Directly related questions
- EXM.1B.SL.TZ0.2c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
- EXM.1B.HL.TZ0.2c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
- EXM.1B.HL.TZ0.2c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
- EXM.1B.HL.TZ0.c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
- EXM.1B.SL.TZ0.2c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
- EXM.1B.SL.TZ0.c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
-
22N.1A.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.1A.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.1A.SL.TZ0.10:
Which molecule is polar?
A. BeH2B. AlH3
C. PH3
D. SiH4
-
22N.1A.SL.TZ0.10:
Which molecule is polar?
A. BeH2B. AlH3
C. PH3
D. SiH4
- 22N.1A.SL.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
- 22N.1A.SL.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
-
22N.2.SL.TZ0.4a.iii:
Explain, with reference to intermolecular forces, why B is more volatile than A.
-
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.
-
20N.2.SL.TZ0.3a:
Outline the two distinct phases of this composite.
-
20N.2.SL.TZ0.a:
Outline the two distinct phases of this composite.
-
20N.2.SL.TZ0.3a:
Outline the two distinct phases of this composite.
-
20N.2.SL.TZ0.a:
Outline the two distinct phases of this composite.
Sub sections and their related questions
Structure 2.2.1—A covalent bond is formed by the electrostatic attraction between a shared pair of electrons and the positively charged nuclei. The octet rule refers to the tendency of atoms to gain a valence shell with a total of 8 electrons. Deduce the Lewis formula of molecules and ions for up to four electron pairs on each atom.
- 22M.2.HL.TZ2.3d(i): Draw the Lewis structure of NO3−.
-
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.1A.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.1A.SL.TZ1.10: What is the explanation for the high melting point of sodium chloride? A. The covalent bond...
- 19M.1A.SL.TZ2.11: Which molecule contains an incomplete octet of electrons? A. NF3 B. BF3 C. BrF D. SF2
-
19M.2.SL.TZ2.5b(iii):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
- 21N.1A.SL.TZ0.11: Which compound contains both ionic and covalent bonds? A. CH3COONa B. CH3COOH C. K2O D. ...
- 19M.1A.HL.TZ1.12: Which species has delocalized electrons? A. OH− B. H2CO C. CO2 D. CO32−
-
19M.2.SL.TZ2.5d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
19M.2.SL.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
-
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.
- 19N.1A.SL.TZ0.13: Which atom does not obey the octet rule? A. C in CO2 B. F in BF3 C. O in H2O D. S in SF6
- 21M.2.SL.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.TZ1.2a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
- 22M.1A.SL.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
- 22M.2.HL.TZ2.3d(i): Draw the Lewis structure of NO3−.
- 22M.2.HL.TZ2.d(i): Draw the Lewis structure of NO3−.
-
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.1A.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.1A.SL.TZ1.10: What is the explanation for the high melting point of sodium chloride? A. The covalent bond...
- 19M.1A.SL.TZ2.11: Which molecule contains an incomplete octet of electrons? A. NF3 B. BF3 C. BrF D. SF2
-
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:
- 21N.1A.SL.TZ0.11: Which compound contains both ionic and covalent bonds? A. CH3COONa B. CH3COOH C. K2O D. ...
- 19M.1A.HL.TZ1.12: Which species has delocalized electrons? A. OH− B. H2CO C. CO2 D. CO32−
-
19M.2.SL.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.d(i):
Identify the type of bonding in sodium hydrogencarbonate.
Between sodium and hydrogencarbonate:
Between hydrogen and oxygen in hydrogencarbonate:
-
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.
-
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.
- 19N.1A.SL.TZ0.13: Which atom does not obey the octet rule? A. C in CO2 B. F in BF3 C. O in H2O D. S in SF6
- 21M.2.SL.TZ1.1d(v): Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.
