Directly related questions
- 17N.1.sl.TZ0.9: The electronegativity values of four elements are given. What is the order of increasing...
- 17N.1.sl.TZ0.9: The electronegativity values of four elements are given. What is the order of increasing...
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17N.3.sl.TZ0.7b.i:
Describe how the structures of LDPE and HDPE affect one mechanical property of the plastics.
-
17N.3.sl.TZ0.7b.i:
Describe how the structures of LDPE and HDPE affect one mechanical property of the plastics.
-
17N.3.sl.TZ0.b.i:
Describe how the structures of LDPE and HDPE affect one mechanical property of the plastics.
-
17N.3.sl.TZ0.9c:
Explain how the inclusion of carbohydrates in plastics makes them biodegradable.
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17N.1.sl.TZ0.10:
Which compound has the shortest C–N bond?
A. CH3NH2
B. (CH3)3CNH2
C. CH3CN
D. CH3CHNH
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17N.3.sl.TZ0.9c:
Explain how the inclusion of carbohydrates in plastics makes them biodegradable.
-
17N.3.sl.TZ0.c:
Explain how the inclusion of carbohydrates in plastics makes them biodegradable.
-
17N.3.sl.TZ0.10a:
Explain, at the molecular level, why vitamin D is soluble in fats. Use section 35 of the data booklet.
-
17N.2.sl.TZ0.2b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
-
17N.1.sl.TZ0.10:
Which compound has the shortest C–N bond?
A. CH3NH2
B. (CH3)3CNH2
C. CH3CN
D. CH3CHNH
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17N.3.sl.TZ0.10a:
Explain, at the molecular level, why vitamin D is soluble in fats. Use section 35 of the data booklet.
-
17N.3.sl.TZ0.a:
Explain, at the molecular level, why vitamin D is soluble in fats. Use section 35 of the data booklet.
- 17N.1.sl.TZ0.11: Which of the following series shows increasing hydrogen bonding with water? A. Propane <...
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17N.3.hl.TZ0.7c:
Identify the type of intermolecular bonding that is responsible for Kevlar®’s strength.
-
17N.3.hl.TZ0.7c:
Identify the type of intermolecular bonding that is responsible for Kevlar®’s strength.
-
17N.3.hl.TZ0.c:
Identify the type of intermolecular bonding that is responsible for Kevlar®’s strength.
- 17N.1.sl.TZ0.11: Which of the following series shows increasing hydrogen bonding with water? A. Propane <...
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17N.2.sl.TZ0.2b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
-
17N.2.sl.TZ0.b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
-
18M.2.hl.TZ2.7e:
Carbon and silicon are elements in group 14.
Explain why CO2 is a gas but SiO2 is a solid at room temperature.
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17N.2.sl.TZ0.3a:
Draw the Lewis (electron dot) structures of PF3 and PF4+ and use the VSEPR theory to deduce the molecular geometry of each species.
-
17N.2.sl.TZ0.3a:
Draw the Lewis (electron dot) structures of PF3 and PF4+ and use the VSEPR theory to deduce the molecular geometry of each species.
-
17N.2.sl.TZ0.a:
Draw the Lewis (electron dot) structures of PF3 and PF4+ and use the VSEPR theory to deduce the molecular geometry of each species.
-
17N.2.sl.TZ0.3b:
Predict with a reason, whether the molecule PF3 is polar or non-polar.
-
17N.2.sl.TZ0.3b:
Predict with a reason, whether the molecule PF3 is polar or non-polar.
-
17N.2.sl.TZ0.b:
Predict with a reason, whether the molecule PF3 is polar or non-polar.
-
17N.2.hl.TZ0.3b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
- 18M.1.hl.TZ1.11: Which metal has the strongest metallic bonding? A. Na B. Mg C. Al D. Ca
- 18M.1.hl.TZ1.11: Which metal has the strongest metallic bonding? A. Na B. Mg C. Al D. Ca
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18M.2.hl.TZ2.7e:
Carbon and silicon are elements in group 14.
Explain why CO2 is a gas but SiO2 is a solid at room temperature.
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18M.2.hl.TZ2.e:
Carbon and silicon are elements in group 14.
Explain why CO2 is a gas but SiO2 is a solid at room temperature.
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18M.1.sl.TZ1.9:
What is the formula of ammonium phosphate?
A. (NH3)3PO4
B. (NH4)3PO4
C. (NH4)2PO4
D. (NH3)2PO3
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18M.2.hl.TZ1.1b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
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18M.2.hl.TZ1.1b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
-
18M.2.hl.TZ1.b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
-
17N.2.hl.TZ0.3b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
-
17N.2.hl.TZ0.b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
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17N.2.hl.TZ0.4a:
Draw the Lewis (electron dot) structures of PF3 and PF5 and use the VSEPR theory to deduce the molecular geometry of each species including bond angles.
-
17N.2.hl.TZ0.4a:
Draw the Lewis (electron dot) structures of PF3 and PF5 and use the VSEPR theory to deduce the molecular geometry of each species including bond angles.
-
17N.2.hl.TZ0.a:
Draw the Lewis (electron dot) structures of PF3 and PF5 and use the VSEPR theory to deduce the molecular geometry of each species including bond angles.
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18M.1.sl.TZ1.9:
What is the formula of ammonium phosphate?
A. (NH3)3PO4
B. (NH4)3PO4
C. (NH4)2PO4
D. (NH3)2PO3
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17N.2.hl.TZ0.4b:
Predict whether the molecules PF3 and PF5 are polar or non-polar.
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17N.2.hl.TZ0.4b:
Predict whether the molecules PF3 and PF5 are polar or non-polar.
-
17N.2.hl.TZ0.b:
Predict whether the molecules PF3 and PF5 are polar or non-polar.
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18M.2.hl.TZ1.1e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
- 18M.1.sl.TZ1.10: Which form of carbon is the poorest electrical conductor? A. Graphite B. Graphene C. ...
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18M.2.hl.TZ1.1e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
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18M.2.hl.TZ1.e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
- 18M.1.sl.TZ1.11: What is the molecular geometry and bond angle in the molecular ion NO3−?
- 18M.1.sl.TZ1.10: Which form of carbon is the poorest electrical conductor? A. Graphite B. Graphene C. ...
- 18M.1.sl.TZ1.11: What is the molecular geometry and bond angle in the molecular ion NO3−?
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18M.2.hl.TZ1.1e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
-
18M.2.hl.TZ1.1e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
-
18M.2.hl.TZ1.e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
- 18M.1.sl.TZ2.9: What is the formula of magnesium nitride? A. MgN B. Mg2N3 C. Mg3N D. Mg3N2
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18M.2.hl.TZ1.2a:
Describe the nature of ionic bonding.
- 18M.1.sl.TZ2.9: What is the formula of magnesium nitride? A. MgN B. Mg2N3 C. Mg3N D. Mg3N2
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18M.2.hl.TZ1.2a:
Describe the nature of ionic bonding.
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18M.2.hl.TZ1.a:
Describe the nature of ionic bonding.
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18M.1.sl.TZ1.12:
What are the strongest intermolecular forces between molecules of propanone, CH3COCH3, in the liquid phase?
A. London (dispersion) forces
B. Covalent bonding
C. Hydrogen bonding
D. Dipole–dipole forces
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18M.1.sl.TZ1.12:
What are the strongest intermolecular forces between molecules of propanone, CH3COCH3, in the liquid phase?
A. London (dispersion) forces
B. Covalent bonding
C. Hydrogen bonding
D. Dipole–dipole forces
- 18M.1.sl.TZ2.10: Which species has the longest carbon to oxygen bond length? A. CO B. CH3OH C. ...
- 18M.1.sl.TZ2.10: Which species has the longest carbon to oxygen bond length? A. CO B. CH3OH C. ...
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18M.2.sl.TZ1.1b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
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18M.2.hl.TZ1.2d.ii:
Outline why solid calcium is a good conductor of electricity.
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18M.1.sl.TZ2.11:
What are the predicted electron domain geometries around the carbon and both nitrogen atoms in urea, (NH2)2CO, applying VSEPR theory?
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18M.2.hl.TZ1.2d.ii:
Outline why solid calcium is a good conductor of electricity.
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18M.2.hl.TZ1.d.ii:
Outline why solid calcium is a good conductor of electricity.
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18M.1.sl.TZ2.11:
What are the predicted electron domain geometries around the carbon and both nitrogen atoms in urea, (NH2)2CO, applying VSEPR theory?
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18M.2.sl.TZ1.1b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
-
18M.2.sl.TZ1.b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
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18M.1.sl.TZ2.12:
The compounds shown below have similar relative molecular masses. What is the correct order of increasing boiling point?
A. CH3COOH < (CH3)2CO < (CH3)2CHOH
B. CH3COOH < (CH3)2CHOH < (CH3)2CO
C. (CH3)2CO < CH3COOH < (CH3)2CHOH
D. (CH3)2CO < (CH3)2CHOH < CH3COOH
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18M.2.sl.TZ2.6a.i:
Explain why the hydrides of group 16 elements (H2O, H2S, H2Se and H2Te) are polar molecules.
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18M.1.sl.TZ2.12:
The compounds shown below have similar relative molecular masses. What is the correct order of increasing boiling point?
A. CH3COOH < (CH3)2CO < (CH3)2CHOH
B. CH3COOH < (CH3)2CHOH < (CH3)2CO
C. (CH3)2CO < CH3COOH < (CH3)2CHOH
D. (CH3)2CO < (CH3)2CHOH < CH3COOH
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18M.2.sl.TZ2.6a.i:
Explain why the hydrides of group 16 elements (H2O, H2S, H2Se and H2Te) are polar molecules.
-
18M.2.sl.TZ2.a.i:
Explain why the hydrides of group 16 elements (H2O, H2S, H2Se and H2Te) are polar molecules.
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18M.2.sl.TZ1.1e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
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18N.2.sl.TZ0.3b:
Draw the Lewis (electron dot) structure for BrO3− that obeys the octet rule.
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18M.2.sl.TZ1.1e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
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18M.2.sl.TZ1.e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
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18M.2.sl.TZ1.1e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
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18M.2.sl.TZ1.1e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
-
18M.2.sl.TZ1.e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
-
18M.2.sl.TZ1.2a:
Describe the nature of ionic bonding.
-
18M.2.sl.TZ1.2a:
Describe the nature of ionic bonding.
-
18M.2.sl.TZ1.a:
Describe the nature of ionic bonding.
-
18M.2.sl.TZ1.2d.ii:
Outline why solid calcium is a good conductor of electricity.
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18M.2.sl.TZ2.6a.ii:
The graph shows the boiling points of the hydrides of group 16 elements.
Explain the increase in the boiling point from H2S to H2Te.
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18M.2.sl.TZ2.6a.ii:
The graph shows the boiling points of the hydrides of group 16 elements.
Explain the increase in the boiling point from H2S to H2Te.
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18M.2.sl.TZ2.a.ii:
The graph shows the boiling points of the hydrides of group 16 elements.
Explain the increase in the boiling point from H2S to H2Te.
-
18M.2.sl.TZ2.6b:
Lewis structures show electron domains and are used to predict molecular geometry.
Deduce the electron domain geometry and the molecular geometry for the NH2− ion.
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18M.2.sl.TZ2.6b:
Lewis structures show electron domains and are used to predict molecular geometry.
Deduce the electron domain geometry and the molecular geometry for the NH2− ion.
-
18M.2.sl.TZ2.b:
Lewis structures show electron domains and are used to predict molecular geometry.
Deduce the electron domain geometry and the molecular geometry for the NH2− ion.
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18M.3.sl.TZ2.1a.i:
Graphene is two-dimensional, rather than three-dimensional, material.
Justify this by using the structure of graphene and information from the table.
-
18M.3.sl.TZ2.1a.i:
Graphene is two-dimensional, rather than three-dimensional, material.
Justify this by using the structure of graphene and information from the table.
-
18M.3.sl.TZ2.a.i:
Graphene is two-dimensional, rather than three-dimensional, material.
Justify this by using the structure of graphene and information from the table.
-
18M.3.sl.TZ2.1a.ii:
Show that graphene is over 1600 times stronger than graphite.
-
18M.3.sl.TZ2.1a.ii:
Show that graphene is over 1600 times stronger than graphite.
-
18M.3.sl.TZ2.a.ii:
Show that graphene is over 1600 times stronger than graphite.
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18M.3.sl.TZ2.1a.iii:
Identify a value from the table which can be used to support the information about graphene given below.
Electrons in a solid are restricted to certain ranges, or bands, of energy (vertical axis). In an insulator or semiconductor, an electron bound to an atom can break free only if it gets enough energy from heat or a passing photon to jump the “band gap”, but in graphene the gap is infinitely small.
-
18M.3.sl.TZ2.1a.iii:
Identify a value from the table which can be used to support the information about graphene given below.
Electrons in a solid are restricted to certain ranges, or bands, of energy (vertical axis). In an insulator or semiconductor, an electron bound to an atom can break free only if it gets enough energy from heat or a passing photon to jump the “band gap”, but in graphene the gap is infinitely small.
-
18M.3.sl.TZ2.a.iii:
Identify a value from the table which can be used to support the information about graphene given below.
Electrons in a solid are restricted to certain ranges, or bands, of energy (vertical axis). In an insulator or semiconductor, an electron bound to an atom can break free only if it gets enough energy from heat or a passing photon to jump the “band gap”, but in graphene the gap is infinitely small.
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18M.3.sl.TZ2.1b:
Diamond, graphene, and graphite are all network solids.
Suggest, giving a reason, the electron mobility of diamond compared to graphene.
-
18M.3.sl.TZ2.1b:
Diamond, graphene, and graphite are all network solids.
Suggest, giving a reason, the electron mobility of diamond compared to graphene.
-
18M.3.sl.TZ2.b:
Diamond, graphene, and graphite are all network solids.
Suggest, giving a reason, the electron mobility of diamond compared to graphene.
-
18M.3.sl.TZ2.1c:
The melting point of diamond at 1 × 106 kPa is 4200 K (in the absence of oxygen).
Suggest, based on molecular structure, why graphene has a higher melting point under these conditions.
-
18M.3.sl.TZ2.1c:
The melting point of diamond at 1 × 106 kPa is 4200 K (in the absence of oxygen).
Suggest, based on molecular structure, why graphene has a higher melting point under these conditions.
-
18M.3.sl.TZ2.c:
The melting point of diamond at 1 × 106 kPa is 4200 K (in the absence of oxygen).
Suggest, based on molecular structure, why graphene has a higher melting point under these conditions.
- 18N.1.sl.TZ0.9: Which species has the same molecular geometry as SO32−? A. BF3 B. SO3 C. PF3 D. CO32−
- 18N.1.sl.TZ0.9: Which species has the same molecular geometry as SO32−? A. BF3 B. SO3 C. PF3 D. CO32−
- 18N.1.sl.TZ0.10: How many lone pairs and bonding pairs of electrons surround the central chlorine atom in...
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18N.2.sl.TZ0.3b:
Draw the Lewis (electron dot) structure for BrO3− that obeys the octet rule.
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18M.2.sl.TZ1.2d.ii:
Outline why solid calcium is a good conductor of electricity.
-
18M.2.sl.TZ1.d.ii:
Outline why solid calcium is a good conductor of electricity.
-
18M.3.sl.TZ1.1a.i:
Part of this molecule is hydrophilic (bonds readily to water) and part hydrophobic (does not bond readily to water). Draw a circle around all of the hydrophilic part of the molecule.
