Dear friends,
I have been quite busy these few days. Sorry for not updating this blog. Thanks to all who have signed up for the email newsletter. I will try to get the first issue out by next week. thank you all for your support. In the meantime, please read through the notes that I have posted up. Remember repetition is the mother of all success.
Happy studying
Zhanrui
Tuesday, July 1, 2008
Sunday, June 29, 2008
E mail newsletter
Dear readers,
I will be creating weekly newsletters which contain information about how to prepare for the A Level chemistry exam including solutions to past year question. If u would like to subscribe to this newsletter, please fill up your email address in the top right hand corner of the blog.
I will be creating weekly newsletters which contain information about how to prepare for the A Level chemistry exam including solutions to past year question. If u would like to subscribe to this newsletter, please fill up your email address in the top right hand corner of the blog.
Saturday, June 28, 2008
Relation of reaction mechanism to reaction kinetics
In a nutshell, the rate equation is made up of reactants which are involved in the rate liming step or the slow step.
When chemists study reaction mechanisms, they are interested to know how the reaction actually occurs, for example which bonds are boken first, which bonds are formed first etc. We cannot determine the reaction mechanism from the balanced chemical reaction.
Let us consider the reaction of propanonne with iodine. From the balanced chemical equation, we can see that there are 2 reactants propanone and iodine. In order to determine the order of reaction w.r.t iodine, we use an excess of propanone and vary the concentration iodine, i.e. iodine will be the limiting reactant. Using the initial rates method, we can determine the order of reaction w.r.t iodine. It was determined that the reaction is zero order w.r.t iodine which means that the rate of reaction is independent of the concentration of iodine. Iodine will not appear in the rate equation
Next we determine the order of reaction w.r.t propanone and it was determined that the order of reaction w.r.t propanone is first order . Hence propanone will appear in the rate equation . Similarly we determine the order of reaction with respect to hydrogen ion concentration to be first order. Hence from the above data, we can write the following rate equation:
The rate equation also indicates that H+ and propanone are reactants in the rate determining step and a reaction mechanism can be proposed based on this data.
Note that the rate equation cannot be used to prove that a particular reaction mechanism is correct, it can only be used to prove that a particular reaction mechanism is wrong. If some one suggests that iodine is a reactant in the rate determining step, we can use the data from the kinetic analysis to say that he is wrong because the rate of reaction is independent of the concentration of iodine and hence iodine cannot be a reactant in the rate determining step.
When chemists study reaction mechanisms, they are interested to know how the reaction actually occurs, for example which bonds are boken first, which bonds are formed first etc. We cannot determine the reaction mechanism from the balanced chemical reaction.
Let us consider the reaction of propanonne with iodine. From the balanced chemical equation, we can see that there are 2 reactants propanone and iodine. In order to determine the order of reaction w.r.t iodine, we use an excess of propanone and vary the concentration iodine, i.e. iodine will be the limiting reactant. Using the initial rates method, we can determine the order of reaction w.r.t iodine. It was determined that the reaction is zero order w.r.t iodine which means that the rate of reaction is independent of the concentration of iodine. Iodine will not appear in the rate equation
Next we determine the order of reaction w.r.t propanone and it was determined that the order of reaction w.r.t propanone is first order . Hence propanone will appear in the rate equation . Similarly we determine the order of reaction with respect to hydrogen ion concentration to be first order. Hence from the above data, we can write the following rate equation:
The rate equation also indicates that H+ and propanone are reactants in the rate determining step and a reaction mechanism can be proposed based on this data.
Note that the rate equation cannot be used to prove that a particular reaction mechanism is correct, it can only be used to prove that a particular reaction mechanism is wrong. If some one suggests that iodine is a reactant in the rate determining step, we can use the data from the kinetic analysis to say that he is wrong because the rate of reaction is independent of the concentration of iodine and hence iodine cannot be a reactant in the rate determining step.
Concentration time graphs
In this section, we would look at how the concentration of reactants vary with time. Using a hypothetical A + B --> C reaction as an example, lets assume that the order of reaction with respect to A is zero order, i.e. the rate of reaction is independent of the [A]. The concentration time graph of A can be represented by a straight line. In other words, [A] decreases at a constant rate with time. This is because the rate of reaction is independent of the [A]. So although the concentration of A decreases with time, the rate of reaction remains the same. If the rate of reaction remains the same, it also means that reactant A is used up at a constant rate, hence the [A] decreases at a constant rate.
Now lets consider reactant B. Lets assume that the order of reaction w.r.t B is first order. Similarly the concentration of B will decrease with time. However the rate of reaction is dependent on the concentration of B, hence as the concentration of B decreases, the rate of reaction also decreases . As a result the concentration of B decreases at a decreasing rate. This can be illustrated by drawing tangents to the curve at different time points. You can see that the gradient of the tangent becomes gentler with time, indicating that the rate of decrease is decreasing.
Now lets consider reactant B. Lets assume that the order of reaction w.r.t B is first order. Similarly the concentration of B will decrease with time. However the rate of reaction is dependent on the concentration of B, hence as the concentration of B decreases, the rate of reaction also decreases . As a result the concentration of B decreases at a decreasing rate. This can be illustrated by drawing tangents to the curve at different time points. You can see that the gradient of the tangent becomes gentler with time, indicating that the rate of decrease is decreasing.
Thursday, June 26, 2008
Reaction kinetics I
Definition of some terms
Rate of reaction is defined as the rate of change of amount or concentration of a particular reactant or product
Rates of most reactions can be related to the concentrations of individual reactants by an equation of the form Rate = k[X]^n, where k is the rate constant, X is the reactant under consideration and n is the order of reaction with respect to X. This expression is known as a rate equation.
Students should note that the rate equation can only be determined experimentally, it is not related to the balanced equation.
The order of a reaction with respect to a given reactant is the power of that reactant's concentration of the experimentally determined rate equation.
The overall order of reaction is the sum of the powers of the concentration terms in the rate equation
The half life of a reaction is the time taken for the concentration of a reactant to fall to half its original value. Students should note that first order reactions have a constant half life. The decay of a radioactive isotope is usually a first order reaction.
Rate of reaction is defined as the rate of change of amount or concentration of a particular reactant or product
Rates of most reactions can be related to the concentrations of individual reactants by an equation of the form Rate = k[X]^n, where k is the rate constant, X is the reactant under consideration and n is the order of reaction with respect to X. This expression is known as a rate equation.
Students should note that the rate equation can only be determined experimentally, it is not related to the balanced equation.
The order of a reaction with respect to a given reactant is the power of that reactant's concentration of the experimentally determined rate equation.
The overall order of reaction is the sum of the powers of the concentration terms in the rate equation
The half life of a reaction is the time taken for the concentration of a reactant to fall to half its original value. Students should note that first order reactions have a constant half life. The decay of a radioactive isotope is usually a first order reaction.
Wednesday, June 25, 2008
Electrolysis in industrial process
Anodising of aluminum
Anodising is the process of increasing the thickness of aluminum oxide layer on the surface of aluminum in order to protect the metal underneath. The aluminum that is to be anodised is made the electrode during the electrolysis of sulphuric acid. Recall that oxygen is evolved at the anode during the electrolysis. the oxygen released combines with aluminum and thickens the oxide layer.
Electrolytic purification of copper
The impure copper rod is made the anode. At the anode, the copper ions is oxidized to Cu2+ ions. The Cu2+ ions is attracted to the cathode where is gains 2 electrons to form the copper metal. The electrolyte is copper sulphate solution. Effectively the copper is transferred from the anode to the cathode.
Anodising is the process of increasing the thickness of aluminum oxide layer on the surface of aluminum in order to protect the metal underneath. The aluminum that is to be anodised is made the electrode during the electrolysis of sulphuric acid. Recall that oxygen is evolved at the anode during the electrolysis. the oxygen released combines with aluminum and thickens the oxide layer.
Electrolytic purification of copper
The impure copper rod is made the anode. At the anode, the copper ions is oxidized to Cu2+ ions. The Cu2+ ions is attracted to the cathode where is gains 2 electrons to form the copper metal. The electrolyte is copper sulphate solution. Effectively the copper is transferred from the anode to the cathode.
Tuesday, June 24, 2008
Calculations related to involving electrolysis
Quantity of charge that passes through during electrolysis
The quantity of charge, Q that passes through during electrolysis in coloumbs is given by the product of current, I in amperes and the time, t in seconds.
Q = It
The syllabus also highlighted that students should know calculations involving the electrolysis of aqueous sulphuric acid and aqueous sodium sulphate. Realise that in both cases of electrolysis , it is actually the electrolysis of water , hence hydrogen and oxygen gas will be evolved. The half equations are presented below.
The effective reaction is the electrolysis of water. Note that the ratio of the volume of hydrogen gas to oxygen gas evolved is 2:1.
The quantity of charge, Q that passes through during electrolysis in coloumbs is given by the product of current, I in amperes and the time, t in seconds.
Q = It
The syllabus also highlighted that students should know calculations involving the electrolysis of aqueous sulphuric acid and aqueous sodium sulphate. Realise that in both cases of electrolysis , it is actually the electrolysis of water , hence hydrogen and oxygen gas will be evolved. The half equations are presented below.
The effective reaction is the electrolysis of water. Note that the ratio of the volume of hydrogen gas to oxygen gas evolved is 2:1.Monday, June 23, 2008
How to determine order of reaction?
How do we deduce the order of reaction?
The syllabus specify that students should know how to deduce the order of reaction with respect to a particular reactant using the initial rates method. Initial rate simply means the rate of reaction at the start of the reaction. For a typical chemical reaction, the initial rate of reaction is the fastest because the rate of reaction slows down as the reactants are being used up. Hence we would use the initial rate of reaction to represent the "true" rate of reaction.
Let's consider the reaction A + B --> C
The initial rates of reaction were determined and the results were presented in the table below
First we would determine the order of reaction with respect to reactant A. To do that we would look at Run(a) and Run(b) because the concentration of the other reactant B is the same. Only the concentration of A is changed. Hence whatever change in rate of reaction observed is due to A. Comparing Run(a) and run(b), we see that [A] is doubled however the reaction rate remains unchanged. Hence the order of reaction w.r.t A is zero order.
Secondly, we would dtermine the order of reaction w.r.t reactant B. Using the same reasoning we compare Run(b) and (c) where the [A] remain constant and [B] changes. We realise that the reaction rate doubles when the [B] doubles. Hence the order of reaction with respect to reactant B is first order.
The syllabus specify that students should know how to deduce the order of reaction with respect to a particular reactant using the initial rates method. Initial rate simply means the rate of reaction at the start of the reaction. For a typical chemical reaction, the initial rate of reaction is the fastest because the rate of reaction slows down as the reactants are being used up. Hence we would use the initial rate of reaction to represent the "true" rate of reaction.
Let's consider the reaction A + B --> C
The initial rates of reaction were determined and the results were presented in the table below
First we would determine the order of reaction with respect to reactant A. To do that we would look at Run(a) and Run(b) because the concentration of the other reactant B is the same. Only the concentration of A is changed. Hence whatever change in rate of reaction observed is due to A. Comparing Run(a) and run(b), we see that [A] is doubled however the reaction rate remains unchanged. Hence the order of reaction w.r.t A is zero order.
Secondly, we would dtermine the order of reaction w.r.t reactant B. Using the same reasoning we compare Run(b) and (c) where the [A] remain constant and [B] changes. We realise that the reaction rate doubles when the [B] doubles. Hence the order of reaction with respect to reactant B is first order.
