Tutorial No.1 +
−
1. Sodium chloride (NaCl) exhibits predominantly ionic bonding. The Na and Cl ions haveelectron structures that are identical to which two inert gases? Solution: The Na+ ion is just a sodium atom (configuration: 2.8.1) that has lost one electron; therefore, it has an electron configuration the same as neon (configuration: 2.8) (Refer to periodic table, text book Figure 2.6).
The Cl−ion is a chlorine atom (configuration: 2.8.7) that h as acquired one extra electron; therefore, it has an electron configuration the same as argon (configuration: 2.8.8). 2. Explain why hydrogen fluoride (HF) has a higher boiling temperature than hydrogen chloride (HCl) (19.4 vs. -85 °C), even though HF has a lower molecular weight. Solution: The intermolecular bonding for HF is hydrogen, whereas for HCl, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher boiling temperature.
3. Compute the percents ionic character of the interatomic bonds for the following compounds:TiO2, ZnTe, CsCl, InSb, and MgCl2. Solution: The percent ionic character is a function of the electron negativities of the ions X A and X B according to text book Equation 2.10 . The electronegativities of the elements are found in text book Figure 2.7 .
For TiO2, X Ti Ti= 1.5 and X O = 3.5, and therefore,
For ZnTe, X Zn Zn = 1.6 and X Te Te= 2.1, and therefore,
For CsCl, X Cs Cs = 0.7 and X Cl Cl= 3.0, and therefore,
For InSb, X In In= 1.7 and X Sb Sb= 1.9, and therefore,
For MgCl2, X Mg = 1.2 and X Cl= 3.0, and therefore,
4. Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C(600 K). Assume an energy for vacancy formation of 0.52 eV/atom. Solution: In order to compute the fraction of atom sites that are vacant in lead at 600 K, we mustemploy text book Equation 4.1 . As stated in the problem, Qv= 0.52 eV/atom. Thus,
-5
= 4.3 x 10
5. Atomic radius, crystal structure, electronegativity, and the most common valence are tabulated in the following table for several elements; for those that are nonmetals, only atomic radii areindicated.
Which of these elements would you expect to form the following with copper: (a) A substitutional solid solution having complete solubility (b) A substitutional solid solution of incomplete solubility (c) An interstitial solid solution Solution: For complete substitutional solubility the following criteria must be met:
i) The difference in atomic radii between Cu and the other element (ΔR %) must be less than ±15%. ii) The crystal structures must be the same. iii) Theelectronegativities must be similar. iv) Thevalencesshould be the same, or nearly the same. Below are tabulated, for the various elements, these criteria.
(a) Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having completesolubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCCcrystal structure, and thus display complete solid solubility at these temperatures. (b) Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii andthat for Cu are greater than ±15%, and/or have a valence different than 2+. (c) C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantlysmaller than the atomic radius of Cu. 6. Calculate the composition, in weight percent, of an alloy that contains 218.0 kg titanium, 15 kgof aluminum, and 10 kg of vanadium. Solution: The concentration, in weight percent, of an element in an alloy may be computed using amodified form of text book Equation 4.3 . For this alloy, the concentration of titanium (C Ti) is just
Similarly, for aluminum
And for vanadium
7. Calculate the unit cell edge length for an 85 wt% Fe – 15 wt% V alloy. All of the vanadium is insolid solution, and, at room temperature the crystal structure for this alloy is BCC. Solution: It is necessary to employ text book Equation 3.5 (chapter 3) ; in this expressiondensity and atomic weight will be averages for the alloy, that is
3
As the unit cell is cubic, then V C = a , hence:
And solving this equation for the unit cell edge length, leads to
Expressions for Aave and ρave are found in text book Equation 4.11a and text book Equation 4.10a, respectively, which,when incorporated into the above expression yields
Since the crystal structure is BCC, the value of n in the above expression is 2 atoms per unitcell. The atomic weights for Fe and V are 55.85 and 50.94 g/mol, respectively (refer to periodic table 3 3 text book Figure 2.6 ), whereasthe densities for the Fe and V are 7.87 g/cm and 6.10 g/cm (from the front section of the text book ).Substitution of these, as well as the concentration values stipulated in the problem statement, into theabove equation gives
8. For a solid solution consisting of two elements (designated as 1 and 2), sometimes it is desirable to determine the number of atoms per cubic centimeter of one element in a solid solution, N 1, given the concentration of that element specified in weight percent, C 1. This computation ispossible using the following expression:
Where N A = Avogadro’s number ρ1 and ρ2 = densities of the two elements A1 = the atomic weight of element 1 Derive the above mentioned equation using text book Equation 4.2 and expressions contained in text book Section 4.4 . Solution: This problem asks that we derive the above mentioned equation, using other equations given in the chapter.The concentration of component 1 in atom percent (C 1′) is just 100C 1′whereC 1′is the atom fractionof component 1. Furthermore, C 1′is defined as C 1′= N 1/ N where N 1 and N are, respectively, thenumber of atoms of component 1 and total number of atoms per cubic centimeter. Thus, from theabove discussion the following holds:
Substitution into this expression of the appropriate form of N from text book Equation 4.2 yields
And, finally, substitution into this equation expressions for C1 ′ (text book Equation 4.6a ), ρave (text book Equation 4.10a ), Aave ( text book Equation 4.11a ), and realizing that C 2 = (100 - C 1), and after some algebraic manipulation weobtain the desired expression:
9. For an ASTM grain size of 8, approximately how many grains would there be per square centimeter at: (a) a magnification of 100, (b) without any magnification?
Solution: (a) To compute the number of grains per square centimeter for anASTM grain size of 8 at a magnification of 100×. N intext book Equation 4.16 :
(b) It is also necessary to compute the value of N ′ for no magnification. In order to solve thisproblem it is necessary to use text book Equation 4.17 :
where N ′ M = the number of grains per square centimeter at magnification M , and n is the ASTM grainsize number. Without any magnification, M in the above equation is 1, and therefore,
( ) = 19.83 grains/cm 1
2
2
’
2
N 1 = 198300 grains/cm 10. Cite the relative mixeddislocations.
Burgers
vector – dislocation
line
orientations
for
edge,
screw,
and
Solution: The Burgers vector and dislocation line are perpendicular for edge dislocations, parallel for screw dislocations, and neither perpendicular nor parallel for mixed dislocations.
