1. Tutorial – 4 1) Calculate the moment of inertia (I) and bond length (r) from microwave spectrum of CO. First line (J = 0 to J=1 transition) in the rotation spectrum of CO is 3.84235 cm -1. Calculate the moment of inertia (I) and bond length (r) of CO. Mass of C = 19.92168 x 10 -27 kg Mass of O = 26.561 x 10 -27 kg Find r(C  O)

2. 2B = 2B = 3.84235 cm -1 B = 1.921175 cm -1 Answer I =  r 2 for COJ=0  J=1 (First line)at3.84235 cm -1  = = = = x 19.92168 x 10 -27 x 26.561 x 10 -27 + 19.92168 x 10 -27 + 26.561 x 10 -27  = = = = x [19.92168 x 26.561] x 10 -27 x 10 -27 + [19.92168 + 26.561] x 10 -27  = = = = 529.14 x 10 -27 46.48  = = = = 11.384 x 10 -27  = = = = 1.1384 x 10 -26 kg  = = = = x m o M c x m o + m o M c + m o

3. Answer I =  r 2 =I = =I = 6.626 x 10 -34 kg m 2 s -1 8 x 9.869 x B m -1 x 2.998×10 8 m s −1 h = 6.626 x 10 -34 Kg m 2 s -1 c = 2.998×10 8 m s −1 B = 1.921175 cm -1 B = 192.1175 m -1 =I = =I = 0.02799 x 10 -42 kg m 2 =I = =I = 1.4569 x 10 -46 kg m 2 192.1175 =r2 = =r2 = I Kg m 2  =r2 = =r2 = m2m2 kg 1.4569 x 10 -46 1.1384 x 10 -26 =r2 = =r2 = m2m2 1.2797 x 10 -20 =r = =r = m 1.131 x 10 -10 =r = =r = 1.131 Å

4. BB’ µ µ’ < > First rotational absorption, For = = Mass of O = 15.9994 Mass of 13 C = ?? The effect of isotopic substitution on the energy levels and hence rotational spectrum of a diatomic molecule such as carbon monoxide

5. Comparison of rotational energy levels of 12 CO and 13 CO Can determine: (i) isotopic masses accurately, to within 0.02% of other methods for atoms in gaseous molecules; (ii) isotopic abundances from the absorption relative intensities. Question 2: First rotational lines in microwave region for both 12 CO and 13 CO are given below: for 12 COJ=0  J=1 (First line)at3.84235 cm -1 for 13 CO J=0  J=1 (First line)at 3.67337 cm -1 Mass of 12 C and 16 O Given : 12 C = 12.0000 amu; 16 O = 15.9994 amu Calculate the atomic weight of 13 C Question-2

6. Answer

7. Ans: Question & Answer 1) Why vibrational (IR) frequency ( ϖ os ) of triple bond (C  C) is higher than C-C ? The vibrational frequency of a particular bond is increasing with:  increasing force constant k ( = increasing bond strength)  decreasing atomic mass and, therefore, k c  c > k c-c [The force constant (k) is proportional to the strength of the covalent bond linking two atoms] Here two atoms in both cases are same (carbon) and hence reduced mass (  ) is same. Thus, ϖ os for triple bond (C  C) is higher than C-C Oscillation Frequency, k is the force constant (bond strength )  = reduced mass of atoms cm -1 Example: cm -1

8. Question & Answer 2) Why vibrational (IR) frequency ( ϖ os ) of N-H (3400 cm -1 ) is higher than P-H (2350 cm -1 ) ? Ans: The vibrational frequency of a particular bond is increasing with:  increasing force constant k ( = increasing bond strength)  decreasing atomic mass Oscillation Frequency, k is the force constant (bond strength )  = reduced mass of atoms cm -1 Here two atoms in both cases are different (NH Vs PH); P is heavier than N and hence reduced mass (  ) for P-H bond is higher than that of N-H bond. Thus, ϖ os for N-H bond (3400 cm -1 ) is higher than P-H (2350 cm -1 ). Example:

9. Question & Answer 3) Why vibrational (IR) frequency ( ϖ os ) of O-H (3600 cm -1 ) is higher than S-H (2570 cm -1 ) ? Ans: The vibrational frequency of a particular bond is increasing with:  increasing force constant k ( = increasing bond strength)  decreasing atomic mass Oscillation Frequency, k is the force constant (bond strength )  = reduced mass of atoms cm -1 Here two atoms in both cases are different (OH Vs SH); S is heavier than O and hence reduced mass (  ) for S-H bond is higher than that of O-H bond. Thus, ϖ os for O-H bond (3600 cm -1 ) is higher than S-H (2570 cm -1 ). Example: cm -1

10. Question & Answer 4) Why vibrational (IR) frequency ( ϖ os ) of F-H (4000 cm -1 ) is higher than Cl-H (2890 cm -1 )? Ans: The vibrational frequency of a particular bond is increasing with:  increasing force constant k ( = increasing bond strength)  decreasing atomic mass Oscillation Frequency, k is the force constant (bond strength)  = reduced mass of atoms cm -1 Here two atoms in both cases are different (FH Vs ClH); Cl is heavier than F and hence reduced mass (  ) for Cl-H bond is higher than that of F-H bond. Thus, ϖ os for F-H bond (4000 cm -1 ) is higher than Cl-H (2890 cm -1 ). Example: cm -1

11. Ans: Question & Answer 5) Why vibrational (IR) frequency (  os ) of C  N triple bond is higher than C-N single bond? The vibrational frequency of a particular bond is increasing with:  increasing force constant k ( = increasing bond strength)  decreasing atomic mass and, therefore, k C  N > k C-N Here two atoms in both cases are same (c and N) and hence reduced mass (  ) is same. Thus, ϖ os for C  N triple bond is higher than C-N single bond. Oscillation Frequency, k is the force constant (bond strength )  = reduced mass of atoms The equation on the above describes the major factors that influence the stretching frequency of a covalent bond between two atoms of mass m 1 and m 2 respectively. The force constant (k) is proportional to the strength of the covalent bond linking m 1 and m 2. In the analogy of a spring, it corresponds to the spring's stiffness. For example, a C=N double bond is about twice as strong as a C-N single bond, and the C≡N triple bond is similarly stronger than the double bond. The infrared stretching frequencies of these groups vary in the same order, ranging from 1100 cm-1 for C-N, to 1660 cm-1 for C=N, to 2220 cm-1 for C≡N. cm -1