Solutions
 Page ID
 11570
Solutions
1.
a. First determine the eigenvalues:
det = 0
(1  l)(2  l)  22 = 0
2 + l  2l + l2  4 = 0
l2  l  6 = 0
(l  3)(l + 2) = 0
l = 3 or l = 2.
Next, determine the eigenvectors. First, the eigenvector associated with eigenvalue 2:
= 2
C11 + 2C21 = 2C11
C11 = 2C21 (Note: The second row offers no new information, e.g. 2C11 + 2C21 = 2C21)
C112 + C212 = 1 (from normalization)
(2C21)2 + C212 = 1
4C212 + C212 = 1
5C212 = 1
C212 = 0.2
C21 = , and therefore C11 = 2.
For the eigenvector associated with eigenvalue 3:
= 3
C12 + 2C22 = 3C12
4C12 = 2C22
C12 = 0.5C22 (again the second row offers no new information)
C122 + C222 = 1 (from normalization)
(0.5C22)2 + C222 = 1
0.25C222 + C222 = 1
1.25C222 = 1
C222 = 0.8
C22 = = 2, and therefore C12 = .
Therefore the eigenvector matrix becomes:
b. First determine the eigenvalues:
det = 0
det det = 0
From 1a, the solutions then become 2, 2, and 3. Next, determine the eigenvectors. First the eigenvector associated with eigenvalue 3 (the third root):
= 3
2 C13 = 3C13 (row one)
C13 = 0
C23 + 2C33 = 3C23 (row two)
2C33 = 4C23
C33 = 2C23 (again the third row offers no new information)
C132 + C232 + C332 = 1 (from normalization)
0 + C232 + (2C23)2 = 1
5C232 = 1
C23 = , and therefore C33 = 2.
Next, find the pair of eigenvectors associated with the degenerate eigenvalue of 2. First, root one eigenvector one:
2C11 = 2C11 (no new information from row one)
C21 + 2C31 = 2C21 (row two)
C21 = 2C31 (again the third row offers no new information)
C112 + C212 + C312 = 1 (from normalization)
C112 + (2C31)2 + C312 = 1
C112 + 5C312 = 1
C11 =
Second, root two eigenvector two:
2C12 = 2C12 (no new information from row one)
C22 + 2C32 = 2C22 (row two)
C22 = 2C32 (again the third row offers no new information)
C122 + C222 + C322 = 1 (from normalization)
C122 + (2C32)2 + C322 = 1
C122 + 5C322 = 1
C12 = (1 5C322)1/2 (Note: again, two equations in three unknowns)
C11C12 + C21C22 + C31C32 = 0 (from orthogonalization)
Now there are five equations with six unknowns.
Arbitrarily choose C11 = 0
(whenever there are degenerate eigenvalues, there are not unique eigenvectors because the degenerate eigenvectors span a 2 or more dimensional space, not two unique directions. One always is then forced to choose one of the coefficients and then determine all the rest; different choices lead to different final eigenvectors but to identical spaces spanned by these eigenvectors).
C11 = 0 =
5C312 = 1
C31 =
C21 = 2
C11C12 + C21C22 + C31C32 = 0 (from orthogonalization)
0 + 2+ C32 = 0
5C32 = 0
C32 = 0, C22 = 0, and C12 = 1
Therefore the eigenvector matrix becomes:
2.
a. K.E. = = = =
K.E. =
K.E. =
K.E. =
b. p = mv = ipx + jpy + kpz
p =
where i, j, and k are unit vectors along the x, y, and z axes.
c. Ly = zpx  xpz
Ly = z  x
3.
First derive the general formulas for , , in terms of r,q, and f, and , , and in terms of x,y, and z. The general relationships are as follows:
x = r Sinq Cosf r2 = x2 + y2 + z2
y = r Sinq Sinf sinq =
z = r Cosq cosq =
tanf =
First , , and from the chain rule:
= y,z + y,z + y,z ,
= x,z + x,z + x,z ,
= x,y + x,y + x,y .
Evaluation of the many "coefficients" gives the following:
y,z = Sinq Cosf , y,z = , y,z =  ,
x,z = Sinq Sinf , x,z = , x,z = ,
x,y = Cosq , x,y =  , and x,y = 0 .
Upon substitution of these "coefficients":
= Sinq Cosf +  ,
= Sinq Sinf + + , and
= Cosq  + 0 .
Next , , and from the chain rule:
= q,f + q,f + q,f ,
= r,f + r,f + r,f , and
= r,q + r,q + r,q .
Again evaluation of the the many "coefficients" results in:
q,f = , q,f = ,
q,f = , r,f = , r,f = ,
r,f =  , r,q = y , r,q = x , and r,q = 0
Upon substitution of these "coefficients":
= +
+
= + 
= y + x + 0 .
Note, these many "coefficients" are the elements which make up the Jacobian matrix used whenever one wishes to transform a function from one coordinate representation to another. One very familiar result should be in transforming the volume element dxdydz to r2Sinqdrdqdf. For example:
=
a. Lx =
Lx =

Lx = 
b. Lz = =  i
Lz =
4.
B dB/dx d2B/dx2
i. 4x4  12x2 + 3 16x3  24x 48x2  24
ii. 5x4 20x3 60x2
iii. e3x + e3x 3(e3x  e3x) 9(e3x + e3x)
iv. x2  4x + 2 2x  4 2
v. 4x3  3x 12x2  3 24x
B(v.) is an eigenfunction of A(i.):
(1x2)  x B(v.) =
(1x2) (24x)  x (12x2  3)
24x  24x3  12x3 + 3x
36x3 + 27x
9(4x3 3x) (eigenvalue is 9)
B(iii.) is an eigenfunction of A(ii.):
B(iii.) =
9(e3x + e3x) (eigenvalue is 9)
B(ii.) is an eigenfunction of A(iii.):
x B(ii.) =
x (20x3)
20x4
4(5x4) (eigenvalue is 4)
B(i.) is an eigenfunction of A(vi.):
 2x B(i) =
(48x2  24)  2x (16x3  24x)
48x2  24  32x4 + 48x2
32x4 + 96x2  24
8(4x4  12x2 + 3) (eigenvalue is 8)
B(iv.) is an eigenfunction of A(v.):
x + (1x) B(iv.) =
x (2) + (1x) (2x  4)
2x + 2x  4  2x2 + 4x
2x2 + 8x  4
2(x2  4x +2) (eigenvalue is 2)
5.
6.
7.
8.
i. In ammonia, the only "core" orbital is the N 1s and this becomes an a1 orbital in C3v symmetry. The N 2s orbitals and 3 H 1s orbitals become 2 a1 and an e set of orbitals. The remaining N 2p orbitals also become 1 a1 and a set of e orbitals. The total valence orbitals in C3v symmetry are 3a1 and 2e orbitals.
ii. In water, the only core orbital is the O 1s and this becomes an a1 orbital in C2v symmetry. Placing the molecule in the yz plane allows us to further analyze the remaining valence orbitals as: O 2pz = a1, O 2py as b2, and O 2px as b1. The (H 1s + H 1s) combination is an a1 whereas the (H 1s  H 1s) combination is a b2.
iii. Placing the oxygens of H2O2 in the yz plane (z bisecting the oxygens) and the (cis) hydrogens distorted slightly in +x and x directions allows us to analyze the orbitals as follows. The core O 1s + O 1s combination is an a orbital whereas the O 1s  O 1s combination is a b orbital. The valence orbitals are: O 2s + O 2s = a, O 2s  O 2s = b, O 2px + O 2px = b, O 2px  O 2px = a, O 2py + O 2py = a, O 2py  O 2py = b, O 2pz + O 2pz = b, O 2pz  O 2pz = a, H 1s + H 1s = a, and finally the H 1s  H 1s = b.
iv. For the next two problems we will use the convention of choosing the z axis as principal axis for the D¥h, D2h, and C2v point groups and the xy plane as the horizontal reflection plane in Cs symmetry.
D¥h D2h C2v Cs
N 1s sg ag a1 a'
N 2s sg ag a1 a'
N 2px pxu b3u b1 a'
N 2py pyu b2u b2 a'
N 2pz su b1u a1 a''
9.
a. Yn(x) = Sin
Pn(x)dx = dx
The probability that the particle lies in the interval 0 £ x £ is given by:
Pn = =
This integral can be integrated to give :
Pn =
Pn =
Pn =  \f(1,4q,2
=
=  Sin
b. If n is even, Sin= 0 and Pn = .
If n is odd and n = 1,5,9,13, ... Sin= 1
and Pn = 
If n is odd and n = 3,7,11,15, ... Sin= 1
and Pn = +
The higher Pn is when n = 3. Then Pn = +
Pn = + = 0.303
c. Y(t) = e= aYne+ bYme
HY = aYnEne+ bYmEme
= a2En + b2Em + a*be
+ b*ae
Since and are zero,
= a2En + b2Em (note the time independence)
d. The fraction of systems observed in Yn is a2. The possible energies measured are En and Em. The probabilities of measuring each of these energies is a2 and b2.
e. Once the system is observed in Yn, it stays in Yn.
f. P(En) = 2 = cn2
cn = x(Lx)dx
= dx
=
These integrals can be evaluated to give:
cn = 60,L6L\b(\f(L2,n2p2
 60,L6\b(\f(2xL2,n2p2
cn = {
 )
 (
 Cos(np)
+ Cos(0))}
cn = L3{ Cos(np) + Cos(np)
+ }
cn =
cn =
cn = )
cn2 = )
If n is even then cn = 0
If n is odd then cn = =
The probability of making a measurement of the energy and obtaining one of the eigenvalues, given by:
En = is:
P(En) = 0 if n is even
P(En) = if n is odd
g. =
=
=
=
= 30\o(h,L\f(x2,2
=
=
= =
10.
= Ci*eeCj
Since = Ejdij
= Cj*CjEje
=
For other properties:
= Ci*eeCj
but, does not necessarily = ajdij because the Yj are not eigenfunctions of A unless [A,H] = 0.
= Ci*Cje
Therefore, in general, other properties are time dependent.
11.
a. The lowest energy level for a particle in a 3dimensional box is when n1 = 1, n2 = 1, and n3 = 1. The total energy (with L1 = L2 = L3) will be:
Etotal = =
Note that n = 0 is not possible. The next lowest energy level is when one of the three quantum numbers equals 2 and the other two equal 1:
n1 = 1, n2 = 1, n3 = 2
n1 = 1, n2 = 2, n3 = 1
n1 = 2, n2 = 1, n3 = 1.