Electrolysis II
Prediction of discharged substances during electrolysis
Redox series
When there is more than one anion or cation ion present in the electrolyte, how do we decide which ion will be discharged. The ion that is preferentially discharged depends on its position in the redox series. The redox series is presented below. Students should take note of the position of the hydrogen ion. In the presence of almost all other ions except copper and silver ions, hydrogen ions will be preferentially discharged.
There is a similar series for anions, albeit a shorter one. Similarly for hydroxide ions, unless the other ion present is bromide or iodide ion, hydroxide ions will be preferentially discharged.
Concentration
The concentration of ions present may also affect which ion will be discharged. Take for example if both lead ion and hydrogen ion are attracted to the cathode and lead ion is present in much higher concentration, lead ion will be preferentially discharged although it is higher in the redox series. In other words, high concentration can promote the discharge of an ion higher in the redox series.
Redox series
When there is more than one anion or cation ion present in the electrolyte, how do we decide which ion will be discharged. The ion that is preferentially discharged depends on its position in the redox series. The redox series is presented below. Students should take note of the position of the hydrogen ion. In the presence of almost all other ions except copper and silver ions, hydrogen ions will be preferentially discharged.
There is a similar series for anions, albeit a shorter one. Similarly for hydroxide ions, unless the other ion present is bromide or iodide ion, hydroxide ions will be preferentially discharged.
Concentration
The concentration of ions present may also affect which ion will be discharged. Take for example if both lead ion and hydrogen ion are attracted to the cathode and lead ion is present in much higher concentration, lead ion will be preferentially discharged although it is higher in the redox series. In other words, high concentration can promote the discharge of an ion higher in the redox series.
Electrolysis I
Relationship between Faraday's Constant and Avagadro's Constant
F: Faraday's Constant = 96500C
L: Avagadro's Constant = 6e23
e: charge of an electron = 1.6e-19
Basically, Faraday's Constant is a quantity that tells us how much charge does a mole of electrons possess. Recall that one mole is 6e23 which is also the Avagadro's number. Thus the relation F=Le
Prediction of the substance liberated during electrolysis
The ion that is discharged at the electrodes during electrolysis is affected by 3 main factors. They are the state of electrolyte (molten or aqueous), position in the redox series and concentration of the ion in the electrolyte.
State of electrolyte
If the electrolyte is aqueous, hydrogen ions (protons) and hydroxide ions are present. Due to the fact that hydrogen ions and hydroxide ions are positioned quite low in the redox series, they are usually preferentially discharged. If the electrolyte is in molten form, hydrogen ions and hydroxide ions are not present.
To illustrate this, let us consider the different products produced during the electrolysis of molten NaCl and aqueous NaCl.
During the electrolysis of molten NaCl, molten Na metal and chlorine gas is produced. Na+ is attracted to the negative electrode (cathode) and chloride ions is attracted to the positive electrode (anode). At the cathode Na+ ion gain an electron to form the molten Na metal. At the anode, Cl- loses an electron to form chlorine gas.
On the other hand, during the electrolysis of aqueous NaCl, hydrogen gas and oxygen gas
are produced. Both hydrogen ions and sodium ions are attracted to the cathode, however hydrogen ion is preferentially discharged. Hydrogen ion gain an electron to form hydrogen gas. Both hydroxide ions and chloride ions are attracted to the anode. Hydroxide ions are preferentially discharged at the anode due to its lower position in the redox series.
Sunday, June 22, 2008
Fuel Cells
What is a fuel cell?
Students should note that in the syllabus, you are not required to know the details about how a fuel cell works. You just need to know the advantages of a fuel cell as a power source. In a nutshell, the hydrogen oxygen fuel cell generates a flow of electron due to the oxidation of hydrogen gas and the reduction of oxygen gas.
Advantages of using a fuel cell
As mentioned in the syllabus, the advantages of using a fuel cell are the fuel cell is smaller in size, has a lower mass and can produce a higher voltage than conventional batteries.
Students should note that in the syllabus, you are not required to know the details about how a fuel cell works. You just need to know the advantages of a fuel cell as a power source. In a nutshell, the hydrogen oxygen fuel cell generates a flow of electron due to the oxidation of hydrogen gas and the reduction of oxygen gas.
Advantages of using a fuel cell
As mentioned in the syllabus, the advantages of using a fuel cell are the fuel cell is smaller in size, has a lower mass and can produce a higher voltage than conventional batteries.
Limitations of the standard cell potentials
We have seen how standard cell potential can be used to predict the feasibility of a redox reaction. However students should also understand the limitations of using the standard cell potentials.
Energetics vs Kinetics
In general, reactions with positive standard cell potentials are energetically feasible. However the standard cell potential does not tell us about the rate of reaction or its kinetic feasibility.
Standard conditions
Standard electrode values are determined under standard conditions such as the concentration of the aqueous metal ion is 1M. Students should not that a concentration of 1M is quite high. In practice, the redox reactions that we are interested in may not have such a high concentration of reacting ions.
Also the standard electrode values vary with the concentration of metal ions. Take for example the reduction of Cu2+. When the concentration of CU2+ is increased beyond 1M, the feasibility of Cu being reduced is increased. Thus the standard electrode potential will become more positive.
Energetics vs Kinetics
In general, reactions with positive standard cell potentials are energetically feasible. However the standard cell potential does not tell us about the rate of reaction or its kinetic feasibility.
Standard conditions
Standard electrode values are determined under standard conditions such as the concentration of the aqueous metal ion is 1M. Students should not that a concentration of 1M is quite high. In practice, the redox reactions that we are interested in may not have such a high concentration of reacting ions.
Also the standard electrode values vary with the concentration of metal ions. Take for example the reduction of Cu2+. When the concentration of CU2+ is increased beyond 1M, the feasibility of Cu being reduced is increased. Thus the standard electrode potential will become more positive.
A simple electrochemical cell
The schematic below illustrates a simple electrochemical cell.
In short an electrochemical cell functions as a battery. Recall that a current is simply a flow of electrons. In an electrochemical cell, the electron transfer that occurs during a redox reaction is made to flow through an external circuit.
How do we identify which species is donating the electron and which species is accepting the electrons? We can make use of the standard electrode values in the data booklet.
From the electrode values we can see that Cu2+ is more likely to be reduced compared to Zn2+. Hence we make Cu2+ the species that is to be reduced and Zn(s) would be the species that will be oxidized, i.e. the electron donating species. If you calculate the standard cell potential, we will see that the standard cell potential is positive, indicating that the reaction is feasible.
Students should note that for an electrochemical cell, the standard cell potential is always positive.
In short an electrochemical cell functions as a battery. Recall that a current is simply a flow of electrons. In an electrochemical cell, the electron transfer that occurs during a redox reaction is made to flow through an external circuit.
How do we identify which species is donating the electron and which species is accepting the electrons? We can make use of the standard electrode values in the data booklet.
From the electrode values we can see that Cu2+ is more likely to be reduced compared to Zn2+. Hence we make Cu2+ the species that is to be reduced and Zn(s) would be the species that will be oxidized, i.e. the electron donating species. If you calculate the standard cell potential, we will see that the standard cell potential is positive, indicating that the reaction is feasible.Students should note that for an electrochemical cell, the standard cell potential is always positive.
Electrochemistry II
How do we measure the the standard electrode potential?
Metals in contact with their ions in aqueous solutions
For example, we want to measure the standard electrode potential of Cu2+(aq)/Cu(s) half cell. The Cu2+(aq) /Cu(s) half cell consists of a copper electrode immersed in a 1M solution of Cu2+.
Ions of the same element in different oxidation state
For example, we want to measure the standard electrode potential of the Fe3+(aq)/Fe2+(aq) half cell. This half cell will consist of a platinum electrode immersed in a solution containing 1M Fe3+ and 1M Fe2+ ions
How to calculate standard cell potential by using standard electrode potential values in the data booklet?
For example if we want to calculate the standard cell potential for the following redox reaction:
We would look up the values for the standard half cell reactions from the data booklet.
Note that the standard electrode potential are always given as reduction potential. In our redox reaction Zn is oxidized, hence we have to reverse the sign of its standard electrode potential. Hence the standard cell potential would be +0.76 + (+0.34) = +1.10V
The standard cell potential is a positive value, indicating that the redox reaction is feasible.
Metals in contact with their ions in aqueous solutions
For example, we want to measure the standard electrode potential of Cu2+(aq)/Cu(s) half cell. The Cu2+(aq) /Cu(s) half cell consists of a copper electrode immersed in a 1M solution of Cu2+.
Ions of the same element in different oxidation state
For example, we want to measure the standard electrode potential of the Fe3+(aq)/Fe2+(aq) half cell. This half cell will consist of a platinum electrode immersed in a solution containing 1M Fe3+ and 1M Fe2+ ions
How to calculate standard cell potential by using standard electrode potential values in the data booklet?
For example if we want to calculate the standard cell potential for the following redox reaction:
We would look up the values for the standard half cell reactions from the data booklet.
Note that the standard electrode potential are always given as reduction potential. In our redox reaction Zn is oxidized, hence we have to reverse the sign of its standard electrode potential. Hence the standard cell potential would be +0.76 + (+0.34) = +1.10VThe standard cell potential is a positive value, indicating that the redox reaction is feasible.
Electrochemistry I
Redox processes
Redox processes can be defined in terms of electron transfer. Oxidation is defined as the loss of electrons. Reduction is defined as the gain of electrons. Redox process can also be defined in terms of changes in oxidation number. Oxidation is defined as an increase in oxidation number and reduction is defined as a decrease in oxidation number.
Standard electrode potential
The standard electrode potential of a half cell is defined as the potential of that half cell relative to a standard hydrogen electrode under standard conditions. The standard conditions are: all solutions have a concentration of 1M. Any gases involved have a pressure of 1 atm. The temperature is 298K.
Standard cell potential
Standard cell potential is the potential of the cell under standard conditions mentioned above.
Description of the standard hydrogen electrode
The standard hydrogen electrode consists of hydrogen gas at 1atm and 25 degrees Celsius bubbling around a platinum electrode. The electrode is immersed in 1M solution of H+ ions.
Redox processes can be defined in terms of electron transfer. Oxidation is defined as the loss of electrons. Reduction is defined as the gain of electrons. Redox process can also be defined in terms of changes in oxidation number. Oxidation is defined as an increase in oxidation number and reduction is defined as a decrease in oxidation number.
Standard electrode potential
The standard electrode potential of a half cell is defined as the potential of that half cell relative to a standard hydrogen electrode under standard conditions. The standard conditions are: all solutions have a concentration of 1M. Any gases involved have a pressure of 1 atm. The temperature is 298K.
Standard cell potential
Standard cell potential is the potential of the cell under standard conditions mentioned above.
Description of the standard hydrogen electrode
The standard hydrogen electrode consists of hydrogen gas at 1atm and 25 degrees Celsius bubbling around a platinum electrode. The electrode is immersed in 1M solution of H+ ions.
Saturday, June 21, 2008
Solubility product Ksp
Solubility product is used to compare the solubility of sparingly soluble ionic solids. I would use AgCl as an example to illustrate solubility product.
Ksp(AgCl) = [Ag+] [Cl-]
The solubility product of AgCl is the concentration of soluble silver ions at equilibrium multiplied by the concentration of soluble chloride ions at equilibrium. Students should note the difference between solubility product and ionic product. Ionic product is simply the product of concentration of ions at a given time, i.e the concentration of ions may not be at equilibrium.