- 21M.2.SL.TZ1.d(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.SL.TZ1.2a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
-
21M.2.SL.TZ1.a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
- 22M.1A.SL.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
[N/A]
NoneStructure 2.2.2—Single, double and triple bonds involve one, two and three shared pairs of electrons respectively. Explain the relationship between the number of bonds, bond length and bond strength.
- 19N.2.SL.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 21N.1A.SL.TZ0.9: Which molecule has the weakest nitrogen to nitrogen bond? A. N2 B. N2H2 C. N2H4 D.
- 21M.2.HL.TZ1.7a(ii): Discuss the relative length of the two O−O bonds in ozone.
- 19N.1A.SL.TZ0.10: Which compound has the shortest C to O bond? A. CH3CHO B. CO C. CO2 D. C2H5OC2H5
-
21M.1A.SL.TZ2.10:
Which compound has the shortest C to N bond?
A. HCN
B. CH3CH2NH2
C. CH3CHNH
D. (CH3)2NH
-
20N.1A.SL.TZ0.9:
Which of these species contains the shortest carbon to oxygen bond length?
A.
B.
C.
D.
-
22N.1A.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
- 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.b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 21N.1A.SL.TZ0.9: Which molecule has the weakest nitrogen to nitrogen bond? A. N2 B. N2H2 C. N2H4 D.
- 21M.2.HL.TZ1.7a(ii): Discuss the relative length of the two O−O bonds in ozone.
- 21M.2.HL.TZ1.a(ii): Discuss the relative length of the two O−O bonds in ozone.
- 19N.1A.SL.TZ0.10: Which compound has the shortest C to O bond? A. CH3CHO B. CO C. CO2 D. C2H5OC2H5
-
21M.1A.SL.TZ2.10:
Which compound has the shortest C to N bond?
A. HCN
B. CH3CH2NH2
C. CH3CHNH
D. (CH3)2NH
-
20N.1A.SL.TZ0.9:
Which of these species contains the shortest carbon to oxygen bond length?
A.
B.
C.
D.
-
22N.1A.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
[N/A]
NoneStructure 2.2.3—A coordination bond is a covalent bond in which both the electrons of the shared pair originate from the same atom. Identify coordination bonds in compounds.
None[N/A]
NoneStructure 2.2.4—The valence shell electron pair repulsion (VSEPR) model enables the shapes of molecules to be predicted from the repulsion of electron domains around a central atom. Predict the electron domain geometry and the molecular geometry for species with up to four electron domains.
- 22M.2.HL.TZ2.3d(ii): Explain the electron domain geometry of NO3−.
-
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...
-
22M.2.HL.TZ2.6c(ii):
Explain the electron domain geometry of SO3.
-
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.TZ2.3d(ii):
State, giving a reason, the shape of the dinitrogen monoxide molecule.
-
21M.2.SL.TZ2.2c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl2.
- 21M.1A.SL.TZ1.9: The Lewis structure of methylamine is shown. What is the molecular geometry around N? A. ...
- 21M.1A.SL.TZ1.12: Along which series is the bond angle increasing? A. NH3 H2O CH4 B. CH4 NH3 H2O C. H2O ...
-
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...
- 21M.1A.SL.TZ2.13: What is the electron domain geometry of Si in SiO2? A. bent B. linear C. square planar D. ...
- 21M.2.SL.TZ1.2a(ii): Predict the shape of the hydrogen sulfide molecule.
-
20N.1A.SL.TZ0.11:
Which combination correctly describes the geometry of the carbonate ion, ?
- 22N.1A.SL.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
- 22M.2.HL.TZ2.3d(ii): Explain the electron domain geometry of NO3−.
- 22M.2.HL.TZ2.d(ii): Explain the electron domain geometry of NO3−.
-
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.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...
-
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.
-
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.TZ2.3d(ii):
State, giving a reason, the shape of the dinitrogen monoxide molecule.
-
19M.2.SL.TZ2.d(ii):
State, giving a reason, the shape of the dinitrogen monoxide molecule.
