a molecle do not remain ood qantm nmbers, instead the projections alon the inter-nclear axis need to be considered. We have also learnt the procedre for determinin the terms of the molecle from the electronic confiration. As in case of atom, in the molecle also the spin orbit interaction takes place and the total qantm nmber of the electrons arise from their projection alon the inter-nclear axis.