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Atomic Structure

Relative electric charge. Position in atom. Relative mass. Atomic Structure. Composition of an atom. Atoms are made up of 3 fundamental subatomic particles :. nucleus + 1 1 nucleus 0 1 around the - 1 1 / 1840 nucleus. Protons (p) Neutrons (n) Electrons (e - ).

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Atomic Structure

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  1. Relative electric charge Position in atom Relative mass Atomic Structure Composition of an atom Atoms are made up of 3 fundamental subatomic particles: nucleus + 1 1 nucleus 0 1 around the - 1 1/1840 nucleus Protons (p) Neutrons (n) Electrons (e-)

  2. In an electric field: Effect of electric fields on beams of p, n & e- - + p n e- The e- is deflected more than p as it has a small mass.

  3. Fleming’s Left-Hand Motor Rule ThuMb : Force, Motion Fore finger : Magnetic Field Centre finger : Current Current flow in the same direction as proton

  4. In a magnetic field: Effect of magnetic fields on beams of p, n & e The magnetic field line is moving perpendicular into the screen p n e- NB: Electrons are deflected to a greater extent than protons because electrons have a much smaller mass.

  5. A E Z Atomic number & Atomic mass where A: mass no. (total no. of p & n) Z: atomic no. (no. of p) N: no. of neutrons ( = A - Z ) NB: 1. Since the atomic no. (no. of p) = no. of electrons, therefore, there is no net charge on an atom 2. Loss of electrons give rise to cations (positively charged ions, no. of p > e-) while the gain of electrons give rise to anions (negatively charged ions, no. of p < e-)

  6. Try these : 29 Al 13 16 13 pne- 13 32 S 2- 16 16 18 pne- 16 + 23 Na 11 12 10 pne- 11

  7. A B C mass no. 55 54 57 no. of neutrons20 21 24 atomic no. Isotopes Isotopes are atoms of the same element with same atomic no. but different mass no. (or different no. of neutrons) For examples: 1H, 2H, 3H are isotopes of hydrogen35Cl, 37Cl are isotopes of chlorine Try this: Which of the following 3 atoms are isotopes of the same element? 35 33 33  

  8. Isotopes • Isotopes of an element have similar chemical propertiesThis is because they have the same no. of electrons and the chemical properties depend upon the transfer and redistribution of electrons. • Isotopes of an element have different physical properties.This is because they havedifferent no. of neutronsand hencedifferent masses.eg. 37Cl has a higher density than 35Cl.

  9. K 19 farthest from nucleus nearest tonucleus Electronic Energy levels or shells Electrons move around the nucleus of an atom in definite energy levels or shells which are identified by numbers called principal quantum numbers (p.q.n.), n. : electrons are arranged in 4 shells ( 2, 8, 8, 1) energy level / shell no. n = 4 n = 3 n = 2 n = 1 no. of electrons ____ ____ ____ ____ Highest energy level(most easily removed)Lowest energy level 1 8 Increasing energy 8 2

  10. Energy levels are split into sublevels / subshells. ( labelled s, p, d, f ) Energy sublevels or subshells Energy level n = 1 n = 2 n = 3 n = 4 no. of subshells 1 2 3 4 types of subshells s s , p s, p, d s, p, d, f labelled as 1s 2s, 2p 3s, 3p, 3d 4s, 4p, 4d, 4f

  11. Electrons & Orbitals type of subshell no. of orbitals s p d f max. no. of electrons 2 6 10 14 Electrons are in constant motion. Within each subshells, electrons can be found in certain regions known as orbitals. Anorbitalis a region or volume of space within which there is a high probability of finding an electron. Each orbital can accommodate two electrons. 1 3 5 7

  12. Relative energies of energy levels & sublevels 4f 4d n = 4 n = 3 n = 2n = 1 4p 3d 4s 3p increasing energy 3s 2p 2s NB: 3d higher energy than 4s 1s

  13. Try this: Shell no. Maximum no. of electrons n = 1 n = 2 n = 3 n = 4 2 2 + 6 = 8 2 + 6 + 10 = 18 2 + 6 + 10 + 14 = 32 General Formula = 2n2

  14. z z   x x y y 1s 2s Shapes of Orbitals Shape: spherical size of s orbitals: 4s > 3s > 2s > 1s

  15. z z z  x x   x y y y px py pz • Shape : dumb-bell • different axes of symmetry • mutually perpendicular to one another • orbitals(px, py, pz) with the same p.q.n. have the same energy i.e. they are degenerate.

