The Covalent Model (DP IB Chemistry: SL): Exam Questions

State the formula of the compound formed between boron and chlorine.

Draw the Lewis (electron dot) structure for the compound formed in part (a).

Explain why the compound formed in part (a) is able to form coordinate bonds.

Group 17 of the Periodic Table contain non-metals that are often referred to as the halogens.

Iodine, I₂, is one of these halogens. At room temperature and pressure it exists as a grey-black solid.

Describe the bonding and forces present in I₂ in the solid state.

The state of the halogens changes down the group, with fluorine being a gas and astatine being a solid.

Explain why the melting point of the halogens increases down the group.

The halogens are all diatomic covalent molecules.

State three physical properties common to all elements in Group 17.

The halogens can also form interhalogen compounds, such as iodine monochloride, ICl.

Predict the state of iodine monochloride at room temperature and pressure. Explain your answer with reference to the intermolecular forces present.

The allotropes of carbon include diamond, graphite, and buckminsterfullerene, C₆₀.

i) State how many atoms each carbon is directly bonded to in each of the allotropes.

[3]

ii) Explain why the carbon atoms in the alltropes form a different number of bonds.

[1]

Compare the overall structure and bonding of diamond with that of buckminsterfullerene.

Describe and explain the differences in electrical conductivity between the three allotropes of carbon.

Graphene is another allotrope consisting of a single layer of graphite.

State one similarity and one difference in structure between graphene and graphite.

Silicon and carbon are elements in the same group of the Periodic Table, that form covalent bonds.

Both silicon and carbon react with oxygen to form dioxides. The structures of silicon dioxide and carbon dioxide are shown below.

Silicon dioxide has a melting point of 1710 ^oC, while carbon dioxide sublimes at -78 ^oC.

Explain this difference with reference to the structure and bonding present in each dioxide.

How many oxygen atoms are bonded to each carbon and to each silicon?

Explain how this links to the formula of each compound.

Predict the O-C-O and O-Si-O bond angles respectively in CO₂ and in SiO₂.

Predict and explain the solubility of both SiO₂ and CO₂ in water.

A student investigates the chlorides of various elements to compare their bonding, structure, and properties.

The student first uses electronegativity to predict bonding type.

Using section 9 of the data booklet, deduce whether the bonding in each compound below is predominantly ionic or covalent. Justify each answer with an electronegativity difference calculation.

i) ICl

ii) SrCl₂

The student knows that sodium chloride, NaCl, is an ionic compound with a high melting point of 801 ^oC.

Explain this property with reference to the structure and bonding in NaCl.

Explain why ionic chlorides, such as NaCl, conduct electricity when molten but not when solid.

The student next investigates two covalent chlorides.

i) Draw the Lewis (electron dot) structure for phosphorus trichloride (PCl₃) and boron trichloride (BCl₃).

[2]

ii) Predict the molecular geometry of PCl₃ and BCl₃.

[2]

Nitrogen hydrides, such as diazene (N₂H₂) and hydrazine (N₂H₄), are highly reactive compounds used in chemical synthesis and as potential fuels. Their structure and bonding determine their properties and reactivity.

Draw the Lewis structures for diazene (N₂H₂) and hydrazine (N₂H₄).

Use VSEPR theory to deduce the molecular geometry around a nitrogen atom in diazene (N₂H₂) and estimate the H-N-N bond angle.

List the compounds dinitrogen (N₂), diimide (N₂H₂), and hydrazine (N₂H₄) in order of increasing nitrogen-nitrogen bond length. Justify the order with reference to the number of shared electron pairs in the bond

Use Section 9 of the Data Booklet to predict which bond in each of the following pairs is more polar:

i)         C–H or C–Cl

[1]

ii)        Si–Li or Si–Cl

[1]

The combustion of ethyne, C₂H₂, is a highly exothermic reaction used in welding torches. The balanced equation for its complete combustion is:

2C₂H₂ (g) + 5O₂ (g) → 4CO₂ (g) + 2H₂O (g) 

Draw the Lewis structure for one molecule of ethyne.