-
18M.3.sl.TZ1.1a.i:
Part of this molecule is hydrophilic (bonds readily to water) and part hydrophobic (does not bond readily to water). Draw a circle around all of the hydrophilic part of the molecule.
-
18M.3.sl.TZ1.a.i:
Part of this molecule is hydrophilic (bonds readily to water) and part hydrophobic (does not bond readily to water). Draw a circle around all of the hydrophilic part of the molecule.
-
18N.2.sl.TZ0.b:
Draw the Lewis (electron dot) structure for BrO3− that obeys the octet rule.
- 18N.2.sl.TZ0.3c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
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18M.3.sl.TZ1.1a.ii:
When a small amount of palmitic acid is placed in water it disperses to form a layer on the surface that is only one molecule thick. Explain, in terms of intermolecular forces, why this occurs.
-
18M.3.sl.TZ1.1a.ii:
When a small amount of palmitic acid is placed in water it disperses to form a layer on the surface that is only one molecule thick. Explain, in terms of intermolecular forces, why this occurs.
-
18M.3.sl.TZ1.a.ii:
When a small amount of palmitic acid is placed in water it disperses to form a layer on the surface that is only one molecule thick. Explain, in terms of intermolecular forces, why this occurs.
- 18N.1.sl.TZ0.10: How many lone pairs and bonding pairs of electrons surround the central chlorine atom in...
- 18N.1.sl.TZ0.11: Which compound has the highest boiling point? A. CH3CHO B. CH3CH2F C. CH3OCH3 D. ...
- 18N.2.sl.TZ0.3c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
- 18N.1.sl.TZ0.11: Which compound has the highest boiling point? A. CH3CHO B. CH3CH2F C. CH3OCH3 D. ...
- 18N.1.sl.TZ0.12: Which molecule is polar? A. BeCl2 B. BCl3 C. NCl3 D. CCl4
- 18N.2.hl.TZ0.8a: Suggest why the three-membered ring in methyloxirane is unstable.
- 18N.2.sl.TZ0.c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
-
18N.2.hl.TZ0.3b.i:
Draw two Lewis (electron dot) structures for BrO3−.
- 18N.2.hl.TZ0.8a: Suggest why the three-membered ring in methyloxirane is unstable.
- 18N.2.hl.TZ0.a: Suggest why the three-membered ring in methyloxirane is unstable.
- 18N.1.sl.TZ0.12: Which molecule is polar? A. BeCl2 B. BCl3 C. NCl3 D. CCl4
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18N.2.hl.TZ0.3b.i:
Draw two Lewis (electron dot) structures for BrO3−.
-
18N.2.hl.TZ0.b.i:
Draw two Lewis (electron dot) structures for BrO3−.
- 18N.2.hl.TZ0.3c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
- 18N.2.hl.TZ0.3c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
- 18N.2.hl.TZ0.c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
- 18N.2.sl.TZ0.4b.ii: State a physical property of sodium oxide.
- 18N.2.sl.TZ0.4b.ii: State a physical property of sodium oxide.
- 18N.2.sl.TZ0.b.ii: State a physical property of sodium oxide.
- 18N.2.sl.TZ0.6b: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
- 18N.2.sl.TZ0.6b: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
- 18N.2.sl.TZ0.b: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
- 18N.2.hl.TZ0.6d: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
- 18N.2.hl.TZ0.6d: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
- 18N.2.hl.TZ0.d: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
-
19M.2.hl.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
-
19M.2.hl.TZ1.2h:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.hl.TZ1.2h:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.hl.TZ1.h:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.hl.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.hl.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.hl.TZ1.c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.hl.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
-
19M.2.hl.TZ1.5c(iii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.hl.TZ1.5c(iii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.hl.TZ1.c(iii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.hl.TZ2.b(i):
Deduce the Lewis (electron dot) structure of ethyne.
-
19M.2.hl.TZ2.1b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.hl.TZ1.5c(iv):
State the type of bond formed when chloramine is protonated.
-
19M.2.hl.TZ1.3c:
Describe the structure and bonding in solid sodium oxide.
-
19M.2.hl.TZ2.1b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.hl.TZ1.5c(iv):
State the type of bond formed when chloramine is protonated.
-
19M.2.hl.TZ1.c(iv):
State the type of bond formed when chloramine is protonated.
-
19M.2.hl.TZ2.b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.hl.TZ2.1b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.hl.TZ1.3c:
Describe the structure and bonding in solid sodium oxide.
-
19M.2.hl.TZ1.c:
Describe the structure and bonding in solid sodium oxide.
-
19M.2.hl.TZ2.1b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.hl.TZ2.b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.hl.TZ2.1d(iii):
Explain why product B is water soluble.
-
19M.2.hl.TZ2.3d(ii):
State, giving a reason, the shape of the dinitrogen monoxide molecule.
-
19M.2.hl.TZ2.1d(iii):
Explain why product B is water soluble.
-
19M.2.hl.TZ2.3d(ii):
State, giving a reason, the shape of the dinitrogen monoxide molecule.
-
19M.2.hl.TZ2.d(ii):
State, giving a reason, the shape of the dinitrogen monoxide molecule.
-
19M.1.hl.TZ1.9:
What is the order of increasing boiling point?
A. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
B. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
C. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
D. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
-
19M.2.hl.TZ2.d(iii):
Explain why product B is water soluble.
- 19M.1.hl.TZ2.9: How does a lithium atom form the most stable ion? A. The atom gains a proton to form a positive...
-
19M.1.hl.TZ1.9:
What is the order of increasing boiling point?
A. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
B. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
C. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
D. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
-
19M.1.hl.TZ1.10:
What is the IUPAC name of NiCO3?
A. nickel(II) carbonate
B. nickel carbonate
C. nickel(I) carbonate
D. nitrogen(I) carbonate
-
19M.2.hl.TZ2.5d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
- 19M.1.hl.TZ2.9: How does a lithium atom form the most stable ion? A. The atom gains a proton to form a positive...
- 19M.1.hl.TZ2.10: Which combination causes the strength of metallic bonding to increase?
-
19M.1.hl.TZ1.10:
What is the IUPAC name of NiCO3?
A. nickel(II) carbonate
B. nickel carbonate
C. nickel(I) carbonate
D. nitrogen(I) carbonate
- 19M.1.hl.TZ1.11: Which combination corresponds to a strong metallic bond?
- 19M.1.hl.TZ2.10: Which combination causes the strength of metallic bonding to increase?
-
19M.2.hl.TZ2.5d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.2.hl.TZ2.d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.2.sl.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
-
19M.2.sl.TZ1.2c:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
- 19M.1.hl.TZ1.11: Which combination corresponds to a strong metallic bond?
-
19M.1.sl.TZ1.9:
What is the order of increasing boiling point?
A. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
B. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
C. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
D. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
-
19M.2.sl.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
-
19M.1.sl.TZ1.9:
What is the order of increasing boiling point?
A. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
B. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
C. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
D. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
- 19M.1.sl.TZ1.10: Which species does not have resonance structures? A. C6H6 B. NH4+ C. CO32− D. O3
- 19M.1.sl.TZ1.10: Which species does not have resonance structures? A. C6H6 B. NH4+ C. CO32− D. O3
- 19M.1.sl.TZ1.11: Which describes an ionic compound?
-
19M.2.sl.TZ1.2c:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.sl.TZ1.c:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.sl.TZ2.b(i):
Deduce the Lewis (electron dot) structure of ethyne.
-
19M.2.sl.TZ2.1b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.sl.TZ1.3a(i):
Describe the structure and bonding in solid sodium oxide.
- 19M.1.sl.TZ1.11: Which describes an ionic compound?
- 19M.1.sl.TZ1.12: Which combination corresponds to a strong metallic bond?
-
19M.2.sl.TZ2.1b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.sl.TZ1.3a(i):
Describe the structure and bonding in solid sodium oxide.
-
19M.2.sl.TZ1.a(i):
Describe the structure and bonding in solid sodium oxide.
-
19M.2.sl.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.sl.TZ2.b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.sl.TZ2.1b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.sl.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.sl.TZ1.c(i):
Draw a Lewis (electron dot) structure of chloramine.
- 19M.1.sl.TZ1.12: Which combination corresponds to a strong metallic bond?
- 19M.1.sl.TZ2.9: How does a lithium atom form the most stable ion? A. The atom gains a proton to form a positive...
-
19M.2.sl.TZ2.1b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
- 19M.1.sl.TZ2.9: How does a lithium atom form the most stable ion? A. The atom gains a proton to form a positive...
-
19M.2.sl.TZ1.5c(ii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.sl.TZ1.5c(ii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.sl.TZ1.c(ii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.sl.TZ2.b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.sl.TZ2.1d(iii):
Explain why product B is water soluble.
- 19M.1.sl.TZ2.10: Which combination causes the strength of metallic bonding to increase?
- 19M.1.sl.TZ2.10: Which combination causes the strength of metallic bonding to increase?
-
19M.2.sl.TZ2.1d(iii):
Explain why product B is water soluble.
- 19M.1.sl.TZ2.11: Which molecule contains an incomplete octet of electrons? A. NF3 B. BF3 C. BrF D. SF2
-
19M.2.sl.TZ2.d(iii):
Explain why product B is water soluble.
-
19M.2.sl.TZ2.5b(iii):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
- 19M.1.sl.TZ2.11: Which molecule contains an incomplete octet of electrons? A. NF3 B. BF3 C. BrF D. SF2
- 19M.1.sl.TZ2.12: Which compound has hydrogen bonds between its molecules? A. CH4 B. CH4O C. CH3Cl D. CH2O
- 19M.1.sl.TZ2.12: Which compound has hydrogen bonds between its molecules? A. CH4 B. CH4O C. CH3Cl D. CH2O
-
19M.2.sl.TZ2.5b(iii):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.2.sl.TZ2.b(iii):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
- 19N.2.sl.TZ0.1a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.2.sl.TZ0.1a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
-
19N.3.sl.TZ0.5b(i):
Determine the percentage of ionic bonding in alumina using sections 8 and 29 of the data booklet.
- 19N.2.sl.TZ0.a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.2.sl.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
-
19N.3.sl.TZ0.5b(i):
Determine the percentage of ionic bonding in alumina using sections 8 and 29 of the data booklet.
-
19N.3.sl.TZ0.b(i):
Determine the percentage of ionic bonding in alumina using sections 8 and 29 of the data booklet.
- 19N.2.sl.TZ0.6a(iv): Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled,...
- 19N.2.sl.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.3.sl.TZ0.9c: Explain why stearic acid has a higher melting point than oleic acid.
- 19N.3.sl.TZ0.9c: Explain why stearic acid has a higher melting point than oleic acid.
- 19N.3.sl.TZ0.c: Explain why stearic acid has a higher melting point than oleic acid.
- 19N.2.sl.TZ0.b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.2.sl.TZ0.1d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
- 19N.3.hl.TZ0.13b: Explain how the double-helical structure of DNA is stabilized once formed.
- 19N.3.hl.TZ0.13b: Explain how the double-helical structure of DNA is stabilized once formed.
- 19N.3.hl.TZ0.b: Explain how the double-helical structure of DNA is stabilized once formed.
- 19N.2.sl.TZ0.6a(iv): Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled,...
- 19N.2.sl.TZ0.a(iv): Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled,...
- 19N.2.sl.TZ0.1d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
- 19N.3.sl.TZ0.10c: Explain why maltose, C12H22O11, is soluble in water.
-
19N.3.sl.TZ0.17b:
Aspirin, C6H4(OCOCH3)COOH, is only slightly soluble in water.
Outline, including an equation, how aspirin can be made more water-soluble. Use section 37 in the data booklet.
- 19N.2.sl.TZ0.d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
- 19N.2.hl.TZ0.1a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
-
19N.3.sl.TZ0.17b:
Aspirin, C6H4(OCOCH3)COOH, is only slightly soluble in water.
Outline, including an equation, how aspirin can be made more water-soluble. Use section 37 in the data booklet.
-
19N.3.sl.TZ0.b:
Aspirin, C6H4(OCOCH3)COOH, is only slightly soluble in water.
Outline, including an equation, how aspirin can be made more water-soluble. Use section 37 in the data booklet.
- 19N.3.sl.TZ0.10c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.3.sl.TZ0.c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.3.hl.TZ0.15c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.3.hl.TZ0.15c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.3.hl.TZ0.c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.2.hl.TZ0.1a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.3.sl.TZ0.4b: Predict, with a reason, whether isotactic or atactic polypropene has the higher melting point.
- 19N.2.hl.TZ0.a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.2.hl.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.3.sl.TZ0.4b: Predict, with a reason, whether isotactic or atactic polypropene has the higher melting point.
- 19N.3.sl.TZ0.b: Predict, with a reason, whether isotactic or atactic polypropene has the higher melting point.
- 19N.1.sl.TZ0.10: Which compound has the shortest C to O bond? A. CH3CHO B. CO C. CO2 D. C2H5OC2H5
- 19N.2.hl.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.1.sl.TZ0.10: Which compound has the shortest C to O bond? A. CH3CHO B. CO C. CO2 D. C2H5OC2H5
- 19N.2.hl.TZ0.b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.2.hl.TZ0.1c: Predict the bond angle in the ozone molecule.
- 19N.1.sl.TZ0.11: Which describes a resonance structure? A. Double bond can be drawn in alternative...
-
19N.2.sl.TZ0.3d(ii):
Explain why the compound C2H6O, produced in (b), has a higher boiling point than compound C2H4O, produced in d(i).
- 19N.2.hl.TZ0.1c: Predict the bond angle in the ozone molecule.
- 19N.2.hl.TZ0.c: Predict the bond angle in the ozone molecule.
- 19N.2.hl.TZ0.1d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
-
19N.2.sl.TZ0.3d(ii):
Explain why the compound C2H6O, produced in (b), has a higher boiling point than compound C2H4O, produced in d(i).
-
19N.2.sl.TZ0.d(ii):
Explain why the compound C2H6O, produced in (b), has a higher boiling point than compound C2H4O, produced in d(i).
- 19N.1.sl.TZ0.11: Which describes a resonance structure? A. Double bond can be drawn in alternative...
- 19N.2.hl.TZ0.1d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
-
19N.1.sl.TZ0.12:
What is the structure and bonding in SiO2 (s)?
- 19N.2.hl.TZ0.d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
-
19N.2.hl.TZ0.3d(ii):
Explain why the compound C3H8O, produced in (a)(iv), has a higher boiling point than compound C3H6O, produced in d(i).
-
19N.1.sl.TZ0.12:
What is the structure and bonding in SiO2 (s)?
- 19N.1.sl.TZ0.9: Which is correct for all solid ionic compounds? A. High volatility B. Poor electrical...
-
19N.2.hl.TZ0.3d(ii):
Explain why the compound C3H8O, produced in (a)(iv), has a higher boiling point than compound C3H6O, produced in d(i).
- 19N.1.sl.TZ0.9: Which is correct for all solid ionic compounds? A. High volatility B. Poor electrical...
-
19N.2.hl.TZ0.d(ii):
Explain why the compound C3H8O, produced in (a)(iv), has a higher boiling point than compound C3H6O, produced in d(i).
-
20N.1.sl.TZ0.9:
Which formula is correct?
A.
B.
C.
D.
-
20N.1.sl.TZ0.9:
Which formula is correct?
A.
B.
C.
D.
- 20N.2.sl.TZ0.2a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
-
20N.1.hl.TZ0.9:
Which of these species contains the shortest carbon to oxygen bond length?
