Trend of Atomic Radius
Defined as half the internuclear distance between 2 similar atoms bonded only by a single bond.
Atomic radius decreases across the period
• Proton no./nuclear charge increases. Hence, drawing electrons closer to nucleus.
• Negligible increase in shielding effect as electrons are added to the valence shell
(no. of inner electron shells remain constant)
Trend of Ionic Radius
Defined as the radius of a spherical ion in an ionic compound.
Na, Mg, Al and Si
• Positive ions decreases from Na+, Mg2+, Al3+ to Si4+
• Nuclear charge increases as no. of proton increases
- Equal no. of electrons
- Negligible increase in screening effect as equal number of electron shells
- Hence, attraction force between the nucleus and the electrons in Si4+ ion is stronger than
that in Na+
P, S and Cl
Negative ions radius decreases from P, S to Cl
• Nuclear charge increases as no. of proton increases
- Equal no. of electrons
- Negligible increase in screening effect as equal no. of electron shells
- Therefore, the attraction force between the nucleus and the electrons in CL ion is
stronger than that in P ion.
Comparison between ionic radius and atomic radius
Size of positive ions : size of parent atoms
• Positive ions are smaller than parent atoms because:
- Cations are formed when a neutral atom loses electon/s. Thus cations have lesser
electrons than parent atoms.
- Cations and parent atoms have equal number of protons.
- nuclear charge is therefore shared among less no of electrons in cations.
- Hence the attraction force between the nucleus and the remaining electrons in cations
is stronger than that of parent atoms.
Size of negative ions : size of parent atoms
• Negative ions are larger than parent atoms
- Anions are formed when a neutral atom gains electron/s. Thus anions have more
electrons than parent atom.
- Anions and parent atoms have equal number of protons.
- Nuclear charge is therefore shared among more no of electrons in cations.
- Hence the attraction force between the nucleus and the remaining electrons in
cations is weaker than that of parent atoms.
Chemical formula of how these elements exist:
Na
Mg
Al
Si
P4
S8
Cl2
Thursday, April 16, 2009
Chemical periodicity part 3
Learning Outcome : Interpret the variation in melting point and electrical conductivity in terms of presence of simple molecular ,giant molecular and metallic bonding in elements
Bonding and structure of elements
Ø Metallic Bonds
Ø Giant Metallic Structure
>Na, Mg, Al
Ø Covalent bonds
Ø Giant covalent structure
>Si
Ø Intramolecular : Covalent bonds
Ø Intermolecular: Van Der Waals forces
Ø Simple molecular structure
>P, S, Cl
Trend of Melting Point
· Na, Mg ,Al
- High Melting Point
Since it is a :
>Giant Metallic structure
» High energy is required to overcome the strong attractive forces between the metal cations and delocalized electrons
>Melting point: Na< Mg
-decrease in cation radius
- increase in no of delocalized electrons
»increase in charge/size ratio, increase in the metallic bond strength
· Si
-Has the highest melting point
>Giant molecular structure
»large amount of energy required to break many strong covalent bonds in the giant molecular structure
· P, S, Cl
-Low melting point
>exists as simple molecular structures
>Intermolecular forces : Van der Waals forces
>Low thermal energy is required to overcome the weak Van der Waals forces
»The melting point of these elements is dependant on the size the electron cloud
»The higher the molecular mass(Mr), the larger the electron cloud,the stronger the VDW forces. Hence,the higher the melting point.
Trend of Electrical Conductitvity
· Na, Mg, Al
-Good electrical conductors
»they are metals with delocalized electrons
-Degree of electrical conductivity :
Al >Mg >Na
»As more valence electrons are contributed to the ‘sea of electrons’,electrical conductivity increases
· Si
-Low electrical conductivity under ordinary conditions
»Conductivity increases with an increase in temperature
»At a higher temperature ,electrons are excited to higher energy levels. Silicon is a semi- conductor .
· P ,S, Cl
- Non-conductor of electricity because absence of delocalized electrons or mobile ions.
Bonding and structure of elements
Ø Metallic Bonds
Ø Giant Metallic Structure
>Na, Mg, Al
Ø Covalent bonds
Ø Giant covalent structure
>Si
Ø Intramolecular : Covalent bonds
Ø Intermolecular: Van Der Waals forces
Ø Simple molecular structure
>P, S, Cl
Trend of Melting Point
· Na, Mg ,Al
- High Melting Point
Since it is a :
>Giant Metallic structure
» High energy is required to overcome the strong attractive forces between the metal cations and delocalized electrons
>Melting point: Na< Mg
-decrease in cation radius
- increase in no of delocalized electrons
»increase in charge/size ratio, increase in the metallic bond strength
· Si
-Has the highest melting point
>Giant molecular structure
»large amount of energy required to break many strong covalent bonds in the giant molecular structure
· P, S, Cl
-Low melting point
>exists as simple molecular structures
>Intermolecular forces : Van der Waals forces
>Low thermal energy is required to overcome the weak Van der Waals forces
»The melting point of these elements is dependant on the size the electron cloud
»The higher the molecular mass(Mr), the larger the electron cloud,the stronger the VDW forces. Hence,the higher the melting point.
