13.2.1 List the characteristics properties of transition elements.
13.2.2 Explain why Sc and Zn are not considered to be transition elements
13.2.3 Explain the existence of variable oxidation number in ions of transition elements.
13.2.4 Define the term ligand
13.2.5 Describe and explain the formation of complexes of d-block elements
13.2.6 Explain why some complexes of d-block elements are coloured
13.2.7 State examples of the catalytic action of transition elements and their compounds.
13.2.8 Outline the economic significance of catalysts in the Contact and Haber processes
Thursday 28 November 2013
Topic 13.1: Trends across period 3
13.1.1 Explain the physical states (under standard conditions) and electrical conductivity (in the molten state) of the chlorides and oxides of the elements in period 3 in terms of their bonding and structure
13.1.2 Describe the reactions of chlorine and the chlorides referred to in 13.1.1 with water
13.1.2 Describe the reactions of chlorine and the chlorides referred to in 13.1.1 with water
Topic 13: Periodicity
Topic 13 of the IB HL Chemistry syllabus is the Periodicity. IBO recommends to spend 4 hours on this topic.
This topic has 2 sub-chapters: "Trends across period 3", and "First-row d-block elements". Each are separated with numerical values in order of mentioned.
These are advanced HL syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.
This topic has 2 sub-chapters: "Trends across period 3", and "First-row d-block elements". Each are separated with numerical values in order of mentioned.
These are advanced HL syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.
Topic 3.3: Chemical properties
3.3.1 Discuss the similarities and differences in the chemical properties of elements in the same group
Group 0: Noble Gases
Group 1: Alkali Metals
The reaction becomes more vigorous as we descend the group.
Group 7: Halogens
Silver Nitrate = Colourless
3.3.2 Discuss the changes in nature, from ionic to covalent and from basic to acidic, of the oxides across period 3
The acid-base properties of the oxides are closely linked to their bonding. Metallic elements, which form ionic oxides, are basic; non-metal oxides, which are covalent are acidic. Aluminium oxide, which can be considered as an ionic oxide with some covalent character, shows amphoteric properties reacting with both acids and bases.
Sodium and Magnesium Oxide are bases
Na2O + HCl →2NaCl + H2O
MgO + H2SO4 → MgSO4 + H2O
Aluminium oxide is amphoteric (alkali and acid) so it reacts with both bases and acids
Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O
Al2O3 + 2NaOH + 3H2O → 2NaAl(OH)4
Phosphorous Oxide, Sulfur Oxide, Dichlorine Oxide reacts with water to form Acid Solutions
P4O10 + 6H2O → 4H3PO4
P4O6 +6H2O → 4 H3PO3
SO3 + H2O → H2SO4
SO2 + H2O → H2SO3
Cl2O7 + H2O → 2HClO4
Cl2O + H2O → 2HClO
Silcon dioxide doesn't react with water, but reacts with concentrated alkalis to form silicates
SiO2 + 2OH → SiO3 + H2O
Group 0: Noble Gases
- Colourless
- Monatomic
- Unreactive
Their lack of reactivity can be explained by the inability of their atoms to lose or gain electrons.
Group 1: Alkali Metals
- Good conductors of electricity
- Low Density
- Shiny surface
- Reactive metals
- Forms ionic compounds with non-metals
They form single charged ions M+. Their low ionization energies give an indication of the ease with which the outer electron is lost.
Reaction with Water
The alkali metals react with water to produce hydrogen and the metal hydroxide.
Group 7: Halogens
- Diatomic molecules
- Coloured
- Gradual change from gases (Fluorine and Chlorine) to Liquid (Bromine) to Solids (Iodine and Astatine)
- Reactive non-metals
- Reactivity decrease down the group
- Form ionic compounds with metals or covalent compounds with other non-metals
Reaction with Group 1 metals
Halogens react with Group 1 metals to form ionic halides. The halogen atoms gains one electron from the Group 1 elements to form a halide ion. The electrostatic force of attraction between the alkali metal ions and halides bonds the ions together.
Once the transfer is complete, the ions are pulled together by the mutual attraction of their opposite charges.