Each of these three states have the same energy:
Etotal = =
Note that these three states are only degenerate if L1 = L2 = L3.
b. ¾ ¾¾ ¾¾ ¾¾ ¾¾
¾
L1 = L2 = L3 L3 ¹ L1 = L2
For L1 = L2 = L3, V = L1L2L3 = L13,
Etotal(L1) = 2e1 + e2
= +
= + =
For L3 ¹ L1 = L2, V = L1L2L3 = L12L3, L3 = V/L12
Etotal(L1) = 2e1 + e2
= +
= +
=
= =
In comparing the total energy at constant volume of the undistorted box (L1 = L2 = L3) versus the distorted box (L3 ¹ L1 = L2) it can be seen that:
£ as long as L3 ³ L1.
c. In order to minimize the total energy expression, take the derivative of the energy with respect to L1 and set it equal to zero. = 0
= 0
But since V = L1L2L3 = L12L3, then L3 = V/L12. This substitution gives:
= 0
= 0
= 0
=
24L16 = 12V2
L16 = V2 = = L14L32
L12 = L32
L3 = L1
d. Calculate energy upon distortion:
cube: V = L13, L1 = L2 = L3 = (V)
distorted: V = L12L3 = L12L1 = L13
L3 = ¹ L1 = L2 =
DE = Etotal(L1 = L2 = L3)  Etotal(L3 ¹ L1 = L2)
= 
=
=
Since V = 8Å3, V2/3 = 4Å2 = 4 x 1016 cm2 , and = 6.01 x 1027 erg cm2:
DE = 6.01 x 1027 erg cm2
DE = 6.01 x 1027 erg cm2
DE = 0.99 x 1011 erg
DE = 0.99 x 1011 erg
DE = 6.19 eV
12.
a. H = (Cartesian coordinates)
Finding andfrom the chain rule gives:
= y + y , = x + x ,
Evaluation of the "coefficients" gives the following:
y = Cosf , y =  ,
x = Sinf , and x = ,
Upon substitution of these "coefficients":
= Cosf  =  ; at fixed r.
= Sinf + = ; at fixed r.
=
= + ; at fixed r.
=
=  ; at fixed r.
+ = + + 
= ; at fixed r.
So, H = (cylindrical coordinates, fixed r)
=
The Schrödinger equation for a particle on a ring then becomes:
HY = EY
= EF
= F
The general solution to this equation is the now familiar expression:
F(f) = C1eimf + C2eimf , where m =
Application of the cyclic boundary condition, F(f) = F(f+2p), results in the quantization of the energy expression: E = where m = 0, ±1, ±2, ±3, ... It can be seen that the ±m values correspond to angular momentum of the same magnitude but opposite directions. Normalization of the wavefunction (over the region 0 to 2p) corresponding to + or  m will result in a value of for the normalization constant.
\ F(f) = eimf
¾¾ ¾¾
¾¾ ¾¾
¾¾ ¾¾
b. = 6.06 x 1028 erg cm2
=
= 3.09 x 1012 erg
DE = (22  12) 3.09 x 1012 erg = 9.27 x 1012 erg
but DE = hn = hc/l So l = hc/DE
l =
= 2.14 x 105 cm = 2.14 x 103 Å
Sources of error in this calculation include:
i. The attractive force of the carbon nuclei is not included in the Hamiltonian.
ii. The repulsive force of the other pelectrons is not included in the Hamiltonian.
iii. Benzene is not a ring.
iv. Electrons move in three dimensions not one.
13.
Y(f,0) = Cos2f.
This wavefunction needs to be expanded in terms of the eigenfunctions of the angular momentum operator, . This is most easily accomplished by an exponential expansion of the Cos function.
Y(f,0) =
=
The wavefunction is now written in terms of the eigenfunctions of the angular momentum operator, , but they need to include their normalization constant, .
Y(f,0) =
=
Once the wavefunction is written in this form (in terms of the normalized eigenfunctions of the angular momentum operator having mas eigenvalues) the probabilities for observing angular momentums of 0, 2, and 2can be easily identified as the squares of the coefficients of the corresponding eigenfunctions.
P2= =
P2= =
P0= =
14.
a. mv2 = 100 eV
v2 =
v = 0.593 x 109 cm/sec
The length of the N2 molecule is 2Å = 2 x 108 cm.
v =
t = = = 3.37 x 1017 sec
b. The normalized ground state harmonic oscillator can be written as:
Y0 = 1/4eax2/2, where a = and x = r  re
Calculating constants;
aN2 =
= 0.48966 x 1019 cm2 = 489.66 Å2
For N2: Y0(r) = 3.53333Åe(244.83Å2)(r1.09769Å)2
aN2+ =
= 0.45823 x 1019 cm2 = 458.23 Å2
For N2+: Y0(r) = 3.47522Åe(229.113Å2)(r1.11642Å)2
c. P(v=0) =
Let P(v=0) = I2 where I = integral:
I= .
(3.53333Åe(244.830Å2)(r1.09769Å)2)dr
Let C1 = 3.47522Å, C2 = 3.53333Å,
A1 = 229.113Å2, A2 = 244.830Å2,
r1 = 1.11642Å, r2 = 1.09769Å,
I = C1C2dr .
Focusing on the exponential:
A1(rr1)2A2(rr2)2 = A1(r2  2r1r + r12)  A2(r2  2r2r + r22)
= (A1 + A2)r2 + (2A1r1 + 2A2r2)r  A1r12  A2r22
Let A = A1 + A2,
B = 2A1r1 + 2A2r2,
C = C1C2, and
D = A1r12 + A2r22 .
I = Cdr
= Cdr
where A(rr0)2 + D' = Ar2 + Br  D
A(r2  2rr0 + r02) + D' = Ar2 + Br  D
such that, 2Ar0 = B
Ar02 + D' = D
and, r0 =
D' = Ar02  D = A D =  D .
I = Cdr
= CeD'dy
= CeD'
Now back substituting all of these constants:
I = C1C2exp
I = (3.47522)(3.53333)
. exp
. exp
I = 0.959
P(v=0) = I2 = 0.92, so there is a 92% probability.
15.
a. En =
DE = En+1  En
= =
=
= 4.27 x 1013 erg
DE =
l = =
= 4.66 x 104 cm
= 2150 cm1
b. Y0 = 1/4eax2/2
=
=
=
=
= 1/2eax2 ½= 0
=
=
=
= 21/2
= 21/21/2
=
Dx = (<x2>  <x>2)1/2.=
=
=
= 3.38 x 1010 cm = 0.0338Å
c. Dx =
The smaller k and m become, the larger the uncertainty in the internuclear distance becomes. Helium has a small m and small attractive force between atoms. This results in a very large Dx. This implies that it is extremely difficult for He atoms to "vibrate" with small displacement as a solid, even as absolute zero is approached.
16.
a. W =
W =
e=
= +
= +
Making this substitution results in the following three integrals:
W = +
+
= + +
a
= 2 + 2 +
a
= + +
W = + a
b. Optimize b by evaluating = 0
=
=  b
So, b= or, b= = ,
and, b = . Substituting this value of b into the expression for W gives:
W = + a
= + a
= 2pam+ 2pam
= am= am
= 0.812889106am1/3 which is in error by only 0.5284% !!!!!
17.
a. H = + kx2
f = a for a < x < a
f = 0 for x ³ a
=
= a5
= a5
+ a5
= a5
+ a5
= a5dx + a5
= a5\o(\s\up10(aa
+ a5a4k,3\o(\s\up10(a
= a5+ a5
= a5
= a5
= a5
= a5= +
b. Substituting a = binto the above expression for E we obtain:
E = +
= km
c. E = +
= + = + = 0
= and 352 = 2mka4
So, a4 = , or a =
Therefore fbest = ,
and Ebest = + = km.
d. =
= = = 0.1952 = 19.52%
18.
a. H0 y= y= Yl,m(q,f)
= 2 l(l+1) Yl,m(q,f)
E= l(l+1)
b. V = eez = eer0Cosq
E= =
= eer0
Using the given identity this becomes:
E= eer0+
eer0
The spherical harmonics are orthonormal, thus = = 0, and E= 0.
E=
= eer0
Using the given identity this becomes:
= eer0+
eer0
= 
This indicates that the only term contributing to the sum in the expression for Eis when l=1, and m=), otherwise vanishes (from orthonormality). In quantum chemistry when using orthonormal functions it is typical to write the term as a delta function, for example dlm,10 , which only has values of 1 or 0; dij = 1 when i = j and 0 when i ¹ j. This delta function when inserted into the sum then eliminates the sum by "picking out" the nonzero component. For example,
= dlm,10 , so
E= =
E= 0(0+1) = 0 and E= 1(1+1) =
Inserting these energy expressions above yields:
E= = 
c. E= E+ E+ E+ ...
= 0 + 0 
= 
a = =
=
d. a =
a = r04 12598x106cm1 = r04 1.2598Å1
aH = 0.0987 Å3
aCs = 57.57 Å3
19.
The above diagram indicates how the SALCAOs are formed from the 1s,2s, and 2p N atomic orbitals. It can be seen that there are 3sg, 3su, 1pux, 1puy, 1pgx, and 1pgy SALCAOs. The Hamiltonian matrices (Fock matrices) are given. Each of these can be diagonalized to give the following MO energies:
3sg; 15.52, 1.45, and 0.54 (hartrees)
3su; 15.52, 0.72, and 1.13
1pux; 0.58
1puy; 0.58
1pgx; 0.28
1pgy; 0.28
It can be seen that the 3sg orbitals are bonding, the 3su orbitals are antibonding, the 1pux and 1puy orbitals are bonding, and the 1pgx and 1pgy orbitals are antibonding.
20.
Using these approximate energies we can draw the following MO diagram:
This MO diagram is not an orbital correlation diagram but can be used to help generate one. The energy levels on each side (C and H2) can be "superimposed" to generate the reactant side of the orbital correlation diagram and the center CH2 levels can be used to form the product side. Ignoring the core levels this generates the following orbital correlation diagram.
21.
a. The two F p orbitals (top and bottom) generate the following reducible representation:
D3h E 2C3 3C2 sh 2S3 3sv
Gp 2 2 0 0 0 2
This reducible representation reduces to 1A1' and 1A2'' irreducible representations.
Projectors may be used to find the symmetryadapted AOs for these irreducible representations.
fa1' =
fa2'' =
b. The three trigonal F p orbitals generate the following reducible representation:
D3h E 2C3 3C2 sh 2S3 3sv
Gp 3 0 1 3 0 1
This reducible representation reduces to 1A1' and 1E' irreducible representations.
Projectors may be used to find the symmetryadapted AOs for these irreducible representations (but they are exactly analogous to the previous few problems):
fa1' =
fe' = (1/6)1/2 (2 f3 – f4 –f5)
fe' = .
c. The 3 P sp2 orbitals generate the following reducible representation:
D3h E 2C3 3C2 sh 2S3 3sv
Gsp2 3 0 1 3 0 1
This reducible representation reduces to 1A1' and 1E' irreducible representations. Again, projectors may be used to find the symmetryadapted AOs for these irreducible representations:
fa1' =
fe' =
fe' = .