A.
B.
C.
D.
-
20N.1.hl.TZ0.9:
Which of these species contains the shortest carbon to oxygen bond length?
A.
B.
C.
D.
-
20N.1.sl.TZ0.10:
Which molecule is most polar?
A.
B.
C.
D.
-
20N.1.sl.TZ0.10:
Which molecule is most polar?
A.
B.
C.
D.
- 20N.2.hl.TZ0.2a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
- 20N.2.sl.TZ0.2a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
- 20N.2.sl.TZ0.a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
-
20N.1.sl.TZ0.11:
Which combination correctly describes the geometry of the carbonate ion, ?
-
20N.1.sl.TZ0.11:
Which combination correctly describes the geometry of the carbonate ion, ?
- 20N.2.sl.TZ0.4c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
- 20N.2.sl.TZ0.4c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
- 20N.2.sl.TZ0.c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
- 20N.2.hl.TZ0.a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
- 20N.2.hl.TZ0.2a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
- 20N.2.hl.TZ0.4c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
-
20N.1.sl.TZ0.12:
Which series shows the correct order of metallic bond strength from strongest to weakest?
A.
B.
C.
D.
- 20N.2.hl.TZ0.4c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
- 20N.2.hl.TZ0.c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
-
20N.1.sl.TZ0.12:
Which series shows the correct order of metallic bond strength from strongest to weakest?
A.
B.
C.
D.
- 20N.2.hl.TZ0.4d(v): Describe the bonding in metals.
- 20N.2.hl.TZ0.4d(v): Describe the bonding in metals.
- 20N.2.hl.TZ0.d(v): Describe the bonding in metals.
- 20N.2.sl.TZ0.4d(iii): Describe the bonding in metals.
-
20N.3.sl.TZ0.1a:
Suggest why a non-polar solvent was needed.
- 20N.2.sl.TZ0.4d(iii): Describe the bonding in metals.
- 20N.2.sl.TZ0.d(iii): Describe the bonding in metals.
- 20N.2.sl.TZ0.4d(iv): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.2.hl.TZ0.4d(vi): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.2.hl.TZ0.4d(vi): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.2.hl.TZ0.d(vi): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
-
20N.3.sl.TZ0.1a:
Suggest why a non-polar solvent was needed.
-
20N.3.sl.TZ0.a:
Suggest why a non-polar solvent was needed.
- 20N.3.sl.TZ0.1c: Non-polar solvents can be toxic. Suggest a modification to the experiment which allows the...
- 20N.2.sl.TZ0.4d(iv): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.2.sl.TZ0.d(iv): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.3.sl.TZ0.1c: Non-polar solvents can be toxic. Suggest a modification to the experiment which allows the...
- 20N.3.sl.TZ0.4b(i): Alloying metals changes their properties. Suggest one property of magnesium that could be...
- 20N.3.sl.TZ0.c: Non-polar solvents can be toxic. Suggest a modification to the experiment which allows the...
- 20N.3.sl.TZ0.11b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
- 20N.3.sl.TZ0.4b(i): Alloying metals changes their properties. Suggest one property of magnesium that could be...
- 20N.3.sl.TZ0.b(i): Alloying metals changes their properties. Suggest one property of magnesium that could be...
-
20N.3.sl.TZ0.5a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
- 20N.3.sl.TZ0.11b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
-
20N.3.sl.TZ0.5a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
-
20N.3.sl.TZ0.a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
-
20N.3.sl.TZ0.6b(ii):
State the most significant intermolecular forces in the phospholipid in b(i).
- 20N.3.sl.TZ0.b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
- 20N.3.hl.TZ0.15b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
-
20N.3.sl.TZ0.6b(ii):
State the most significant intermolecular forces in the phospholipid in b(i).
-
20N.3.sl.TZ0.b(ii):
State the most significant intermolecular forces in the phospholipid in b(i).
-
20N.3.hl.TZ0.6a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
-
20N.3.hl.TZ0.4c(i):
Alloying metals changes their properties. Suggest one property of magnesium that could be improved by making a magnesium–CNT alloy.
-
20N.3.hl.TZ0.4c(i):
Alloying metals changes their properties. Suggest one property of magnesium that could be improved by making a magnesium–CNT alloy.
-
20N.3.hl.TZ0.c(i):
Alloying metals changes their properties. Suggest one property of magnesium that could be improved by making a magnesium–CNT alloy.
- 20N.3.hl.TZ0.15b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
-
20N.3.hl.TZ0.6a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
-
20N.3.hl.TZ0.a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
- 20N.3.hl.TZ0.b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
- 21M.1.sl.TZ2.9: Which compound has the greatest volatility under the same conditions? A. SO2 B. SiO2 C. ...
- 21M.1.sl.TZ1.9: The Lewis structure of methylamine is shown. What is the molecular geometry around N? A. ...
- 21M.1.sl.TZ1.24: Which series is in order of increasing boiling point? A. CH2CH2CH3OH CH3COCH3 ...
- 21M.1.sl.TZ2.9: Which compound has the greatest volatility under the same conditions? A. SO2 B. SiO2 C. ...
- 21M.1.sl.TZ1.9: The Lewis structure of methylamine is shown. What is the molecular geometry around N? A. ...
-
21M.1.sl.TZ1.10:
Which compound contains both ionic and covalent bonds?
A.
B.
C.
D.
-
21M.1.sl.TZ2.10:
Which compound has the shortest C to N bond?
A. HCN
B. CH3CH2NH2
C. CH3CHNH
D. (CH3)2NH
-
21M.1.sl.TZ2.10:
Which compound has the shortest C to N bond?
A. HCN
B. CH3CH2NH2
C. CH3CHNH
D. (CH3)2NH
- 21M.1.sl.TZ1.24: Which series is in order of increasing boiling point? A. CH2CH2CH3OH CH3COCH3 ...
-
21M.1.sl.TZ2.11:
What is the formula of the compound formed from Ca2+ and PO43−?
A. CaPO4
B. Ca3(PO4)2
C. Ca2(PO4)3
D. Ca(PO4)2
- 21M.1.hl.TZ2.12: Which atom has an expanded octet? A. C in CO2 B. S in SCl4 C. O in H2O2 D. P in PCl3
-
21M.1.sl.TZ2.11:
What is the formula of the compound formed from Ca2+ and PO43−?
A. CaPO4
B. Ca3(PO4)2
C. Ca2(PO4)3
D. Ca(PO4)2
-
21M.1.sl.TZ1.10:
Which compound contains both ionic and covalent bonds?
A.
B.
C.
D.
- 21M.1.sl.TZ1.11: Which substance is most likely to be ionic?
- 21M.1.sl.TZ2.12: Which is the correct order based on increasing strength? A. covalent bonds < hydrogen bonds...
- 21M.1.hl.TZ2.12: Which atom has an expanded octet? A. C in CO2 B. S in SCl4 C. O in H2O2 D. P in PCl3
- 21M.1.sl.TZ2.12: Which is the correct order based on increasing strength? A. covalent bonds < hydrogen bonds...
- 21M.1.sl.TZ1.11: Which substance is most likely to be ionic?
- 21M.1.sl.TZ1.12: Along which series is the bond angle increasing? A. NH3 H2O CH4 B. CH4 NH3 H2O C. H2O ...
-
21M.2.sl.TZ1.2a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
- 21M.1.sl.TZ1.12: Along which series is the bond angle increasing? A. NH3 H2O CH4 B. CH4 NH3 H2O C. H2O ...
- 21M.2.sl.TZ1.1a: Outline why metals, like iron, can conduct electricity.
-
21M.2.sl.TZ1.2a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
- 21M.2.sl.TZ1.1a: Outline why metals, like iron, can conduct electricity.
- 21M.2.sl.TZ1.a: Outline why metals, like iron, can conduct electricity.
-
21M.2.sl.TZ1.a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
- 21M.2.sl.TZ1.2a(ii): Predict the shape of the hydrogen sulfide molecule.
- 21M.2.sl.TZ1.2a(ii): Predict the shape of the hydrogen sulfide molecule.
- 21M.2.sl.TZ1.1c(i): Describe the bonding in this type of solid.
- 21M.2.hl.TZ1.1a: Outline why metals, like iron, can conduct electricity.
- 21M.2.sl.TZ1.a(ii): Predict the shape of the hydrogen sulfide molecule.
- 21M.2.sl.TZ1.3a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21M.2.hl.TZ1.1a: Outline why metals, like iron, can conduct electricity.
- 21M.2.hl.TZ1.a: Outline why metals, like iron, can conduct electricity.
- 21M.2.sl.TZ1.1c(i): Describe the bonding in this type of solid.
- 21M.2.sl.TZ1.c(i): Describe the bonding in this type of solid.
- 21M.2.sl.TZ1.3a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21M.2.sl.TZ1.1c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.1d(i): Describe the bonding in this type of solid.
- 21M.2.hl.TZ1.1d(i): Describe the bonding in this type of solid.
- 21M.2.hl.TZ1.d(i): Describe the bonding in this type of solid.
- 21M.2.sl.TZ1.a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21M.2.hl.TZ1.3a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21M.2.hl.TZ1.1d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.sl.TZ1.1c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.sl.TZ1.c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.3a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21M.2.sl.TZ1.1e: Explain why the addition of small amounts of carbon to iron makes the metal harder.
- 21M.2.hl.TZ1.1d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
-
21M.2.hl.TZ1.7a(i):
Draw a Lewis (electron dot) structure for ozone.
- 21M.2.hl.TZ1.1f: Explain why the addition of small amounts of carbon to iron makes the metal harder.
- 21M.2.sl.TZ1.e: Explain why the addition of small amounts of carbon to iron makes the metal harder.
- 21M.2.sl.TZ1.1e: Explain why the addition of small amounts of carbon to iron makes the metal harder.
-
21M.2.hl.TZ1.7a(i):
Draw a Lewis (electron dot) structure for ozone.
- 21M.2.hl.TZ1.1f: Explain why the addition of small amounts of carbon to iron makes the metal harder.
- 21M.2.hl.TZ1.f: Explain why the addition of small amounts of carbon to iron makes the metal harder.
-
21M.2.hl.TZ1.a(i):
Draw a Lewis (electron dot) structure for ozone.
- 21M.2.hl.TZ2.2b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
- 21M.2.hl.TZ2.2b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
- 21M.2.sl.TZ2.2b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
- 21M.2.hl.TZ2.b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
-
21M.2.hl.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF4, and sulfur dichloride, SCl2.
-
21M.2.hl.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF4, and sulfur dichloride, SCl2.
- 21M.2.sl.TZ2.2b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
- 21M.2.sl.TZ2.b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
- 21N.1.sl.TZ0.9: Which molecule has the weakest nitrogen to nitrogen bond? A. N2 B. N2H2 C. N2H4 D.
-
21M.2.hl.TZ2.c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF4, and sulfur dichloride, SCl2.
- 21M.2.hl.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21N.1.sl.TZ0.9: Which molecule has the weakest nitrogen to nitrogen bond? A. N2 B. N2H2 C. N2H4 D.
- 21M.2.hl.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21N.1.sl.TZ0.10: Which combination would create the strongest ionic bond?
-
21M.2.sl.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl2.
- 21M.2.hl.TZ2.d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
-
21N.2.sl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
-
21M.2.sl.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl2.
-
21M.2.sl.TZ2.c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl2.
- 21M.2.sl.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21N.1.sl.TZ0.10: Which combination would create the strongest ionic bond?
- 21N.1.sl.TZ0.11: Which compound contains both ionic and covalent bonds? A. CH3COONa B. CH3COOH C. K2O D. ...
-
21N.2.sl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
- 21N.1.sl.TZ0.11: Which compound contains both ionic and covalent bonds? A. CH3COONa B. CH3COOH C. K2O D. ...
- 21N.1.sl.TZ0.12: The following compounds have similar relative molecular masses. What is the order of increasing...
- 21N.1.sl.TZ0.12: The following compounds have similar relative molecular masses. What is the order of increasing...
- 21M.2.sl.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21M.2.sl.TZ2.d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
-
21N.2.sl.TZ0.a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
- 21N.2.sl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
-
21N.2.hl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
- 21N.1.sl.TZ0.13: Which alcohol is least soluble in water? A. CH3OH B. CH3CH2OH C. CH3CH2CH2OH D. ...
- 21N.1.sl.TZ0.13: Which alcohol is least soluble in water? A. CH3OH B. CH3CH2OH C. CH3CH2CH2OH D. ...
- 21N.2.sl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
-
21N.2.hl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
-
21N.2.hl.TZ0.a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
- 21N.2.sl.TZ0.b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
- 21N.2.sl.TZ0.3b(ii): Explain the polarity of PCl3.
- 21N.2.hl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
-
21N.2.hl.TZ0.6c(v):
Suggest a reason that the Winkler Method used to measure biochemical oxygen demand (BOD) must be done at constant temperature.
- 21N.2.sl.TZ0.3b(ii): Explain the polarity of PCl3.
-
21N.2.hl.TZ0.c(v):
Suggest a reason that the Winkler Method used to measure biochemical oxygen demand (BOD) must be done at constant temperature.
-
21N.2.hl.TZ0.6c(v):
Suggest a reason that the Winkler Method used to measure biochemical oxygen demand (BOD) must be done at constant temperature.
- 21N.2.hl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
- 21N.2.hl.TZ0.b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
- 21N.2.sl.TZ0.b(ii): Explain the polarity of PCl3.
-
22M.1.sl.TZ2.9:
Which statement best describes the intramolecular bonding in HCN (l)?
A. Electrostatic attractions between H+ and CN− ions
B. Hydrogen bonding
C. Van der Waals forces and hydrogen bonding
D. Electrostatic attractions between pairs of electrons and positively charged nuclei
- 21N.2.hl.TZ0.3b(ii): Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.
- 21N.2.hl.TZ0.3b(ii): Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.
- 21N.2.hl.TZ0.b(ii): Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.
- 22M.1.hl.TZ1.13: What are the electron domain and molecular geometries of the XeF4 molecule?
- 22M.1.sl.TZ1.9: A compound consists of the ions Ca2+ and PO43–. What are the name and formula of the compound?
-
22M.1.sl.TZ2.9:
Which statement best describes the intramolecular bonding in HCN (l)?
A. Electrostatic attractions between H+ and CN− ions
B. Hydrogen bonding
C. Van der Waals forces and hydrogen bonding
D. Electrostatic attractions between pairs of electrons and positively charged nuclei
- 22M.1.sl.TZ1.9: A compound consists of the ions Ca2+ and PO43–. What are the name and formula of the compound?
- 22M.1.sl.TZ1.10: What is the explanation for the high melting point of sodium chloride? A. The covalent bond...
- 22M.1.hl.TZ1.13: What are the electron domain and molecular geometries of the XeF4 molecule?
- 22M.1.sl.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
- 22M.1.hl.TZ2.9: In which of the following compounds does ionic bonding predominate? A. HCl B. NaF C. ...
- 22M.1.sl.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
- 22M.1.hl.TZ2.9: In which of the following compounds does ionic bonding predominate? A. HCl B. NaF C. ...
-
22M.1.sl.TZ2.11:
What is the name of the compound with formula Ti3(PO4)2?
A. Titanium phosphate
B. Titanium(II) phosphate
C. Titanium(III) phosphate
D. Titanium(IV) phosphate
- 22M.1.sl.TZ1.10: What is the explanation for the high melting point of sodium chloride? A. The covalent bond...