This would be the equation
Displacement reactions
The relative reactivity of the elements can also be seen by placing them in direct competition for an extra electron.
A chlorine nucleus has a stronger attraction for an electron than a bromine nucleus because of its smaller atomic radius and so takes the electron from the bromide ion. The chlorine has gained an electron and so forms the chloride ion. The bromide ion loses an electron to form bromine.
Reaction with Silver
The halogens form insoluble salts with silver. Adding a solution containing the halide to a solution containing silver ions produces a precipitate which is useful in identifying the halide
Silver Chloride = White
Silver Bromide = Cream
Silver Iodide = YellowSilver Nitrate = Colourless
3.3.2 Discuss the changes in nature, from ionic to covalent and from basic to acidic, of the oxides across period 3
The acid-base properties of the oxides are closely linked to their bonding. Metallic elements, which form ionic oxides, are basic; non-metal oxides, which are covalent are acidic. Aluminium oxide, which can be considered as an ionic oxide with some covalent character, shows amphoteric properties reacting with both acids and bases.
Na2O + HCl →2NaCl + H2O
MgO + H2SO4 → MgSO4 + H2O
Aluminium oxide is amphoteric (alkali and acid) so it reacts with both bases and acids
Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O
Al2O3 + 2NaOH + 3H2O → 2NaAl(OH)4
Phosphorous Oxide, Sulfur Oxide, Dichlorine Oxide reacts with water to form Acid Solutions
P4O10 + 6H2O → 4H3PO4
P4O6 +6H2O → 4 H3PO3
SO3 + H2O → H2SO4
SO2 + H2O → H2SO3
Cl2O7 + H2O → 2HClO4
Cl2O + H2O → 2HClO
Silcon dioxide doesn't react with water, but reacts with concentrated alkalis to form silicates
SiO2 + 2OH → SiO3 + H2O
Wednesday 27 November 2013
Topic 3.2: Physical properties
3.2.1 Define the terms first ionization energy and electronegativity
First Ionization Energy is the minimum energy required to remove an electron from a neutral gaseous atom in its ground state.
Key words:
Minimum Energy, Neutral Gaseous Atom and Ground State
Electro-negativity defined as the relative attraction that an atom has for the shared pair of electrons in a covalent
Key Words:
Relative Attraction, Shared Pair and Covalent
3.2.2 Describe and explain the trends in atomic radii, ionic radii, first ionization energies, electro-negativities and melting points for alkali metals (Li to Cs) and the halogens (F to I)
Atomic Radii
Since the outer electrons are difficult to locate, in practise, the atomic radius is measured as half the distance between two bonded atoms. For this reason noble gases are given no value as they do not bond with other atoms.
On descending a group, the atomic radius increase. This is because the outer electrons are getting further from the nucleus. This applies for both alkali metals and halogens.
It is defined by the distance between nucleus and outer most electrons of a positive metal cation or negative non-metal anion.
Ionization Energy
It is defined by the minimum energy required to remove an electron from a neutral gaseous atom in its ground state.
Electronegativity
Electronegativity is defined as the relative attraction that an atom has for the shared pair of electrons in a covalent bond.
Melting points
3.2.3 Describe and explain the trends in atomic radii, ionic radii, first ionization energies and electro-negativities for elements across period 3
Atomic radii
Across a period, the atomic radius decreases. This is because of the increase in effective nuclear charge and no increase shielding, therefore the electrons get pulled closer towards the nucleus
Ionic radii
Across a period, the metal cation and the non-metal anion get smaller. This is because of an increase in protons (Effective Nuclear Charge) whilst all the ions have the same electronic configuration
First Ionization Energy is the minimum energy required to remove an electron from a neutral gaseous atom in its ground state.
Key words:
Minimum Energy, Neutral Gaseous Atom and Ground State
Electro-negativity defined as the relative attraction that an atom has for the shared pair of electrons in a covalent
Key Words:
Relative Attraction, Shared Pair and Covalent
3.2.2 Describe and explain the trends in atomic radii, ionic radii, first ionization energies, electro-negativities and melting points for alkali metals (Li to Cs) and the halogens (F to I)
Atomic Radii
Since the outer electrons are difficult to locate, in practise, the atomic radius is measured as half the distance between two bonded atoms. For this reason noble gases are given no value as they do not bond with other atoms.
