Directly related questions
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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
-
17N.1.sl.TZ0.10:
Which compound has the shortest C–N bond?
A. CH3NH2
B. (CH3)3CNH2
C. CH3CN
D. CH3CHNH
-
21M.1.sl.TZ1.10:
Which compound contains both ionic and covalent bonds?
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.TZ2.10:
Which compound has the shortest C to N bond?
A. HCN
B. CH3CH2NH2
C. CH3CHNH
D. (CH3)2NH
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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.
-
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.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.
- 21N.1.sl.TZ0.9: Which molecule has the weakest nitrogen to nitrogen bond? A. N2 B. N2H2 C. N2H4 D.
- 21N.1.sl.TZ0.9: 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.1.sl.TZ0.11: Which compound contains both ionic and covalent bonds? A. CH3COONa B. CH3COOH C. K2O D. ...
- 22M.1.sl.TZ1.10: What is the explanation for the high melting point of sodium chloride? A. The covalent bond...
- 22M.1.sl.TZ1.10: 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.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(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
- 22M.2.sl.TZ2.a(i): Outline one difference between the bonding of carbon atoms in C60 and diamond.
-
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.TZ2.1b(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.TZ2.5d(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:
- 19N.2.hl.TZ0.1c: Predict the bond angle in the ozone molecule.
- 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.1.sl.TZ0.10: Which compound has the shortest C to O bond? A. CH3CHO B. CO C. CO2 D. C2H5OC2H5
- 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.TZ0.1d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
- 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.d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
- 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.TZ0.2c: 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.TZ0.2d: 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.TZ0.9: The electronegativity values of four elements are given. What is the order of increasing...
- 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. ...
-
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.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.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.
-
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.TZ2.1b(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.TZ2.5b(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.TZ0.1b: 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...
- 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.TZ0.1d: Discuss how the different bond strengths between the oxygen atoms in O2 and O3 in the ozone layer...
- 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.TZ0.1a: 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.
- 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.TZ0.1b: Outline why both bonds in the ozone molecule are the same length and predict the bond length in...
- 19N.2.hl.TZ0.b: 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.TZ0.5b(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.
-
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.
- 21M.1.sl.TZ2.12: Which is the correct order based on increasing strength? A. covalent bonds < hydrogen bonds...
- 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.TZ1.1c(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.TZ1.1d(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.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.2.sl.TZ2.4a(ii): Explain why C60 and diamond sublime at different temperatures and pressures.
- 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.TZ2.8a(i):
Outline two differences between the bonding of carbon atoms in C60 and diamond.
-
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.TZ0.13:
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.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
- 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.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.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