1. Describe what the following symmetry operations do and give the symbols that represent them: (a) proper rotation - simple rotation about an axis passing through the molecule by an angle 2p/n where n is an integer, and is symbolized by "Cn". (b) reflection - reflection of all atoms through a plane that passes through the molecule, and is symbolized by "s". (c) inversion - reflection of all atoms through a point in the molecule, and is symbolized by "i". (d) improper rotation - combination of a rotation (2p/n) and reflection of all atoms through a plane that is perpendicular to the axis of rotation, and is symbolized by "Sn". 2. Which uppercase letters in the Roman alphabet contain a symmetry axis of proper rotation? We can also add the letter "K" if written in the following manner: 3. Which of the following molecules have a three-fold (C3)State where the axis (or axes) lies. CH4 has 4 independent C3 axes, one along

PF5 is a trigonal bipyramid (two pyramids sharing a triangular base) molecule. There are two types of F in this structure: 2(axial) and 3 (equatorial). There is one C3 axis, it lies along the Faxial-P-Faxial HCl, being a linear molecule contains a rotational axis ofparts to home ac unit infinite order along the internuclear axis.ac unit to small for house IF5 has six pairs of valence electrons surroundingac unit suddenly stopped working the I atom, one of these pairs is unshared while the other five are shared with the 5 F atoms. IF5 has a square pyramidalThere are two types of F atoms: 1 (axial) and 4 (base). IF5 has a four-fold (C4) rotational symmetry. Its axis lies along the I-Faxial bond.

does not have three-fold rotational symmetry. 4. Which of the following molecules are disymmetric? the symmetry operation that is present which makes the molecule not disymmetric (e.g. reflection, inversion, improper rotation) and show where this symmetry element (plane for reflection, point for inversion, axis for rotation) resides in the molecule. (a) CH2Cl2 -not disymmetric since it contains two mirror planes. One of these planes contain CH2, the other contains CCl2. (b) CHCl3 - not disymmetric since it contains threeThere is a mirror plane for each Cl with the C-H (c) glycine - not disymmetric since it contains one mirrorThis plane contains the central carbon, the carboxyl atoms and the amino group. (d) alanine - disymmetric or chiral. 5. Enumerate all the symmetry elements present in the molecule C3 axis perpendicular to the plane of the molecule. Mirror plane containing all atoms. 3 mirror planes (all coincidental with the C3 axis

and one for each B-F bond) 3 C2 axis perpendicular to the C3 axis (one on each B-F bond) S3 axis (parallel and same position as C3) 6. Calculate the net number of spheres in (a) a primitive cubic(b) a body-centered-cubic unit cell; (a) primitive cubic unit cell - lattice points are cornersA cube has eight corners, eight lattice points per cell, eight corner spheres per cell. Each corner sphere is shared by eight unit cells. Thus, only 1/8 of a corner sphere can be assigned exclusively to a unit cell. 8 corner spheres, each one giving 1/8, equal one sphere per unit cell. (b) body-centered-cubic unit cell - similar to primitive cell except for body centered sphere which belongs fully to the unitThus, 2 spheres per unit cell. (c) face-centered-cubic unit cell - similar to primitive cell except for lattice points at the center of each face of the cube. A cube has 6 faces, thus, a unit cell will have 6 face-centered

One half of a face-centered sphere can be regarded as belonging to a unit cell. Thus, there are 4 (3 from the face-centered spheres, 1 from the corner spheres) spheres per unit cell. 7. Copper crystallizes in a face-centered cubic unit cell that has an edge length of 3.61 Angstroms. Assuming that the atoms touch each other along the face diagonal, calculate the atomic radius of copper and the density of copper metal. The length of a face diagonal (from Pythagorean theorem) is 1.414 a where a is the length of the edge of the cube. touch each other along the face diagonal, thus the length of a face diagonal is equal to (radius of one corner atom plus twice the radius of the face-centered atom plus the radius of the atom on the opposite corner) 4 r. 4 r = 1.414 a r = 0.3535 a = 0.3535 (3.61 A) = 1.28 Angstroms The density of the metal can be obtained by finding the number of atoms per unit cell, multiplying this number by the mass of

each atom (A.W. / Avogadro's number), and dividing the result by the volume of the unit cell (a3): density (g/mL) = n (A.W. g/mole) / (NA)(a3) where n = atoms per unit cell, A.W. is in g/mole, NA is Avogadro's number and a is the length of an edge in centimeter. Density (g/mL) = 4 atoms (63.546 g/mole) / (6.02 x 1023 atoms/mole) (3.61 x 10-8 cm)3 8. Clausthalite is a mineral composed of lead selenide, PbSe. The mineral adopts a NaCl-type structure. The density of PbSe at 298 K is 8.27 g/cm3. Calculate the length of an edge of the PbSe unit cell. In the NaCl-type structure, there are 4 cations and 4 anionsThus, in lead selenide, there will be 4 PbSe perFrom the previous problem, one can see that: a3 = n (F.W. g/mole) / (NA) (density where F.W. stands for the formula weight of the salt. a3 = 4 PbSe units (286.2 g/mole) / (6.02 x 1023 PbSe units / mole) (8.27 g/mL) 9. One form of crystalline iron has a body-centered cubic latice

with an iron atom located at every lattice point. at 25oC is 7.86 g/cm3. The length of the edge of the unit cell is 2.87 Angstroms. Use these facts to estimate Once again, we begin with the following: NA = n (A.W. g/mole) / (density(g/mL))(a3) = 2 atoms (55.847 g/mole) / (7.86 g/mL) (2.87 x 10-8 10. In a typical X-ray crystallography experiment, X-rays of wavelength 0.71 Angstroms are generated by bombarding molybdenum metal with an energetic beam of electrons. Why are these X-rays more effectively diffracted by crystals than is visible light. Remember the diffraction grating that you used last semester in studying the spectra of various elements (hydrogen, nitrogen,The most effective diffraction of light occurs when the distances between slits are comparable to the wavelength ofThis comes directly from Bragg's diffraction law: nl = 2d sin q where l is the wavelength and d is the separation between slits.