On descending a group, the atomic radius increase. This is because the outer electrons are getting further from the nucleus. This applies for both alkali metals and halogens.
Ionic Radii
It is defined by the distance between nucleus and outer most electrons of a positive metal cation or negative non-metal anion.
- Both metal cation and non-metal anion increase in size on descending the group
- Metal cations tend to be smaller than their atom as they have lost their outer most electrons
- Non-metal anions tend to be larger than their atom as they have gained electrons into their outer energy levels
Ionization Energy
It is defined by the minimum energy required to remove an electron from a neutral gaseous atom in its ground state.
- Ionization energy decreases on descending a group
- The outer most electrons are easier to remove as they are a greater distance from the nucleus shielding from the positive nucleus charge by the core electrons
Electronegativity
Electronegativity is defined as the relative attraction that an atom has for the shared pair of electrons in a covalent bond.
- On descending a group, electronegativity decreases. While the atomic radius and the nuclear charge increases, the level of shielding increases and the effective nuclear charge decrease.
- The most electronegative elements are in the top right of the periodic table. (F, O and N). Which are involved in hydrogen bonding.
Melting points
- The variation in melting points is related to the strength of the inter-molecular forces holding the atoms together
- Melting point of metals decreases on descending the group. As the atoms become larger, the metallic bonds holding them together gets weaker. The valence electrons are further from the nucleus.
- Melting point of non-metals increases on descending the group. This is due to the increase in the strength of the Van der Waal's forces between the molecules. The strength of these forces is determined by the total number of electrons in the atom.
- Across a period, the metals in groups 1, 2 and 3 display an increase in melting point. This is due to an increase in strength of the metallic bonds as the number of valence electrons increases
- The Group 4 element will have the highest melting point as it will be a covalently bonded macro-molecule. This contains only strong covalent bonds in the solid phase.
- The non-metal elements in Groups 5, 6 and 7 will have much lower melting point. Theses are also covalently bonded but are simple molecular structures with weaker Van der Waal's forces holding them together.
- The Group 0 noble gases will have the lowest melting point due to it being monoatomic. It will therefore have the weakest of Van der Waal's forces.
3.2.3 Describe and explain the trends in atomic radii, ionic radii, first ionization energies and electro-negativities for elements across period 3
Atomic radii
Across a period, the atomic radius decreases. This is because of the increase in effective nuclear charge and no increase shielding, therefore the electrons get pulled closer towards the nucleus
Ionic radii
Across a period, the metal cation and the non-metal anion get smaller. This is because of an increase in protons (Effective Nuclear Charge) whilst all the ions have the same electronic configuration
First Ionization Energies
Ionization energy generally increases across a period. This is due to the increase effective nuclear charge as the number of protons increases with no increase in shielding.
Electronegativity
Across a period, electronegativity increases. This is the result of an increased number of protons and thus an increased effective nuclear charge with no greater level of shielding
3.2.4 Compare the relative electro-negativity values of two or more elements based on their positions in the periodic table
We can tell whether a molecule has covalent or ionic bonds
Ionization energy generally increases across a period. This is due to the increase effective nuclear charge as the number of protons increases with no increase in shielding.
Electronegativity
Across a period, electronegativity increases. This is the result of an increased number of protons and thus an increased effective nuclear charge with no greater level of shielding
3.2.4 Compare the relative electro-negativity values of two or more elements based on their positions in the periodic table
By looking at the difference in electro-negativity values
We can tell whether a molecule has covalent or ionic bonds
Topic 3.1: The periodic table
3.1.1 Describe the arrangement of elements in the periodic table in order of increasing atomic number
The elements in a periodic table is placed in order of increasing atomic number (Z), which we know is a fundamental property of the element - number of protons in the nucleus of its atom. To read the periodic table, simply start from the top left and read across. You will find that the atomic number increases.
3.1.2 Distinguish between the terms group and period
Group is the columns or the vertical elements. Period is the row or the horizontal elements.
3.1.3 Apply the relationship between the electron arrangement of elements and their position in the periodic table up to Z=20.
Groups state how many valence electrons there are
Periods state how many electron shells there are.
Valence electron is an electron in the outer shell of an atom
Example,
Carbon has an electron arrangement of 2, 6
It is in the 2nd period and 6th group, hence its position on the periodic table
Calcium has an electron arrangement of 2, 8, 8, 2
It is in the 4th period and 2nd group, hence its position as an alkali earth metal.
3.1.4 Apply the relationship between the number of electrons in the highest occupied energy level for an element and its position in the periodic table
An atom with one valence electron will be in group 1
2, 1 would be in group 1
An atom with two valence electron will be in group 2
2, 2 would be in group 2
...etc
An atom with a full outer shell will be in group 0
2, 0 would be in group 0
The elements in a periodic table is placed in order of increasing atomic number (Z), which we know is a fundamental property of the element - number of protons in the nucleus of its atom. To read the periodic table, simply start from the top left and read across. You will find that the atomic number increases.
3.1.2 Distinguish between the terms group and period
Group is the columns or the vertical elements. Period is the row or the horizontal elements.
3.1.3 Apply the relationship between the electron arrangement of elements and their position in the periodic table up to Z=20.
Groups state how many valence electrons there are
Periods state how many electron shells there are.
Valence electron is an electron in the outer shell of an atom
Example,
Carbon has an electron arrangement of 2, 6
It is in the 2nd period and 6th group, hence its position on the periodic table
Calcium has an electron arrangement of 2, 8, 8, 2
It is in the 4th period and 2nd group, hence its position as an alkali earth metal.
3.1.4 Apply the relationship between the number of electrons in the highest occupied energy level for an element and its position in the periodic table
An atom with one valence electron will be in group 1
2, 1 would be in group 1
An atom with two valence electron will be in group 2
2, 2 would be in group 2
...etc
An atom with a full outer shell will be in group 0
2, 0 would be in group 0
Topic 3: Periodicity
Topic 3 of the IB HL Chemistry syllabus is the Periodicity. IBO recommends to spend 6 hours on this topic.
This topic has 3 sub-chapters: "The periodic table", "Physical properties" and "Chemical Properties". Each are separated with numerical values in order of mentioned.
These are advanced basic syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.
This topic has 3 sub-chapters: "The periodic table", "Physical properties" and "Chemical Properties". Each are separated with numerical values in order of mentioned.
These are advanced basic syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.
Sunday 17 November 2013
Topic 14.3: Delocalization of electrons
14.3.1 Describe the delocalization of π electrons and explain how this can account for structures of some species.
Instead of being confined to one location, electrons show a tendency to be shared between more than one bonding position, and are said to be delocalized. Free from the constriants of a single bonding position, delocalized electron spread themselves out.
Delocalization is a characteristic of electrons in pi bonds when there is more than one possible position for a double bond within a molecule. For example, if we consider the structure of the nitrate ion.
It can inter-change from any of the different Lewis structures.This suggests that the pi electrons have delocalized and spread themselves equally between all three possible bonding positions. This is the Lewis structure normally drawn.
Resonance is introduced because it can't be represented in a single diagram. The actual structure of the species is a composite or average of the number of Lewis structures that can be drawn, each known as a resonance structure.
Other resonance structures include carbonate, ozone and methanoate.
Benzene is a particular interesting case as all the carbons have made sp^2 bonds.
Because the p orbitals are so close together, they form a huge cloud instead.
This allows the electron to move freely around the benzene hexagon. This is easily portrayed in a Lewis Structure.
Delocalized electrons gives special properties to the structures in which they are found.
Intermediate bond lengths and strengths
By spreading the electrons between more than one bonding position, delocalization causes the affected bonds to be equal each other in length and strength, with values in between those of single and double bonds. The concept of bond order is sometimes used to describe the electron density within the bond. It is calculated by how many bonds divided by number of bonding positions. The higher the bond order, the greater the electron density
Greater Stability
Species with delocalized electrons are more stable than related species with all electrons localized in bonds. This is because delocalization spreads electrons as far apart as possible and so minimizes the repulsion between them. They require extra energy called resonance energy to disrupt the delocalized pi electron cloud.
Electrical conductivity in metal and graphite
Structures that have delocalized electrons spread through the entire structure (graphite). This enables them to move in response to a potential difference applied. In other words, conduct electricity.
Instead of being confined to one location, electrons show a tendency to be shared between more than one bonding position, and are said to be delocalized. Free from the constriants of a single bonding position, delocalized electron spread themselves out.
Delocalization is a characteristic of electrons in pi bonds when there is more than one possible position for a double bond within a molecule. For example, if we consider the structure of the nitrate ion.
It can inter-change from any of the different Lewis structures.This suggests that the pi electrons have delocalized and spread themselves equally between all three possible bonding positions. This is the Lewis structure normally drawn.
Resonance is introduced because it can't be represented in a single diagram. The actual structure of the species is a composite or average of the number of Lewis structures that can be drawn, each known as a resonance structure.
Other resonance structures include carbonate, ozone and methanoate.
Benzene is a particular interesting case as all the carbons have made sp^2 bonds.
Because the p orbitals are so close together, they form a huge cloud instead.
This allows the electron to move freely around the benzene hexagon. This is easily portrayed in a Lewis Structure.
Delocalized electrons gives special properties to the structures in which they are found.
Intermediate bond lengths and strengths
By spreading the electrons between more than one bonding position, delocalization causes the affected bonds to be equal each other in length and strength, with values in between those of single and double bonds. The concept of bond order is sometimes used to describe the electron density within the bond. It is calculated by how many bonds divided by number of bonding positions. The higher the bond order, the greater the electron density
Greater Stability
Species with delocalized electrons are more stable than related species with all electrons localized in bonds. This is because delocalization spreads electrons as far apart as possible and so minimizes the repulsion between them. They require extra energy called resonance energy to disrupt the delocalized pi electron cloud.
Electrical conductivity in metal and graphite
Structures that have delocalized electrons spread through the entire structure (graphite). This enables them to move in response to a potential difference applied. In other words, conduct electricity.
Topic 14.2: Hybridization
14.2.1 Describe σ and π bonds
When two atomic orbitals overlaps along the bond axis - an imaginary line between the two nuclei - the bond is described as a sigma bond. Denoted as σ. This type of bond forms by the overlaps of s orbitals, p orbitals and hybrid orbitals.
14.2.2 Explain hybridization in terms of the mixing of atomic orbitals to form new orbitals for bonding.
Carbon forms four covalent bonds. Yet if we consider the electron configuration in the carbon atom, we would not predict this as it has only two singly occupied orbitals available for bonding.
Because carbon can have 4 covalent bonds, this indicates that this lowest energy or ground state electron configuration changes during bonding. The electron in 2s is promoted to 2p by the process of excitation. The atoms now have 4 singly occupied orbitals available for bonding.
14.2.3 Identify and explain the relationships between Lewis structures, molecular shapes and types of hybridization (sp, sp^2 and sp^3)
Though this process can explain the four covalent bonds, the three p orbitals should have a slightly higher energy than the s orbital. So if there were used in bonding, we would expect unequal bonds. However methane has four identical bonds, so we can safely assume that the orbitals have been made equal during the bonding process. They form new hybrid atomic orbitals which are the same as each other, but different from the original orbitals.This mixing of orbitals is called hybridization.
sp^3
When carbon forms four single bonds, it under goes sp^3 hybridization, producing 4 equal orbitals. Note that it should have a tetrahedral arrangement.
sp^2
When carbon forms a double bond, it undergoes sp^2 hybridization, producing three equal orbtials. This typically forms, double bonds as the remaining p orbital becomes part of the pi bond. Note that it should have a planar triangular arrangement.
sp
When carbon forms a triple bond, it under goes sp hybridization, producing two equal orbitals. Note that it should have a linear arrangement.
These coalesce into a cylinder of negative charge around the atom, making the molecule susceptible to attack by electrophilic reagents.
Non-bonding pairs can also take part in hybridization. The non-bonding pair in ammonia, NH3, resides in a sp^3 orbital.
Note: Look at the number of charge centres, then the shape of the molecule. Then you would be able to find the type of hybridization.
When two atomic orbitals overlaps along the bond axis - an imaginary line between the two nuclei - the bond is described as a sigma bond. Denoted as σ. This type of bond forms by the overlaps of s orbitals, p orbitals and hybrid orbitals.
When two p orbitals overlaps sideways, the electron density of the molecular orbital is concentrated in two regions, above and below the plane of the bond angle. This type of bond only forms by the overlap of p orbitals alongside the formation of a sigma bond. In other words, pi bonds only form within double bond or triple bond.
14.2.2 Explain hybridization in terms of the mixing of atomic orbitals to form new orbitals for bonding.
Carbon forms four covalent bonds. Yet if we consider the electron configuration in the carbon atom, we would not predict this as it has only two singly occupied orbitals available for bonding.
Because carbon can have 4 covalent bonds, this indicates that this lowest energy or ground state electron configuration changes during bonding. The electron in 2s is promoted to 2p by the process of excitation. The atoms now have 4 singly occupied orbitals available for bonding.
14.2.3 Identify and explain the relationships between Lewis structures, molecular shapes and types of hybridization (sp, sp^2 and sp^3)
Though this process can explain the four covalent bonds, the three p orbitals should have a slightly higher energy than the s orbital. So if there were used in bonding, we would expect unequal bonds. However methane has four identical bonds, so we can safely assume that the orbitals have been made equal during the bonding process. They form new hybrid atomic orbitals which are the same as each other, but different from the original orbitals.This mixing of orbitals is called hybridization.
sp^3
When carbon forms four single bonds, it under goes sp^3 hybridization, producing 4 equal orbitals. Note that it should have a tetrahedral arrangement.
sp^2
When carbon forms a double bond, it undergoes sp^2 hybridization, producing three equal orbtials. This typically forms, double bonds as the remaining p orbital becomes part of the pi bond. Note that it should have a planar triangular arrangement.
sp
When carbon forms a triple bond, it under goes sp hybridization, producing two equal orbitals. Note that it should have a linear arrangement.
These coalesce into a cylinder of negative charge around the atom, making the molecule susceptible to attack by electrophilic reagents.
Non-bonding pairs can also take part in hybridization. The non-bonding pair in ammonia, NH3, resides in a sp^3 orbital.
Note: Look at the number of charge centres, then the shape of the molecule. Then you would be able to find the type of hybridization.
Topic 14.1: Shapes of molecules and ions
14.1.1 Predict the shape and bond angles for species with five and six negative charge centres using the VSEPR theory.
Expanded Octet is when compounds have more than eight electrons around the central atom is possible because the d orbitals avaliable in the valence shell of these atoms have energy values relatively close to those of the p orbitals. So promtions of electrons from 3p to 3d orbitals will allow additional electron pairs to form. This is how elements such as phosphorus and sulfur expand their octets forming species with five or six charge centers.
This table shows all the shapes you need to know. Remember the names for all of the different shapes.
A is central atom
X are other atoms
E are lone pairs
Five negative charge centres.
Triangular bipyramidal
Consists of 90 degrees, 120 degrees and 180 degree bonds.
Seesaw
With one pair of lone pair, it reduces the degree is reduced from 120 to 117 between the angles
T-shaped
With two pairs of lone pair, the bond angles are 90 degrees and 180 degrees.
Linear
It is just 180 degrees
Square Pyramidal
All degrees are <90 degrees or <180 degrees.
Square Planar
All angles are 90 degrees or 180 degrees
Summary
2 charge centers are linear
3 charge centers are planar triangular
With lone pair V-shaped
4 charge centers are tetrahedral
With lone pair Pyramidal
With 2 lone pair V-shaped
5 charge centers are triangular bipyramidal
With lone pair See-saw
With 2 lone pair T-shaped
With 3 lone pairs Linear
6 charge centers are octahedral
With lone pair Square pyramidal
With 2 lone pair Square planar
Expanded Octet is when compounds have more than eight electrons around the central atom is possible because the d orbitals avaliable in the valence shell of these atoms have energy values relatively close to those of the p orbitals. So promtions of electrons from 3p to 3d orbitals will allow additional electron pairs to form. This is how elements such as phosphorus and sulfur expand their octets forming species with five or six charge centers.
This table shows all the shapes you need to know. Remember the names for all of the different shapes.
A is central atom
X are other atoms
E are lone pairs
Five negative charge centres.
Triangular bipyramidal
Consists of 90 degrees, 120 degrees and 180 degree bonds.
Seesaw
With one pair of lone pair, it reduces the degree is reduced from 120 to 117 between the angles
T-shaped
With two pairs of lone pair, the bond angles are 90 degrees and 180 degrees.
Linear
It is just 180 degrees
Six negative charged centres
Octahedral
All 90 degrees or 180 degrees bond.
Square Pyramidal
All degrees are <90 degrees or <180 degrees.
Square Planar
All angles are 90 degrees or 180 degrees
Summary
2 charge centers are linear
3 charge centers are planar triangular
With lone pair V-shaped
4 charge centers are tetrahedral
With lone pair Pyramidal
With 2 lone pair V-shaped
5 charge centers are triangular bipyramidal
With lone pair See-saw
With 2 lone pair T-shaped
With 3 lone pairs Linear
6 charge centers are octahedral
With lone pair Square pyramidal
With 2 lone pair Square planar
Topic 14: Bonding
Topic 14 of the IB HL Chemistry syllabus is the Bonding. IBO recommends to spend 5 hours on this topic.
This topic has 3 sub-chapters: "Shapes of molecules and ions", "Hybridization" and "Delocalization of electrons". Each are separated with numerical values in order of mentioned.
These are advanced HL syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.
This topic has 3 sub-chapters: "Shapes of molecules and ions", "Hybridization" and "Delocalization of electrons". Each are separated with numerical values in order of mentioned.
These are advanced HL syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.
Topic 4.5: Physical Properties
4.5.1 Compare and explain the properties of substances resulting from different types of bonding.
Melting and boiling points
Melting and boiling points increases with increasing molecular size and the extent of polarity within the bonds of molecules.
Solubility
Consider an ionic compound being placed in water. At the contact surface, partial charges in the water molecules are attracted to ions of opposite charge in the lattice which may cause them to dislodge from their position. Ions separated from the lattice, which may cause them to dislodge from their positions. This is called hydrated.
It works the same for non-polar substances as well. Except it is covalent bonds which are dissolved instead of ionic bonds.
Anything that doesn't dissolve in water can dissolve in non-polar solvents, like lipids or hexane.
Electrical conductivity
Ability to conduct electricity depends on whether it contains electrons that are able to move. Ionic compounds are not able to conduct electricity in the solid state as the ions are firmly held within the lattice and can't move. Ionic compounds in aqueous or liquid state will be able to conduct electricity as there are free electrons.
This is a simple test to test whether it is a covalent or ionic compound. If the light doesn't light up, its a covalent bond.
Melting and boiling points
Melting and boiling points increases with increasing molecular size and the extent of polarity within the bonds of molecules.
Solubility
Consider an ionic compound being placed in water. At the contact surface, partial charges in the water molecules are attracted to ions of opposite charge in the lattice which may cause them to dislodge from their position. Ions separated from the lattice, which may cause them to dislodge from their positions. This is called hydrated.
It works the same for non-polar substances as well. Except it is covalent bonds which are dissolved instead of ionic bonds.
Anything that doesn't dissolve in water can dissolve in non-polar solvents, like lipids or hexane.
Electrical conductivity
Ability to conduct electricity depends on whether it contains electrons that are able to move. Ionic compounds are not able to conduct electricity in the solid state as the ions are firmly held within the lattice and can't move. Ionic compounds in aqueous or liquid state will be able to conduct electricity as there are free electrons.
This is a simple test to test whether it is a covalent or ionic compound. If the light doesn't light up, its a covalent bond.
Topic 4.4: Metallic Bonding
4.4.1 Describe the metallic bond as the electrostatic attraction between a lattice of positive ions and delocalized electrons.
In the elemental state, when there is no other element present to accept the electron and form an ionic compound, the outer electrons held only loosely by atom's nucleus tend to 'wander off' or, more correctly, become delocalized. This means that in metals they will no longer be associated closely with any one atomic nucleus but instead can spread themselves through the metals structure. The metal atoms without these electrons become positively charged ions and form a regular lattice structure which electrons can move freely.
4.4.2 Explain the electrical conductivity and malleability of metals
Metals are good conductors of electricity because the delocalized electrons are highly mobile and can move through the metal structure in response to an applied voltage. This mobility of electrons is responsible for the fact that they are also very good conductors. The unduly change in the confirmation of the metal through applied pressure. This property means that metals can be shaped under pressure so they are said to be malleable. A related property is that metals are ductile, meaning that they can be drawn out into threads.
In the elemental state, when there is no other element present to accept the electron and form an ionic compound, the outer electrons held only loosely by atom's nucleus tend to 'wander off' or, more correctly, become delocalized. This means that in metals they will no longer be associated closely with any one atomic nucleus but instead can spread themselves through the metals structure. The metal atoms without these electrons become positively charged ions and form a regular lattice structure which electrons can move freely.
4.4.2 Explain the electrical conductivity and malleability of metals
Metals are good conductors of electricity because the delocalized electrons are highly mobile and can move through the metal structure in response to an applied voltage. This mobility of electrons is responsible for the fact that they are also very good conductors. The unduly change in the confirmation of the metal through applied pressure. This property means that metals can be shaped under pressure so they are said to be malleable. A related property is that metals are ductile, meaning that they can be drawn out into threads.
Topic 4.3: Intermolecular Forces
4.3.1 Describe the types of intermolecular forces (attractions between molecules that have temporary dipoles, permanent dipoles or hydrogen bonding) and explain how they arise from the structural features of the molecule
Van der Waal's forces
Electron behave somewhat like mobile clouds of negative charge, the density of this cloud may at any one moment be greater over one atom that the other. This is known as a temporary dipole or instantaneous dipole. The strength increases as number of electron increases.
Dipole-Dipole
Molecules which have a polarity has a permanent dipole. It results in opposite charges on neighboring molecules attracting each other, generating a force known as the dipole-dipole.
Hydrogen Bond
When a molecule contains hydrogen covalently bonded to a very electronegative atom (F, N, O), these molecules are attracted to each other by a particularly strong type of intermolecular force called a hydrogen bond. The hydrogen bond is in essence a particular case of dipole-dipole. Given its small size, hydrogen can only exert a strong attractive force on a lone pair in the electronegative atom of a neighboring molecule.
4.3.2 Describe and explain how intermolecular forces affect the boiling points of substances.
The strength of intermolecular forces will play a particular important role indetermining the volatility of a substance. Change state from solid to liquid and from liquid to gas both involve seperating particles by overcoming the forces between them. It follows that the stronger the intermolecular forces, the more energy will be required to do this and so the higher will be the substance's melting and boiling points.
Van der Waal's forces
Electron behave somewhat like mobile clouds of negative charge, the density of this cloud may at any one moment be greater over one atom that the other. This is known as a temporary dipole or instantaneous dipole. The strength increases as number of electron increases.
Dipole-Dipole
Molecules which have a polarity has a permanent dipole. It results in opposite charges on neighboring molecules attracting each other, generating a force known as the dipole-dipole.
Hydrogen Bond
When a molecule contains hydrogen covalently bonded to a very electronegative atom (F, N, O), these molecules are attracted to each other by a particularly strong type of intermolecular force called a hydrogen bond. The hydrogen bond is in essence a particular case of dipole-dipole. Given its small size, hydrogen can only exert a strong attractive force on a lone pair in the electronegative atom of a neighboring molecule.
4.3.2 Describe and explain how intermolecular forces affect the boiling points of substances.
The strength of intermolecular forces will play a particular important role indetermining the volatility of a substance. Change state from solid to liquid and from liquid to gas both involve seperating particles by overcoming the forces between them. It follows that the stronger the intermolecular forces, the more energy will be required to do this and so the higher will be the substance's melting and boiling points.
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