Answer to Question #163924 in General Chemistry for Cliff

Question

A certain molecule has the fundamental and the first overtone centered at 2143.26cm–¹ and 4260.04cm–¹. Determine the vibrational frequency and first anharmonicity constant

Answer 1

The energy levels for the Morse potential are: 

G(v) = (v + ½)ω_e - (v + ½)²ω_ex_e(in cm^-1) 

The fundamental corresponds to the transition between v = 0 and v = 1. This occurs at: 

G(1) - G(0) = [(1 + ½)ω_e - (1 + ½)²ω_ex_e] – [(0 + ½)ω_e - (0 + ½)²ω_ex_e] 

 = [(3/2)ω_e– (9/4)ω_ex_e] – [(1/2)ω_e -(1/4)ωe_ex_e] = ω_e – 2ω_ex_e

The first overtone corresponds to the transition between v = 0 and v = 2. This occurs at: 

G(2) - G(0) = [(2 + ½)ω_e - (2 + ½)²ω_ex_e] – [(0 + ½)ω_e - (0 + ½)²ω_ex_e] 

 = [(5/2)ω_e – (25/4)ω_ex_e] – [(1/2)ω_e – (1/4)ω_ex_e] = 2ω_e – 6ω_ex_e

As these occur at 3034 cm-1 and 5941 cm-1 respectively; 

ω_e – 2ω_ex_e = 2143.26 cm^-1 (1) 

2ω_e – 6ω_ex_e = 4260.04 cm^-1 (2) 

These are simply two simultaneous equations to solve. Taking 2 × (1) - (2) gives: 

 2 × [ω_e – 2ω_ex_e] - [2ω_e – 6ω_ex_e] = (2 × 2143.26 – 4260.04) cm-1

 2 ω_ex_e = 26.48 cm^-1 or ω_ex_e = 13.24 cm^-1

Substituting this value into (1) gives: 

ω_e – (26.48 cm^-1) = 2143.26 cm^-1

 or ω_e = 2169.74 cm^-1