Details of the three Sub-atomic (fundamental) Particles Particle Position Relative Mass Relative Charge Proton Nucleus 1 +1 Neutron Nucleus 1 0 Electron Orbitals 1/1840 -1 An atom of Lithium (Li) can be represented as follows: 7 3 Li Atomic Number Atomic Symbol Mass Number 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.
184.108.40.206 Fundamental particles
Appreciate that knowledge and understanding of atomic structure has evolved over time. Protons, neutrons and electrons: relative charge and relative mass. An atom consists of a nucleus containing protons and neutrons surrounded by electrons.
220.127.116.11 Mass number and isotopes
Mass number (A) and atomic (proton) number (Z). Students should be able to: • determine the number of fundamental particles in atoms and ions using mass number, atomic number and charge • explain the existence of isotopes.
The Time of Flight Mass Spectrometer The mass spectrometer can be used to determine all the isotopes present in a sample of an element and to therefore identify elements. The following are the essential 4 steps in a mass spectrometer. 1. Ionisation The sample can be ionised in a number of ways. Two of these techniques are electron impact and electrospray ionisation Electron impact •A Vaporised sample is injected at low pressure •An electron gun fires high energy electrons at the sample •This Knocks out an outer electron •Forming positive ions with different charges E.g. Ti Ti+ + e– Electro Spray Ionisation • The sample is dissolved in a volatile, polar solvent • injected through a fine hypodermic needle giving a fine mist or aerosol • the tip of needle has high voltage • at the tip of the needle the sample molecule, M, gains a proton, H+ , from the solvent forming MH+ • M(g) + H+ MH+ (g) • The solvent evaporates away while the MH+ ions move towards a negative plate Learn all these steps carefully! It needs to be under a vacuum otherwise air particles would ionise and register on the detector There are various models for atomic structure 1 Ionisation area Acceleration area Ion drift area Detection area Ion detector Heavy ions Light ions Time measurement Electron impact is used for elements and substances with low formula mass. Electron impact can cause larger organic molecules to fragment Electro Spray Ionisation is used preferably for larger organic molecules. The softer conditions of this technique mean fragmentation does not occur 2. Acceleration •Positive ions are accelerated by an electric field •To a constant kinetic energy KE = ½ KE = kinetic energy of particle (J) m = mass of the particle (kg) = velocity of the particle (ms–1 ) 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 = / t = time of flight (s) 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 You don’t need to learn these equations but may be asked to use them in a calculation Given that all the particles have the same kinetic energy, the velocity of each particle depends on its mass. Lighter particles have a faster velocity, and heavier particles have a slower velocity. Combining the two equations gives you Rearranged gives 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
The relative atomic mass quoted on the periodic table is a weighted average of all the isotopes R.A.M = (isotopic mass x % abundance) 100 24 25 26 20 40 60 80 100 % abundance m/z 78.70% 10.13% 11.17% Fig: spectra for Magnesium from mass spectrometer Use these equations to work out the R.A.M For above example of Mg R.A.M = [(78.7 x 24) + (10.13 x 25) + (11.17 x 26)] /100 = 24.3 3 If asked to give the species for a peak in a mass spectrum then give charge and mass number e.g. 24Mg+ 24Mg+ 25Mg+ 26Mg+ R.A.M = (isotopic mass x relative abundance) total relative abundance If relative abundance is used instead of percentage abundance use this equation N Goalby chemrevise.org 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 = [(124×2) + (126×4) + (128×7) + (130×6)] 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%
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 70 72 74 m/z relative abundance Cl35Cl35 +Cl35Cl37 +Cl37Cl37 + 158 160 162 m/z relative abundance Br79Br79 + Br79Br81 + Br81Br79 + Br81Br81 + 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 If a molecule is put through a mass spectrometer with an Electron impact ionisation stage it will often break up and give a series of peaks caused by the fragments. The peak with the largest m/z, however, will be due to the complete molecule and will be equal to the relative molecular mass , Mr ,of the molecule. This peak is called the parent ion or molecular ion Molecular ion C4H10 + Spectra for C4H10 58 Mass spectrum for butane 43 29 Mass spectra for Cl2 and Br2 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. 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 Br79-Br81 and Br81-79