- 22M.1.sl.TZ1.11: Which molecule is most polar? A. CF4 B. CCl4 C. CHF3 D. CClF3
-
22M.1.sl.TZ2.11:
What is the name of the compound with formula Ti3(PO4)2?
A. Titanium phosphate
B. Titanium(II) phosphate
C. Titanium(III) phosphate
D. Titanium(IV) phosphate
- 22M.1.sl.TZ1.11: Which molecule is most polar? A. CF4 B. CCl4 C. CHF3 D. CClF3
- 22M.1.sl.TZ1.12: For which species can resonance structures be drawn? A. HCOOH B. HCOO– C. CH3OH D. H2CO3
-
22M.2.sl.TZ1.1g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
-
22M.1.sl.TZ2.12:
What is the main interaction between liquid CH4 molecules?
A. London (dispersion) forces
B. Dipole–dipole forces
C. Hydrogen bonding
D. Covalent bonding
-
22M.1.sl.TZ2.12:
What is the main interaction between liquid CH4 molecules?
A. London (dispersion) forces
B. Dipole–dipole forces
C. Hydrogen bonding
D. Covalent bonding
- 22M.1.sl.TZ1.12: For which species can resonance structures be drawn? A. HCOOH B. HCOO– C. CH3OH D. H2CO3
- 22M.2.sl.TZ2.3d(i): Draw the Lewis structure of NO3−.
- 22M.2.hl.TZ1.5a(ii): Compound A and Compound B are both liquids at room temperature and pressure. Identify the...
- 22M.2.hl.TZ1.5a(ii): Compound A and Compound B are both liquids at room temperature and pressure. Identify the...
- 22M.2.hl.TZ1.a(ii): Compound A and Compound B are both liquids at room temperature and pressure. Identify the...
-
22M.2.sl.TZ1.1g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
-
22M.2.sl.TZ1.g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
- 22M.2.sl.TZ2.3d(i): Draw the Lewis structure of NO3−.
-
22M.2.sl.TZ1.2b:
Draw the Lewis (electron dot) structure of the ammonia molecule.
-
22M.2.hl.TZ1.6a(ii):
Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO3, that obeys the octet rule, showing any non-zero formal charges on the atoms.
-
22M.2.hl.TZ1.6a(ii):
Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO3, that obeys the octet rule, showing any non-zero formal charges on the atoms.
-
22M.2.hl.TZ1.a(ii):
Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO3, that obeys the octet rule, showing any non-zero formal charges on the atoms.
- 22M.2.sl.TZ2.d(i): Draw the Lewis structure of NO3−.
- 22M.2.sl.TZ2.3d(ii): Explain the electron domain geometry of NO3−.
- 22M.2.sl.TZ2.3d(ii): Explain the electron domain geometry of NO3−.
-
22M.2.hl.TZ2.5c:
Describe the bonding in iron, Fe (s).
-
22M.2.sl.TZ1.2b:
Draw the Lewis (electron dot) structure of the ammonia molecule.
-
22M.2.sl.TZ1.b:
Draw the Lewis (electron dot) structure of the ammonia molecule.
- 22M.2.sl.TZ2.d(ii): Explain the electron domain geometry of NO3−.
- 22M.2.sl.TZ2.4a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
- 22M.2.sl.TZ1.3c(ii): Identify the strongest force between the molecules of Compound B.
-
22M.2.hl.TZ2.5c:
Describe the bonding in iron, Fe (s).
-
22M.2.hl.TZ2.c:
Describe the bonding in iron, Fe (s).
- 22M.2.hl.TZ2.6c(i): Draw the Lewis structure of SO3.
- 22M.2.sl.TZ2.4a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
- 22M.2.hl.TZ2.6c(i): Draw the Lewis structure of SO3.
- 22M.2.hl.TZ2.c(i): Draw the Lewis structure of SO3.
- 22M.2.sl.TZ1.3c(ii): Identify the strongest force between the molecules of Compound B.
- 22M.2.sl.TZ1.c(ii): Identify the strongest force between the molecules of Compound B.
- 22M.2.sl.TZ2.a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
- 22M.2.sl.TZ2.4a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.6c(ii):
Explain the electron domain geometry of SO3.
-
22M.2.hl.TZ2.6c(ii):
Explain the electron domain geometry of SO3.
-
22M.2.hl.TZ2.c(ii):
Explain the electron domain geometry of SO3.
- 22M.2.sl.TZ2.4a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.7c:
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
-
22M.2.hl.TZ1.1g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
- 22M.2.sl.TZ2.a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
- 22N.1.sl.TZ0.9: Which molecule can be represented by resonance structures? A. H2S B. HNO3 C. H2O2 D. HClO
-
22M.2.hl.TZ1.1g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
-
22M.2.hl.TZ1.g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
-
22M.2.hl.TZ2.7c:
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
-
22M.2.hl.TZ2.c:
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
-
22M.2.hl.TZ2.8a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
- 22N.1.sl.TZ0.9: Which molecule can be represented by resonance structures? A. H2S B. HNO3 C. H2O2 D. HClO
-
22M.2.hl.TZ2.8a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
22M.2.hl.TZ2.a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
22M.2.hl.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22N.1.sl.TZ0.10:
Which molecule is polar?
A. BeH2B. AlH3
C. PH3
D. SiH4
-
22N.2.sl.TZ0.1c:
Deduce the Lewis (electron dot) structure for the nitrate anion.
-
22N.1.sl.TZ0.10:
Which molecule is polar?
A. BeH2B. AlH3
C. PH3
D. SiH4
-
22N.2.sl.TZ0.1c:
Deduce the Lewis (electron dot) structure for the nitrate anion.
-
22N.2.sl.TZ0.c:
Deduce the Lewis (electron dot) structure for the nitrate anion.
-
22N.2.sl.TZ0.1f.iv:
Deduce the Lewis (electron dot) structure and shape for dinitrogen monoxide showing nitrogen as the central atom.
- 22N.1.sl.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
-
22N.2.sl.TZ0.1f.iv:
Deduce the Lewis (electron dot) structure and shape for dinitrogen monoxide showing nitrogen as the central atom.
-
22N.2.sl.TZ0.f.iv:
Deduce the Lewis (electron dot) structure and shape for dinitrogen monoxide showing nitrogen as the central atom.
- 22N.1.sl.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
- 22N.2.sl.TZ0.2c: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
-
22N.2.hl.TZ0.1f.v:
Deduce the Lewis (electron dot) structure, including formal charges, and shape for dinitrogen monoxide showing nitrogen as the central atom.
- 22N.1.sl.TZ0.12: Alloying a metal with a metal of smaller atomic radius can disrupt the lattice and make it more...
-
22N.2.hl.TZ0.1f.v:
Deduce the Lewis (electron dot) structure, including formal charges, and shape for dinitrogen monoxide showing nitrogen as the central atom.
-
22N.2.hl.TZ0.f.v:
Deduce the Lewis (electron dot) structure, including formal charges, and shape for dinitrogen monoxide showing nitrogen as the central atom.
- 22N.2.sl.TZ0.2c: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
- 22N.2.sl.TZ0.c: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
- 22N.1.sl.TZ0.12: Alloying a metal with a metal of smaller atomic radius can disrupt the lattice and make it more...
-
22N.2.sl.TZ0.4a.iii:
Explain, with reference to intermolecular forces, why B is more volatile than A.
- 22N.2.hl.TZ0.2d: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
-
22N.1.sl.TZ0.13:
Chlorofluorocarbons (CFCs) contain bonds of the following lengths:
C—C = 1.54 × 10−10 m
C—F = 1.38 × 10−10 m
C—Cl = 1.77 × 10−10 m
What is the order of increasing bond strength in the CFC molecule?
A. C—C < C—F < C—Cl
B. C—C < C—Cl < C—F
C. C—Cl < C—C < C—F
D. C—F < C—C < C—Cl
- 22N.2.hl.TZ0.2d: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
- 22N.2.hl.TZ0.d: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
-
22N.1.sl.TZ0.13:
Chlorofluorocarbons (CFCs) contain bonds of the following lengths:
C—C = 1.54 × 10−10 m
C—F = 1.38 × 10−10 m
C—Cl = 1.77 × 10−10 m
What is the order of increasing bond strength in the CFC molecule?
A. C—C < C—F < C—Cl
B. C—C < C—Cl < C—F
C. C—Cl < C—C < C—F
D. C—F < C—C < C—Cl
-
22N.2.sl.TZ0.4a.iii:
Explain, with reference to intermolecular forces, why B is more volatile than A.
-
22N.2.sl.TZ0.a.iii:
Explain, with reference to intermolecular forces, why B is more volatile than A.
- 23M.2.HL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
-
23M.1.SL.TZ1.22:
What is the formula of copper (I) sulfide?
A. CuSB. Cu2S
C. CuSO3
D. Cu2SO3
- 23M.2.SL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
- 23M.2.HL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
-
23M.1.SL.TZ1.22:
What is the formula of copper (I) sulfide?
A. CuSB. Cu2S
C. CuSO3
D. Cu2SO3
- 23M.2.HL.TZ1.i: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
- 23M.2.SL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
- 23M.2.SL.TZ1.i: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
-
23M.2.HL.TZ1.6ai:
Determine the standard enthalpy of reaction (), in kJ mol−1, for the oxidation of SO2 to SO3.
Substance Enthalpy of formation, (), in kJ mol−1 SO2 −296.8 SO3 −395.8 -
23M.2.HL.TZ1.6ai:
Determine the standard enthalpy of reaction (), in kJ mol−1, for the oxidation of SO2 to SO3.
Substance Enthalpy of formation, (), in kJ mol−1 SO2 −296.8 SO3 −395.8 -
23M.2.HL.TZ1.i:
Determine the standard enthalpy of reaction (), in kJ mol−1, for the oxidation of SO2 to SO3.
Substance Enthalpy of formation, (), in kJ mol−1 SO2 −296.8 SO3 −395.8 - 23M.2.HL.TZ1.9a: Explain why a colorimeter set at a wavelength of 500 nm is not suitable to investigate reactions...
- 23M.2.HL.TZ1.9a: Explain why a colorimeter set at a wavelength of 500 nm is not suitable to investigate reactions...
- 23M.2.HL.TZ1.a: Explain why a colorimeter set at a wavelength of 500 nm is not suitable to investigate reactions...
-
23M.2.HL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.SL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.HL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.HL.TZ2.a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.SL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.SL.TZ2.a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.HL.TZ2.3a:
An electrolytic cell was set up using inert electrodes and a dilute aqueous solution of magnesium chloride, MgCl2 (aq).
-
23M.2.HL.TZ2.3a:
An electrolytic cell was set up using inert electrodes and a dilute aqueous solution of magnesium chloride, MgCl2 (aq).
-
23M.2.HL.TZ2.a:
An electrolytic cell was set up using inert electrodes and a dilute aqueous solution of magnesium chloride, MgCl2 (aq).
-
23M.2.HL.TZ2.4a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
-
23M.2.SL.TZ2.4a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
-
23M.2.HL.TZ2.4a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
-
23M.2.HL.TZ2.a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
-
23M.2.SL.TZ2.4a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
-
23M.2.SL.TZ2.a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
-
23M.1.HL.TZ1.12:
What are the formal charges on the atoms in this molecular ion?
S C N A. −1 0 0 B. 0 0 −1 C. −1 +1 0 D. 0 +1 −1 -
23M.1.HL.TZ1.12:
What are the formal charges on the atoms in this molecular ion?
S C N A. −1 0 0 B. 0 0 −1 C. −1 +1 0 D. 0 +1 −1 - 23M.1.HL.TZ1.13: Which types of hybridization are present in...
- 23M.1.HL.TZ1.13: Which types of hybridization are present in...
-
23M.2.SL.TZ2.3a:
An electrolytic cell was set up using inert electrodes and molten magnesium chloride, MgCl2 (l).
-
23M.2.SL.TZ2.3a:
An electrolytic cell was set up using inert electrodes and molten magnesium chloride, MgCl2 (l).
-
23M.2.SL.TZ2.a:
An electrolytic cell was set up using inert electrodes and molten magnesium chloride, MgCl2 (l).
-
23M.1.SL.TZ2.1:
Which is the correct equation for the electrolysis of molten sodium chloride?
A. 2NaCl (l) → 2Na (l) + Cl2 (g)
B. 2NaCl (s) → 2Na (s) + Cl2 (g)
C. 2NaCl (l) → 2Na (s) + Cl2 (g)
D. 2NaCl (aq) → 2Na (s) + Cl2 (g)
-
23M.1.SL.TZ2.1:
Which is the correct equation for the electrolysis of molten sodium chloride?
A. 2NaCl (l) → 2Na (l) + Cl2 (g)
B. 2NaCl (s) → 2Na (s) + Cl2 (g)
C. 2NaCl (l) → 2Na (s) + Cl2 (g)
D. 2NaCl (aq) → 2Na (s) + Cl2 (g)
Sub sections and their related questions
4.1 Ionic bonding and structure
-
18M.2.hl.TZ1.2a:
Describe the nature of ionic bonding.
-
18M.1.sl.TZ1.9:
What is the formula of ammonium phosphate?
A. (NH3)3PO4
B. (NH4)3PO4
C. (NH4)2PO4
D. (NH3)2PO3
-
18M.2.sl.TZ1.2a:
Describe the nature of ionic bonding.
- 18M.1.sl.TZ2.9: What is the formula of magnesium nitride? A. MgN B. Mg2N3 C. Mg3N D. Mg3N2
- 18N.2.sl.TZ0.4b.ii: State a physical property of sodium oxide.
-
19M.2.hl.TZ1.3c:
Describe the structure and bonding in solid sodium oxide.
-
19M.2.hl.TZ2.5d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.1.hl.TZ1.10:
What is the IUPAC name of NiCO3?
A. nickel(II) carbonate
B. nickel carbonate
C. nickel(I) carbonate
D. nitrogen(I) carbonate
- 19M.1.hl.TZ2.9: How does a lithium atom form the most stable ion? A. The atom gains a proton to form a positive...
-
19M.2.sl.TZ1.3a(i):
Describe the structure and bonding in solid sodium oxide.
-
19M.2.sl.TZ2.5b(iii):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
- 19M.1.sl.TZ2.9: How does a lithium atom form the most stable ion? A. The atom gains a proton to form a positive...
-
19N.3.sl.TZ0.17b:
Aspirin, C6H4(OCOCH3)COOH, is only slightly soluble in water.
Outline, including an equation, how aspirin can be made more water-soluble. Use section 37 in the data booklet.
- 19N.1.sl.TZ0.9: Which is correct for all solid ionic compounds? A. High volatility B. Poor electrical...
-
20N.1.sl.TZ0.9:
Which formula is correct?
A.
B.
C.
D.
-
21M.1.sl.TZ1.10:
Which compound contains both ionic and covalent bonds?
A.
B.
C.
D.
- 21M.1.sl.TZ1.11: Which substance is most likely to be ionic?
-
21M.1.sl.TZ2.11:
What is the formula of the compound formed from Ca2+ and PO43−?
A. CaPO4
B. Ca3(PO4)2
C. Ca2(PO4)3
D. Ca(PO4)2
- 21M.1.hl.TZ2.12: Which atom has an expanded octet? A. C in CO2 B. S in SCl4 C. O in H2O2 D. P in PCl3
- 21M.2.sl.TZ1.1c(i): Describe the bonding in this type of solid.
