b) Drawing the methane molecule as suggested in the exercise, we get the following figure:

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Kimberly Floyd
5 years ago
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1 Exersise 1: Moment of inertia a) As shown at the lecture, sulfur hexafluoride SF 6 has an octahedral structure, where sulfur atom is in the origin (the center of inertia point). Directing three principal axes along the S-F bonds, we can draw SF 6 molecule as: It is easy to see that the moment of inertia about each of the three principal axes Z, Y, and X is the same. Therefore, SF 6 molecule is a spherical top. The moment of inertia can be calculated as: I = 4 m F R SF 2, where m F = 19 m p is the F atom mass, m p is the proton mass, and R SF is the length of the S-F bond. Substituting the numerical values, we get I 3, kg m 2 b) Drawing the methane molecule as suggested in the exercise, we get the following figure: Where only two principal axes Z and Y are shown for simplicity. It is clear from the figure symmetry that all three moments of inertia are the same. Therefore, CH 4 is a spherical top. Let us calculate the moment of inertia about axis Y. Using the Pythagorean theorem, we notice, that the distance from each of the four H atoms to this axis r H = r/2 is the same, where r is the diagonal of a profile plane. Therefore, the total moment of inertia about Y axis is given by: 1

2 I = 4 m H r H 2 = 4 m H (r /2 ) 2 = m H r 2 = 2 m H a 2 = 8 / 3 m H R CH 2, where R CH is the length of the C-H bond. Substituting numerical values, we get: I 5, kg m 2 c) Let us draw CHCl 3 molecule. It is clear from the symmetry of the molecule that three chlorine atoms are in the corners of a regular triangle and that C and H atoms are placed on the altitude descended to the triangle center which we denote 0. The cross in figure X indicates the center of mass (c.m.) of the CHCl 3 molecule and the distance 0X we denote d. It is clear that the principle axis Z coincides with the altitude and goes through the C and H atoms. The two other axes also go through the c.m., Y is parallel to the Cl-Cl bond and X is parallel to the altitude of the regular triangle Cl-Cl-Cl. The length of Cl-Cl bond we denote as a. In order to prove that the moments of inertia about X and Y axes have the same value, we draw the plane containing these axes (which is the regular triangle) and use the Pythagorean theorem. 2

3 Denoting a side of the triangle as a, we get: I X = 2 M a 2 /4 = M a 2 /2 I Y = 2 M h M h 2 2 = 2 M (a 2 -a 2 /4)/9 + M 4(a 2 -a 2 /4)/9 = M a 2 /6 + Ma 2 /3 = M a 2 /2 Therefore, we see that I X = I Y which means that CHCl 3 is a symmetric top. Let us calculate the moment of inertia around the Z axis: I Z = 3 m Cl d 1 2 = 3 m Cl (R CCl ) 2 sin 2 α = 3 35 m p (R CCl ) 2 sin Substituting the numerical values, we get: I Z = 5, kg m 2 Exersise 2: a) The expression for the rotational constant was given at the lecture: B = ħ/(4 π c I) The moment of inertia I for CO molecule is equal to I = M 1 R M 2 R 2 2, where R 1 and R 2 are the distance from the molecular center of mass to the C and O nuclear, respectively. One can fount the position of the center of mass considering the equilibrium of the molecule on the sharp holder: 3

4 M 1 g R 1 = M 2 g R 2 R 1 = R M 2 /(M 1 + M 2 with the solution R 1 + R 2 = R = R CO R 2 = R M 2 /(M 1 + M 2 ) Substituting R 1 and R 2 into the expression for I, we get: I = M 1 R 2 M 2 2 / (M 1 +M 2 ) 2 + M 2 R 2 M 1 2 / (M 1 +M 2 ) 2 = M 1 M 2 /(M 1 +M 2 ) 2 [M 1 +M 2 ]R 2 I = µ R 2, where µ= M 1 M 2 /(M 1 +M 2 ) is the reduced mass. Of cause, the obtained result is valid for any diatomic molecule and also for linear configuration molecule CO 2 where the center of mass is on the C atom and does not contribute to the moment of inertia. Therefore, for CO we have: B CO = ħ/(4 π c I CO ) = ħ/(4 π c µ CO R 2 CO) for CO 2 we have: B CO2 = ħ/(4 π c I CO2 ) = ħ/(4 π c µ OO 4 R 2 CO) Substituting numerical values, we get: B CO 1,59 cm -1, B CO2 0,34 cm -1 b) CH 3 I is a symmetric top molecule. In order to find both rotational constants we first have to find the center of mass of the molecule. It is evidently on the principal axis between C and I nuclei, see the figure, where L 1 =R IC is the length of the I-C bond and L 2 is the distance between the C nucleus and the H-atom plane: M I g X = M C g (L 1 -X) + 3M H g (L 1 +L 2 -X) (M I +M C +3M H ) X = M C L 1 +3M H (L 1 +L 2 ) X = [M C L 1 +3M H (L 1 +L 2 )]/M = where M = M I +M C + 3M H = L 1 (M C +3M H )/M + L 2 3M H /M (1) The position of the center of mass is now known and we are calculating the moments of inertia I and I, see figure below where all three axes X, Y, and Z are shown. The moments of inertia about axes X and Y is the same and equal to I, while the moment of inertia about axis Z is equal to I. L 2 is calculated as: 4

