Chemistry 4110-5110 (Fall 2010) Name__________________

Problem Set #1 (due Monday, Sept. 13th) (1) (12 points) i) Give the electronic configuration for the iridium cation, Ir+.

ii) Neatly sketch the d valence orbital which is oriented in the xz plane, including all radial and angular nodes and wavefunction sign changes, and iii) Indicate the number of unpaired electrons in the ground state.

i) Neutral Ir: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d7 or [Xe] 6s24f145d7 Ir+: be sure to remove 6s electron, not 5d!!: [Xe] 6s14f145d7 ii) Remember the lobes are between axes and for a 5d AO there are two radial nodes: iii) write out the valence orbitals: 6s1 is ___, 5d7 is ___ ___ ___ ___ ___ so, there are 4 unpaired electrons. Remember to count the unpaired 6s electron: all unpaired electrons, no mater what subshell, count! (2) (18 points) Using Slater’s rules, calculate Zeff for the following electrons: a) A valence electron in germanium Electronic configuration: 1s22s22p63s23p24s23d104p2. Slater grouping: (1s2)(2s2,2p6)(3s2,3p6)(3d10)(4s2,4p2), Z = 32 4p is the valence electron shell; σ = 3(0.35) + 18(0.85) + 10(1.0) = 26.35; Zeff = 32 – 26.35 = 5.65 b) A 3d electron in nickel Electronic configuration: 1s22s22p63s23p64s23d8. Slater grouping: (1s2)(2s2,2p6)(3s2,3p6)(3d8)(4s2); Z = 28 For 3d, σ = 7(0.35) + 18(1.00) = 20.45 (remember: for a d shell, all e-s to the left shield 1.0). Zeff = 28 – 20.45 = 7.55

c) A 4f electron in Ta3+ Electronic configuration (Ta): 1s22s22p63s23p64s23d104p65s24d105p66s24f145d3. Z = 73 Electronic configuration (Ta3+): 1s22s22p63s23p64s23d104p65s24d105p66s04f145d2. (NOTE: again, remove 6s electrons first, then 5d electrons) Slater grouping: (1s2)(2s2,2p6)(3s2,3p6)(3d10)(4s2,4p6)(4d10)(4f14)(5s2,5p6)(5d2)(6s0,6p0); For 4f, σ = 13(0.35) + 46(1.00) = 50.55 (remember: for a f shell, all e-s to the left shield 1.0). Zeff = 73 – 50.55 = 22.45. Note: Don’t subtract 3 electrons from Z!!! (3) (12 points) Give a relative ordering of the following series of atoms for

(a) ionization energy, IE, and (b) electron affinity, EA. Give your reasoning for each ordering (don’t just look up these values in tables!!).

He Se Ba As Cl K a) IE ordering, using the horizontal and vertical trends would be: IE(He) > IE(Cl) > IE(Se) > IE(As) > IE(K) ~ IE(Ba) But it is actually: IE(He) > IE(Cl) > IE(As) > IE(Se) > IE(K) ~ IE(Ba) Arsenic has a higher IE than selenium, due to the greater ease of removing a 3p4 electron than a 3p3 electron (electron pair repulsion and the half-filled shell stability effects). Potassium and barium are a tough call: Ba is to the right, but two rows lower, so this is conflicting. If you say the horizontal trend is more important we’d conclude IE(Ba) > IE(K), but if the vertical trend magnified by two rows is more important we’d conclude IE(K) > IE(Ba). I’ll accept either conclusion, as long as you discuss the issue. b) EA follows the IE trend with some key differences: He has no place to put an extra electron (other than the next principle quantum shell, which is much higher in energy), so it has the lowest value of EA. Also, adding an electron to As disrupts a half-filled p subshell and introduces electron-pair repulsion. Regarding EA(K) versus EA(Ba), BA must place an extra electron into a higher energy d subshell, so it should have a low EA. Not as low as He though, since the energy gap in helium to place an electron in the n=2 principle shell is much greater than the subshell gap between 6s and 5d. EA(Cl) > EA(Se) > EA(As) > EA(K) > EA(Ba) > EA(He)

(4) (12 points) Consider the reaction of hydrazine to produce ammonia and H2, as shown below. Using what you know about bond strength trends, predict whether this reaction would be enthalpically favored. Explain your answer carefully.