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. Electrons are arranged on: The A-level model Principle energy levels numbered 1,2,3,4.. 1 is closest to nucleus Sub energy levels labelled s , p, d and f s holds up to 2 electrons p holds up to 6 electrons d holds up to 10 electrons f holds up to 14 electrons Split into Split into Orbitals which hold up to 2 electrons of opposite spin Shapes of orbitals Orbitals represent the mathematical probabilities of finding an electron at any point within certain spatial distributions around the nucleus. Each orbital has its own approximate, three dimensional shape. It is not possible to draw the shape of orbitals precisely. Principle level 1 2 3 4 Sub-level 1s 2s, 2p 3s, 3p, 3d 4s, 4p, 4d, 4f An atom fills up the sub shells in order of increasing energy (note 3d is higher in energy than 4s and so gets filled after the 4s) 1s2s2p3s3p 4s3d4p5s4d5p Writing electronic structure using letters and numbers For Calcium 1s2 2s2 2p6 3s2 3p6 4s2 Number of main energy level Name of type of sub-level Number of electrons in sub-level 2s 2p 1s Using spin diagrams For fluorine An arrow is one electron The arrows going in the opposite direction represents the different spins of the electrons in the orbital Box represents one orbital •s sublevels are spherical • p sublevels are shaped like dumbbells 2p When filling up sub levels with several orbitals, fill each orbital singly before starting to pair up the electrons. 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 3s 1 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 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 exp. When a positive ion is formed electrons are lost from the outermost shell Mg is 1s2 2s2 2p6 3s2 becomes Mg2+ is 1s2 2s2 2p6 When a negative ion is formed electrons are gained O is 1s2 2s2 2p4 becomes O2- is 1s2 2s2
Ionisation Energies Definition :First ionisation energy 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 This is represented by the equation: H(g) H+ (g) + e- Always gaseous Remember these definitions very carefully The equation for 1st ionisation energy always follows the same pattern. It does not matter if the atom does not normally form a +1 ion or is not gaseous 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 Definition :Second ionisation energy 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) 2. The distance of the electrons from the nucleus (The bigger the atom the further the outer electrons are from the nucleus and the 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) Many questions can be answered by application of these factors The patterns in successive ionisation energies for an element give us important information about the electronic structure for that element. Successive ionisation energies 1 2 3 4 5 6 Ionisation energy No of electrons removed Notice the big jump between 4 and 5. Explanation The fifth electron is in a inner shell closer to the nucleus and therefore attracted much more strongly by the nucleus than the fourth electron. It also does not have any shielding by inner complete shells of electron 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? 1 2 3 4 5 Ionisation energy kJ mol-1 590 1150 4940 6480 8120 Here there is a big jump between the 2nd and 3rd ionisations energies which means that this element must be in group 2 of the periodic table as the 3rd electron is removed from an electron shell closer to the nucleus with less. The first Ionisation energy of the elements The shape of the graph for periods two and three is similar. A repeating pattern across a period is called periodicity. The pattern in the first ionisation energy gives us useful information about electronic structure You need to carefully learn the patterns Many questions can be answered by application of the 3 factors that control ionisation energy 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 energy than H as it has one more proton Q. Why do first ionisation energies decrease down a group? 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 these two small drops as they are different to the usual factors Q. Why is there a small drop from P to S? 3s 3p With sulphur there are 4 electrons in the 3p sub shell and the 4th is starting to doubly fill the first 3p orbital. When the second electron is added to a 3p orbital there is a slight repulsion between the two negatively charged electrons which makes the second electron easier to remove. 3s 3p Two electrons of opposite spin in the same orbital phosphorus 1s2 2s2 2p63s23p3 sulphur 1s2 2s2 2p63s23p4
18.104.22.168 Electron configuration
Students should be able to: • define first ionisation energy • write equations for first and successive ionisation energies • explain how first and successive ionisation energies in Period 3 (Na–Ar) and in Group 2 (Be–Ba) give evidence for electron configuration in sub-shells and in shells.