r , 0 Y Y Y T T T > > > >

Practice Problems Set I

MIME260

P1. Calculate the fraction of bonding of MgO that is ionic (use the figure below).

Solution: 2 % ionic character = {1  exp  0.25  X A  X B   } 100   Electronegativities of Mg and O are 1.2 and 3.5 respectively. 2 ionic component: {1  exp  0.25  3.5  1.2   } 100 = 73.4%  

P2. Consider the atomic bonding in the following materials,  MgO  NiAl  SiC Whose bonding is directional? Solution: SiC P3. Consider three materials, A, B and C. The atomic bonding energies within these materials are denoted as EA(r), EB(r), and EC(r) respectively with r being the atomic separation. Denoting the corresponding equilibrium atomic separation (i.e., the separation at which dE / dr  0 ) within these materials as

r0 A , r0 B , and r0 C , respectively, we

have

dFA dr

 r  r0

A

dFC dr

 r  r0

C

dFB dr

r  r0 B

, with F   dE dr

EB ( r0 B )  EA ( r0 A )  EC ( r0C ) Please range the values of Young’s modulus (Y) and melting temperature (Tm) for these materials. Solution:

YA  YC  YB Tm B  Tm A  TmC

1

Practice Problems Set I

MIME260

P4. The net potential energy between two adjacent ions, EN, may be represented as follows, A B EN =   n r r

(E.1)

Calculate the bonding energy E0 in terms of the parameters A, B, and n using the following procedure: a.

Differentiate EN with respect to r, and then set the resulting expression equal to zero, since the curve of EN  versus r is a minimum at E0.

b.

Solve for r in terms of A, B, and n, which yields r0, the equilibrium interionic spacing.

c.

Determine the expression for E0 by substitution of r0 into Equation (E.1).

Solution (a) Differentiation of Equation 2.11 yields

dEN A nB = (1 + 1)  (n + 1) dr r r (b) 1/(1 - n )

dEN  A  0  r  r0 =   dr  nB  (c)

E0 = E ( r0 ) = 

A 1/(1 - n )

 A    nB 

+

B  A    nB 

n /(1 - n )

P5. For a K+–Cl– ion pair, attractive and repulsive energies EA and ER, respectively, depend on the distance between the ions r, according to

EA  

1.436 , r

ER 

5.8 106 r9

For these expressions, energies are expressed in electron volts per K+–Cl– pair, and r is the distance in nanometers. The net energy EN is just the sum of the two expressions above. (a) Superimpose on a single plot EN, ER, and EA versus r up to 1.0 nm. (b) On the basis of this plot, determine (i) the equilibrium spacing r0 between the K+ and Cl– ions, and (ii) the magnitude of the bonding energy E0 between the two ions. (c) Mathematically determine the r0 and E0 values using the solutions to Problem 1 and compare these with the graphical results from part (b).

2

Practice Problems Set I

MIME260

Solution (a) Curves of EA, ER, and EN are shown on the plot below.

(b) From this plot r0 = 0.28 nm,

E0 = – 4.6 eV

(c) Using the solutions in Problem #1,

r0  0.279 nm

E0  ? 4.57 eV

P6. The net potential energy EN between two adjacent ions is sometimes represented by the expression

EN  

 r C  D exp    r  

(E.3)

in which r is the interionic separation and C, D, and ρ are constants whose values depend on the specific material. (a) Derive an expression for the bonding energy E0 in terms of the equilibrium interionic separation r0 and the constants D and ρ using the following procedure: 1). Differentiate EN with respect to r and set the resulting expression equal to zero. 2). Solve for C in terms of D, ρ, and r0. 3). Determine the expression for E0 by substitution for C in Equation (E.3). (b) Derive another expression for E0 in terms of r0, C, and ρ using a procedure analogous to the one in part (a).

3

Practice Problems Set I

MIME260

Solution (a) dE/dr = 0 

C De r /  De (r0 / ) 2   0  C = r0 r2   Substitution into Equation (E.3) yields an expression for E0 as

 r  E0 = De ( r0 / ) 1  0    (b)

D=

C  e ( r0 / ) , r02

E0 =

 C  1  r0  r0 

P7. Make a plot of bonding energy versus melting temperature for the metals listed in the Table below. Using this plot, approximate the bonding energy for copper, which has a melting temperature of 1084C.

Solution: Below is plotted the bonding energy versus melting temperature for these four metals. From this plot, the bonding energy for copper (melting temperature of 1084C) should be approximately 3.6 eV. 4

Practice Problems Set I

MIME260

P8. What type(s) of bonding would be expected for each of the following materials: brass (a copper-zinc alloy), rubber, barium sulfide (BaS), solid xenon, bronze, nylon, and aluminum phosphide (AlP)? a. For brass, the bonding is metallic since it is a metal alloy. b. For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.) c. For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table. d. For solid xenon, the bonding is van der Waals since xenon is an inert gas. e. For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin). f. For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.) g. For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table.

5