Happy Christmas

Wishing you all a very Happy Christmas

See if you can work out all the Chemistry involved with this simple demonstration of copper in silver nitrate solution. 

Reactivity Series of Metals

Note that Al displays much less reactivity that it actually has, due to an oxide coating over its surface which is unreactive. That's why it can be used for mountain bikes and window frames. The reactivity series of metals is as follows:

Melting Point Trends in Period Three

The following link shows a graph of the trend and gives some explanation. Notice that it gives a different value for phosphorous than the worksheet that I gave. Where's the mistake? Check with the data booklet and also the interactive periodic table site.

Melting and boiling points across period three



Periodic Table

Periodicity Video Clips

some video links

Homework. Period Three Trends

Period Three Trends

Don’t panic we will do bonding next and the trends can be simplified:

Sodium, magnesium and Aluminium are metals. Therefore, they have typical metallic properties i.e. shiny, malleable, and conduct electricity. The delocalized electrons can move and therefore, maintain the metallic bond even if the shape is changed, in addition they can move and so an electric current can flow. Si, P S Cl and Ar are non- metals. Ar is a noble gas and so exits as atoms but the others have covalent bonds, either as a giant covalent structure i.e. Si or as simple covalent molecules. The melting point will depend very much on the type of covalent bonding. Giant covalent substances will have very high melting points, as the strong covalent bonds need to be broken. However, simple molecular substances will have relatively low melting points due to the weak forces of attraction between the molecules (Van der Waal’s). 

The metallic oxides are ionic. These are basic and produce hydroxide solutions. Al₂O₃ is an exception as it has both kinds of bonding, ionic and covalent and so can react with acids and bases. It is said to be amphoteric. However, in water in does not dissolve and so won't affect the pH.

The covalent oxides are acidic; they react with water to form an acid. 

The ionic oxides are neutral or slightly acidic. Any slight acidity is just due to protons being pulled off water molecules, releasing H⁺ ions. The smaller and more highly charged the ion, the more this happens.

The covalent chlorides react with water to form two acids (usually), one of them being HCl.  ACl₃ is strange in that it is actually a covalent compound and is made up of a metal and a non-metal. It produces HCl when it reacts with water. 

Your homework is to: 

• Produce a neat table of results from your experimental observations
• Write an equation for all of the reactions that occurred. (include the oxides and chlorides that we didn’t have)
• Make conclusions from your results about the trends of the elements, oxides and chlorides of period three that you saw
• Explain the conclusions that you made and relate them to the type of bonding

Use your notes form the lesson, the notes / links on the blog and your text book

Period Three Trends

The following is a link to the IB website and it has good summary tables for the reactions of the oxides and chlorides. Remember to think about the type of bonding when you are explaining the trends of the period three elements, oxides and chlorides.

Period Three Trends

Chem Guide Period Three

Questions on the Halogens

Group (VII) (The Halogens)

1.      Explain the trend in atomic size in descending the group.

Each time the group is descended by one, a new quantum level is added. Therefore, the distance from the nuclear charge to the outer electrons increases and so does the screening effect, therefore, the attraction on the outer electrons is reduced and the atoms get bigger.

2.      What trend in electron affinities would you predict?

As the atomic size (distance from outer electrons to the nuclear charge) and the screening effect increases going down the group, the attraction on an added electron from the nuclear charge will decrease. Therefore, I would expect the electron affinity to decrease going down the group.

3.      Based on number 2, what trends in reactivity would you expect? Give reasons.

The halogens have seven electrons in their outer shell and so react by obtaining one more (either by sharing in covalent bonding or gaining one in ionic bonding). One of the factors influencing their reactivity will be their ability to gain an electron. This will be greatest when they are smallest with the least screening  (greatest attraction). Therefore, the reactivity should decrease going down the group or increase going up the group.

4.      The actual values for the electron affinity are; F_(g)   -328 (KJ/mol), Cl_(g)   

-349, Br_(g) -324 and I_(g) -245

What is surprising is that Cl has a higher value (more heat released) than F. This is because that F is such a small atom that the electrons in the outer quantum level are closely packed. This causes there to be more repulsion for an incoming electron and therefore, the overall heat energy released is actually less than for Cl. However, F is still more reactive than Cl and this is because in a reaction other things are occurring, such as: Atomisation and hydration energy (if its in a solution). Flourine has the lowest atomization energy and the greatest hydration energy. This compensates for the slightly lower electron affinity value. Therefore, overall fluorine is the most reactive of the halogens.