The leftover P pz orbital generate the following irreducible representation:
D3h E 2C3 3C2 sh 2S3 3sv
Gpz 1 1 1 1 1 1
This irreducible representation is A2''
fa2'' = f9.
Drawing an energy level diagram using these SALCAOs would result in the following:
22.
a. For nondegenerate point groups, one can simply multiply the representations (since only one representation will be obtained):
a1 Ä b1 = b1
Constructing a "box" in this case is unnecessary since it would only contain a single row. Two unpaired electrons will result in a singlet (S=0, MS=0), and three triplets (S=1, MS=1; S=1, MS=0; S=1, MS=1). The states will be: 3B1(MS=1), 3B1(MS=0), 3B1(MS=1), and 1B1(MS=0).
b. Remember that when coupling nonequivalent linear molecule angular momenta, one simple adds the individual Lz values and vector couples the electron spin. So, in this case (1pu12pu1), we have ML values of 1+1, 11, 1+1, and 11 (2, 0, 0, and 2). The term symbol D is used to denote the spatially doubly degenerate level (ML=±2) and there are two distinct spatially nondegenerate levels denoted by the term symbol S (ML=0) Again, two unpaired electrons will result in a singlet (S=0, MS=0), and three triplets (S=1, MS=1;S=1, MS=0;S=1, MS=1). The states generated are then:
1D (ML=2); one state (MS=0),
1D (ML=2); one state (MS=0),
3D (ML=2); three states (MS=1,0, and 1),
3D (ML=2); three states (MS=1,0, and 1),
1S (ML=0); one state (MS=0),
1S (ML=0); one state (MS=0),
3S (ML=0); three states (MS=1,0, and 1), and
3S (ML=0); three states (MS=1,0, and 1).
c. Constructing the "box" for two equivalent p electrons one obtains:
ML
MS
2
1
0
1
p1ap1a
0
p1ap1b
p1ap1b,
p1ap1b
From this "box" one obtains six states:
1D (ML=2); one state (MS=0),
1D (ML=2); one state (MS=0),
1S (ML=0); one state (MS=0),
3S (ML=0); three states (MS=1,0, and 1).
d. It is not necessary to construct a "box" when coupling nonequivalent angular momenta since vector coupling results in a range from the sum of the two individual angular momenta to the absolute value of their difference. In this case, 3d14d1, L=4, 3, 2, 1, 0, and S=1,0. The term symbols are: 3G, 1G, 3F, 1F, 3D, 1D, 3P, 1P, 3S, and 1S. The L and S angular momenta can be vector coupled to produce further splitting into levels:
J = L + S ... L  S.
Denoting J as a term symbol subscript one can identify all the levels and subsequent (2J + 1) states:
3G5 (11 states),
3G4 (9 states),
3G3 (7 states),
1G4 (9 states),
3F4 (9 states),
3F3 (7 states),
3F2 (5 states),
1F3 (7 states),
3D3 (7 states),
3D2 (5 states),
3D1 (3 states),
1D2 (5 states),
3P2 (5 states),
3P1 (3 states),
3P0 (1 state),
1P1 (3 states),
3S1 (3 states), and
1S0 (1 state).
e. Construction of a "box" for the two equivalent d electrons generates (note the "box" has been turned side ways for convenience):
MS
ML
1
0
4
d2ad2b
3
d2ad1a
d2ad1b, d2bd1a
2
d2ad0a
d2ad0b, d2bd0a, d1ad1b
1
d1ad0a, d2ad1a
d1ad0b, d1bd0a, d2ad1b, d2bd1a
0
d2ad2a, d1ad1a
d2ad2b, d2bd2a, d1ad1b, d1bd1a, d0ad0b
The term symbols are: 1G, 3F, 1D, 3P, and 1S. The L and S angular momenta can be vector coupled to produce further splitting into levels:
1G4 (9 states),
3F4 (9 states),
3F3 (7 states),
3F2 (5 states),
1D2 (5 states),
3P2 (5 states),
3P1 (3 states),
3P0 (1 state), and
1S0 (1 state).
23.
a. Once the spatial symmetry has been determined by multiplication of the irreducible representations, the spin coupling gives the result:
b. There are three states here :
1.) 3a1a1b1a,
2.) , and
3.) 3a1b1b1b
c. 3a1a3a1b
24.
a. All the Slater determinants have in common the 1sa1sb2sa2sb "core" and hence this component will not be written out explicitly for each case.
3P(ML=1,MS=1) = p1ap0a
= a(pz)a
=
3P(ML=0,MS=1) = p1ap1a
= aa
=
=
=
= ipxapya
3P(ML=1,MS=1) = p1ap0a
= a(pz)a
=
As you can see, the symmetries of each of these states cannot be labeled with a single irreducible representation of the C2v point group. For example, pxapza is xz (B1) and pyapza is yz (B2) and hence the 3P(ML=1,MS=1) state is a combination of B1 and B2 symmetries. But, the three 3P(ML,MS=1) functions are degenerate for the C atom and any combination of these three functions would also be degenerate. Therefore, we can choose new combinations that can be labeled with "pure" C2v point group labels.
3P(xz,MS=1) = pxapza
= = 3B1
3P(yx,MS=1) = pyapxa
= = 3A2
3P(yz,MS=1) = pyapza
= = 3B2
Now, we can do likewise for the five degenerate 1D states:
1D(ML=2,MS=0) = p1ap1b
= ab
=
1D(ML=2,MS=0) = p1ap1b
= ab
=
1D(ML=1,MS=0) =
=
=
1D(ML=1,MS=0) =
=
=
1D(ML=0,MS=0) =
=
+ ab)
=
+
+ )
= )
Analogous to the three 3P states, we can also choose combinations of the five degenerate 1D states which can be labeled with "pure" C2v point group labels:
1D(xxyy,MS=0) = pxapxb  pyapyb
= = 1A1
1D(yx,MS=0) = pxapyb + pyapxb
= = 1A2
1D(zx,MS=0) = pzapxb  pzbpxa
= = 1B1
1D(zy,MS=0) = pzapyb  pzbpya
= = 1B2
1D(2zz+xx+yy,MS=0) = )
= 1D(ML=0,MS=0) = 1A1
The only state left is the 1S:
1S(ML=0,MS=0) =
=
 ab)
=

 )
= )
Each of the components of this state are A1 and hence this state has
A1 symmetry.
b. Forming symmetryadapted AOs from the C and H atomic orbitals would generate the following:
The bonding, nonbonding, and antibonding orbitals of CH2 can be illustrated in the following manner:
c.
d.  e. It is necessary to determine how the wavefunctions found in part a. correlate with states of the CH2 molecule:
3P(xz,MS=1); 3B1 = sg2s2pxpz ¾¾® s2n2pps*
3P(yx,MS=1); 3A2 = sg2s2pxpy ¾¾® s2n2pps
3P(yz,MS=1); 3B2 = sg2s2pypz ¾¾® s2n2ss*
1D(xxyy,MS=0); 1A1 ¾¾® s2n2pp2  s2n2s2
1D(yx,MS=0); 1A2 ¾¾® s2n2spp
1D(zx,MS=0); 1B1 ¾¾® s2n2s*pp
1D(zy,MS=0); 1B2 ¾¾® s2n2s*s
1D(2zz+xx+yy,MS=0); 1A1 ¾¾® 2s2n2s*2 + s2n2pp2 + s2n2s2
Note, the C + H2 state to which the lowest 1A1 (s2n2s2) CH2 state decomposes would be sg2s2py2. This state (sg2s2py2) cannot be obtained by a simple combination of the 1D states. In order to obtain pure sg2s2py2 it is necessary to combine 1S with 1D. For example,
sg2s2py2 =  .
This indicates that a configuration correlation diagram must be drawn with a barrier near the 1D asymptote to represent the fact that 1A1 CH2 correlates with a mixture of 1D and 1S carbon plus hydrogen. The C + H2 state to which the lowest 3B1 (s2ns2pp) CH2 state decomposes would be sg2spy2px.
f. If you follow the 3B1 component of the C(3P) + H2 (since it leads to the groundstate products) to 3B1 CH2 you must go over an approximately 20 Kcal/mole barrier. Of course this path produces 3B1 CH2 product. Distortions away from C2v symmetry, for example to Cs symmetry, would make the a1 and b2 orbitals identical in symmetry (a'). The b1 orbitals would maintain their different symmetry going to a'' symmetry. Thus 3B1 and 3A2 (both 3A'' in Cs symmetry and odd under reflection through the molecular plane) can mix. The system could thus follow the 3A2 component of the C(3P) + H2 surface to the place (marked with a circle on the CCD) where it crosses the 3B1 surface upon which it then moves and continues to products. As a result, the barrier would be lowered.
You can estimate when the barrier occurs (late or early) using thermodynamic information for the reaction (i.e. slopes and asymptotic energies). For example, an early barrier would be obtained for a reaction with the characteristics:
and a late barrier would be obtained for a reaction with the characteristics:
This relation between reaction endothermicity or exothermicity and the character of the transition state is known as the Hammond postulate. Note that the C(3P1) + H2 > CH2 reaction of interest here has an early barrier.
g. The reaction C(1D) + H2 > CH2 (1A1) should have no symmetry barrier (this can be recognized by following the 1A1 (C(1D) + H2) reactants down to the 1A1 (CH2) products).
25.
This problem in many respects is analogous to problem 24.
The 3B1 surface certainly requires a two configuration CI wavefunction; the s2s2npx (p2py2spx) and the s2n2pxs* (p2s2pxpz). The 1A1 surface could use the s2s2n2 (p2s2py2) only but once again there is no combination of 1D determinants which gives purely this configuration (p2s2py2). Thus mixing of both 1D and 1S determinants are necessary to yield the required p2s2py2 configuration. Hence even the 1A1 surface would require a multiconfigurational wavefunction for adequate description.
Configuration correlation diagram for the reaction C2H2 + C > C3H2.
26.
a. CCl4 is tetrahedral and therefore is a spherical top. CHCl3 has C3v symmetry and therefore is a symmetric top. CH2Cl2 has C2v symmetry and therefore is an asymmetric top.
b. CCl4 has such high symmetry that it will not exhibit pure rotational spectra because it has no permanent dipole moment. CHCl3 and CH2Cl2 will both exhibit pure rotation spectra.
27.
NH3 is a symmetric top (oblate). Use the given energy expression,
E = (A  B) K2 + B J(J + 1),
A = 6.20 cm1, B = 9.44 cm1, selection rules DJ = ±1, and the fact that lies along the figure axis such that DK = 0, to give:
DE = 2B (J + 1) = 2B, 4B, and 6B (J = 0, 1, and 2).
So, lines are at 18.88 cm1, 37.76 cm1, and 56.64 cm1.
