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ADVANCED INORGANIC CHEMISTRY QUIZ 4 November 29, 2012 INSTRUCTIONS: PRINT YOUR NAME ————> NAME . WORK all 4 problems USE THE CORRECT NUMBER OF SIGNIFICANT FIGURES YOUR SUPPPLEMENTAL MATERIALS CONTAIN: A PERIODIC TABLE Water MO diagram C 2 v and D 5 h POINT GROUP CHARACTER TABLEs Quiz 4 R = 0.08206 lit-atm/mol-K 1 (30) R = 8.3145 J/mol-K 2 (30) 3 (20) c = 2.9979 X 10 8 m/s 4 (20) h = 6.626 X 10 -34 J-s TOTAL(100)

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Page 1: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

ADVANCED INORGANIC CHEMISTRY QUIZ 4

November 29, 2012

INSTRUCTIONS:

PRINT YOUR NAME ————> NAME .

WORK all 4 problems

USE THE CORRECT NUMBER OF SIGNIFICANT FIGURES

YOUR SUPPPLEMENTAL MATERIALS CONTAIN:

A PERIODIC TABLE

Water MO diagram

C2v and D5h POINT GROUP CHARACTER TABLEs

Quiz 4

R = 0.08206 lit-atm/mol-K 1 (30)

R = 8.3145 J/mol-K 2 (30)

3 (20)

c = 2.9979 X 108 m/s 4 (20)

h = 6.626 X 10-34 J-s TOTAL(100)

Page 2: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

1. Give a short answer or explanation for the following.

(a) Explain why s-p mixing must be considered for early second row diatomics such as B2

and C2, but s-p mixing is ignored for late second row diatomics such as O2 and F2.

(b) Give at least two reasons how the use of symmetry helps in molecular orbital

construction.

The molecular orbital diagram for water is included in the quiz. Refer to it for (c) – (e)

(c) Using the diagram explain why a proton asttached to oxygen instead of hysrogen

(d) The 1 b2 orbital is called a pure non-bonding orbital. Explain what this means.

(e) How does the molecular orbital diagram differ from the valence bond (Lewis structure )

description of water.

.

Page 3: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

2 . Consider the species, NO+,

NO (neutral), and NO-. (these species have s-p mixing)

(a) Sketch the energy levels of the molecular orbitals (just draw one diagram and populat it with

the electrons for neutrsl NO. ).

(b) Calculate the bond order of each species.

(c) Rank the species in order of increasing bond length

< <

(d) Rank the species in order of increasing bond strength.

< <

(e) Which if any of the species is paramagnetic?

(f) Sketch the lowest energy orbital’s shape. Sketch the shape of one of the π* anti-

bonding orbitals.

Page 4: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

3. This problem runs over two pages.You will construct the symmetry adapted orbitals for BrF3.

Point group C2v

(a) Construct reducible representastions for the group orbitals by applying the symmetry

operations.

E C2(z) σv(xz) σv(yz)

s

Γ= 3 1 3 1

E C2(z) σv(xz) σv(yz)

Γ= 3 1 3 1 P (σ bond)

E C2(z) σv(xz) σv(yz)

Γ= ____ ______ ______ _____

p(π-bond(1))

E C2(z) σv(xz) σv(yz)

Γ= ____ ______ ______ _____

p(π-bond(2))

(continue on next page)

Page 5: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

(b) Reduce your representation to a sum of irreducible representations, and identify thre central

aton that interacts with the orbital. (I did the first two)

s or p σ Γ = 2 A1 = s, pz + B1 central atom orbitals s, pz px p(π-bond(1)) E C2(z) σv(xz) σv(yz)

from previous page Γ= ____ ______ ______ _____

Reduce and identify the central atom orbitals.

#A1 =

#A2 =

#B1 =

#B2 =

p(π-bond(1)) Γ = _______ + ________ + __________ central atom orbitals _______ ______ __________ ______________________________________________________________________________ p(π-bond(1)) E C2(z) σv(xz) σv(yz)

from previous page Γ= ____ ______ ______ _____

Reduce and identify the central atom orbitals.