-
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.1A.SL.TZ1.9: The Lewis structure of methylamine is shown. What is the molecular geometry around N? A. ...
- 21M.1A.SL.TZ1.12: Along which series is the bond angle increasing? A. NH3 H2O CH4 B. CH4 NH3 H2O C. H2O ...
-
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.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...
- 21M.1A.SL.TZ2.13: What is the electron domain geometry of Si in SiO2? A. bent B. linear C. square planar D. ...
- 21M.2.SL.TZ1.2a(ii): Predict the shape of the hydrogen sulfide molecule.
- 21M.2.SL.TZ1.a(ii): Predict the shape of the hydrogen sulfide molecule.
-
20N.1A.SL.TZ0.11:
Which combination correctly describes the geometry of the carbonate ion, ?
- 22N.1A.SL.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
[N/A]
NoneStructure 2.2.5—Bond polarity results from the difference in electronegativities of the bonded atoms. Deduce the polar nature of a covalent bond from electronegativity values.
- 22N.1A.SL.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
- 22N.1A.SL.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
[N/A]
NoneStructure 2.2.6—Molecular polarity depends on both bond polarity and molecular geometry. Deduce the net dipole moment of a molecule or ion by considering bond polarity and molecular geometry.
- 22M.1A.SL.TZ1.11: Which molecule is most polar? A. CF4 B. CCl4 C. CHF3 D. CClF3
- 21N.2.SL.TZ0.3b(ii): Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.
- 21M.2.SL.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21N.2.SL.TZ0.3b(ii): Explain the polarity of PCl3.
-
20N.1A.SL.TZ0.10:
Which molecule is most polar?
A.
B.
C.
D.
-
22N.1A.SL.TZ0.10:
Which molecule is polar?
A. BeH2B. AlH3
C. PH3
D. SiH4
- 22M.1A.SL.TZ1.11: Which molecule is most polar? A. CF4 B. CCl4 C. CHF3 D. CClF3
- 21N.2.SL.TZ0.3b(ii): Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.
- 21N.2.SL.TZ0.b(ii): Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.
- 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.TZ0.3b(ii): Explain the polarity of PCl3.
- 21N.2.SL.TZ0.b(ii): Explain the polarity of PCl3.
-
20N.1A.SL.TZ0.10:
Which molecule is most polar?
A.
B.
C.
D.
-
22N.1A.SL.TZ0.10:
Which molecule is polar?
A. BeH2B. AlH3
C. PH3
D. SiH4
[N/A]
NoneStructure 2.2.7—Carbon and silicon form covalent network structures. Describe the structures and explain the properties of silicon, silicon dioxide and carbon’s allotropes: diamond, graphite, fullerenes and graphene.
-
22M.2.SL.TZ2.8a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
22M.2.SL.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
19N.1A.SL.TZ0.12:
What is the structure and bonding in SiO2 (s)?
- 22M.1A.SL.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
-
22M.2.SL.TZ2.8a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
22M.2.SL.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
-
22M.2.SL.TZ2.a(i):
Outline two differences 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.
-
19N.1A.SL.TZ0.12:
What is the structure and bonding in SiO2 (s)?
- 22M.1A.SL.TZ2.10: What is the type of bonding in a compound that has high boiling and melting points, poor...
[N/A]
NoneStructure 2.2.8—The nature of the force that exists between molecules is determined by the size and polarity of the molecules. Intermolecular forces include London (dispersion), dipole-induced dipole, dipole–dipole and hydrogen bonding. Deduce the types of intermolecular force present from the structural features of covalent molecules.
- 22M.2.SL.TZ1.5a(ii): Compound A and Compound B are both liquids at room temperature and pressure. Identify the...
- 22M.2.SL.TZ1.3c(ii): Identify the strongest force between the molecules of Compound B.
-
22M.1A.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
- 19M.1A.SL.TZ2.12: Which compound has hydrogen bonds between its molecules? A. CH4 B. CH4O C. CH3Cl D. CH2O
-
19M.2.HL.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.