28.
To convert between cm1 and energy, multiply by hc = (6.62618x1034J sec)(2.997925x1010cm sec1) = 1.9865x1023 J cm.
Let all quantities in cm1 be designated with a bar,
e.g. = 1.78 cm1.
a. hc=
Re = ,
m = = x 1.66056x1027 kg
= 1.0824x1026 kg.
hc= hc(1.78 cm1) = 3.5359x1023 J
Re =
Re = 1.205x1010 m = 1.205 Å
De = , = = = 6.35x106 cm1
wexe = , = = = 13.30 cm1.
D= D + , =  +
= 66782.2  +
= 65843.0 cm1 = 8.16 eV.
ae = +
= +
= + = 0.0175 cm1.
B0 = Be  ae(1/2) , =  = 1.78  0.0175/2
= 1.77 cm1
B1 = Be  ae(3/2) , =  = 1.78  0.0175(1.5)
= 1.75 cm1
b. The molecule has a dipole moment and so it should have a pure rotational spectrum. In addition, the dipole moment should change with R and so it should have a vibrationrotation spectrum.
The first three lines correspond to J = 1 ® 0, J = 2 ® 1, J = 3 ® 2
E = we(v + 1/2)  wexe(v + 1/2)2 + BvJ(J + 1)  DeJ2(J + 1)2
DE = we  2wexe  B0J(J + 1) + B1J(J  1)  4DeJ3
=  2 J(J + 1) + J(J  1)  4J3
= 1885  2(13.3)  1.77J(J + 1) + 1.75J(J  1)  4(6.35x106)J3
= 1858.4  1.77J(J + 1) + 1.75J(J  1)  2.54x105J3
= 1854.9 cm1
= 1851.3 cm1
= 1847.7 cm1
29.
The C2H2Cl2 molecule has a sh plane of symmetry (plane of molecule), a C2 axis (^ to the molecular plane), and inversion symmetry, this results in C2h symmetry. Using C2h symmetry, the modes can be labeled as follows: n1, n2, n3, n4, and n5 are ag, n6 and n7 are au, n8 is bg, and n9, n10, n11, and n12 are bu.
30.
Molecule I Molecule II
RCH = 1.121 Å RCH = 1.076 Å
ÐHCH = 104° ÐHCH = 136°
yH = R Sin (q/2) = ±0.8834 yH = ±0.9976
zH = R Cos (q/2) = 0.6902 zH = 0.4031
Center of Mass(COM):
clearly, X = Y = 0,
Z = = 0.0986 Z = 0.0576
a. Ixx =  M(Y2 + Z2)
Ixy =  MXY
Ixx = 2(1.121)2  14(0.0986)2 Ixx = 2(1.076)2  14(0.0576)2
= 2.377 = 2.269
Iyy = 2(0.6902)2  14(0.0986)2 Iyy = 2(0.4031)2  14(0.0576)2
= 0.8167 = 0.2786
Izz = 2(0.8834)2 Izz = 2(0.9976)2
= 1.561 = 1.990
Ixz = Iyz = Ixy = 0
b. Since the moment of inertia tensor is already diagonal, the principal moments of inertia have already been determined to be
(Ia < Ib < Ic):
Iyy < Izz < Ixx Iyy < Izz < Ixx
0.8167 < 1.561 < 2.377 0.2786 < 1.990 < 2.269
Using the formula: A = = X
A = cm1
similarly, B = cm1, and C = cm1.
So,
Molecule I Molecule II
y Þ A = 20.62 y Þ A = 60.45
z Þ B = 10.79 z Þ B = 8.46
x Þ C = 7.08 x Þ C = 7.42
c. Averaging B + C:
B = (B + C)/2 = 8.94 B = (B + C)/2 = 7.94
A  B = 11.68 A  B = 52.51
Using the prolate top formula:
E = (A  B) K2 + B J(J + 1),
Molecule I Molecule II
E = 11.68K2 + 8.94J(J + 1) E = 52.51K2 + 7.94J(J + 1)
Levels: J = 0,1,2,... and K = 0,1, ... J
For a given level defined by J and K, there are MJ degeneracies given by: (2J + 1) x
d.
Molecule I Molecule II
e. Assume molecule I is CH2 and molecule II is CH2. Then,
DE = EJj(CH2)  EJi(CH2), where:
E(CH2) = 52.51K2 + 7.94J(J + 1), and E(CH2) = 11.68K2 + 8.94J(J + 1)
For Rbranches: Jj = Ji + 1, DK = 0:
DER = EJj(CH2)  EJi(CH2)
= 7.94(Ji + 1)(Ji + 1 + 1)  8.94Ji(Ji + 1)
= (Ji + 1){7.94(Ji + 1 + 1)  8.94Ji}
= (Ji + 1){(7.94 8.94)Ji + 2(7.94)}
= (Ji + 1){Ji + 15.88}
For Pbranches: Jj = Ji  1, DK = 0:
DEP = EJj(CH2)  EJi(CH2)
= 7.94(Ji  1)(Ji  1 + 1)  8.94Ji(Ji + 1)
= Ji{7.94(Ji  1)  8.94(Ji + 1)}
= Ji{(7.94 8.94)Ji  7.94  8.94}
= Ji{Ji  16.88}
This indicates that the R branch lines occur at energies which grow closer and closer together as J increases (since the 15.88  Ji term will cancel). The P branch lines occur at energies which lie more and more negative (i.e. to the left of the origin). So, you can predict that if molecule I is CH2 and molecule II is CH2 then the Rbranch has a band head and the Pbranch does not. This is observed, therefore our assumption was correct: molecule I is CH2 and molecule II is CH2.
f. The band head occurs when = 0.
= [(Ji + 1){Ji + 15.88}] = 0
= = 0
= 2Ji + 14.88 = 0
\ Ji = 7.44, so J = 7 or 8.
At J = 7.44:
DER = (J + 1){J + 15.88}
DER = (7.44 + 1){7.44 + 15.88} = (8.44)(8.44) = 71.2 cm1 above the origin.
31.
a.
D6h
E
2C6
2C3
C2
3C2'
3C2"
i
2S3
2S6
sh
3sd
3sv
A1g
1
1
1
1
1
1
1
1
1
1
1
1
x2+y2,z2
A2g
1
1
1
1
1
1
1
1
1
1
1
1
Rz
B1g
1
1
1
1
1
1
1
1
1
1
1
1
B2g
1
1
1
1
1
1
1
1
1
1
1
1
E1g
2
1
1
2
0
0
2
1
1
2
0
0
Rx,Ry
(xz,yz)
E2g
2
1
1
2
0
0
2
1
1
2
0
0
(x2y2,xy)
A1u
1
1
1
1
1
1
1
1
1
1
1
1
A2u
1
1
1
1
1
1
1
1
1
1
1
1
z
B1u
1
1
1
1
1
1
1
1
1
1
1
1
B2u
1
1
1
1
1
1
1
1
1
1
1
1
E1u
2
1
1
2
0
0
2
1
1
2
0
0
(x,y)
E2u
2
1
1
2
0
0
2
1
1
2
0
0
GCH
6
0
0
0
0
2
0
0
0
6
2
0
b. The number of irreducible representations may be found by using the following formula:
nirrep = ,
where g = the order of the point group (24 for D6h).
nA1g =
= {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1)
+(3)(0)(1)+(3)(2)(1)+(1)(0)(1)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(1)+(3)(2)(1)+(3)(0)(1)}
= 1
nA2g = {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1)
+(3)(0)(1)+(3)(2)(1)+(1)(0)(1)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(1)+(3)(2)(1)+(3)(0)(1)}
= 0
nB1g = {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1)
+(3)(0)(1)+(3)(2)(1)+(1)(0)(1)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(1)+(3)(2)(1)+(3)(0)(1)}
= 0
nB2g = {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1)
+(3)(0)(1)+(3)(2)(1)+(1)(0)(1)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(1)+(3)(2)(1)+(3)(0)(1)}
= 0
nE1g = {(1)(6)(2)+(2)(0)(1)+(2)(0)(1)+(1)(0)(2)
+(3)(0)(0)+(3)(2)(0)+(1)(0)(2)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(2)+(3)(2)(0)+(3)(0)(0)}
= 0
nE2g = {(1)(6)(2)+(2)(0)(1)+(2)(0)(1)+(1)(0)(2)
+(3)(0)(0)+(3)(2)(0)+(1)(0)(2)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(2)+(3)(2)(0)+(3)(0)(0)}
= 1
nA1u = {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1)
+(3)(0)(1)+(3)(2)(1)+(1)(0)(1)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(1)+(3)(2)(1)+(3)(0)(1)}
= 0
nA2u = {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1)
+(3)(0)(1)+(3)(2)(1)+(1)(0)(1)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(1)+(3)(2)(1)+(3)(0)(1)}
= 0
nB1u = {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1)
+(3)(0)(1)+(3)(2)(1)+(1)(0)(1)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(1)+(3)(2)(1)+(3)(0)(1)}
= 0
nB2u = {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1)
+(3)(0)(1)+(3)(2)(1)+(1)(0)(1)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(1)+(3)(2)(1)+(3)(0)(1)}
= 1
nE1u = {(1)(6)(2)+(2)(0)(1)+(2)(0)(1)+(1)(0)(2)
+(3)(0)(0)+(3)(2)(0)+(1)(0)(2)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(2)+(3)(2)(0)+(3)(0)(0)}
= 1
nE2u = {(1)(6)(2)+(2)(0)(1)+(2)(0)(1)+(1)(0)(2)
+(3)(0)(0)+(3)(2)(0)+(1)(0)(2)+(2)(0)(1)
+(2)(0)(1)+(1)(6)(2)+(3)(2)(0)+(3)(0)(0)}
= 0
We see that GCH = A1gÅE2gÅB2uÅE1u
c. x and y Þ E1u , z Þ A2u , so, the ground state A1g level can be excited to the degenerate E1u level by coupling through the x or y transition dipoles. Therefore E1u is infrared active and ^ polarized.
d. (x2 + y2, z2) Þ A1g, (xz, yz) Þ E1g, (x2  y2, xy) Þ E2g ,so, the ground state A1g level can be excited to the degenerate E2g level by coupling through the x2  y2 or xy transitions or be excited to the degenerate A1g level by coupling through the xz or yz transitions. Therefore A1g and E2g are Raman active..
e. The B2u mode is not IR or Raman active.
32.
a. Evaluate the zcomponent of mfi:
mfi = <2pze r Cosq1s>, where y1s = e , and y2pz = r Cosq e .
mfi = <r Cosq e e r Cosqe >
= <r Cosq e e r Cosqe >
= Cos2q
= 2p
= 2p Cos3q\s\up15(p0
= 2p
= = = 0.7449
b. Examine the symmetry of the integrands for <2pz e x 1s> and <2pz e y 1s>. Consider reflection in the xy plane:
Function
Symmetry
2pz
1
x
+1
1s
+1
y
+1
Under this operation, the integrand of <2pz e x 1s> is (1)(1)(1) = 1 (it is antisymmetric) and hence <2pz e x 1s> = 0.
