NAME:_____SOLUTIONS_______ Chem 51, fall, 2002, third midterm exam
A periodic table of the elements with useful constants is included as an insert. You may use the back for scratch work but enter all work to be graded in the space provided with each question. Show work in all problems involving computation to receive credit. This examination has 6 questions on four pages and is worth a total of 150 points..
1) (45 points) This problem deals with isomers with the molecular formula CH3NO2.
There are many structural isomers of compounds with this molecular formula. Two solutions are provided on Figure 1.
a) Draw an acceptable Lewis electron dot structure (show the dots) of one of the isomers. Provide the formal charge and hybridization of each heavy atom in your structure.
b) Provide estimates of the bond angles of the isomer whose structure you have provided.
c) Select two bonds in your structure that do not involve hydrogen and describe them in terms of overlap of orbitals. Are they sigma and/or pi bonds?
Solution I on Figure 1.
Two resonance structures are possible. One resonance structure is shown on the figure. The C-N single bond is a sigma type bond which is an overlap of a C sp3 hybridized orbital with a N sp2 hybridized orbitals. In the Lewis structure shown, there is a sigma type N-O single bond which is an overlap of a N sp2 hybridized orbital with an O orbital, either an unhybridized orbital or a sp3 hybridized orbital. The N-O double bond consists of a sigma type bond which is an overlap of a N sp2 hybridized orbital and a N sp2 hybridized orbital and of a pi type bond which is an overlap of an O unhybridized p orbital with a N unhybridized p orbital. If one combines the other resonance structure, one concludes that the two N-O bonds are equal and are halfway between a single and double bond.
Solution II on Figure 1
Only Lewis structure with zero formal charges on the atoms is shown. A second resonance structure with formal charges is possible. This is not shown but contributes to the full picture. This second resonance structure has the consequence that the C-N bond has partial double bond character and all heavy atoms lie in the same plane. The following description applies to the structure shown. There are two C-O bonds. The C-O single bond is a sigma type which is an overlap of a C sp2 hybridized orbital with an O sp3 hybridized orbital. The C-O double bond consists of a sigma type bond, an overlap of a C sp3 hybridized orbital with an O sp2 hybridized orbital, plus a pi type bond, an overlap of a C unhybridized p orbital and an O unhybridized p orbital. The C-N single bond is a sigma type bond, an overlap of a C sp2 hybridized orbital with a N sp3 hybridized orbital.
2) (20 points) The dividing line between metals and non-metals is a zig-zag line at the right hand of the periodic table. Use your knowledge of electronic structure to justify the location and orientation of the line (i.e. it is tilted and is neither horizontal nor vertical).
Atoms are described by several atomic parameters such as electronegativity and electron affinity but the ionization energy is most instructive as elements with low ionization energies are metals. There is a general trend of increasing ionization energy as one moves across a row. Metals are found to the left and non-metals to the right. However, the ionization energy decreases as one goes down a column. Consequently, the transition from metals (low ionization energy) to non-metals (high ionization energy) occurs at higher atomic number as one moves down the table. These two effects account for the zig-zag nature of the boundary.
3) (15 points) There is a general trend in ionization energies across a row in a periodic table. The first ionization energy of boron is anomalous.
a) Use your knowledge of electronic structure to explain why.
The electron added last in the Aufbau procedure and the first to be ionized is in a 2p orbital. Although the nuclear charge increases by one unit from Be to B, the 2p electron sees less of the nuclear charge than the 2s because of the centrifugal force resulting from the angular momentum of the 2p electron.
b) Suggest an element whose third ionization energy is anomalous for the same reason.
The third ionization energy is the energy required to remove an electron from the cation M+2. The M+2 species must have the configuration (1s)2(2s)2(2p) and therefore M would have the configuration (1s)2(2s)2(2p)3. The element is nitrogen. Other elements in the nitrogen family such as phosphorus would also be acceptable.
c) What consequences might your explanation have for the electron affinity of beryllium?
The electron affinity of beryllium is the negative of the ionization energy of Be-. Be- and B are isoelectronic. If it is relatively easy to ionize B, it will be even easier to ionize Be-. A greatly reduced electron affinity is expected. In fact, it is positive.
4) (20 points) What type of bonding (ionic or covalent) would you expect in manganese(VII) oxide? Briefly support your answer and predict the physical properties of the compound.
A simplistic understanding of chemistry based on rules of thumb argues that all compounds of metals and non-metals are ionic. However, many but not all are. Ionic bonding is only possible with a favorable energy balance. The energetic cost to remove seven electrons to achieve an oxidation number is prohibitively high and an ionic bond is not possible. The compound does exist so the bonding must be covalent. Although the intramolecular forces are strong, the intermolecular forces are weak. The compound will have a low melting and boiling points and will be soluble in a non-polar solvent such as hexane. The liquid will be a non-conductor of electricity as the mobile species have zero net charge.
5) (20 points) A highly excited helium atom decays from a state with an energy of 3.87862 x 10-18 J/atom to one with an energy of 3.80283 x 10-18 J/atom. Calculate the wavelength and momentum of the photon emitted in this decay process.
The energy loss of the atom as it decays between the two states equals the energy of the photon produced by the transition.
D E = Eg = (3.87862 – 3.80283) x 10-18 J = 7.579 x 10-20 J/photon
From the relation E = hn , one obtains n = (7.579 x 10-20 J)/(6.626 x 10-34 J-s) =
1.1428 x 1014 Hz (1 Hz = 1/s).
From the relation l n = c, one obtains l = (2.99792 x 108 m/s)/(1.1428 x 1014 1/s) = 2.6210 x 10-6 m.
The momentum of the photon is obtained from de Broglie’s relation: l = h/p so p = (6.626 x 10-34 J-s)/(2.62610 x 10-6 m) = 2.528 x 10-28 Kg-m/s.
6) (30 points) This question deals with the electron with the highest orbital energy in the element arsenic, atomic number 33.
a) The ionization energy of this electron is 9.82 eV (947 kJ/mole). Calculate the effective nuclear charge of this electron and discuss the physical significance of your result.
From its position in the periodic table, the electron with the highest energy is a 4p electron so n =4.
IE = -E = Rze2/n2 so ze = [n2IE/R]0.5 =
[(42)(9.82 eV)/(13.6 eV)]0.5 = 3.40
The atomic number of arsenic is 33. The 28 core electrons will effectively screen the nuclear charge and by themselves reduce the effective nuclear charge by 28 to 5. The actual effective nuclear charge is less. The remaining decrease is achieved by the 5 valence electrons.
b) Will the electron ever be found at the nucleus? Briefly discuss.
The l quantum number for the 4p electron is one. The electron has orbital angular momentum. The associated centrifugal force will prevent the electron from ever reaching the nucleus. In addition, the angular node for each of the p orbitals goes through the nucleus. Both factors lead to the same result. A p electron will not be found at the nucleus.
c) Calculate an estimate in Ångstrom of the average distance of this electron from the nucleus.
<r> = a0n2/ze = (0.529Å)(42)/(3.40) = 2.5 Å.
d) An atom of arsenic is placed in a strong electric field that serves to define the z axis. Are their surfaces where this electron will never be found? If your answer to this question is yes, describe the geometrical characteristics (e.g. an ellipsoid?) of these surfaces. Give the number of each type of surface.
The electron under consideration is a 4p electron. From the value of n, 4, there will be a total of n – 1 = 3 nodal surfaces. The number of angular nodes is given by the l quantum number which is one. The angular nodes for this case, a p orbital, are planes. The remaining two nodes are radial nodes, spherical surfaces.
14 November 2002, WES