Note that the solubility product concept applies to only sparingly soluble ionic compounds, i.e. it cannot be used for soluble compunds like NaCl. The solubility product is a modified equilibrium constant hence it is affected by temperature.
Common ion effect
The solubility of a sparing soluble ionic compound AB can be reduced by the presence of A+ or B- from a second source. For example we can compare the solubility of AgCl in water and in NaCl solution. The common ion is the chloride ion. Intuitively we know that AgCl is more soluble in water than in NaCl solution. Why is this so? Consider the solubility equilibria of AgCl:
In a solution of pure water, there are no chloride ions, so AgCl dissolves until the ionic product equals to Ksp. In a solution of NaCl, less AgCl can dissolve, as the ionic product will reach Ksp sooner as there are chloride ions already present in solution.
Its quite hard to explain in words but in the A level examinations, students will be asked to do some calculations. Based on the calculations, the effect of the common ion will be more apparent.
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Ksp(AgCl) = [Ag+] [Cl-]
The solubility product of AgCl is the concentration of soluble silver ions at equilibrium multiplied by the concentration of soluble chloride ions at equilibrium. Students should note the difference between solubility product and ionic product. Ionic product is simply the product of concentration of ions at a given time, i.e the concentration of ions may not be at equilibrium.
Note that the solubility product concept applies to only sparingly soluble ionic compounds, i.e. it cannot be used for soluble compunds like NaCl. The solubility product is a modified equilibrium constant hence it is affected by temperature.
Common ion effect
The solubility of a sparing soluble ionic compound AB can be reduced by the presence of A+ or B- from a second source. For example we can compare the solubility of AgCl in water and in NaCl solution. The common ion is the chloride ion. Intuitively we know that AgCl is more soluble in water than in NaCl solution. Why is this so? Consider the solubility equilibria of AgCl:
In a solution of pure water, there are no chloride ions, so AgCl dissolves until the ionic product equals to Ksp. In a solution of NaCl, less AgCl can dissolve, as the ionic product will reach Ksp sooner as there are chloride ions already present in solution.
Its quite hard to explain in words but in the A level examinations, students will be asked to do some calculations. Based on the calculations, the effect of the common ion will be more apparent.
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Friday, June 20, 2008
Bicarbonate buffer: Controlling blood pH
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The buffer in blood is made up of carbonic acid (weak acid) and hydrogen carbonate (conjugate base).
When protons is released into the blood, the protons combine with hydrogen carbonate (conjugate base) to form carbonic acid. As a result, the concentration of protons in the blood only increases slightly and the resulting pH change is small.
When hydroxide ions are released into the blood, the alkali that is added reacts with carbonic acid to form salt an dwater. As a result, the hydroxide ions are removed from blood, and the pH of blood only changes slightly.
The explanation that I have given above is actually a oversimplified one. For a more physiological and thorough explanation, please refer to the following website on Blood, Sweat and Buffers. For the purpose of the A Level examinations, my explanation will suffice.
Link to Contents Page
The buffer in blood is made up of carbonic acid (weak acid) and hydrogen carbonate (conjugate base).
When protons is released into the blood, the protons combine with hydrogen carbonate (conjugate base) to form carbonic acid. As a result, the concentration of protons in the blood only increases slightly and the resulting pH change is small.
When hydroxide ions are released into the blood, the alkali that is added reacts with carbonic acid to form salt an dwater. As a result, the hydroxide ions are removed from blood, and the pH of blood only changes slightly.
The explanation that I have given above is actually a oversimplified one. For a more physiological and thorough explanation, please refer to the following website on Blood, Sweat and Buffers. For the purpose of the A Level examinations, my explanation will suffice.
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Buffers and pH control
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The pH of a buffer solution can be calculated using the Handerson-Hasselbach equation.
Buffers are solutions which resist changes in pH on addition of small amount of acid or alkali. Buffers usually consist of a weak acid and its conjugate base (acid buffer) or a weak base and its conjugate acid (basic buffer).
How does an acid buffer work?
As mentioned above, an acid buffer consists of a weak acid and its conjugate base. An example is ethanoic acid and its conjugate base ethanoate ion. The conjugate base is usually supplied as a salt. The buffer is most effective at resisting pH changes when the concentration of the undissociated acid molecule (ethanoic acid) is similar to the concentration of the conjugate base (ethanoate ion).
When a small amount of acid is added, the protons combine with the conjugate base (ethanoate ions) to form the weak acid (ethanoic acid). The H+ that is added is removed, hence the concentration of H+ in solution and thus the pH changes only slightly.
When a small amount of base is added, the base reacts with the weak acid to form salt and water. The hydroxide ions are removed from the solution in the form of water and the pH of the solution only changes slightly.
Calculating the pH of buffer solutionThe pH of a buffer solution can be calculated using the Handerson-Hasselbach equation.
Students should note that this equation can only be used for calculating the pH of buffer solutions. Also the equation above can only be used to calculate the pH of acid buffers. For base buffers, use the equation below
Thursday, June 19, 2008
Acids and bases II
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3) pH, dissociation constants (Ka, Kb) and ionic product of water (Kw)
The pH of a solution is the negative logarithm to base ten of the molar concentration of hydrogen ion. Students should note that pH only tells you how acidic or basic a solution is, it does not tell you about the strength of the acid or base. A concentrated solution of a weak acid will have a low pH , but it does not mean that the acid is a strong acid
In order to measure the strength of the acid , chemists use the acid dissociation constant. The acid dissociation constant is an equilibrium constant. Basically it is the ratio of concentration of protons and conjugate base to the concentration of undissociated acid molecules. Since acid dissociation constants are equilibrium constants, they are not affected by concentration, unlike pH which is concentration dependent. The greater the Ka, the stronger the acid.
Dissociation of water
Water dissociates to a very small extent to form hydrogen ions and hydroxide ions. This is represented by the ionic product for water Kw, which is the [H+] X [OH-], 10e-14 for pure water at 25 degree celsius. Note that like acid dissociation constant, the ionic product for water is effectively an equilibrium constant, hence it is affected by temperature, i.e. Kw will vary with temperature.
4) Indicators for acid base titration
The most common indicators are litmus, methyl orange and phenolpthalein. Methyl orange is red at acidic pH, orange at pH 3 to 4 and yellow at pH 5 onwards. Litmus paper is red at acidic pH and blue at pH values of 7 and above. Phenolpthalein is colourless at at acidic pH (pH 0 to 7 ) and red at alkalike pH (pH 9 and above). Students shouldnote that not all indicators change colour at pH 7.
In a titration of strong acid against a strong alkali, at equivalence point when the acid is completely neutralized, the pH changes by a large extent from around pH 4 to 10. Thus both methyl orange and phenolpthalein can be used to indicate that equivalenc epoint has been reached.
In the titration of a strong acid with a weak base, the pH change is less drastic, from around pH 4 to 8, hence phenolpthalein is not a suitable indicator as it changes colour at around pH 8. Methy orange will be a suitable indicator.
In the titration of a weak acid with a strong base, the pH change at equivalence point is from around pH 7 to 11. Phenolpthalein will be a suitable indicator as it changes colour at around pH 8.
In the titration of a weak acid with weak base, the pH change at equivalence point is very small. hence there are no suitable indicators. You can probably detect the pH change using a pH meter.
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3) pH, dissociation constants (Ka, Kb) and ionic product of water (Kw)
The pH of a solution is the negative logarithm to base ten of the molar concentration of hydrogen ion. Students should note that pH only tells you how acidic or basic a solution is, it does not tell you about the strength of the acid or base. A concentrated solution of a weak acid will have a low pH , but it does not mean that the acid is a strong acid
In order to measure the strength of the acid , chemists use the acid dissociation constant. The acid dissociation constant is an equilibrium constant. Basically it is the ratio of concentration of protons and conjugate base to the concentration of undissociated acid molecules. Since acid dissociation constants are equilibrium constants, they are not affected by concentration, unlike pH which is concentration dependent. The greater the Ka, the stronger the acid.
Dissociation of water
Water dissociates to a very small extent to form hydrogen ions and hydroxide ions. This is represented by the ionic product for water Kw, which is the [H+] X [OH-], 10e-14 for pure water at 25 degree celsius. Note that like acid dissociation constant, the ionic product for water is effectively an equilibrium constant, hence it is affected by temperature, i.e. Kw will vary with temperature.
4) Indicators for acid base titration
The most common indicators are litmus, methyl orange and phenolpthalein. Methyl orange is red at acidic pH, orange at pH 3 to 4 and yellow at pH 5 onwards. Litmus paper is red at acidic pH and blue at pH values of 7 and above. Phenolpthalein is colourless at at acidic pH (pH 0 to 7 ) and red at alkalike pH (pH 9 and above). Students shouldnote that not all indicators change colour at pH 7.
In a titration of strong acid against a strong alkali, at equivalence point when the acid is completely neutralized, the pH changes by a large extent from around pH 4 to 10. Thus both methyl orange and phenolpthalein can be used to indicate that equivalenc epoint has been reached.
In the titration of a strong acid with a weak base, the pH change is less drastic, from around pH 4 to 8, hence phenolpthalein is not a suitable indicator as it changes colour at around pH 8. Methy orange will be a suitable indicator.
In the titration of a weak acid with a strong base, the pH change at equivalence point is from around pH 7 to 11. Phenolpthalein will be a suitable indicator as it changes colour at around pH 8.
In the titration of a weak acid with weak base, the pH change at equivalence point is very small. hence there are no suitable indicators. You can probably detect the pH change using a pH meter.
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Acids and bases I
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1) Bronsted-Lowry Theory of Acids and Bases
According to the Bronsted Lowry theory, an acid is defined as a substance that can donate proton to another substance, i.e. a proton donor. An example of an acid is hydrochloric acid. Students also need to know what a conjugate base is. An acid whhich has donated its proton becomes a conjugate base. Using hydrochloric acid as an example, hydrochloric acid is the acid and the chloride ion is the conjugate base. Using ethanol as an example, if ethanol behaves as a Bronsted acid, it loses its proton and becomes ethanoate ion. Therefore the conjugate base is the ethanoate ion.
A base is a substance which accepts a proton from another substance, i.e. a proton acceptor. An example is ammonia. Ammonia is a Bronsted base, it accepts a proton and becomes the ammonium ion. In this case ammonia is the base and the ammonium ion is the conjugate acid. Note that a bronsted base is related to its conjugate acid and a bronsted acid is related to its conjugate base.
Note that certain acids and bases do not conform to the bronsted acid and base definition. Take for example sodium hydroxide, it is well know that sodium hydroxide is a base. However NaOH is not a proton acceptor hence it is not a Bronsted base.
There is another definition of acids and bases proposed by Arrhenius. According to the Arrhenius definition, an acid dissociates in water to form hydrgen ions and a base dissociates in water to produce hydroxide ions. NaOH will fit the Arrhenius definiton of a base. There are other definitions such as the Lewis acid definition. For more information students can refer to the wikipedia entry.
2) Differences between a weak and strong acid.
A strong acid such as hydrochloric acid dissociates fully in water to form hydrogen ions and chloride ions. A weak acid dissociates partially in water. An example is ethanoic acid. When ethanoic acid is dissolved in water, some of the ethanoic acid molecules will dissociate to form the ethanoate ion and hydrogen ions. However not all of the ethanoic acid molecules dissociate so there will be ethanoic acid molecules in the wtaer in addition to ethanoate ions and protons.