#A1 =

#A2 =

#B1 =

#B2 =

p(π-bond(1)) Γ = _______ + ________ + __________ central atom orbitals _______ ______ __________

Page 6: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

FF

F FFU

O

O

z

x

y

3. A ion of UO2F55-

has pentagonal bipyramidal geometry. The point group of the ion is D5h

(b) The group orbitals for the O atoms (using the s orbitals) are shown below with their

symmetries. To the right of the figures indicate which U orbital will interact with the group

orbital to form a molecular orbital. Consider the s, p, and d orbitals on U. Sketch the shape of

the bonding molecular orbital formed by this interaction.

Group U Bonding

Orbital orbital Molecular orbital (s)

U

U

A1'

A2"

Continues on next page.

Page 7: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

U

U

U

U

U

y

A1'

E1'

E1"

(b) The group orbitals for the F atoms (using the s orbitals) are shown below with their

symmetries. The ion has been rotated to show the view perpendicular to the xy plane. To the

right of the figures indicate which U orbital will interact with the group orbital to form a

molecular orbital. Consider the s, p, and d orbitals on U. Sketch the shape of the bonding

molecular orbital formed by this interaction.

Group U Bonding

Orbital orbital Molecular orbital(s)

Page 8: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered
Page 9: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

Character table for D5h (x axis coincident with C'2 axis)

D5h E 2C5 2(C5)2 5C'2 h 2S5 2(S5)

3 5 v linear

functions,

rotations

quadratic

functions

A'1 +1 +1 +1 +1 +1 +1 +1 +1 - x2+y

2, z

2

A'2 +1 +1 +1 -1 +1 +1 +1 -1 Rz -

E'1 +2 +2cos(2

/5)

+2cos(4

/5) 0 +2

+2cos(2

/5) +2cos(4

/5) 0 (x, y) -

E'2 +2 +2cos(4

/5)

+2cos(2

/5) 0 +2

+2cos(4

/5) +2cos(2

/5) 0 -

(x2-y

2,

xy)

A''1 +1 +1 +1 +1 -1 -1 -1 -1 - -

A''2 +1 +1 +1 -1 -1 -1 -1 +1 z -

E''1 +2 +2cos(2

/5)

+2cos(4

/5) 0 -2

-2cos(2

/5) -2cos(4 /5) 0 (Rx, Ry) (xz, yz)

E''2 +2 +2cos(4

/5)

+2cos(2

/5) 0 -2

-2cos(4

/5) -2cos(2 /5) 0 - -

Character table for point group D5h

Page 10: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

3. This problem runs over two pages. You will construct the symmetry adapted group orbitals

for BrF5. Point group C4v

(a) Construct reducible representations for the group orbitals by applying the symmetry

operations. Hint: they are all odd numbers.

E 2 C4(z) C2(z) 2 σv(xz) 2σd(bisectror

of xz and yz)

Γ = ____ ______ ______ _____ ________

E 2 C4(z) C2(z) 2 σv(xz) 2σd(bisectror of xz/yz)

When you finish bring it to me to grade so you can do the next part.

Page 11: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

Reproduce your representation here and reduce it.

Γ = ____ ______ ______ _____ ________

E 2 C4(z) C2(z) 2 σv(xz) 2σd(bisectror of xz/yz)

Reduce and identify the central atom orbitals.

#A1 =

#A2 =

#B1 =

#B2 =

#E =

(s-bond(1)) Γ = _______ + ________ + __________ + ________ +_____________ Not all blanks A1 A2 B1 B2 E may be needed central atom orbitals _______ ________ __________ ________ __________ (Use s, p, and d) A1 A2 B1 B2 E

Page 12: ADVANCED INORGANIC CHEMISTRY EXAM Iwhitener/courses/fa2012/... · ADVANCED INORGANIC CHEMISTRY QUIZ 4 November ... m/s 4 (20) h = 6.626 X 10 ... Explain why s-p mixing must be considered

4. The following involves drawing the pi molecular orbitals for a square planar complex such as

XeF4 . The group orbital symmetry is as shown. Identify the central atom orbital that interacts,

and draw the bonding combination for each set.