1.1-revision-guide-atomic-structure-aqa.pdf

1.1 Atomic Structure Details of the three sub-atomic (fundamental) particles Particle Position Relative Mass Relative Charge proton nucleus 1 +1 neutron nucleus...

1.1 Atomic Structure Details of the three sub-atomic (fundamental) particles Particle Position Relative Mass Relative Charge proton nucleus 1 +1 neutron nucleus 1 0 There are various electron orbitals 1/1840 -1 models for atomic structure An atom of Lithium (Li) can be represented as follows: Mass number 7 Li Atomic symbol Atomic number 3 The atomic number, Z, is the number of protons in the nucleus. The mass number ,A, is the total number of protons and neutrons in the atom. Number of neutrons = A - Z Isotopes Isotopes are atoms with the same number of protons, but different numbers of neutrons. Isotopes have similar chemical properties because they have the same electronic structure. They may have slightly varying physical properties because they have different masses. The time of flight mass spectrometer Ionisation Acceleration Detection The mass spectrometer can be Ion drift area area area area used to determine all the isotopes present in a sample of an element and to therefore Ion identify elements. detector It needs to be under a vacuum otherwise air particles would Heavy Light ionise and register on the ions ions detector Time measurement The following are the essential 4 steps in a mass spectrometer. Learn all these steps 1. Ionisation carefully! The sample can be ionised in a number of ways. Two of these techniques are electron impact and electrospray ionisation Electron impact Electron impact is used for elements A vaporised sample is injected at low pressure and substances with low formula mass. An electron gun fires high energy electrons at the sample Electron impact can cause larger organic This knocks out an outer electron molecules to fragment. Forming positive ions with different charges e.g. Ti (g) Ti+ (g)+ e– Electro spray Ionisation The sample is dissolved in a volatile, polar solvent Electro spray ionisation is used injected through a fine needle giving a fine mist or aerosol preferably for larger organic the tip of needle has high voltage molecules. The ‘softer’ conditions at the tip of the needle the sample molecule, M, gains a proton, H+, from the of this technique mean solvent forming MH+ fragmentation does not occur. M(g) + H+  MH+(g) The solvent evaporates away while the MH+ ions move towards a negative plate N Goalby chemrevise.org 1 2. Acceleration Given that all the particles have the same kinetic energy, the Positive ions are accelerated by an electric field velocity of each particle depends on its mass. Lighter particles To a constant kinetic energy have a faster velocity, and heavier particles have a slower KE = ½ velocity. KE = kinetic energy of particle (J) m = mass of the particle (kg) You do not need to learn these equations but = velocity of the particle (ms–1) may be asked rearrange them and use them in a calculation. Rearranged gives 3. Flight Tube The positive ions with smaller m/z values will have the same kinetic energy as those with larger m/z and will move faster. The heavier particles take longer to move through the drift area. The ions are distinguished by different flight times = / Combining the two equations t = time of flight (s) gives you d = length of flight tube (m) = velocity of the particle (m s–1) 4. Detection The ions reach the detector and generate a small current, which is fed to a computer for analysis. The current is produced by electrons transferring from the detector to the positive ions. The size of the current is proportional to the abundance of the species For each isotope the mass spectrometer can Sometimes two electrons may be measure a m/z (mass/charge ratio) and an removed from a particle forming a 2+ ion. 24Mg2+ with a 2+ charge would have a abundance m/z of 12 Example A sample of nickel was analysed and one of the isotopes found was 59Ni. The ions were accelerated to have 1.000 x 10-16 J of kinetic energy and travelled through a flight tube that was 0.8000 m long. How long would one ion of 