Students must be careful not to mix up concentration and strength of acid. A concentrated acid might not be a strong acid. A 5 M ethanoic acid is a concentrated acid solution but ethanoic acid is not a strong acid.
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1) Bronsted-Lowry Theory of Acids and Bases
According to the Bronsted Lowry theory, an acid is defined as a substance that can donate proton to another substance, i.e. a proton donor. An example of an acid is hydrochloric acid. Students also need to know what a conjugate base is. An acid whhich has donated its proton becomes a conjugate base. Using hydrochloric acid as an example, hydrochloric acid is the acid and the chloride ion is the conjugate base. Using ethanol as an example, if ethanol behaves as a Bronsted acid, it loses its proton and becomes ethanoate ion. Therefore the conjugate base is the ethanoate ion.
A base is a substance which accepts a proton from another substance, i.e. a proton acceptor. An example is ammonia. Ammonia is a Bronsted base, it accepts a proton and becomes the ammonium ion. In this case ammonia is the base and the ammonium ion is the conjugate acid. Note that a bronsted base is related to its conjugate acid and a bronsted acid is related to its conjugate base.
Note that certain acids and bases do not conform to the bronsted acid and base definition. Take for example sodium hydroxide, it is well know that sodium hydroxide is a base. However NaOH is not a proton acceptor hence it is not a Bronsted base.
There is another definition of acids and bases proposed by Arrhenius. According to the Arrhenius definition, an acid dissociates in water to form hydrgen ions and a base dissociates in water to produce hydroxide ions. NaOH will fit the Arrhenius definiton of a base. There are other definitions such as the Lewis acid definition. For more information students can refer to the wikipedia entry.
2) Differences between a weak and strong acid.
A strong acid such as hydrochloric acid dissociates fully in water to form hydrogen ions and chloride ions. A weak acid dissociates partially in water. An example is ethanoic acid. When ethanoic acid is dissolved in water, some of the ethanoic acid molecules will dissociate to form the ethanoate ion and hydrogen ions. However not all of the ethanoic acid molecules dissociate so there will be ethanoic acid molecules in the wtaer in addition to ethanoate ions and protons.
Students must be careful not to mix up concentration and strength of acid. A concentrated acid might not be a strong acid. A 5 M ethanoic acid is a concentrated acid solution but ethanoic acid is not a strong acid.
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Tuesday, June 17, 2008
Updates
I have added a navigation bar near the top of the blog. You can access the sitemap via this bar
For the chapter on Chemical periodicity, I have also added a question from the actual A Levels Examinations. Students are encouraged to try the question to get an idea of how this chapter can be tested in the examinations.
The chapters on Chemical Periodicity has been rewritten. Hope it is more clear now. The textbook by Hill and Holman, Chemistry in Context has a good chapter written on this topic.
For the chapter on Chemical periodicity, I have also added a question from the actual A Levels Examinations. Students are encouraged to try the question to get an idea of how this chapter can be tested in the examinations.
The chapters on Chemical Periodicity has been rewritten. Hope it is more clear now. The textbook by Hill and Holman, Chemistry in Context has a good chapter written on this topic.
Monday, June 16, 2008
Chemical Equilibria II
9）The equilibrium in terms of partial pressure Kp
Kp is generally used when the reaction involves gaseous reactants and products. For example, reaction between hydrogen and iodine to form hydrogen iodide.
10）Haber Process
Haber process involves the production of ammonia from nitrogen and hydrogen gas. Recall that the reaction is exothermic and that 1 mol of nitrogen gas reaacts with 3 mol of hydrogen gas to produce 2 mol of ammonia gas. By Le Chatelier's Princple high pressure and low temperature will increase the proportion of ammonia at equilibrium.
However industrially a high temperature of 450 degrees and a pressure of 250 atm is used. Although the yield of ammonia is lower at higher temperature, using a low temeprature reduces the rate of reaction making the process uneconomical. Fe is also used as a catalyst to speed up the rate of reaction.
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Kp is generally used when the reaction involves gaseous reactants and products. For example, reaction between hydrogen and iodine to form hydrogen iodide.
10）Haber Process
Haber process involves the production of ammonia from nitrogen and hydrogen gas. Recall that the reaction is exothermic and that 1 mol of nitrogen gas reaacts with 3 mol of hydrogen gas to produce 2 mol of ammonia gas. By Le Chatelier's Princple high pressure and low temperature will increase the proportion of ammonia at equilibrium.
However industrially a high temperature of 450 degrees and a pressure of 250 atm is used. Although the yield of ammonia is lower at higher temperature, using a low temeprature reduces the rate of reaction making the process uneconomical. Fe is also used as a catalyst to speed up the rate of reaction.
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Sunday, June 15, 2008
Chemical Equilibria I
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1)Reversible reaction
A reversible reaction is a chemical reaction which can take place in both directions, i.e. reactant react to form products and products can also react to form the original reactants.
2)Dynamic equilibrium
A system is said to be in dynamic equilibrium occurs when the rate of forward reaction is equal to the rate of reverse reaction and the concentration of products and reactants remain constant. It is dynamic because reactants are being continuously converted to products and products are being continuously converted back to reactants.
3)Le Chatelier's Principle
Le Chatelier's Principle states that in a system in equilibrium, when a change is made to some external factor, the position of equilibrium shifts to oppose the change.
I will use the production of ammonia as an example.
4)Effects of changes in concentration
What happens when you increase the concentration of nitrogen gas. According to Le Chatelier's Principle the equilibrium reacts to remove the extra nitrogen gas that is added. Thus the equilibrium shifts to the right and produces more ammonia.
5)Effects of changes in pressure
Changes in pressure only affects reactions involving gases. From the chemical equation, we can see that 1 mol of nitrogen gas reacts with 3 mol of hydrogen gas to form 2 mol of ammonia gas, i.e. 4 mol of gaseous reactants react to form 2 mol of gaseous products. What happens when the pressure of the system is increased? According to Le Chatelier's Principle, the equilibrium shifts to the right to reduce the pressure.
6)Effects of changes in temperature
The formation of ammonia from nitrogen and hydrogen is an exothermic reaction, i.e. the forward reaction is exothermic . What happens when the temperature is increased? Based on Le Chatelier's Principle, the equilibrium will react to oppose the increase in temperature and the equilibrium shifts to the left.
7)Effects of a catalyst
A catalyst does not change the position of the equilibrium. It only increases the rate at which the system reaches equilibrium.
8)Equilibrium Constant Kc
If a reversible reaction is allowed to reach equilibrium, the product of the concentration of products divided by the product of the concentrations of reactants has a constant value at a particular temperature.
Changes in concentration and pressure and the presence of a catalyst does not change the equilibrium constant. Only temperature change will cause a change in the equilibrium constant. For an exothermic reaction, the equilibrium constant decreases as temperature is increased . For an endothermic reaction, the equilibrium constant increase with increasing temperature.
To be continued
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1)Reversible reaction
A reversible reaction is a chemical reaction which can take place in both directions, i.e. reactant react to form products and products can also react to form the original reactants.
2)Dynamic equilibrium
A system is said to be in dynamic equilibrium occurs when the rate of forward reaction is equal to the rate of reverse reaction and the concentration of products and reactants remain constant. It is dynamic because reactants are being continuously converted to products and products are being continuously converted back to reactants.
3)Le Chatelier's Principle
Le Chatelier's Principle states that in a system in equilibrium, when a change is made to some external factor, the position of equilibrium shifts to oppose the change.
I will use the production of ammonia as an example.
4)Effects of changes in concentration
What happens when you increase the concentration of nitrogen gas. According to Le Chatelier's Principle the equilibrium reacts to remove the extra nitrogen gas that is added. Thus the equilibrium shifts to the right and produces more ammonia.
5)Effects of changes in pressure
Changes in pressure only affects reactions involving gases. From the chemical equation, we can see that 1 mol of nitrogen gas reacts with 3 mol of hydrogen gas to form 2 mol of ammonia gas, i.e. 4 mol of gaseous reactants react to form 2 mol of gaseous products. What happens when the pressure of the system is increased? According to Le Chatelier's Principle, the equilibrium shifts to the right to reduce the pressure.
6)Effects of changes in temperature
The formation of ammonia from nitrogen and hydrogen is an exothermic reaction, i.e. the forward reaction is exothermic . What happens when the temperature is increased? Based on Le Chatelier's Principle, the equilibrium will react to oppose the increase in temperature and the equilibrium shifts to the left.
7)Effects of a catalyst
A catalyst does not change the position of the equilibrium. It only increases the rate at which the system reaches equilibrium.
8)Equilibrium Constant Kc
If a reversible reaction is allowed to reach equilibrium, the product of the concentration of products divided by the product of the concentrations of reactants has a constant value at a particular temperature.
Changes in concentration and pressure and the presence of a catalyst does not change the equilibrium constant. Only temperature change will cause a change in the equilibrium constant. For an exothermic reaction, the equilibrium constant decreases as temperature is increased . For an endothermic reaction, the equilibrium constant increase with increasing temperature.
To be continued
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The Periodic Table: Chemical periodicity II
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7) Variation of oxidation number in chlorides and oxides
The oxidation numbers of elements in their oxides is always positive because oxygen is a very electronegative atom. The maximum oxidation number of each element is the same as its group number. The group number of the element corresponds to the number of electrons in its outermost shell.
Most of the elements have their usual oxidation states. Just take note that phosphorous has an oxidation state of +5 in phosphorous oxide and +5 in phosphorous chloride. Sulphur has an oxidation state of +4 in sulphur dioxide and +6 in sulphur trioxide.
The oxidation number becomes more positive across the period because the number of valence electrons increase across the period. These valence electrons can take part in bonding with chlorine and oxygen to form chlorides and oxides.
8) Reaction of oxides with water
Metallic oxides reacts with water to form alkaline solutions.
Sodium oxide react vigourously with water to form an alkaline solution of sodium hydroxide.
Magnesium oxide reacts less readily with water to form magnesium hydroxide. Its low reactivity with wtaer is due to the high charge density of Mg ions which holds the oxygen ion more firmly.
Aluminum oxide does not react with water.
Silicon oxide does not react with water.
Non metalllic oxides react with water to form strong acids.
Phosphorous (V) oxide react with water to form phosphoric (V) acid.
Sulphur dioxide react with water to form sulphurous acid.
Sulphur trioxide react with water to form sulphuric acid.
9) Reaction of oxides and hydroxides with acid and NaOH
Metal oxides and hydroxides (Na, Mg) are ionic compounds. They are basic in nature and reacts with acid to form salt and water.
Non metal oxides and hydroxides (Si, P, S, Cl) are covalent compounds. They acidic in nature and reacts with NaOH to form salt and water.
Aluminum oxide is not souble in water but it is amphoteric in nature; i.e. it reacts with both acid and base. Aluminium oxide reacts with HCl to form aluminum chloride and water. Aluminium oxide reacts with NaOH to from an a complex ion (aluminate).
10) Reaction of chlorides with water
Simple ionic chlorides like NaCl and magnesium chloride simply dissolve in water. The solutions of ionic chlorides are neutral.
Aluminum chloride reacts with water to form a complex ion. The aluminum complex can polarize water molecule due to its high charge density. The highly charged aluminium ion draws electrons away from surrounding water molecules, causing them to give up H+.
Silicon chloride reacts with water to form silicon dioxide and HCl.
Phosphorous (V) chloride reacts with water to form phosphoric (V) acid and HCl.
