EXAM 1 - CHEM 231 - 134 points total

1.  Identify the point group of an object in the cases below. Note that more symmetry may be required to exist besides what is specifically listed in each case.

a) Case 1: you find C4, C3 and C2 axes but no mirror planes and no inversion center.

The presence of both a C4 and a C3 *requires* that there are actually 3 x C4 axes and 4 x C3 axes, and this immediately leads to an 'O' class of poitn groups. Since there is no inversion and no mirror planes identified, the point group is simply 'O'.

b) Case 2: you find a C4 axis and, on a different axis, a C2 axis, and a mirror plane parallel to C4.

The presence of one C4 and a C2 on a different axis is only possible if the C2 is perpendicular to the C4. There must then be 4 such C2 axes, puttign this item in the 'D' class of groups. The presence of a mirror plane parallel (but not perpendicular) to the C4 makes this the D4d point group.

c) Case 3: you find three mutually perpendicular mirror planes and an inversion center

The presence of three mutually mirror planes occurs in the D2h point group. Some of your drew 3 such planes and deduced an Oh point group, but three mirrors can exist (as they do in a D2h point group) without requiring any C4 or C3 axes.

d) Case 4: you find a C5 axis and a C3 axis and note the object has inversion symmetry.

The presence of both a C5 and a C3 axis, in addition to inversion *only * happens in one point group -> Ih

e) Case 5: you find a C3 axis and a perpendicular mirror plane

The presence of both one C3 and a horizontal mirror is all that is required to define the C3h point group.

2. This question concerns the hypothetical molecule PF2(CH3)3.  You may consider that there is free rotation around the methyl groups (so that they behave as big spheres).

a) How many isomers are possible for this molecule?

Three of symmetries D3h, C2v and Cs. (click to see structures)

b) Demonstrate whether infrared spectroscopy could be used to distinguish these isomers by observing the P-F vibration n(P-F).

The D3h isomer is predicted to exhibit just one absorption in the IR, while the other isomers will both show 2 P-F absorptiokns. So th D3h isomer could be identofoed conclusively, but not the C2v vs Cs isomers. (click to see details)

3.The H-C-H angles in CH4 and  CH2F2 ,  are 109.5, and 112 deg respectively. 
a) What do these angles suggest about the % s- and p-orbital used in the C-H bonds as compared to C-F bonds in each compound? 

The larger H-C-H angle in CH2F2 as compared to CH4 implies the C hybrids used to make the C-H bonds have a greater amount (%) s character and less % p orbital than in CH4. Therefore it must also be true that the F-C-F angle is smaller than the H-C-H angles because it has less (%) s orbital and more p-orbital.

b) There is a formula used to estimate the % s- and p-orbital composition in hybrid orbitals of atom A with two bonds to other atoms X, like this, X-A-X.  The formula is based on the angle, f, of the unit X-A-X.  The formula specifies the percentage of s-orbital used to hybridize atom A to produce the angle, f, can be found using this formula:

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

Using this formula to calculate the % s- and p-orbital used in the C-H bonds in CH4 and CH2F2 given that CH4 has a H-C-H angle of 109.5 and  CH2F2 has a H-C-H angle of 112 deg.  Comment on whether the results agrees with your answer from (a).

The cos 109.5 = %s-orbital/ (% s orbital -100%) solves to give 25% s, as expected for a perfectly tetrahredrally hybridized C atom in CH4.

The cos 112 = %s-orbital/ (% s orbital -100%) solves to give 27.5% s, a larger amount, in agreement with the prediction in part (a)

4. Find your Table of Valence Orbital Energies (one of the hand-outs I gave in class). Use this to predict what element will make the most covalent bonds with Ge, germanium (other than to itself, of course).  Explain your choice of “most covalent element” making bonds to Ge.

Covalency is correlated to a close match in the energies of the atomic orbitals that make moelcular orbitals. Silicon has 3s and 3p energies of -14.9 and -7.7 eV, respecively , closest to the 4s adn 4p orbitals of Ge (-15.6 adn -7.6 eV) and by this criterion, is expected to make the most covalent bonds.

5. In a few brief sentences, explain concisely the connection between the following pairs of terms:

a) MOT and electronic (UV/vis) spectroscopy

Only MOT creates empty anti-bonding orbitals that can be used to "explain" where an electron goes when excited to higher energies by absorption of light energy in electronic (UV/vis) spectroscopy.

b) VSEPR and resonance

The localized nature of bonding described by VSEPR is unable to be used to describe molecules with delocalized electronic structures. Resonance structures are multiple equivalent ways of drawing bonds according to VSEPR which when used together, approximate a delocalized, or partial double bonded character, for certain molecules like benzene or nitrate.

c) % orbital composition and conservation of orbitals

The conservation of orbitals requires that when atomic orbitals are used as a basis set of functions to make new hybrid oirbitals or molecular orbitals, the number of new hybrids or MO's must equal the dimension, i.e. the number, of AO's. It also means that for any one AO, it must be used completely. The important consequence is that if one hybrid or MO has 80% s used, then only 20% s orbital remains for the other hybrids. This explains the different and unequal compositions of MOs and hybrid orbitals in less symmetrical molecules.

d) ionization energy and orbital energies

Atomic orbital energies = - ionization energies.

The more electronegative an element, the lower in energy will be its AO's.

e) orbital energies and covalency

The maximum covalency results from mixing atomic orbitals of similar energy.

f) molecular angles and % orbital composition

The angles in molecules AXn of related geometries can be adjusted by the proportions of s and p (and d) orbitals. Generally, addition of a greater % s character serves to increase X-A-X angles, while conversiely, increasing the % p orbital decreases the X-A-X angles.

6. In this (very lengthy) question, you are asked to complete the preliminary steps towards generating the MO diagram for nitrite using the procedure from class (used for the water molecule).  Note that you are not asked to generate the MO diagram however. 

Nitrite is an anion,  [NO2]-.   Your goal is to determine:

  • how many MO’s will be formed (using only the valence atomic orbitals)
  • what are the symmetries of these MO’s
  • the general expressions for the LCAO’s that will form these MO’s.  You do not need to provide any coefficients!
  • Sketches of the MO’s described by the LCAOs in (c).  For (c) and (d), note that some of the LCAOs are straightforward, other are difficult because they mix ≥ 3 orbitals. I do not expect you to get all of these but try to specify as much as possible.
  • click here and here for two diagrams of answers to the above questions

    For the MO’s (i through viii) that were calculated for nitrate using the program Gaussian (assuming NO2- was oriented in the yz plane. If yours is oriented in the xz plane, B1 and B2 labels are switched):

    i. sigma bonding, A1

    ii. sigma bonding, B2

    iii. sigma non-bonding, A1

    iv. pi bidning, B1

    v. pi non-bonding, A2

    vi. pi anti-bonding, B1

    vii. sigma non-bodning, B2

    viii. sigma anti-bonding, B2