-
22N.2.SL.TZ0.4a.iii:
Explain, with reference to intermolecular forces, why B is more volatile than A.
- EXM.1B.HL.TZ0.2c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
- EXM.1B.SL.TZ0.2c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
- 22M.2.SL.TZ1.5a(ii): Compound A and Compound B are both liquids at room temperature and pressure. Identify the...
- 22M.2.SL.TZ1.a(ii): Compound A and Compound B are both liquids at room temperature and pressure. Identify the...
- 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.1A.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
- 19M.1A.SL.TZ2.12: Which compound has hydrogen bonds between its molecules? A. CH4 B. CH4O C. CH3Cl D. CH2O
-
19M.2.HL.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.HL.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.
-
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.
- EXM.1B.HL.TZ0.2c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
- EXM.1B.HL.TZ0.c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
- EXM.1B.SL.TZ0.2c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
- EXM.1B.SL.TZ0.c: Suggest why water was chosen to extract ascorbic acid from the spinach leaves with reference to...
[N/A]
NoneStructure 2.2.9—Given comparable molar mass, the relative strengths of intermolecular forces are generally: London (dispersion) forces < dipole–dipole forces < hydrogen bonding. Explain the physical properties of covalent substances to include volatility, electrical conductivity and solubility in terms of their structure.
-
22M.2.SL.TZ2.7c:
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
- 21N.1A.SL.TZ0.13: Which alcohol is least soluble in water? A. CH3OH B. CH3CH2OH C. CH3CH2CH2OH D. ...
-
19M.2.SL.TZ1.10a:
The melting points of cocoa butter and coconut oil are 34 °C and 25 °C respectively.
Explain this in terms of their saturated fatty acid composition.
-
19M.2.SL.TZ2.11c:
Lecithin aids the body’s absorption of vitamin E.
Suggest why vitamin E is fat-soluble.
-
19M.2.SL.TZ1.9a:
The melting points of cocoa butter and coconut oil are 34 °C and 25 °C respectively.
Explain this in terms of their saturated fatty acid composition.
-
19M.2.SL.TZ2.7d:
Lecithin aids the body’s absorption of vitamin E.
Suggest why vitamin E is fat-soluble.
-
19M.1A.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.1A.SL.TZ2.24:
Which compound has the lowest boiling point?
A. CH3CH2CH2CH2CH2CH3
B. CH3CH2CH2CH2CH3
C. CH3CH(CH3)CH2CH3
D. CH3C(CH3)2CH3
-
19M.1A.SL.TZ2.32:
Which compound has the lowest boiling point?
A. CH3CH2CH2CH2CH2CH3
B. CH3CH2CH2CH2CH3
C. CH3CH(CH3)CH2CH3
D. CH3C(CH3)2CH3
- 21N.1A.SL.TZ0.12: The following compounds have similar relative molecular masses. What is the order of increasing...
-
19M.2.SL.TZ1.10:
Ascorbic acid and retinol are two important vitamins.
Explain why ascorbic acid is soluble in water and retinol is not. Use section 35 of the data booklet.
-
19M.1A.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
- 21M.2.SL.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21M.1A.SL.TZ2.9: Which compound has the greatest volatility under the same conditions? A. SO2 B. SiO2 C. ...
- 21M.1A.SL.TZ2.12: Which is the correct order based on increasing strength? A. covalent bonds < hydrogen bonds...
- 21M.1A.SL.TZ1.24: Which series is in order of increasing boiling point? A. CH2CH2CH3OH CH3COCH3 ...
-
22N.2.SL.TZ0.4a.iii:
Explain, with reference to intermolecular forces, why B is more volatile than A.
-
22M.2.SL.TZ2.7c:
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
-
22M.2.SL.TZ2.c:
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
- 21N.1A.SL.TZ0.13: Which alcohol is least soluble in water? A. CH3OH B. CH3CH2OH C. CH3CH2CH2OH D. ...