59Ni+ take to travel along the flight tube? The Avogadro constant L = 6.022 × 1023 mol–1 Mass of one ion of 59Ni+ = mass of one mole of 59Ni+ The Avogadro constant = 59/ 6.022 × 1023 = 9.797X10-23 g = 9.797X10-26 kg t= 0.8000 √( 9.797X10-26/(2x 1.000 x 10-16)) t=1.771X10-5 s N Goalby chemrevise.org 2 Calculating Relative Atomic Mass The relative atomic mass quoted on the periodic table is a weighted average of all the isotopes Fig: spectra for 100 magnesium from mass 80 78.70% spectrometer % abundance 60 24Mg+ If asked to give the species for a peak 40 in a mass spectrum then give charge 25Mg+ 26Mg+ and mass number e.g. 24Mg+ 20 10.13% 11.17% m/z 24 25 26 R.A.M =  (isotopic mass x % abundance) Use these equations to work out the R.A.M 100 For above example of Mg R.A.M = [(78.7 x 24) + (10.13 x 25) + (11.17 x 26)] /100 = 24.3 R.A.M =  (isotopic mass x relative abundance) If relative abundance is used instead of total relative abundance percentage abundance use this equation Example: Calculate the relative atomic mass of tellurium from the following abundance data: 124-Te relative abundance 2; 126-Te relative abundance 4; 128-Te relative abundance 7; 130-Te relative abundance 6 R.A.M = [(124x2) + (126x4) + (128x7) + (130x6)] 19 = 127.8 Example: Copper has two isotopes 63-Cu and 65-Cu. The relative atomic mass of copper is 63.5. Calculate the percentage abundances of these two isotopes. 63.55 = yx63 + (1-y)x65 63.55 = 63y +65 -65y 63.55 = 65 -2y 2y = 1.45 y = 0.725 %abundance 63-Cu =72.5% %abundance 65-Cu = 27.5% N Goalby chemrevise.org 3  Mass spectra for Cl2 and Br2 Cl has two isotopes Cl35 (75%) and Cl37(25%) Br has two isotopes Br79 (50%) and Br81(50%) These lead to the following spectra caused by the diatomic molecules Br79Br81 + Br81Br79 + Cl35Cl35 + relative relative abundance abundance Cl35Cl37 + Br79Br79 + Br81Br81 + Cl37Cl37 + 70 72 74 m/z m/z 158 160 162 The 160 peak has double the abundance of the other two peaks because there is double the probability of 160 Br79-Br81 + as can be both Br79-Br81 and Br81-Br79 Mass spectrometers have been included in planetary space probes so that elements on other planets can be identified. Elements on other planets can have a different composition of isotopes. Measuring the Mr of a molecule Spectra for C4H10 If a molecule is put through a mass spectrometer with an Electron impact ionisation stage it will often break Mass spectrum for butane up and give a series of peaks caused by the fragments. The peak with the largest m/z, however, 43 will be due to the complete molecule and will be Molecular ion equal to the relative molecular mass , Mr ,of the C4H10+ molecule. This peak is called the parent ion or molecular ion 29 58 If a molecule is put through a mass spectrometer with electro spray ionisation then fragmentation will not occur. There will be one peak that will equal the mass of the MH+ ion. It will therefore be necessary to subtract 1 to get the Mr of the molecule. So if a peak at 521.1 is for MH+, the relative molecular mass of the molecule is 520.1. N Goalby chemrevise.org 4 Electronic Structure Models of the atom An early model of the atom was the Bohr model (GCSE model) (2 electrons in first shell, 8 in second etc.) with electrons in spherical orbits. Early models of atomic structure predicted that atoms and ions with noble gas electron arrangements should be stable. The A-level model Electrons are arranged on: Sub energy levels labelled s , p, d and f Principle energy levels Split s holds up to 2 electrons Split Orbitals which hold up numbered 1,2,3,4.. to 2 electrons of into p holds up to 6 electrons into 1 is closest to nucleus opposite spin d holds up to 10 electrons f holds up to 14 electrons Shapes of orbitals Orbitals represent the Principle level 1 2 3 4 mathematical probabilities of finding an electron at any point within certain spatial Sub-level distributions around the 1s 2s, 2p 3s, 3p, 3d 4s, 4p, 4d, 4f nucleus. Each orbital has its own approximate, three An atom fills up the sub shells in order of increasing energy (note 3d is dimensional shape. higher in energy than 4s and so gets filled after the 4s) It