CYU5301 Unit III Session 1 PDF

Summary

This document covers electronic configurations and electronic states of diatomic molecules. It delves into topics such as molecular orbitals and electronic states, providing a foundational understanding of quantum chemistry.

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Session 1 Electronic configurations and electronic states of diatomic molecules Contents Introduction 1.1 Electronic orbitals of atoms. 1.2 Homonuclear diatomic molecules 1.2.1 Shapes and symmetry properties of molecular orbitals. 1.2.2 Nomenclature of molecular orbitals. 1.2.3 Electronic configura...

Session 1 Electronic configurations and electronic states of diatomic molecules Contents Introduction 1.1 Electronic orbitals of atoms. 1.2 Homonuclear diatomic molecules 1.2.1 Shapes and symmetry properties of molecular orbitals. 1.2.2 Nomenclature of molecular orbitals. 1.2.3 Electronic configuration of a molecule. 1.2.4 Spin multiplicity. 1.3 Heteronuclear diatomic molecules. 1.3.1 Orbitals have no inversion symmetry. 1.3.2 Order of molecular orbital energy levels. 1.4 Electronic states arising from electronic configurations of diatomic molecules. 1.4.1 Total orbital angular momentum along the internuclear axis. 1.4.2 Inversion symmetry of the electronic wave function of a homonuclear diatomic molecule. 1.4.3 Symmetry of the electronic wave function with respect to reflection through a vertical plane in a diatomic molecule. 1.4.4 Identification of all possible electronic states generated by a given electronic configuration. 1.5 Hund’s rules for the identification of the ground state. 1.6 Selection rules in electronic spectroscopy. Summary Learning outcomes Introduction In a spectrum, the positions (frequency) of the peaks depend on the energy level separations in the transitions involved. The intensities (heights) of the peaks depend on the characteristics of the wave functions of the states involved in the transitions. The said characteristics also determine the selection rules. In the case of rotational spectroscopy, you used simple 1 Unit III − An Introduction to Electronic Spectroscopy of Molecules models, viz. rigid rotor and non-rigid rotor, in determining the energy levels and the characteristics of the wave functions. Similarly, in the case of vibrational spectroscopy you used harmonic oscillator and Morse oscillator as the models. In both of the above cases you obtained simple algebraic expressions for the positions of peaks. The specific selection rules involved changes in quantum numbers which were deduced from the said models. Such simple models are not available for the calculation of electronic energy levels and wave functions of molecules. Extensive computer programmes are utilised in calculating electronic energy levels. However, careful study of the nature (various symmetry properties) of the electronic wave functions gives the specific selections rules in terms of the changes of various quantum numbers (or quantised properties) without extensive calculations. (There is no gross selection rule since any molecule can give an electronic spectrum.) In this session you study the nature of electronic wave functions of diatomic molecules. You will learn the specific selection rules in terms of the associated quantum numbers or quantised properties. As you have learned in Unit II of CYU3300, electrons in a molecule reside in molecular orbitals. Molecular orbitals may be imagined to be formed due to the overlap of atomic orbitals. As such you begin the study by refreshing your memory on the orbitals of atoms. Symmetry properties of molecular orbitals decide the symmetry properties of electronic wave functions of molecules. As such, a thorough study of various symmetry properties of molecular orbitals of homonuclear diatomic molecules is undertaken next. Thereafter, what is learned there is extended to heteronuclear diatomic molecules. In CYU3300 you learned that, analogous to an atom, a molecule consists of electronic shells. An electronic configuration of a molecule is a description of populations of electrons in these shells. A molecular orbital describes the state of a single electron in a molecule. The electronic wave function of a molecule, which is used in deriving the selection rules, describes the collective behaviour all the electrons in a molecule. Such a wave function describes an electronic state of a molecule. A given electronic configuration may generate a number of such electronic states. You will learn to deduce symmetry properties of such electronic states which are important in studying electronic spectra of diatomic molecules. At the end of the session you will learn the specific selection rules in electronic spectroscopy of a diatomic molecule. 1.1 Electronic orbitals of atoms You have studied the electronic structure of atoms in Unit I of CYU3300. There you learned that the electrons in an atom reside in atomic orbitals. An atomic orbital gives the electronic density distribution in space. You learned that different atomic orbitals have different shapes; see Figure 1.1. 2 Session 1 − Electronic configurations and electronic states of diatomic molecules z z z + − + − + y y y − + x x x s pz py px Figure 1.1. s− and p−orbitals of atoms An s-orbital is spherical and the corresponding wave function is positive everywhere as indicated in the above figure. A p z orbital is concentrated along the z-axis. The signs in the diagrams refer to the sign of the wave function. (these + and – signs do not refer to a charge.) For example, the wave function corresponding to a p z orbital is positive for positive z-values and negative for negative z-values. You have learned that an electron in an atomic orbital is described by four quantum numbers, n, l , ml and s. No two electrons in an atom can have the same set of four quantum numbers (at least one quantum number is different). ml (called the magnetic quantum number), is quantum number for the z- component of the orbital angular momentum. Following table summarises the possible values of ml of an electron in different orbitals. Orbital Possible values of ml s 0 p −1, 0, + 1 d −2, − 1, 0, + 1, + 2 f −3, − 2, − 1, 0, + 1, + 2, + 3 s is the spin quantum number of an electron which takes the values + 1 2 or − 12. You also learned that the energy of an electron in an atom depends on the orbital it is in. For example, energy of an electron in the 1s orbital in a lithium atom is different from that in the 2s orbital in the same atom. The order of these orbital energies is 1s  2s  2p  3s  3p  4s  3d ....... Such an orbital energy level is also called a shell. The number of orbitals associated with an orbital energy level is called the degeneracy of that orbital energy level. For example, s-levels are non- degenerate; a p-level is 3-fold degenerate. Note that the degeneracy of an atomic orbital energy level is equal to the number different ml values associated with it. 3 Unit III − An Introduction to Electronic Spectroscopy of Molecules The maximum number of electrons an orbital can accommodate is two; one with s = + 1 2 and the other with s = − 1 2. Using this fact and the degeneracy you can determine the maximum number of electrons an orbital energy level can have. Q1. Determine the maximum number of electrons a p-level can have. A1. Each orbital can have only two electrons. Hence three orbitals can have 2 + 2 + 2 = 2  3 = 6. You also learned that the electronic configuration of an atom indicates the distribution of electrons among the atomic orbital energy levels. For example the electronic configuration of ground state of C is 1s2 2s2 2p2. As indicated above, in carbon atom, the 1s and 2s orbitals are fully occupied since an s-level is non-degenerate. However, 2p-level is partially filled. This leads to a number of possible p-orbital occupations; viz. ( p x ) , ( py ) , 2 2 ( pz )2 , ( px )1 ( py ) , ( py ) ( pz )1 and ( p ) ( p ). From earlier studies you know 1 1 1 1 z x that the last three occupations have lower energy than the first three. Activity 1.1 1 By examining Figure 1.1, state the sign of p x and p y orbitals in different regions of space. 