Trend of Electrical Conductitvity
· Na, Mg, Al
-Good electrical conductors
»they are metals with delocalized electrons
-Degree of electrical conductivity :
Al >Mg >Na
»As more valence electrons are contributed to the ‘sea of electrons’,electrical conductivity increases
· Si
-Low electrical conductivity under ordinary conditions
»Conductivity increases with an increase in temperature
»At a higher temperature ,electrons are excited to higher energy levels. Silicon is a semi- conductor .
· P ,S, Cl
- Non-conductor of electricity because absence of delocalized electrons or mobile ions.
Chemical Equilibria
a.) A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products. For a reaction involving two reactants and two products this can be expressed symbolically as
aA + bB cC + dD
A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is distinct from reversible process in thermodynamics.Dynamic Equilibrium refers to a reversible reaction in which the rates of forward and reverse reactions become equal and there is no change in the concentration of the reactants and products.b.) Le Chatelier's Principle can be used to predict the effect of a change in conditions on a chemical equilibrium.
c.) If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or total pressure, then the equilibrium shifts to partially counter-act the imposed change.
Concentration
Changing the concentration of an ingredient will shift the equilibrium to the side that would reduce that change in concentration. The chemical system will attempt to partially oppose the change affected to the original state of equilibrium. In turn, the rate of reaction, extent and yield of products will be altered corresponding to the impact on the system.
This can be illustrated by the equilibrium of carbon monoxide and hydrogen gas, reacting to form methanol.
CO + 2 H2 ⇌ CH3OH
Suppose we were to increase the concentration of CO in the system. Using Le Châtelier's principle we can predict that the amount of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to fill the “gap” and favor the side where the species was reduced. This observation is supported by the "collision theory". As the concentration of CO is increased, the frequency of collisions of that reactant would increase also, allowing for an increase in forward reaction, and generation of the product. Even if a desired product is not thermodynamically favored, the end product can be obtained if it is continuously removed from the solution.
Temperature
Let us take for example the reaction of nitrogen gas with hydrogen gas. This is a reversible reaction, in which the two gases react to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
This is an exothermic reaction when producing ammonia. If we were to lower the temperature, the equilibrium would shift in such a way as to produce heat. Since this reaction is exothermic to the right, it would favour the production of more ammonia. In practice, in the Haber process the temperature is instead increased to speed the reaction rate at the expense of producing less ammonia.
Changes in pressure owing to changes in volume
The equilibrium concentrations of the products and reactants do not directly depend on the pressure subjected to the system. However, a change in pressure due to a change in volume of the system will shift the equilibrium.
Once again, let us refer to the reaction of nitrogen gas with hydrogen gas to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
Note the number of moles of gas on the left hand side, and the number of moles of gas on the right hand side. When the volume of the system is changed, the partial pressures of the gases change. Because there are more moles of gas on the reactant side, this change is more significant in the denominator of the equilibrium constant expression, causing a shift in equilibrium.
Thus, an increase in pressure due to decreasing volume causes the reaction to shift to the side with the fewer moles of gas.[2] A decrease in pressure due to increasing volume causes the reaction to shift to the side with more moles of gas. There is no effect on a reaction where the number of moles of gas is the same on each side of the chemical system (or equation)
g.) The Haber synthesis was developed into an industrial process.Conditions:medium temperature (~500oC) very high pressure (~250 atmospheres, ~351kPa) a catalyst (a porous iron catalyst prepared by reducing magnetite, Fe3O4).Osmium is a much better catalyst for the reaction but is very expensive.
This process produces an ammonia, NH3(g), yield of approximately 10-20%.
The reaction between nitrogen gas and hydrogen gas to produce ammonia gas is exothermic, releasing 92.4kJ/mol of energy at 298K (25oC).
N2(g)nitrogen
+
3H2(g)hydrogen
heat, pressure, catalyst
2NH3(g) ammonia
H = -92.4 kJ mol-1
By Le Chetalier's Principle:
Increasing the pressure causes the equilibrium position to move to the right resulting in a higher yeild of ammonia since there are more gas molecules on the left hand side of the equation (4 in total) than there are on the right hand side of the equation. Increasing the pressure means the system adjusts to reduce the effect of the change, that is, to reduce the pressure by having fewer gas molecules.
Decreasing the temperature causes the equilibrium position to move to the right resulting in a higher yield of ammonia since the reaction is exothermic (releases heat). Reducing the temperature means the system will adjust to minimise the effect of the change, that is, it will produce more heat since energy is a product of the reaction, and will therefore produce more ammonia gas as wellHowever, the rate of the reaction at lower temperatures is extremely slow, so a higher temperature must be used to speed up the reaction which results in a lower yield of ammonia.
The equilibrium expression for this reaction is:
Keq =
[NH3]2
[N2][H2]3
h.) Chain Disproportionation occurs when two radicals meet, instead of coupling, they exchange a proton. That gives two terminated chains, one saturated and the other with a terminal double bond.
2R-(CH2CH2)n-CH2CH2• → R-(CH2CH2)n-CH=CH2 + R-(CH2CH2)n-CH2CH3
http://www.scribd.com/doc/6618373/InDepth-Overview-and-Comparison-of-2006-2007-AL-Chem-Syllabus
aA + bB cC + dD
A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is distinct from reversible process in thermodynamics.Dynamic Equilibrium refers to a reversible reaction in which the rates of forward and reverse reactions become equal and there is no change in the concentration of the reactants and products.b.) Le Chatelier's Principle can be used to predict the effect of a change in conditions on a chemical equilibrium.
c.) If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or total pressure, then the equilibrium shifts to partially counter-act the imposed change.
Concentration
Changing the concentration of an ingredient will shift the equilibrium to the side that would reduce that change in concentration. The chemical system will attempt to partially oppose the change affected to the original state of equilibrium. In turn, the rate of reaction, extent and yield of products will be altered corresponding to the impact on the system.
This can be illustrated by the equilibrium of carbon monoxide and hydrogen gas, reacting to form methanol.
CO + 2 H2 ⇌ CH3OH
Suppose we were to increase the concentration of CO in the system. Using Le Châtelier's principle we can predict that the amount of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to fill the “gap” and favor the side where the species was reduced. This observation is supported by the "collision theory". As the concentration of CO is increased, the frequency of collisions of that reactant would increase also, allowing for an increase in forward reaction, and generation of the product. Even if a desired product is not thermodynamically favored, the end product can be obtained if it is continuously removed from the solution.
Temperature
Let us take for example the reaction of nitrogen gas with hydrogen gas. This is a reversible reaction, in which the two gases react to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
This is an exothermic reaction when producing ammonia. If we were to lower the temperature, the equilibrium would shift in such a way as to produce heat. Since this reaction is exothermic to the right, it would favour the production of more ammonia. In practice, in the Haber process the temperature is instead increased to speed the reaction rate at the expense of producing less ammonia.
Changes in pressure owing to changes in volume
The equilibrium concentrations of the products and reactants do not directly depend on the pressure subjected to the system. However, a change in pressure due to a change in volume of the system will shift the equilibrium.
Once again, let us refer to the reaction of nitrogen gas with hydrogen gas to form ammonia:
N2 + 3 H2 ⇌ 2 NH3 ΔH = −92kJ mol-1
Note the number of moles of gas on the left hand side, and the number of moles of gas on the right hand side. When the volume of the system is changed, the partial pressures of the gases change. Because there are more moles of gas on the reactant side, this change is more significant in the denominator of the equilibrium constant expression, causing a shift in equilibrium.
Thus, an increase in pressure due to decreasing volume causes the reaction to shift to the side with the fewer moles of gas.[2] A decrease in pressure due to increasing volume causes the reaction to shift to the side with more moles of gas. There is no effect on a reaction where the number of moles of gas is the same on each side of the chemical system (or equation)
g.) The Haber synthesis was developed into an industrial process.Conditions:medium temperature (~500oC) very high pressure (~250 atmospheres, ~351kPa) a catalyst (a porous iron catalyst prepared by reducing magnetite, Fe3O4).Osmium is a much better catalyst for the reaction but is very expensive.
This process produces an ammonia, NH3(g), yield of approximately 10-20%.
The reaction between nitrogen gas and hydrogen gas to produce ammonia gas is exothermic, releasing 92.4kJ/mol of energy at 298K (25oC).
N2(g)nitrogen
+
3H2(g)hydrogen
heat, pressure, catalyst
2NH3(g) ammonia
H = -92.4 kJ mol-1
By Le Chetalier's Principle:
Increasing the pressure causes the equilibrium position to move to the right resulting in a higher yeild of ammonia since there are more gas molecules on the left hand side of the equation (4 in total) than there are on the right hand side of the equation. Increasing the pressure means the system adjusts to reduce the effect of the change, that is, to reduce the pressure by having fewer gas molecules.
Decreasing the temperature causes the equilibrium position to move to the right resulting in a higher yield of ammonia since the reaction is exothermic (releases heat). Reducing the temperature means the system will adjust to minimise the effect of the change, that is, it will produce more heat since energy is a product of the reaction, and will therefore produce more ammonia gas as wellHowever, the rate of the reaction at lower temperatures is extremely slow, so a higher temperature must be used to speed up the reaction which results in a lower yield of ammonia.
The equilibrium expression for this reaction is:
Keq =
[NH3]2
[N2][H2]3
h.) Chain Disproportionation occurs when two radicals meet, instead of coupling, they exchange a proton. That gives two terminated chains, one saturated and the other with a terminal double bond.
2R-(CH2CH2)n-CH2CH2• → R-(CH2CH2)n-CH=CH2 + R-(CH2CH2)n-CH2CH3
http://www.scribd.com/doc/6618373/InDepth-Overview-and-Comparison-of-2006-2007-AL-Chem-Syllabus
Chemical periodicity part 2
Periodic Table
1st ionization energy is the minimum energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous cations.
It is a endothermic process.
Across Periods:
There are more protons in each nucleus so the nuclear charge in each element increases.
Therefore the force of attraction between the nucleus and outer electron is increased.
There is a negligible increase in shielding because each successive electron enters the same energy level.
More energy is needed to remove the outer electron.
1st I.E increases
First ionization energy generally increases going across Period 3.
The first ionization energy drops between magnesium and aluminium before increasing again.
The first ionization energy drops between phosphorus and sulphur before increasing again.
Irregularities:
1st I.E of Magnesium is greater than Aluminium.
Magnesium: 1s2 2s2 2p6 3s2
Aluminium: 1s2 2s2 2p6 3s2 3p1
3p1 electron is at a higher energy level. Thus it is much easier to be removed than 3s2 electrons.
1st I.E of Phosphorus is greater than Sulphur.
Phosphorus: 1s2 2s2 2p6 3s2 3p3
Sulphur: 1s2 2s2 2p6 3s2 3p4
The 3p electrons in phosphorus are all unpaired. In sulphur, there is one group of paired electrons and 2 unpaired ones.
There is some repulsion between paired electrons (interelectronic repulsion) in the same orbital.
Less energy needed to remove one electron from 3p4 than 3p3.
Down the Group:
There will be an additional shell of electrons
There is an increase in screening/shielding effect which outweighs the increase in effective nuclear charge.
Valence electrons is further away from the nucleus.
Less energy needed to remove electrons
1st I.E decreases
1st ionization energy is the minimum energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous cations.
It is a endothermic process.
Across Periods:
There are more protons in each nucleus so the nuclear charge in each element increases.
Therefore the force of attraction between the nucleus and outer electron is increased.
There is a negligible increase in shielding because each successive electron enters the same energy level.
More energy is needed to remove the outer electron.
1st I.E increases
First ionization energy generally increases going across Period 3.
The first ionization energy drops between magnesium and aluminium before increasing again.
The first ionization energy drops between phosphorus and sulphur before increasing again.
Irregularities:
1st I.E of Magnesium is greater than Aluminium.
Magnesium: 1s2 2s2 2p6 3s2
Aluminium: 1s2 2s2 2p6 3s2 3p1
3p1 electron is at a higher energy level. Thus it is much easier to be removed than 3s2 electrons.
1st I.E of Phosphorus is greater than Sulphur.
Phosphorus: 1s2 2s2 2p6 3s2 3p3
Sulphur: 1s2 2s2 2p6 3s2 3p4
The 3p electrons in phosphorus are all unpaired. In sulphur, there is one group of paired electrons and 2 unpaired ones.
There is some repulsion between paired electrons (interelectronic repulsion) in the same orbital.
Less energy needed to remove one electron from 3p4 than 3p3.
Down the Group:
There will be an additional shell of electrons
There is an increase in screening/shielding effect which outweighs the increase in effective nuclear charge.
Valence electrons is further away from the nucleus.
Less energy needed to remove electrons
1st I.E decreases
Chemcial Bonding part 3
Chemical Bonding
Learning outcomes:
(a) Describe ionic bonding, as in sodium chloride and magnesium oxide, including the use of ‘dot and cross’ diagrams.
Answer: NaCl
x
x
xx
x
x
x
xxNa + Cl x [ Na ] + [ x Cl ]-
x
x x
Sodium atom has one valence electron, it loses one to become a positive 1+ charged ion to achieve the stable octet arrangement.
Chlorine atom has seven valence electrons, it gains one electron from sodium atom to form a negative -1 charged ion to achieve the stable octet arrangement.
Being oppositely charged, they attract each other and are bonded in a single ionic bond to form the lattice structure, NaCl.
Answer: MgO
xx
xx
xx
xx Mg + x O à [ Mg ] 2+ [ O ] 2-
Magnesium atom has two valence electrons, it loses two electrons to become a positive 2+ charged ion to achieve the stable octet arrangement.
Oxygen atom has six valence electrons, it gains two electrons from Magnesium atom to form a negative 2- charged ion to achieve the stable octet arrangement.
Being oppositely charged, they attracted each other by electrostatic force and are bonded in a double ionic bond.
(i) Describe metallic bonding in terms of a lattice of positive ions surrounded by mobile electrons.
Answer: The valence electrons from metals are far away from the nucleus hence they are not strongly attracted to the positive charged of the protons in the nucleus. These valence electrons can easily leave the atom and becomes free mobile delocalized electrons. Metals are structures held up by metallic bonding which is the attraction force between the positive cations and the “sea of negative delocalized electrons”. The more the delocalized electrons, the stronger the metallic bonding, the higher the melting point of the metal.
(m) Describe, in simple terms , the lattice structure of a crystalline solid which is:
(i) ionic, as in sodium chloride, magnesium oxide.
Answer: NaCl
The lattice structure of sodium chloride consists of large number of Na+ and Cl- ions arranged in an orderly manner. One sodium ion is bonded three dimensionally to six chloride ions, each by a single ionic bond. The ions are arranged in straight rows and forms a large structures with flat sides, a crystal structure.The ions of Na+ and Cl- will be repelled when a force is applied to the lattice, bringing the like charged ions together, hence NaCl is hard but brittle.
Answer: MgO
The lattice structure of magnesium oxide consists of large number of Mg2+ and O2- ions arranged in an orderly manner. One magnesium ion is bonded three dimensionally with six oxygen ions, each by a double ionic bond. The ions of Mg2+ and O2- will be repelled when a force is applied to the lattice, bringing the like charged ions together, hence MgO is hard but brittle. Magnesium oxide also has a higher melting point than sodium chloride because the electrostatic force of attraction between the oppositely charged ions in magnesium oxide is twice that of sodium chloride’s.
(ii) Simple molecular, as in iodine.
Answer: Iodine is made up of small discrete iodine molecules. Each molecule is made up of two iodine atoms covalently bonded to one another. These iodine molecules attracted each other by weak Van Der Waal’s intermolecular forces to form a lattice structure which has very low melting point and boiling point.
