Question

Calculate the expected wavenumber for a N----D stretch given that N---H stretches are normally observed at 3400 cm −1 .

show detailed steps. From Inorganci Chemistry.

Answer 1

we have formula strech in wavenumber = ( 1/ 2pi x c) sqrt ( k/u)

c= speed of light , K = force constant    = constant

strech of N-D / strech of N-H =   sqrt [ ( u of N-H) / u of N-D) ]

u of N-H is reduced mass of N-H = ( mN x mH) / ( mN + mH)

where mN = atomic mass of N = 14amu , mH = 1 amu ,    , mD = 2amu

now u N-H = ( 14 x 1 ) / ( 14+1) = 14/15 = 0.933

reduced mass u-N-D = ( 14 x 2) / ( 14+2) = 28 /16 = 1.75

now   wavenumberof N-D / ( 3400) = sqrt ( 0.933 /1.75)

wavenumber of N-D = 2483 cm-1

( since force constants of N-D and N-H not given we assume force constant is same for both )