2 State the degeneracies of d- and f-levels? [Unit I of CYU3300] 3 Determine the maximum number of electrons that can be in s-, p-, d- and f-levels. 4 Write down, in standard notation, the ground state electronic configurations of Li, Be, B, C, N, O, F and Ne atoms. Comment on the maximum number of electrons each of the involved shells can have. 5 The ground state electronic configuration of Cr is 1s2 2s2 2p6 3s2 3p6 3d4 4s2. Indicate, in standard notation, the possible occupations of electrons in the 3p- and 3d-levels in the above configuration. [see Unit I of CYU3300 to refresh your memory on 3d orbitals] 1.2 Homonuclear diatomic molecules 1.2.1 Shapes and symmetry properties of molecular orbitals In Unit II of CYU3300 you learned that molecular orbitals are formed due to the overlap of atomic orbitals. Shape of the resulting molecular orbitals depends on the shape of the overlapping atomic orbitals and their relative signs. (See Unit II of CYU3300). The molecular orbitals formed by the 4 Session 1 − Electronic configurations and electronic states of diatomic molecules overlap of two s-orbitals in a homonuclear diatomic molecule are shown in Figure 1.2. 1sa − 1s b *u1s + − Nucleus a Nucleus b ⚫ ⚫ z−axis Internuclear Nodal Internuclear ⚫ ⚫ plane axis axis + + 1sa 1s b + ⚫ ⚫ z−axis g 1s 1sa + 1s b Figure 1.2: Formation of molecular orbitals by overlapping 1s atomic orbitals. As in atomic orbitals, the + and – signs in molecular orbitals represent regions where the wave function is positive and negative, respectively. Direct overlap of two 1s atomic orbitals (labelled as 1s a and 1s b ) generates two molecular orbitals, denoted by g1s and *u 1s. One of them is a bonding orbital, g1s , formed by overlapping 1sa + 1sb and the other one is an anti bonding orbital, *u 1s , formed by the overlapping 1sa − 1s b. In bonding orbitals, the region of highest electron density occurs in between the nuclei. In anti bonding orbitals, there is a plane between nuclei with zero electron density (nodal plane). Note that, traditionally, the internuclear axis is taken to be the z-axis. For a diatomic molecule it is an infinite order rotational axis and is the principal axis of symmetry. This axis is an infinite order rotational axis for both g1s and *u 1s orbitals as well. In this sense, we say that g1s and *u 1s are rotationally symmetric about the internuclear axis. (Note that, in studying symmetry of molecules in CYU4300 we referred to the nuclear frame of the molecule. We neglected the electrons there. However, here we are talking about various symmetry properties of orbitals.) As you have learned in CYU4300, any symmetry plane containing the principal axis is a vertical plane,  v. (Do not get confused with the sigma orbitals with the reflection plane. Both are represented using the letter sigma.) Figure 1.3 indicates the result of a reflection operation on the * orbital through the xz-plane (which is a  v ). 5 Unit III − An Introduction to Electronic Spectroscopy of Molecules y y x x ++ −− ^ v ++ −− ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ z z Figure 1.3: Reflection of *u 1s through the xz-plane You observe that the orbital remains unchanged due to the reflection operation. Mathematically it may be written as ˆ v *u1s = + *u1s. Here, ( ) ( ) ( ) ̂ v indicates the reflection operation and *u 1s indicates the anti bonding orbital. The mathematical expression indicates that ̂ v operation on the orbital gives the orbital back. Such an orbital is said to be of “+” symmetry with respect to a ̂ v operation. Same is true for the bonding sigma orbital as well. Hence the bonding sigma orbital has “+” reflection symmetry. [You will learn that reflection symmetry plays an important role in determining the specific selection rules in electronic spectroscopy.] Q2. Why have you used subscripts u and g with the above molecular orbitals? A2. The bond midpoint is an inversion centre of the bonding molecular orbital. Such a molecular orbital is said to have gerade symmetry. This fact is represented using a subscript “g”. When inverted through the bond midpoint the shape of the anti bonding orbital does not change. However, the sign changes during inversion. Such an orbital is said to have ungerade symmetry. This fact is indicated with a subscript “u”. See Figure 1.4 where î represents the inversion operation through the bond midpoint. + − ^ i − + *u1s ⚫ ⚫ ⚫ ⚫ Bond midpoint ^ i g 1s ⚫ ⚫ ⚫ ⚫ + + Figure 1.4: Inversion of bonding and anti bonding sigma orbitals through the bond midpoint. 6 Session 1 − Electronic configurations and electronic states of diatomic molecules Q3. How do you represent the above facts mathematically? ( ) ( A3. î *u1s = − *u1s ) î ( g1s ) = + ( g1s ) [You will learn that inversion symmetry plays an important role in determining the selection rules in the electronic spectroscopy of homonuclear diatomic molecules.] In contrast to s-orbitals, p – orbitals centred on two atoms can overlap directly as well as laterally. The p z – orbitals lying along the internuclear axis, overlap directly. p x and py orbitals that are on a plane perpendicular to the internuclear axis and overlap laterally. See Figures 1.5, 1.6 and 1.7. y y x x + + + + − + − + z z pz pz − − − − px py px py Atom a Atom b Figure 1.5: 2p – orbitals of an atom. Lobes of p x , py and p z lie along the x, y, z axes − ⚫ + − ⚫ + − ⚫ + − ⚫ + z-axis z-axis pz − atom a pz − atom b *u2pz − antibonding molecular orbital − ⚫ + + ⚫ − − ⚫ + ⚫ − z-axis z-axis pz − atom a pz − atom b  g2pz − bonding molecular orbital Figure 1.6: Direct overlap of 2p z orbitals on two atoms produces a g 2pz bonding molecular orbital and a *u 2p z antibonding molecular orbital 7 Unit III − An Introduction to Electronic Spectroscopy of Molecules + − + − ⚫ ⚫ ⚫ ⚫ z-axis z-axis − + − + py py *g 2py − antibonding molecular orbital Atom a Atom b + + + ⚫ ⚫ ⚫ ⚫ z-axis − z-axis − − py py  u 2py− bonding molecular orbital Atom a Atom b Figure 1.7: Lateral overlap of 2p y – orbitals forms u 2p y bonding molecular orbital and a *g 2py antibonding molecular orbital. As in figure 1.7, overlap of 2p x − orbitals form u 2p x and *g 2px molecular orbitals. The lateral overlap of p x and p y atomic orbitals of two atoms generate two bonding molecular orbitals with same energy and two antibonding molecular orbitals with same energy. Orbitals with same energy are called degenerate orbitals. Therefore, lateral overlap of p x and p y orbitals form doubly degenerate the u 2p and *g 2p orbital energy levels. Each molecular orbital can accommodate two electrons as in the case of an atomic orbital. It can be seen from Figure 1.7 that in  molecular orbitals, high electron density regions occur laterally. Unlike  and * molecular orbitals,  and * -orbitals are not rotationally symmetric.  -orbitals have ungerade inversion symmetry and * -orbitals have gerade inversion symmetry. Q4. Determine the reflection symmetry of the bonding and antibonding orbitals in Figure 1.7 with respect to reflection through a plane perpendicular to the plane of this paper and passing through nuclei. A4. See Figure 1.8. y y x x + − − + ++ −− ^ xz ++ −− ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ ⚫ z z − + + − Figure 1.8: Reflection of *g through the xz-plane 8 Session 1 − Electronic configurations and electronic states of diatomic molecules From Figure 1.8 it is clear that the orientation of the orbital remains the same in the mirror image. However, the signs of the lobes have changed. This means that,  ( ) ( ˆ xz *g 2py = − *g 2p y. We say that ) *g 2py has “−” symmetry with respect to reflection through the xz-plane. Activity 1.2 1 Briefly explain what is meant by the statement that “ g1s and u* 1s orbitals are rotationally symmetric about the internuclear axis”. 2 Using a diagram similar to that in Figure 1.3, determine the symmetry of g1s with respect to reflection through xz-plane. 3. Determine the symmetry of g1s and u* 1s with respect to reflection through yz- plane. 4 Determine the symmetry of u 2py with respect to reflection through xz-plane. 5 Determine the symmetry of u 2py and *g 2py with respect to reflection through yz-plane. 6 Unlike in the case of  orbitals, symmetry of u 2py and *g 2py with respect to reflection through xz-plane is different from reflection through yz-plane. Why? 7 The inversion operation, î , of a function, f, of the position coordinates (x,y,z), through the origin of the coordinate system, is defined by î f(x,y,z) = f( − x, − y, −z). Using this definition, identify the gerade and ungerade functions in the following set. (i) f1(x,y,z) = x + y + z (ii) f2 (x,y,z) = x 2 + y 2 + z 2 (iii) f3 (x,y,z) = x + y 2 + z3 (iv) f4 (x,y,z) = cos(x + y + z) 8 The reflection operation,  ˆ xz , of a function, f, of the position coordinates (x,y,z) through the xz-plane is defined by ˆ xz f(x,y,z) = f(x, −y,z). Using this definition, identify the type of reflection symmetry of the functions in the following set. (i) f1(x,y,z) = x + y + z (ii) f2 (x,y,z) = x 2 + y 2 + z 2 (iii) f3 (x,y,z) = x + y 2 + z3 (iv) f4 (x,y,z) = y ( x + z ) 1.2.2 Nomenclature of molecular orbitals You have already seen that a molecular orbital is designated with a lower case Greek letter followed by the symbol indicating the atomic orbitals that generated it. In addition, a superscript “*” indicates an antibonding molecular orbital. A subscript “g” or “u” indicates the type of its symmetry with respect to inversion through the bond midpoint. As in the case of an atom, the component of orbital angular momentum of an electron in a molecular orbital along the z-axis is equal to ml where 9 Unit III − An Introduction to Electronic Spectroscopy of Molecules the quantum number ml = 0,1, 2,3...... For a given molecular orbital, you may take ml to be equal to that of the atomic orbitals used in generating it. Q5. Can you explain meaning of the sentence in italics using examples? A5. Let us find out the value of ml for the bonding and antibonding orbitals described in Figure 1.2. ml = 0 for 1s orbitals. Hence, ml = 0 for g 1s and *u 1s. ml = 0 for a 2p z orbitals. Hence ml = 0 for g 2p and *u 2p. In fact the Greek letter used in designating a molecular orbital is chosen based on the value of ml for the orbital. You define  = ml. Then the Greek letter is chosen as shown in the following table.  0 1 2 3 4 Letter      Q6. What are the possible values of ml of an electron in a  ,  ,  and  orbital, respectively? A6. ml = 1,  2,  3,  4 , respectively. The energy of an electron in a molecular orbital is determined by . Hence, the degeneracy of a molecular orbital energy level is due to different values of the quantum number ml. Q7. What are the degeneracies  ,  ,  ,  ,.... molecular orbital energy levels? A7. As in the case of atoms, the degeneracy of a molecular orbital energy level is equal to the number of ml values associated with it. As indicated in A6, each energy level is associated with two possible values of ml ( =  ). Hence, any molecular orbital energy level, except a  -level, is doubly degenerate. A  -level is non-degenerate. A molecular orbital energy level may also be called a molecular electronic shell. Activity 1.3 1. How many molecular orbitals are associated with each of the, , ,  and  molecular electronic shells? 2. What is the degeneracy of a molecular electronic shell with  = 10 ? 10 Session 1 − Electronic configurations and electronic states of diatomic molecules 3. The magnetic quantum number of an electron in an antibonding molecular orbital,  , of a homonuclear diatomic molecule is found to be equal to −3. The orbital  has the property, î  = −  , with respect to inversion, î , through the bond midpoint. Deduce the symbol for the molecular orbital . 1.2.3 Electronic configuration of a molecule Only a summary of the important concepts are given here since you have studied this topic, extensively, in Unit II of CYU3300. A molecular electronic configuration indicates the populations of molecular orbital energy levels of a molecule. For this purpose, you should first know the order of molecular orbital energy levels (same as you had to know the order of atomic orbital energy levels before start filling them in obtaining electronic configurations of atoms.) In general, the order of increasing energy of homonuclear diatomic molecular orbital energies, except for O 2 and F2 , is given by the following sequence. g1s  *u 1s  g 2s  *u 2s  u 2p  g 2p  *g 2p  *u 2p  g 3s  u 3p  g 3d In O 2 and F2 the order is g1s  *u 1s  g 2s  *u 2s  g 2p  u 2p  *g 2p  *u 2p  g 3s  u 3p  g 3d Ground electronic configurations of homonuclear diatomic molecules can be obtained by feeding electrons to the above sequence of orbital energy levels. A  -level can hold up to two electrons and any other level can hold up to four electrons (since the latter are doubly degenerate). As with atoms, in obtaining the ground configuration, you first add one electron (with the same spin quantum number s) into each of the two orbitals in a degenerate level and then pair each electron with an electron with opposite spin. See Figure 1.9. 