- 21M.2.sl.TZ1.1c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.sl.TZ1.3a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21M.2.hl.TZ1.1d(i): Describe the bonding in this type of solid.
- 21M.2.hl.TZ1.1d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.3a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21N.1.sl.TZ0.10: Which combination would create the strongest ionic bond?
- 21N.1.sl.TZ0.11: Which compound contains both ionic and covalent bonds? A. CH3COONa B. CH3COOH C. K2O D. ...
- 22M.1.sl.TZ1.9: A compound consists of the ions Ca2+ and PO43–. What are the name and formula of the compound?
- 22M.1.sl.TZ1.10: What is the explanation for the high melting point of sodium chloride? A. The covalent bond...
-
22M.1.sl.TZ2.9:
Which statement best describes the intramolecular bonding in HCN (l)?
A. Electrostatic attractions between H+ and CN− ions
B. Hydrogen bonding
C. Van der Waals forces and hydrogen bonding
D. Electrostatic attractions between pairs of electrons and positively charged nuclei
- 22M.1.sl.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
-
22M.1.sl.TZ2.11:
What is the name of the compound with formula Ti3(PO4)2?
A. Titanium phosphate
B. Titanium(II) phosphate
C. Titanium(III) phosphate
D. Titanium(IV) phosphate
- 22M.1.hl.TZ2.9: In which of the following compounds does ionic bonding predominate? A. HCl B. NaF C. ...
-
22M.2.sl.TZ1.1g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
-
22M.2.hl.TZ1.1g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
-
23M.1.SL.TZ2.1:
Which is the correct equation for the electrolysis of molten sodium chloride?
A. 2NaCl (l) → 2Na (l) + Cl2 (g)
B. 2NaCl (s) → 2Na (s) + Cl2 (g)
C. 2NaCl (l) → 2Na (s) + Cl2 (g)
D. 2NaCl (aq) → 2Na (s) + Cl2 (g)
-
23M.1.SL.TZ1.22:
What is the formula of copper (I) sulfide?
A. CuSB. Cu2S
C. CuSO3
D. Cu2SO3
-
19M.2.hl.TZ1.3c:
Describe the structure and bonding in solid sodium oxide.
-
19M.2.hl.TZ1.c:
Describe the structure and bonding in solid sodium oxide.
-
19M.2.hl.TZ2.5d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.2.hl.TZ2.d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.1.hl.TZ1.10:
What is the IUPAC name of NiCO3?
A. nickel(II) carbonate
B. nickel carbonate
C. nickel(I) carbonate
D. nitrogen(I) carbonate
- 19M.1.hl.TZ2.9: How does a lithium atom form the most stable ion? A. The atom gains a proton to form a positive...
-
19M.2.sl.TZ1.3a(i):
Describe the structure and bonding in solid sodium oxide.
-
19M.2.sl.TZ1.a(i):
Describe the structure and bonding in solid sodium oxide.
-
19M.2.sl.TZ2.5b(iii):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.2.sl.TZ2.b(iii):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
- 19M.1.sl.TZ2.9: How does a lithium atom form the most stable ion? A. The atom gains a proton to form a positive...
-
19N.3.sl.TZ0.17b:
Aspirin, C6H4(OCOCH3)COOH, is only slightly soluble in water.
Outline, including an equation, how aspirin can be made more water-soluble. Use section 37 in the data booklet.
-
19N.3.sl.TZ0.b:
Aspirin, C6H4(OCOCH3)COOH, is only slightly soluble in water.
Outline, including an equation, how aspirin can be made more water-soluble. Use section 37 in the data booklet.
- 19N.1.sl.TZ0.9: Which is correct for all solid ionic compounds? A. High volatility B. Poor electrical...
-
20N.1.sl.TZ0.9:
Which formula is correct?
A.
B.
C.
D.
-
21M.1.sl.TZ1.10:
Which compound contains both ionic and covalent bonds?
A.
B.
C.
D.
- 21M.1.sl.TZ1.11: Which substance is most likely to be ionic?
-
21M.1.sl.TZ2.11:
What is the formula of the compound formed from Ca2+ and PO43−?
A. CaPO4
B. Ca3(PO4)2
C. Ca2(PO4)3
D. Ca(PO4)2
- 21M.1.hl.TZ2.12: Which atom has an expanded octet? A. C in CO2 B. S in SCl4 C. O in H2O2 D. P in PCl3
- 21M.2.sl.TZ1.1c(i): Describe the bonding in this type of solid.
- 21M.2.sl.TZ1.1c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.sl.TZ1.c(i): Describe the bonding in this type of solid.
- 21M.2.sl.TZ1.c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.sl.TZ1.3a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21M.2.sl.TZ1.a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21M.2.hl.TZ1.1d(i): Describe the bonding in this type of solid.
- 21M.2.hl.TZ1.1d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.d(i): Describe the bonding in this type of solid.
- 21M.2.hl.TZ1.d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.3a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21M.2.hl.TZ1.a: Deduce the ratio of Fe2+:Fe3+ in Fe3O4.
- 21N.1.sl.TZ0.10: Which combination would create the strongest ionic bond?
- 21N.1.sl.TZ0.11: Which compound contains both ionic and covalent bonds? A. CH3COONa B. CH3COOH C. K2O D. ...
- 22M.1.sl.TZ1.9: A compound consists of the ions Ca2+ and PO43–. What are the name and formula of the compound?
- 22M.1.sl.TZ1.10: What is the explanation for the high melting point of sodium chloride? A. The covalent bond...
-
22M.1.sl.TZ2.9:
Which statement best describes the intramolecular bonding in HCN (l)?
A. Electrostatic attractions between H+ and CN− ions
B. Hydrogen bonding
C. Van der Waals forces and hydrogen bonding
D. Electrostatic attractions between pairs of electrons and positively charged nuclei
- 22M.1.sl.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
-
22M.1.sl.TZ2.11:
What is the name of the compound with formula Ti3(PO4)2?
A. Titanium phosphate
B. Titanium(II) phosphate
C. Titanium(III) phosphate
D. Titanium(IV) phosphate
- 22M.1.hl.TZ2.9: In which of the following compounds does ionic bonding predominate? A. HCl B. NaF C. ...
-
22M.2.sl.TZ1.1g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
-
22M.2.sl.TZ1.g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
-
22M.2.hl.TZ1.1g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
-
22M.2.hl.TZ1.g:
State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.
-
18M.2.hl.TZ1.2a:
Describe the nature of ionic bonding.
-
18M.2.hl.TZ1.a:
Describe the nature of ionic bonding.
-
23M.1.SL.TZ2.1:
Which is the correct equation for the electrolysis of molten sodium chloride?
A. 2NaCl (l) → 2Na (l) + Cl2 (g)
B. 2NaCl (s) → 2Na (s) + Cl2 (g)
C. 2NaCl (l) → 2Na (s) + Cl2 (g)
D. 2NaCl (aq) → 2Na (s) + Cl2 (g)
-
23M.1.SL.TZ1.22:
What is the formula of copper (I) sulfide?
A. CuSB. Cu2S
C. CuSO3
D. Cu2SO3
-
18M.1.sl.TZ1.9:
What is the formula of ammonium phosphate?
A. (NH3)3PO4
B. (NH4)3PO4
C. (NH4)2PO4
D. (NH3)2PO3
-
18M.2.sl.TZ1.2a:
Describe the nature of ionic bonding.
-
18M.2.sl.TZ1.a:
Describe the nature of ionic bonding.
- 18M.1.sl.TZ2.9: What is the formula of magnesium nitride? A. MgN B. Mg2N3 C. Mg3N D. Mg3N2
- 18N.2.sl.TZ0.4b.ii: State a physical property of sodium oxide.
- 18N.2.sl.TZ0.b.ii: State a physical property of sodium oxide.
4.2 Covalent bonding
- 17N.1.sl.TZ0.9: The electronegativity values of four elements are given. What is the order of increasing...
-
17N.1.sl.TZ0.10:
Which compound has the shortest C–N bond?
A. CH3NH2
B. (CH3)3CNH2
C. CH3CN
D. CH3CHNH
- 18M.1.sl.TZ2.10: Which species has the longest carbon to oxygen bond length? A. CO B. CH3OH C. ...
-
18M.1.sl.TZ2.11:
What are the predicted electron domain geometries around the carbon and both nitrogen atoms in urea, (NH2)2CO, applying VSEPR theory?
-
18M.2.sl.TZ2.6a.i:
Explain why the hydrides of group 16 elements (H2O, H2S, H2Se and H2Te) are polar molecules.
-
18M.2.sl.TZ2.6a.ii:
The graph shows the boiling points of the hydrides of group 16 elements.
Explain the increase in the boiling point from H2S to H2Te.
-
19M.2.hl.TZ2.1b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.hl.TZ2.5d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.2.sl.TZ2.1b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.sl.TZ2.5b(iii):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
- 19N.2.sl.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.2.sl.TZ0.1d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
- 19N.2.hl.TZ0.1a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.2.hl.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.2.hl.TZ0.1c: Predict the bond angle in the ozone molecule.
- 19N.2.hl.TZ0.1d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
-
19N.3.sl.TZ0.5b(i):
Determine the percentage of ionic bonding in alumina using sections 8 and 29 of the data booklet.
- 19N.1.sl.TZ0.10: Which compound has the shortest C to O bond? A. CH3CHO B. CO C. CO2 D. C2H5OC2H5
-
20N.1.hl.TZ0.9:
Which of these species contains the shortest carbon to oxygen bond length?
A.
B.
C.
D.
-
21M.1.sl.TZ1.10:
Which compound contains both ionic and covalent bonds?
A.
B.
C.
D.
-
21M.1.sl.TZ2.10:
Which compound has the shortest C to N bond?
A. HCN
B. CH3CH2NH2
C. CH3CHNH
D. (CH3)2NH
- 21M.1.sl.TZ2.12: Which is the correct order based on increasing strength? A. covalent bonds < hydrogen bonds...
- 21M.2.sl.TZ1.1c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.1d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21N.1.sl.TZ0.9: Which molecule has the weakest nitrogen to nitrogen bond? A. N2 B. N2H2 C. N2H4 D.
- 21N.1.sl.TZ0.11: Which compound contains both ionic and covalent bonds? A. CH3COONa B. CH3COOH C. K2O D. ...
- 21N.2.hl.TZ0.3b(ii): Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.
- 22M.1.sl.TZ1.10: What is the explanation for the high melting point of sodium chloride? A. The covalent bond...
- 22M.1.hl.TZ2.9: In which of the following compounds does ionic bonding predominate? A. HCl B. NaF C. ...
- 22M.2.sl.TZ2.4a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
- 22M.2.sl.TZ2.4a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.8a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
22M.2.hl.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22N.1.sl.TZ0.13:
Chlorofluorocarbons (CFCs) contain bonds of the following lengths:
C—C = 1.54 × 10−10 m
C—F = 1.38 × 10−10 m
C—Cl = 1.77 × 10−10 m
What is the order of increasing bond strength in the CFC molecule?
A. C—C < C—F < C—Cl
B. C—C < C—Cl < C—F
C. C—Cl < C—C < C—F
D. C—F < C—C < C—Cl
- 22N.2.sl.TZ0.2c: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
- 22N.2.hl.TZ0.2d: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
-
23M.2.HL.TZ1.6ai:
Determine the standard enthalpy of reaction (), in kJ mol−1, for the oxidation of SO2 to SO3.
Substance Enthalpy of formation, (), in kJ mol−1 SO2 −296.8 SO3 −395.8 -
23M.2.HL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
- 23M.1.HL.TZ1.13: Which types of hybridization are present in...
-
23M.2.SL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
19M.2.hl.TZ2.1b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.hl.TZ2.b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.hl.TZ2.5d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.2.hl.TZ2.d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.2.sl.TZ2.1b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.sl.TZ2.b(ii):
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
-
19M.2.sl.TZ2.5b(iii):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.2.sl.TZ2.b(iii):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
- 19N.2.sl.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.2.sl.TZ0.1d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
- 19N.2.sl.TZ0.b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.2.sl.TZ0.d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
- 19N.2.hl.TZ0.1a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.2.hl.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.2.hl.TZ0.1c: Predict the bond angle in the ozone molecule.
- 19N.2.hl.TZ0.1d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
- 19N.2.hl.TZ0.a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.2.hl.TZ0.b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.2.hl.TZ0.c: Predict the bond angle in the ozone molecule.
- 19N.2.hl.TZ0.d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
-
19N.3.sl.TZ0.5b(i):
Determine the percentage of ionic bonding in alumina using sections 8 and 29 of the data booklet.
-
19N.3.sl.TZ0.b(i):
Determine the percentage of ionic bonding in alumina using sections 8 and 29 of the data booklet.
- 19N.1.sl.TZ0.10: Which compound has the shortest C to O bond? A. CH3CHO B. CO C. CO2 D. C2H5OC2H5
-
20N.1.hl.TZ0.9:
Which of these species contains the shortest carbon to oxygen bond length?
A.
B.
C.
D.
-
21M.1.sl.TZ1.10:
Which compound contains both ionic and covalent bonds?
A.
B.
C.
D.
-
21M.1.sl.TZ2.10:
Which compound has the shortest C to N bond?
A. HCN
B. CH3CH2NH2
C. CH3CHNH
D. (CH3)2NH
- 21M.1.sl.TZ2.12: Which is the correct order based on increasing strength? A. covalent bonds < hydrogen bonds...
- 21M.2.sl.TZ1.1c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.sl.TZ1.c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.1d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21N.1.sl.TZ0.9: Which molecule has the weakest nitrogen to nitrogen bond? A. N2 B. N2H2 C. N2H4 D.
- 21N.1.sl.TZ0.11: Which compound contains both ionic and covalent bonds? A. CH3COONa B. CH3COOH C. K2O D. ...
- 21N.2.hl.TZ0.3b(ii): Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.
- 21N.2.hl.TZ0.b(ii): Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.
- 22M.1.sl.TZ1.10: What is the explanation for the high melting point of sodium chloride? A. The covalent bond...
- 22M.1.hl.TZ2.9: In which of the following compounds does ionic bonding predominate? A. HCl B. NaF C. ...
- 22M.2.sl.TZ2.4a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
- 22M.2.sl.TZ2.4a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
- 22M.2.sl.TZ2.a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
- 22M.2.sl.TZ2.a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.8a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
22M.2.hl.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
22M.2.hl.TZ2.a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22N.1.sl.TZ0.13:
Chlorofluorocarbons (CFCs) contain bonds of the following lengths:
C—C = 1.54 × 10−10 m
C—F = 1.38 × 10−10 m
C—Cl = 1.77 × 10−10 m
What is the order of increasing bond strength in the CFC molecule?
A. C—C < C—F < C—Cl
B. C—C < C—Cl < C—F
C. C—Cl < C—C < C—F
D. C—F < C—C < C—Cl
- 22N.2.sl.TZ0.2c: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
- 22N.2.sl.TZ0.c: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
- 22N.2.hl.TZ0.2d: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
- 22N.2.hl.TZ0.d: Compare, giving a reason, the length of the carbon-nitrogen bond in the ring to the length of the...
- 17N.1.sl.TZ0.9: The electronegativity values of four elements are given. What is the order of increasing...
-
17N.1.sl.TZ0.10:
Which compound has the shortest C–N bond?
A. CH3NH2
B. (CH3)3CNH2
C. CH3CN
D. CH3CHNH
-
23M.2.HL.TZ1.6ai:
Determine the standard enthalpy of reaction (), in kJ mol−1, for the oxidation of SO2 to SO3.
