Unit 1 AS Chemistry Atomic sructure and Periodicity QUICK PEEK

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Brief overview (quick peek) of AQA Unit 1 AS Chemistry Atomic sructure and Periodicity, including interactive exam based questions

Brief overview (quick peek) of AQA Unit 1 AS Chemistry Atomic sructure and Periodicity, including interactive exam based questions

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  • 1. Unit 1 Atomic Structure (& mass spec) Periodicity (Period 3 and IEs) Quick Peek
  • 2. Define the terms mass number and atomic number. Use these terms to explain the difference between the terms element and isotope. Isotope: atoms of an element with the same atomic number (proton number) but different mass number (due to different numbers of neutrons). Mass number = The total number of neutrons and protons in an atom of an element Atomic number = The number of protons in an atom of an element What does a mass spectrometer do?
  • 3. THE MASS SPECTROMETER = an instrument which : (a) converts neutral atoms (or molecules) into positively charged ions by removing an electron (b) then separates these ions according to their relative mass (m) to relative charge (z) ratio, (m/z) (c) then measures "m/z" and the % abundance for each ion. (d) then produces a MASS SPECTRUM = graph of “m/z” versus % abundance
  • 4. A Mass Spectrometer to form SEPARATE atoms / molecules to remove air molecules which would also be measured Order of actions = V I A D D R = Victory Is A Definite Detectable Result V I A D D R
  • 5. Ionisation Electron gun fires high-energy electrons at the minimum energy on to the sample  Minimum energy: so no more than 1 electron is knocked out, so reducing the risk of 2+, 3+ ion formation A(g) A+(g) + e-  produces a positive ion of the atom (A+) or molecule (M+) High energy electrons Positive ions repelled Electron gun – electrically heated coil Electron trap (+) Vapourised sample Ion repeller (+) M2+ ions may also be formed if 2 e- knocked off – very rare Ions of molecules can fragment M(g) M+(g) + e-
  • 6. Acceleration High speed beam of ionised sample Ionisation chamber at +10000 volts Final plate at 0 voltsIntermediate plate +ve ions repelled by high +ve potential  accelerates the +ve ions through a slit  narrow beam of fast moving +ve ions. Sample needs to be ionised and fast moving to make subsequent separation and detection possible
  • 7. Deflection Electromagnet Ion stream C Ion stream A Ion stream B  Mixed ion stream from accelerating unit High speed beam of +ve ions is deflected by a strong, variable magnetic field. Deflection is GREATER for : (a) lighter (m lower) ions and (b) more charged (z higher) ions i.e. ions with LOWER m/z ratio Does ion stream A, B or C have lowest m/z ratio? : A = Lowest m/z C = Highest m/z Note : deflection also greater for faster ions This field strength is steadily increased causing ions of INCREASING m/z value to be deflected in turn onto the detector. Which ion stream, A, B or C, is being detected in the diagram? B
  • 8. Detection and measurement Ion stream B Metal box Wire to amplifier Each ion reaching the detector takes an e- from the metal box  tiny current produced  current measures abundance of that ion. Output from recorder is called a ‘mass spectrum’ showing abundance (or detector current) against m/z ratio for each isotope. Relative abundance m/z
  • 9. Ar(Kr) = 84.06= (82 x 12/100) + (83 x 12/100) + (84 x 55/100) + (86 x 21/100) Minor peaks are also observed at m/z 41, 41.5, 42 and 43. Why ? These are caused by Kr2+ particles produced by the rare event of 2 electrons being knocked off the atom during ionisation. The % distribution of isotopes is the same for the Kr2+ particles as it is for the Kr+ particles, even though their abundance is considerably lower than the major m/z peaks
  • 10. Explain the terms ionisation, fragmentation, acceleration, deflection and detection as used in mass spectrometry. Ionisation Fragmentation Acceleration Deflection Detection A stream of high speed electrons bombards gaseous sample and knocks off an electron to form a gaseous cation Bond(s) break in the gaseous cation and creates a smaller molecule (or atom) cation and a free radical molecule (or atom) Beam of gaseous cations passes through holes / slits in two negatively charged plates with a potential difference across them Beams of cations are deflected by magnetic field. The smaller the m/z the greater the deflection. Gaseous cations with a particular m/z ratio hit the detector and acquire electrons thereby generating a transient electrical current. The amount of current is proportional to the abundance of cations colliding at the detector.
  • 11. Define Ar of element X Mr of molecule Y Average* mass of one atom of X 1/12th mass of one atom of 12C isotope *weighted average related to abundance of its naturally occurring isotopes Average mass of one molecule of Y 1/12th mass of one atom of 12C isotope
  • 12. Calculate the relative mass of an 16O atom [Ar(O)] Calculate the relative mass of a 12C16O2 molecule [Mr(CO2)] = Mass of one O atom 1/12 x Mass of one 12C atom = 2.656 x 10-26 1/12 x 1.992 x 10-26 kg kg = 16 (g/mol) = Mass of one CO2 molecule 1/12 x Mass of one 12C atom = 7.304 x 10-26 1/12 x 1.992 x 10-26 kg kg = 44 (g/mol)