You may find the atomisation data for fluorine and chlorine surprising as they are both gases at room temperature. Therefore, atomisation involves breaking the bond between the atoms. Look at the following link which includes bond energy data, and all should be clear to you.

Chem Guide Group 7 properties

Homework The Halogens

The Halogens

Please make sure that all your observations for the reaction with iron wool and with the solutions; KI_(aq), KBr_(aq) and KCl_(aq) are recorded. Please write equations for the reactions and explain the trend in reactivity.

E.g. Cl₂ + 2I^- →  I₂ +  2Cl^-

Or showing the two half equations:  Cl₂ + 2e^- → 2Cl^-    (reduction)

                                                                  2I^- → I₂ + 2e^-        (oxidation)

This illustrates that chlorine can oxidise iodide ions and that chlorine is more reactive than iodine.

Period three reactions with water

Period three reactions with water

The oxidising ability of the halogens

look at the following link:

The Oxidising power of the halogens

Electron Affinity Trends

look at this link on electronegativity trends. Notice how these trends sometimes fall down, e.g. the value for F(g) is lower than the trend would suggest. Can you think why this is so?

Electron Affinity Trends


Note: This is not the same as electronegativity. Electronegativity values relate to the ability of an atom within a covalent bond to pull the bonding electrons towards itself. A significant difference in electronegativity values leads to a polar bond. This may or may not lead to a polar molecule, depending if the dipoles cancel due to symmetry in the molecule. If they cancel the molecule will be overall non polar, if they don't it will be polar.

Periodic Patterns

This link gives you a menu to select physical and chemical properties and the trends in the Periodic Table:

Trends in the Periodic Table

Ionisation Energy Trends

This is a useful link that explains the trend in ionisation values in the Periodic Table:

Trends in ionisation energy

Group 1 Metals Reacting With Water

Can you spot the "interesting science" in this video?

Explain carefully the trend in reactivity.

Notes from Atomic Structure Test

Notes from atomic Structure test:

The following are points that you need to clarify in order to improve your understanding and marks on this unit:

·      Copper has the electronic configuration of [Ar] 3d¹⁰ 4s¹ (remember also that Cr has a half filled d subshell)

·      Although the 4s orbital fills before the 3d. The 4s electrons are removed first.

·      The definition of RAM! The weighted average of an atom relative to a 1/12 of carbon-12

·      The sequence in a mass spectrometer is: vaporization, ionization, acceleration, deflection and detection

·      Acceleration is caused by a very high positive potential difference

·      Deflection is caused by an electromagnet around the main tube, causing the positive ions to move in a curved path

·      The amount of deflection depends on the mass to charge ratio (m/z). The higher the mass and lower the charge the bigger the m/z ratio and the less the deflection. The lower the m/z ratio, the smaller the mass and or the bigger the charge, the greater the deflection

·      Big jumps in ionization energy relate to breaking into the next quantum level. This value is much higher as its nearer the nucleus and the quantum level is full

Atomic Structure Summary notes

This is a useful summary and links to the syllabus. You can download it via facebook, or just see it through this link:

Atomic Structure Summary

Video clip on orbitals

You may find this useful. here is a video clip of another teachers's lecture on this. What do you think? Does it make more sense than mine? The clip below shows the orientation of the orbitals.

Internal Assessments

You need to look carefully at the assessment criteria given in the interactive IB site. A link to this is on this blog. In addition you may find the following link useful.

Internal assessment

Link to Assessment criteria

Ferromagnetic, Dimagnetic and Paramagnetic

These are easily confused, here are some definitions:


Diamagnetic materials have a weak, negative susceptibility to magnetic fields. Diamagnetic materials are slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed. In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom. Diamagnetic properties arise from the realignment of the electron paths under the influence of an external magnetic field. Most elements in the periodic table, including copper, silver, and gold, are diamagnetic.

Paramagnetic materials have a small, positive susceptibility to magnetic fields. These materials are slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed. Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field. Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum.

