Question

If an Fe-¹⁶O stretch is suspected to correspond to a vibrational feature at 630 cm^-1, where should the feature appear if the compound is prepared with ¹⁸O?

Answer 1

Solution:

The stretching frequency (v) decreases with increasing reduced mass (),

v = (1/2πc) √(k/)

For, Fe-O16 stretch, reduced mass is calculated as,

(Fe-O 16) = m1m2/m1+m2 = 56 x 16 / 56 +16 = 12.44

Thus,

630 cm-1 = (1/2πc) √(k/12.44) ----(1)

For Fe-O18 stretch, reduced mass is calculated as,

(Fe-O 18) = m1 m2/m1+m2 = 56 x 18 / 56 + 18 = 13.62

Thus,

v = (1/2πc) √(k/ 13.62) ------(2)

From equation 1 and 2:

v / 630 cm-1 = √ 12.44 / √13.62

v = 630 cm-1 x √ (12.44 / 13.62)

v = 630 cm-1 x √ 0.9134 = 630 cm-1 x 0.96 = 604.8 cm-1

v = 605 cm-1

Thus, stretch is shifted towards lower frequency region.