  • 13. Define the term first ionisation energy and write an equation to represent the first ionisation energy of chlorine. State whether the process in b) is exothermic or endothermic and explain your choice. State and explain the trend in first ionisation energies down a group of the Periodic Table. The minimum energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of singly charged gaseous cations Cl(g)  Cl+ (g) + e- Endothermic because energy must be supplied to overcome the electrostatic force of attraction between negatively charged electron and the positively charged nucleus First ionisation energies decrease down a group. Although nuclear charge increases down a group, atomic radius increases, as does electron shielding by occupied orbital shells, hence the outer shell electron(s) are less attracted to the nucleus and require less energy to remove. What is the second ionisation energy?
  • 14. (c) Which of these three isotopes of sulphur would you expect to have the: highest first ionisation energy? . highest melting point? greatest chemical reactivity? (d) Use the following % composition data to calculate the relative atomic mass of sulphur. Naturally occurring sulphur contains 95.0% sulphur-32, 0.8% sulphur-33 and 4.2% sulphur-34. All identical because all have same electron configuration Sulphur-34 because higher mass atoms create stronger van der Waal forces All identical because all have same electron configuration Ar = (32 x 95/100) + (33 x 0.8/100) + (34 x4.2/100) = 32.092 Naturally occurring sulphur contains 95.0% sulphur-32, 0.76% sulphur-33 and 4.2% sulphur-34. S 32 16 S 33 16 S 34 16
  • 15. Electronegative trend is valid for 1st, 2nd, 3rd periods, but d-block elements fluctuate EN values (effect upon 4th, 5th periods etc.) FrF Fr 0.7 and F 4.0 electronegativities . What is electronegativity? Which two elements, if combined in a compound, would produce a compound with the greatest difference in electronegativity? What happens to electronegativity, ionization energy and atomic radius as you go DOWN A GROUP? Ability of an atom to attract electron density (or e- or –ve charge) in a covalent bond or shared pair
  • 16. IncreasingAtomicSizeWhy? Across a period: effective nuclear charge acting on electrons increases (electrons occupy existing shells, no additional shielding, however increased proton nuclear charge) Down a group: effective nuclear charge acting on electrons decreases (increasing electron shells creates increased electron shielding) Decreasing Atomic Size
  • 17. 3 5 4 n log10 In VARIATIONS IN IONISATION ENERGIES FOR SILICON Hence, electron arrangement is :2,8,4 Si X X X X XX X X X X XX X X  4 electrons FURTHEST from and MOST SHIELDED from the nucleus  EASIEST to remove  2 electrons VERY CLOSE to and NOT SHIELDED from the nucleus  MOST DIFFICULT to remove  8 electrons at INTERMEDIATE DISTANCE, with INTERMEDIATE SHIELDING from the nucleus  INTERMEDIATE DIFFICULTY to remove This model of atomic structure is associated with Niels Bohr (1915)
  • 18. 3 5 4 n log10 In VARIATIONS IN IONISATION ENERGIES FOR ELEMENT C Click on the group number of element suggested by these data 1 2 3 4 5 6 7 0
  • 19. VARIATIONS IN THE SUCCESSIVE IONISATION ENERGIES OF THE SODIUM ATOM 2.50 3.00 3.50 4.00 4.50 5.00 5.50 1 2 3 4 5 6 7 8 9 10 11 log In Number of electrons removed
  • 20. (a) Element X has an atomic number of 9. Figure 1 shows its mass spectrum and Figure 2 is a graph of its successive lg ionisation energies. (b) (i) Write down the electronic configuration of the element X. (ii) To which group of the Periodic Table does X belong? (iii) What is the relative atomic mass of X? iv) Describe and explain the trend shown by the ionisation graph in Figure 2 Fig 2 logIn n Fig 1 Rel.Abundance m : z ratio 19 38 x x x x x x x x x 1s2 2s2 2p5 since Fig 2 shows basic e- configuration to be 2,7 Group 7 since Fig 2 shows 7 e- in outer e- shell 19 19=F+ and 38=F2 +
  • 21. As successive electrons are removed, the ion which remains has an increasing positive charge so more energy is needed to remove further electrons The large increase between IE7 and IE8 occurs because this is the transition between inner shell (closer to the nucleus, less shieldung, greater effective nuclear attraction) and outer shell (further from the nucleus, greater shielding, less effective nuclear attraction) electrons. Written answer: explaining the observed trend
  • 22. FIRST IONISATION ENERGY (I1) vs ATOMIC NUMBER 0 500 1000 1500 2000 2500 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Atomic Number (Z) I1/kJpermole Na LARGE DECREASE in IE1 between periods because e- removed from more distant and more shielded higher energy electron orbital shells, which overcomes the increased nuclear attraction due to increased proton numberExplained by transition between s and higher energy/better shielded p sub-level Explained by electron removed from spin-paired p suborbital where there is repulsion between paired e which reduces required IE Si Mg Al P S Cl Ar O B
  • 23. N N/2 N/4 N/8 0 Radioactive decay (e.g. 14C) Number of atoms TimeSecond Half-lifeFirst Half-life ThirdHalf-life X X X X X X X X X X X X t½ = the time taken for N atoms to decay to N/2 atoms (e.g. 5730y for 14C) Nt=No e-λt ln Nt = -λt No