Tutorial – 4

1. Tutorial – 4 1) Calculate the moment of inertia (I) and bond length (r) from microwave spectrum of CO. Find r(CO) 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

2. Answer I = r2 for CO J=0 J=1 (First line)at 3.84235 cm-1 2B = 3.84235 cm-1 B = 1.921175 cm-1 19.92168 x 10-27 x 26.561 x 10-27 Mcx mo kg  =  = 19.92168 x 10-27 + 26.561 x 10-27 Mc+mo [19.92168 x 26.561] x 10-27 x 10-27 kg  = [19.92168 + 26.561] x 10-27 529.14 x 10-27 kg  = 46.48 11.384 x 10-27 kg  = kg 1.1384 x 10-26  =

3. Answer I = r2 B = 1.921175 cm-1 B = 192.1175 m-1 h = 6.626 x 10-34 Kg m2s-1 c = 2.998×108 m s−1 6.626 x 10-34 kg m2 s-1 I = 8 x 9.869 x B m-1 x 2.998×108 m s−1 I Kg m2 r2 = 0.02799 x 10-42  kg kg m2 I = 192.1175 1.4569 x 10-46 1.4569 x 10-46 I = r2 = m2 kg m2 1.1384 x 10-26 1.2797 x 10-20 m2 r2 = m 1.131 x 10-10 r = 1.131 Å r =

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

5. Question-2 Comparison of rotational energy levels of 12CO and 13CO 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 12CO and 13CO are given below: for 12CO J=0 J=1 (First line)at 3.84235 cm-1 for 13CO J=0  J=1 (First line) at 3.67337 cm-1 Mass of 12C and 16O Given : 12C = 12.0000 amu; 16O = 15.9994 amu Calculate the atomic weight of 13C

6. Answer

7. Question & Answer 1) Why vibrational (IR) frequency (ϖos) of triple bond (CC) is higher than C-C ? Ans: k is the force constant (bond strength ) = reduced mass of atoms Oscillation Frequency, • The vibrational frequency of a particular bond is increasing with: • increasing force constant k ( = increasing bond strength) • decreasing atomic mass cm-1 and, therefore, k cc> kc-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, ϖos for triple bond (CC) is higher than C-C cm-1 Example:

8. Question & Answer 2) Why vibrational (IR) frequency (ϖos) of N-H (3400 cm-1) is higher than P-H (2350 cm-1)? Ans: k is the force constant (bond strength ) = reduced mass of atoms Oscillation Frequency, • The vibrational frequency of a particular bond is increasing with: • increasing force constant k ( = increasing bond strength) • decreasing atomic mass 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, ϖos for N-H bond (3400 cm-1) is higher than P-H (2350 cm-1). cm-1 Example:

9. Question & Answer 3) Why vibrational (IR) frequency (ϖos) of O-H (3600 cm-1) is higher than S-H (2570 cm-1) ? Ans: k is the force constant (bond strength ) = reduced mass of atoms Oscillation Frequency, • The vibrational frequency of a particular bond is increasing with: • increasing force constant k ( = increasing bond strength) • decreasing atomic mass 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, ϖos for O-H bond (3600 cm-1) is higher than S-H (2570 cm-1). cm-1 Example:

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

11. Question & Answer 5) Why vibrational (IR) frequency (os) of CN triple bond is higher than C-N single bond? Ans: Oscillation Frequency, k is the force constant (bond strength ) = reduced mass of atoms • The vibrational frequency of a particular bond is increasing with: • increasing force constant k ( = increasing bond strength) • decreasing atomic mass cm-1 The equation on the above describes the major factors that influence the stretching frequency of a covalent bond between two atoms of mass m1and m2respectively. The force constant (k) is proportional to the strength of the covalent bond linking m1 and m2. 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 and, therefore, k CN > kC-N Here two atoms in both cases are same (c and N) and hence reduced mass () is same. Thus, ϖos for CN triple bond is higher than C-N single bond.