o e ccccc aa bb cc dd aa bb cc dd 12 34 56alpha.chem.umb.edu/chemistry/grouptheory/chp05ans.pdf ·...

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Chapter 5Answers to Problems

25.1 P(E)F

3 3 3 3 3 3 3 3 2 2 2O E C C C C C C C C C C C2 2 2 2

label aa bb cc dd aa bb cc dd 12 34 56

j 2 2 6 4 5 3 4 5 3 6 2 1 1RF F F F F F F F F F F F F

E 2 -1 -1 -1 -1 -1 -1 -1 -1 2 2 2

i j 2 2 6 4 5 3 4 5 3 6 2 1 1 P RF 2F -F -F -F -F -F -F -F -F 2F 2F 2FR

2 1 2 3 4 5 6 1 2 3 4 5 6P(E)F % 4F + 4F – 2F – 2F – 2F – 2F % 2F + 2F – F – F – F – F

1This is the same result as P(E)F , and therefore the companion function can be found in thesame way as shown in the text, either by another projection or a transformation by asymmetry operation.

4P(E)F

3 3 3 3 3 3 3 3 2 2 2O E C C C C C C C C C C C2 2 2 2



E 2 -1 -1 -1 -1 -1 -1 -1 -1 2 2 2

i j 4 4 6 5 1 5 2 2 6 1 3 4 3 P RF 2F -F -F -F -F -F -F -F -F 2F 2F 2FR

4 1 2 3 4 5 6 1 2 3 4 5 6P(E)F % –2F – 2F + 4F + 4F – 2F – 2F % –F – F + 2F + 2F – F – F

3This is the same result as P(E)F , and the companion function can be obtained by theidentical addition shown in the text (p. 145).

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5P(E)F

3 3 3 3 3 3 3 3 2 2 2O E C C C C C C C C C C C2 2 2 2



E 2 -1 -1 -1 -1 -1 -1 -1 -1 2 2 2

i j 5 5 2 3 1 1 4 2 4 6 6 5 P RF 2F -F3 -F -F -F -F -F -F -F 2F 2F 2FR

5 1 2 3 4 5 6 1 2 3 4 5 6P(E)F % –2F – 2F – 2F – 2F + 4F + 4F % –F – F – F – F + 2F + 2F

1 5To get the partner add P(E)F + 2P(E)F :

1 1 2 3 4 5 6P(E)F % 2F + 2F – F – F – F – F

5 1 2 3 4 5 62P(E)F % –2F – 2F – 2F – 2F + 4F + 4F------------------------------------------------------

3 4 5 6 3 4 5 6– 3F – 3F + 3F + 3F % –F – F + F + F

This is the negative of the same function as previously obtained in the text.

4 2 15.2 Effect of C (56) on G (E) % P(E)F

4C (56) transforms the initial functions as follows:

1 2 3 4 5 6Before F F F F F F

4 3 1 2 5 6After F F F F F F

1 2 3 4 5 6 4 3 1 2 5 62F + 2F – F – F – F – F 6 2F + 2F – F – F – F – F

1 2 3 4 5 6 3 4 = F – F + 2@ + 2F – F – F % P(E)F = P(E)F

The partner function can be generated from this as shown in the text.

4 2 1Effect of C (34) on G (E) % P(E)F



5 6 3 4 2 1After F F F F F F

1 2 3 4 5 6 5 6 3 4 2 12F + 2F – F – F – F – F 6 2F + 2F – F – F – F – F

1 2 3 4 5 6 5 6 = –F – F – @ – F + 2F + 2F % P(E)F = P(E)F

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This can be used to obtain the partner by the addition shown in the answer to 5.1 givenabove.

4 2 1Effect of C (12) on G (E) % P(E)F



1 2 6 5 3 4After F F F F F F

1 2 3 4 5 6 1 2 6 5 3 42F + 2F – F – F – F – F 6 2F + 2F – F – F – F – F

1 2 3 4 5 6 1 2 2F + 2F – F – F – F – F % P(E)F = P(E)F

This just gives back the initial function.

4 5 1 2 3 4 4 6 5 65.3 IE E dJ = I(F – F )(F – F )dJ = 0 Clearly, the other two combinations E E and E Ewill give 0.

