Atomic and Molecular Structure
(01:160:421/16:160:521)
Answers to Homework Problems, Chapter 13
13.1
(a) False....De is always greater than Do. (b) I surely hope so... True (c) False. E is
the total energy. See first paragraph in Summary, page 430.
13.2
H2(g)  2H(g)
Hoo
1 eV = 23.0605 kcal/mol
Hoo = 2H(H(g)) - H(H2(g)) = Do = 4.4781eV =
4.4781 eV x 23.0605 kcal/mol/eV = 103.26 kcal/mol =103.26 kcal/mol x 4.184 kJ/kcal =
432.07 kJ/mol = bond strength H2(g) = H2 bond dissociation energy
13.3
(A) H2(g)  2H(g)
Ho = Do = 4.478 eV
(B) H2+(g)  H(g) + H+(g) Ho = Do = 2.651 eV
(C) H2(g)  H2+(g) + e-
Ho = IP(H2) = ?
We will use "Hess's Law" (remember that one?). Looking at equations (A) - (C), we see
that sooner or later we will need to bring H(g) and H+(g) into the mix. Hence we will also
need IP for H
(D) H(g)  H+(g) + e-
Ho = IP(H) = E1 = 13.598 eV (Example, page 146)
1
Now
(C) = (A) - (B) + (D)
so that
? = 4.478 eV - 2.651 eV + 13.598 eV = 15.425 eV
(C) H2(g) -> H2+(g) + e- Ho = 15.425 eV = IP1 for H2 (quite large, actually)
13.4
De = Do + ZPE = Do + 0.5h
(a) De(1H35Cl) = Do(1H35Cl) + 0.5h(1H35Cl) =
4.43 eV + 0.5 x 6.626x10-34Js x 8.65x1013s-1 x (1 eV/1.602x10-19 J)) =
4.43 eV + 0.18 eV = 4.61 eV
(b) De(2H 35Cl) = Do(2H 35Cl) + 0.5h(2H35Cl)
De(2H 35Cl) = De(1H 35Cl) because a potential energy surface is independent of the
nuclear masses! This is an important conclusion!
Look at the electronic Hamiltonian and you will see no mention of nuclear masses
there...nuclear charges and coordinates, yes, but no nuclear masses.
The nuclear mass independence furthermore means that vibrational force
constants (the k's) are independent of nuclear masses:
(2H 35Cl) ≠  (1H 35Cl)
k(2H 35Cl) = k(1H 35Cl) but
 = (1/2)(k/)1/2
(1H 35Cl) = (1/2)(k(1H 35Cl)/( 1H 35Cl))1/2
(2H 35Cl) = (1/2)(k(2H 35Cl)/( 2H 35Cl))1/2 = (1/2)(k(1H 35Cl)/( 2H 35Cl))1/2
2
Hence form ratio
(1H 35Cl)/(2H 35Cl) =
[(1/2)(k(1H 35Cl)/( 1H 35Cl))1/2]/{(1/2)(k(2H 35Cl)/( 2H 35Cl))1/2} =
(2H 35Cl)/(1H 35Cl)]1/2 =
[m(2H)m(35Cl)/m(2H) +m(35Cl)] 1/2/ [m(1H)m(35Cl)/m(1H) +m(35Cl)]1/2
= (1.892)1/2/(0.972)1/2 = 1.395 so that
(2H 35Cl) = (1H 35Cl)/1.395 = 8.65x1013s-1/1.395 = 6.20 x1013 s-1
Do(2H 35Cl) = De(2H 35Cl) - 0.5h(2H35Cl) =
4.61 eV - 0.5 x 6.626x10-34Js x 8.65x1013s-1 x (1 eV/1.602x10-19 J))
4.61 eV - 0.13 eV = 4.48 eV
Do is isotope-dependent, even though De is not.
13.5
Evib = (v + 0.5)he
harmonic oscillator approximation
Evib = (v + 0.5)he - (v + 0.5)2hxe
anharmonic oscillator approximation,
incl. first anharmonicity correction term.
In general, anharmonicity lowers the vibrational frequencies and frequency
spacings relative to the harmonic oscillator approximation (i.e. xe is generally positive)
(a) De = Do + ZPE = Do + 0.5h in harm osc app
De = Do + ZPE = (including anharmonicity) = Do + 0.5h- (0.5)2hxe =
Do + 0.5h- 0.25hxe
3
(b) De = Do + 0.5h- 0.25hxe = 2.4287 eV + 0.5x1405.65 cm-1- 0.25x23.20 cm-1 =
2.4287 eV + 0.5x1405.65 cm-1x(eV/8066 cm-1) - 0.25x23.20 cm-1xeV/8066 cm-1) =
2.4287 eV + 0.0871 eV - 0.0007 eV = 2.5151 eV
13.14 (a) proton mass = mp = 1836.15 me = 1836.15 (au)
(b) charge of electron = -e = -1 (au)
(c) Planck's constant = h = ħ x (2 = 2 (au)
(d) E(1s)He+ = (-(Z=2)2/(n=1)2) au = -22/12 au = -4 (au)
(e) Planck's constant has units of (energy)x(time) = Js, so
one au of time = 1ħ/1Hartree = (h/2)Hartree =
6.62608x10-34s/[2x3.141592 x(4.35975x10-18)] = 2.41696 x 10-17 s
so 1 s = 2.416 x 10-17 (au) = 4.134 x 1016 (au)
I think 2.41696 x 10-17 s is the time it takes Bohr's electron to travel one full orbit in the
H-atom.
(f) unit of speed = (length/time) = 0.529177x10-10/2.41696 x 10-17 = 0.21894 x 107
speed of light = c = 2.99792458 x 108 /0.21894 x 107 = 136.92 ~ 137 (au)
(g) E(1s)H = -13.598 eV (1 au/27.2114 eV) = -0.4997 au
(h) 1 D = 10-18 statC x 1cm = 3.33564 x 10-30 C m = (charge x length)
unit of dipole moment = 1e x 1ao = 1.602 x 10-19C x 0.529177x10-10 m =
8.47836x10-30 Cm
Hence 1 D = 3.33564 x 10-30 C m/8.47836x10-30 Cm = 0.3934 (au)
4