% ionic character = [ ]{ } 100))(25.0(exp1 2!"""

BAXX

={ } 100)5.28.0(25.0exp[1 2!"!""

= 51.45

K: 162622433221 spspss

+K : 62622

33221 pspss

I: 510261026262254543433221 pdspdspspss

:!I 6102610262622

54543433221 pdspdspspss

+K has an electron structure the same as argon. !I has the electron structure the same as xenon.

2. Callister 2.15

For an !+

!ClNa ion pair, attractive and repulsive energies AE and R

E , respectively,

depend on the distance between the ions r, according to rEA

436.1!=

,

8

61032.7

rER

!"

=

For these expressions, energies are expressed in electron volts per !+

!ClNa pair, and

r is the distance in nanometers. The net energy NE is just the sum of the two

expressions above.

(a) Superimpose on a single plot NE , R

E and AE versus r up to 1.0 nm.

(b) On the basis of this plot, determine (i) the equilibrium spacing 0r between the

+Na and !

Cl ions, and (ii) the magnitude of the bonding energy 0E between the two

ions.

(c) Mathematically determine the 0r and

0E values using the solutions to Problem

2.14 and compare these with the graphical results from part (b).

3. Callister 2.18a Briefly cite the main differences between ionic, covalent, and metallic bonding.

4. Callister 2.23 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.

5. Core 2.28 (A) Species B has the higher interatomic spacing, because

rA< rB.

(B) Species A has the higher melting temperature, because the depth of its energy well is greater.

(C) Species C has the higher coefficient of thermal expansion, because the slope of B

U (Brr > )

is less than that of )(AArrU >

rB

rA

6. Core 3.02

22162 Ra = ! Ra 22=

3

233

)22(

)/98.26)(10022.6

1)(4(

/7.2R

molegatoms

mole

cell

atoms

cmgv

m !==="

3

321

3

3

243 00293.0)

10

1)(

/7.2

1092.7( nm

nm

cm

cmgR =

!="

#

nmR 143.0=!

[001]

[100]

[010]

a

R

7. Callister 3.21 Sketch a unit cell for the face-centered orthorhombic crystal structure.

aR

aR

4

3

34

=

=

833.02

16

32

2

)4

3)(2(

2

2

2

)14

14(

2

2

2

2]110[

2

=

!

===!

+!

=

"""

"

a

a

a

R

aa

R

Farea

833.0589.016

3)

4

3()

4

14(

2

2

]100[

2

<===!

!

="

""

a

a

aa

R

Farea

[001]

]111[]111[)1,0,1()0,1,0( or=!

(0,1,1)-(1,0,0)= ]111[]111[ or [100]

[010]

a2 a3

R

5855.0495.0587.0285.0

4)1425.0(3

4 3

=!!

!!!

==nmnmnm

atomsnm

V

MAPF

"

10. Callister 3.41 Determine the Miller indices for the planes shown in the following unit cell:

2

1

2

1 +y

+x

2

1 +z