(iii) giant molecular, as in graphite; diamond.
Answer: Graphite
Graphite is a carbon allotrope in which large number of carbon atoms are arranged in flat, hexagonal parallel layers to form a giant covalent structure. The carbon atoms in each layer are arranged in rings of six atoms with each carbon atom covalently bonded strongly to three other atoms in its layers. Weak Van Der Waal’s forces are present between the layers. The layers of atoms can slide over each other easily making graphite soft and slippery.
Answer: Diamond
Diamond is a carbon allotrope in which each carbon atom forms four covalent bonds with four other carbon atoms in a three dimensional tetrahedral arrangement. The structure is a giant network of carbon atoms held together by covalent bonds. Each carbon atom in diamond is at the centre of a tetrahedron.
(iv) Hydrogen-bonded, as in ice.
Answer: Ice has an open regular structure, with the O atom in H2O molecule forms two hydrogen bonds with two water molecules. The two H atoms form two more hydrogen bonds with two other water molecules. Each H2O molecule is therefore hydrogen bonded to four other H2O molecules in a tetrahedral arrangement.
(v) Metallic, as in copper.
Answer: Metals have giant structures. In metal like copper , the atoms are packed together closely in regular three-dimensional patterns to form giant lattice structures. In copper, it consisted of positive Cu2+ ions bonded to a sea of delocalized electrons.
(n) Outline the importance of hydrogen bonding to the physical properties of substances, including ice and water.
Answer: Each H2O molecule is hydrogen –bonded to four other H2O molecules in a tetrahedral arrangement. The open structure explains the fact that ice is less dense than liquid water at 0 oC . Hence, during winter, only the top layer of water is frozen. The bottom part of water is relatively warmer and keeps the aquatic life alive.
Learning outcomes:
(a) Describe ionic bonding, as in sodium chloride and magnesium oxide, including the use of ‘dot and cross’ diagrams.
Answer: NaCl
x
x
xx
x
x
x
xxNa + Cl x [ Na ] + [ x Cl ]-
x
x x
Sodium atom has one valence electron, it loses one to become a positive 1+ charged ion to achieve the stable octet arrangement.
Chlorine atom has seven valence electrons, it gains one electron from sodium atom to form a negative -1 charged ion to achieve the stable octet arrangement.
Being oppositely charged, they attract each other and are bonded in a single ionic bond to form the lattice structure, NaCl.
Answer: MgO
xx
xx
xx
xx Mg + x O à [ Mg ] 2+ [ O ] 2-
Magnesium atom has two valence electrons, it loses two electrons to become a positive 2+ charged ion to achieve the stable octet arrangement.
Oxygen atom has six valence electrons, it gains two electrons from Magnesium atom to form a negative 2- charged ion to achieve the stable octet arrangement.
Being oppositely charged, they attracted each other by electrostatic force and are bonded in a double ionic bond.
(i) Describe metallic bonding in terms of a lattice of positive ions surrounded by mobile electrons.
Answer: The valence electrons from metals are far away from the nucleus hence they are not strongly attracted to the positive charged of the protons in the nucleus. These valence electrons can easily leave the atom and becomes free mobile delocalized electrons. Metals are structures held up by metallic bonding which is the attraction force between the positive cations and the “sea of negative delocalized electrons”. The more the delocalized electrons, the stronger the metallic bonding, the higher the melting point of the metal.
(m) Describe, in simple terms , the lattice structure of a crystalline solid which is:
(i) ionic, as in sodium chloride, magnesium oxide.
Answer: NaCl
The lattice structure of sodium chloride consists of large number of Na+ and Cl- ions arranged in an orderly manner. One sodium ion is bonded three dimensionally to six chloride ions, each by a single ionic bond. The ions are arranged in straight rows and forms a large structures with flat sides, a crystal structure.The ions of Na+ and Cl- will be repelled when a force is applied to the lattice, bringing the like charged ions together, hence NaCl is hard but brittle.
Answer: MgO
The lattice structure of magnesium oxide consists of large number of Mg2+ and O2- ions arranged in an orderly manner. One magnesium ion is bonded three dimensionally with six oxygen ions, each by a double ionic bond. The ions of Mg2+ and O2- will be repelled when a force is applied to the lattice, bringing the like charged ions together, hence MgO is hard but brittle. Magnesium oxide also has a higher melting point than sodium chloride because the electrostatic force of attraction between the oppositely charged ions in magnesium oxide is twice that of sodium chloride’s.
(ii) Simple molecular, as in iodine.
Answer: Iodine is made up of small discrete iodine molecules. Each molecule is made up of two iodine atoms covalently bonded to one another. These iodine molecules attracted each other by weak Van Der Waal’s intermolecular forces to form a lattice structure which has very low melting point and boiling point.
(iii) giant molecular, as in graphite; diamond.
Answer: Graphite
Graphite is a carbon allotrope in which large number of carbon atoms are arranged in flat, hexagonal parallel layers to form a giant covalent structure. The carbon atoms in each layer are arranged in rings of six atoms with each carbon atom covalently bonded strongly to three other atoms in its layers. Weak Van Der Waal’s forces are present between the layers. The layers of atoms can slide over each other easily making graphite soft and slippery.