Take home message:
Compounds of Na and Mg are ionic compounds Thus they have high melting points and form basic oxides. Ionic compounds do not react with water, they simply dissolve.
Aluminum compounds are ionic with strong covalent character and form amphoteric oxides. Aluminum compounds react with water to form acidic solutions.
Si compounds has a giant molecular structure and have high melting points. They form acidic oxides.
Phosphorus and sulphur compounds have simple molecular structures and low melting points. They form acidic oxides that react with water giving rise to acidic solutions.
Side note: The syllabus for this section looks very demanding and difficult. However this section isnt really being asked in long answer type of questions. So dont panic
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7) Variation of oxidation number in chlorides and oxides
The oxidation numbers of elements in their oxides is always positive because oxygen is a very electronegative atom. The maximum oxidation number of each element is the same as its group number. The group number of the element corresponds to the number of electrons in its outermost shell.
Most of the elements have their usual oxidation states. Just take note that phosphorous has an oxidation state of +5 in phosphorous oxide and +5 in phosphorous chloride. Sulphur has an oxidation state of +4 in sulphur dioxide and +6 in sulphur trioxide.
The oxidation number becomes more positive across the period because the number of valence electrons increase across the period. These valence electrons can take part in bonding with chlorine and oxygen to form chlorides and oxides.
8) Reaction of oxides with water
Metallic oxides reacts with water to form alkaline solutions.
Sodium oxide react vigourously with water to form an alkaline solution of sodium hydroxide.
Magnesium oxide reacts less readily with water to form magnesium hydroxide. Its low reactivity with wtaer is due to the high charge density of Mg ions which holds the oxygen ion more firmly.
Aluminum oxide does not react with water.
Silicon oxide does not react with water.
Non metalllic oxides react with water to form strong acids.
Phosphorous (V) oxide react with water to form phosphoric (V) acid.
Sulphur dioxide react with water to form sulphurous acid.
Sulphur trioxide react with water to form sulphuric acid.
9) Reaction of oxides and hydroxides with acid and NaOH
Metal oxides and hydroxides (Na, Mg) are ionic compounds. They are basic in nature and reacts with acid to form salt and water.
Non metal oxides and hydroxides (Si, P, S, Cl) are covalent compounds. They acidic in nature and reacts with NaOH to form salt and water.
Aluminum oxide is not souble in water but it is amphoteric in nature; i.e. it reacts with both acid and base. Aluminium oxide reacts with HCl to form aluminum chloride and water. Aluminium oxide reacts with NaOH to from an a complex ion (aluminate).
10) Reaction of chlorides with water
Simple ionic chlorides like NaCl and magnesium chloride simply dissolve in water. The solutions of ionic chlorides are neutral.
Aluminum chloride reacts with water to form a complex ion. The aluminum complex can polarize water molecule due to its high charge density. The highly charged aluminium ion draws electrons away from surrounding water molecules, causing them to give up H+.
Silicon chloride reacts with water to form silicon dioxide and HCl.
Phosphorous (V) chloride reacts with water to form phosphoric (V) acid and HCl.
Take home message:
Compounds of Na and Mg are ionic compounds Thus they have high melting points and form basic oxides. Ionic compounds do not react with water, they simply dissolve.
Aluminum compounds are ionic with strong covalent character and form amphoteric oxides. Aluminum compounds react with water to form acidic solutions.
Si compounds has a giant molecular structure and have high melting points. They form acidic oxides.
Phosphorus and sulphur compounds have simple molecular structures and low melting points. They form acidic oxides that react with water giving rise to acidic solutions.
Side note: The syllabus for this section looks very demanding and difficult. However this section isnt really being asked in long answer type of questions. So dont panic
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The Periodic Table: Chemical Periodicity I
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In this chapter we would concentrate on the elements found in Period 3 of the Periodic Table. They are : Na, Mg, Al, Si, P, S, Cl and Ar
1) Variation of atomic radii across Period 3
Atomic radii decreases across the period. Across the period, electron are added to same electron shell (Principal quantum number =3). Shielding effect is similar across the period. However nuclear charge increases across the period due to the increase in number of protons. Hence effective nuclear charge increases and atomic radii decreases.
2) Variation of ionic radii across Period 3
Ionic radii decreases across the period from Na+ to Si4+. Across the period, the outer most electron of the ion is in the 2p subshell, hence the outer most electron experiences the same shielding effect. However nuclear charge increases across the period due to the increase in number of protons. Hence effective nuclear charge increases and ionic radii decreases.
Note that we have to leave out P, S, Cl and Ar in the comparison as their outermost electron is found in the 3p subshell.
3) Variation of melting point across the period
The melting point of an element depends on the bonds present and the structure of the element.
Metals
Na, Mg and Al are metals. The bonds present are metallic bonds defined as the electrostatic forces of attraction between the positive metal ions and sea of delocalised electrons. The metal ions are pack together to form a metallic lattice. Hence the melting point of metals are high and increases from Na to Al as the nuber of delocalized electrons increase.
Si
The atoms of Si are bonded to each other by covalent bonds in a giant molecular structure. When Si melts, all the covlent bonds have to be broken. This is unlike metals where some metallic bonding still remains in the liquid metal. Hence Si has a melting point that is very much higher than metals
P, S, Cl and Ar
Phosphorous, sulphur and chlorine has simple molecular structures. Within the molecule, the atoms are bonded together by strong covalent bonds. Van der Waals interactions exists between molecules. These intermolecular interactions are weak, thus P,S and Cl have low melting points.
4) Variation in the first ionization energy
First IE is the energy required to remove the outermost electron from one mole of atoms in the gas phase. First IE of the elements in period 3 increases across the period. Across the period, electron are removed from the same electron shell (Principal quantum number =3). Shielding effect is similar across the period. However nuclear charge increases across the period due to the increase in number of protons. Hence effective nuclear charge increases and first IE increases.
5) Reaction with oxygen
Na reacts very vigorously with oxygen to form sodium oxide.
Mg reacts very vigorously with oxygen to form MgO.
Al reacts vigorously with oxygen to form aluminium oxide.
P reacts vigourously with oxygen to form phosphorous (V) oxide.
S reacts slowly with oxygen to form sulphur dioxide and sulphur trioxide.
6) Reaction with chlorine
Na reacts very vigorously to giveNaCl.
Mg reacts vigorously to form magnesium chloride.
Al reacts vigorously to form aluminium chloride.
Si reacts slowly to form silicon chloride.
P reacts slowly to form phosphorous (V) chloride.
The Cambridge International Examinations Website has put up the questions for A Levels Nov 2006 Paper 2. These questions are available for download. Students can download the question paper and try question 3. This question is about chemical periodicity. You can try it and check your answers against the mark scheme provided.
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In this chapter we would concentrate on the elements found in Period 3 of the Periodic Table. They are : Na, Mg, Al, Si, P, S, Cl and Ar
1) Variation of atomic radii across Period 3
Atomic radii decreases across the period. Across the period, electron are added to same electron shell (Principal quantum number =3). Shielding effect is similar across the period. However nuclear charge increases across the period due to the increase in number of protons. Hence effective nuclear charge increases and atomic radii decreases.
2) Variation of ionic radii across Period 3
Ionic radii decreases across the period from Na+ to Si4+. Across the period, the outer most electron of the ion is in the 2p subshell, hence the outer most electron experiences the same shielding effect. However nuclear charge increases across the period due to the increase in number of protons. Hence effective nuclear charge increases and ionic radii decreases.
Note that we have to leave out P, S, Cl and Ar in the comparison as their outermost electron is found in the 3p subshell.
3) Variation of melting point across the period
The melting point of an element depends on the bonds present and the structure of the element.
Metals
Na, Mg and Al are metals. The bonds present are metallic bonds defined as the electrostatic forces of attraction between the positive metal ions and sea of delocalised electrons. The metal ions are pack together to form a metallic lattice. Hence the melting point of metals are high and increases from Na to Al as the nuber of delocalized electrons increase.
Si
The atoms of Si are bonded to each other by covalent bonds in a giant molecular structure. When Si melts, all the covlent bonds have to be broken. This is unlike metals where some metallic bonding still remains in the liquid metal. Hence Si has a melting point that is very much higher than metals
P, S, Cl and Ar
Phosphorous, sulphur and chlorine has simple molecular structures. Within the molecule, the atoms are bonded together by strong covalent bonds. Van der Waals interactions exists between molecules. These intermolecular interactions are weak, thus P,S and Cl have low melting points.
4) Variation in the first ionization energy
First IE is the energy required to remove the outermost electron from one mole of atoms in the gas phase. First IE of the elements in period 3 increases across the period. Across the period, electron are removed from the same electron shell (Principal quantum number =3). Shielding effect is similar across the period. However nuclear charge increases across the period due to the increase in number of protons. Hence effective nuclear charge increases and first IE increases.
5) Reaction with oxygen
Na reacts very vigorously with oxygen to form sodium oxide.
Mg reacts very vigorously with oxygen to form MgO.
Al reacts vigorously with oxygen to form aluminium oxide.
P reacts vigourously with oxygen to form phosphorous (V) oxide.
S reacts slowly with oxygen to form sulphur dioxide and sulphur trioxide.
6) Reaction with chlorine
Na reacts very vigorously to giveNaCl.
Mg reacts vigorously to form magnesium chloride.
Al reacts vigorously to form aluminium chloride.
Si reacts slowly to form silicon chloride.
P reacts slowly to form phosphorous (V) chloride.
The Cambridge International Examinations Website has put up the questions for A Levels Nov 2006 Paper 2. These questions are available for download. Students can download the question paper and try question 3. This question is about chemical periodicity. You can try it and check your answers against the mark scheme provided.
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Group II Elements
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The Group II elements included in the syllabus are Magnesium(Mg), Calcium (Ca), Strontium (Sr) and Barium (Ba)
1) Reactions of elements with oxygen
Group II elements react with oxygen to form oxides
2) Reactions with water
Reactivity of group II elements with water increases down the group
Mg reacts slowly with cold water to form MgO and H2. Mg reacts rapidly with steam.
Calcium reacts steadily with water to form calcium hydroxide and hydrogen.
Strontium and barium reacts explosively with water to form hydroxides and hydrogen.
The syllabus only require students to describe the reaction. If you are interested to know why reactivity of group II elements increases down the group, chemguide.co.uk has a good explanation.
3) Behaviour of oxides with water
Reactivity of group II oxides with water increases down the group.
Group II oxides react with water to form hydroxides.
4) Thermal stability of Group II nitrates
Thermal stability of Group II nitrates increases down the group
The thermal stability of Group II nitrates can be explained in terms of the charge density of the metal cation and the polarisability of the nitrate ion.
Down the group, the ionic radii of the Group II elements increases. The charge of the metal ion remains the same, hence charge density decreases. The ion with a lower charge density is less able to polarize the nitrate ion. Hence more heat have to be supplied to decompose the nitrate to metal oxide and NO2.
5) Make predictions
There would e some questions in the exam that would require you to make some predictions. For example, thye woould say an unknown metal X is a group II element. How would its metal hydroxide react with water? So you are expected to know that it would react to form a metal hydroxide.
End of notes for Group II elements
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The Group II elements included in the syllabus are Magnesium(Mg), Calcium (Ca), Strontium (Sr) and Barium (Ba)
1) Reactions of elements with oxygen
Group II elements react with oxygen to form oxides
2) Reactions with water
Reactivity of group II elements with water increases down the group
Mg reacts slowly with cold water to form MgO and H2. Mg reacts rapidly with steam.