4 1 1 2 1 2 3 4 5 6 1IE E dJ = I(F – F )(F + F + F + F + F + F )dJ = 0 Other combinations with E willhave the same result.

4 2 1 2 1 2 3 4 5 6IE E dJ = I(F – F )(2F + 2F – F – F – F – F )dJ = 2 – 2 =0

5 2 3 4 1 2 3 4 5 6IE E dJ = I(F – F )(2F + 2F – F – F – F – F )dJ = –1 + 1 =0

6 2 5 6 1 2 3 4 5 6IE E dJ = I(F – F )(2F + 2F – F – F – F – F )dJ = –1 + 1 =0

4 3 1 2 3 4 5 6IE E dJ = I(F – F )(F + F – F – F )dJ = 0

5 3 3 4 3 4 5 6IE E dJ = I(F – F )(F + F – F – F )dJ = 1 – 1 =0

6 3 5 6 3 4 5 6IE E dJ = I(F – F )(F + F – F – F )dJ = –1 + 1 = 0

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2 A 35.4 The terms of the full P(T )s operator are shown below. The 8C operations have zerocharacters and can be skipped.

d 2 2 2 4 4 4 4 4 4T E ... C C C S S S S S S3 3 3

Label x y z x y z x y z

j A A C D B B C C D B DR s s s s s s s s s s s

2T 3 -1 -1 -1 -1 -1 -1 -1 -1 -1

i j A A C D B B C C D B DPR s 3s -s -s -s -s -s -s -s -s -s

d d d d d dF F F F F F

bd ac bc ad ab cd

C A D A A Bs s s s s s

1 1 1 1 1 1

C A D A A Bs s s s s s

Gathering the terms:

2 A A B C D A B C DP(T )s % 6s – 2s – 2s – 2s % 3s – s – s – s

AThis is the result obtained from the P(T)s operator in the subgroup T.

5.5 (a)

2h 2 2 2D E C (z) C (y) C (x) i F(xy) F(xz) F(yz)

H' 4 0 0 0 0 0 0 4

H g 3g 1u 2u' = A + B + B + B

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By inspection

g A B C DP(A ) % s + s + s + s

1 g A B C DM (A ) = ½(s + s + s + s )

3gProjection for B :


j A A D B C C B D AR s s s s s s s s s

3gB 1 -1 -1 1 1 -1 -1 1

i j A A D B C C B D AP R s s -s -s s s -s -s s

3g A A B C D 2 3g A B C DP(B )s % s – s + s – s Y M (B ) = ½(s – s + s – s )

1 gThis is clearly orthogonal to M (A ).

1uProjection for B :



1uB 1 1 -1 -1 -1 -1 1 1

i j A A D B C C B D AP R s s s -s -s -s -s s s

1u A A B C D 3 1u A B C DP(B )s % s – s – s + s Y M (B ) = ½(s – s – s + s )

1 g 2 3gThis is clearly orthogonal to both M (A ) and M (B ) .

2uProjection for B :



2uB 1 -1 1 -1 -1 1 -1 1

i j A A D B C C B D AP R s s -s s -s -s s -s s

2u A A B C D 3 2u A B C DP(B )s % s + s – s – s Y M (B ) = ½(s + s – s – s )

1 g 2 3g 3 1uThis is clearly orthogonal to both M (A ), M (B ) , and M (B ).

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5.5 (b)

2d 4 2 2 dD E 2S C 2C ' 2F

H' 4 0 0 0 2

H 1 2' = A + B + E

2If we do the work in the subgroup D , we will not only reduce the number of terms, but moreimportantly we will lift the E degeneracy, allowing separate projections for the two

2d 2 1 2 1 2 3degenerate SALCs. In the descent from D to D , A 6 A, B 6 B , and E 6 B + B .