is not possible to draw the 1s2s2p3s3p 4s3d4p5s4d5p shape of orbitals precisely. Writing electronic structure using letters and numbers s sublevels are Number of electrons spherical in sub-level For oxygen 1s2 2s2 2p4 Number of main Name of energy level type of p sublevels are shaped sub-level like dumbbells For calcium 1s2 2s2 2p6 3s2 3p6 4s2 Using spin diagrams For fluorine An arrow is one electron 2p Box represents one 2s orbital The arrows going in the opposite direction represents 1s the different spins of the electrons in the orbital When filling up sub levels with several orbitals, fill each orbital singly before starting to pair up the electrons 2p 5 N Goalby chemrevise.org  The periodic table is split into blocks. A s block element is one whose outer electron is filling a s-sub shell e.g. sodium 1s2 2s2 2p6 3s1 A p block element is one whose outer electron is filling a p-sub shell e.g. chlorine 1s2 2s2 2p6 3s2 3p5 A d block element is one whose outer electron is filling a d-sub shell e.g. vanadium 1s22s22p63s23p6 4s23d3 Electronic structure for ions When a positive ion is formed electrons are lost When a negative ion is formed electrons are gained from the outermost shell. O is 1s2 2s2 2p4 becomes O2- is 1s2 2s2 2p6 Mg is 1s2 2s2 2p6 3s2 becomes Mg2+ is 1s2 2s2 2p6 Electronic structure of d-block elements The electronic structure of the d-block has some complications. As mentioned earlier, conventionally we say that 4s fills before 3d and so we write them in that order. There is, however, disagreement in the scientific community about whether this is true. If you look at the electronic structures below you will see both chromium and copper have an unusual arrangement in having a half filled 4s sub shell. You will also see that when d-block elements form ions they lose the 4s electrons first. You may find if you research different reasons for these observations. It may well be many of the reasons are false and we have to accept that some things in chemistry don’t neatly follow patterns we can explain. You do need to learn these electronic structure though! Sc 1s22s22p63s23p6 4s23d1 Sc3+ [Ar] 4s03d0 Ti 1s22s22p63s23p6 4s23d2 Ti3+ [Ar] 4s03d1 V 1s22s22p63s23p6 4s23d3 V3+ [Ar] 4s03d2 Cr 1s22s22p63s23p6 4s13d5 Cr3+ [Ar] 4s03d3 Mn 1s22s22p63s23p6 4s23d5 Mn2+ [Ar] 4s03d5 Fe 1s22s22p63s23p6 4s23d6 When Fe3+ [Ar] 4s03d5 forming ions Co 1s22s22p63s23p6 4s23d7 lose 4s Co2+ [Ar] 4s03d7 Ni 1s22s22p63s23p6 4s23d8 before 3d Ni2+ [Ar] 4s03d8 Cu 1s22s22p63s23p6 4s13d10 Cu2+ [Ar] 4s03d9 Zn 1s22s22p63s23p6 4s23d10 Zn2+ [Ar] 4s03d10 N Goalby chemrevise.org 6  Ionisation Energies Definition :First ionisation energy Remember these definitions very carefully The first ionisation energy is the enthalpy change when one mole of gaseous atoms forms one mole of gaseous ions with a single positive charge The equation for 1st ionisation This is represented by the equation: H(g)  H+ (g) + e- energy always follows the same pattern. It does not matter if the atom does Always gaseous not normally form a +1 ion or is not gaseous Definition :Second ionisation energy The second ionisation energy is the enthalpy change when one mole of gaseous ions with a single positive charge forms one mole of gaseous ions with a double positive charge This is represented by the equation: Ti+ (g)  Ti2+(g) + e- Factors that affect ionisation energy There are three main factors 1.The attraction of the nucleus (The more protons in the nucleus the greater the attraction) Many questions can be 2. The distance of the electrons from the nucleus answered by application (The bigger the atom the further the outer electrons are from the nucleus and the of these factors weaker the attraction to the nucleus) 3. Shielding of the attraction of the nucleus (An electron in an outer shell is repelled by electrons in complete inner shells, weakening the attraction of the nucleus) Successive ionisation energies The patterns in successive ionisation energies for an element give us important information about the electronic structure for that element. Why are successive ionisation energies always larger? The second ionisation energy