Calcium reacts steadily with water to form calcium hydroxide and hydrogen.
Strontium and barium reacts explosively with water to form hydroxides and hydrogen.
The syllabus only require students to describe the reaction. If you are interested to know why reactivity of group II elements increases down the group, chemguide.co.uk has a good explanation.
3) Behaviour of oxides with water
Reactivity of group II oxides with water increases down the group.
Group II oxides react with water to form hydroxides.
4) Thermal stability of Group II nitrates
Thermal stability of Group II nitrates increases down the group
The thermal stability of Group II nitrates can be explained in terms of the charge density of the metal cation and the polarisability of the nitrate ion.
Down the group, the ionic radii of the Group II elements increases. The charge of the metal ion remains the same, hence charge density decreases. The ion with a lower charge density is less able to polarize the nitrate ion. Hence more heat have to be supplied to decompose the nitrate to metal oxide and NO2.
5) Make predictions
There would e some questions in the exam that would require you to make some predictions. For example, thye woould say an unknown metal X is a group II element. How would its metal hydroxide react with water? So you are expected to know that it would react to form a metal hydroxide.
End of notes for Group II elements
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Site Map
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Electrochemistry I
Electrochemsistry II
Simple Electrochemical Cell
Limitations of Standard Cell Potentials
Fuel Cells
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Electrochemistry I
Electrochemsistry II
Simple Electrochemical Cell
Limitations of Standard Cell Potentials
Fuel Cells
Electrolysis I
Electrolysis II
Calculations involving electrolysis
Electrolysis in industrial process
Reaction kinetics I
How to determine the order of reaction?
Chemical Equilibria I
Chemical Equilibria II
Acids and Bases I
Acids and Bases II
Buffers and pH control
Bicarbonate buffer: Controlling blood pH
Solubility Product: Ksp
The Periodic Table: Chemical Periodicity I
The Periodic Table: Chemical Periodicity II
Group II Elements
Group VII: The Halogens
Chemistry of Transition Elements I
Chemistry of Transition Elements II
Worked examples
N2007/II/5
N2007/III/1
N2007/III/2
N2007/III/5
Exam Advice
A Level chemistry Syllabus
Cambridge International Examinations Website
How to use the worked examples?
Saturday, June 14, 2008
Group VII: The Halogens
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The halogens refer to chlorine, bromine and iodine.
1) Colour and volatility
Chlorine is a dense green gas.
Bromine is a brown volatile liquid.
Iodine is a shiny black solid
Down the group, the number of electrons in the halogen molecule increases, hence the strength of van der Waals forces between the halogen molecules increases. This explains why chlorine exists as a gas, bromine exists as a liquid and iodine exists as a solid.
2) Relative reactivity of halogens as oxidizing agents
Chlorine is the strongest oxidizing agent followed by bromine and iodine. Recall that an oxidizing agent causes oxidation and it itself is reduced, i.e. an oxidizing agent has a tendency to gain electrons.
The electrode potential values become less positive down the group, indicating that down the group the halogens has a reduced tendency to gain electrons, hence the oxidizing power of halogen decreases.
3) Reaction of halogens with hydrogen
Chlorine reacts explosively in sunlight with hydrogen. It also reacts slowly in the dark with hydrogen.
Bromine reacts with hydrogen only at high temperatures
Iodine reacts with hydrogen to form an equilibrium mixture of hydrogen, iodine and HI.
4) Relative thermal stability of hydrides (hydrogen halides)
Thermal stability of hydrogen halides decreases down the group.
Down the group, the atomic radii of the halogen increases and the bond dissociation enthalpy of hydrogen halides decreases.
Bond energy of H--Cl : 431 kJ/mol
Bond energy of H--Br: 366 kJ/mol
Bond energy of H--I: 299 kJ/mol
Note: In the exam, students may be asked to quote bond energy values from the Data booklet. You do not have to memorize these bond energies. they are in the data booklet.
5) Reaction of halide ions with aqueous silver ions followed by aqueous ammonia
Chloride ions form a white precipitate with silver ions. White ppt is soluble in aqueous NH3
Bromide ions form a pale yellow ppt with silver ions. Yellow ppt is sparingly soluble in aqueous ammonia.
Iodide ions form a deep yellow ppt with silver ions. Deep yellow ppt is insoluble in aqueous ammonia.
Silver ions is usually supplied as silver nitrate. This is a common test for halides. Aqueous ammonia is added to differentiate between silver chloride and silveer bromide as it is very hard to differentiate between the white and pale yellow ppt.
Aqueous ammonia has to be added slowly as the pale yellow silver bromide is soluble in excess aqueous ammonia. Aqueous ammonia forms a complex with silver ions, shifting the equlibrium to the left, and increases the solubility of AgCl. AgCl is most soluble followed by AgBr and AgI, thus the addition of ammonia can fully solubilise AgCl. AgBr is only soluble in concentrated ammonia while ammonia has no effect on the solubility of AgI.
6) Reaction of halide ions with concentrated sulphuric acid
Concentrated sulphuric acid can act as a proton donor, i.e. it supplies H+ ions which reacts with the halide ion to form hydrides (HCl, HBr and HI). The hydrides appear as steamy white fumes
Concentrated sulphuric acid can also act as an oxidizing agent.
Recall that iodine is the weakest oxidizing agent, hence it is the most easily oxidized. Concentrated sulphuric acid can oxidize iodide ions to iodine, hence you will see purple vapour
Concentrated sulphuric acid can also oxidize bromide ions to bromine gas.
Concentrated sulphuric acid cannot oxidize chloride ions to chlorine gas.
7) Reaction of chlorine with NaOH
Chlorine reacts with cold NaOH to form NaCl and NaClO (sodium hypochlorate)
Note that in this reaction one atom of chlorine is undergoing oxidation and the other atom of chlorine is undergoing reduction.
Chlorine is oxidized. The oxidation number of chlorine increases from 0 in Chlorine gas to +1 in NaClO. Chlorine ia also reduced. The oxidation number of chlorine decreases from 0 in Chlorine gas to -1 in NaCl.
Chlorine reacts with hot NaOH to form NaCl and NaClO3 (sodium chlorate)
In this reaction one atom of chlorine is undergoing oxidation and the other atom of chlorine is undergoing reduction.
Chlorine is oxidized. The oxidation number of chlorine increases from 0 in Chlorine gas to +5 in NaClO3. Chlorine ia also reduced. The oxidation number of chlorine decreases from 0 in Chlorine gas to -1 in NaCl.
End of group VII notes
Link to Contents Pages
The halogens refer to chlorine, bromine and iodine.
1) Colour and volatility
Chlorine is a dense green gas.
Bromine is a brown volatile liquid.
Iodine is a shiny black solid
Down the group, the number of electrons in the halogen molecule increases, hence the strength of van der Waals forces between the halogen molecules increases. This explains why chlorine exists as a gas, bromine exists as a liquid and iodine exists as a solid.
2) Relative reactivity of halogens as oxidizing agents
Chlorine is the strongest oxidizing agent followed by bromine and iodine. Recall that an oxidizing agent causes oxidation and it itself is reduced, i.e. an oxidizing agent has a tendency to gain electrons.
The electrode potential values become less positive down the group, indicating that down the group the halogens has a reduced tendency to gain electrons, hence the oxidizing power of halogen decreases.
3) Reaction of halogens with hydrogen
Chlorine reacts explosively in sunlight with hydrogen. It also reacts slowly in the dark with hydrogen.
Bromine reacts with hydrogen only at high temperatures
Iodine reacts with hydrogen to form an equilibrium mixture of hydrogen, iodine and HI.
4) Relative thermal stability of hydrides (hydrogen halides)
Thermal stability of hydrogen halides decreases down the group.
Down the group, the atomic radii of the halogen increases and the bond dissociation enthalpy of hydrogen halides decreases.
Bond energy of H--Cl : 431 kJ/mol
Bond energy of H--Br: 366 kJ/mol
Bond energy of H--I: 299 kJ/mol
Note: In the exam, students may be asked to quote bond energy values from the Data booklet. You do not have to memorize these bond energies. they are in the data booklet.
5) Reaction of halide ions with aqueous silver ions followed by aqueous ammonia
Chloride ions form a white precipitate with silver ions. White ppt is soluble in aqueous NH3
Bromide ions form a pale yellow ppt with silver ions. Yellow ppt is sparingly soluble in aqueous ammonia.
Iodide ions form a deep yellow ppt with silver ions. Deep yellow ppt is insoluble in aqueous ammonia.
Silver ions is usually supplied as silver nitrate. This is a common test for halides. Aqueous ammonia is added to differentiate between silver chloride and silveer bromide as it is very hard to differentiate between the white and pale yellow ppt.
Aqueous ammonia has to be added slowly as the pale yellow silver bromide is soluble in excess aqueous ammonia. Aqueous ammonia forms a complex with silver ions, shifting the equlibrium to the left, and increases the solubility of AgCl. AgCl is most soluble followed by AgBr and AgI, thus the addition of ammonia can fully solubilise AgCl. AgBr is only soluble in concentrated ammonia while ammonia has no effect on the solubility of AgI.
6) Reaction of halide ions with concentrated sulphuric acid
Concentrated sulphuric acid can act as a proton donor, i.e. it supplies H+ ions which reacts with the halide ion to form hydrides (HCl, HBr and HI). The hydrides appear as steamy white fumes
Concentrated sulphuric acid can also act as an oxidizing agent.
Recall that iodine is the weakest oxidizing agent, hence it is the most easily oxidized. Concentrated sulphuric acid can oxidize iodide ions to iodine, hence you will see purple vapour
Concentrated sulphuric acid can also oxidize bromide ions to bromine gas.
Concentrated sulphuric acid cannot oxidize chloride ions to chlorine gas.
7) Reaction of chlorine with NaOH
Chlorine reacts with cold NaOH to form NaCl and NaClO (sodium hypochlorate)
Note that in this reaction one atom of chlorine is undergoing oxidation and the other atom of chlorine is undergoing reduction.
Chlorine is oxidized. The oxidation number of chlorine increases from 0 in Chlorine gas to +1 in NaClO. Chlorine ia also reduced. The oxidation number of chlorine decreases from 0 in Chlorine gas to -1 in NaCl.
Chlorine reacts with hot NaOH to form NaCl and NaClO3 (sodium chlorate)
In this reaction one atom of chlorine is undergoing oxidation and the other atom of chlorine is undergoing reduction.
Chlorine is oxidized. The oxidation number of chlorine increases from 0 in Chlorine gas to +5 in NaClO3. Chlorine ia also reduced. The oxidation number of chlorine decreases from 0 in Chlorine gas to -1 in NaCl.
End of group VII notes
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Friday, June 13, 2008
Chemistry of transition elements II
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7) Transition elements in redox systems
These transition metals were highlighted in the syllabus. For e.g. Fe3+ can act as a reducing agent while it itself is oxized to Fe2+.
8) Transition metals react with ligands to form complexes
A ligand possess one or more lone pairs of electrons and donates its electrons to the transition metal ion
A complex is formed by the coordination of lone pairs of electrons from a ligand to a cation which has empty orbitals to accomodate them. Transition metal ions have incompletely filled 3d subshells to accomodate the lone pair donated by the ligand.
A coordinate bond is a covalent bond in which the shared paired of electrons is provided by only one of the bonded atoms. In the case of transition ion complexes, the shared paired of electrons is provided by the ligand
9) Ligand exchange
A stronger ligand can displace a weaker ligand from a complex ion.
