ENGR321: Midterm One Winter 2002

1) (30 points) Use the section of the periodic table given below to help answer the following questions:

TEST VERSION ONE a) What is the electronic configuration of sulfur (S)? The atomic number of S is Z = 16, so the electronic configuration must hold 16 electrons: 1s2 2s2 2p6 3s2 3p4 b) What is the valence of S? The valence is the number of s and p electrons in the largest n-value shell, in this case the n = 3 shell. The valence is 6. c) Name two IONS that have the same valence as S. There are lots of choices, but the trick is that the atoms must be either negatively or positiviely charged; neutral atoms are not ions. Some choices are: P-1, Si-2, Al -3, Cl+, Ar+2, N-1, C -2, etc... d) Are these elements, as a whole, electropositive or electronegative? As a whole, this group, on the right hand side of the periodic table, are electronegative. e) Which of these elements have a valence of 7? F, Cl, Br f) Will S prefer to give up or accept electrons in order to bond to another atom? With a valence of 6, S is more than half full, and therefore will prefer to accept additional electrons to complete the n = 3 shell.

TEST VERSION TWO a) What is the electronic configuration of calcium (Ca)? The atomic number of Ca is Z = 20, so the electronic configuration must hold 20 electrons: 1s2 2s2 2p6 3s2 3p6 4s2 b) What is the valence of Ca? The valence is the number of s and p electrons in the largest n-value shell, in this case the n = 4shell. The valence is 2. c) Name two IONS that have the same valence as Ca. There are lots of choices, but the trick is that the atoms must be either negatively or positiviely charged; neutral atoms are not ions. Some choices are: K-1, Na-1, Sc+1,Ti+2, etc... d) Are these elements, as a whole, electropositive or electronegative? As a whole, this group, on the left hand side of the periodic table, are electropositive. e) Which of these elements have a valence of 2? Be, Mg, Ca (plus Sc, Ti and V, since the 3d sub-shell fills before the 4p sub-shell) f) Will Ca prefer to give up or accept electrons in order to bond to another atom? With a valence of 2, Ca is less than half full, and therefore will prefer to give up electrons to close the n = 3 shell.

TEST VERSION ONE: First, calculate the percent ionicity using the Pauling ionicity equation: We know it will NOT BE IONIC. Now we must check the valence: Al: valence = 3, and Mg: valence = 2. Both are less than half full, so they will both tend to bond metallically. Thus our final answer is: METALLIC

TEST VERSION TWO: First, calculate the percent ionicity using the Pauling ionicity equation: Only partly ionic. What is the rest? Check the valences: Be: valence = 2, and S: valence = 6. Since S is more than half full, it will prefer to bond covalently, so the final answer is: About 22% ionic and 78% covalent.

TEST VERSION ONE: The coefficient of thermal expansion is related to the symmetry of the energy well. Deeper, more symmetric wells will have a smaller COTE. For a lerger COTE, we want a shallower and more asymmetric well.

TEST VERSION TWO: Elastic modulus is related to the bonding force curve as . To get a smaller modulus, we need a smaller curvature of the bottom of the well; a less sharp well (this also usually means the well is not as deep).

TEST VERSION ONE:

TEST VERSION TWO:

TEST VERSION ONE: Calculate and compare the linear density along the [111] direction in FCC and BCC crystals. Along the [111] direction in BCC the atoms are close packed (they touch one another) so that the linear density is 100%. For FCC, it helps to draw out the diagonal plane: The [111] direction in FCC has two atoms lying on it, for a total atomic length of 2R. The length of the [111] direction is the length of the body diagonal = sqrt(3) a = sqrt(3) sqrt(2) 2R. The linear density is therefore: The linear density along [111] in BCC is much higher than along [111] in FCC.

TEST VERSION TWO: Calculate and compare the linear density along the [011] direction in FCC and BCC crystals. Along the [ 011] direction in FCC the atoms are close packed, so that the linear density is 100%. For BCC, it helps to draw out a face plane: The [011] direction in BCC is a face diagonal with two atoms lying on it, for a total atomic length of 2R. The length of the [011] direction is the length of the face diagonal = sqrt(2)a = 4Rsqrt(2)/sqrt(3). The linear density is therefore: The linear density along the [011] in FCC is much larger than along [011] in BCC.

7) (30 points)

TEST VERSION ONE: Ge (germanium) crystallizes in a cubic structure called the “diamond-cubic” structure. Given the atomic weight of Ge is 72.6, the atomic radius is 122pm, the lattice parameter a = 566pm, and the atomic packing factor is 34%, calculate the density of Ge. To find the density, we need to know the number of atoms per unit cell inthe diamond-cubic crystal. Everything else is given in the problem. We can find the n from the APF equation.

TEST VERSION TWO: Si (silicon) crystallizes in a cubic structure called the “diamond-cubic” structure. Given the density of Si is 2.33 g/cm^3, the atomic weight is 28.09, the atomic radius is 117pm, and the lattice parameter a = 543pm, determine the atomic packing factor for Si. To find the APF, we need to know the number of atoms per unit cell in diamond-cubic crystals. We can find this by inverting the density equation: