4.3 Covalent structures

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Directly related questions

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...
20N.2.hl.TZ0.2a: Predict the electron domain and molecular geometries around the oxygen atom of molecule A using...
17N.2.hl.TZ0.4b:
Predict whether the molecules PF₃ and PF₅ are polar or non-polar.
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.
21M.1.sl.TZ1.12: Along which series is the bond angle increasing? A. NH3 H2O CH4 B. CH4 NH3 H2O C. H2O ...
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.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.hl.TZ2.c:
Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF₄, and sulfur dichloride, SCl₂.
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.TZ1.1b:
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.1.sl.TZ1.10: Which form of carbon is the poorest electrical conductor? A. Graphite B. Graphene C. ...
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.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.TZ1.1b:
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.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.TZ2.6b:
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.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.
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.
21N.2.sl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P₄ molecule, containing only single bonds.
21N.2.sl.TZ0.3a(i):
Sketch the Lewis (electron dot) structure of the P₄ molecule, containing only single bonds.
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.sl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
21N.2.sl.TZ0.b(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.TZ0.3a(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.hl.TZ0.3b(i): Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl...
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...
22M.1.sl.TZ1.12: For which species can resonance structures be drawn? A. HCOOH B. HCOO– C. CH3OH D. H2CO3
22M.1.sl.TZ1.12: For which species can resonance structures be drawn? A. HCOOH B. HCOO– C. CH3OH D. H2CO3
22M.1.sl.TZ1.11: Which molecule is most polar? A. CF4 B. CCl4 C. CHF3 D. CClF3
22M.1.sl.TZ1.11: Which molecule is most polar? A. CF4 B. CCl4 C. CHF3 D. CClF3
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.TZ0.3c: Predict, using the VSEPR theory, the geometry of the BrO3− ion and the O−Br−O bond angles.
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.sl.TZ0.3b:
Draw the Lewis (electron dot) structure for BrO₃⁻ that obeys the octet rule.
18N.2.sl.TZ0.3b:
Draw the Lewis (electron dot) structure for BrO₃⁻ that obeys the octet rule.
18N.2.sl.TZ0.b:
Draw the Lewis (electron dot) structure for BrO₃⁻ that obeys the octet rule.
18N.2.hl.TZ0.8a: Suggest why the three-membered ring in methyloxirane is unstable.
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.
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.TZ2.10: 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.TZ1.2b:
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.
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.TZ1.5c(iv):
State the type of bond formed when chloramine is protonated.
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.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.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
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.TZ2.3d(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.TZ2.1b(i):
Deduce the Lewis (electron dot) structure of ethyne.
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.TZ2.11: Which molecule contains an incomplete octet of electrons? A. NF3 B. BF3 C. BrF D. SF2
19M.1.sl.TZ2.11: Which molecule contains an incomplete octet of electrons? A. NF3 B. BF3 C. BrF D. SF2
19N.1.sl.TZ0.12:
What is the structure and bonding in SiO₂(s)?
19N.1.sl.TZ0.12:
What is the structure and bonding in SiO₂(s)?
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...
22N.1.sl.TZ0.11: Which structure of CF2Cl2 is shown with correct bond and molecular dipoles?
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.TZ0.1c:
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.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.
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.TZ0.4a:
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.
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.TZ2.7e:
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.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.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.
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...
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.1.sl.TZ0.12: Which molecule is polar? A. BeCl2 B. BCl3 C. NCl3 D. CCl4
18N.2.hl.TZ0.3b.i:
Draw two Lewis (electron dot) structures for BrO₃⁻.
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.
19M.2.sl.TZ1.5c(i):
Draw a Lewis (electron dot) structure of chloramine.
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.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.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
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.2.sl.TZ0.a: Draw the Lewis structures of oxygen, O2, and ozone, O3.
19N.1.sl.TZ0.11: Which describes a resonance structure? A. Double bond can be drawn in alternative...
19N.1.sl.TZ0.11: Which describes a resonance structure? A. Double bond can be drawn in alternative...
20N.1.sl.TZ0.10:
Which molecule is most polar?

A.   ${CHF}_{3}$

B.   ${CF}_{4}$

C.   ${CClF}_{3}$

D.   ${CCl}_{4}$
20N.1.sl.TZ0.10:
Which molecule is most polar?

A.   ${CHF}_{3}$

B.   ${CF}_{4}$

C.   ${CClF}_{3}$

D.   ${CCl}_{4}$
20N.1.sl.TZ0.11:
Which combination correctly describes the geometry of the carbonate ion, ${CO}_{3}^{2 -}$ ?
20N.1.sl.TZ0.11:
Which combination correctly describes the geometry of the carbonate ion, ${CO}_{3}^{2 -}$ ?
20N.2.sl.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...
21M.1.sl.TZ1.9: The Lewis structure of methylamine is shown. What is the molecular geometry around N? A. ...
21M.1.sl.TZ1.9: The Lewis structure of methylamine is shown. What is the molecular geometry around N? A. ...
21M.2.sl.TZ1.2a(i):
Draw the Lewis (electron dot) structure of hydrogen sulfide.
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.
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.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.TZ1.7a(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.TZ2.2c:
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.
21N.2.sl.TZ0.3b(ii): Explain the polarity of PCl3.
21N.2.sl.TZ0.3b(ii): Explain the polarity of PCl3.
21N.2.sl.TZ0.b(ii): Explain the polarity of PCl3.
22M.1.hl.TZ1.13: What are the electron domain and molecular geometries of the XeF4 molecule?
22M.1.hl.TZ1.13: What are the electron domain and molecular geometries of the XeF4 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.TZ2.3d(i): Draw the Lewis structure of NO3−.
22M.2.sl.TZ2.3d(i): Draw the Lewis structure of NO3−.
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.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(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.
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.TZ2.6c(i): Draw the Lewis structure of SO3.
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.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.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.TZ0.9: Which molecule can be represented by resonance structures? A. H2S B. HNO3 C. H2O2 D. HClO
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. BeH₂

B. AlH₃

C. PH₃

D. SiH₄
22N.1.sl.TZ0.10:
Which molecule is polar?

A. BeH₂

B. AlH₃

C. PH₃

D. SiH₄
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.TZ0.1f.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.
23M.2.HL.TZ1.2ai: 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.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.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.TZ1.9a: 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.TZ2.3a:
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.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).

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4.3 Covalent structures

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