  16. Electronic Configuration Electrons are arranged in orbitals according to a set of rules. 1. Pauli Exclusion Principle: An atomic orbital can be occupied by only two electrons and their spins are always opposite.    

  17. 6s ... 5s 5p ... 4s 4p 4d 4f 3s 3p 3d 2s 2p 1s Increasing energy Electronic Configuration 2. Aufbau (or building up) Principle: Electrons fill the orbitals of lowestenergy first and then a higher energy.

  18. 2p 2p Electronic Configuration 3. Hund’s Rule: When electrons are added successively to a set of orbital of the same energy level, they occupy them singly first before any pairing occurs.  

  19. M(g) M+(g) + eH = + value kJ mol -1 Ionisation Energies Information about the arrangement of electrons has been obtained by bombarding the atoms with streams of fast-moving electrons and so removing electrons from the atom. This method allows us to measure approximately the energy required to remove an electron, called the ionisation energy. First ionisation energy of an element is the amount of energy required to remove one mole of electrons from one mole of gaseous atoms.

  20. Examples: Na(g) Na+(g) + e H = + 494 kJ mol -1 Mg(g) Mg+(g) + e H = + 736 kJ mol -1 Second ionisation energy of an element is the amount of energy required to remove one mole of electrons from one mole of gaseous ions. M +(g) M 2+(g) + eH = + value kJ mol -1

  21. Example: Na +(g) Na 2+(g) + e H = + 4560 kJ mol -1 Mg +(g) Mg 2+(g) + e H = + 1450 kJ mol -1 NB:1st IE < 2nd IE < 3rd IE < ..........< .........

  22. Factors affecting the first IE of elements Atomic radius(distance of outermost electron from the nucleus) As atomic radius increases, outer electrons experience less nuclear attraction, hence 1st IE decreases. Size of nuclear charge(no. of p in nucleus) As size of nuclear charge increases, outer electrons experience greater nuclear attraction, hence 1st IE increases.

  23. Screening or shielding effect by inner-shells electrons Outer electrons are shielded from nuclear attraction by repelling effect of inner-shells electrons. As no. of inner shells increases, the shielding effect by inner-shells electron increases, and hence 1st IE decreases. NB: Electrons in the same shell exert anegligible shielding effecton each other.

  24. Try this: Compare the 1st IE of 5B & 6C. 5B : 1s22s2 2p16C : 1s22s2 2p2 C has a greater nuclear charge than B (because C has a greater no. of proton) atomic radius of C is smaller than that of B.  outer electron of C would experience greaternuclear attraction. Both B & C have  the same shielding effect by inner-shell electrons (because they have the same no. of inner principal quantum shell) As a result, 1st IE of C > 1st IE of B.

  25. Trend of 1st IE across a period 1st IE / kJmol -1 Ne ( 2,3,3 pattern) F N O Period 3 Be C Li B Mg Period 2 Na 3 4 5 6 7 8 9 10 11 12 atomic number

  26. Across any period, there is a general increase in the 1st IE due to  the increase in the nuclear charge &  the decrease in atomic radius across the period. (the shielding effect remains almost the same from one element to the other because electrons are added successively to the same shell)

  27. However, • the 1st IE of B is lower than that of Be. • 5B : 1s2 2s2 2p14Be : 1s2 2s2 2pelectron of B is of ahigher energythan the2s electron of Be. Hence, a lower amount of energy is required to remove the 2p electron from B than to remove the 2s electron from Be.

