Name /(PJStudent number_V'- _Section

Chemistry 483 Exam 4

1. (50 points) define and/or characterize

a. The 5 symmetry elements A "

s: J v) : J S.J c;

December 5, 2008

c. Rotation reflection operator

;1 ~~-SN J'

~ 6h~.

e. Trace of a matrix

1

/

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f. Centrifugal distortion

g. Selection rule

~tn7tk~a4.~i~

~~~b?~.L~'

F<n.. ~ 4v-="1: I ) f~ A<Te. r/

h. Transition dipole

i. Rotation matrix

~t-./.,. ~1(J.~~~ ~~ z{~h1~~~~~< FVLzivod~td (&:J9 ~9 )

- Ah1 g.. CJ#> p.Jj. Basis for an irreducible representation

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2. (25 points) The S02 molecule has C2vsymmetry. Suppose the molecule is in the yz

plane with S at the origin and the two 0 atoms at (O,±Yo,zo)' Use the Px orbitals on S

and the two 0 atoms as a basis and

a. determine the representation matrices of C2v.

b. what are the characters of this representation?

c. what irreducible representations are contained in this reducible representation?

d. if we denote the P; orbitals on the 0 atoms as PL and PR for left and right

a.

respectively what irreducible representations does PL +PR belong to?

e. what irreducible representation does the P, orbital on S belong to.

!£= ( / Ob)010 ./o 0 J (

0 10I 0 0o C) I

)

,. I ("'dOO)-.I- af'V -=- 0-1 0

o cs r- ]

b, X - 1 1 -3:3-No-tL )' ~ we. ~ h tl<JM? a... (XL) ~F

~~ I~=O o I ). tiTV ::. (/00)(' 0 0 ) (0 0 - I ) 0-1 0

lE= o J 0 J (£2= 0,-1 0 ..J I 0 j

o 0-1o 01 -I 0 0 o 0 '

t X ~.3 -j

Jo~~~~

() -= _.' (:5 - 1/. (C2) -I- X, (OV) ,- ""3X). (Oy))~ 4 ~ ~ ,a I (:3 - 1-I-1 - 3) :::::0 a = -L (3+ / -f / +3) =- 2.

A-- 73} 4-J - 4-

a Az :. ~ (:3- I - 1 -+ 3) z: / (j( 132 ~ 0 3

1

.!

Name. 1<'Student;:-:-nu=m"i:':'ber=---!!j7f----Section

A.

d. E (f'L + fk) ~+(!:7L +'8?JA

C2 (fl-l- fJZ- J:= - ((L -I-{I<)1\

(Tv (fL +-TIZ J =- -i (1/L -I- ff7.)

r - / -/ l-J

4

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3. (25 points) Formaldehyde, H2CO, has C2vsymmetry. How many normal modes of

vibration does formaldehyde have? What are their symmetries? Which are IR active?

b.

:3 N- /; - 12-6 :.60II

/+ ..e -,{-IA '" " J

C~ ir; OV-2- 2 4

E12

tlp-. ~ f (12 - Z -';2 - 1- ) -:: .t

A ...L (J2. +2 .-1--2- - 4 ) = .3G(" 8,.:: -1

(182

= ~ (12 i 2 - '2 -+ 4 ) = 4-

rrzrn1 -c~ ~ ~ 4A-L

tfJ ~ /lA.-t. AtJ~') 13~,

~~~V~

j:3A,) r e.: 282Jm=

A\ -r::7 B'~I} loJ/J ••••.

I..

..5