Question

If an Fe-160 stretch is suspected to correspond to a vibrational feature at 630 cm-1, where should the feature appear if the compound is prepared with 180?

Answer 1

The given question is based on a formula for the calculation of vibration wavenumber .Further it is considered that force constant(k) remains same for the isotopic molecules(molecules labelled with different isotopes of same element,like here the two FeO^* molecules are isotopic as they are labelled with different isotopes of oxygen-16,18).

With these points,question is solved as shown in the pic attached.
ra Ginen 1 1680 cant bunen for fo-160, vibrational wanenumber 3 06 o fe- 180 oro w ie õ for Od to To find i a solution :- = (un nibrational wanenumber ome know 3x10 10 cms where tun w ww . ca velocity of light & 3x10 %om u= Force constant (Nm) MS M = Reduced mass of the Molecule (kg) 7 160 and pe 180 e constant abe when & is also constant ie the case with isotopic molecules. Isotopic molecules here Pe-160 and Pe-180 have same na sou K is also considered to be constant. astong de 20c, u all are constant so abone relations oo (1) can be applied of Wymi w z - MF-160 (2) en Dre-180 Je-160 Mere 180 u = MIM2 x 103 , м, е м, @ka Matula» олай а the inolued elements & No is avogadro's no.(-6.023 x 1023, mely we know (MI+M2) No [MR = 55.841) gmail - ng К [ M6 = 16 ам. sobre 55.84x16 X10 3 (55.84 16) No o 55.8t 16) No on [ Meso= 18g mot Mpe-160 = Mpex MiGO x 10 nuoma (MPC + M160) No Mre- 100 = Mre & M180 x 10 3 (MFe + M180) No - 55.84x 18 x 10 3 (55.84+ 18) No X (55.84+18) Do = 16 (73.84) 18 (7184 Mife-160 Mre- 100 - 88.84x 168 10% (55.84+ 16) to S884x 18 x 10

Putting the values in equation (2), we have Jre-180 = Ofe-160 . V 19 x 71.84 16 X 73.8 4 0 NO - Tre -1800 96 Jre-160 =) o fe- 18,0 = 0.96X & Fe-160 ron =0.9,6x630 cm opanoolCoMND I [ OXO 2 fe-160 = 630 an, genen = 6.02. 18 un arxa Aus 8- The up brational feature for the required compound (Re-180) would appear at 602.18 an o r