Answer: Diamond
Diamond is a carbon allotrope in which each carbon atom forms four covalent bonds with four other carbon atoms in a three dimensional tetrahedral arrangement. The structure is a giant network of carbon atoms held together by covalent bonds. Each carbon atom in diamond is at the centre of a tetrahedron.
(iv) Hydrogen-bonded, as in ice.
Answer: Ice has an open regular structure, with the O atom in H2O molecule forms two hydrogen bonds with two water molecules. The two H atoms form two more hydrogen bonds with two other water molecules. Each H2O molecule is therefore hydrogen bonded to four other H2O molecules in a tetrahedral arrangement.
(v) Metallic, as in copper.
Answer: Metals have giant structures. In metal like copper , the atoms are packed together closely in regular three-dimensional patterns to form giant lattice structures. In copper, it consisted of positive Cu2+ ions bonded to a sea of delocalized electrons.
(n) Outline the importance of hydrogen bonding to the physical properties of substances, including ice and water.
Answer: Each H2O molecule is hydrogen –bonded to four other H2O molecules in a tetrahedral arrangement. The open structure explains the fact that ice is less dense than liquid water at 0 oC . Hence, during winter, only the top layer of water is frozen. The bottom part of water is relatively warmer and keeps the aquatic life alive.
Chemical Bonding part 2
(f) Hydrogen bonding – a specialized form of permanent dipole –dipole interaction.
Hydrogen bonding is formed between the hydrogen atom of an ammonia or water molecule (with a small positive charge) in one molecule and the lone pair of electrons of the electronegative atom such as nitrogen, oxygen or fluorine in another neighbouring molecule.
(g) Bond length – the distance between the nuclei of the two atoms in the covalent bond or sum of covalent (atomic) radii.
The shorter the bond length, the stronger the covalent bond.
Bond energy – the amount of energy required to break one mole of covalent bond at gaseous state.
The greater the bond energy, the stronger the bond.
Bond polarity – the dipole-dipole intermolecular forces between the slightly positively-charged ends of one molecule to the negative end of another or the same molecule.
(h) Van der waals’ forces – the induced dipoles are due to the random movement of electrons around the atoms. At any point of time, the electron distribution may be slightly displaced towards one side of an atom or molecule, making that side slightly negative, and the opposite side slightly positive.
Permanent dipole-dipole attraction – it is the attraction between molecules that have permanent dipole moments (polar molecules) in liquid and solid states.
(j) Ionic bonding – they are formed when electrons are transferred from the valence shell of atoms of one element to other atoms.
1. high melting and boiling point
2. soluble in water, insoluble in organic solvents (exceptions)
3. poor electrical conductors in solid state but good conductors in molten and aqueous state.
4. hard and brittle
Covalent bonding – the sharing (pairing) of electrons between atoms.
Hydrogen bonding – a specialized form of permanent dipole –dipole interaction.
Metallic bonding – the force of attraction between positive metal ions and the negative delocalized electrons.
1. high degree of electrical conductivity
2. high degree of thermal conductivity
3. high melting and boiling point
4. shiny metal lustre
5. insoluble in water
6. malleable
Van der waals’ forces – the induced dipoles are due to the random movement of electrons around the atoms. At any point of time, the electron distribution may be slightly displaced towards one side of an atom or molecule, making that side slightly negative, and the opposite side slightly positive.
Hydrogen bonding is formed between the hydrogen atom of an ammonia or water molecule (with a small positive charge) in one molecule and the lone pair of electrons of the electronegative atom such as nitrogen, oxygen or fluorine in another neighbouring molecule.
(g) Bond length – the distance between the nuclei of the two atoms in the covalent bond or sum of covalent (atomic) radii.
The shorter the bond length, the stronger the covalent bond.
Bond energy – the amount of energy required to break one mole of covalent bond at gaseous state.
The greater the bond energy, the stronger the bond.
Bond polarity – the dipole-dipole intermolecular forces between the slightly positively-charged ends of one molecule to the negative end of another or the same molecule.
(h) Van der waals’ forces – the induced dipoles are due to the random movement of electrons around the atoms. At any point of time, the electron distribution may be slightly displaced towards one side of an atom or molecule, making that side slightly negative, and the opposite side slightly positive.
Permanent dipole-dipole attraction – it is the attraction between molecules that have permanent dipole moments (polar molecules) in liquid and solid states.
(j) Ionic bonding – they are formed when electrons are transferred from the valence shell of atoms of one element to other atoms.
1. high melting and boiling point
2. soluble in water, insoluble in organic solvents (exceptions)
3. poor electrical conductors in solid state but good conductors in molten and aqueous state.
4. hard and brittle
Covalent bonding – the sharing (pairing) of electrons between atoms.
Hydrogen bonding – a specialized form of permanent dipole –dipole interaction.
Metallic bonding – the force of attraction between positive metal ions and the negative delocalized electrons.
1. high degree of electrical conductivity
2. high degree of thermal conductivity
3. high melting and boiling point
4. shiny metal lustre
5. insoluble in water
6. malleable
Van der waals’ forces – the induced dipoles are due to the random movement of electrons around the atoms. At any point of time, the electron distribution may be slightly displaced towards one side of an atom or molecule, making that side slightly negative, and the opposite side slightly positive.