To qualitatively assess whether this reaction is exothermic (not quantitative- I’m not asking for you to look up actual thermodynamic data, but to use the general trends discussed in class), look at the bonds broken versus the bonds made. If the bonds made are stronger than the bonds broken, then the reaction will be exothermic and visa versa. Catalog the bonds: Bonds broken: two N-H bonds Bonds made: one N-N bond, one H-H bond. The vertical trend predicts that H-H is stronger than N-H, so from this the reaction might be exothermic, but N-N bonds are anomalously weak due to lone pair repulsion between two closely-bonded N atoms, and should be much weaker than an N-H bond. We predict that the difference between N-H and N-N will be quite large, so overall the reaction should be endothermic.

(5) (14 points) For the following molecules, give at least two reasonable

Lewis structures (not necessarily resonance structures). Assign formal charges for each atom, and state which structure is most likely. Show all valence electrons!

a) ONSi - The most reasonable framework is with Si in the center, since it is in the carbon group and is to the left of N andO. Total valence e-‘s: 6 + 5 + 4 + 1 = 16, or 8 EP’s. Form a framwork, subtract 2 EP’s for the bonds, leaving 6 LP’s. Add 3 LP’s to the terminal N and O. The center atom Si has only 4 e-‘s, so it will need to form two extra bonds: and Calculate the FC’s: For Si, 4 – (4 + 0) = 0, O(left): 6 – (2 +2(2)) = 0; O(right): 6 – (1 + 3(2)) = -1. N(left): 5 – (2 + 2(2)) = -1; N(right): 5 – (3 + 2(1)) = 0 So: and Since oxygen is more electronegative than N, the Lewis structure that places a negative formal charge on oxygen, the right one, is most favorable. b) S4N3

+ (a cyclic compound) The simplest cyclic compound is a ring of atoms, with 7 bonds connecting them in the framework. Why not use it? Total valence e-‘s: 6(4) + 3(5) - 1 = 38, or 19 EP’s. Subtract 7 BP’s, leaving 12 LP’s to distribute on 7 atoms. Place 2 LP’s on 5 of the atoms (doesn’t matter which- we’ll look at FC’s afterwards), and one LP on two atoms. We’ll need to share LP’s from adjacent atoms to make those atoms have an octet. For FC’s, use the shortcut: atoms making their “normal” number of bonds are neutral, those making an extra bond are +1, and those making one less bond are -1. S “needs” two bonds, N “needs” three bonds. I’ll make two rings with the atoms in a different order; there are many options:

0

+1

N

S

S S

S

NN

+1

-1

0 0

0

0

+1

N

S

N N

S

SS

0

0 0

0

0

The Lewis structure on the right has lower FC’s, so it is more favorable.

O Si NO Si N

O Si N

0 0 -1O Si N

0 0-1

(6) (16 points) For the following molecules, use VSEPR to predict the ideal geometry, draw the molecular geometry, and describe any expected distortions.

a) SeIBrClF Total valence: 6 + 4(7) = 34, or 17 EP’s. Place Se in the center, make 4 bonds to the halides. 17 – 4 = 13 LPs, 3 on each terminal halide, leaving one for the central Se. This is an AX4E ideal trigonal pyramidal geometry. Place the largest groups in the equatorial positions: the LP, the I, and the Br. Distort away from the equatorial LP: distort: b) BrXeF4

+

Total valence: 8 + 5(7) - 1 = 42, or 21 EP’s. What to put in the center? This is an odd case, since Xe, a noble gas, is furthest to the right but it is furthest down and can make more than one bond. Make 5 bonds to the halides. 21 – 5 = 16 LPs, 3 on each terminal halide, leaving one for the central Xe. This is an AX5E ideal octahedral geometry. Place the largest groups trans to each other: the LP and the Br. Distort away from the LP:

XeF

F F

FBr distort:

Xe

F

F F

FBr

SeBr

I

F

Cl

SeBr

I

F

Cl

(7) (16 points) It is a fact that if you fill a subshell for any principle quantum number n (s, p, d, or f), the filled subshell will have spherical symmetry. Consider the total wave function for the 2p orbitals, ψ2(total) = ψ2(2px) + ψ2(2py) + ψ2(2pz), and prove this statement mathematically in this particular case (HINT: use the angular parts of the equations given in Table 2.3, and recall what defines a function with spherical symmetry)

To prove that the shape of the total subshell is spherical, you must show that the angular dependence on both θ and φ drops out of ψ2(total). Since atomic wave functions can be separated into radial and angular portions, we need only consider the angular wavefunctions for p orbitals. The polar coordinate functions which are listed in Table 2.2 for the px, py, pz orbitals have the following forms: ψ(2px) = sin(θ)cos(φ), ψ(2py) = sin(θ)sin(φ), ψ(2pz) = cos(θ), ψ2(total, angular) = ψ(2px)2 + ψ(2py)2 + ψ(2pz)2 = {sin(θ)2cos(φ)2 + sin(θ)2sin(φ)2 + cos(θ)2} = {sin(θ)2[cos(φ)2 + sin(φ)2 = 1] + cos(θ)2} = {sin(θ)2 + cos(θ)2 = 1} Since all dependence on angle drops out, this is equivalent to saying that the sum of the squares is a spherical function.

Point Group And Character Table Practice Problems

Here are some molecular symmetry practice problems. Notice that you actually need to remember how to derive the shapes of some of these molecules- an essential skill that I insist on harping about!

1) Determine the point group for the following molecules:

a) NBr3

Just like NH3’s shape:

A C3 primary axis, vertical mirrors, no h or C2’s, so C3v

b) SeCl6

Do VSEPR, determine that this is a AX6 molecule, so octahedral: Oh

c) CCl4

Do VSEPR, determine that this is a AX4 molecule, so tetrahedral: Td

d) H2C=CH2

Draw the molecule:

You should see a C2 axis, and then find one of the two perpendicular C2’s.

This is a flat molecule, so the plane of the page is a mirror, and it is h (not

v!). Thus, this is D2h

e) H2C=CF2

Draw it:

Now we don’t have C2’s, just one C2 and vertical mirror planes: C2v

f) ozone

Do a VSEPR analysis of ozone (O3), determine the central oxygen is AX2E, so a bent molecule:

A C2 axis bisects the O-O bonds and runs through the central oxygen. O3 is flat, so there is (are!) vertical mirror planes: C2v

g) SCN-

This is a linear molecule with different ends, so it is Cv

h) PCl5

Do a VSEPR analysis: this is AX5, a trigonal bipyramidal molecule:

The equatorial PCl3 plane is a h, we

have C2’s, so D3h

i) Cl3PO.

Do a VSEPR analysis: the central P is AX4, so based on a tetrahedral ideal geometry:

A C3, vertical mirros, so once again C3v

2) Find the point group for the following molecules or ions:

a) FeBr3Cl34- (two isomers!)

Only two isomers: both sets of Three ligands meridional, or both facial. The point groups are C2v and C3v, respectively

b) (ignore H atoms)

Trim away the H’s and look at this: This molecule is D4h

c)

A C4, C2’s, a h, so D4h

d) (ignore Ph groups)

A C2, no C2’s, a v, so C2v

e)

Hard to see unless you look down the right direction: there is a C3 axis,

C2’s, but no mirrors, so D3

f)

The plane of the page is a mirror plane, and there is a C2 perpendicular to

that: C2h

g) a cube:

Once you realize there are multiple C4 axes, it is a special point group: Oh

h)

No perpendicular C2’s, so C3h

i)

As drawn there are no symmetry elements: C1

If you rotate about the C-C bond, you will add a mirror plane between the

two ends, and the point group symmetry is Cs:

j)

As drawn you might miss the mirror plane. Simply rotate so that the unique

Cl atom is in the plane of projection and you’ll see that this is Cs

3) Determine the point group for the following objects:

a) The peace sign: Yet another C2v object

b) The Mercedes-Benz sign:

If the symbol is the same on the backside, there will be a

horizontal mirror, C2’s, and the point group will be D3h. If not, we will

have the C3 and some v’s so the point group would be C3v. When in

doubt, ask about the object in question.