5 2 2 1/ /2 L = [(R CH ) cos (α/2)-h1 2 2 ] = [(RCH) cos (α/2)- (RCH) sin (α/2)/3] = R CH [(1+2cosα)/3] 1/2 where h 1 2 = (a 2 -a 2 /4)/9 = a 2 /(12) = (R CH ) 2 sin 2 (α/2)/3 We calculate the moment of inertia about axis Y: I = M I X 2 + M C (L 1 -X) M H R M H R 2 2 = = M I X 2 + M C (L 1 -X) M H [h (L 1 +L 2 -X) 2 ] + M H [(2h 1 ) 2 +(L 1 +L 2 -X) 2 ] = Now we substitute X from eq.(1) = M I [L 1 (M C +3M H )/M + L 2 3M H /M] 2 + M C [L 1 - L 1 (M C +3M H )/M+L 2 3M H /M] M H h M H (L 1 +L 2 -X) 2 ] + M H (2h 1 ) 2 +M H (L 1 +L 2 -X) 2 = = M I [L 1 (M C +3M H )/M + L 2 3M H /M] 2 + M C [L 1 - L 1 (M C +3M H )/M - L 2 3M H /M] 2 + 3M H (L 1 +L 2 -X) 2 + 6M H (h 1 ) 2 = = M I [L 1 (M C +3M H )/M + L 2 3M H /M] 2 + M C [L 1 - L 1 (M C +3M H )/M - L 2 3M H /M] 2 + 3M H [L 1 +L 2 - L 1 (M C +3M H )/M - L 2 3M H /M] 2 + 6M H (h 1 ) 2 = = M I [L 1 (M C +3M H )/M + L 2 3M H /M] 2 + M C [L 1 (M I )/M - L 2 3M H /M] 2 + 3M H [L 1 (M I )/M + L 2 (M I +M C )/M] 2 + 6M H (h 1 ) 2 = Now we take the term 1/(M 2 ) out of the brackets =M I /(M 2 ) [L 1 (M C +3M H ) + L 2 3M H ] 2 + M C /(M 2 ) [L 1 (M I ) - L 2 3M H ] 2 + 3M H /(M 2 ) [L 1 (M I ) + L 2 (M I +M C )] 2 + 6M H (h 1 ) 2 = 5

6 Now we take a square in all brackets using the formula (a±b) 2 =a 2 +b 2 ±2ab =M I /(M 2 ) [(L 1 (M C +3M H )) 2 + (L 2 3M H ) 2 +2L 1 L 2 (Mc+3M H )3M H ] + +M C /(M 2 ) [(L 1 (M I )) 2 + (L 2 3M H ) 2-2L 1 L 2 M I 3M H ]+ + 3M H /(M 2 ) [(L 1 (M I )) 2 + (L 2 (M I +M C )) 2 +2L 1 L 2 M I (M I +M C )] + 6M H (h 1 ) 2 = Now we join the terms with L 1 2, L 2 2, and 2L 1 L 2 =L 1 2 /(M 2 ) [M I (M C +3M H ) 2 +M C M I 2 +3M H M I 2 ] + + L 2 2 /(M 2 ) [M I (3M H ) 2 +M C (3M H ) 2 +3M H (M I +M C ) 2 ] + +2L 1 L 2 /(M 2 )[M I 3M H (M C +3M H )-M C M I 3M H +3M H M I (M I +M C )] + 6M H (h 1 ) 2 = =L 1 2 M I (M C +3M H )/M + L 2 2 3M H (M I +M C )/M + 2L 1 L 2 M I 3M H /M + 6M H (h 1 ) 2 Substituting L 1, L 2, and h 1 2 we obtain the resulting formula: I = (R IC ) 2 M I (M C +3M H )/M + (R CH ) 2 (1+2cosα)M H (M I +M C )/M R IC R CH [3(1+2cosα)] 1/2 M I M H /M + M H (R CH ) 2 (1-cos α) The moment of inertia I is calculated according to: I = 3 M H h 2 2 = 3 M H 4(a 2 -a 2 /4)/9 = M H a 2 = 4M H (R CH ) 2 sin 2 (α/2) Thus, I = 2M H (R CH ) 2 (1-cosα) The rotational constants are given by: A = ħ /(4πcI ); B = ħ /(4πcI ) Substituting the numerical values we get: I 1, kg m 2, I 1, kg m 2 A 5, m -1 = 5,8 cm -1 B 25 m -1 = 0,25 cm -1 It is clear from the calculation that the major contribution to the moment of inertia I is given by the I and C atoms, while the contribution from three H atoms is very small and can be neglected. Therefore, the moment of inertia I is large. The moment of inertia I is produced exclusively by 3 light H atoms, therefore it is small. c) The rotational constant A is larger for CH 3 Cl than for CD 3 Br, because D atom is heavier than the H atom and therefore the moment of inertia I is smaller for CH 3 Cl. 6

7 The rotational constant B is also larger for CH 3 Cl than for CD 3 Br, because Br atom is heavier than the Cl atom and therefore the moment of inertia I is smaller for CH 3 Cl. Therefore, the rotational energy levels more closely spaced for CD 3 Br, than for CH 3 Cl. 7

Lewis structures show the number and type of bonds between atoms in a molecule or polyatomic ion.

VSEPR & Geometry Lewis structures show the number and type of bonds between atoms in a molecule or polyatomic ion. Lewis structures are not intended to show the 3-dimensional structure (i.e. shape or geometry)