Substance Enthalpy of formation, (), in kJ mol−1 SO2 −296.8 SO3 −395.8 -
23M.2.HL.TZ1.i:
Determine the standard enthalpy of reaction (), in kJ mol−1, for the oxidation of SO2 to SO3.
Substance Enthalpy of formation, (), in kJ mol−1 SO2 −296.8 SO3 −395.8 -
23M.2.HL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.HL.TZ2.a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
- 23M.1.HL.TZ1.13: Which types of hybridization are present in...
-
23M.2.SL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.SL.TZ2.a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
- 18M.1.sl.TZ2.10: Which species has the longest carbon to oxygen bond length? A. CO B. CH3OH C. ...
-
18M.1.sl.TZ2.11:
What are the predicted electron domain geometries around the carbon and both nitrogen atoms in urea, (NH2)2CO, applying VSEPR theory?
-
18M.2.sl.TZ2.6a.i:
Explain why the hydrides of group 16 elements (H2O, H2S, H2Se and H2Te) are polar molecules.
-
18M.2.sl.TZ2.6a.ii:
The graph shows the boiling points of the hydrides of group 16 elements.
Explain the increase in the boiling point from H2S to H2Te.
-
18M.2.sl.TZ2.a.i:
Explain why the hydrides of group 16 elements (H2O, H2S, H2Se and H2Te) are polar molecules.
-
18M.2.sl.TZ2.a.ii:
The graph shows the boiling points of the hydrides of group 16 elements.
Explain the increase in the boiling point from H2S to H2Te.
4.3 Covalent structures
-
17N.2.sl.TZ0.3a:
Draw the Lewis (electron dot) structures of PF3 and PF4+ and use the VSEPR theory to deduce the molecular geometry of each species.
-
17N.2.sl.TZ0.3b:
Predict with a reason, whether the molecule PF3 is polar or non-polar.
-
17N.2.hl.TZ0.4a:
Draw the Lewis (electron dot) structures of PF3 and PF5 and use the VSEPR theory to deduce the molecular geometry of each species including bond angles.
-
17N.2.hl.TZ0.4b:
Predict whether the molecules PF3 and PF5 are polar or non-polar.
-
18M.2.hl.TZ1.1b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
-
18M.2.hl.TZ2.7e:
Carbon and silicon are elements in group 14.
Explain why CO2 is a gas but SiO2 is a solid at room temperature.
- 18M.1.sl.TZ1.10: Which form of carbon is the poorest electrical conductor? A. Graphite B. Graphene C. ...
- 18M.1.sl.TZ1.11: What is the molecular geometry and bond angle in the molecular ion NO3−?
-
18M.2.sl.TZ1.1b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
- 18M.1.sl.TZ2.10: Which species has the longest carbon to oxygen bond length? A. CO B. CH3OH C. ...
-
18M.1.sl.TZ2.11:
What are the predicted electron domain geometries around the carbon and both nitrogen atoms in urea, (NH2)2CO, applying VSEPR theory?
-
18M.2.sl.TZ2.6b:
Lewis structures show electron domains and are used to predict molecular geometry.
Deduce the electron domain geometry and the molecular geometry for the NH2− ion.
-
18M.3.sl.TZ2.1a.i:
Graphene is two-dimensional, rather than three-dimensional, material.
Justify this by using the structure of graphene and information from the table.
-
18M.3.sl.TZ2.1a.ii:
Show that graphene is over 1600 times stronger than graphite.
-
18M.3.sl.TZ2.1a.iii:
Identify a value from the table which can be used to support the information about graphene given below.
Electrons in a solid are restricted to certain ranges, or bands, of energy (vertical axis). In an insulator or semiconductor, an electron bound to an atom can break free only if it gets enough energy from heat or a passing photon to jump the “band gap”, but in graphene the gap is infinitely small.
-
18M.3.sl.TZ2.1b:
Diamond, graphene, and graphite are all network solids.
Suggest, giving a reason, the electron mobility of diamond compared to graphene.
-
18M.3.sl.TZ2.1c:
The melting point of diamond at 1 × 106 kPa is 4200 K (in the absence of oxygen).
Suggest, based on molecular structure, why graphene has a higher melting point under these conditions.
- 18N.1.sl.TZ0.9: Which species has the same molecular geometry as SO32−? A. BF3 B. SO3 C. PF3 D. CO32−
- 18N.1.sl.TZ0.10: How many lone pairs and bonding pairs of electrons surround the central chlorine atom in...
- 18N.1.sl.TZ0.12: Which molecule is polar? A. BeCl2 B. BCl3 C. NCl3 D. CCl4
-
18N.2.sl.TZ0.3b:
Draw the Lewis (electron dot) structure for BrO3− that obeys the octet rule.
- 18N.2.sl.TZ0.3c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
-
18N.2.hl.TZ0.3b.i:
Draw two Lewis (electron dot) structures for BrO3−.
- 18N.2.hl.TZ0.3c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
- 18N.2.hl.TZ0.8a: Suggest why the three-membered ring in methyloxirane is unstable.
-
19M.2.hl.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.hl.TZ1.5c(iii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.hl.TZ1.5c(iv):
State the type of bond formed when chloramine is protonated.
-
19M.2.hl.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
-
19M.2.hl.TZ2.3d(ii):
State, giving a reason, the shape of the dinitrogen monoxide molecule.
-
19M.2.sl.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.sl.TZ1.5c(ii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.sl.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
- 19M.1.sl.TZ1.10: Which species does not have resonance structures? A. C6H6 B. NH4+ C. CO32− D. O3
- 19M.1.sl.TZ2.11: Which molecule contains an incomplete octet of electrons? A. NF3 B. BF3 C. BrF D. SF2
- 19N.2.sl.TZ0.1a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.2.sl.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.1.sl.TZ0.11: Which describes a resonance structure? A. Double bond can be drawn in alternative...
-
19N.1.sl.TZ0.12:
What is the structure and bonding in SiO2 (s)?
-
20N.1.sl.TZ0.10:
Which molecule is most polar?
A.
B.
C.
D.
-
20N.1.sl.TZ0.11:
Which combination correctly describes the geometry of the carbonate ion, ?
- 20N.2.sl.TZ0.2a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
- 20N.2.hl.TZ0.2a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
- 21M.1.sl.TZ1.9: The Lewis structure of methylamine is shown. What is the molecular geometry around N? A. ...
- 21M.1.sl.TZ1.12: Along which series is the bond angle increasing? A. NH3 H2O CH4 B. CH4 NH3 H2O C. H2O ...
-
21M.2.sl.TZ1.2a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
- 21M.2.sl.TZ1.2a(ii): Predict the shape of the hydrogen sulfide molecule.
-
21M.2.hl.TZ1.7a(i):
Draw a Lewis (electron dot) structure for ozone.
-
21M.2.sl.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl2.
-
21M.2.hl.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF4, and sulfur dichloride, SCl2.
-
21N.2.sl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
- 21N.2.sl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
- 21N.2.sl.TZ0.3b(ii): Explain the polarity of PCl3.
-
21N.2.hl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
- 21N.2.hl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
- 22M.1.sl.TZ1.11: Which molecule is most polar? A. CF4 B. CCl4 C. CHF3 D. CClF3
- 22M.1.sl.TZ1.12: For which species can resonance structures be drawn? A. HCOOH B. HCOO– C. CH3OH D. H2CO3
- 22M.1.hl.TZ1.13: What are the electron domain and molecular geometries of the XeF4 molecule?
- 22M.1.sl.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
-
22M.2.sl.TZ1.2b:
Draw the Lewis (electron dot) structure of the ammonia molecule.
-
22M.2.hl.TZ1.6a(ii):
Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO3, that obeys the octet rule, showing any non-zero formal charges on the atoms.
- 22M.2.sl.TZ2.3d(i): Draw the Lewis structure of NO3−.
- 22M.2.sl.TZ2.3d(ii): Explain the electron domain geometry of NO3−.
- 22M.2.sl.TZ2.4a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
- 22M.2.sl.TZ2.4a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
- 22M.2.hl.TZ2.6c(i): Draw the Lewis structure of SO3.
-
22M.2.hl.TZ2.6c(ii):
Explain the electron domain geometry of SO3.
-
22M.2.hl.TZ2.8a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
22M.2.hl.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
- 22N.1.sl.TZ0.9: Which molecule can be represented by resonance structures? A. H2S B. HNO3 C. H2O2 D. HClO
-
22N.1.sl.TZ0.10:
Which molecule is polar?
A. BeH2B. AlH3
C. PH3
D. SiH4
- 22N.1.sl.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
-
22N.2.sl.TZ0.1c:
Deduce the Lewis (electron dot) structure for the nitrate anion.
-
22N.2.sl.TZ0.1f.iv:
Deduce the Lewis (electron dot) structure and shape for dinitrogen monoxide showing nitrogen as the central atom.
-
22N.2.hl.TZ0.1f.v:
Deduce the Lewis (electron dot) structure, including formal charges, and shape for dinitrogen monoxide showing nitrogen as the central atom.
- 23M.2.HL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
- 23M.2.HL.TZ1.9a: Explain why a colorimeter set at a wavelength of 500 nm is not suitable to investigate reactions...
-
23M.2.HL.TZ2.3a:
An electrolytic cell was set up using inert electrodes and a dilute aqueous solution of magnesium chloride, MgCl2 (aq).
- 23M.2.SL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
-
23M.2.SL.TZ2.3a:
An electrolytic cell was set up using inert electrodes and molten magnesium chloride, MgCl2 (l).
-
19M.2.hl.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.hl.TZ1.5c(iii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.hl.TZ1.5c(iv):
State the type of bond formed when chloramine is protonated.
-
19M.2.hl.TZ1.c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.hl.TZ1.c(iii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.hl.TZ1.c(iv):
State the type of bond formed when chloramine is protonated.
-
19M.2.hl.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
-
19M.2.hl.TZ2.b(i):
Deduce the Lewis (electron dot) structure of ethyne.
-
19M.2.hl.TZ2.3d(ii):
State, giving a reason, the shape of the dinitrogen monoxide molecule.
-
19M.2.hl.TZ2.d(ii):
State, giving a reason, the shape of the dinitrogen monoxide molecule.
-
19M.2.sl.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.sl.TZ1.5c(ii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.sl.TZ1.c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
19M.2.sl.TZ1.c(ii):
Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.
Molecular geometry:
H–N–H bond angle:
-
19M.2.sl.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
-
19M.2.sl.TZ2.b(i):
Deduce the Lewis (electron dot) structure of ethyne.
- 19M.1.sl.TZ1.10: Which species does not have resonance structures? A. C6H6 B. NH4+ C. CO32− D. O3
- 19M.1.sl.TZ2.11: Which molecule contains an incomplete octet of electrons? A. NF3 B. BF3 C. BrF D. SF2
- 19N.2.sl.TZ0.1a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.2.sl.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.2.sl.TZ0.a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.2.sl.TZ0.b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.1.sl.TZ0.11: Which describes a resonance structure? A. Double bond can be drawn in alternative...
-
19N.1.sl.TZ0.12:
What is the structure and bonding in SiO2 (s)?
-
20N.1.sl.TZ0.10:
Which molecule is most polar?
A.
B.
C.
D.
-
20N.1.sl.TZ0.11:
Which combination correctly describes the geometry of the carbonate ion, ?
- 20N.2.sl.TZ0.2a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
- 20N.2.sl.TZ0.a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
- 20N.2.hl.TZ0.2a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
- 20N.2.hl.TZ0.a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
- 21M.1.sl.TZ1.9: The Lewis structure of methylamine is shown. What is the molecular geometry around N? A. ...
- 21M.1.sl.TZ1.12: Along which series is the bond angle increasing? A. NH3 H2O CH4 B. CH4 NH3 H2O C. H2O ...
-
21M.2.sl.TZ1.2a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
- 21M.2.sl.TZ1.2a(ii): Predict the shape of the hydrogen sulfide molecule.
-
21M.2.sl.TZ1.a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
- 21M.2.sl.TZ1.a(ii): Predict the shape of the hydrogen sulfide molecule.
-
21M.2.hl.TZ1.7a(i):
Draw a Lewis (electron dot) structure for ozone.
-
21M.2.hl.TZ1.a(i):
Draw a Lewis (electron dot) structure for ozone.
-
21M.2.sl.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl2.
-
21M.2.sl.TZ2.c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl2.
-
21M.2.hl.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF4, and sulfur dichloride, SCl2.
-
21M.2.hl.TZ2.c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF4, and sulfur dichloride, SCl2.
-
21N.2.sl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
- 21N.2.sl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
- 21N.2.sl.TZ0.3b(ii): Explain the polarity of PCl3.
-
21N.2.sl.TZ0.a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
- 21N.2.sl.TZ0.b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
- 21N.2.sl.TZ0.b(ii): Explain the polarity of PCl3.
-
21N.2.hl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
- 21N.2.hl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
-
21N.2.hl.TZ0.a(i):
Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.
- 21N.2.hl.TZ0.b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
- 22M.1.sl.TZ1.11: Which molecule is most polar? A. CF4 B. CCl4 C. CHF3 D. CClF3
- 22M.1.sl.TZ1.12: For which species can resonance structures be drawn? A. HCOOH B. HCOO– C. CH3OH D. H2CO3
- 22M.1.hl.TZ1.13: What are the electron domain and molecular geometries of the XeF4 molecule?
- 22M.1.sl.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
-
22M.2.sl.TZ1.2b:
Draw the Lewis (electron dot) structure of the ammonia molecule.
-
22M.2.sl.TZ1.b:
Draw the Lewis (electron dot) structure of the ammonia molecule.
-
22M.2.hl.TZ1.6a(ii):
Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO3, that obeys the octet rule, showing any non-zero formal charges on the atoms.
-
22M.2.hl.TZ1.a(ii):
Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO3, that obeys the octet rule, showing any non-zero formal charges on the atoms.
- 22M.2.sl.TZ2.3d(i): Draw the Lewis structure of NO3−.
- 22M.2.sl.TZ2.3d(ii): Explain the electron domain geometry of NO3−.
- 22M.2.sl.TZ2.d(i): Draw the Lewis structure of NO3−.
- 22M.2.sl.TZ2.d(ii): Explain the electron domain geometry of NO3−.
- 22M.2.sl.TZ2.4a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
- 22M.2.sl.TZ2.4a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
- 22M.2.sl.TZ2.a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
- 22M.2.sl.TZ2.a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
- 22M.2.hl.TZ2.6c(i): Draw the Lewis structure of SO3.
-
22M.2.hl.TZ2.6c(ii):
Explain the electron domain geometry of SO3.
- 22M.2.hl.TZ2.c(i): Draw the Lewis structure of SO3.
-
22M.2.hl.TZ2.c(ii):
Explain the electron domain geometry of SO3.
-
22M.2.hl.TZ2.8a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
22M.2.hl.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
22M.2.hl.TZ2.a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
- 22N.1.sl.TZ0.9: Which molecule can be represented by resonance structures? A. H2S B. HNO3 C. H2O2 D. HClO
-
22N.1.sl.TZ0.10:
Which molecule is polar?
A. BeH2B. AlH3
C. PH3
D. SiH4
- 22N.1.sl.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
-
22N.2.sl.TZ0.1c:
Deduce the Lewis (electron dot) structure for the nitrate anion.
-
22N.2.sl.TZ0.1f.iv:
Deduce the Lewis (electron dot) structure and shape for dinitrogen monoxide showing nitrogen as the central atom.
