Computational Chemistry

A deuterium atom is heavier than a hydrogen atom, but the real point is not its
weight, which involves gravity, but its mass, which does not. The vibrational
frequency of a bond depends on its stiffness (the force constant) and on the masses
of the atoms involved. For a diatomic molecule A–B the vibrational frequency (in
wavenumbers) is governed by the simple formula

~v¼

1

2 pc

k

m

 1 = 2

wherecis the velocity of light,kis the force constant, andm(mu) is the reduced
mass of the two atoms,M 1 m 2 /(M 1 þm 2 ). IfM 1 is big compared tom 2 , this equation
devolves to

~n¼constant

k

m 2

 1 = 2

as expected, since essentially the big mass does not move. Thus substituting D for
H in a sizable molecule reduces the C-Hydrogen stretch wavenumber by a factor
of 21/2= 1.4. With polyatomic molecules, accounting for mass is a bit more
complicated. The force constant matrix must be “mass weighted” and diagonalized
to give a matrix with the displacement vectors of the vibrations, and a matrix with
the frequencies [1].

Reference

1. Details of how this is done in a computational chemistry program are given in http://www.
   gaussian.com/g_whitepap/white_pap.htm, “Vibrational analysis in Gaussian”

Chapter 2, Harder Questions, Answers

Q4

We assumed that the two bond lengths of water are equal.Mustan acyclic molecule
AB 2 have equal A–B bond lengths? What about a cyclic molecule AB 2?
Intuitively, there is no reason why acyclic or cyclic AB 2 should have unequal
A–B bond lengths: one A–B bond seems just as good as the other. Butprovingthis
is another matter.
Consider a molecule AB 2 , linear, bent, or cyclic. Each of the two A–B bonds has
the same force constant – we can’t have one, say, single and one double, because
this on-paper arrangement would correspond to a resonance hybrid with each bond
the same ca. 1.5 in bond order:

596 Answers