-
19M.2.SL.TZ1.a:
The melting points of cocoa butter and coconut oil are 34 °C and 25 °C respectively.
Explain this in terms of their saturated fatty acid composition.
-
19M.2.SL.TZ2.c:
Lecithin aids the body’s absorption of vitamin E.
Suggest why vitamin E is fat-soluble.
-
19M.2.SL.TZ1.a:
The melting points of cocoa butter and coconut oil are 34 °C and 25 °C respectively.
Explain this in terms of their saturated fatty acid composition.
-
19M.2.SL.TZ2.d:
Lecithin aids the body’s absorption of vitamin E.
Suggest why vitamin E is fat-soluble.
-
19M.1A.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.1A.SL.TZ2.24:
Which compound has the lowest boiling point?
A. CH3CH2CH2CH2CH2CH3
B. CH3CH2CH2CH2CH3
C. CH3CH(CH3)CH2CH3
D. CH3C(CH3)2CH3
-
19M.1A.SL.TZ2.32:
Which compound has the lowest boiling point?
A. CH3CH2CH2CH2CH2CH3
B. CH3CH2CH2CH2CH3
C. CH3CH(CH3)CH2CH3
D. CH3C(CH3)2CH3
- 21N.1A.SL.TZ0.12: The following compounds have similar relative molecular masses. What is the order of increasing...
-
19M.2.SL.TZ1.10:
Ascorbic acid and retinol are two important vitamins.
Explain why ascorbic acid is soluble in water and retinol is not. Use section 35 of the data booklet.
-
19M.1A.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
- 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.1A.SL.TZ2.9: Which compound has the greatest volatility under the same conditions? A. SO2 B. SiO2 C. ...
- 21M.1A.SL.TZ2.12: Which is the correct order based on increasing strength? A. covalent bonds < hydrogen bonds...
- 21M.1A.SL.TZ1.24: Which series is in order of increasing boiling point? A. CH2CH2CH3OH CH3COCH3 ...
-
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.
[N/A]
NoneStructure 2.2.10—Chromatography is a technique used to separate the components of a mixture based on their relative attractions involving intermolecular forces to mobile and stationary phases. Explain, calculate and interpret the retardation factor values, RF.
None[N/A]
NoneStructure 2.2.11—Resonance structures occur when there is more than one possible position for a double bond in a molecule. Deduce resonance structures of molecules and ions.
- 22M.2.HL.TZ2.3d(i): Draw the Lewis structure of NO3−.
- 22M.2.HL.TZ2.3d(ii): Explain the electron domain geometry of NO3−.
- 19N.2.HL.TZ0.1a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 22M.1A.HL.TZ1.12: For which species can resonance structures be drawn? A. HCOOH B. HCOO– C. CH3OH D. H2CO3
- 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.
- 19M.1A.HL.TZ1.10: Which species does not have resonance structures? A. C6H6 B. NH4+ C. CO32− D. O3
-
19M.2.HL.TZ2.3d(i):
State what the presence of alternative Lewis structures shows about the nature of the bonding in the molecule.
-
21M.2.HL.TZ1.7a(i):
Draw a Lewis (electron dot) structure for ozone.
- 21M.2.HL.TZ1.7a(ii): Discuss the relative length of the two O−O bonds in ozone.
- 19N.1A.HL.TZ0.11: Which describes a resonance structure? A. Double bond can be drawn in alternative...
- 21M.1A.HL.TZ1.12: Which contain delocalised electrons? I. C6H5OHII. CH3COO−III. CO32− A. I and II only B. I...
- 22M.2.HL.TZ2.3d(i): Draw the Lewis structure of NO3−.
- 22M.2.HL.TZ2.3d(ii): Explain the electron domain geometry of NO3−.
- 22M.2.HL.TZ2.d(i): Draw the Lewis structure of NO3−.
- 22M.2.HL.TZ2.d(ii): Explain the electron domain geometry of NO3−.