  28. the 1st IE of O is lower than that of N. • 8O : 1s2 2s2 2p47N : 1s2 2s2 2p3 Themutual repulsionof thepaired electronsin a2porbital of O makes the removal of one electron from that orbital easier compared to a 2p electron of N which does not experience such repulsion.

  29. Down a group, the 1st IE decreases. This is due to  the increase in atomic radius  the increase in shielding effect by inner-shell electrons. These two factors outweigh the effect of the increasing nuclear charge.

  30. Try this: Explain why the 1. the 1st IE of Al < 1st IE of Mg 2. the 1st IE of S < 1st IE of P 1st IE / kJmol -1 Ar Cl P S Mg Si Na Al 11 12 13 14 15 16 17 18 atomic number

  31. Relevance of successive IEs to electronic configuration of atom. A plot of successive IEs of Mg ( 1s2 2s2 2p6 3s2) Log 10 IE 1 2 3 4 5 6 7 8 9 1o 11 12 no. of electrons removed

  32. A plot of successive IEs of Mg: 1s2 2s2 2p6 3s2 n = 3 n = 1 Gradual increase  electrons in the same p.q. shell Log 10 IE 1s2 n = 2 Closest to the nucleus 2p62s2 drastic “jump” change of p.q. shell. Furthest away from the nucleus 3s2 1 2 3 4 5 6 7 8 9 10 11 12 no. of electrons removed

  33. Try this: Log 10 IE 1 2 3 4 5 6 7 8 9 1o 11 12 13 14 no. of electrons removed

  34. A plot of successive log10 IEs of Element X against the no. of electrons removed. Log 10 IE n = 1 n = 2 Electrons in the outermost shell Drastic “jump” n = 3 1 2 3 4 5 6 7 8 9 1o 11 12 13 14 no. of electrons removed

  35. n = 1 n = 2 n = 3 Element X belongs to Group IV because it has 4 valence electrons. It has 3 principal quantum shells and its electronic configuration is 1s2 2s2 2p6 3s2 3p2 Valence electrons

  36. Sketch a graph of the first seven successive log10 IEs of Al against the no. of electrons removed. 13Al: 1s2 2s2 2p6 3s2 3p1 Log 10 IE 2p4 Drastic “jump” 3p13s2 1 2 3 4 5 6 7 no. of electrons removed

  37. A plot of successive IEs of Al against the no. of electrons removed. IE / kJ mol -1 1s2 13Al: 1s2 2s2 2p6 3s2 3p1 Moderate “jump” presence of subshells 2s2 2p6 Drastic “jump” 3s2 3p1 1 2 3 4 5 6 7 8 9 10 11 12 13 no. of electrons removed

  38. 8833 928 1239 3253 3547 The first six successive IEs of an element X are as follows: In which group does the element belong to? 578 1817 2745 11578 14831 18378 kJmol -1     There is adrastic increasein the4thIE  the4thelectron to be removed is in aninner and more stable principal quantum shellthat is closer to the nucleus.  there are3valence electrons.  Element X belongs to groupIII.

  39. The successive IEs of an element Y are as follows: In which group does element Y belong? 740 1500 7700 10500 13600 18000 21700 2800 6200 760 3100 4400 3700    Ans: Element Y belongs to Group II.

  40. N92P1Q1 (a) Write down the electronic configuration of the selenium atom. (b) How would you expect the first ionisation energy of selenium to compare with that of (i) sulphur (ii) bromine Give your reasoning.

  41. (a)34Se:1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4 Gp VI (b)16S: 1s2 2s2 2p6 3s2 3p4 Gp VI The 4p electron of Se to be removed is i) more shielded by inner-shells’ electrons and ii) further away from the nucleus compared with the 3p electron to be removed from S. Therefore, the 1st IE of Se is expected to be lower than that of S.

  42. same period (a)34Se:1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4 Gp VI (b)35Br: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 Gp VII Se has a i) smaller nuclear charge and ii) larger atomic radius than Br Therefore, the 1st IE of Se is expected to be lower than that of Br.

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