1 The Electronic Structure of Atoms 4.1The Electromagnetic Spectrum 4.2Deduction of Electronic Structure from Ionization Enthalpies 4.3The Wave-mechanical Model of the Atom 4.4 Atomic Orbitals 4

2 Line spectrum of hydrogen Fig.4-5(b) 4.1 The electromagnetic spectrum (SB p.83)

3 How Do Atoms Emit Light ? H 2 (g) Electric discharge Or Heating H(g) Hydrogen atom in ground state means its electron has the lowest energy

4 Ionisation energy of hydrogen

5 Energy Atom in ground state H(g) Electric discharge Or Heating H*(g) ExcitedGround Atoms in excited state -This is not stable state so electron drops back giving out energy in the form of light - The energy given out is difference between the energy of electron in in outer orbit and the energy of the orbit into which it drops

6 H*(g) H(g) Ground Excited Energy Atom in ground state Atoms in excited state Not Stable photon Atom returns to ground state photon E = energy of the emitted light = E 2 – E 1 E2E2 E1E1

7 Q.2 (a) The spectral lines come closer at higher frequency and eventually merge into a continuum (b)n =   n = 2

8 (c)The electron has been removed from the atom. I.e. the atom has been ionized. Q.2 e1e1 H(g)  H + (g) + e 

9 Balmer series caused by electron dropping back to 2 nd orbit

10 Lyman series is Obtained when electron in hydrogen atom drops from exited state to ground state which is in 1 st orbit

11 Interpretation of the atomic hydrogen spectrum 4.1 The electromagnetic spectrum (SB p.84)

12 Niels Bohr ( ) Interpretation of the atomic hydrogen spectrum Bohr’s Atomic Model of Hydrogen Nobel Prize Laureate in Physics, 1922

13 Interpretation of the atomic hydrogen spectrum Bohr’s model of H atom

14 Interpretation of the atomic hydrogen spectrum 1.The electron can only move around the nucleus of a hydrogen atom in certain circular orbits with fixed radii. Each allowed orbit is assigned an integer, n, known as the principal quantum number.

15 Interpretation of the atomic hydrogen spectrum 2. Different orbits have different energy levels. An orbit with higher energy is further away from the nucleus.

16 Interpretation of the atomic hydrogen spectrum 3.Spectral lines arise from electron transitions from higher orbits to lower orbits. E2E2 E1E1 n = 2 n = 1

17 Interpretation of the atomic hydrogen spectrum 5.The theory failed when applied to elements other than hydrogen (multi- electron systems)

18 Convergence Limits and Ionization Enthalpies Ionization enthalpy is the energy needed to remove one mole of electrons from one mole of gaseous atoms in ground state to give one moles of gaseous ions (n =  ) X(g)  X + (g) + e  Units : kJ mol  1

19 Na Li He n = 3  n =  n = 2  n =  n = 1  n =  n = 3 n = 2 n = 1 H Electron transition for ionization of atom Energy Level of e  to be removed in ground state of atom Atom Lyman series ? Balmer series ? Paschen series ?

20 EE E2E2 E1E1 E2’E2’ E1’E1’ HeH Ionization enthalpy : He > H Relative positions of energy levels depend on the nuclear charge of the atom.

Deduction of Electronic Structure from Ionization Enthalpies

22 Ionization energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms in ground state to give one moles of gaseous ions (n =  ) X(g)  X + (g) + e  Ionization enthalpy

23 X(g) + e -  X - (g)  H 1 st E.A. X  (g) + e -  X 2  (g)  H 2 nd E.A. 2.Electron affinity The enthalpy change when one mole of electrons are added to one mole of atoms or negative ions in gaseous state. Electron affinities can be positive or negative 1 st electron affinity is negative 2 nd electron affinity is positive as energy is requred to add electron toan already negative atom.

24 Q.11(a) Be(g)  Be + (g) + e  IE 1 Be + (g)  Be 2+ (g) + e  IE 2 Be 2+ (g)  Be 3+ (g) + e  IE 3 Be 3+ (g)  Be 4+ (g) + e  IE 4

25 Q.11(b) IE 1 < IE 2 < IE 3 < IE 4 Positive ions with higher charges (which have less radius) attract electrons more strongly. Thus, more energy is needed to remove an electron from positive ions with higher charges.

26 Q.11(c) IE 4 (kJ mol  1 ) IE 3 (kJ mol  1 ) IE 2 (kJ mol  1 ) IE 1 (kJ mol  1 )

27 Q.11(c) (i)The first two electrons are relatively easy to be removed.  they experience less attraction from the nucleus,  they are further away from the nucleus and occupy the n = 2 electron shell.

28 Q.11(c) (ii)The last two electrons are very difficult to be removed.  they experience stronger attraction from the nucleus,  they are close to the nucleus and occupy the n = 1 electron shell.