1 A B C D 2 1 AThe totally symmetric SALC is clearly M (A) = ½(s + s + s + s ). In D , the P(B )sprojection is found as follows:

2 2 2 2D E C (z) C (y) C (x)

j A A D B CR s s s s s

1B 1 1 -1 -1

i j A A D B CPR s s s -s -s

2 1 A B C D 2 2 A B C D 2dM (B ) = ½(s – s – s + s ) Y M (B ) = ½(s – s – s + s ) in D

2 A 2The first of the degenerate functions is found as the projection P(B )s in D :

2 2 2 2D E C (z) C (y) C (x)


2B 1 -1 1 -1

i j A A D B CPR s s -s s -s

3 2 A B C D 3 A B C D 2dM (B ) = ½(s + s – s – s ) Y M (E ) = ½(s + s – s – s ) in Da

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3 A 2The second of the degenerate functions is found as the projection P(B )s in D :

2 2 2 2D E C (z) C (y) C (x)


3B 1 -1 -1 1

i j A A D B CPR s s -s -s s

4 3 A B C D 4 A B C D 2dM (B ) = ½(s – s + s – s ) Y M (E ) = ½(s – s + s – s ) in Db

The orthogonality of all four functions is easily shown.

2d AIf the doubly degenerate projection is carried out in D on s , we obtain an initial projection

A D 2P(E) % s – s , because only the operations E and C have nonzero characters. The

3 A D 1normalized function M '(E ) = 1//2 (s – s ) is orthogonal to our previously obtained M anda

2 1 A 2 A 2dM expressions, which are the same functions generated by P(A )s and P(B )s in D . The

4 3companion can be generated by applying the operation S (for example) to M '(E ) to obtain3 a

4 B C 3 4 1 1M '(E ) = 1//2 (s – s ). Both M '(E ) and M '(E ) are orthogonal to each other and to M (A )b a b

2 2 3 3and M (B ). They are related to the previously obtained functions as M (E ) % M '(E ) +a a

4 4 3 4M '(E ) and M (E ) % M '(E ) – M '(E ). The distinction is a matter of definition, with neitherb b a b

pair of degenerate SALCs being "right" and the other "wrong".

5.6 For all of these planar cyclic systems, the projections can be read off from the characters for

n nhthe appropriate representations in the C subgroup of the molecule's true D point group (n= 3, 4, 5).

3 3(a) C H

3h 3' = A' + E' in D Y ' = A + E in C

The A projection and resulting normalized wave function are

1 1 2 3 1 1 2 3P(A)p % p + p + p Y A 1//3 (p + p + p )

From the complex conjugate pair of E representations we obtain the following twoprojections:

1 1 2 3P(E )p % p + ,p + ,*pa

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1 1 2 3P(E )p % p + ,*p + ,pa

We can add and subtract these two to obtain real functions, realizing that

, + ,* = 2 cos 2B/3 = 2(–½) = –1, – ,* = 2i sin 2B/3 and ,* – , = –2i sin 2B/3

The imaginary constant 2i sin 2B/3 can be factored out of the subtractive combination priorto normalization. Thus, we obtain the following real SALCs:

1 1 1 2 3P(E )p + P(E )p % 2p – p – pa b

1 1 2 3P(E )p – P(E )p % p – pa b

These give the normalized functions

2 1 2 3A = 1//6 (2p – p – p )

3 2 3A = 1/2 (p – p )

4 4(b) C H

g 2u 2u 4h z 4' = E + A + B in D Y ' = A + B + E in C

4By inspection of the C character table, the A and B functions are

1 1 2 3 4A (A) = 1/2 (p + p + p + p )

4 1 2 3 4A (B) = 1/2 (p – p + p – p )

The two E projections are

1 1 2 3 4P(E )p % (p + ip – p – ip )a

1 1 2 3 4P(E )p % (p – ip – p + ip )b

By addition and subtraction, these give the two normalized functions

2 1 3A (E ) = 1//2 (p – p )"

3 2 4A (E ) = 1//2 (p – p )$

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5 5(c) C H

2 1 2 5h 1 2 5' = A " + E " + E " in D Y ' = A + E + E in C

1 1 2 3 4 5 5By inspection, A (A) = 1//5 (p + p + p + p + p ). From the C character table we can

1 2read off the two E projections and the two E projections. The following relationships willbe useful in transforming these into real functions:

Let T = 2B/5 = 72 and 2T = 4B/5 = 144o o

, + ,* = 2 cos T = 2 (0.3090), – ,* = –2i sin T = –2i (0.9511)

