Structure 2.4. From models to materials

The electronegativity of antimony is 2.0 and that of chlorine is 3.2.

In which region of the bonding triangle diagram would SbCl₅ be found?

Answer: Region C 

EN_agv = `frac{2.0+3.2}{2}`= 2.6

 ΔEN =∣ 3.2 − 2.0 ∣= 1.2 

Locate on the bonding triangle 

• x-axis (average EN) → 2.6 

• y-axis (difference in EN) → 1.2 

This point lies roughly in region C of the diagram.

Two elements that have very similar electronegativity values chemically combine. The compound they form is plotted on a bonding triangle diagram. Which statement about the compound must be correct? 

A. The compound has a low conductivity compared to a metal. 

B. The covalent character is high. 

C. The ionic character is high. 

D. The metallic character is low. 

Answer: B. The covalent character is high. 

A. Incorrect: 

This statement is likely true for a covalent compound, as covalent compounds  generally do not conduct electricity well in the solid or molten state. However, it is not  the most direct or certain conclusion from the initial premise. Some covalent network  solids (like graphite) can have high conductivity.  

B. Correct: 

The premise states that the two elements have "very similar electronegativity values."  The difference in electronegativity (ΔEN) between two atoms is the primary factor in  determining bond type. A small ΔEN means the electrons are shared relatively equally,  which is the definition of a covalent bond. Therefore, the covalent character of the  bond is high.  

C. Incorrect: 

Ionic bonds form when there is a large difference in electronegativity between the two  atoms, causing one atom to essentially donate an electron to the other. Since the  problem states the electronegativity values are "very similar," the ionic character is  low, not high. 

D. Incorrect: 

Metallic character is a property of a single element (or a mixture of metals, forming an  alloy). It is not a characteristic of a compound formed between two elements. The  problem describes a compound, not a metallic element or alloy. 

Which is the correct composition of the alloy stainless steel?

A. Iron, carbon and silver.

B. Iron, carbon and lead.

C. Iron, carbon and chromium.

D. Iron and carbon only.

Answer: C. Iron, carbon and chromium. 

Stainless steel is an alloy (a mixture of metals) designed to resist rusting and corrosion. It is mainly composed of: 

• Iron (Fe) – the base metal 

• Carbon (C) – increases strength and hardness 

• Chromium (Cr) – provides corrosion resistance by forming a protective layer of  chromium oxide (Cr₂O₃) on the surface 

Which statement about polymers is correct? 

A. All addition polymers are biodegradable. 

B. All condensation polymers can be hydrolysed to release amino acids.

C. The formation of all addition polymers produces only one product.

D. All condensation polymers are naturally occurring. 

Answer: C. The formation of all addition polymers produces only one product.

A. Incorrect: 

Most addition polymers (like polyethylene, polystyrene, PVC) are not biodegradable – they have strong C–C backbones that resist breakdown. 

B. Incorrect: 

Only polyamides (like proteins or nylon) yield amino acids on hydrolysis. Other  condensation polymers like polyesters yield alcohols and carboxylic acids, not amino  acids. 

C. Correct: 

In addition polymerization, no small molecule (like H₂O or HCl) is eliminated – only the polymer is formed. 

D. Incorrect: 

Some condensation polymers (like nylon, Terylene) are synthetic, while others (like  proteins, cellulose) are natural. Not all are naturally occurring.

The diagram shows three repeat units in the structure of an addition polymer.

Which alkene monomer is used to make this polymer? 

Answer: D. 

1. Identify the polymer’s repeating unit 

From the diagram, each carbon chain unit looks like: 

−CH₂ − CHF − 

That means the repeating unit has: 

o One carbon bonded to two hydrogens (CH₂) 

o The next carbon bonded to one hydrogen and one fluorine (CHF)

2. Reverse to the monomer 

To form this polymer, the monomer must be an alkene (double bond between the two carbons). 

Therefore, replace the single C–C bond with a double bond: 

CH2 = CHF 

In which row are the monomer and polymer chain correctly matched? 

Answer: A. Monomer: CH₃CH=CHCH₃, Polymer: –CH(CH₃)–CH(CH₃)–CH(CH₃)–CH(CH₃)–

A. Correct:

Polymerizing but-2-ene gives a backbone where each carbon bears a CH₃ substituent: –CH(CH₃)–CH(CH₃)–

B. Incorrect:

CH₂=CHF should give –CH₂–CHF–..., not –CHC–CHF–...

C. Incorrect:

CH₃CH=CH₂ (propene) should give –CH₂–CH(CH₃)–...; the shown chain places extra –CH₃– units in the backbone.

D. Incorrect:

CH₂=CHCH₂CH₃ (1-butene) should give –CH₂–CH(CH₂CH₃)–..., not –CH₂–CH₂CH₂–CH(CH₂CH₃)–...

a. Define the term electronegativity. 

b. Explain why elements with low electronegativities can conduct electricity.

c. 

i. Explain, in terms of electronegativity, why ionic bonding results from the transfer of  one or more electrons from one atom to another. 

ii. Explain, in terms of electronegativities and ionization energy, why this transfer takes  place. 

a. Define the term electronegativity. 

Electronegativity is the ability of an atom to attract the shared pair of electrons in a  covalent bond toward itself. 

b. Explain why elements with low electronegativities can conduct electricity. Elements with low electronegativities (typically metals) lose their valence electrons  easily, forming a lattice of positive ions surrounded by delocalized electrons. These free-moving electrons allow the metal to conduct electricity. c. 

i. Explain in terms of electronegativity why ionic bonding results from the transfer of  one or more electrons from one atom to another.  