- 19N.2.HL.TZ0.1a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 19N.2.HL.TZ0.a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
- 22M.1A.HL.TZ1.12: For which species can resonance structures be drawn? A. HCOOH B. HCOO– C. CH3OH D. H2CO3
- 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.
- 19M.1A.HL.TZ1.10: Which species does not have resonance structures? A. C6H6 B. NH4+ C. CO32− D. O3
-
19M.2.HL.TZ2.3d(i):
State what the presence of alternative Lewis structures shows about the nature of the bonding in the molecule.
-
19M.2.HL.TZ2.d(i):
State what the presence of alternative Lewis structures shows about the nature of the bonding in the molecule.
-
21M.2.HL.TZ1.7a(i):
Draw a Lewis (electron dot) structure for ozone.
- 21M.2.HL.TZ1.7a(ii): Discuss the relative length of the two O−O bonds in ozone.
-
21M.2.HL.TZ1.a(i):
Draw a Lewis (electron dot) structure for ozone.
- 21M.2.HL.TZ1.a(ii): Discuss the relative length of the two O−O bonds in ozone.
- 19N.1A.HL.TZ0.11: Which describes a resonance structure? A. Double bond can be drawn in alternative...
- 21M.1A.HL.TZ1.12: Which contain delocalised electrons? I. C6H5OHII. CH3COO−III. CO32− A. I and II only B. I...
[N/A]
NoneStructure 2.2.12—Benzene, C6H6, is an important example of a molecule that has resonance. Discuss the structure of benzene from physical and chemical evidence.
-
19M.2.HL.TZ2.6d(ii):
The minor product, C6H5–CH2–CH2Br, can exist in different conformational forms (isomers).
Outline what this means.
-
19M.2.HL.TZ1.1b:
Draw the structure of one other isomer of xylene which retains the benzene ring.
-
19M.2.HL.TZ1.1b:
Draw the structure of one other isomer of xylene which retains the benzene ring.
- 19M.1A.HL.TZ1.36: Which statement is not correct regarding benzene? A. It is planar. B. The ring contains...
-
19M.2.HL.TZ1.2a:
Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
-
19M.2.HL.TZ1.2b(i):
The pH of an aqueous solution of benzoic acid at 298 K is 2.95. Determine the concentration of hydroxide ions in the solution, using section 2 of the data booklet.
-
19M.2.HL.TZ1.2b(ii):
Formulate the equation for the complete combustion of benzoic acid in oxygen using only integer coefficients.
-
19M.2.HL.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.HL.TZ2.6d(ii):
The minor product, C6H5–CH2–CH2Br, can exist in different conformational forms (isomers).
Outline what this means.
-
19M.2.HL.TZ2.d(ii):
The minor product, C6H5–CH2–CH2Br, can exist in different conformational forms (isomers).
Outline what this means.
-
19M.2.HL.TZ1.1b:
Draw the structure of one other isomer of xylene which retains the benzene ring.
-
19M.2.HL.TZ1.b:
Draw the structure of one other isomer of xylene which retains the benzene ring.
-
19M.2.HL.TZ1.1b:
Draw the structure of one other isomer of xylene which retains the benzene ring.
-
19M.2.HL.TZ1.b:
Draw the structure of one other isomer of xylene which retains the benzene ring.
- 19M.1A.HL.TZ1.36: Which statement is not correct regarding benzene? A. It is planar. B. The ring contains...
-
19M.2.HL.TZ1.2a:
Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
-
19M.2.HL.TZ1.2b(i):
The pH of an aqueous solution of benzoic acid at 298 K is 2.95. Determine the concentration of hydroxide ions in the solution, using section 2 of the data booklet.
-
19M.2.HL.TZ1.2b(ii):
Formulate the equation for the complete combustion of benzoic acid in oxygen using only integer coefficients.