Similarly, under this operation the integrand of <2pz e y 1s> is
(1)(1)(1) = 1 (it is also antisymmetric) and hence <2pz e y 1s> = 0.
c. tR = ,
Ei = E2pz = Z2
Ef = E1s = Z2
Ei  Ef = Z2
Making the substitutions for Ei  Ef and mfi in the expression for tR we obtain:
tR = ,
= ,
= ,
Inserting e2 = we obtain:
tR = =
= 25.6289
= 25,6289 x
= 1.595x109 sec x
So, for example:
Atom
tR
H
1.595 ns
He+
99.7 ps
Li+2
19.7 ps
Be+3
6.23 ps
Ne+9
159 fs
33.
a. H = H0 + lH'(t), H'(t) = Vq(t), H0jk = Ekjk, wk = Ek/
i= Hy
let y(r,t) = iand insert into the Schrödinger equation:
ieiwjtjj = ijj
eiwjtjj = 0
eiwjt = 0
im eiwmt = eiwjt
So,
m = ei(wjm)t
Going back a few equations and multiplying from the left by jk instead of jm we obtain:
eiwjt = 0
ik eiwkt = eiwjt
So,
k = ei(wjk)t
Now, let:
cm = cm(0) + cm(1)l + cm(2)l2 + ...
ck = ck(0) + ck(1)l + ck(2)l2 + ...
and substituting into above we obtain:
m(0) + m(1)l + m(2)l2 + ... = lH'mj ei(wjm)t
first order:
m(0) = 0 Þ cm(0) = 1
second order:
m(1) =
(n+1)st order:
m(n) =
Similarly:
first order:
k(0) = 0 Þ ck¹m(0) = 0
second order:
k(1) =
(n+1)st order:
k(n) =
So,
m(1) = cm(0) H'mm ei(wmm)t = H'mm
cm(1)(t) = =
and similarly,
k(1) = cm(0) H'km ei(wmk)t = H'km ei(wmk)t
ck(1)(t) = Vkm =
m(2) =
m(2) = H'mj ei(wjm)t + H'mm
cm(2) = ei(wjm)t' 
= 
= 
= + t 
Similarly,
k(2) =
= H'kj ei(wjk)t +
H'km ei(wmk)t
ck(2)(t) = ei(wjk)t'
 ei(wmk)t'
=
 h,ei(wmk
=
+ h,ei(wmk
=
+
So, the overall amplitudes cm, and ck, to second order are:
cm(t) = 1 + + t +

ck(t) = +
+ ei(wmk)t +
b. The perturbation equations still hold:
m(n) = ; k(n) =
So, cm(0) = 1 and ck(0) = 0
m(1) = H'mm
cm(1) = Vmm =
k(1) = H'km ei(wmk)t
ck(1) = Vkm =
=
m(2) = ei(wmj+h)t Vmj eht ei(wjm)t +
Vmm eht
cm(2) = 
= e2ht  e2ht
k(2) = ei(wmj+h)t H'kj ei(wjk)t +
H'km ei(wmk)t
ck(2) = 
= 
Therefore, to second order:
cm(t) = 1 + + e2ht
ck(t) =
+
c. In part a. the c(2)(t) grow linearly with time (for Vmm = 0) while in part b. they remain finite for h > 0. The result in part a. is due to the sudden turning on of the field.
d. ck(t)2 =
=
=
ck(t)2 =
Now, look at the limit as h ® 0+:
ck(t)2 ¹ 0 when Em = Ek
limhÆ0+a d(EmEk)
So, the final result is the 2nd order golden rule expression:
ck(t)2 = d(EmEk)limhÆ0+
34.
a. Tnm »
evaluating <1sV2s> (using only the radial portions of the 1s and 2s wavefunctions since the spherical harmonics will integrate to unity) where V = (e2/r), the change in Coulomb potential when tritium becomes He:
<1sV2s> = e e r2dr
<1sV2s> =
=
<1sV2s> =
<1sV2s> = =
Now,
En = , E1s = , E2s = , E2s  E1s =
So,
Tnm = = = = 0.312 (for Z = 1)
b. jm(r) = j1s = 2e Y00
The orthogonality of the spherical harmonics results in only sstates having nonzero values for Anm. We can then drop the Y00 (integrating this term will only result in unity) in determining the value of A1s,2s.
yn(r) = y2s = e
Remember for j1s Z = 1 and for y2s Z = 2
Anm = e e r2dr
Anm = e r2dr
Anm =
We obtain:
Anm =
Anm =
Anm =
Anm = 2
The transition probability is the square of this amplitude:
Tnm = = = 0.25 (for Z = 1).
The difference in these two results (parts a. and b.) will become negligible at large values of Z when the perturbation becomes less significant than in the case of Z = 1.
35.
is along Z (lab fixed), and is along z (the CI molecule fixed bond). The angle between Z and z is b:
.= emCosb = emD
So,
I = <D.D> = Sinbdbdgda
= emSinbdbdgda.
Now use:
DD= *,
to obtain:
I = em*Sinbdbdgda.
Now use:
Sinbdbdgda = dJjdMmdKn,
to obtain:
I = em*dJjdMmdKn
= em<J'M'10JM><JKJ'K'10>.
We use:
<JKJ'K'10> =
and,
<J'M'10JM> =
to give:
I = em
= em8p2(i)(J'1+M+J'1+K)
= em8p2(i)(M+K)
The 3J symbols vanish unless: K' + 0 = K and M' + 0 = M.
So,
I = em8p2(i)(M+K)dM'MdK'K.
b. and vanish unless J' = J + 1, J, J  1
\ DJ = ±1, 0
The K quantum number can not change because the dipole moment lies along the molecule's C3 axis and the light's electric field thus can exert no torque that twists the molecule about this axis. As a result, the light can not induce transitions that excite the molecule's spinning motion about this axis.
36.
a. B atom: 1s22s22p1, 2P ground state L = 1, S = , gives a degeneracy ((2L+1)(2S+1)) of 6.
O atom: 1s22s22p4, 3P ground state L = 1, S = 1, gives a degeneracy ((2L+1)(2S+1)) of 9.
The total number of states formed is then (6)(9) = 54.
b. We need only consider the p orbitals to find the low lying molecular states:
Which, in reality look like this:
This is the correct ordering to give a 2S+ ground state. The only lowlying electron configurations are 1p35s2 or 1p45s1. These lead to 2P and 2S+ states, respectively.
c. The bond orders in both states are 2.5.
d. The 2S is + but g/u symmetry cannot be specified since this is a heteronuclear molecule.
e. Only one excited state, the 2P, is spinallowed to radiate to the 2S+. Consider symmetries of transition moment operators that arise in the electric dipole contributions to the transition rate z ® S+, x,y ® P, \ the 2P ® 2S+ is electric dipole allowed via a perpendicular band.
f. Since ionization will remove a bonding electron, the BO+ bond is weaker than the BO bond.
g. The ground state BO+ is 1S+ corresponding to a 1p4 electron configuration. An electron configuration of 1p3 5s1 leads to a 3P and a 1P state. The 3P will be lower in energy. A 1p2 5s2 configuration will lead to higher lying states of 3S, 1D, and 1S+.
h. There should be 3 bands corresponding to formation of BO+ in the 1S+, 3P, and 1P states. Since each of these involves removing a bonding electron, the FranckConden integrals will be appreciable for several vibrational levels, and thus a vibrational progression should be observed.
37.
a. The bending (p) vibration is degenerate.
b. HCºN
Ý
bending fundamental
c. HCºN
Ý
stretching fundamental
d. CH stretch (n3 in figure) is s, CN stretch is s, and HCN (n2 in figure) bend is p.
e. Under z (s) light the CN stretch and the CH stretch can be excited, since y0 = s, y1 = s and z = s provides coupling.
f. Under x,y (p) light the HCN bend can be excited, since y0 = s, y1 = p and x,y = p provides coupling.
g. The bending vibration is active under (x,y) perpendicular polarized light. DJ = 0, ±1 are the selection rules for ^ transitions. The CH stretching vibration is active under (z)  polarized light. DJ = ±1 are the selection rules for  transitions.
38.
F fi = ei fj = h fi + fi
Let the closed shell Fock potential be written as:
Vij = , and the 1e component as:
hij = fi  Ñ2  fj , and the delta as:
dij = , so that: hij + Vij = dijei.
using: fi = , fj = , and fk = , and transforming from the MO to AO basis we obtain:
Vij = CmiCgkCnjCkk
=
= Vmn where,
Vmn = Pgk, and Pgk = ,
hij = hmn , where
hmn = cm  Ñ2  cn , and
dij = = .
So, hij + Vij = dijej becomes:
hmn + Vmn = ej ,
ej  hmn  Vmn = 0 for all i,j
CmiCnj = 0 for all i,j
Therefore,
Cnj = 0
This is FC = SCE in the AO basis.
39.
The Slater Condon rule for zero (spin orbital) difference with N electrons in N spin orbitals is:
E = = +
= +
= +
If all orbitals are doubly occupied and we carry out the spin integration we obtain:
E = 2+ ,
where i and j now refer to orbitals (not spinorbitals).
40.
If the occupied orbitals obey Ffk = ekfk , then the expression for E in problem 39 can be rewritten as.
E = +
We recognize the closed shell Fock operator expression and rewrite this as:
E = + =
41.
I will use the QMIC software to do this problem. Lets just start from the beginning. Get the starting "guess" MO coefficients on disk. Using the program MOCOEFS it asks us for the first and second MO vectors. We input 1, 0 for the first mo (this means that the first MO is 1.0 times the He 1s orbital plus 0.0 times the H 1s orbital; this bonding MO is more likely to be heavily weighted on the atom having the higher nuclear charge) and 0, 1 for the second. Our beginning LCAOMO array looks like: and is placed on disk in a file we choose to call "mocoefs.dat". We also put the AO integrals on disk using the program RW_INTS. It asks for the unique one and two electron integrals and places a canonical list of these on disk in a file we choose to call "ao_integrals.dat". At this point it is useful for us to step back and look at the set of equations which we wish to solve: FC = SCE. The QMIC software does not provide us with a socalled generalized eigenvalue solver (one that contains an overlap matrix; or metric), so in order to use the diagonalization program that is provided we must transform this equation (FC = SCE) to one that looks like (F'C' = C'E). We do that in the following manner:
Since S is symmetric and positive definite we can find an Ssuch that SS= 1, SS = S, etc.
rewrite FC = SCE by inserting unity between FC and multiplying the whole equation on the left by S. This gives:
SFSSC = SSCE = SCE.