Explain why the carbon-carbon triple bond in ethyne is significantly stronger and shorter than the oxygen-oxygen double bond in oxygen.

The products of the combustion, carbon dioxide and water, have different molecular polarities.

i) Using section 9 of the data booklet, determine the electronegativity difference (Δχ) for a C=O bond and an O-H bond.

[1]

ii) Both CO₂ and H₂O molecules contain polar bonds, but only H₂O is a polar molecule. Explain this difference with reference to molecular geometry.

[3]

The combustion of ethyne is highly exothermic. Explain this in terms of the energy absorbed to break bonds and the energy released when forming bonds.

Ammonia, NH₃, and boron trifluoride, BF₃, are simple covalent molecules.

State the shape and the H-N-H bond angle in ammonia.

Ammonia reacts with a hydrogen ion, H⁺, to form the ammonium ion, NH₄⁺.

i) State the type of bond formed between the ammonia molecule and the hydrogen ion.

[1]

ii) The H-N-H bond angle in the ammonium ion is 109.5^o. Explain why this value is different from the bond angle in the ammonia molecule.

[2]

i) State the shape and bond angle of a boron trifluoride molecule.

[1]

ii) Explain why a molecule of boron trifluoride is non-polar, even though it contains polar bonds.

[2]

Draw the Lewis structure of the product formed when one molecule of ammonia reacts with one molecule of boron trifluoride.

Phosphorus tribromide and sulfur tetrafluoride are two colourless compounds which both react with water to form toxic products. 

Deduce the Lewis structure of both molecules.

Predict the shapes of phosphorus tribromide and sulfur tetrafluoride.

Explain why both phosphorus tribromide and sulfur tetrafluoride are polar.

The isomers of formulas C₄H₁₀O and C₄H₈O, and the alkane C₄H₁₀, provide a good basis for comparing the effects of molecular structure on intermolecular forces and physical properties.

Explain why butan-1-ol (CH₃CH₂CH₂CH₂OH) has a significantly higher boiling point than its isomer, ethoxyethane (CH₃CH₂OCH₂CH₃).

Explain why butan-1-ol has a higher boiling point than its structural isomer, 2-methylpropan-2-ol ((CH₃)₃COH).

Place the following C₄ compounds in order of increasing boiling point (lowest first) and explain your reasoning: butan-1-ol, butanal (CH₃CH₂CH₂CHO), and butane (CH₃CH₂CH₂CH₃).

Explain the difference in solubility in water between butan-1-ol and ethoxyethane.

The hydrides of elements show a wide range of properties depending on their bonding. This question compares the hydride of a Group 1 metal with the hydrides of Group 15 elements.

Sodium reacts with hydrogen to form sodium hydride, NaH, an ionic compound. Predict, with a reason, whether potassium hydride (KH) would have a higher or lower melting point than sodium hydride.

Nitrogen, from Group 15, forms a simple covalent hydride, ammonia (NH₃). Explain why ammonia has a very low boiling point and is a gas at room temperature.

Phosphorus is directly below nitrogen in Group 15. Explain why the boiling point of phosphine (PH₃) is significantly lower than that of ammonia (NH₃), despite its greater molar mass.

Carbon and silicon are both in Group 14 and form covalent network structures, both as elements and in their common oxides.

i) Draw the structure of silicon dioxide.

[1]

ii) State the type of bonding present in silicon dioxide.

[1]

Compare the structures and resulting physical properties of silicon and diamond.

Explain the large differences in the boiling points of carbon dioxide (–78.5 ^oC), silicon dioxide (2230 ^oC), and diamond (3550 ^oC) with reference to their structure and bonding.

In 1996 the Nobel prize in Chemistry was awarded for the discovery of a new carbon allotrope, known as fullerenes.

Outline the structure of buckminsterfullerene.

Graphite is a covalent substance with a melting point of around 3600 °C.

i) Describe the structure and bonding of graphite.

[3]

ii) Explain why graphite has a very high melting point.

[2]

Explain the difference in electrical conductivity between graphite and diamond.

Graphite is soft enough to be used as a lubricant, whereas diamond is one of the hardest known substances.

Explain this difference in hardness with reference to the structure and bonding in the two allotropes.