-
22N.2.sl.TZ0.c:
Deduce the Lewis (electron dot) structure for the nitrate anion.
-
22N.2.sl.TZ0.f.iv:
Deduce the Lewis (electron dot) structure and shape for dinitrogen monoxide showing nitrogen as the central atom.
-
22N.2.hl.TZ0.1f.v:
Deduce the Lewis (electron dot) structure, including formal charges, and shape for dinitrogen monoxide showing nitrogen as the central atom.
-
22N.2.hl.TZ0.f.v:
Deduce the Lewis (electron dot) structure, including formal charges, and shape for dinitrogen monoxide showing nitrogen as the central atom.
-
17N.2.sl.TZ0.3a:
Draw the Lewis (electron dot) structures of PF3 and PF4+ and use the VSEPR theory to deduce the molecular geometry of each species.
-
17N.2.sl.TZ0.3b:
Predict with a reason, whether the molecule PF3 is polar or non-polar.
-
17N.2.sl.TZ0.a:
Draw the Lewis (electron dot) structures of PF3 and PF4+ and use the VSEPR theory to deduce the molecular geometry of each species.
-
17N.2.sl.TZ0.b:
Predict with a reason, whether the molecule PF3 is polar or non-polar.
-
17N.2.hl.TZ0.4a:
Draw the Lewis (electron dot) structures of PF3 and PF5 and use the VSEPR theory to deduce the molecular geometry of each species including bond angles.
-
17N.2.hl.TZ0.4b:
Predict whether the molecules PF3 and PF5 are polar or non-polar.
-
17N.2.hl.TZ0.a:
Draw the Lewis (electron dot) structures of PF3 and PF5 and use the VSEPR theory to deduce the molecular geometry of each species including bond angles.
-
17N.2.hl.TZ0.b:
Predict whether the molecules PF3 and PF5 are polar or non-polar.
-
18M.2.hl.TZ1.1b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
-
18M.2.hl.TZ1.b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
-
18M.2.hl.TZ2.7e:
Carbon and silicon are elements in group 14.
Explain why CO2 is a gas but SiO2 is a solid at room temperature.
-
18M.2.hl.TZ2.e:
Carbon and silicon are elements in group 14.
Explain why CO2 is a gas but SiO2 is a solid at room temperature.
- 23M.2.HL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
- 23M.2.HL.TZ1.i: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
- 23M.2.HL.TZ1.9a: Explain why a colorimeter set at a wavelength of 500 nm is not suitable to investigate reactions...
- 23M.2.HL.TZ1.a: Explain why a colorimeter set at a wavelength of 500 nm is not suitable to investigate reactions...
-
23M.2.HL.TZ2.3a:
An electrolytic cell was set up using inert electrodes and a dilute aqueous solution of magnesium chloride, MgCl2 (aq).
-
23M.2.HL.TZ2.a:
An electrolytic cell was set up using inert electrodes and a dilute aqueous solution of magnesium chloride, MgCl2 (aq).
- 23M.2.SL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
- 23M.2.SL.TZ1.i: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
-
23M.2.SL.TZ2.3a:
An electrolytic cell was set up using inert electrodes and molten magnesium chloride, MgCl2 (l).
-
23M.2.SL.TZ2.a:
An electrolytic cell was set up using inert electrodes and molten magnesium chloride, MgCl2 (l).
- 18M.1.sl.TZ1.10: Which form of carbon is the poorest electrical conductor? A. Graphite B. Graphene C. ...
- 18M.1.sl.TZ1.11: What is the molecular geometry and bond angle in the molecular ion NO3−?
-
18M.2.sl.TZ1.1b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
-
18M.2.sl.TZ1.b:
The structural formula of urea is shown.
Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.
- 18M.1.sl.TZ2.10: Which species has the longest carbon to oxygen bond length? A. CO B. CH3OH C. ...
-
18M.1.sl.TZ2.11:
What are the predicted electron domain geometries around the carbon and both nitrogen atoms in urea, (NH2)2CO, applying VSEPR theory?
-
18M.2.sl.TZ2.6b:
Lewis structures show electron domains and are used to predict molecular geometry.
Deduce the electron domain geometry and the molecular geometry for the NH2− ion.
-
18M.2.sl.TZ2.b:
Lewis structures show electron domains and are used to predict molecular geometry.
Deduce the electron domain geometry and the molecular geometry for the NH2− ion.
-
18M.3.sl.TZ2.1a.i:
Graphene is two-dimensional, rather than three-dimensional, material.
Justify this by using the structure of graphene and information from the table.
-
18M.3.sl.TZ2.1a.ii:
Show that graphene is over 1600 times stronger than graphite.
-
18M.3.sl.TZ2.1a.iii:
Identify a value from the table which can be used to support the information about graphene given below.
Electrons in a solid are restricted to certain ranges, or bands, of energy (vertical axis). In an insulator or semiconductor, an electron bound to an atom can break free only if it gets enough energy from heat or a passing photon to jump the “band gap”, but in graphene the gap is infinitely small.
-
18M.3.sl.TZ2.1b:
Diamond, graphene, and graphite are all network solids.
Suggest, giving a reason, the electron mobility of diamond compared to graphene.
-
18M.3.sl.TZ2.1c:
The melting point of diamond at 1 × 106 kPa is 4200 K (in the absence of oxygen).
Suggest, based on molecular structure, why graphene has a higher melting point under these conditions.
-
18M.3.sl.TZ2.a.i:
Graphene is two-dimensional, rather than three-dimensional, material.
Justify this by using the structure of graphene and information from the table.
-
18M.3.sl.TZ2.a.ii:
Show that graphene is over 1600 times stronger than graphite.
-
18M.3.sl.TZ2.a.iii:
Identify a value from the table which can be used to support the information about graphene given below.
Electrons in a solid are restricted to certain ranges, or bands, of energy (vertical axis). In an insulator or semiconductor, an electron bound to an atom can break free only if it gets enough energy from heat or a passing photon to jump the “band gap”, but in graphene the gap is infinitely small.
-
18M.3.sl.TZ2.b:
Diamond, graphene, and graphite are all network solids.
Suggest, giving a reason, the electron mobility of diamond compared to graphene.
-
18M.3.sl.TZ2.c:
The melting point of diamond at 1 × 106 kPa is 4200 K (in the absence of oxygen).
Suggest, based on molecular structure, why graphene has a higher melting point under these conditions.
- 18N.1.sl.TZ0.9: Which species has the same molecular geometry as SO32−? A. BF3 B. SO3 C. PF3 D. CO32−
- 18N.1.sl.TZ0.10: How many lone pairs and bonding pairs of electrons surround the central chlorine atom in...
- 18N.1.sl.TZ0.12: Which molecule is polar? A. BeCl2 B. BCl3 C. NCl3 D. CCl4
-
18N.2.sl.TZ0.3b:
Draw the Lewis (electron dot) structure for BrO3− that obeys the octet rule.
- 18N.2.sl.TZ0.3c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
-
18N.2.sl.TZ0.b:
Draw the Lewis (electron dot) structure for BrO3− that obeys the octet rule.
- 18N.2.sl.TZ0.c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
-
18N.2.hl.TZ0.3b.i:
Draw two Lewis (electron dot) structures for BrO3−.
- 18N.2.hl.TZ0.3c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
-
18N.2.hl.TZ0.b.i:
Draw two Lewis (electron dot) structures for BrO3−.
- 18N.2.hl.TZ0.c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
- 18N.2.hl.TZ0.8a: Suggest why the three-membered ring in methyloxirane is unstable.
- 18N.2.hl.TZ0.a: Suggest why the three-membered ring in methyloxirane is unstable.
4.4 Intermolecular forces
- 17N.1.sl.TZ0.11: Which of the following series shows increasing hydrogen bonding with water? A. Propane <...
-
17N.2.sl.TZ0.2b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
-
17N.2.hl.TZ0.3b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
-
17N.3.sl.TZ0.7b.i:
Describe how the structures of LDPE and HDPE affect one mechanical property of the plastics.
-
17N.3.sl.TZ0.9c:
Explain how the inclusion of carbohydrates in plastics makes them biodegradable.
-
17N.3.sl.TZ0.10a:
Explain, at the molecular level, why vitamin D is soluble in fats. Use section 35 of the data booklet.
-
17N.3.hl.TZ0.7c:
Identify the type of intermolecular bonding that is responsible for Kevlar®’s strength.
-
18M.2.hl.TZ1.1e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
-
18M.2.hl.TZ1.1e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
-
18M.2.hl.TZ2.7e:
Carbon and silicon are elements in group 14.
Explain why CO2 is a gas but SiO2 is a solid at room temperature.
-
18M.1.sl.TZ1.12:
What are the strongest intermolecular forces between molecules of propanone, CH3COCH3, in the liquid phase?
A. London (dispersion) forces
B. Covalent bonding
C. Hydrogen bonding
D. Dipole–dipole forces
-
18M.2.sl.TZ1.1e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
-
18M.2.sl.TZ1.1e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
-
18M.3.sl.TZ1.1a.i:
Part of this molecule is hydrophilic (bonds readily to water) and part hydrophobic (does not bond readily to water). Draw a circle around all of the hydrophilic part of the molecule.
-
18M.3.sl.TZ1.1a.ii:
When a small amount of palmitic acid is placed in water it disperses to form a layer on the surface that is only one molecule thick. Explain, in terms of intermolecular forces, why this occurs.
-
18M.1.sl.TZ2.12:
The compounds shown below have similar relative molecular masses. What is the correct order of increasing boiling point?
A. CH3COOH < (CH3)2CO < (CH3)2CHOH
B. CH3COOH < (CH3)2CHOH < (CH3)2CO
C. (CH3)2CO < CH3COOH < (CH3)2CHOH
D. (CH3)2CO < (CH3)2CHOH < CH3COOH
- 18N.1.sl.TZ0.11: Which compound has the highest boiling point? A. CH3CHO B. CH3CH2F C. CH3OCH3 D. ...
- 18N.2.sl.TZ0.6b: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
- 18N.2.hl.TZ0.6d: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
-
19M.2.hl.TZ1.2h:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.hl.TZ2.1b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.hl.TZ2.1d(iii):
Explain why product B is water soluble.
-
19M.1.hl.TZ1.9:
What is the order of increasing boiling point?
A. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
B. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
C. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
D. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
-
19M.2.sl.TZ1.2c:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.sl.TZ2.1b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.sl.TZ2.1d(iii):
Explain why product B is water soluble.
-
19M.1.sl.TZ1.9:
What is the order of increasing boiling point?
A. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
B. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
C. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
D. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
- 19M.1.sl.TZ1.11: Which describes an ionic compound?
- 19M.1.sl.TZ2.12: Which compound has hydrogen bonds between its molecules? A. CH4 B. CH4O C. CH3Cl D. CH2O
-
19N.2.hl.TZ0.3d(ii):
Explain why the compound C3H8O, produced in (a)(iv), has a higher boiling point than compound C3H6O, produced in d(i).
- 19N.2.sl.TZ0.6a(iv): Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled,...
- 19N.3.sl.TZ0.9c: Explain why stearic acid has a higher melting point than oleic acid.
- 19N.3.hl.TZ0.13b: Explain how the double-helical structure of DNA is stabilized once formed.
- 19N.3.sl.TZ0.10c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.3.hl.TZ0.15c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.3.sl.TZ0.4b: Predict, with a reason, whether isotactic or atactic polypropene has the higher melting point.
-
19N.2.sl.TZ0.3d(ii):
Explain why the compound C2H6O, produced in (b), has a higher boiling point than compound C2H4O, produced in d(i).
- 20N.2.sl.TZ0.4c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
- 20N.2.hl.TZ0.4c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
-
20N.3.sl.TZ0.1a:
Suggest why a non-polar solvent was needed.
- 20N.3.sl.TZ0.1c: Non-polar solvents can be toxic. Suggest a modification to the experiment which allows the...
-
20N.3.sl.TZ0.5a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
-
20N.3.sl.TZ0.6b(ii):
State the most significant intermolecular forces in the phospholipid in b(i).
- 20N.3.sl.TZ0.11b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
-
20N.3.hl.TZ0.6a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
- 20N.3.hl.TZ0.15b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
- 21M.1.sl.TZ1.24: Which series is in order of increasing boiling point? A. CH2CH2CH3OH CH3COCH3 ...
- 21M.1.sl.TZ2.9: Which compound has the greatest volatility under the same conditions? A. SO2 B. SiO2 C. ...
- 21M.1.sl.TZ2.12: Which is the correct order based on increasing strength? A. covalent bonds < hydrogen bonds...
- 21M.2.sl.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21M.2.hl.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21N.1.sl.TZ0.12: The following compounds have similar relative molecular masses. What is the order of increasing...
- 21N.1.sl.TZ0.13: Which alcohol is least soluble in water? A. CH3OH B. CH3CH2OH C. CH3CH2CH2OH D. ...
-
21N.2.hl.TZ0.6c(v):
Suggest a reason that the Winkler Method used to measure biochemical oxygen demand (BOD) must be done at constant temperature.
-
22M.1.sl.TZ2.9:
Which statement best describes the intramolecular bonding in HCN (l)?
A. Electrostatic attractions between H+ and CN− ions
B. Hydrogen bonding
C. Van der Waals forces and hydrogen bonding
D. Electrostatic attractions between pairs of electrons and positively charged nuclei
-
22M.1.sl.TZ2.12:
What is the main interaction between liquid CH4 molecules?
A. London (dispersion) forces
B. Dipole–dipole forces
C. Hydrogen bonding
D. Covalent bonding
- 22M.2.sl.TZ1.3c(ii): Identify the strongest force between the molecules of Compound B.
- 22M.2.hl.TZ1.5a(ii): Compound A and Compound B are both liquids at room temperature and pressure. Identify the...
- 22M.2.sl.TZ2.4a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.7c:
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
-
22M.2.hl.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22N.2.sl.TZ0.4a.iii:
Explain, with reference to intermolecular forces, why B is more volatile than A.
-
19M.2.hl.TZ1.2h:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.hl.TZ1.h:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.hl.TZ2.1b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.hl.TZ2.1d(iii):
Explain why product B is water soluble.
-
19M.2.hl.TZ2.b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.hl.TZ2.d(iii):
Explain why product B is water soluble.
-
19M.1.hl.TZ1.9:
What is the order of increasing boiling point?
A. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
B. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
C. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
D. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
-
19M.2.sl.TZ1.2c:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.sl.TZ1.c:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
-
19M.2.sl.TZ2.1b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.sl.TZ2.1d(iii):
Explain why product B is water soluble.
-
19M.2.sl.TZ2.b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
-
19M.2.sl.TZ2.d(iii):
Explain why product B is water soluble.
-
19M.1.sl.TZ1.9:
What is the order of increasing boiling point?
A. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
B. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
C. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
D. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
- 19M.1.sl.TZ1.11: Which describes an ionic compound?
- 19M.1.sl.TZ2.12: Which compound has hydrogen bonds between its molecules? A. CH4 B. CH4O C. CH3Cl D. CH2O
-
19N.2.hl.TZ0.3d(ii):
Explain why the compound C3H8O, produced in (a)(iv), has a higher boiling point than compound C3H6O, produced in d(i).
-
19N.2.hl.TZ0.d(ii):
Explain why the compound C3H8O, produced in (a)(iv), has a higher boiling point than compound C3H6O, produced in d(i).
- 19N.2.sl.TZ0.6a(iv): Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled,...