Ferromagnetic materials have a large, positive susceptibility to an external magnetic field. They exhibit a strong attraction to magnetic fields and are able to retain their magnetic properties after the external field has been removed. Ferromagnetic materials have some unpaired electrons so their atoms have a net magnetic moment. They get their strong magnetic properties due to the presence of magnetic domains. In these domains, large numbers of atom's moments (10¹² to 10¹⁵) are aligned parallel so that the magnetic force within the domain is strong. When a ferromagnetic material is in the unmagnitized state, the domains are nearly randomly organized and the net magnetic field for the part as a whole is zero. When a magnetizing force is applied, the domains become aligned to produce a strong magnetic field within the part. Iron, nickel, and cobalt are examples of ferromagnetic materials. Components with these materials are commonly inspected using the magnetic particle method.

http://www.ndt-ed.org/EducationResources/CommunityCollege/MagParticle/Physics/MagneticMatls.htm

Electronic Configuration and Quantum Numbers

The following gives a summary:

Traditional nomenclature

Many different models have been proposed throughout the history of quantum mechanics, but the most prominent system of nomenclature spawned from the Hund-Mulliken molecular orbital theory of Friedrich Hund, Robert S. Mulliken, and contributions from Schrödinger, Slater and John Lennard-Jones. This system of nomenclature incorporated Bohr energy levels, Hund-Mulliken orbital theory, and observations on electron spin based on spectroscopyand Hund's rules.

This model describes electrons using four quantum numbers, n, ℓ, m_ℓ, m_s. It is also the common nomenclature in the classical description of nuclear particle states (e.g. protons and neutrons).

• The first, n, describes the electron shell, or energy level.
  • The value of n ranges from 1 to "n", where "n" is the shell containing the outermost electron of that atom. For example, in cesium (Cs), the outermost valence electron is in the shell with energy level 6, so an electron in cesium can have an n value from 1 to 6. This is known as the principal quantum number.
• The second, ℓ, describes the subshell (0 = s orbital, 1 = p orbital, 2 = d orbital, 3 = f orbital, etc.).
  • The value of ℓ ranges from 0 to n − 1. This is because the first p orbital (ℓ = 1) appears in the second electron shell (n = 2), the first d orbital (ℓ = 2) appears in the third shell (n = 3), and so on. A quantum number beginning in 3, 0, … describes an electron in the s orbital of the third electron shell of an atom.
• The third, m_ℓ, describes the specific orbital (or "cloud") within that subshell.*
  • The values of m_ℓ range from −ℓ to ℓ. The s subshell (ℓ = 0) contains only one orbital, and therefore the m_ℓ of an electron in an s subshell will always be 0. The p subshell (ℓ = 1) contains three orbitals (in some systems, depicted as three "dumbbell-shaped" clouds), so the m_ℓ of an electron in a p subshell will be −1, 0, or 1. The d subshell (ℓ = 2) contains five orbitals, with m_ℓ values of −2, −1, 0, 1, and 2.
• The fourth, m_s, describes the spin of the electron within that orbital.*
  • An electron can have a spin of ±½, m_s will be either, corresponding with "spin" and "opposite spin." Each electron in any individual orbital must have different spins, therefore, an orbital never contains more than two electrons.

* Note that, since atoms and electrons are in a state of constant motion, there is no universal fixed value for m_ℓ and m_s values. Therefore, the m_ℓ and m_s values are defined somewhat arbitrarily. The only requirement is that the naming schematic used within a particular set of calculations or descriptions must be consistent (e.g. the orbital occupied by the first electron in a p subshell could be described as m_ℓ = −1 or m_ℓ = 0, or m_ℓ = 1, but the m_ℓ value of the other electron in that orbital must be the same, and the m_ℓ assigned to electrons in other orbitals must be different).

These rules are summarized as follows:

name	symbol	orbital meaning	range of values	value example
principal quantum number	n	shell	1 ≤ n	n = 1, 2, 3, …
azimuthal quantum number (angular momentum)	ℓ	subshell (s orbital is listed as 0, p orbital as 1 etc.)	0 ≤ ℓ ≤ n − 1	for n = 3: ℓ = 0, 1, 2 (s, p, d)
magnetic quantum number, (projection of angular momentum)	mℓ	energy shift (orientation of the subshell's shape)	−ℓ ≤ mℓ ≤ ℓ	for ℓ = 2: mℓ = −2, −1, 0, 1, 2
spin projection quantum number	ms	spin of the electron (−½ = counter-clockwise, ½ = clockwise)	−½, ½	for an electron, either: −½, ½

Example: The quantum numbers used to refer to the outermost valence electrons of the Carbon (C) atom, which are located in the 2p atomic orbital, are; n = 2 (2nd electron shell), ℓ = 1 (p orbital subshell), m_ℓ = 1, 0 or −1, m_s = ½ (parallel spins).