of an element is always bigger than the first ionisation energy. When the first electron is removed a positive ion is formed. The ion increases the attraction on the remaining electrons and so the energy required to remove the next electron is larger. How are ionisation energies linked to electronic structure? Explanation Ionisation The fifth electron is in a inner energy shell closer to the nucleus and therefore attracted much more Notice the big strongly by the nucleus than the jump between 4 fourth electron. and 5. It also does not have any shielding by inner complete shells 1 2 3 4 5 6 of electron No of electrons removed Example: What group is this element in? Here there is a big jump between the 2nd and 3rd ionisations energies which means that this 1 2 3 4 5 element must be in group 2 of the periodic table Ionisation 590 1150 4940 6480 8120 as the 3rd electron is removed from an electron energy kJ mol-1 shell closer to the nucleus with less shielding and so has a larger ionisation energy N Goalby chemrevise.org 7 The first Ionisation energy of the elements The shape of the graph for periods two and three is similar. A repeating pattern across a Ionisation energy kJ mol-1 period is called periodicity. 2000 1500 The pattern in the first ionisation energy gives us useful information about 1000 electronic structure 500 You need to carefully learn the 0 patterns 5 10 15 20 Atomic number Q. Why has helium the largest first ionisation energy? A. Its first electron is in the first shell closest to the nucleus and has no shielding effects from inner shells. He has a bigger first ionisation Many questions can be energy than H as it has one more proton answered by application of the 3 factors that control Q. Why do first ionisation energies decrease down a group? ionisation energy A. As one goes down a group, the outer electrons are found in shells further from the nucleus and are more shielded so the attraction of the nucleus becomes smaller Q. Why is there a general increase in first ionisation energy across a period? A. As one goes across a period the electrons are being added to the same shell which has the same distance from the nucleus and same shielding effect. The number of protons increases, however, making the effective attraction of the nucleus greater. Q. Why has Na a much lower first ionisation energy than neon? This is because Na will have its outer electron in a 3s shell further from the nucleus and is more shielded. So Na’s outer electron is easier to remove and has a lower ionisation energy. Q. Why is there a small drop from Mg to Al? Al is starting to fill a 3p sub shell, whereas Mg has its outer electrons in the 3s sub shell. The electrons in the 3p subshell are slightly easier to remove because the 3p electrons are higher in energy and are also slightly shielded by the 3s electrons Learn carefully the explanations for Q. Why is there a small drop from P to S? these two small With sulfur there are 4 electrons in the 3p sub shell and the 4th is starting to doubly fill drops as they are the first 3p orbital. different to the When the second electron is added to a 3p orbital there is a slight repulsion between usual factors the two negatively charged electrons which makes the second electron easier to remove. 3p 3p 3s 3s Two electrons of opposite spin in the same orbital phosphorus 1s2 2s2 2p63s23p3 sulfur 1s2 2s2 2p63s23p4 N Goalby chemrevise.org 8 Patterns in the second ionisation energy. If the graph of second ionisation or each successive element is plotted then a similar pattern to the first ionisation energy is observed but all the elements will have shifted one to the left. 5000 Na 2nd Ionisation energy 4500 4000 (kJ/mol) 3500 3000 Ar 2500 S P 2000 Al Cl 1500 Mg Si 1000 10 12 14 16 18 20 Atomic Number The group 1 elements are now at the peaks of the graph Lithium would now have the second largest ionisation of all elements as its second electron would be removed from the first 1s shell closest to the nucleus and has no shielding effects from inner shells. Li has a bigger second ionisation energy than He as it has more protons. N Goalby chemrevise.org 9