Ethene, C₂H₄, and hydrazine, N₂H₄, are hydrides of adjacent elements in the periodic table. 

State and explain the H一C一H bond angle in ethene and the H一N一H bond angle in hydrazine.

Hydrazine can be oxidised to form diimide, which is a useful compound used in organic synthesis. 

Deduce the molecular geometry of diimide, N₂H₂, and estimate its H–N–N bond angle.

Explain whether ethene and hydrazine are polar or non-polar.

Hydrazine forms a cation with an ethane-like structure called hydrazinediium,  N₂H₆²⁺.

Predict the value of the H–N–H bond angle in N₂H₆²⁺.

The oxides of elements in Periods 2 and 3 display a wide range of bonding, structural, and physical properties.

Explain why the melting point of sodium oxide, Na₂O, is significantly higher than that of phosphorus(V) oxide, P₄O₁₀. In your answer, you should refer to the structure and bonding of each compound.

The bonding in carbon monoxide, CO, is more complex than in many other oxides as it contains a coordinate (dative) covalent bond. Describe the bonding in a molecule of carbon monoxide.

Using section 11 and section 12 of the data booklet, compare the length and strength of the bond in carbon monoxide with the bond in molecular nitrogen, N₂. Justify your answer with reference to the number of shared electrons.

Alcohols are a homologous series of organic compounds whose physical properties, such as boiling point and solubility, are determined by the nature of their intermolecular forces.

Explain why methanol is soluble in water.

The boiling points of the first three primary alcohols are shown below.

Describe and explain this trend.

The boiling points of longer-chain primary alcohols continue the trend.

Describe and explain this trend.

Predict, with a reason, whether ethanol or ethane-1,2-diol will have the higher boiling point.

C₂H₆, C₃H₈ and C₄H₁₀ are the first three unbranched alkanes.

i) Place these alkanes in order of increasing boiling point.

[1]

ii) Explain the trend in boiling point with reference to the intermolecular forces present.

[2]

Pentane, C₅H₁₂, has several structural isomers, including pentane and 2,2-dimethylpropane.

i) Draw the skeletal formula for pentane and 2,2-dimethylpropane.

[2]

ii) Predict and explain which of these two isomers would have the lower boiling point.

[3]

The molecular formula C₂H₆O corresponds to two isomers with different functional groups, ethanol and methoxymethane.

i) Draw the skeletal formulae of both isomers.

[2]

ii) Identify the strongest type of intermolecular force present in each isomer.

[2]

iii) Explain which isomer would have the significantly higher boiling point.

[2]

The peptide bond that links amino acids together to form proteins is an amide linkage. Understanding the geometry of the simple amide, methanamide (HCONH₂), is key to understanding protein structure.

Draw the Lewis structure for this molecule.

Predict the electron domain geometry and the molecular geometry around the carbon and nitrogen atoms in methanamide.

Predict and explain the bond angles around:

i) the carbon atom.

[3]

ii) the nitrogen atom.

[3]

State, with a reason, whether HCONH₂ is a polar molecule.

Tetrafluoroethene, C₂F₄, and tetrafluorohydrazine, N₂F₄, are fluorides of adjacent elements in the Periodic Table.

Draw the Lewis (electron dot) structures for C₂F₄ and N₂F₄ showing all valance electrons.

Predict and explain the F-C-F bond angle in tetrafluoroethene and the F-N-F bond angle in tetrafluorohydrazine.

Tetrafluorohydrazine is a polar molecule but tetrafluoroethene is not.

Explain the difference in molecular polarity.

Diamond and graphite are two allotropes of carbon with distinct properties. Some of the physical and structural properties of diamond and graphite are shown below:

Suggest why the melting point of graphite is higher than that of diamond, using the information in the table.

Graphene has the structure of a single layer of graphite. 

Suggest, giving a reason, the electron mobility of graphene compared to graphite.

Graphite is a layered giant structure, containing London dispersion forces between the layers, whereas diamond has covalent bonds across all planes.

Describe and explain, based on structure and bonding, the differences expected when each of graphite and diamond are moved across a paper surface.