Ionic bonding occurs because there is a large difference in electronegativity between two atoms. 

The more electronegative atom attracts electrons away from the less electronegative  atom, resulting in electron transfer and formation of ions. 

ii. Explain in terms of electronegativities and ionization energy why this transfer takes place. 

The atom with low electronegativity and low ionization energy loses electrons easily,  while the atom with high electronegativity gains electrons readily. This makes electron transfer energetically favorable. 

The following substances are all hard materials used in body armour: AlN, B₄C, SiC,  Si₃N₄, and WC. 

a. Plot labelled points for each compound on the bonding triangle diagram and deduce the type of bonding they are predicted to exhibit.

b. Compare the degree of metallic bonding in silicon carbide (SiC) with silicon nitride  (Si₃N₄). 

c. Suggest one limitation of using the bonding triangle diagram to predict bonding type

a. Plot & bonding type 

Use Pauling electronegativities: 

Al 1.6, B 2.0, C 2.55, Si 1.9, N 3.0, W 2.36. 

For each binary pair (X–Y): 

Average EN = `frac{chi_X+chi_Y}{2}`,  ΔEN = ∣ `""chi_X - chi_Y` ∣

(When plotted, these points fall from near the covalent bottom edge (SiC, B₄C, WC) up toward the ionic apex (AlN), with Si₃N₄ in between.) 

b. Compare metallic bonding in SiC vs Si₃N₄ 

SiC shows slightly more metallic character (smaller ΔEN, more delocalization / semiconducting). 

Si₃N₄ is less metallic / more ionic-polar covalent (larger ΔEN), so it is a better insulator. c. One limitation of the bonding triangle 

It uses only average EN and ΔEN, so it ignores structure and multi-center bonding (e.g.,  giant networks, delocalization, coordination/oxidation state). Hence, real compounds  often have mixed bonding not captured by a single point. 

Which row in the table describes the formation of a polymer? 

Answer: B. Monomer: Ethene; Polymer: poly(ethene). 

A. Incorrect: Ethane (C₂H₆) is saturated – no double bond → can’t polymerize.

B. Correct: Ethene (C₂H₄) is an alkene with a C=C double bond; under polymerization,  the double bond opens to form long chains. 

C. Incorrect: Monomer and polymer names don’t correspond. 

D. Incorrect: Wrong polymer name — should be poly(ethene), not poly(ethane). 

Describe how, in general, alloying improves the usefulness of metals and how strength  is increased in terms of structure. Use diagrams in your answers. No specific alloys need  to be mentioned. 

1. Structure of pure metals (diagram 1)

• In a pure metal, atoms are arranged in layers of identical atoms in a regular, close packed lattice. 

• These layers of atoms can slide easily over each other when a force is applied →  Metals are malleable and ductile. 

2. Structure of an alloy (diagram 2) 

• When another element (metal or non-metal) is added, the atoms of the added  element have a different size. 

• This distorts the regular layers in the metal lattice, making it harder for the layers  to slide over each other. 

3. Explanation of increased strength 

• The distortion caused by atoms of different sizes prevents dislocation  movement in the lattice. 

• Hence, alloys are harder and stronger than pure metals. 

• They are also less malleable and less ductile because atoms cannot easily slide. 4. Improved usefulness 

Alloying allows metals to have enhanced properties such as: 

• Greater strength and hardness 

• Resistance to corrosion 

• Improved durability and tensile strength 

• Better suitability for specific applications (e.g., construction, tools, machinery)

State the name of the linkage present in polylactic acid whose repeating unit is shown below. 

A. Ester. 

B. Amide. 

C. Peptide. 

D. Carbonyl.

Answer: A. Ester 

The repeating unit: –O–CH(CH₃)–C(=O)– 

This segment contains a –COO– linkage, which is characteristic of an ester functional  group. 

A. Correct: Present between the lactic acid monomers after condensation. This is what  forms the polymer chain in polylactic acid (PLA). 

B. Incorrect: Requires nitrogen; no N atom here. 

C. Incorrect: A specific type of amide linkage found in proteins, not in PLA.

D. Incorrect: A carbonyl group is part of the structure but not the linking group  between monomers. 

The equation below shows the synthesis of Kevlar. 

Which row describes Kevlar? 

Answer: B. Condensation polymerization; Polyamide. 

• Type of polymerization: 

A small molecule (H₂O) is eliminated during each linkage formation → This is  condensation polymerization. It is not addition polymerization because no double bonds  are involved. 

• Type of polymer: 

The linkage formed is –CONH–, called an amide (peptide) linkage. Therefore, Kevlar  is a polyamide (similar to nylon). 

Nylon-6,6 is a condensation polymer synthesized from hexanedioic acid and 1,6- diamino hexane. What type of linkage is present? 

A. Carboxyl. 

B. Amine. 

C. Ester. 

D. Amide.

Answer: D. Amide. 

Nylon-6,6 is made from: 

• Hexanedioic acid (adipic acid) — contains –COOH groups 

• 1,6-diaminohexane (hexamethylenediamine) — contains –NH₂ groups When these react together, the –COOH and –NH₂ groups combine to form –CONH– linkages and release water (H₂O). This is the characteristic linkage of a polyamide.