1. Normal modes of vibration of a molecule – An Exercise by Dr.D.UTHRA Head Department of Physics DG Vaishnav College Chennai-106

2. This presentation has been designed to serve as a self- study material for Postgraduate Physics students pursuing their programme under Indian Universities, especially University of Madras and its affiliated colleges. If this aids the teachers too who deal this subject, to make their lectures more interesting, the purpose is achieved. -D.Uthra

3. I acknowledge my sincere gratitude to my teacher Dr.S.Gunasekaran, for teaching me group theory with so much dedication and patience & for inspiring me and many of my friends to pursue research. My acknowledgement to all my students who inspired me to design this presentation. - D.Uthra

4. By understanding this presentation you can find  the no. of vibration that a molecule can execute  the type of vibration, whether it is stretching or bending or torsion or out-of plane bending  the no. of symmetric and asymmetric vibration the molecule executes and to which species it belongs to

5.  3N-6 degrees of freedom are required to describe vibration of a non-linear molecule  3N-5 degrees of freedom are required to describe vibration of a linear molecule

6. An exercise! Search Ur Allied Chemistry book!! For the following molecules, Find the molecule is linear or non-linear Calculate the number of normal modes of vibration CO 2, H 2 O, NH 3, BF 3, HCl, CH 4, CO, H 2 S, C 2 H 2, PCl 3, BCl 3, N 2 F 2, B 2 H 6, C 6 H 6, SF 6, C 3 H 4, COS, CH 3 Cl, H 2 O 2 This is usually a Part A question!

7. Internal Coordinates  Changes in the bond lengths and in the inter-bond angles. Types of internal coordinates and their notation bond stretching ( Δ r ) angle deformation ( Δ Φ) torsion ( Δτ ) out of plane deformation ( Δ Φ')

8. How to calculate number of vibrations using internal coordinates?  n r = b  n Φ = 4b-3a+a 1  n τ = b-a 1  n Φ’ = no.of linear subsections in a molecule where, b - no.of bonds (ignoring the type of bond) a - no.of atoms a 1 - no.of atoms with multiplicity one Multiplicity means the number of bonds meeting the atom, ignoring the bond type. In H 2 O, multiplicity of H is 1 and that of O is 2. In NH 3, multiplicity of N is 3, while H is 1. In C 2 H 2, it is 1 for H and 2 for C (not 4, b’coz you ignore bond type). In CH 4, it is 1 for H and 4 for C. In CO 2, it is 2 for C and 1 for O.

9. Types of vibrations  n r - no.of stretching vibrations  n Φ - deformations  n τ - torsional vibrations  n Φ’ - out of plane bending

10. Exercise Again!  Now using the internal coordinates, calculate the normal modes of vibration again for the molecules CO 2, H 2 O, NH 3, BF 3, HCl, CH 4, CO, H 2 S, C 2 H 2, PCl 3, BCl 3, N 2 F 2, B 2 H 6, C 6 H 6, SF 6, C 3 H 4, COS, CH 3 Cl, H 2 O 2

11. How to calculate normal modes of vibration using character table of the molecule ? Let us try to calculate normal modes of vibration using the formula  n i ' - number of normal modes of vibration  h - order of the point group  ℓ i - dimension of the i th species  Ψ j (R) – from the character table  χ j ' (R) =(N R -2)(1+2cosθ) (for proper rotations) =N R (-1+2cosθ) (for other symmetry operations) N R - number of atoms that undergo No change due to operation R θ - angle of rotation for R n i '= (1/h)∑ R ℓ Ψ j (R)χ j ' (R)

12. For XY 2 bent molecule, calculate normal modes of vibration by all three methods

13. Using 3N-6 formula For an XY 2 bent molecule  N=3 ; 3N-6 = 3 modes of vibration

14. Using internal coordinates In XY 2 bent molecule,  b = 2 (no.of bonds)  a = 3 (no.of atoms)  a 1 = 2 (no.of atoms with multiplicity one) Hence, n r = 2 n Φ = 4*2-3*3+2 =1 n τ = 2-2 = 0 3N-6 = n r +n Φ +n τ = 2+1= 3

15.  For a XY 2 bent molecule, n r = 2 n Φ = 1 implies, this molecule has  two bond stretching vibrations and  one angle deformation (bending)vibration

16. Using Character table XY 2 bent molecule belongs to C 2v point group From character table, h - order of the point group = 4 ℓ i - dimension of each species =1 C 2v EC2C2 σvσv σv ’σv ’ A1A1 1111 A2A2 11 B1B1 1 1 B2B2 1 1

17. To calculate χ j ' (R) C 2v E C 2 σvσv σv ’σv ’ A1A1 1111 A2A2 11 B1B1 1 1 B2B2 1 1 NRNR 3113 Θ (in degrees) 018000 2cosθ 2-222 (1+2cosθ) 3-- (-1+2cosθ) --11 (N R -2)*(1+2cosθ) 31-- N R *(-1+2cosθ) --13

18. To find n i ‘( A 1 ) h = 4 ; ℓ =1; Ψ (E) = 1; Ψ ( C 2 ) = 1; Ψ ( σ v ) = 1; Ψ ( σ v ’ ) = 1; χ ' (E) = 3; χ ' ( C 2 ) = 1; χ ' ( σ v ) = 1; χ ' ( σ v ’ ) = 3. To find n i ‘( A 2 ) h = 4 ; ℓ =1; Ψ (E) = 1; Ψ ( C 2 ) = 1; Ψ ( σ v ) = -1; Ψ ( σ v ’ ) = -1; χ ' (E) = 3; χ ' ( C 2 ) = 1; χ ' ( σ v ) = 1; χ ' ( σ v ’ ) = 3. n i ‘( A 1 ) = (1/4) [ 1*1*3 + 1*1*1 + 1*1*1 + 1*1*3] =(1/4)*8=2 n i ‘( A 2 ) = (1/4) [ 1*1*3 + 1*1*1 + 1*-1*1 + 1*-1*3]=(1/4)*0=0 n i ‘( B 1 ) = (1/4) [ 1*1*3 + 1*-1*1 + 1*1*1 + 1*-1*3]=(1/4)*0=0 n i ‘( B 2 ) = (1/4) [ 1*1*3 + 1*-1*1 + 1*-1*1 + 1*1*3]=(1/4)*4=1 n i '= (1/h)∑ R ℓ Ψ (R)χ ' (R) h - order of the C 2v point group = 4 ℓ i - dimension of each species in C 2v point group=1 Ψ j (R) – from the character table

19. For an XY 2 bent molecule  There are 3 normal modes of vibration as 3N-6 = 3  Using internal coordinates, the nature of these 3 vibrations are found i.e.,  two bond stretching vibrations and  one angle deformation (bending) vibration  Further, using Character table,  n i ‘(A 1 )=2  n i ‘(B 2 )=1 indicate, these 3 vibrations are distributed among A 1 and B 2 species.  From the knowledge of group theory, as the nature of A species is to have only symmetric vibrations and B is to have only asymmetric vibrations,  There are two symmetric vibrations and one asymmetric vibration Thus, knowledge of group theory systematically unfolds you the nature of vibration of any molecule!!

20. Tabulate and Tally! S.No.Molecule (formula) Linear/ Non-Linear N Normal modes of vibration by using Tally! 3N-5/3N-6internal coordinates Character table

21. Congrats!! You have learnt to calculate normal modes of vibration. Try for many more molecules. Make it a fun game, a mental exercise,…

22. All the best and C u in the next presentation!! -uthra mam