Problem Set #2

Problems Set #2
(Level 1 problems)

**PLEASE NOTE!!!**

Due to problems with accessing ChemDraw on my home computer, the answers posted 2-24-08 at 10PM will not have any graphics to illustrate the answers. These will be added later and you will be notified.

1. For each of the molecules or ions below, determine the geometry of the molecule and assign the point symmetry.

a) TeCl4 -- (AX5 type with one lone pair) molecular geometry = see saw, point group = C2v

       b)   TeF3 - (AX5 type with one lone pair, one unpaired e-) molecular geometry = T-shaped , C2v or trigonal pyramidal ), C3v depending on where you put the electrons
       c)   XeF4 -- (AX6 type with two lone pair) molecular geometry = sq. planar, D4h
       d)   XeO3 -- (AX4 type with one lone pair) molecular geometry = trigonal pyramidal, C3v
       e)   ClO2F3 -- (AX5 type with no lone pair) molecular geometry = trigonal pyramidal, either D3h (trans =O) or C2v (cis =O in eq. plane, best choice)

2. a) Draw the Lewis structure (including any resonance structures) for sulfite, SO32-. This should show an AX4 type with one lone pair having trigonal pyramidal molecular geometry. 3 resonance structures delocalize one S=O bond and 2 S-O bonds.
      b) Assign the point group to sulfite. = C3v
      c) Orient sulfite in a Cartesian coordinate system.The z-axis is parallel to C3.
      d) What hybridization at the central atom in sulfite will be used (according to VSEPR) to produce the correct geometry? The AX4 type shape requires sp3 hybridiziation using VSEPR theory.
      e) What valence atomic orbitals on Sand O were not used in (d) ? Will these contribute to any part of bonding in these molecules and if so, how? Thus Sulfur uses its 3s and 3px,py,pz orbitals to make S-O bonds. Leftover on S are the 3d orbitals. On O, it can use simple one 2p orbital to make the S-O bond, leaving 2s adn two 2p's unused. THe Sd adn the unused O p's can be used to form the pi bonds.

3. For each of the three molecules: NO2- , AsF3(CH3)2 , PF3

Draw its structure
Assign its point symmetry
Orient it in Cartesian space.

NO2-, nitrile ion: bent planar, C2v, z aligned with C2, (best putting molecule in yz plane)

AsF3(CH3)2: trigonal bipyramidal, cis eq. methyls, C2v, z aligned with C2.

PF3: trigonal pyramidal, C3v, z aligned with C3.

4. Using the MO diagram for N2, determine the following for nitric oxide, NO:

number valence electrons = 11
bond order = 2.5
predict its chemical reactivity : high because it has one unpaired electron in an antibonding orbital. It is a radical.

5. What are the point groups of the molybdenum molecules below?

MoOCl4 = C4v

Mo(CO)6 = Oh

Mo(benzenedithiolate) = C2v

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Problems Set #2
(Level 2 problems)

6. A hybrid orbital Y1 is described by the linear combination below:

Phi = 0.77 s + 0.45 px + 0.45 py

What is the percent atomic orbital contribution to this hybrid?

60% s and 20% each px and py

The two hybrids Phi(a) and Phi(b) :

Phi(a) = 0.45 s + 0.63 px + 0.63 py
Phi(b) = 0.45 s + 0.63 px - 0.63 py

would be suitable for making bonds in which molecule below? (circle it)

Both Phi(a) and Phi(b) have 20% s and 40% each px and py, therefore they describe a bent pair of hybrid orbitals with an acute angle.

The molecule AB2 on the far left is the molecule whose drawn shape best matches these hybrids on A.

7. There is a “rule of thumb” that says the angle, f, formed by any three atoms X-A-X in a molecule, the percentage of s orbital used to hybridize atom A can be found using this formula:

cos f (in deg) = % s / (% s-100%)

Using this formula in the questions below.

A molecule AB3 has C3v point symmetry. The angle B-A-B is 107.5.
What is the %s and %p in the hybrid orbitals used to make A-B bonds?

Solving: cos 107.5 = %s / (%s-100%) gives 23.5 % s. Therfore, the remainder of the "sp3" type hybridization in a trigonal pyramidal molecule is 76.5 %

Which of the following molecules is most likely to be AB3 ?

BCl3 CBr3 PF3 or NH3

Since and ideal sp3 hybrid would be 25%s and 75% p, this suggests a more acute B-A-B angle. This best fits NH3. The other choices would not fit: BCl3 is planar (not trig. pyr), CBr3 does not exist, PF3 likely has even more acute angles (based on what was observed with SH2)

c) The H-C-H angles in CH4 , CH3F , CH2F2 , are 109.5, 111 and 112 deg respectively. What is the % of s- and p-orbitals used in the C-H bonds as compared to C-F bonds in each compound? Speculate a reason why this is so.

Using the same approach:

C-H bond %s . . . C-H bond %p . . . C-F bond %s . . . C-H bond %p . .

CH4 . . . . . . 25% . . . . . . . . . . 75% . . . . . . . . . . .0% . . . . . . . . . 0.%

CH3F . . . . . 26.3% . . . . . . . . .73.6% . . . . . . . . 21.1% . . . . . . . 78.9%

CH2F2 . . . . 27.3% . . . . . . . . .72.7% . . . . . . . . 22.7% . . . . . . . .77.3%

As more F is added to CH4, the HCH angles increase; therefore the F-C-H(F) angles must decrease.

In both F-compounds, there is a larger % of p (and less %s) in the C-F bonds than in the C-H bonds. This makes sense with the angles, since more %p causes more acute angles.