Curriculum
Y1 · I · #26Chemical bonds

Testing

Testing Your Knowledge: Chemical Bonds

You've spent this unit exploring ionic, covalent, and metallic bonding — how atoms hold together, why molecules have the shapes they do, and how bonding explains physical properties like melting point, solubility, and conductivity. This lesson is a chance to check how well those ideas have stuck, before we move on. No new content today — just five quick questions covering the key skills from the unit.

Quick Recap Before You Start

  • Simple molecular (covalent) compounds tend to have low melting points and don't conduct electricity when molten, since they contain no free ions (Talbot, p. 167).
  • Solubility depends on the type of intermolecular force: compounds relying mainly on London (dispersion) forces dissolve better in non-polar solvents, while those capable of hydrogen bonding often dissolve well in water (Talbot, p. 167).
  • Polarity of molecules depends on both bond polarity and molecular shape — dipoles can cancel out if they are arranged symmetrically, giving an overall non-polar molecule even though individual bonds are polar (Talbot, p. 159).

Keep these ideas in mind as you tackle the questions below.

Practice Questions

1. Explain why iodine (I₂) does not conduct electricity even when melted.

2. A student has two solids: one dissolves easily in water, the other dissolves easily in hexane (a non-polar solvent). Which type of intermolecular force is most likely dominant in each solid, and why?

3. Consider 1,2-difluorobenzene and 1,4-diflu

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Source excerpts

From Chemistry for the IB Diploma 3e · Talbot

p.167relevance 7.4

S2.2 The covalent model 155 ■Physical properties of simple molecular compounds ■ Simple molecular compounds are gases, liquids or soft solids with low melting points. ■ Most simple covalent compounds whose intermolecular forces are mainly London (dispersion) f orces (for example, iodine and the halogenoalkanes) are poorly soluble in water, but are soluble in less polar or non-p…

p.642relevance 5.3

630 R3: What are the mechanisms of chemical change? 630 H3C H3C H3C H3C CH3 CH3 O 2NR3 OH OHO + 2NR3+ hf ■Figure R3.88 Formation of radicals from the photoinitiator mixture in a dental composite The radicals then undergo a radical addition reaction with an alkene on the end of ‘ bis- GMA ’ monomer (Figure R3.89). This starts the polymerization process which hardens the composit…

p.159relevance 5.1

S2.2 The covalent model 147 Deduce which of the difluorobenzene molecules shown is non-polar (has zero dipole moment). 03_29 Cam/Chem AS&A2 Barking Dog Art F F F F F FA B C Answer Compound C, 1,4-difluorobenzene. In compound C, the C–F dipoles are on opposite sides of the benzene ring so their two equal dipoles oppose each other and cancel. In molecules A and B, the C–F dipoles…

p.307relevance 4.7

early 20th century that chemists started to investigate how infrared radiation interacted with matter. The first commercial infrared spectrometers were manufactured in the USA during the 1940s. Infrared spectroscopy is used in modern breath testing machines to identify individuals who drive while under the influence of alcohol. Greenhouse gases The absorption of infrared radiat…

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + ATP
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