Source: Wikipedia

Introduction to the wave nature of electrons and orbitals

Orbitals

Wave nature of electrons

Atomic Orbitals

Atomic Orbitals (chemguide)

Electronic Structure and Atomic Orbitals

Electrons in Atoms

The Hydrogen Spectrum

The Hydrogen Spectrum

The mass spectrometer

The Mass Spectrometer

Mass Spectra Data

Periodic Table

link to Periodic Table

Periodic Table

Homework

Write up the Charle's law experiment following the IB critirea for DCP and CE.

Homework and test date

Please finish the questions: "The Ideal Way" and "Some Ideal Questions"

You need to revise for your test on Quantative chemistry on Thursday 6th Oct.

Notes on The Gas Laws

Boyle’s Law:     Pressure is inversely proportional to the volume   P x V = constant

                          P₁V₁=P₂V₂ 

You need to know the graphs, pressure against volume and pressure against (1/V)

Charle’s Law:    Volume is proportional to the temperature. V/T = constant

                            V₁/T₁ = V₂/T₂

                                      You need to know the graph of Volume against temperature (⁰C) and how it can be    

                             extrapolated to give a value of absolute zero. Should be -273 ^oC (0K)

The Ideal Gas Law and the gas constant:

The volume of a gas depends on:

·         The temperature

·         The number of moles

·         The pressure

The volume of the gas particles themselves is insignificant compared to the space between them. Therefore, it doesn’t matter which gas you are referring to, one mole of gas at a fixed temperature and pressure will occupy the same volume. One mole of any gas at 1atm and 273K will occupy 22.4 dm³ (litres).

So the gas constant for one mole of a gas under standard conditions (STP, 1atm, 273K) can be easily worked out. However, its value depends on if you use atmosphere (atm) for pressure or pascals (Pa) or KPa.

For 1atm, 273 K. Note: if you are using atm use dm³  as your volume units

PV/T = constant        = 1 x 22.4/273   = 0.082 atm dm³ k^-1 mol^-1

1 atm = 101317 Pa  Note: 1Pa = 1 N/m². Remember pressure is force/area. So you have to use m³ for the units of volume if you are using pascals and dm³   if you are using kPa.

PV/T = constant     = (101325 x 0.0224)/273  = 8.314 Pa m³ k^-1 mol^-1   

Note: 1 N/m = J  so it can be written as: 8.314 J K^-1 mol^-1

So for n moles PV/T = nR    where R is the gas constant.

So PV = nRT

Unit conversions:

1atm = 101325Pa = 101.325kPa = 760 mm Hg

Homework

Homework PG

Complete the uncertainty and % error calculations for your titration experiment.

Make sure that you have tried the virtual labs on Boyle’s and Charle’s law. Write notes about these gas laws

Boyle's Law

Charle's Law

Error, Uncertainty and Significant figures

Error and uncertainty in practical work

The error is the difference between the result obtained and the generally accepted 'correct' result found in the data book or other literature. If the 'correct' result is available it should be recorded and the percentage error calculated and commented upon in your conclusion. Without the 'correct ' value no useful comment on the error can be made.

The percentage error is equal to:

the difference between the value obtained and the literature value    x 100
                              the literature value

Uncertainty occurs due to the limitations of the apparatus itself and the taking of readings from scientific apparatus. For example during a titration there are generally four separate pieces of apparatus, each of which contributes to the uncertainty.

When making a single measurement with a piece of apparatus then the absolute uncertainty and the percentage uncertainty can both be stated relatively easily. For example consider measuring 25.0 cm³ with a 25 cm³ pipette which measures to + 0.1 cm³. The absolute uncertainty is 0.1 cm³ and the percentage uncertainty is equal to:

                        0.1  x 100   =  0.4%

                       25.0

If two volumes or two masses are simply added or subtracted then the absolute uncertainties are added. For example suppose two volumes of 25.0 cm³ + 0.1 cm³ are added. In one extreme case the first volume could be 24.9 cm³ and the second volume 24.9 cm³ which would give a total volume of 48.8 cm³. Alternatively the first volume might have been 25.1 cm³ which when added to a second volume of 25.1 cm³ gives a total volume of 50.2 cm³. The final answer therefore can be quoted between 48.8 cm³ and 50.2 cm³, that is, 50.0 cm³ + 0.2 cm³.

 

When using multiplication, division or powers then percentage uncertainties should be used during the calculation and then converted back into an absolute uncertainty when the final result is presented. For example, during a titration there are generally four separate pieces of apparatus, each of which contributes to the uncertainty.

e.g. when using a balance that weighs to + 0.001 g the uncertainty in weighing 2.500 g will equal

                               0.001 x 100 = 0.04%

                               2.500

Similarly a pipette measures 25.00 cm³ + 0.04 cm³.