- 19N.2.sl.TZ0.a(iv): Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled,...
- 19N.3.sl.TZ0.9c: Explain why stearic acid has a higher melting point than oleic acid.
- 19N.3.sl.TZ0.c: Explain why stearic acid has a higher melting point than oleic acid.
- 19N.3.hl.TZ0.13b: Explain how the double-helical structure of DNA is stabilized once formed.
- 19N.3.hl.TZ0.b: Explain how the double-helical structure of DNA is stabilized once formed.
- 19N.3.sl.TZ0.10c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.3.sl.TZ0.c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.3.hl.TZ0.15c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.3.hl.TZ0.c: Explain why maltose, C12H22O11, is soluble in water.
- 19N.3.sl.TZ0.4b: Predict, with a reason, whether isotactic or atactic polypropene has the higher melting point.
- 19N.3.sl.TZ0.b: Predict, with a reason, whether isotactic or atactic polypropene has the higher melting point.
-
19N.2.sl.TZ0.3d(ii):
Explain why the compound C2H6O, produced in (b), has a higher boiling point than compound C2H4O, produced in d(i).
-
19N.2.sl.TZ0.d(ii):
Explain why the compound C2H6O, produced in (b), has a higher boiling point than compound C2H4O, produced in d(i).
- 20N.2.sl.TZ0.4c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
- 20N.2.sl.TZ0.c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
- 20N.2.hl.TZ0.4c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
- 20N.2.hl.TZ0.c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
-
20N.3.sl.TZ0.1a:
Suggest why a non-polar solvent was needed.
- 20N.3.sl.TZ0.1c: Non-polar solvents can be toxic. Suggest a modification to the experiment which allows the...
-
20N.3.sl.TZ0.a:
Suggest why a non-polar solvent was needed.
- 20N.3.sl.TZ0.c: Non-polar solvents can be toxic. Suggest a modification to the experiment which allows the...
-
20N.3.sl.TZ0.5a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
-
20N.3.sl.TZ0.a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
-
20N.3.sl.TZ0.6b(ii):
State the most significant intermolecular forces in the phospholipid in b(i).
-
20N.3.sl.TZ0.b(ii):
State the most significant intermolecular forces in the phospholipid in b(i).
- 20N.3.sl.TZ0.11b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
- 20N.3.sl.TZ0.b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
-
20N.3.hl.TZ0.6a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
-
20N.3.hl.TZ0.a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
- 20N.3.hl.TZ0.15b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
- 20N.3.hl.TZ0.b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest why...
- 21M.1.sl.TZ1.24: Which series is in order of increasing boiling point? A. CH2CH2CH3OH CH3COCH3 ...
- 21M.1.sl.TZ2.9: Which compound has the greatest volatility under the same conditions? A. SO2 B. SiO2 C. ...
- 21M.1.sl.TZ2.12: Which is the correct order based on increasing strength? A. covalent bonds < hydrogen bonds...
- 21M.2.sl.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21M.2.sl.TZ2.d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21M.2.hl.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21M.2.hl.TZ2.d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21N.1.sl.TZ0.12: The following compounds have similar relative molecular masses. What is the order of increasing...
- 21N.1.sl.TZ0.13: Which alcohol is least soluble in water? A. CH3OH B. CH3CH2OH C. CH3CH2CH2OH D. ...
-
21N.2.hl.TZ0.6c(v):
Suggest a reason that the Winkler Method used to measure biochemical oxygen demand (BOD) must be done at constant temperature.
-
21N.2.hl.TZ0.c(v):
Suggest a reason that the Winkler Method used to measure biochemical oxygen demand (BOD) must be done at constant temperature.
-
22M.1.sl.TZ2.9:
Which statement best describes the intramolecular bonding in HCN (l)?
A. Electrostatic attractions between H+ and CN− ions
B. Hydrogen bonding
C. Van der Waals forces and hydrogen bonding
D. Electrostatic attractions between pairs of electrons and positively charged nuclei
-
22M.1.sl.TZ2.12:
What is the main interaction between liquid CH4 molecules?
A. London (dispersion) forces
B. Dipole–dipole forces
C. Hydrogen bonding
D. Covalent bonding
- 22M.2.sl.TZ1.3c(ii): Identify the strongest force between the molecules of Compound B.
- 22M.2.sl.TZ1.c(ii): Identify the strongest force between the molecules of Compound B.
- 22M.2.hl.TZ1.5a(ii): Compound A and Compound B are both liquids at room temperature and pressure. Identify the...
- 22M.2.hl.TZ1.a(ii): Compound A and Compound B are both liquids at room temperature and pressure. Identify the...
- 22M.2.sl.TZ2.4a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
- 22M.2.sl.TZ2.a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.7c:
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
-
22M.2.hl.TZ2.c:
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
-
22M.2.hl.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.hl.TZ2.a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22N.2.sl.TZ0.4a.iii:
Explain, with reference to intermolecular forces, why B is more volatile than A.
-
22N.2.sl.TZ0.a.iii:
Explain, with reference to intermolecular forces, why B is more volatile than A.
- 17N.1.sl.TZ0.11: Which of the following series shows increasing hydrogen bonding with water? A. Propane <...
-
17N.2.sl.TZ0.2b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
-
17N.2.sl.TZ0.b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
-
17N.2.hl.TZ0.3b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
-
17N.2.hl.TZ0.b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
-
17N.3.sl.TZ0.7b.i:
Describe how the structures of LDPE and HDPE affect one mechanical property of the plastics.
-
17N.3.sl.TZ0.b.i:
Describe how the structures of LDPE and HDPE affect one mechanical property of the plastics.
-
17N.3.sl.TZ0.9c:
Explain how the inclusion of carbohydrates in plastics makes them biodegradable.
-
17N.3.sl.TZ0.c:
Explain how the inclusion of carbohydrates in plastics makes them biodegradable.
-
17N.3.sl.TZ0.10a:
Explain, at the molecular level, why vitamin D is soluble in fats. Use section 35 of the data booklet.
-
17N.3.sl.TZ0.a:
Explain, at the molecular level, why vitamin D is soluble in fats. Use section 35 of the data booklet.
-
17N.3.hl.TZ0.7c:
Identify the type of intermolecular bonding that is responsible for Kevlar®’s strength.
-
17N.3.hl.TZ0.c:
Identify the type of intermolecular bonding that is responsible for Kevlar®’s strength.
-
18M.2.hl.TZ1.1e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
-
18M.2.hl.TZ1.1e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
-
18M.2.hl.TZ1.e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
-
18M.2.hl.TZ1.e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
-
18M.2.hl.TZ2.7e:
Carbon and silicon are elements in group 14.
Explain why CO2 is a gas but SiO2 is a solid at room temperature.
-
18M.2.hl.TZ2.e:
Carbon and silicon are elements in group 14.
Explain why CO2 is a gas but SiO2 is a solid at room temperature.
-
18M.1.sl.TZ1.12:
What are the strongest intermolecular forces between molecules of propanone, CH3COCH3, in the liquid phase?
A. London (dispersion) forces
B. Covalent bonding
C. Hydrogen bonding
D. Dipole–dipole forces
-
18M.2.sl.TZ1.1e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
-
18M.2.sl.TZ1.1e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
-
18M.2.sl.TZ1.e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
-
18M.2.sl.TZ1.e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
-
18M.3.sl.TZ1.1a.i:
Part of this molecule is hydrophilic (bonds readily to water) and part hydrophobic (does not bond readily to water). Draw a circle around all of the hydrophilic part of the molecule.
-
18M.3.sl.TZ1.1a.ii:
When a small amount of palmitic acid is placed in water it disperses to form a layer on the surface that is only one molecule thick. Explain, in terms of intermolecular forces, why this occurs.
-
18M.3.sl.TZ1.a.i:
Part of this molecule is hydrophilic (bonds readily to water) and part hydrophobic (does not bond readily to water). Draw a circle around all of the hydrophilic part of the molecule.
-
18M.3.sl.TZ1.a.ii:
When a small amount of palmitic acid is placed in water it disperses to form a layer on the surface that is only one molecule thick. Explain, in terms of intermolecular forces, why this occurs.
-
18M.1.sl.TZ2.12:
The compounds shown below have similar relative molecular masses. What is the correct order of increasing boiling point?
A. CH3COOH < (CH3)2CO < (CH3)2CHOH
B. CH3COOH < (CH3)2CHOH < (CH3)2CO
C. (CH3)2CO < CH3COOH < (CH3)2CHOH
D. (CH3)2CO < (CH3)2CHOH < CH3COOH
- 18N.1.sl.TZ0.11: Which compound has the highest boiling point? A. CH3CHO B. CH3CH2F C. CH3OCH3 D. ...
- 18N.2.sl.TZ0.6b: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
- 18N.2.sl.TZ0.b: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
- 18N.2.hl.TZ0.6d: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
- 18N.2.hl.TZ0.d: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
4.5 Metallic bonding
-
17N.2.sl.TZ0.2b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
-
17N.2.hl.TZ0.3b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
- 18M.1.hl.TZ1.11: Which metal has the strongest metallic bonding? A. Na B. Mg C. Al D. Ca
-
18M.2.hl.TZ1.2d.ii:
Outline why solid calcium is a good conductor of electricity.
-
18M.2.sl.TZ1.2d.ii:
Outline why solid calcium is a good conductor of electricity.
- 19M.1.hl.TZ1.11: Which combination corresponds to a strong metallic bond?
- 19M.1.hl.TZ2.10: Which combination causes the strength of metallic bonding to increase?
- 19M.1.sl.TZ1.12: Which combination corresponds to a strong metallic bond?
- 19M.1.sl.TZ2.10: Which combination causes the strength of metallic bonding to increase?
-
20N.1.sl.TZ0.12:
Which series shows the correct order of metallic bond strength from strongest to weakest?
A.
B.
C.
D.
- 20N.2.sl.TZ0.4d(iii): Describe the bonding in metals.
- 20N.2.sl.TZ0.4d(iv): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.2.hl.TZ0.4d(v): Describe the bonding in metals.
- 20N.2.hl.TZ0.4d(vi): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.3.sl.TZ0.4b(i): Alloying metals changes their properties. Suggest one property of magnesium that could be...
-
20N.3.hl.TZ0.4c(i):
Alloying metals changes their properties. Suggest one property of magnesium that could be improved by making a magnesium–CNT alloy.
- 21M.2.sl.TZ1.1a: Outline why metals, like iron, can conduct electricity.
- 21M.2.sl.TZ1.1c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.sl.TZ1.1e: Explain why the addition of small amounts of carbon to iron makes the metal harder.
- 21M.2.hl.TZ1.1a: Outline why metals, like iron, can conduct electricity.
- 21M.2.hl.TZ1.1d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.1f: Explain why the addition of small amounts of carbon to iron makes the metal harder.
- 21M.2.sl.TZ2.2b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
- 21M.2.hl.TZ2.2b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
-
22M.2.hl.TZ2.5c:
Describe the bonding in iron, Fe (s).
- 22N.1.sl.TZ0.12: Alloying a metal with a metal of smaller atomic radius can disrupt the lattice and make it more...
- 23M.2.HL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
-
23M.2.HL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.HL.TZ2.4a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
- 23M.2.SL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
-
23M.2.SL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.SL.TZ2.4a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
- 19M.1.hl.TZ1.11: Which combination corresponds to a strong metallic bond?
- 19M.1.hl.TZ2.10: Which combination causes the strength of metallic bonding to increase?
- 19M.1.sl.TZ1.12: Which combination corresponds to a strong metallic bond?
- 19M.1.sl.TZ2.10: Which combination causes the strength of metallic bonding to increase?
-
20N.1.sl.TZ0.12:
Which series shows the correct order of metallic bond strength from strongest to weakest?
A.
B.
C.
D.
- 20N.2.sl.TZ0.4d(iii): Describe the bonding in metals.
- 20N.2.sl.TZ0.4d(iv): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.2.sl.TZ0.d(iii): Describe the bonding in metals.
- 20N.2.sl.TZ0.d(iv): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.2.hl.TZ0.4d(v): Describe the bonding in metals.
- 20N.2.hl.TZ0.4d(vi): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.2.hl.TZ0.d(v): Describe the bonding in metals.
- 20N.2.hl.TZ0.d(vi): Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the...
- 20N.3.sl.TZ0.4b(i): Alloying metals changes their properties. Suggest one property of magnesium that could be...
- 20N.3.sl.TZ0.b(i): Alloying metals changes their properties. Suggest one property of magnesium that could be...
-
20N.3.hl.TZ0.4c(i):
Alloying metals changes their properties. Suggest one property of magnesium that could be improved by making a magnesium–CNT alloy.
-
20N.3.hl.TZ0.c(i):
Alloying metals changes their properties. Suggest one property of magnesium that could be improved by making a magnesium–CNT alloy.
- 21M.2.sl.TZ1.1a: Outline why metals, like iron, can conduct electricity.
- 21M.2.sl.TZ1.1c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.sl.TZ1.1e: Explain why the addition of small amounts of carbon to iron makes the metal harder.
- 21M.2.sl.TZ1.a: Outline why metals, like iron, can conduct electricity.
- 21M.2.sl.TZ1.c(iv): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.sl.TZ1.e: Explain why the addition of small amounts of carbon to iron makes the metal harder.
- 21M.2.hl.TZ1.1a: Outline why metals, like iron, can conduct electricity.
- 21M.2.hl.TZ1.1d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.1f: Explain why the addition of small amounts of carbon to iron makes the metal harder.
- 21M.2.hl.TZ1.a: Outline why metals, like iron, can conduct electricity.
- 21M.2.hl.TZ1.d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.hl.TZ1.f: Explain why the addition of small amounts of carbon to iron makes the metal harder.
- 21M.2.sl.TZ2.2b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
- 21M.2.sl.TZ2.b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
- 21M.2.hl.TZ2.2b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
- 21M.2.hl.TZ2.b(ii): Describe metallic bonding and how it contributes to electrical conductivity.
-
22M.2.hl.TZ2.5c:
Describe the bonding in iron, Fe (s).
-
22M.2.hl.TZ2.c:
Describe the bonding in iron, Fe (s).
- 22N.1.sl.TZ0.12: Alloying a metal with a metal of smaller atomic radius can disrupt the lattice and make it more...
-
17N.2.sl.TZ0.2b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
-
17N.2.sl.TZ0.b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
-
17N.2.hl.TZ0.3b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
-
17N.2.hl.TZ0.b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
- 18M.1.hl.TZ1.11: Which metal has the strongest metallic bonding? A. Na B. Mg C. Al D. Ca
-
18M.2.hl.TZ1.2d.ii:
Outline why solid calcium is a good conductor of electricity.
-
18M.2.hl.TZ1.d.ii:
Outline why solid calcium is a good conductor of electricity.
- 23M.2.HL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
- 23M.2.HL.TZ1.i: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
-
23M.2.HL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.HL.TZ2.a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.HL.TZ2.4a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
-
23M.2.HL.TZ2.a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
- 23M.2.SL.TZ1.2ai: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
- 23M.2.SL.TZ1.i: Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.
-
23M.2.SL.TZ2.2a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.SL.TZ2.a:
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
-
23M.2.SL.TZ2.4a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
-
23M.2.SL.TZ2.a:
Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
-
18M.2.sl.TZ1.2d.ii:
Outline why solid calcium is a good conductor of electricity.
-
18M.2.sl.TZ1.d.ii:
Outline why solid calcium is a good conductor of electricity.