, + , * = 2 cos 2T = 2 (–0.8090)2 2

, – , * = –2i sin 2T = –2i (0.5878)2 2

1The two E projections are

1 1 1 2 3 4 5Pp (E ) % (p + ,p + , p + , p + , p )a 2 2* *

1 1 1 2 3 4 5Pp (E ) % (p + , p + , p + , p + ,p )b * 2* 2

Adding:

1 1 1 1 1 2 3 4 5Pp (E ) + Pp (E ) % 2p + 2p cos T + 2p cos 2T + 2p cos 2T + 2p cos Ta b

1 2 3 4 5% p + p cos T + p cos 2T + p cos 2T + p cos T

1 2 3 4 5 2 1% p + 0.3090p – 0.8090p – 0.8090p + 0.3090p % A (E )

Normalizing:

2N IA dJ = 1 + 0.0955 + 0.6545 + 0.6545 + 0.0955 = 2.52 2

N = (1/2.5) = (2/5)½ ½

Normalized function:

2 1 1 2 3 4 5A (E ) = (2/5) {p + p cos T + p cos 2T + p cos 2T + p cos T}½

Subtracting:

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1 1 1 1 2 3 4 5Pp (E ) – Pp (E ) % –2ip sin T – 2ip sin 2T + 2ip sin 2T + 2ip sin Ta b

2 3 4 5% p sin T + p sin 2T – p sin 2T – p sin T

2 3 4 5 3 1% 0.9511p + 0.5878p – 0.5878p – 0.9511p % A (E )

Normalizing:

3N IA dJ = 0.9045 + 0.3455 + 0.3455 + 0.9045 = 2.52 2

N = (1/2.5) = (2/5)½ ½


3 1 2 3 4 5A (E ) = (2/5) {p sin T + p sin 2T – p sin 2T – p sin T}½

2The two E projections are

1 2 1 2 3 4 5Pp (E ) % (p + ,p + , p + , p + , p )a 2 2* *

1 2 1 2 3 4 5Pp (E ) % (p + , p + , p + , p + ,p )b * 2* 2

Adding:

1 2 2 2 1 2 3 4 5Pp (E ) + Pp (E )% 2p + 2p cos 2T + 2p cos T + 2p cos T + 2p cos 2Ta b

1 2 3 4 5% p + p cos 2T + p cos T + p cos T + p cos 2T

1 2 3 4 5 4 2% p – 0.8090p + 0.3090p + 0.3090p – 0.8090p % A (E )

Normalizing:

4N IA dJ = 1 + 0.6545 + 0.0955 + 0.0955 + 0.6545 = 2.52 2

N = (1/2.5) = (2/5)½ ½


4 2 1 2 3 4 5A (E ) = (2/5) {p + p cos 2T + p cos T + p cos T + p cos 2T}½

Subtracting:

1 2 1 2 2 3 4 5Pp (E ) – Pp (E ) % –2ip sin 2T – 2ip sin T + 2ip sin T + 2ip sin 2Ta b

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2 3 4 5% p sin 2T + p sin T – p sin T – p sin 2T

2 3 4 5 5 2% 0.5878p + 0.9511p – 0.9511p – 0.5878p % A (E )

Normalizing:

5N IA dJ = 0.34555 + 0.9045 + 0.9045 + 0.3455 = 2.52 2

N = (1/2.5) = (2/5)½ ½


5 2 2 3 4 5A (E ) = (2/5) {p sin 2T + p sin T – p sin T – p sin 2T}½

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3h5.7 The three basis sets, their reducible representations in D , and the species of which they arecomposed are shown below.

3h 3 2 h 3 vD E 2C 3C F 2S 3F

F' 3 0 1 3 0 1

F 1' = A ' + E'

3h 3 2 h 3 vD E 2C 3C F 2S 3F

z' 3 0 -1 -3 0 1

z 2' = A " + E'

3h 3 2 h 3 vD E 2C 3C F 2S 3F

2' 3 0 -1 -3 0 1

2 2' = A ' + E'

3If we carry out the projections in the subgroup C , all three reducible representations will be' = A + E. Therefore, once we find the forms of the SALCs for one case, the other two

3cases will have the same forms. In addition, working in C lifts the double degeneracy,allowing us to read off the pair of projections from the characters for each species. Takingthe sigma case as the model, the A projection and resulting normalized wave function are