Chemical Bonding
b) Describe, including the use of ‘dot-and-cross’ diagrams,
i) covalent bonding in H2, O2, Cl2, HCl, CO2, CH4, C2H6
Covalent bond is formed from sharing of electrons between atoms. A covalent bond is the electrostatic force of attraction between the nuclei of 2 atoms for the shared pair of electron.
Molecule
‘dot-and-cross’ diagram
Description(to achieve stability)
H2
2 hydrogen atoms shared 1 pair of e- in the valence shell
O2
2 oxygen atoms shared 2 pairs of e- in the valence shell
Cl2
2 chlorine atoms shared 1 pair of e- in the valence shell
HCl
1 hydrogen & 1 chlorine atoms shared 1 pair of e- in the valence shell
CO2
1 carbon & 2 oxygen atoms shared 2 pairs of e- in the valence shell
CH4
1 carbon & 4 hydrogen atoms shared 4 pairs of e- in the valence shell
C2H6
2 carbon & 6 hydrogen atoms shared 7 pair of e- in the valence shell
ii) dative bond, in BF3->NH3
Dative bond is a type of covalent bond where the shared paired of electrons comes from one donor only. The donor must possess 1 lone pair of electron in its valence shell. The acceptor must have an empty orbital in its outer quantum shell to accommodate the lone pair.
NH3 + BF3----à BF3->NH3
NH3’s 1 lone pair donates to BF3. Now both N and B have 8e- in valence shell.
c) Explain shapes and bond angles of molecules by, using qualitative model of electron pair repulsion, using as simple examples: BF3 (trigonal), CO2 (linear), CH4 (tetrahedral), NH3 (pyramidal), H2O(V-shape), SF6(octahedral), by using VSEPR Theory.
Valence Shell Electron Pair Repulsion Theory can be use to predict the shape f simple covalent molecules.
I. These b.p & l.p. electrons around central atom determine the shapes of the molecule.
II. The electrons repel each other as far as possible.
III. Degree of repulsion
l.p.-l.p > l.p-b.p. > b.p.-b.p.
105’ 107’ 109’
Molecule
Structural formula
Kind of repulsion
Bond angle
Shape
Description
BF3
l.p.-l.p.
360’/3=120’
Trigonal planar
(because it is a triangle in 2D.)
B is in Grp III F is in Grp VII. There are 3F atoms. The strength of repulsion is equal because all are l.p.
CO2
l.p.-l.p.
360’/2=180’
Linear (because it is in a straight line)
There is repulsion between 2 O atoms, so the repulsion is the largest.
CH4
b.p-b.p.
109’
Tetrahedral (because there are 4 faces)
There is repulsion between H atoms.
NH3
l.p.-b.p.
107’
Trigonal pyramidal (because it is a triangle in 3D.
There is repulsion between N and H atom. So the angle will be pushed smaller.
H2O
l.p.-l.p.
105’
V-shape
There is repulsion causing the H atoms to be pushed closer to each other.
SF6
b.p-b.p
90’
Octahedral (because it has two 4 sided pyramid)
This is a molecule with more than 8e-.
Such case can be seen in period 3 and 4. There is no l.p. repulsion, so there is equal bond angle.
Factors affecting Bond Size:
I. Total number of e- pairs.
The more the number of e- pairs the central atom has, the smaller the bond angle.
II. Number of lone pairs.
L.p. electrons cause more repulsion than the b.p. electrons. Bond pairs are pushed closer to one another.
III. Size of central atom.
As size of central atom increases, the l.p. region increases, resulting in greater repulsion between the l.p.-l.p. & b.p-l.p.. The b.p. are pushed closer to one another, bond angle decrease.
IV. Electronegativity of central atom
The greater the electronegativity of a central atom, the closer the b.p. are towards the central atom, increasing b.p.-b.p. repulsion, hence bond angle increases.
d) Describe covalent bonding in terms of orbital overlap, giving σ and π bonds
A sigma bond is formed from the head-on overlapping of orbitals.
A Pi bond is formed from the side-on overlapping of p orbitals.
e) Predict the shapes and bond angles of molecules analogous to those specified in c).
1) Identify the central atom.
2) Determine the no. of electrons surrounding the central atom.
· For molecules with no charge,
No. of e- surrounding central atom = No. of valence e- of central atom + No. of surrounding atoms
· For charged particles,
No. surrounding atom = No. of valence e- of central atom ± magnitude of charge + No. of surrounding atoms
3) Determine the no. of electron pairs.
· No. of electron pairs = ½ x No. of e- surrounding central atom
4) Determine no. of b.p. electron and l.p. electron.
· No. of b.p. = No. of atoms around the central atoms
· No. of l.p. = No. of electron pairs – No. of b.p.
5) Identify shape of the molecule using the table below.
No. of Electron pairs
No. of Bond pairs
No. of Lone pairs
Geometrical shape
Bond angle
2
2
0
Linear
180’
3
3
0
Trigonal planar
equal or near 120’
2
1
V-shape/bent
4
4
0
Tetrahedral
equal or near 108’
3
1
Trigonal pyramidal
2
2
V-shape/bent
5
5
0
Trigonal bipyramidal
90’, 120’
4
1
Distorted tetrahedron
-
3
2
T-shape
90’
2
3
Linear
180’
6
6
0
Octahedral
90’
5
1
Square pyramidal
4
2
Square planar
i) covalent bonding in H2, O2, Cl2, HCl, CO2, CH4, C2H6
Covalent bond is formed from sharing of electrons between atoms. A covalent bond is the electrostatic force of attraction between the nuclei of 2 atoms for the shared pair of electron.