11 Unit III − An Introduction to Electronic Spectroscopy of Molecules Energy *u 2p *g2p g2p u2p *u 2s g2s *u 1s g1s Li 2 Be2 B2 C2 N2 O2 F2 Figure 1.9: A schematic representation of electronic configurations of some diatomic molecules in their ground states Q8. What is the standard notation for the ground electronic configurations of B 2 and C 2 molecules? A8. You use a notation similar to that used in representing the electronic configuration of atoms. ( g1s ) ( *u1s ) ( g 2s ) ( *u 2s ) ( u 2p)2 2 2 2 2 B2 ( g1s ) ( *u1s ) ( g 2s ) ( *u 2s ) ( u 2p)4 2 2 2 2 C2 A term within brackets indicates a molecular orbital energy level or a shell and the superscript indicates the number of electrons in that energy level. Note that, in designating an energy level, you use the same symbol used in designating the associated molecular orbitals. When all the orbitals in a shell are fully occupied you say that the shell is closed or filled. For example in B 2 , g1s, *u1s, g 2s and *u 2s -shells are closed. u 2p -shell is partially filled or open. 12 Session 1 − Electronic configurations and electronic states of diatomic molecules 1.2.4 Spin multiplicity The total spin, denoted by S, of a set of N electrons is defined by N S= sj (1) j =1 where s j is the spin of the jth electron. s j =  1 2 depending on whether the jth electron is in spin up state or spin down state, respectively. Note that S is a positive quantity. Then the spin multiplicity of that set of N electrons is defined by Spin multiplicity = 2S + 1 (2) As you will see, spin multiplicity plays an important role in determining selection rules in electronic spectroscopy. When many orbital occupations are possible for a given electronic configuration then the orbital occupations with the highest spin multiplicity has the lowest energy. Q9. Write down the electronic configuration of the ground state of N 2 and determine its spin multiplicity. A9. As indicated in Figure 1.9, the ground electronic configuration of N 2 ( ) ( *u1s ) (g 2s) (*u 2s ) ( u 2p)4 (g 2p) 2 2 2 2 2 is g1s You know that the spins are paired in a filled molecular orbital (as with atomic orbitals). Hence, the contribution of electrons in a filled orbital towards the total spin is zero. All the involved orbitals are filled in the above configuration. Hence S = 0  Spin multiplicity = 2S + 1 = 2  0 + 1 = 1. The spin multiplicity of the ground state of N 2 is one. You call it a singlet state. Q10. Write down the ground electronic configuration of O 2 and determine its spin multiplicity. A10. As indicated in Figure 1.9, the electronic configuration of O 2 is ( g1s ) ( *u1s ) ( g 2s ) (*u 2s ) (g 2p) ( u 2p)4 ( *g 2p ) 2 2 2 22 2 Only partially filled orbitals contribute to the total spin. Therefore, in this calculation you need to consider only the electrons in the *g 2p level. We can have two situations here. (i) Spins are not paired, S = + 1 2 + 1 2 = − 1 2 − 1 2 = 1 Spins are paired, S = + 1 2 − 1 2 = 0 Spin multiplicity in case (i) is 2S + 1 = 3. We call it a triplet state, It gives the lower energy state which is indicated in Figure 1.9 with parallel spins. Spin multiplicity in case (ii) is 2S + 1 = 0 ; This gives a singlet state. 13 Unit III − An Introduction to Electronic Spectroscopy of Molecules Q11. Denote the two molecular orbitals in *g 2p shell by *+1 g 2p and *−1 g 2p where the value of ml of each orbital is indicated as a ( ) 2 subscript. Write down the possible orbital occupations for *g 2p in triplet and singlet states of O 2 in the configuration described in A10. A11. In the triplet state, both electrons have the same spin state. Hence they cannot be in the same orbital. Hence1 for the triplet state the ( )( ) 1 orbital occupation is −1 g 2p +1 g 2p. The spins are paired in the * * singlet state. Hence there is no such a restriction. Thus2the possible ( )( ) ( ) 1 1 orbital occupations are −1 g 2p +1 g 2p , −1 g 2p and * * * ( ). * 2 +1 g 2p In general the two orbitals in a molecular energy level with the z-component of angular momentum,  , are designated by − and + , where  = g or u and  represent the Greek letter corresponding to the value of  ; e.g.  when  = 0. Activity 1.4 1. Show that the contribution towards total spin, S, from the electrons in a closed shell is zero. 2. Indicate the two orbitals in a molecular orbital energy level with  = 3 , with gerade inversion symmetry, in standard notation. Indicate all possible orbital occupations of the partial configuration (  u ). 2 3. 4. Determine possible values of spin multiplicities of the following electronic configurations. 1 ,  / , 2 and 4. [Here,  / indicates a configuration with one electron each in two different  levels.] 5. Determine the maximum possible spin multiplicity of a configuration of a molecule having the following number of partially filled shells. (i) Only one (ii) Only two 6. Assume that the order of molecular orbital energy levels indicated in section 1.2.3 remains the same when ions are formed and do the following for each ion, C 22 + and N2+. (i) Write down the ground state electronic configuration in standard notation. (ii) Write down the possible orbital occupations within the above mentioned configuration. (iii) Determine the spin multiplicity of possible states within the above mentioned configuration. 14 Session 1 − Electronic configurations and electronic states of diatomic molecules 1.3 Heteronuclear diatomic molecules All the discussion in section 1.2 is valid for heteronuclear diatomic molecules as well, except for the two points indicated below. 1.3.1 Orbitals have no inversion symmetry A heteronuclear diatomic molecule has no centre of symmetry. As such the molecular orbitals of these molecules have no inversion symmetry. The orbitals and the corresponding energy levels are written using the same symbols as described in section 1.2 except that they do not have subscripts “g” and “u”. 1.3.2 Order of molecular orbital energy levels The following order of energy levels can be used to construct the electronic configurations of most of the common heteronuclear diatomic molecules such as NO, CO, CN. SO. 1s  *1s  2s  * 2s  2p  2p  * 2p  * 2p  3s  3p  3d Q12. Write down the electronic configuration of the ground state of the molecules NO, CO, CN and PO. Write down the possible orbital occupations of the partially filled shells and determine the possible spin multiplicities. A12. NO Total number of electrons in the molecule = 7 + 8 = 15. ( 1s )2 ( *1s ) ( 2s )2 ( * 2s ) ( 2p )2 ( 2p )4 ( * 2p ) 2 2 1 Configuration: ( ) ( ) 1 1 Possible orbital occupations: * −1 2p and *+1 2p. Total spin of any of the above configurations = S =  1 2 = 1 2 Spin multiplicity = 2S+1 = 2; This is a doublet. CO Total number of electrons in the molecule = 6 + 8 = 14. ( 1s )2 ( *1s ) ( 2s )2 ( * 2s ) ( 2p )2 ( 2p )4 2 2 Configuration: There are no partially filled shells. Total spin = S = 0 Spin multiplicity = 2S+1 = 1 15 Unit III − An Introduction to Electronic Spectroscopy of Molecules CN Total number of electrons in the molecule = 6 + 7 = 13. ( 1s )2 ( *1s ) ( 2s )2 ( * 2s ) ( 2p )2 ( 2p )3 2 2 Configuration: Possible orbital occupations: ( −1 2p )2 ( +1 2p )1 and ( −1 2p )1 ( +1 2p )2. Total spin of any of the above configurations = S = + 12 − 12 − 12 = − 12 + 12 + 12 = 1 2 Spin multiplicity = 2S+1 = 2 PO Total number of electrons in the molecule = 15 + 8 = 23. Configuration: ( 1s )2 ( *1s ) ( 2s )2 ( * 2s ) ( 2p )2 ( 2p )4 ( * 2p ) 2 2 4 (  2p ) ( 3s )2 ( 3p )1 * 2 Possible orbital occupations: ( −13p )1 and ( +13p )1. Total spin = S =  1 2 = 1 2 Spin multiplicity = 2S+1 = 2 1.4 Electronic states arising from electronic configurations of diatomic molecules In CYU4301 you learned that the selection rules in spectroscopy are determined by the transition moment. The transition moment is determined by the wave functions of the two states involved in the transition. Whether the transition moment is zero or not can be determined just by observing some properties of these wave functions. For light diatomic molecules, the spin-orbit interaction, i.e. the interaction between the magnetic moment created by orbital motion of electrons and the magnetic moment created by the spin of the electrons, is weak and is neglected. This is referred to as Hund’s case (a). Following properties of electronic wave functions are used in determining the selection rules in Hund’s case (a). The component of the total orbital angular momentum along the internuclear axis. Spin multiplicity. 16 Session 1 − Electronic configurations and electronic states of diatomic molecules Inversion symmetry in the case of homonuclear diatomic molecules. Reflection symmetry with respect to reflection through a vertical plane. A symbol has been developed in indicating the above four properties associated with a total electronic wave function of a molecule. It is of the form, 2S+1 Q where 2S+1 is the spin multiplicity, Q is an upper case Greek letter depending on the component of the total orbital angular momentum along the internuclear axis,  is the “+” or “–” sign depending on the nature of reflection symmetry of the wave function and  is “g” or “u” depending on the nature of inversion symmetry of the wave function. 2S+1 Q is called the electronic term symbol or the electronic state symbol. You have learned to calculate the spin multiplicity in section 1.2.4. Now let us learn how to determine the other quantities appearing in the state symbol so that you can determine the possible terms generated for a given electronic configuration. 1.4.1 Total orbital angular momentum along the internuclear axis Define a quantum number, M z , corresponding to the component of the total orbital angular momentum of the electrons in a molecule, having N electrons, along the internuclear axis (in the z-direction) by N M z =  ml j (3) j =1 where ml j is the quantum number corresponding to the component of the orbital angular momentum of the jth electron along the internuclear axis. [Then the z-component of total orbital angular momentum is M z.] Analogous to  for a single electron in a molecular orbital, you define  = M z. Then, Q for the state symbol is defined depending on the value of  as shown below.  0 1 2 3 4 Q      The upper case Greek letters used for state symbols are;  = sigma,  = pi,  = delta,  = phi and  = gamma. 17 Unit III − An Introduction to Electronic Spectroscopy of Molecules Q13. What are the possible Q for the molecular electronic states arising from the ground state electronic configuration of NO, viz. ( 1s )2 ( *1s ) ( 2s )2 ( * 2s ) ( 2p )2 ( 2p )4 ( * 2p ). 2 2 1 A13. In calculating  it is desirable to write down the possible orbital occupations since the value of ml of an electron depends on the molecular orbital it is in. For NO you can write them, based on the discussion in A12. (i) ( 1s ) *1s ( ) ( 2s )2 ( * 2s ) ( 2p )2 ( −1 2p )2 ( +1 2p )2 ( *−1 2p ) 2 2 2 1 ( 1s )2 ( *1s ) ( 2s )2 ( * 2s ) ( 2p )2 ( −1 2p )2 ( +1 2p )2 ( *+1 2p ) 2 2 1 (ii) Remembering that ml = 0 for an electron in a sigma orbital you obtain (i)  = M z = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 − 1 − 1 + 1 + 1 − 1 = 1 (ii)  = M z = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 − 1 − 1 + 1 + 1 + 1 = 1 Hence, only Q =  is possible for NO in its ground state. Note that the contribution from electrons in closed shells towards M z is zero (see activity 1.5.1). Hence,  = 0 and Q =  for a configuration that involves only closed shells; e.g. CO described in A12. Only the electrons in partially filled shells have to be considered in determining Q. Q14. What are the possible Q for the molecular electronic states arising from the ground state electronic configuration of CN, viz. ( 1s )2 ( *1s ) ( 2s )2 ( * 2s ) ( 2p )2 ( 2p )3. 2 2 A14. You have to consider only the partially filled shell/s. From A12 you learn that the possible occupation patterns for the partially filled shell are ( −1 2p ) ( +1 2p ) and ( −1 2p ) ( +1 2p ). Hence 2 1 1 2  = −1 − 1 + 1 = 1 or  = −1 + 1 + 1 = 1. Hence, only Q =  is possible for CN in its ground state. Activity 1.5 1. Show that the contribution from electrons in a closed shell towards Mz is zero. 2. Determine Q of all possible states associated with the ground state electronic configuration of O 2 (see A11). 18 Session 1 − Electronic configurations and electronic states of diatomic molecules 1.4.2 Inversion symmetry of the electronic wave function of a homonuclear diatomic molecule Following rules can be easily derived (see activity 1.6.1) for products of functions having inversion symmetry. 1. A product of even or odd number of functions with gerade symmetry has gerade symmetry. 2. A product of even number of functions with ungerade symmetry has gerade symmetry. 3. A product of odd number of functions with ungerade symmetry has ungerade symmetry. 4. A product of a function with gerade symmetry with a function of ungerade symmetry has ungerade symmetry. It can be shown that the electronic wave function of a molecule is a sum of a set of terms where each term is a product of all the occupied orbitals. This fact together with the above four rules can be used in determining the inversion symmetry of the wave function associated with a given orbital occupation. Q15. Determine the inversion symmetry of the wave function corresponding to the ground state electronic configuration of B 2 and B+2. ( ) ( *u1s ) ( g 2s ) ( *u 2s ) ( u 2p)2. 2 2 2 2 A15. Configuration of B 2 is g1s The orbital product is g1s g1s *u1s *u1s g 2s g 2s *u 2s *u 2s  u 2p  u 2p It is a product of 4 functions with g-symmetry and 6 functions with u-symmetry. According to the above rules the product has g inversion symmetry. Hence the electronic wave function corresponding to the ground state configuration of B 2 has g inversion symmetry. ( ) ( *u1s ) (g 2s ) (*u 2s ) ( u 2p)1 2 2 2 2 Configuration of B+2 is g1s The orbital product is g1s g1s *u1s *u1s g 2s g 2s *u 2s *u 2s  u 2p It is a product of 4 functions with g-symmetry and 5 functions with u-symmetry. According to the above rules the product has u inversion symmetry. Hence the electronic wave function corresponding to the ground state configuration of B+2 has u inversion symmetry. Note that the contribution towards the inversion symmetry by closed shells is gerade. (See activity 1.6.2) Hence, the inversion symmetry of the total 19 Unit III − An Introduction to Electronic Spectroscopy of Molecules wave function may be determined by a product of a gerade function and the orbital product of the partially filled shells. Q16. Determine the inversion symmetry of the wave function corresponding to the ground state electronic configuration of C+2. ( )( A16. Configuration of C+2 is g1s *u 1s g 2s *u 2s ( u 2p ) )( )( ) 2 2 2 2 3 Partially filled shell is u 2p. Hence the relevant orbital product is g  u 2p  u 2p  u 2p where g represents the gerade contribution coming from all the closed shells. It is a product of 1 function with g inversion symmetry and 3 functions with u inversion symmetry. According to the above rules the product has u-symmetry. Hence the electronic wave function corresponding to the ground state configuration of C+2 has u inversion symmetry. Note that, multiplication of a function,  , by a function having gerade symmetry does not change the inversion symmetry of . Hence it is sufficient to consider only the orbital occupations in partially filled shells in determining the inversion symmetry of a term. Activity 1.6 1. Using the definition given for the inversion operation in activity 1.2.7 derive the 4 rules given above for the determination of inversion symmetry of a product of functions with inversion symmetry. 