Letting: F' = SFS
C' = SC, and inserting these expressions above give:
F'C' = C'E
Note, that to get the next iteration’s MO coefficients we must calculate C from C':
C' = SC, so, multiplying through on the left by Sgives:
SC' = SSC = C
This will be the method we will use to solve our fock equations.
Find Sby using the program FUNCT_MAT (this program generates a function of a matrix). This program will ask for the elements of the S array and write to disk a file (name of your choice ... a good name might be "shalf") containing the Sarray. Now we are ready to begin the iterative Fock procedure.
a. Calculate the Fock matrix, F, using program FOCK which reads in the MO coefficients from "mocoefs.dat" and the integrals from "ao_integrals.dat" and writes the resulting Fock matrix to a user specified file (a good filename to use might be something like "fock1").
b. Calculate F' = SFSusing the program UTMATU which reads in F and Sfrom files on the disk and writes F' to a user specified file (a good filename to use might be something like "fock1p"). Diagonalize F' using the program DIAG. This program reads in the matrix to be diagonalized from a user specified filename and writes the resulting eigenvectors to disk using a user specified filename (a good filename to use might be something like "coef1p"). You may wish to choose the option to write the eigenvalues (Fock orbital energies) to disk in order to use them at a later time in program FENERGY. Calculate C by using. C = SC'. This is accomplished by using the program MATXMAT which reads in two matrices to be multiplied from user specified files and writes the product to disk using a user specified filename (a good filename to use might be something like "mocoefs.dat").
c. The QMIC program FENERGY calculates the total energy:
2<khk> + 2<klkl>  <kllk> + , and
ek + <khk> + .
This is the conclusion of one iteration of the Fock procedure ... you may continue by going back to part a. and proceeding onward.
d. and e. Results for the successful convergence of this system using the supplied QMIC software are as follows (this data is provided to give the student assurance that they are on the right track; alternatively one could switch to the QMIC program SCF and allow that program to iteratively converge the Fock equations):
The oneelectron AO integrals:
The twoelectron AO integrals:
1 1 1 1 1.054700
2 1 1 1 0.4744000
2 1 2 1 0.5664000
2 2 1 1 0.2469000
2 2 2 1 0.3504000
2 2 2 2 0.6250000
The "initial" MOAO coefficients:
AO overlap matrix (S):
S
**************
ITERATION 1
**************
The charge bond order matrix:
The Fock matrix (F):
S F S
The eigenvalues of this matrix (Fock orbital energies) are:
Their corresponding eigenvectors (C' = S * C) are:
The "new" MOAO coefficients (C = S * C'):
The oneelectron MO integrals:
The twoelectron MO integrals:
1 1 1 1 0.9779331
2 1 1 1 0.1924623
2 1 2 1 0.5972075
2 2 1 1 0.1170838
2 2 2 1 0.0007945194
2 2 2 2 0.6157323
The closed shell Fock energy from formula:
2<khk> + 2<klkl>  <kllk> + = 2.84219933
from formula:
ek + <khk> + = 2.80060530
the difference is: 0.04159403
**************
ITERATION 2
**************
The charge bond order matrix:
The Fock matrix:
S F S
The eigenvalues of this matrix (Fock orbital energies) are:
Their corresponding eigenvectors (C' = S * C) are:
The "new" MOAO coefficients (C = S * C'):
The oneelectron MO integrals:
The twoelectron MO integrals:
1 1 1 1 0.9626070
2 1 1 1 0.1949828
2 1 2 1 0.6048143
2 2 1 1 0.1246907
2 2 2 1 0.003694540
2 2 2 2 0.6158437
The closed shell Fock energy from formula:
2<khk> + 2<klkl>  <kllk> + = 2.84349298
from formula:
ek + <khk> + = 2.83573675
the difference is: 0.00775623
**************
ITERATION 3
**************
The charge bond order matrix:
The Fock matrix:
S F S
The eigenvalues of this matrix (Fock orbital energies) are:
Their corresponding eigenvectors (C' = S * C) are:
The "new" MOAO coefficients (C = S * C'):
The oneelectron MO integrals:
The twoelectron MO integrals:
1 1 1 1 0.9600707
2 1 1 1 0.1953255
2 1 2 1 0.6060572
2 2 1 1 0.1259332
2 2 2 1 0.004475587
2 2 2 2 0.6158972
The closed shell Fock energy from formula:
2<khk> + 2<klkl>  <kllk> + = 2.84353018
from formula:
ek + <khk> + = 2.84225941
the difference is: 0.00127077
**************
ITERATION 4
**************
The charge bond order matrix:
The Fock matrix:
S F S
The eigenvalues of this matrix (Fock orbital energies) are:
Their corresponding eigenvectors (C' = S * C) are:
The "new" MOAO coefficients (C = S * C'):
The oneelectron MO integrals:
The twoelectron MO integrals:
1 1 1 1 0.9596615
2 1 1 1 0.1953781
2 1 2 1 0.6062557
2 2 1 1 0.1261321
2 2 2 1 0.004601604
2 2 2 2 0.6159065
The closed shell Fock energy from formula:
2<khk> + 2<klkl>  <kllk> + = 2.84352922
from formula:
ek + <khk> + = 2.84332418
the difference is: 0.00020504
**************
ITERATION 5
**************
The charge bond order matrix:
The Fock matrix:
S F S
The eigenvalues of this matrix (Fock orbital energies) are:
Their corresponding eigenvectors (C' = S * C) are:
The "new" MOAO coefficients (C = S * C'):
The oneelectron MO integrals:
The twoelectron MO integrals:
1 1 1 1 0.9595956
2 1 1 1 0.1953862
2 1 2 1 0.6062872
2 2 1 1 0.1261639
2 2 2 1 0.004621811
2 2 2 2 0.6159078
The closed shell Fock energy from formula:
2<khk> + 2<klkl>  <kllk> + = 2.84352779
from formula:
ek + <khk> + = 2.84349489
the difference is: 0.00003290
**************
ITERATION 6
**************
The charge bond order matrix:
The Fock matrix:
S F S
The eigenvalues of this matrix (Fock orbital energies) are:
Their corresponding eigenvectors (C' = S * C) are:
The "new" MOAO coefficients (C = S * C'):
The oneelectron MO integrals:
The twoelectron MO integrals:
1 1 1 1 0.9595859
2 1 1 1 0.1953878
2 1 2 1 0.6062925
2 2 1 1 0.1261690
2 2 2 1 0.004625196
2 2 2 2 0.6159083
The closed shell Fock energy from formula:
2<khk> + 2<klkl>  <kllk> + = 2.84352827
from formula:
ek + <khk> + = 2.84352398
the difference is: 0.00000429
**************
ITERATION 7
**************
The charge bond order matrix:
The Fock matrix:
S F S
The eigenvalues of this matrix (Fock orbital energies) are:
Their corresponding eigenvectors (C' = S * C) are:
The "new" MOAO coefficients (C = S * C'):
The oneelectron MO integrals:
The twoelectron MO integrals:
1 1 1 1 0.9595849
2 1 1 1 0.1953881
2 1 2 1 0.6062936
2 2 1 1 0.1261697
2 2 2 1 0.004625696
2 2 2 2 0.6159083
The closed shell Fock energy from formula:
2<khk> + 2<klkl>  <kllk> + = 2.84352922
from formula:
ek + <khk> + = 2.84352827
the difference is: 0.00000095
**************
ITERATION 8
**************
The charge bond order matrix:
The Fock matrix:
S F S
The eigenvalues of this matrix (Fock orbital energies) are:
Their corresponding eigenvectors (C' = S * C) are:
The "new" MOAO coefficients (C = S * C'):
The oneelectron MO integrals:
The twoelectron MO integrals:
1 1 1 1 0.9595841
2 1 1 1 0.1953881
2 1 2 1 0.6062934
2 2 1 1 0.1261700
2 2 2 1 0.004625901
2 2 2 2 0.6159081
The closed shell Fock energy from formula:
2<khk> + 2<klkl>  <kllk> + = 2.84352827
from formula:
ek + <khk> + = 2.84352827
the difference is: 0.00000000
f. In looking at the energy convergence we see the following:
Iter
Formula 1
Formula 2
1
2.84219933
2.80060530
2
2.84349298
2.83573675
3
2.84353018
2.84225941
4
2.84352922
2.84332418
5
2.84352779
2.84349489
6
2.84352827
2.84352398
7
2.84352922
2.84352827
8
2.84352827
2.84352827
If you look at the energy differences (SCF at iteration n  SCF converged) and plot this data versus iteration number, and do a 5th order polynomial fit, we see the following:
In looking at the polynomial fit we see that the convergence is primarily linear since the coefficient of the linear term is much larger than those of the cubic and higher terms.
g. The converged SCF total energy calculated using the result of problem 40 is an upper bound to the ground state energy, but, during the iterative procedure it is not. Only at convergence does the expectation value of the Hamiltonian for the Hartree Fock determinant become equal to that given by the equation in problem 40.
h. Yes, the 1s2 configuration does dissociate properly because at at R®¥ the lowest energy state is He + H+, which also has a 1s2 orbital occupancy (i.e., 1s2 on He and 1s0 on H+).
42.
2. At convergence the MO coefficients are:
f1 = f2 =
and the integrals in this MO basis are:
h11 = 2.615842 h21 = 0.1953882 h22 = 1.315354
g1111 = 0.9595841 g2111 = 0.1953881 g2121 = 0.6062934
g2211 = 0.1261700 g2221 = 004625901 g2222 = 0.6159081
a. H = =
=
=
b. The eigenvalues are E1 = 4.279131 and E2 = 2.007770. The corresponding eigenvectors are:
C1 = , C2 =
c.
=
=
= a b.
d. The third configuration 1s2s = ,
Adding this configuration to the previous 2x2 CI results in the following 3x3 'full' CI:
H =
=
Evaluating the new matrix elements:
H13 = H31 = *(0.1953882 + 0.1953881) = 0.0
H23 = H32 = *(0.1953882 + 0.004626) = 0.269778
H33 = 2.615842  1.315354 + 0.606293 + 0.126170
= 3.198733
=
e. The eigenvalues are E1 = 4.279345, E2 = 3.256612 and E3 = 1.949678. The corresponding eigenvectors are:
C1 = , C2 = , C3 =
f. We need the nonvanishing matrix elements of the dipole operator in the MO basis. These can be obtained by calculating them by hand. They are more easily obtained by using the TRANS program. Put the 1e AO integrals on disk by running the program RW_INTS. In this case you are inserting z11 = 0.0, z21 = 0.2854, and z22 = 1.4 (insert 0.0 for all the 2e integrals) ... call the output file "ao_dipole.ints" for example. The converged MOAO coefficients should be in a file ("mocoefs.dat" is fine). The transformed integrals can be written to a file (name of your choice) for example "mo_dipole.ints". These matrix elements are:
z11 = 0.11652690, z21 = 0.54420990, z22 = 1.49117320
The excitation energies are E2  E1 = 3.256612  4.279345 = 1.022733, and
E3  E1 = 1.949678. 4.279345 = 2.329667.