-
19M.2.HL.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.HL.TZ1.a:
Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
-
19M.2.HL.TZ1.b(i):
The pH of an aqueous solution of benzoic acid at 298 K is 2.95. Determine the concentration of hydroxide ions in the solution, using section 2 of the data booklet.
-
19M.2.HL.TZ1.b(ii):
Formulate the equation for the complete combustion of benzoic acid in oxygen using only integer coefficients.
-
19M.2.HL.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.
[N/A]
NoneStructure 2.2.13—Some atoms can form molecules in which they have an expanded octet of electrons. Visually represent Lewis formulas for species with five and six electron domains around the central atom. Deduce the electron domain geometry and the molecular geometry for these species using the VSEPR model.
- 22M.1A.HL.TZ2.12: What is the molecular geometry of SF4? A. Tetrahedral B. Trigonal bipyramidal C. ...
- 19M.1A.HL.TZ2.11: Which species has a square planar molecular geometry? A. SF4 B. XeF4 C. CF4 D. PF4+
- 22M.1A.HL.TZ1.13: What are the electron domain and molecular geometries of the XeF4 molecule?
-
20N.1A.HL.TZ0.11:
Which combination correctly describes the geometry of ?
- 21M.1A.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
- 22N.1A.HL.TZ0.10: Which elements are capable of forming expanded octets? I. NitrogenII. PhosphorusIII....
-
22N.1A.SL.TZ0.10:
Which molecule is polar?
A. BeH2B. AlH3
C. PH3
D. SiH4
- 22N.1A.HL.TZ0.11: Which molecule has a tetrahedral molecular geometry? A. HNO3 B. SF4 C. XeF4 D. XeO4
- 22M.1A.HL.TZ2.12: What is the molecular geometry of SF4? A. Tetrahedral B. Trigonal bipyramidal C. ...
- 19M.1A.HL.TZ2.11: Which species has a square planar molecular geometry? A. SF4 B. XeF4 C. CF4 D. PF4+
- 22M.1A.HL.TZ1.13: What are the electron domain and molecular geometries of the XeF4 molecule?
-
20N.1A.HL.TZ0.11:
Which combination correctly describes the geometry of ?
- 21M.1A.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
- 22N.1A.HL.TZ0.10: Which elements are capable of forming expanded octets? I. NitrogenII. PhosphorusIII....
-
22N.1A.SL.TZ0.10:
Which molecule is polar?
A. BeH2B. AlH3
C. PH3
D. SiH4
- 22N.1A.HL.TZ0.11: Which molecule has a tetrahedral molecular geometry? A. HNO3 B. SF4 C. XeF4 D. XeO4
[N/A]
NoneStructure 2.2.14—Formal charge values can be calculated for each atom in a species and used todetermine which of several possible Lewis formulas is preferred. Apply formal charge to determine a preferred Lewis formula from different Lewis formulas for a species.
-
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.
[N/A]
NoneStructure 2.2.15—Sigma bonds σ form by the head-on combination of atomic orbitals where the electron density is concentrated along the bond axis. Pi bonds π form by the lateral combination of p-orbitals where the electron density is concentrated on opposite sides of the bond axis. Deduce the presence of sigma bonds and pi bonds in molecules and ions.
-
22M.2.HL.TZ1.5a(iii):
State the number of (sigma) and (pi) bonds in Compound A.
-
22M.2.HL.TZ2.7b(i):
Sketch the shape of one sigma () and one pi () bond.
-
22M.2.HL.TZ2.7b(ii):
Identify the number of sigma and pi bonds in HCN.
-
19M.1A.HL.TZ2.12:
How many sigma (σ) and pi (π) bonds are present in hydrogen cyanide, HCN?
-
21N.1A.HL.TZ0.12:
Which molecules contain two pi () bonds?
I. HCN
II. H2CO3
III. H2C2O4
A. I and II onlyB. I and III only
C. II and III only
D. I, II and III
-
22M.2.HL.TZ1.5a(iii):
State the number of (sigma) and (pi) bonds in Compound A.