c) An interlaced Jewish star:

The interlacing “destroys” the mirrors and perpendicular C2’s, so all we

have left is the C6 rotational axis: C6.

d) Adamantane (might need a model):

If you look down the 4 points with tertiary carbons, you’ll see they are

actually C3 axes. The only point group that has multiple C3’s (and no

C4’s) is Td

e) The Campanile (Berkeley campus) (ignore the clock faces):

This clock tower has a pointy end (the top) and four sides. If we

ignore the clocks we can see each side has a vertical mirror, so

C4v

Here is a nice website with practice problems, where the molecules are

linked to a 3D graphics program so you can rotate things around to look

them more easily:

http://www.scs.uiuc.edu/~chem315/pointgrouppractice.htm

4) Using character tables to determine the symmetry properties of

molecular bonds, vibrational modes, etc., takes practice. There are

some problems in the textbook to practice on; here’s another one.

Follow the procedure from the lecture notes, step by step.

A new molecule EH3 is prepared. It’s structure (either trigonal planar

or pyramidal) is not known. An infrared spectrum, however shows

two bands for the E-H bond vibrations. From a symmetry analysis of

these two possible geometries, which structure does this molecule

adopt? (HINT: don’t do a full 3N-6 analysis- just consider the

vibrating E-H bonds).

We have two geometries to consider- need to know their point groups. A planar

EH3 is D3h, and a pyramidal EH3 is C3v. Do vibrational analysis in D3h first:

Consider the arrows, representing E-H bond stretches.

Look at D3h character table and consider the action

of each operation.

E: all 3 arrows stay the same

C3: all arrows move, so you get zero for the character

C2: each C2 is along one arrow, so 1 stays the same

h: all three arrows are in the mirror plane, so all stay the same

S3: all arrows still move; the extra reflection doesn’t change this.

v: each vertical mirror contains one arrow, so 1 stays the same.

Thus, vib = 3 0 1 3 0 1

Use vib = 3 0 1 3 0 1, the character table, and the decomposition formula

to decompose vib into its irreducible representations:

n(A1’): 1/12[1(1)(3) + 2(1)(0) + 3(1)(1) + 1(1)(3) + 2(1)(0) + 3(1)(1)] = 1

n(A2’): 1/12[1(1)(3) + 2(1)(0) + 3(-1)(1) + 1(1)(3) + 2(1)(0) + 3(-1)(1)] = 0

n(E’): 1/12[1(2)(3) + 2(-1)(0) + 3(0)(1) + 1(2)(3) + 2(-1)(0) + 3(0)(1)] = 1

n(A1”): 1/12[1(1)(3) + 2(1)(0) + 3(1)(1) + 1(-1)(3) + 2(-1)(0) + 3(-1)(1)] = 0

n(A2”): 1/12[1(1)(3) + 2(1)(0) + 3(-1)(1) + 1(-1)(3) + 2(-1)(0) + 3(1)(1)] = 0

n(E”): 1/12[1(2)(3) + 2(-1)(0) + 3(0)(1) + 1(-2)(3) + 2(1)(0) + 3(0)(1)] = 0

So, vib = 1A1’ + 1E’. Look at the table, see that only E’ is IR-active. Thus, we

conclude that there would be only one E-H stretching IR band.

Now look at pyramidal EH3:

Consider the arrows, representing E-H bond stretches.

Look at C3v character table and consider the action

of each operation.

E: all 3 arrows stay the same

C3: all arrows move, so you get zero for the character

v: each vertical mirror contains one arrow, so 1 stays the same.

Thus, vib = 3 0 1

decompose vib into its irreducible representations:

n(A1): 1/6[1(1)(3) + 2(1)(0) + 3(1)(1)] = 1

n(A2): 1/6[1(1)(3) + 2(1)(0) + 3(-1)(1)] = 0

n(E): 1/6[1(2)(3) + 2(-1)(0) + 3(0)(1)] = 1

So, vib = 1A1 + 1E. Look at the C3v table, see that both are IR-active. Thus, we

conclude that there would be two E-H stretching IR bands. Thus, we decide that

this molecule is pyramidal.