29 Electron Diagram of Beryllium Second Shell First Shell

30 Energy Level Diagram of Beryllium n =2 n= 1

31 EE IE 1 = E  - E 2 E2E2 E1E1 Be

32 EE IE 1 E2E2 E1E1 Be E2’E2’ E1’E1’ Be + IE 2 IE 1 < IE 2

33 EE IE 1 E2E2 E1E1 Be E2’E2’ E1’E1’ Be + E 2 ’’ E 1 ’’ Be 2+ IE 2 IE 1 < IE 2 << IE 3 IE 3

34 EE IE 1 E2E2 E1E1 Be E2’E2’ E1’E1’ Be + E 2 ’’ E 1 ’’ Be 2+ E 2 ’’’ E 1 ’’’ Be 3+ IE 1 < IE 2 << IE 3 < IE 4 IE 3 IE 2 IE 4

35 The Concept of Spin Spin is the angular momentum intrinsic to a body. E.g.Earth’s spin is the angular momentum associated with Earth’s rotation about its own axis.

36 Paired electrons in an energy level should have opposite spins. Electrons with opposite spins are represented by arrows in opposite directions. Q.12

37 4 groups of electrons Q.12

38 n = 4 n = 3 n = 2 n = 1 Which group of electrons is in the first shell ? Q.12

39 2, 8, 8, Q.12

40 n = 4 n = 2 n = 3 n = 1 Q.12

41 Evidence of the Existence of Shells 4.2 Deduction of electronic structure from ionization enthalpies (p.91) 2, 8, 1

42 n = 2 n = 3 n = 1 Na

43 Variation of IE 1 with Atomic Number Evidence for Subshell

44 Only patterns across periods are discussed

45 1.A general  in IE 1 with atomic number across Periods 2 and (a)

46 2 – (a)

47 2.IE 1 value : Group 2 > Group 3 3. IE 1 value : Group 5 > Group 6 13.(a)

48 2.Peaks appear at Groups 2 & 5 3. Troughs appear at Groups 3 & 6 13.(a)

49 2,1 2,2 2,3 2,4 2,5 2,6 2,7 2,8 2,8,1 2,8,2 2,8,3 2,8,4 2,8,5 2,8,6 2,8,7 2,8,8 On moving across a period from left to right, 1. the nuclear charge of the atoms  (from +3 to +10 or +11 to +18) 2. electrons are being removed from the same shell  e  s removed experience stronger attraction from the nucleus. 13.(b)

50 IE 1 : B(2,3) < Be(2,2) 13.(b) The value for Beryllium is higher than for Lithium due to the increased nuclear charge. There is no extra shielding There is a DROP in the value for Boron. This is because the extra electron has gone into one of the 2p orbitals. The increased shielding makes the electron easier to remove It was evidence such as this that confirmed the existence of sub-shells. If there hadn’t been any sub-shell, the value would have been higher than that of Beryllium.

51 2s 2p 1s n =  IE 2s IE 2p IE 2s (Be) > IE 2p (B) IE 1 : Be > B 13.(b)

52 It is more difficult to remove an electron from a half-filled p subshell as half filled subshells are more stable IE 1 : N(2,5) > O(2,6) ; P(2,8,5) > S(2,8,6) There is a DROP in the value for Oxygen. The extra electron has paired up with one of the electrons already in one of the 2p orbitals. The repulsive force beteen the two paired-up electrons means that less energy is required to remove one of them.

53 O(2,6)  O + (2,5) 2p Is the 2p energy level of O + lower or higher than that of O ? Electrons in O + experience a stronger attractive force from the nucleus. (O + has more effective nuclear charge) Not because of the half-filled 2p subshell O + is always less stable than O.

54 Quantum Numbers 2.Subsidiary quantum number ( ) = 0, 1, 2,…,n-1 = 0  spherical s subshell= 1  dumb-bell p subshell = 2  d subshell = 3  f subshell complicated shapes

55 Q.14(a) 32n = 4 18n = = =9 1+3=4 4s, 4p, 4d, 4f 3s, 3p, 3d 2s, 2pn = 2 21(1s)1sn = 1 Total no. of electrons Total no. of orbitals Subshells Principal quantum shell

56 14(b) The total number of electrons in a principal quantum shell = 2n 2

57 In practice, a boundary surface is chosen such that within which there is a high probability (e.g. 90%) of finding the electron. An orbital is the path of an electron

58 s Orbitals 4.4 Atomic orbitals (p.98)

59 nodal surface There is no chance of finding the electron on the nodal surface.

60

61 Probabilities of finding the electron at A or B > 0 A B Probability of finding the electron at C = 0 C How can the electron move between A & B ?

62 p Orbitals Two lobes along an axis 4.4 Atomic orbitals (p.100)

63 For each 2p orbital, there is a nodal plane on which the probability of finding the electron is zero. 4.4 Atomic orbitals (p.100) yz planexz plane xy plane

Atomic orbitals (p.101) d Orbitals Four lobes along two axes Two lobes & one belt Four lobes between two axes

65 The END