-
22M.2.HL.TZ1.a(iii):
State the number of (sigma) and (pi) bonds in Compound A.
-
22M.2.HL.TZ2.7b(i):
Sketch the shape of one sigma () and one pi () bond.
-
22M.2.HL.TZ2.7b(ii):
Identify the number of sigma and pi bonds in HCN.
-
22M.2.HL.TZ2.b(i):
Sketch the shape of one sigma () and one pi () bond.
-
22M.2.HL.TZ2.b(ii):
Identify the number of sigma and pi bonds in HCN.
-
19M.1A.HL.TZ2.12:
How many sigma (σ) and pi (π) bonds are present in hydrogen cyanide, HCN?
-
21N.1A.HL.TZ0.12:
Which molecules contain two pi () bonds?
I. HCN
II. H2CO3
III. H2C2O4
A. I and II onlyB. I and III only
C. II and III only
D. I, II and III
[N/A]
NoneStructure 2.2.16—Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals for bonding. Analyse the hybridization and bond formation in molecules and ions. Identify the relationships between Lewis formulas, electron domains, molecular geometry and type of hybridization. Predict the geometry around an atom from its hybridization, and vice versa.
- 22M.2.HL.TZ1.5a(iv): Deduce the hybridization of the central carbon atom in Compound A.
- 22M.2.HL.TZ2.7b(iii): State the hybridization of the carbon atom in HCN.
- 19M.1A.HL.TZ1.13: How many carbon atoms are sp3, sp2 and sp hybridized in the molecule?
- 19M.1A.HL.TZ2.13: What is the hybridization of carbon and oxygen in methanol?
-
19M.2.HL.TZ2.3d(iii):
Deduce the hybridization of the central nitrogen atom in the molecule.
- 22M.1A.HL.TZ1.12: In which compound are all carbon atoms sp3 hybridized? A. C2H2 B. C2H2Cl2 C. C2Cl4 D. C2Cl6
- 19N.1A.HL.TZ0.12: Which atom is sp2 hybridized? A. C in H2CO B. C in CO2 C. N in CH3NH2 D. O in H2O
- 21M.1A.HL.TZ1.13: In which series are all carbon atoms sp2 hybridized? A. C2H2 H2CO HCOOH B. C2H4 ...
- 21N.1A.HL.TZ0.13: What is the hybridization of nitrogen and chlorine in NCl3?
- 22M.2.HL.TZ1.5a(iv): Deduce the hybridization of the central carbon atom in Compound A.
- 22M.2.HL.TZ1.a(iv): Deduce the hybridization of the central carbon atom in Compound A.
- 22M.2.HL.TZ2.7b(iii): State the hybridization of the carbon atom in HCN.
- 22M.2.HL.TZ2.b(iii): State the hybridization of the carbon atom in HCN.
- 19M.1A.HL.TZ1.13: How many carbon atoms are sp3, sp2 and sp hybridized in the molecule?
- 19M.1A.HL.TZ2.13: What is the hybridization of carbon and oxygen in methanol?
-
19M.2.HL.TZ2.3d(iii):
Deduce the hybridization of the central nitrogen atom in the molecule.
-
19M.2.HL.TZ2.d(iii):
Deduce the hybridization of the central nitrogen atom in the molecule.
- 22M.1A.HL.TZ1.12: In which compound are all carbon atoms sp3 hybridized? A. C2H2 B. C2H2Cl2 C. C2Cl4 D. C2Cl6
- 19N.1A.HL.TZ0.12: Which atom is sp2 hybridized? A. C in H2CO B. C in CO2 C. N in CH3NH2 D. O in H2O
- 21M.1A.HL.TZ1.13: In which series are all carbon atoms sp2 hybridized? A. C2H2 H2CO HCOOH B. C2H4 ...
- 21N.1A.HL.TZ0.13: What is the hybridization of nitrogen and chlorine in NCl3?