2. Show that the contribution towards inversion symmetry of a state generated from the closed shell part of a configuration is gerade. 3. Determine the inversion symmetry of the electronic wave function of a state corresponding to the ground state configuration of C 2+ , C 22 + and N2+. 1.4.3 Symmetry of the electronic wave function with respect to reflection through a vertical plane in a diatomic molecule Determination of this reflection symmetry is an involved process where you have to carefully study the spin and spatial parts of the total electronic wave function, which is beyond the scope of this lesson material. Here we only present four simple rules for the determination of reflection symmetry for states arising from electronic configurations with a total of zero, one and two electrons in partially filled shells. In order to achieve this goal easily we define equivalent electrons as electrons in the same shell. Non-equivalent electrons reside in different shells. 20 Session 1 − Electronic configurations and electronic states of diatomic molecules Q17. Identify the equivalent and non-equivalent electrons in the ground state configuration of Li 2. ( )( )( ) 2 2 2 A17. Ground state configuration of Li 2 is g1s *u1s g 2s Two electrons in the g 1s shell are equivalent to each other. Same is true for the two electrons in each of the shells *u 1s and g 2s. Electrons in g 1s shell are non-equivalent to the electrons in either *u 1s or g 2s shell. Similarly the electrons in *u 1s are non- equivalent to those in g 1s or g 2s shell and the electrons in g 2s are non-equivalent to those in g 1s or *u 1s shell. You use the symbols  2 ,  2 , 2 ,… to identify two equivalent electrons in  ,  ,  ,… shells, respectively. You use the symbols / ,  / ,  / ,… to identify two non- equivalent electrons in (two different)  ,  ,  ,… type shells, respectively. Note that in the above notation you have dropped the part of the orbital (or shell) identification symbol that indicates the atomic orbital from which the molecular orbital originates and the superscript that indicates whether the orbital is bonding or anti-bonding. This information is not necessary in determining the electronic states. The rules for the determination of reflection symmetry are summarised as follows. 1. Reflection symmetry exists only for electronic states with Q = . 2. A  state of a configuration having only completely filled shells (e.g.  2 ,  4 , 4 ,…) and/or partially filled  shells (e.g.  , / ), has + reflection symmetry. 3. When two non-equivalent electrons in molecular orbitals other than sigma (e.g.  / ,  / ,..) are involved, equal numbers of  states with + symmetry and – symmetry are generated in both singlet and triplet spin multiplicity. 4. When two equivalent electrons are involved, – symmetry is coupled with the  state of triplet and the + symmetry is associated with the  state of the singlet. (e.g.  2 generates 1 + and 3  − ; 2 generates 1 +  and 3  −.) You learn to use the above rules by working out examples. 21 Unit III − An Introduction to Electronic Spectroscopy of Molecules 1.4.4 Identification of all possible electronic states generated by a given electronic configuration Identification of electronic states is done using the state symbol, 2S+1 Q. Hence, the identification of all possible electronic states for a given electronic configuration reduces to the determination of all possible state symbols. Q18. Derive the electronic state symbol for the ground state of H 2. A18. Configuration of H 2 is ( g1s ) 2 The molecule has no partially filled shells. For such a configuration =0  Q= S=0  2S + 1 = 1 Inversion symmetry: gerade [Only a completely filled shell is involved] Reflection symmetry: + [All shells are filled or only  orbitals involved] Hence the only possible state is 1 g+. Application of the rules you learned in this section reveals that the only possible state for a configuration of any homonuclear diatomic molecule with no partially filled shells is 1 g+. The only possible state for any heteronuclear diatomic molecule with no partially filled shells is 1 +. Q19. Derive the electronic state symbol for the ground state of He+2. ( )( ) 2 1 A19. Ground state configuration of He+2 is g1s *u1s You have to consider only the electrons in partially filled shells. Only one electron in the partially filled shell and its orbital ( ) 1 occupation is *u 1s  = ml = 0  Q =  [since ml = 0 for an electron in a  orbital] S =  12 = 1 2  2S + 1 = 2 Inversion symmetry: ungerade [Since the partially filled orbital has ungerade symmetry] Reflection symmetry: + [Since only  orbitals are involved] Hence the only possible state is 2  +u. 22 Session 1 − Electronic configurations and electronic states of diatomic molecules Q20. Derive the possible electronic state symbols for the ground state electronic configuration of O 2. A20. Ground state configuration of O 2 is ( )( )( )( )( ) ( u 2p)4 ( *g 2p ) 2 2 2 2 2 2 g1s *u1s g 2s *u 2s g 2p ( ) 2 The configuration of the partially filled shell is *g 2p. A *g 2p shell has two orbitals, *+1g 2p and *−1g 2p with ml = +1 and ml = −1 , respectively. ( ) ( ) 2 2 Thus the possible orbital occupations are *+1g 2p , *−1g 2p and ( ) ( ) * 1 * 1 −1g 2p +1g 2p. Now you find the possible electronic states for each of the occupations. ( ) 2 Occupation *+1g 2p  = +1 + 1 = 2  Q =  S = + 1 2 − 1 2 = 0  2S + 1 = 1 [Spins are paired since the electrons are in the same orbital] Inversion symmetry: gerade [multiple of gerade orbitals ( *+1g 2p  *+1g 2p ] ) ( ) Reflection symmetry: Not relevant since the state in not . Hence the state symbol: 1 g ( ) 2 Occupation *−1g 2p  = −1 − 1 = 2  Q =  S = + 1 2 − 1 2 = 0  2S + 1 = 1 [Spins are paired since the electrons are in the same orbital] Inversion symmetry: gerade [multiple of gerade orbitals ( *−1g 2p  *−1g 2p ] ) ( ) Reflection symmetry: Not relevant since the state in not . Hence the state symbol: 1 g ( ) ( ) 1 1 Occupation *−1g 2p * +1g 2p  = −1 + 1 = 0  Q =  23 Unit III − An Introduction to Electronic Spectroscopy of Molecules The spins can be parallel or anti-parallel and you have to consider these two cases separately. Parallel spins S = + 1 2 + 1 2 = − 1 2 − 1 2 = 1  2S + 1 = 3 Inversion symmetry: gerade [multiple of gerade orbitals ( * −1g 2p )( * +1g 2p )] Reflection symmetry: − [Two equivalent electrons in a non-sigma orbital in triplet state] Hence the state symbol: 3 g− Anti-parallel spins S = + 1 2 − 1 2 = 0  2S + 1 = 1 Inversion symmetry: gerade [multiple of gerade orbitals ( * −1g 2p )( * +1g 2p )] Reflection symmetry: + [Two equivalent electrons in a non-sigma orbital in singlet state] Hence the state symbol: g+ 1 Hence the possible states are 1 g+ , 3 g− and 1 g. In determining the electronic states only the partially filled shells matter. Hence, it is sufficient to give only the configuration of partially filled shells in a problem. Q21. Derive the possible electronic state symbols for an electronic configuration with two non-equivalent  -electrons, i.e.  /. A21.  and  / are two different  -shells. Each has two orbitals;  −1 &  +1 and  −/ 1 and  +/ 1. ( ) Possible orbital occupations are ( −1 ) −/ 1 , ( +1 ) +/ 1 , ( ) 1 1 1 1 ( −1 )1 ( +/ 1 ) ( ) and ( +1 ) −/ 1. 1 1 1 Now you find the possible electronic states for each of the occupation patters. Occupation ( −1 ) −/ 1 ( ) 1 1  = −1 − 1 = 2  Q =  24 Session 1 − Electronic configurations and electronic states of diatomic molecules The spins can be parallel or anti-parallel and the two cases are considered separately. Parallel spins S = + 1 2 + 1 2 = − 1 2 − 1 2 = 1  2S + 1 = 3 Inversion symmetry: Not relevant. Reflection symmetry: Not relevant since the state is not . Hence the state symbol: 3  Anti-parallel spins S = + 1 2 − 1 2 = 0  2S + 1 = 1 Inversion symmetry: Not relevant. Reflection symmetry: Not relevant since the state is not . Hence the state symbol: 1  ( ) Occupation ( +1 ) +/ 1 1 1 The only difference from the previous occupation is that both electrons have ml = +1 instead of −1. However, this does not change the determination of parameters in state symbol. Hence this occupation also leads to same states as above. Hence state symbols: 3  and 1  ( ) Occupation ( −1 ) +/ 1 1 1  = −1 + 1 = 0  Q =  The spins can be parallel or anti-parallel and the two cases are considered separately. Parallel spins S = + 1 2 + 1 2 = − 1 2 − 1 2 = 1  2S + 1 = 3 Inversion symmetry: Not relevant Reflection symmetry:  [Two non-equivalent electrons in non-sigma orbitals in triplet state] Hence the state symbols: 3  + and 3 − Anti-parallel spins S = + 1 2 − 1 2 = 0  2S + 1 = 1 Inversion symmetry: Not relevant Reflection symmetry:  [Two non-equivalent electrons in a non-sigma orbitals in singlet state] 25 Unit III − An Introduction to Electronic Spectroscopy of Molecules Hence the state symbols: 1 + and 1 − Occupation ( +1 ) −/ 1 ( ) 1 1 This occupation also leads to same states as above. Hence state symbols: 1 + , 1 − , 3  + and 3  −. Hence the possible states are 1 + , 1 − , 3  + , 3  − , 1  and 3 . Table 1.1 indicates all possible electronic states that can be generated from configurations with all shells filled except the ones shown in the first column. Table 1.1: Electronic configurations and corresponding electronic states Configuration State symbols  2 ,  4 , 4 , 1 + / 1 + , 3  +  1  , 3  / 1 + , 1 − , 3  + , 3  − , 1  , 3   1  , 3  , 1 , 3   1 +  , 3 2  2 1 + , 3  − , 1   , 3 2  2 1 + , 3  − , 1  Activity 1.7 1. Giving reasons state whether the electrons in the following configurations are equivalent or not. (i)  (ii)  / (iii)  3 (iv)  (v)  2  2. Find electronic states generated by the configurations 4 ,  , / and 2. 3. Determine the following properties of the electronic states (of diatomic molecules) represented by the symbols 1  g+ , 3 g− , 3  u and 1 . (i) Spin multiplicity (ii) z-component of the total orbital angular momentum (iii) Reflection symmetry (iv) Whether the molecule is homonuclear or heteronuclear. 4. For a heteronuclear diatomic molecule, the total set of states arising from / configuration is 1  + , 1  − , 3  + , 3  − , 1  and 3 . Deduce the total sets of states 26 Session 1 − Electronic configurations and electronic states of diatomic molecules for a homonuclear diatomic molecule in   configuration where  and are / either g and/or u. 5. Determine all possible molecular orbital occupations in configurations, not having more than two unfilled shells, which can generate a 3  state. 1.5 Hund’s rules for the identification of the ground state The electronic state with the lowest energy is called the ground state. Under normal conditions, molecules are in their ground electronic states. Hund’s rules can be used to identify the ground electronic state generated from a given configuration. Hunds rules For a given electronic configuration, 1. the state with highest spin multiplicity has the lowest energy. 2. for a given spin multiplicity, the state with the highest  has the lowest energy. They can be applied to a diatomic molecule in finding the ground electronic state. Q22. Find the lowest energy state arising from / electronic configuration. A22. The possible states are 1 + , 1 − , 3  + , 3  − , 1  and 3  (see A21). According to Hund’s rule 1 the triplet states have lower energy. Hence, the ground state may be 3  + , 3  − or 3 . Hund’s rule 2 says that the state with highest  has the lowest energy. Hence the ground state must be 3 . The ground electronic state (minimum energy state) is labelled with the letter X in front of the electronic state symbol. Other states with the same spin multiplicity as the ground state are prefixed with A, B, C,…. respectively in the order of increasing energy. The states with spin multiplicities different from the ground state are labelled with a, b, c, … etc. Exceptions to this rule are N 2 and C 2 where the singlet ground state 1 g+ is prefixed with X and excited triplet states are prefixed with A,B,C. 27 Unit III − An Introduction to Electronic Spectroscopy of Molecules 1.6 Selection rules in electronic spectroscopy During an electronic transition, a molecule goes from one electronic state to another. Selection rules presented here apply to molecules of Hund’s case (a) where total orbital angular momentum and total spin angular momentum are not coupled. 1.  = 0,  1 Difference in the quantum number  of the final and initial states must be zero or 1. E.g. Transitions    and    are allowed but transitions    and    are not allowed. 2. S = 0 Total spin quantum number of the initial and final states must be the same. This rule is violated when spin-orbit interaction is substantial. This is especially true when molecules with heavy atoms are involved. 3. Transitions between  electronic states having different reflection symmetry are forbidden. +  − (forbidden) +  + (allowed) −  − (allowed) 4. Transitions between states having similar inversion symmetry are forbidden in homonuclear diatomic molecules. g  u (allowed) g  g (forbidden) u  u (forbidden) Q23. State whether the following electronic transitions are allowed. If not, state the selection rule/s that forbid the transition. (i) B 3−u  X 3g− (ii) b 1g+  X 3g− (iii) A 3g  X 3g− A23. (i) Allowed. (ii) Not allowed because, g  u symmetry, +  − symmetry and the spin selection rule ( S = 0 ) are violated. (iii) Not allowed because, orbital angular momentum selection (  = 0,  1 ), rule and g  u symmetry are violated. 