Molecule
‘dot-and-cross’ diagram
Description(to achieve stability)
H2
2 hydrogen atoms shared 1 pair of e- in the valence shell
O2
2 oxygen atoms shared 2 pairs of e- in the valence shell
Cl2
2 chlorine atoms shared 1 pair of e- in the valence shell
HCl
1 hydrogen & 1 chlorine atoms shared 1 pair of e- in the valence shell
CO2
1 carbon & 2 oxygen atoms shared 2 pairs of e- in the valence shell
CH4
1 carbon & 4 hydrogen atoms shared 4 pairs of e- in the valence shell
C2H6
2 carbon & 6 hydrogen atoms shared 7 pair of e- in the valence shell
ii) dative bond, in BF3->NH3
Dative bond is a type of covalent bond where the shared paired of electrons comes from one donor only. The donor must possess 1 lone pair of electron in its valence shell. The acceptor must have an empty orbital in its outer quantum shell to accommodate the lone pair.
NH3 + BF3----à BF3->NH3
NH3’s 1 lone pair donates to BF3. Now both N and B have 8e- in valence shell.
c) Explain shapes and bond angles of molecules by, using qualitative model of electron pair repulsion, using as simple examples: BF3 (trigonal), CO2 (linear), CH4 (tetrahedral), NH3 (pyramidal), H2O(V-shape), SF6(octahedral), by using VSEPR Theory.
Valence Shell Electron Pair Repulsion Theory can be use to predict the shape f simple covalent molecules.
I. These b.p & l.p. electrons around central atom determine the shapes of the molecule.
II. The electrons repel each other as far as possible.
III. Degree of repulsion
l.p.-l.p > l.p-b.p. > b.p.-b.p.
105’ 107’ 109’
Molecule
Structural formula
Kind of repulsion
Bond angle
Shape
Description
BF3
l.p.-l.p.
360’/3=120’
Trigonal planar
(because it is a triangle in 2D.)
B is in Grp III F is in Grp VII. There are 3F atoms. The strength of repulsion is equal because all are l.p.
CO2
l.p.-l.p.
360’/2=180’
Linear (because it is in a straight line)
There is repulsion between 2 O atoms, so the repulsion is the largest.
CH4
b.p-b.p.
109’
Tetrahedral (because there are 4 faces)
There is repulsion between H atoms.
NH3
l.p.-b.p.
107’
Trigonal pyramidal (because it is a triangle in 3D.
There is repulsion between N and H atom. So the angle will be pushed smaller.
H2O
l.p.-l.p.
105’
V-shape
There is repulsion causing the H atoms to be pushed closer to each other.
SF6
b.p-b.p
90’
Octahedral (because it has two 4 sided pyramid)
This is a molecule with more than 8e-.
Such case can be seen in period 3 and 4. There is no l.p. repulsion, so there is equal bond angle.
Factors affecting Bond Size:
I. Total number of e- pairs.
The more the number of e- pairs the central atom has, the smaller the bond angle.
II. Number of lone pairs.
L.p. electrons cause more repulsion than the b.p. electrons. Bond pairs are pushed closer to one another.
III. Size of central atom.
As size of central atom increases, the l.p. region increases, resulting in greater repulsion between the l.p.-l.p. & b.p-l.p.. The b.p. are pushed closer to one another, bond angle decrease.
IV. Electronegativity of central atom
The greater the electronegativity of a central atom, the closer the b.p. are towards the central atom, increasing b.p.-b.p. repulsion, hence bond angle increases.
d) Describe covalent bonding in terms of orbital overlap, giving σ and π bonds
A sigma bond is formed from the head-on overlapping of orbitals.
A Pi bond is formed from the side-on overlapping of p orbitals.
e) Predict the shapes and bond angles of molecules analogous to those specified in c).
1) Identify the central atom.
2) Determine the no. of electrons surrounding the central atom.
· For molecules with no charge,
No. of e- surrounding central atom = No. of valence e- of central atom + No. of surrounding atoms
· For charged particles,
No. surrounding atom = No. of valence e- of central atom ± magnitude of charge + No. of surrounding atoms
3) Determine the no. of electron pairs.
· No. of electron pairs = ½ x No. of e- surrounding central atom
4) Determine no. of b.p. electron and l.p. electron.
· No. of b.p. = No. of atoms around the central atoms
· No. of l.p. = No. of electron pairs – No. of b.p.
5) Identify shape of the molecule using the table below.
No. of Electron pairs
No. of Bond pairs
No. of Lone pairs
Geometrical shape
Bond angle
2
2
0
Linear
180’
3
3
0
Trigonal planar
equal or near 120’
2
1
V-shape/bent
4
4
0
Tetrahedral
equal or near 108’
3
1
Trigonal pyramidal
2
2
V-shape/bent
5
5
0
Trigonal bipyramidal
90’, 120’
4
1
Distorted tetrahedron
-
3
2
T-shape
90’
2
3
Linear
180’
6
6
0
Octahedral
90’
5
1
Square pyramidal
4
2
Square planar
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