The order of ligand strength is CN->NH3>Cl->H2O
CO and oxygen are ligands for Fe in haemoglobin
Haemoglobin contains an Fe2+ ion. Oxygen can act as a ligand and form coordinate bonds with Fe2+. This bonding is reversible to allow haemoglobin to release oxygen where it is needed. Carbon monoxide is a stronger ligand than oxygen. It bonds strongly to Fe2+ . This prevents Fe2+ and thus haemoglobin from bonding with oxygen. This prevents the transport of oxygen in the body eventually leading to death.
10) Transition metals can act as catalyst
Transition metals can act as heterogenous catalysts. Reactant molecules are adsorbedon the surface of the catalyst . The 3d electrons of transition metals enable the transition metal to form temporary bonds with reactant molecules, facilitating the breaking of "old" bonds and the formation of "new" bonds leading to the product formation. Examples include iron (catalyst for the Haber process), Ni (catalyst for hydrogenation).
Transition metals can also act as homogenous catalysts due to the variable oxidation states of transition metals.
For example, the oxidation of iodide ions by peroxodisulphate ions is energetically favourable but in reality the reaction takes place very slowly without a catalyst due to the fact that both reactants are negative ions thus they repel each other.
Fe2+ can act as a catalyst for this reaction. First Fe2+ reduces peroxodisulphate ions to sulphate ions and it itself is oxidized to Fe3+.
Subsequently, Fe3+ oxidizes iodide ions to iodine and Fe2+ is regenerated.
11) Transition metal complexes are coloured
In an isolated tansition metal aton, the 3d orbitals are at the same energy level (degenerate) . However in a complex ion , the d orbitals have slightly different energy due to the presence of ligands (d orbital splitting). An electron can be promoted to a higher energy level by absorbing a photon of light. The color of the complex metal ion is the frequencies of light that are not absorbed.
End of chemistry of transition metals
Link to Chemistry of transition elements I
Link to Contents Page
7) Transition elements in redox systems
These transition metals were highlighted in the syllabus. For e.g. Fe3+ can act as a reducing agent while it itself is oxized to Fe2+.
8) Transition metals react with ligands to form complexes
A ligand possess one or more lone pairs of electrons and donates its electrons to the transition metal ion
A complex is formed by the coordination of lone pairs of electrons from a ligand to a cation which has empty orbitals to accomodate them. Transition metal ions have incompletely filled 3d subshells to accomodate the lone pair donated by the ligand.
A coordinate bond is a covalent bond in which the shared paired of electrons is provided by only one of the bonded atoms. In the case of transition ion complexes, the shared paired of electrons is provided by the ligand
9) Ligand exchange
A stronger ligand can displace a weaker ligand from a complex ion.
The order of ligand strength is CN->NH3>Cl->H2O
CO and oxygen are ligands for Fe in haemoglobin
Haemoglobin contains an Fe2+ ion. Oxygen can act as a ligand and form coordinate bonds with Fe2+. This bonding is reversible to allow haemoglobin to release oxygen where it is needed. Carbon monoxide is a stronger ligand than oxygen. It bonds strongly to Fe2+ . This prevents Fe2+ and thus haemoglobin from bonding with oxygen. This prevents the transport of oxygen in the body eventually leading to death.
10) Transition metals can act as catalyst
Transition metals can act as heterogenous catalysts. Reactant molecules are adsorbedon the surface of the catalyst . The 3d electrons of transition metals enable the transition metal to form temporary bonds with reactant molecules, facilitating the breaking of "old" bonds and the formation of "new" bonds leading to the product formation. Examples include iron (catalyst for the Haber process), Ni (catalyst for hydrogenation).
Transition metals can also act as homogenous catalysts due to the variable oxidation states of transition metals.
For example, the oxidation of iodide ions by peroxodisulphate ions is energetically favourable but in reality the reaction takes place very slowly without a catalyst due to the fact that both reactants are negative ions thus they repel each other.
Fe2+ can act as a catalyst for this reaction. First Fe2+ reduces peroxodisulphate ions to sulphate ions and it itself is oxidized to Fe3+.
Subsequently, Fe3+ oxidizes iodide ions to iodine and Fe2+ is regenerated.
11) Transition metal complexes are colouredIn an isolated tansition metal aton, the 3d orbitals are at the same energy level (degenerate) . However in a complex ion , the d orbitals have slightly different energy due to the presence of ligands (d orbital splitting). An electron can be promoted to a higher energy level by absorbing a photon of light. The color of the complex metal ion is the frequencies of light that are not absorbed.
End of chemistry of transition metals
Link to Chemistry of transition elements I
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Chemistry of Transition elements I
Note: My notes based on the order of learning outcomes presented in the syllabus.
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1) Definition of transtion elements
Transition elements are d block elements which form one or more stable ions with incompletely filled subshell of d electrons
Zn for example is not a transition element because the electronic configuration of Zn2+ is [Ar]3d10, i.e. its 3d subshells are completely filled
2) Electronic configuration of transition elements
The syllabus requires that students know the electronic configuration of transition metals from Sc to Zn. Remember that the 4s subshell is filled first before the 3d subshell. There are 5 orbitals in the 3d subshell. All 5 have to be singly filled before the electrons are paired
Special cases:
Cr: [Ar]3d5 4s1
Cu:[Ar]3d10 4s1
In both cases the 4s orbital is not completely filled due to the stability associated with a fully filled d shell and a half filled d shell
3) The atomic radii, ionic radii and first ionization energy of transition elements are similar.
4)Comparison between s block and transition element
Transition elements have higher melting points, higher densities and better conductivity than s block elements. This is due to the stronger metallic bonds present in transition elements. In transition elements, the d electrons and s electrons are available to take part in delocalisation. In contrast only the s electrons are able to delocalize in the s block element.
The atomic radius, ionic radius and first ionisation energy of d block elements are similar to s block element
5) Transition elements tend to have variable oxidation states
The difference in energy between the 3d and 4s electrons is much smaller than the difference in energy between the 3s and 3p electrons, hence the 3d and 4s electrons may be lost in chemical reactions.
Link to Contents Page
1) Definition of transtion elements
Transition elements are d block elements which form one or more stable ions with incompletely filled subshell of d electrons
Zn for example is not a transition element because the electronic configuration of Zn2+ is [Ar]3d10, i.e. its 3d subshells are completely filled
2) Electronic configuration of transition elements
The syllabus requires that students know the electronic configuration of transition metals from Sc to Zn. Remember that the 4s subshell is filled first before the 3d subshell. There are 5 orbitals in the 3d subshell. All 5 have to be singly filled before the electrons are paired
Special cases:
Cr: [Ar]3d5 4s1
Cu:[Ar]3d10 4s1
In both cases the 4s orbital is not completely filled due to the stability associated with a fully filled d shell and a half filled d shell
3) The atomic radii, ionic radii and first ionization energy of transition elements are similar.
4)Comparison between s block and transition element
Transition elements have higher melting points, higher densities and better conductivity than s block elements. This is due to the stronger metallic bonds present in transition elements. In transition elements, the d electrons and s electrons are available to take part in delocalisation. In contrast only the s electrons are able to delocalize in the s block element.
The atomic radius, ionic radius and first ionisation energy of d block elements are similar to s block element
5) Transition elements tend to have variable oxidation states
The difference in energy between the 3d and 4s electrons is much smaller than the difference in energy between the 3s and 3p electrons, hence the 3d and 4s electrons may be lost in chemical reactions.
6) Prediction of likely oxidation states given a particular electronic configuration
E.g. Fe is likely to have a +2 oxidation state due to the loss of the 2 outer most 4s electrons and a +3 oxidation state due to a loss of the 4s electrons and another 3d electrons giving rise to a stable half filled 3 d configuration
Note that some transition elements have oxidation state which are less obvious to predict. In the case of Ti, it is not easy to see why the main oxidation states of Ti are +3 and +4.
Continued in Chemistry of transition elements II
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Continued in Chemistry of transition elements IILink to Contents Page
N2007/III/5
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5ai)
The requirements needed for 2 molecules to form a hydrogen bond between them are:
one molecule must have a hydrogen atom bonded to either oxygen, nitrogen, or fluorine and
the second molecule must have a lone pair on an oxygen, nitrogen, or fluorine which can then interact with the hydrogen on the first molecule.
For example, hydrogen bonding can occur between 2 water molecules. The hydrogen atom bonded to oxygen in one water molecule can interact with the oxygen atom of the other water molecule.
5aii)
5ai)
The requirements needed for 2 molecules to form a hydrogen bond between them are:
one molecule must have a hydrogen atom bonded to either oxygen, nitrogen, or fluorine and
the second molecule must have a lone pair on an oxygen, nitrogen, or fluorine which can then interact with the hydrogen on the first molecule.
For example, hydrogen bonding can occur between 2 water molecules. The hydrogen atom bonded to oxygen in one water molecule can interact with the oxygen atom of the other water molecule.
5aii)
Carboxylic acids consist of a non polar hydrocarbon chain and a polar carboxylate group. Hydrogen bonding occurs between the carboxylate groups of acid molecules. Also Van Der Waals forces exists between the non polar hydrocarbon chains . Water molecules interact with each other through hydrogen bonding.
The first 4 members of the carboxylic acid series has short hydrocarbon chains and the predominant interactions between acid molecules is H bonding which is similar to the H bonding that exists between the water molecules hence they are soluble in water
For members of the the carbocxylic acid series with longer side chains, the predominant interaction between acid molecules is Van Der Waals forces due to the long hydrocarbon chain. This interaction is weaker than the hydrogen bonding that exists between water molecules, hence it is not favourable for the carboxylic acid to dissolve in water.
Side note: this question is surprisingly hard to answer. If anyone has a better answer please post it in the comments. chemguide.co.uk has a reasonably good explanation. in fact i can't find any other explanation on Google. My answer is mainly based on their explanation.
Wednesday, June 11, 2008
N2007/III/2
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2ei)
This question maylook intimidating at the first glance. The way to approach this question is to look for clues in the question itself. First you would notice that the ratio of carbon atoms to hydrogen atoms is close to 1:1. Most of the time, this 1:1 ratio suggests that a benzene ring is present. This is supported by the structure shown in C which indeed contains a benzene ring.
A is able to react with thionyl chloride SOCl2. This suggest that an alcohol function group or a carboxylic acid functional group is present. Alcohols react with thionyl chloride to form alkyl chlorides and carboxylic acids react with thionyl chloride to form acyl chlorides
However the alkyl chloride product does not have any oxygen atoms. This suggest that the functional group present is carboxylic acid .
2eii)
Reaction I is an acid base reaction
OH- abstracts a proton from the amide linkage forming water, N- and Na+.
Reaction II is a Nucleophilic substitution reaction.
The N- generated acts as a nucleophile and substitutes bromine.
Recall that the C-Br bond is polar, with carbon carrying a slight positive charge. The slightly positive carbon attracts the N-.
Reaction III is a hydrolysis reaction.
The amide linkage is hydrolyzed to carboxylic acid and amine functional groups
2eiii)
Reagent: HCl
Condition:Heat
Side note: You would notice that for essay questions, the examiners like to mix 2 different topics in one question. In this case the first portion of the question is about inorganic chemistry (which i have not solved) and the second portion is about organic chemistry. This style of questioning is especially detrimental to students who like to "spot" topics. My advice is dont spot topics. Moreover for organic chemistry the topics only make sense if you study all of them together as they are all interlinked.
