SCS 241 Materials Science for Engineers
Homework #1 (Due: Sun 16/02/2025)
1. A periodic table is a good collection of elements easily translated into properties. Discuss
why group IA-IIA elements are likely to be positively charged while group VIA-VIIA
elements are likely to be negatively charged.
Ans,
●​ Group IA-IIA (Metals): These elements easily lose electrons to become positively
charged ions (cations). Group IA loses 1 electron (+1 charge), and Group IIA loses 2
electrons (+2 charge).
●​ Group VIA-VIIA (Nonmetals): These elements easily gain electrons to become
negatively charged ions (anions). Group VIA gains 2 electrons (-2 charge), and Group
VIIA gains 1 electron (-1 charge).
2. Show electron configurations of the following atoms and indicate if the atoms are stable
or reactive.
Na, Ne, Cl, Ti
●​ Sodium (Na): [Ne] 3s¹. Reactive because it has just 1 electron in its outer shell.
●​ Neon (Ne): [He] 2s² 2p⁶. Stable because its outer shell is full (8 electrons).
●​ Chlorine (Cl): [Ne] 3s² 3p⁵. Reactive because it needs just 1 more electron to fill its
outer shell.
●​ Titanium (Ti): [Ar] 4s² 3d². Reactive because it has only a few electrons in its outer
shells
3. Briefly discuss the intermolecular forces you expect to see in the following molecules:
CO2, H2O, CO, NH3, Nylon
●​ CO2 (Carbon Dioxide): London dispersion forces are the only intermolecular forces
present in CO22. CO2 is a non-polar molecule because it is symmetrical, and the
dipoles cancel out.
●​ H2O (Water): Hydrogen bonding, dipole-dipole forces, and London dispersion forces
are all present in water1. Hydrogen bonding is particularly significant in water due to
the high electronegativity of oxygen.
●​ CO (Carbon Monoxide): Dipole-dipole interactions and London dispersion forces are
present in CO.
●​ NH3 (Ammonia): Hydrogen bonding, dipole-dipole forces, and London dispersion
forces are present.
●​ Nylon: Nylon can form hydrogen bonds between N-H and C=O groups on adjacent
chains, contributing significantly to its strength.
4. Electronegativity (EN) is the ability of an atom to pull electrons towards itself. Discuss
how EN determines the type of bonding (covalent, ionic, and metallic bonds) in the
materials or molecules.
●​ Big EN Difference = Ionic Bond: When atoms have a large difference in EN
(typically > 1.7), the more electronegative atom steals electrons from the other. This
creates ions (charged particles) that attract each other, forming an ionic bond (e.g.,
NaCl).
●​ Small EN Difference = Covalent Bond: When atoms have a small difference in EN
(typically < 1.7), they share electrons. This sharing forms a covalent bond.
●​ Equal Sharing (Nonpolar Covalent): If ENs are nearly identical, the sharing is
equal (e.g., H₂).
●​ Unequal Sharing (Polar Covalent): If there's a slight EN difference, the
sharing is unequal, creating a polar covalent bond (e.g., H₂O).
●​ Metallic Bond: Metallic bonding involves the sharing of electrons within a "sea" of
electrons. EN differences do not play a major role in determining the metallic bonds.
5. Calculate the packing efficiency of the Body-centered cubic (BCC) closed packs (show
the steps).
6. Packing in solid state is a very important concept that determines several properties of the
materials. Consider a very simple 2-dimension case shown below. The four white circles
touch one another, forming a square. If the radius of the white circles is R, calculate the
radius of the black circle that can fit perfectly inside the four white circles.
+
the packing efficiency of the ions in this unit
cell.
7. From question #6, imagine that the white
circles are atoms or ions. A unit cell is
represented on the right as a square. Calculate
8. Determine a coordination number, a radius (r), and a packing efficiency of metal M,
which exhibits a face-centered cubic structure with a unit cell (a = 0.5 nm).