The uncertainty due to the pipette is thus 0.04 x 100 = 0.16%

                                                                  25.00

Assuming the uncertainty due to the burette and the volumetric flask is 0.50% and 0.10% respectively the overall uncertainty is obtained by summing all the individual uncertainties:

Overall uncertainty = 0.04 + 0.16 + 0.50 + 0.10 = 0.80% ~ 1.0%

Hence if the answer is 1.87 mol dm^-3 the uncertainty is 1.0% or 0.0187 mol dm^-3

The answer should be given as 1.87 + 0.02 mol dm^-3.

 

If the generally accepted ‘correct’ value (obtained from the data book or other literature) is known then the total error in the result is the difference between the literature value and the experimental value divided by the literature value expressed as a percentage. For example, if the ‘correct’ concentration for the concentration determined above is 1.90 mol dm^-3 then:

the total error = (1.90 – 1.87)  x 100  = 1.6%.

                             1.9

Significant figures

Whenever a measurement of a physical quantity is taken there will be uncertainty in the reading. The measurement quoted should include the first figure that is uncertain. This should include zero if necessary. Thus a reading of 25.30^oC indicates that the temperature was taken with a thermometer that is accurate to + or - 0.01^oC. If a thermometer accurate to only + or - 0.1^oC was used the temperature should be recorded as 25.3^oC.

Zero can cause problems when determining the number of significant figures. Essentially zero only becomes significant when it comes after a non-zero digit (1,2,3,4,5,6,7,8,9).

000123.4        0.0001234                                1.0234         1.2340

              zero not a significant figure               zero is a significant figure

              values quoted to 4 sig. figs.              values quoted to 5 sig. figs.

Zeros after a non-zero digit but before the decimal point may or may not be significant depending on how the measurement was made. For example 123 000 might mean exactly one hundred and twenty three thousand or one hundred and twenty three thousand to the nearest thousand. This problem can be neatly overcome by using scientific notation.

            1.23000 x 10⁶  quoted to six significant figures

            1.23 x 10⁶        quoted to three significant figures.

Calculations.

1. When adding or subtracting it is the number of decimal places that is important.

e.g.      7.10 g          +          3.10 g          =         10.20 g

          3 sig. figs.                3 sig. figs.                4 sig. figs.

This answer can be quoted to four significant figures since the balance used in both cases was accurate to + or - .01g.

2.  When multiplying or dividing it is the number of significant figures that is important. The number with the least number of significant figures used in the calculation determines how many significant figures should be used when quoting the answer.

e.g. When the temperature of 0.125 kg of water is increased by 7.2^oC the heat required =

       0.125 kg  x  7.2^oC  x 4.18 kJ kg^{-1 o}C^-1 = 3.762 kJ.

Since the temperature was only recorded to two significant figures the answer should strictly be given as 3.8 kJ.

In practice the IB does not tend to penalise in exams if the number of significant figures in an answer differs by one from the correct number (unless the question specifically asks for them) but will penalise if they are grossly wrong.

 Source: Old file, however I think it's from the Geoff Nuess book, Chemistry Companion

Homework

Homework for Tuesday:

Write up the lab

Method – Use the past tense. You should include the names of the apparatus that you used: Burette, volumetric pipette (25.0 cm3), burette, phenolphthalein indicator solution etc

Results- Use a clear table with units and uncertainties

Calculations- Show the step by step calculations in order to calculate the concentration of the unknown

Remember the step by step guide:

Step1. Write a balanced equation

Step2. Calculate the number of moles in the solution with a known concentration (the standard solution)

Step3. Deduce from the ratio in the equation the number of moles in the solution with unknown concentration. The sulphuric acid in this case

Step4. Calculate the concentration of the unknown

Step5. Remember that the solution of unknown concentration was diluted by a factor of four. Therefore, you need to multiply by four to get the concentration of the original acid solution

Evaluation

You need to list the sources of error and evaluate each one in terms of how it could have affected your result

You need to suggest realistic improvements that relate to the weaknesses that you have identified

The Gas Laws

Here is a good site where you can carry out virtual experiments in order to find out the relationship between pressure and volume and between volume and temperature. You can put the data in excel and plot the appropriate graphs: pressure against 1/V  and Volume against T (kelvin). To see the relationships.

The Gas laws virtual site

This is also a good site
Gas Laws

Homework

The first four questions from the concentrate on concentration package.