Using the SlaterConden rules to obtain the matrix elements between configurations we obtain:
Hz =
=
=
Now, <Y1zY2> = C1THzC2, (this can be accomplished with the program UTMATU)
= T
= .757494
and, <Y1zY3> = C1THzC3
= T
= 0.014322
g. Using the converged coefficients the orbital energies obtained from solving the Fock equations are e1 = 1.656258 and e2 = 0.228938. The resulting expression for the PT firstorder wavefunction becomes:
1s2>(1) =  2s2>
1s2>(1) =  2s2>
1s2>(1) = 0.04419822s2>
h. As you can see from part c., the matrix element <1s2H1s2s> = 0 (this is also a result of the Brillouin theorem) and hence this configuration does not enter into the firstorder wavefunction.
i. 0> = 1s2>  0.04419822s2>. To normalize we divide by:
= 1.0009762
0> = 0.9990251s2>  0.0441552s2>
In the 2x2 CI we obtained:
0> = 0.998451231s2>  0.055634392s2>
j. The expression for the 2nd order RSPT is:
E(2) =  = 
= 0.005576 au
Comparing the 2x2 CI energy obtained to the SCF result we have:
4.279131  (4.272102) = 0.007029 au
43. STO total energy: 2.8435283
STO3G total energy 2.8340561
321G total energy 2.8864405
The STO3G orbitals were generated as a best fit of 3 primitive Gaussians (giving 1 CGTO) to the STO. So, STO3G can at best reproduce the STO result. The 321G orbitals are more flexible since there are 2 CGTOs per atom. This gives 4 orbitals (more parameters to optimize) and a lower total energy.
44.
R
HeH+ Energy
H2 Energy
1.0
2.812787056
1.071953297
1.2
2.870357513
1.113775015
1.4
2.886440516
1.122933507
1.6
2.886063576
1.115567684
1.8
2.880080938
1.099872589
2.0
2.872805595
1.080269098
2.5
2.856760263
1.026927710
10.0
2.835679293
0.7361705303
Plotting total energy vs. geometry for HeH+:
Plotting total energy vs. geometry for H2:
For HeH+ at R = 10.0 au, the eigenvalues of the converged Fock matrix and the corresponding converged MOAO coefficients are:
.1003571E+01
.4961988E+00
.5864846E+00
.1981702E+01
.4579189E+00
.8245406E05
.1532163E04
.1157140E+01
.6572777E+00
.4580946E05
.6822942E05
.1056716E+01
.1415438E05
.3734069E+00
.1255539E+01
.1669342E04
.1112778E04
.7173244E+00
.1096019E+01
.2031348E04
Notice that this indicates that orbital 1 is a combination of the s functions on He only (dissociating properly to He + H+).
For H2 at R = 10.0 au, the eigenvalues of the converged Fock matrix and the corresponding converged MOAO coefficients are:
.2458041E+00
.1456223E+00
.1137235E+01
.1137825E+01
.1977649E+00
.1978204E+00
.1006458E+01
.7903225E+00
.5632566E+00
.5628273E+00
.8179120E+00
.6424941E+00
.1976312E+00
.1979216E+00
.7902887E+00
.1006491E+01
.5629326E+00
.5631776E+00
.6421731E+00
.8181460E+00
Notice that this indicates that orbital 1 is a combination of the s functions on both H atoms (dissociating improperly; equal probabilities of H2 dissociating to two neutral atoms or to a proton plus hydride ion).
45. The H2 CI result:
R
1Sg+
3Su+
1Su+
1Sg+
1.0
1.074970
0.5323429
0.3997412
0.3841676
1.2
1.118442
0.6450778
0.4898805
0.1763018
1.4
1.129904
0.7221781
0.5440346
0.0151913
1.6
1.125582
0.7787328
0.5784428
0.1140074
1.8
1.113702
0.8221166
0.6013855
0.2190144
2.0
1.098676
0.8562555
0.6172761
0.3044956
2.5
1.060052
0.9141968
0.6384557
0.4530645
5.0
0.9835886
0.9790545
0.5879662
0.5802447
7.5
0.9806238
0.9805795
0.5247415
0.5246646
10.0
0.980598
0.9805982
0.4914058
0.4913532
For H2 at R = 1.4 au, the eigenvalues of the Hamiltonian matrix and the corresponding determinant amplitudes are:
determinant
1.129904
0.722178
0.544035
0.015191
1sga1sgb
0.99695
0.00000
0.00000
0.07802
1sgb1sua
0.00000
0.70711
0.70711
0.00000
1sga1sub
0.00000
0.70711
0.70711
0.00000
1sua1sub
0.07802
0.00000
0.00000
0.99695
This shows, as expected, the mixing of the first 1Sg+ (1sg2) and the 2nd 1Sg+ (1su2) determinants in the first and fourth states, and the
3Su+ = (),
and 1Su+= ()
states as the second and third states.
Also notice that the first 1Sg+ state has coefficients (0.99695  0.07802) (note specifically the +  combination) and the second 1Sg+ state has the opposite coefficients with the same signs (note specifically the + + combination). The + + combination always gives a higher energy than the +  combination.
46.
F atoms have 1s22s22p5 2P ground electronic states that are split by spinorbit coupling into 2P3/2 and 2P1/2 states that differ by only 0.05 eV in energy.
a.
The degeneracy of a state having a given J is 2J+1, and the J=3/2 state is lower in energy because the 2p orbital shell is more than half filled (I learned this in inorganic chemistry class), so
qel = 4 exp(0/kT) + 2 exp(0.05 eV/kT).
0.05 eV is equivalent to k(500 K), so 0.05/kT = 500/T, hence
qel = 4 exp(0/kT) + 2 exp(500/T).
b.
Q = qN/N!
so, ln Q = N lnq – lnN!
E =kT2 ∂lnQ/∂T = NkT2 ∂lnq/∂T = Nk{1000 exp(500/T)/[4 + 2 exp(500/T)]}
c. Using the fact that kT=0.03eV at T=300°K, make a (qualitative) graph of /N vs T for T ranging from 100°K to 3000°K.
At T = 100 K, E/N is small and equal to 1000k exp(5)/(4 + 2 exp(5)).
At T = 3000 K, E/N has grown to 1000k exp(1/6)/(4 + 2 exp(1/6)) which is
approximately 1000k/6.
47.
a.
The difference between a linear and bent transition state would arise in the vibrational and rotational partition functions. For the linear TS, one has 3N6 vibrations (recall that one loses one vibration as a reaction coordinate), but for the bent TS, one has 3N7 vibrations. For the linear TS, one has 2 rotational axes, and for the bent TS, one has 3.
So the ratio of rate constants will reduce to ratios of vibration and rotation partition functions. In particular, one will have
klinear/kbent = (qvib3N6 qrot2/qvib3N7qrot3) = (qvib/qrot).
b. Using
qt ~ 108, qr ~ 102, qv ~ 1,
I would expect klinear/kbent to be of the order of 1/102 = 102.
48.
Constructing the Slater determinant corresponding to the "state" 1s(a)1s(a) with the rows labeling the orbitals and the columns labeling the electron gives:
1sa1sa =
=
= 0
49.
Starting with the MS=1 3S state (which in a "box" for this ML=0, MS=1 case would contain only one product function; 1sa2sa) and applying S gives:
S 3S(S=1,MS=1) = 3S(S=1,MS=0)
= 3S(S=1,MS=0)
= 1sa2sa
= S(1)1sa2sa + S(2)1sa2sa
= 1sb2sa
+ 1sa2sb
=
So, 3S(S=1,MS=0) =
3S(S=1,MS=0) =
The three triplet states are then:
3S(S=1,MS=1)= 1sa2sa,
3S(S=1,MS=0) = , and
3S(S=1,MS=1) = 1sb2sb.
The singlet state which must be constructed orthogonal to the three singlet states (and in particular to the 3S(S=1,MS=0) state) can be seen to be:
1S(S=0,MS=0) = .
Applying S2 and Sz to each of these states gives:
Sz 1sa2sa = 1sa2sa
= Sz(1)1sa2sa + Sz(2))1sa2sa
= 1sa2sa + 1sa2sa
= 1sa2sa
S2 1sa2sa = (SS+ + Sz2 + Sz) 1sa2sa
= SS+1sa2sa + Sz21sa2sa + Sz1sa2sa
= 0 + 2 1sa2sa + 21sa2sa
= 22 1sa2sa
Sz =
= 1sb2sa
+ 1sa2sb
= 1sb2sa
+ 1sa2sb
= 0
S2 = (SS+ + Sz2 + Sz)
= SS+
=
=
= 2
= 2
= 2 2
Sz 1sb2sb = 1sb2sb
= Sz(1)1sb2sb + Sz(2))1sb2sb
= 1sb2sb + 1sb2sb
= 1sb2sb
S2 1sb2sb = (S+S + Sz2  Sz) 1sb2sb
= S+S1sb2sb + Sz21sb2sb  Sz1sb2sb
= 0 + 2 1sb2sb + 21sb2sb
= 22 1sb2sb
Sz =
= 1sb2sa
 1sa2sb
= 1sb2sa
 1sa2sb
= 0
S2 = (SS+ + Sz2 + Sz)
= SS+
=
=
= 0
= 0
= 0 2
50.
As shown in problem 22c, for two equivalent p electrons one obtains six states:
1D (ML=2); one state (MS=0),
1D (ML=2); one state (MS=0),
1S (ML=0); one state (MS=0), and
3S (ML=0); three states (MS=1,0, and 1).
By inspecting the "box" in problem 22c, it should be fairly straightforward to write down the wavefunctions for each of these:
1D (ML=2); p1ap1b
1D (ML=2); p1ap1b
1S (ML=0);
3S (ML=0, MS=1); p1ap1a
3S (ML=0, MS=0);
3S (ML=0, MS=1); p1bp1b
51.
We can conveniently couple another s electron to the states generated from the 1s12s1 configuration:
3S(L=0, S=1) with 3s1(L=0, S=) giving:
L=0, S=, ; 4S (4 states) and 2S (2 states).
1S(L=0, S=0) with 3s1(L=0, S=) giving:
L=0, S=; 2S (2 states).