For those who are interested, the diagram below illustrates how the cyclic amide linkage is formed.
2ei)
This question maylook intimidating at the first glance. The way to approach this question is to look for clues in the question itself. First you would notice that the ratio of carbon atoms to hydrogen atoms is close to 1:1. Most of the time, this 1:1 ratio suggests that a benzene ring is present. This is supported by the structure shown in C which indeed contains a benzene ring.
A is able to react with thionyl chloride SOCl2. This suggest that an alcohol function group or a carboxylic acid functional group is present. Alcohols react with thionyl chloride to form alkyl chlorides and carboxylic acids react with thionyl chloride to form acyl chlorides
However the alkyl chloride product does not have any oxygen atoms. This suggest that the functional group present is carboxylic acid .
2eii)
Reaction I is an acid base reaction
OH- abstracts a proton from the amide linkage forming water, N- and Na+.
Reaction II is a Nucleophilic substitution reaction.
The N- generated acts as a nucleophile and substitutes bromine.
Recall that the C-Br bond is polar, with carbon carrying a slight positive charge. The slightly positive carbon attracts the N-.
Reaction III is a hydrolysis reaction.
The amide linkage is hydrolyzed to carboxylic acid and amine functional groups
2eiii)
Reagent: HCl
Condition:Heat
Side note: You would notice that for essay questions, the examiners like to mix 2 different topics in one question. In this case the first portion of the question is about inorganic chemistry (which i have not solved) and the second portion is about organic chemistry. This style of questioning is especially detrimental to students who like to "spot" topics. My advice is dont spot topics. Moreover for organic chemistry the topics only make sense if you study all of them together as they are all interlinked.
For those who are interested, the diagram below illustrates how the cyclic amide linkage is formed.
Tuesday, June 10, 2008
Mark scheme for Nov 2006 Paper 2
I chanced upon this on the cambridge international examinations website.
http://www.cie.org.uk/docs/dynamic/4619.pdf
There are also specimen papers with mark schemes. Do check out their website.
Update:
I think the CIE website is extremely useful. Other than the mark scheme, they also provide examiners report. As a student i never knew that such reports were available online. In addition, due to the recent changes in syllabus the website also provides specimen papers and answers for these specimen papers. I would strongly encourage students to do the questions by themselves first before refering to the mark scheme. I emphasize again, do the questions by yourself first. Then compare your answers to the ones given by the examiners, it would be highly beneficial. If there are any teachers reading this blog, please tell your students about this website.
Based on personal experience, the actual exam is likely to be very similar to the specimen paper in terms of the style of the questions, so familiarize yourself with the specimen paper. Dont just do it once, do it twice or thrice.
Links to Contents page
http://www.cie.org.uk/docs/dynamic/4619.pdf
There are also specimen papers with mark schemes. Do check out their website.
Update:
I think the CIE website is extremely useful. Other than the mark scheme, they also provide examiners report. As a student i never knew that such reports were available online. In addition, due to the recent changes in syllabus the website also provides specimen papers and answers for these specimen papers. I would strongly encourage students to do the questions by themselves first before refering to the mark scheme. I emphasize again, do the questions by yourself first. Then compare your answers to the ones given by the examiners, it would be highly beneficial. If there are any teachers reading this blog, please tell your students about this website.
Based on personal experience, the actual exam is likely to be very similar to the specimen paper in terms of the style of the questions, so familiarize yourself with the specimen paper. Dont just do it once, do it twice or thrice.
Links to Contents page
Monday, June 9, 2008
N2007/III/1
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1ci)
Reaction II involves the oxidation of aldehyde to carboxylic acid
Reagent: Potassium dichromate (aq)
Condition: r.t.p
Reaction III involves the reduction of aldehyde to alcohol
Reagents: LiAlH4 in organic sovent
Conditions: Heat
1cii)
butan-1-ol --> 1-bromobutane
this reaction involves the conversion of alcohol to alkyl halides
Reagent: PBr3
Condition: r.t.p
Side note: Alkyl halides are very useful in organic synthesis. For example, you can add carbon atoms to alkyl halides by reaction with CN
1di)
The rate of hydrolysis of 2-bromobutane will be slower than 2 iodobutane. This is because iodine is less electronegative than bromine, thus it forms the weakest bond to carbon and the iodide ion is a better leaving group than bromide ion.
1dii)
Alkyl halides can undergo nucleophilic substitution. The cyanide ion can be used to extend the carbon chain.
LiAlH4 is a reducing agent. It can reduce nitriles to amines
When alkyl halides react with ethanolic NaOH, dehydrohalogenation occurs. As a result an alkene is formed. This is in contrast to aqueous NaOH. Alkyl halides undergo hydrolysis in the presence of NaOH(aq). Subsequently in the presence of cold KMnO4, the alkene is oxidized to a diol
e)
In the presence of UV light , butane undergoes free radical substitution with bromine. There are 6 hydrogen atoms that can be abstracted from the terminal carbon atoms compared to the 4 hydrogen atoms that can be abstracted from the middle 2 carbon atoms. Hence 1 bromobutane should be the more abundant product statistically.
Side note: Experimentally, it has been determined that 2 bromobutane is the more abundant product because of the greater stability of the secondary radical compared to the primary radical.
I am not very sure about the correct answer to this question. This is because i think that the stability of radicals is not in the A level syllabus. Based on the syllabus, students are required to describe the free radical sustitution mechanism. I dont think that includes the stability of radicals.
Links to Contents page
1ci)
Reaction II involves the oxidation of aldehyde to carboxylic acid
Reagent: Potassium dichromate (aq)
Condition: r.t.p
Reaction III involves the reduction of aldehyde to alcohol
Reagents: LiAlH4 in organic sovent
Conditions: Heat
1cii)
butan-1-ol --> 1-bromobutane
this reaction involves the conversion of alcohol to alkyl halides
Reagent: PBr3
Condition: r.t.p
Side note: Alkyl halides are very useful in organic synthesis. For example, you can add carbon atoms to alkyl halides by reaction with CN
1di)
The rate of hydrolysis of 2-bromobutane will be slower than 2 iodobutane. This is because iodine is less electronegative than bromine, thus it forms the weakest bond to carbon and the iodide ion is a better leaving group than bromide ion.
1dii)
Alkyl halides can undergo nucleophilic substitution. The cyanide ion can be used to extend the carbon chain.
LiAlH4 is a reducing agent. It can reduce nitriles to amines
When alkyl halides react with ethanolic NaOH, dehydrohalogenation occurs. As a result an alkene is formed. This is in contrast to aqueous NaOH. Alkyl halides undergo hydrolysis in the presence of NaOH(aq). Subsequently in the presence of cold KMnO4, the alkene is oxidized to a diol
e)
In the presence of UV light , butane undergoes free radical substitution with bromine. There are 6 hydrogen atoms that can be abstracted from the terminal carbon atoms compared to the 4 hydrogen atoms that can be abstracted from the middle 2 carbon atoms. Hence 1 bromobutane should be the more abundant product statistically.
Side note: Experimentally, it has been determined that 2 bromobutane is the more abundant product because of the greater stability of the secondary radical compared to the primary radical.
I am not very sure about the correct answer to this question. This is because i think that the stability of radicals is not in the A level syllabus. Based on the syllabus, students are required to describe the free radical sustitution mechanism. I dont think that includes the stability of radicals.
Links to Contents page
How to make full use of the worked examples?
It is important to try to do the questions by yourself first. You will benefit most by thinking through the question first and writing down your answers. By writing down your answers, you will remember them. During revision, i would encourage students to refer to their lecture notes while attempting the questions. As you do more questions, you will realize that you can remember more stuff and you would be less dependent on your notes.
Also if anyone has any questions that you cant solve, please leave a comment, I will try to solve them as soon as possible.
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Also if anyone has any questions that you cant solve, please leave a comment, I will try to solve them as soon as possible.
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Sunday, June 8, 2008
First worked example: N2007/II/5
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Due to copyright issues, I cannot put up the question online. So i will refer to the question using the standard notation. E.g N2004/II/5a refers to the Nov 2004 Paper II Question 5a
First worked example : N2007/II/5
Look at the different compounds and identify functional groups present in each compound
Compound A contains the alkene functional group and aldehyde functional group
Compound B contains the alkene and ester linkage functional group. Be aware that the ester functional group OCO is different from the carbonyl functional group C=O
Compound C contains the alcohol(phenol) and alkene functional group
Compound D contains the alkene functional group
5a)
the reagent given is a acyl chloride
acyl chloride + alcohol --> ester + HCl
I still haven figure out the best way to draw structural formula and upload online. Once i think of something i will upload the structural formula
Ans: C
dilute HNO3 is able to catalyze the hydrolysis of esters
Esters are hydrolyzed to give carboxylic acids and alcohol
Ans: B
Na is able to react with phenol functional group to form the phenoxide ion
Ans: C
Tollen's reagent is actually an oxidizing agent which can oxidize aldehydes to carboxylic acid. However the reagent is alkaline hence the carboxylate ion is formed. Just be careful when you draw the structural formula.
Ans: A
5bi)
As mentioned earlier compound C has 2 functional groups: the alkene and phenol functional group. Only the alkene functional group will react with Br(aq) . Thus Br and OH will be added across the double bond
5bii)
reflux with hot concentrated acidified manganate will result in the oxidative cleavage of the double bond
As a result 2 carboxylic acids will be formed
5ci)
Pardon my bad drawing. Basically the aldehyde is being reduced to a primary alcohol
5cii)
Stage I
reagent: LiAlH4
Conditions: organic solvent, heat
StageII
reagent: ethanoic acid
Condition: dil HCl, heat
Ok this is the end of the first worked example. Please feel free to leave comments.
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Due to copyright issues, I cannot put up the question online. So i will refer to the question using the standard notation. E.g N2004/II/5a refers to the Nov 2004 Paper II Question 5a
First worked example : N2007/II/5
Look at the different compounds and identify functional groups present in each compound
Compound A contains the alkene functional group and aldehyde functional group
Compound B contains the alkene and ester linkage functional group. Be aware that the ester functional group OCO is different from the carbonyl functional group C=O
Compound C contains the alcohol(phenol) and alkene functional group
Compound D contains the alkene functional group
5a)
the reagent given is a acyl chloride
acyl chloride + alcohol --> ester + HCl
I still haven figure out the best way to draw structural formula and upload online. Once i think of something i will upload the structural formula
Ans: C
dilute HNO3 is able to catalyze the hydrolysis of esters
Esters are hydrolyzed to give carboxylic acids and alcohol
Ans: B
Na is able to react with phenol functional group to form the phenoxide ion
Ans: C
Tollen's reagent is actually an oxidizing agent which can oxidize aldehydes to carboxylic acid. However the reagent is alkaline hence the carboxylate ion is formed. Just be careful when you draw the structural formula.
Ans: A
5bi)
As mentioned earlier compound C has 2 functional groups: the alkene and phenol functional group. Only the alkene functional group will react with Br(aq) . Thus Br and OH will be added across the double bond
5bii)
reflux with hot concentrated acidified manganate will result in the oxidative cleavage of the double bond
As a result 2 carboxylic acids will be formed
5ci)
Pardon my bad drawing. Basically the aldehyde is being reduced to a primary alcohol5cii)
Stage I
reagent: LiAlH4
Conditions: organic solvent, heat
StageII
reagent: ethanoic acid
Condition: dil HCl, heat
Ok this is the end of the first worked example. Please feel free to leave comments.
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