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1 1 2 3 1 1 2 3P(A)N % N + N + N Y M (F) = 1//3 (N + N + N )

From the complex conjugate pair of E representations we obtain the following twoprojections:

1 1 2 3P(E )N % N + ,N + ,*Na

1 1 2 3P(E )N % N + ,*N + ,Na

We can add and subtract these two to obtain real functions, realizing that

, + ,* = 2 cos 2B/3 = 2(–½) = –1, – ,* = 2i sin 2B/3 and ,* – , = –2i sin 2B/3

The imaginary constant 2i sin 2B/3 can be factored out of the subtractive combination priorto normalization. Thus, we obtain the following real SALCs:

1 1 1 2 3P(E )N + P(E )N % 2N – N – Na b

1 1 2 3P(E )N – P(E )N % N – Na b

These give the normalized functions

2 1 2 3M (F) = 1//6 (2N – N – N )

3 2 3M (F) = 1/2 (N – N )

It follows that the other two sets have SALCs with the same forms:

4 4 5 6M (z) = 1//3 (N + N + N )

5 4 5 6M (z) = 1//6 (2N – N – N )

6 5 6M (z) = 1/2 (N – N )

7 7 8 9M (2) = 1//3 (N + N + N )

8 7 8 9M (2) = 1//6 (2N – N – N )

7 8 9M (2) = 1/2 (N – N )

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4h5.8 The four basis sets, their reducible representations in D , and the species of which they arecomposed are shown below.

4h 4 2 2 2 4 h v dD E 2C C 2C ' 2C " i 2S F 2F 2F

F' 4 0 0 2 0 0 0 4 2 0

F 1g 1g u' = A + B + E


z' 4 0 0 -2 0 0 0 -4 2 0

z 2u 2u g' = A + B + E


2' 4 0 0 -2 0 0 0 4 -2 0

2 2g 2g u' = A + B + E

4If we carry out the projections in the subgroup C , all three reducible representations will be ' = A + B + E. Therefore, once we find the forms of the SALCs for one case, the other two

4cases will have the same forms. In addition, working in C lifts the double degeneracy,allowing us to read off the pair of projections from the characters for each species. Takingthe sigma case as a model, the A projection and resulting normalized wave function are

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1 1 2 3 4 1 1 2 3 4P(A)N % N + N + N + N Y M (A) = 1/2 (N + N + N + N )

Likewise, for the B projection and resulting normalized wave function we have

1 1 2 3 4 2 1 2 3 4P(B)N % N – N + N – N Y M (B) = 1/2 (N – N + N – N )

The two E projections are

1 1 2 3 4P(E )N % N + iN – N + iNa

1 1 2 3 4P(E )N % N – iN – N – iNb

By adding and subtracting we obtain the following two normalized functions

3 1 3 4 2 4M (E) = 1//2 (N – N ) and M (E) = 1//2 (N – N )

1 3h 35.9 (a) For trigonal planar sp hybrids, ' = A ' + E in D . Use the rotational subgroup C to2

x yread off the projections for the s = A and (p , p ) = E AOs as SALCs of the hybrids. Fromthe A projection we obtain the normalized function

1 2 3s = 1//3 (Q + Q Q )

For the two complex conjugate E projections we have

1 2 3P(E ) % Q + ,Q + ,*Qa

1 2 3P(E ) % Q + ,*Q + ,Qb

Add and subtract these two imaginary functions to obtain real functions, realizing that

, + ,* = 2 cos 2B/3 = 2(–½) = –1, – ,* = 2i cos 2B/3 and ,* – , = –2i cos 2B/3

The real functions, then, are

1 2 3P(E ) + P(E ) % 2Q – Q – Qa b

2 3P(E ) – P(E ) % Q – Qa b

The normalized expressions for the AOs are

x 1 2 3p = 1//6 (2Q – Q – Q )

y 2 3p = 1/2 (Q – Q )

In matrix form the three AO expressions become

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Taking the transpose of the 3 x 3 matrix A to obtain the B matrix, we obtain an expressionfor the hybrids as functions of the AOs:

The three individual sp functions are2

1 xQ = 1//3 s + 2//6 p

2 x yQ = 1//3 s – 1//6 p + 1/2 p

3 x yQ = 1//3 s – 1//6 p – 1/2 p

1g 1g u 4(b) For square planar dsp hybrids, ' = A + B + E . Use the rotational subgroup C to2

x y x -yread off the projections for the s = A, (p , p ) = E, and d 2 2 = B AOs as SALCs of thehybrids. From the A and B projections we obtain the normalized functions

1 2 3 4s = 1/2 (Q +Q + Q + Q )

x -y 1 2 3 4d 2 2 = 1/2 (Q – Q + Q – Q )

For the E projections we obtain

1 2 3 4P(E ) % Q + iQ – Q – iQa

1 2 3 4P(E ) % Q – iQ – Q + iQa

Adding and subtracting these give the following two normalized functions:

x 1 3p = 1//2 (Q – Q )

y 2 4p = 1//2 (Q – Q )

Gathering all four equations into matrix form gives

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Using the transpose of the 4 x 4 A matrix to obtain the B matrix, we obtain

The four dsp function are2

1 x x -yQ = 1/2 s + 1//2 p + 1/2 d 2 2

2 y x -yQ = 1/2 s + 1//2 p – 1/2 d 2 2

3 x x -yQ = 1/2 s – 1//2 p + 1/2 d 2 2

4 y x -yQ = 1/2 s – 1//2 p – 1/2 d 2 2

1g 1u(c) For octahedral d sp hybrids, ' = A + E + T . We can use the previously obtained 2 3

expressions for F-SALCs developed in section 5.1 to write equations for the AOs asfunctions of the hybrids.

1 2 3 4 5 6s(A) = 1//6 (Q + Q + Q + Q + Q + Q )

x 5 6p (T) = 1//2 (Q – Q )

y 3 4p (T) = 1//2 (Q – Q )

z 1 2p (T) = 1//2 (Q – Q )

z 1 2 3 4 5 6d 2(E) = 1//12 (2Q + 2Q – Q – Q – Q – Q )

x -y 3 4 5 6d 2 2(E) = 1/2 (Q + Q – Q – Q )

In matrix form these are

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Using the transpose of the 6 x 6 A matrix to obtain the B matrix, we obtain

The six hybrid functions are

1 z zQ = 1//6 s + 1//2 p + 1//3 d 2

2 z zQ = 1//6 s – 1//2 p + 1//3 d 2

3 y z x -yQ = 1//6 s + 1//2 p – 1//12 d 2 + 1/2 d 2 2

4 y z x -yQ = 1//6 s – 1//2 p – 1//12 d 2 + 1/2 d 2 2

5 x z x -yQ = 1//6 s + 1//2 p – 1//12 d 2 – 1/2 d 2 2

6 x z x -yQ = 1//6 s – 1//2 p – 1//12 d 2 – 1/2 d 2 2

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5.10 The matrix equation for the AOs as functions of the hybrids is

Taking the transpose of the A matrix to obtain the B matrix gives the following matrixequation for the hybrids as functions of the AOs:

The resulting individual functions are

1 x zQ = 1//5 s + 2//6 p – 2//30 d 2

2 x y zQ = 1//5 s – 1//6 p + 1//2 p – 2//30 d 2

3 x y zQ = 1//5 s – 1//6 p – 1//2 p – 2//30 d 2

4 z zQ = 1//5 s + 1//2 p + 3//30 d 2

5 z zQ = 1//5 s – 1//2 p + 3//30 d 2

These hybrids might be used to describe bonding in a transition metal tbp complex in whichthe axial and equatorial positions made equal contributions to the bonding. The

zinvolvement of the d 2 orbital in the hybrid set makes this mode of hybridization, in general,inappropriate for describing the bonding in tbp structures in which the central atom is a p-block element.