28 Session 1 − Electronic configurations and electronic states of diatomic molecules Activity 1.8 1. In a hypothetical molecule, the order of energies of the states arising out of the configuration / is 3  < 3  + < 1  < 1  + < 3  − < 1  −. Giving reasons, assign letters A, B, C…. and a, b, c,… to these states. 2. State whether each of the following electronic transitions is allowed. If not allowed, give the selection rule/s that forbids the transition. (i) B 1u+  X 1g+ (ii) B 3 u−  X 2g− (iii) A 1g+  X 1u+ (iv) a 3 g  X 1g− (v) A 3 u+  X 3 g− 3. In an experiment a researcher synthesised a heteronuclear diatomic molecule in the ground state of an exotic electronic configuration P. The only open shell part of P is /. All the other shells are closed. In order to carry out a reaction he needs to have the molecule in an electronic configuration W. The only difference between P and W is that, instead of / , it has  as the only open shell part. It is known that all the states associated with P have lower energy than those associated with W. Determine whether the student can obtain the molecule in configuration W by exposing a sample of synthesised molecules to electromagnetic radiation. Summary You use simple models in understanding rotational and vibrational spectra of molecules. Such simple models are not available in studying electronic spectra of molecules. Extensive numerical calculation is necessary in determining the electronic energies of molecules. However, the selection rules can be expressed using a number of symmetry properties of electronic wave functions of molecules which can be determined easily without using a computer. In this session you studied these symmetry properties in detail for diatomic molecules and learned to express the selection rules in terms of them. As with atoms, an electronic configuration of a molecule is a description of the number of electrons in molecular electronic shells. A shell is named according to the magnitude of the z-component of orbital angular momentum of an electron in it.  -shells are non-degenerate. All the other shells are doubly degenerate. A given molecular electronic configuration gives rise to a number of electronic states, each of which is described by a wave function. Different electronic states (even those arising from the same configuration) have different energy. In electronic spectroscopy a molecule goes from one electronic state into another with the absorption of a photon. The selection rules for a transition between two molecular electronic states of a light diatomic molecule depend on the z-component of the orbital angular momentum, spin multiplicity and reflection symmetry of the two 29 Unit III − An Introduction to Electronic Spectroscopy of Molecules electronic states involved in the transition. Inversion symmetry is also important in determining the feasibility of a transition in a homonuclear diatomic molecule. The term symbol used in designating an electronic state indicates all these properties. Determination of term symbols of possible states associated with an electronic configuration is done using only the part of configuration involving partially filled shells. In this determination, first you determine the possible orbital occupations in partially filled shells. For each occupation you then determine the possible spin multiplicities. For each of these spin multiplicities you determine the possible reflection symmetries. In addition, with homonuclear diatomic molecules you determine the inversion symmetry of the electronic state using the inversion symmetry of the molecular orbitals. Using selection rules you can identify allowed transitions between electronic states once all possible term symbols are determined for the configurations you are interested in. Learning outcomes After a successful study of this session you should be able to do the following. Sketch s and p atomic orbitals and on them, indicate the signs of the corresponding wave function. State the degeneracy of s, p, d and f atomic energy levels and determine the maximum number of electrons in each of them. Define electronic configuration of an atom and write down the ground state electronic configurations of atoms in the 1st and 2nd periods of the periodic table. Write down possible occupations of orbitals in a given electronic configuration of an atom. Sketch the shapes of g 1s , *u 1s , g 2pz , *u 2p z , u 2p y and *g 2py molecular orbitals, with + and − , signs for a homonuclear diatomic molecule. Explain what a molecular orbital which is rotationally symmetric about the internuclear axis is and identify such orbitals. Describe what is meant by inversion symmetry of a function of position coordinates, ( x, y,z ) , define the terms gerade and ungerade, and determine the inversion symmetry of functions and orbitals. Explain what is meant by symmetry with respect to reflection of a function of position coordinates ( x, y,z ) through a vertical plane, define + and – reflection symmetry and determine the reflection symmetry of functions and orbitals. 30 Session 1 − Electronic configurations and electronic states of diatomic molecules Define the terms, molecular electronic configuration, molecular orbital energy level, molecular electronic shell and electronic state of a molecule. Explain how the symbols used in designating a molecular orbital and a molecular electronic shell are derived and deduce such symbols for orbitals using data. Explain and deduce the degeneracy of various molecular electronic shells. Write down possible occupations of orbitals in a given molecular electronic configuration. Write down the order of molecular orbital energies for light diatomic molecules. Write down the ground state electronic configuration of a given light diatomic molecule. Define and determine the spin multiplicity of a given electronic configuration of a diatomic molecule. Define the electronic state symbol. Determine M Z of all possible electronic states associated with an electronic configuration of a diatomic molecule. Determine the inversion symmetry of electronic states associated with an electronic configuration of a diatomic molecule. Determine the reflection symmetry of electronic states associated with an electronic configuration of a diatomic molecule. Define and determine equivalent electrons in an electronic configuration. Determine all possible electronic states associated with a given configuration of a light diatomic molecule. State the rules used in labelling electronic states of a diatomic molecule using letters a,b,c,… and A,B,C,… and label such states using them. State Hund’s rules for the determination of the ground electronic state and use them in problem solving. State the selection rules in electronic spectroscopy in terms of spin multiplicity, M Z , inversion symmetry and reflection symmetry of electronic states. Apply the knowledge in selection rules in electronic spectroscopy in problem solving. 31 Unit III − An Introduction to Electronic Spectroscopy of Molecules 32

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