Constructing a "box" for this case would yield:
ML
MS
0
1sa2sa3sa
1sa2sa3sb, 1sa2sb3sa, 1sb2sa3sa
One can immediately identify the wavefunctions for two of the quartets (they are single entries):
4S(S=,MS=): 1sa2sa3sa
4S(S=,MS=): 1sb2sb3sb
Applying S to 4S(S=,MS=) yields:
S4S(S=,MS=) = 4S(S=,MS=)
= 4S(S=,MS=)
S1sa2sa3sa =
So, 4S(S=,MS=) =
Applying S+ to 4S(S=,MS=) yields:
S+4S(S=,MS=) = 4S(S=,MS=)
= 4S(S=,MS=)
S+1sb2sb3sb =
So, 4S(S=,MS=) =
It only remains to construct the doublet states which are orthogonal to these quartet states. Recall that the orthogonal combinations for systems having three equal components (for example when symmetry adapting the 3 sp2 hybrids in C2v or D3h symmetry) give results of + + +, +2  , and 0 + . Notice that the quartets are the + + + combinations and therefore the doublets can be recognized as:
2S(S=,MS=) =
2S(S=,MS=) =
2S(S=,MS=) =
2S(S=,MS=) =
52.
As illustrated in problem 24, a p2 configuration (two equivalent p electrons) gives rise to the term symbols: 3P, 1D, and 1S. Coupling an additional electron (3d1) to this p2 configuration will give the desired 1s22s22p23d1 term symbols:
3P(L=1,S=1) with 2D(L=2,S=) generates;
L=3,2,1, and S=, with term symbols 4F, 2F,4D, 2D,4P, and 2P,
1D(L=2,S=0) with 2D(L=2,S=) generates;
L=4,3,2,1,0, and S=with term symbols 2G, 2F, 2D, 2P, and 2S,
1S(L=0,S=0) with 2D(L=2,S=) generates;
L=2 and S=with term symbol 2D.
53. The notation used for the Slater Condon rules will be as follows:
(a.) zero (spin orbital) difference;
= +
= +
(b.) one (spin orbital) difference (fp ¹ fp');
= +
= fpp' +
(c.) two (spin orbital) differences (fp ¹ fp' and fq ¹ fq');
= 
= gpqp'q'  gpqq'p'
(d.) three or more (spin orbital) differences;
= 0
i. 3P(ML=1,MS=1) = p1ap0a
= < 10 H  10>
Using the Slater Condon rule (a.) above (I will denote these SCaSCd):
= f11 + f00 + g1010  g1001
ii. 3P(ML=0,MS=0) =
=
+ + )
Evaluating each matrix element gives:
= f1a1a + f1b1b + g1a1b1a1b  g1a1b1b1a (SCa)
= f11 + f11 + g1111  0
= g1a1b1b1a  g1a1b1a1b (SCc)
= 0  g1111
= g1b1a1a1b  g1b1a1b1a (SCc)
= 0  g1111
= f1b1b + f1a1a + g1b1a1b1a  g1b1a1a1b (SCa)
= f11 + f11 + g1111  0
Substitution of these expressions give:
=
+ f11 + f11 + g1111)
= f11 + f11 + g1111  g1111
iii. 1S(ML=0,MS=0);
=
 
+ +
 +
+ )
Evaluating each matrix element gives:
= f0a0a + f0b0b + g0a0b0a0b  g0a0b0b0a (SCa)
= f00 + f00 + g0000  0
=
= g0a0b1a1b  g0a0b1b1a (SCc)
= g0011  0
=
= g0a0b1a1b  g0a0b1b1a (SCc)
= g0011  0
= f1a1a + f1b1b + g1a1b1a1b  g1a1b1b1a (SCa)
= f11 + f11 + g1111  0
=
= g1a1b1a1b  g1a1b1b1a (SCc)
= g1111  0
= f1a1a + f1b1b + g1a1b1a1b  g1a1b1b1a (SCa)
= f11 + f11 + g1111  0
Substitution of these expressions give:
=
+ g1111 + g1111  g0011 + g1111 + f11 + f11 + g1111)
=
iv. 1D(ML=0,MS=0) =
Evaluating we note that all the Slater Condon matrix elements generated are the same as those evaluated in part iii. (the signs for the wavefunction components and the multiplicative factor of two for one of the components, however, are different).
=
+ f11 + g1111 + g1111 + 2g0011 + g1111 + f11 + f11
+ g1111)
=
54.
i. 1D(ML=2,MS=0) = p1ap1b
=
= f1a1a + f1b1b + g1a1b1a1b  g1a1b1b1a (SCa)
= f11 + f11 + g1111  0
= 2f11 + g1111
ii. 1S(ML=0,MS=0) =
=
 + )
Evaluating each matrix element gives:
= f1a1a + f1b1b + g1a1b1a1b  g1a1b1b1a (SCa)
= f11 + f11 + g1111  0
= g1a1b1b1a  g1a1b1a1b (SCc)
= 0  g1111
= g1b1a1a1b  g1b1a1b1a (SCc)
= 0  g1111
= f1b1b + f1a1a + g1b1a1b1a  g1b1a1a1b (SCa)
= f11 + f11 + g1111  0
Substitution of these expressions give:
=
= f11 + f11 + g1111+ g1111
iii. 3S(ML=0,MS=0) =
= f11 + f11 + g1111  0
= g1a1b1b1a  g1a1b1a1b (SCc)
= 0  g1111
= g1b1a1a1b  g1b1a1b1a (SCc)
= 0  g1111
= f1b1b + f1a1a + g1b1a1b1a  g1b1a1a1b (SCa)
= f11 + f11 + g1111  0
Substitution of these expressions give:
=
= f11 + f11 + g1111 g1111
55.
The order of the answers is J, I, G. K, B, D, E, A, C, H, F
56.
p = N/(VNb) – N2 a/(kTV2)
but p/kT = (∂lnQ/∂V)T,N
so we can integrate to obtain ln Q
lnQ = ò (p/kT) dV = ò [N/(VNb) – N2 a/(kTV2)] dV
= N ln(VNb) + N2a/kT (1/V)
So,
Q = {(VNb)exp[(a/kT) (N/V)]}N
57.
a.
MD because you need to keep track of how far the molecule moves as a function of time and MC does not deal with time.
b.
MC is capable of doing this although MD is also. However, MC requires fewer computational steps, so I would prefer to use it.
c.
MC can do this, as could MD. Again, because MC needs fewer computational steps, I’d use it.
Suppose you are carrying out a MonteCarlo simulation involving 1000 Ar atoms. Further suppose that the potentials are pairwise additive and that your computer requires approximately 50 floating point operations (FPO's) (e.g. multiply, add, divide, etc.) to compute the interaction potential between any pair of atoms
d.
For each MC move, we must compute only the change in potential energy. To do this, we need to compute only the change in the pair energies that involve the atom that was moved. This will require 999x50 FPOs (the 99 being the number of atoms other than the one that moved). So, for a million MC steps, I would need 106 x 999 x 50 FPOs. At 100 x106 FPOs per second, this will require 495 seconds, or a little over eight minutes.
e.
Because the statistical fluctuations in MC calculations are proportional to (1/N)1/2, where N is the number of steps taken, I will have to take 4 times as many steps to cut the statistical errors in half. So, this will require 4 x 495 seconds or 1980 seconds.
f.
If we have one million rather than one thousand atoms, the 495 second calculation of part d would require
999,999/999
times as much time. This ratio arises because the time to compute the change in potential energy accompanying a MC move is proportional to the number of other atoms. So, the calculation would take 495 x (999,999/999) seconds or about 500,000 seconds or about 140 hours.
g.
We would be taking 109s/(1015 s per step) = 106 MD steps.
Each step requires that we compute all forces(∂V∂RI,J) between all pairs of atoms. There are 1000x999/2 such pairs. So, to compute all the forces would require
(1000x999/2)x 50 FPOs = 2.5 x107 FPOs. So, we will need
2.5 x107 FPOs/step x 106 steps/(100 FPOs per second)
= 2.5 x105 seconds or about 70 hours.
h.
The graduate student is 108 times slower than the 100 Mflop computer, so it will take her/him 108 times as long, so 495 x108 seconds or about 1570 years.
58.
First, Na has a 2S ground state term symbol whose degeneracy is 2S + 1 = 2.
Na2 has a 1S ground state whose degeneracy is 1.
The symmetry number for Na2 is s = 2.
The D0 value given is 17.3 kcal mol1.
The Kp equilibrium constant would be given in terms of partial pressures as (and then using pV=NkT)
Kp = pNa2/pNa2 = (kT)1 (qNa/V)2/(qNa2/V)
in terms of the partition functions.
a.
qNa = (2pmkT/h2)3/2 V qel
qNA2 = (2pm’kT/h2)3/2 V (8p2IkT/h2) 1/2 [ exphn/2kT) (1 exphn/kT))1 exp(De/kT)
We can combine the De and the –hn/2kT to obtain the D0 which is what we were given.
b. For Na (I will use cgs units in all cases):
q/V = (2p 23 1.66x1024 1.38 x1016 1000)3/2 2
= (6.54 x1026) x 2 = 1.31 x1027
For Na2:
q/N = 23/2 x (6.54 x1026) (1000/0.221) (1/2) (1exp(229/1000))1 exp(D0/kT)
= 1.85 x1027 (2.26 x103) (4.88) (5.96 x103)
= 1.22 x1035
So,
Kp = [1.22 x1035]/[(1.38 x1016)(1000) (1.72 x1054)
= 0.50 x106 dynes cm2 = 0.50 atm1.
59.
The differences in krate will arise from differences in the number of translational, rotational, and vibrational partition functions arising in the adsorbed and gasphase species. Recall that
krate = (kT/h) exp(E*/kT) [qTS/V]/[(qNO/V) (qCl2/V)]
In the gas phase,
NO has 3 translations, two rotations, and one vibration
Cl2 has 3 translations, two rotations, and one vibration
the NOCl2 TS, which is bent, has 3 translations, three rotations, and five vibrations (recall that one vibration is missing and is the reaction coordinate)
In the adsorbed state,
NO has 2 translations, one rotation, and three vibrations
Cl2 has 2 translations, one rotation, and three vibrations
the NOCl2 TS, which is bent, has 2 translations, one rotation, and eight vibrations (again, one vibration is missing and is the reaction coordinate).
So, in computing the partition function ratio:
[qTS/V]/[(qNO/V) (qCl2/V)]
for the adsorbed and gasphase cases, one does not obtain the same number of translational, rotational, and vibrational factors. In particular, the ratio of these factors for the adsorbed and gasphase cases gives the ratio of rate constants as follows:
kad/kgas = (qtrans/V)/qvib
which should be of the order of 108 (using the ratio of partition functions as given).
Notice that this result suggests that reaction rates can be altered by constraining the reacting species to move freely in lower dimensions even if one does not alter the energetics (e.g., activation energy or thermochemistry).
Contributions
Jack Simons (Henry Eyring Scientist and Professor of Chemistry, U. Utah) Telluride Schools on Theoretical Chemistry
Integrated by Tomoyuki Hayashi (UC Davis)