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5.11 The representation for the F-SALCs is based on the following vector model:

The reducible representation and its component species are

4v 4 2 v dC E 2C C 2F 2F

' 5 1 1 3 1

1 1' = 2A + B + E

If we apply projection operators to any reference function, we will project LCAOs that are

1expressions only of the members of the same set, basal or axial. The basal functions are F ,

2 3 4 5 5 1F , F , F , and the axial function is simply F . The axial function F clearly has Asymmetry, so the symmetries of the SALCs that can be formed from the basal set are theremaining species in '. The symmetries of the two sets, then, are

basal 1 1 axial 1' = A + B + E and ' = A

4We can carry out the basal projections in the subgroup C , from which it is immediatelyapparent that nondegenerate functions are

1 1 1 2 3 4M (A ) = 1/2 (F + F + F + F )

2 1 1 2 3 4M (B ) = 1/2 (F – F + F – F )

4The two E projections take advantage of the complex conjugate representations in C :

1 1 2 3 4P(E )F % F + iF – F – iFa

1 1 2 3 4P(E )F % F – iF – F + iFb

By addition and subtraction, we obtain the following two real functions:

3 1 3M (E) = 1//2 (F – F )

4 2 4M (E) = 1//2 (F – F )

5For the axial position, the SALC is simply F :

5 1 5M (A ) = F

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The AOs on a central atom that would match with these SALCs are

1 z z 1 x -y x yA = s, p , d 2 B = d 2 2 E = (p , p )

z x -yWe will ignore the d 2 initially, because we are assuming d 2 2sp hybrids on the central3

atom. If we accept the SALCs we obtained from the projection operator approach, wewould make the following matches between SALCs and central-atom AOs:

1Looking at the matches between AOs and SALCs, it appears that the B and E SALCs arecorrectly formulated, despite the basal-axial segregation inherent in our approach.

1However, the two A SALCs are artificially exclusive with regard to the two kinds of

1 1 5positions. The M (A ) SALC would be improved by including the F function, which would

1in no way change its A symmetry. This results in the following match:

If the axial and equatorial positions made equal contributions to the bonding (not a typicalresult) the redefined SALC would be

1 1 1 2 3 4 5M' (A ) = 1//5 (F + F + F + F +F )

More generally, this would be

1 1 basal 1 2 3 4 axial 5M' (A ) = c (F + F + F + F ) + c F

-85-

If the angle between axial and basal positions were exactly 90 , there would be noo

5 1possibility of basal participation in M (A ), because the four basal functions would fall on a

znodal plane of the matching p orbital. In most real examples of square pyramidalgeometry, the central atom is not coplanar with the basal positions. Such departures fromperfect 2 = 90 geometry would allow some measure of constructive overlap between theo

basal functions. The modified function to include such minor basal participation would be

5 1 axial 5 basal 1 2 3 4M' (A ) = c' F ± c' (F + F + F + F )

where the negative sign would be appropriate for 2 > 90 , and the positive sign would beo

z zappropriate for 2 < 90 . If the d 2 orbital were to replace the p orbital as the major axialo

AO on the central atom, the basal positions would make a significant contribution and theSALC would be

5 1 axial 5 basal 1 2 3 4M" (A ) = c" F – c" (F + F + F + F )

zThe match between d 2 and this modified SALC would be as follows:

zOverlap between the annular part of the d 2 orbital and the basal functions of the SALCwould be most effective at 2 = 90 .o

5.12 The purpose of this question is to explore an alternative approach to constructing SALCsand hybrids in general, and for the tbp case in specific. In the Dahr method, an expression foreach SALC that would match with the various central-atom AOs is written down as the sum of

ithe projections of each pendant atom function, N , on the reference axes of the central atom AO,

iwith proper adjustment of the mathematical signs of the orbital. Each projection is given by N

i i icos 2 , where 2 is the angle between the AB bond of each pendant atom and the reference axisof the central AO. The d orbital functions require combining projections. In the case of the

zSALC to match with d 2, this approach yields a function with unequal contributions in theequatorial plane, which is intuitively incorrect. Recognizing this, Dahr corrects the function

3 3empirically to maintain C symmetry. Beyond addressing the C dissymmetry, Dahr's correctionimplicitly gives equal weight to the equatorial and axial positions, although this is not explicitlystated in the paper. The final set of functions is comparable to Eqs. (5.26a - 5.26e), shown onpage 161 of the text. As such, SALCs represent a special case in which axial and equatorialpositions are made equivalent. Dahr does not note the lack of generality in these results. Theadvantages and disadvantages of each approach are primarily a matter of personal preference.

o e ccccc aa bb cc dd aa bb cc dd 12 34 56alpha.chem.umb.edu/chemistry/grouptheory/chp05ans.pdf ·...

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