The Electronic Structure of Atoms

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

Chapter 4 The electronic structure of atoms (SB p.80) The electronic structure of atoms Niels Bohr Bohr’s model of H atom

4.1 The electromagnetic spectrum (SB p.82) The electromagnetic spectrum

4.1 The electromagnetic spectrum (SB p.82) Continuous spectrum of white light Fig.4-5(a)

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

4.1 The electromagnetic spectrum (SB p.83) The emission spectrum of atomic hydrogen UV Visible IR Let's Think 1

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

4.1 The electromagnetic spectrum (SB p.85) Bohr proposed for a hydrogen atom: 1. An electron in an atom can only exist in certain states characterized by definite energy levels (called quantum). 2. Different orbits have different energy levels. An orbit with higher energy is further away from the nucleus. 3. When an electron jumps from a higher energy level (of energy E1) to a lower energy level (of energy E2), the energy emitted is related to the frequency of light recorded in the emission spectrum by:E = E1 - E2 = h

4.1 The electromagnetic spectrum (SB p.86) How can we know the energy levels are getting closer and closer together?

4.1 The electromagnetic spectrum (SB p.87) Frequency of light emitted Planck ’s constant E = E1 - E2 = h

4.1 The electromagnetic spectrum (SB p.87) dark background(photographic plate) bright lines bright background(photographic plate) dark lines Emission spectrum of hydrogen Absorption spectrum of hydrogen

4.1 The electromagnetic spectrum (SB p.87) bright background(photographic plate) dark lines Production of the absorption spectrum Absorption spectrum of hydrogen

4.1 The electromagnetic spectrum (SB p.87) H (g) H+(g) + e- Convergence limits and ionization What line in the H spectrum corresponds to this electron transition (n= ∞  n=1)? Last line in the Lyman Series For n=∞  n=1:

4.1 The electromagnetic spectrum (SB p.87) Let's Think 2 Example 4-1B Example 4-1A

4.1 The electromagnetic spectrum (SB p.89) Check Point 4-1 The uniqueness of atomic emission spectra No two elements have identical atomic spectra atomic spectra can be used to identify unknown elements.

4.2 Deduction of Electronic Structure from Ionization Enthalpies

4.2 Deduction of electronic structure from ionization enthalpies (p.91) Ionization enthalpy Ionization enthalpy(ionization energy) of an atom is the energy required to remove one mole of electrons fromone mole of its gaseous atoms to form one mole of gaseous positive ions. The first ionization enthalpy M(g)  M+(g) + e- H = 1st I.E. The second ionization enthalpy M+(g)  M2+(g) + e- H = 2nd I.E.

4.2 Deduction of electronic structure from ionization enthalpies (p.91) Evidence of shells  shells

4.2 Deduction of electronic structure from ionization enthalpies (p.91) 2,8 2,7 2,5 2,2 2,4 2,6 2,3 2,1 Check Point 4-2 Evidence of sub-shells  subshells

4.3 The Wave-mechanical Model of the Atom

4.3 The Wave-mechanical model of the atom (p.94) Bohr’s atomic model and its limitations Bohr considered the electron in the H atom (a one-electron system) moves around the nucleus in circular orbits. Basing on classical mechanics, Bohr calculated values of frequencies of light emitted for electron transitions between such ‘orbits’. The calculated values for the frequencies of light matched with the data in the emission spectrum of H.

4.3 The Wave-mechanical model of the atom (p.94) Bohr tried to apply similar models to atoms of other elements (many-electron system), e.g. Na atom. Basing on classical mechanics, Bohr calculated values of frequencies of light emitted for electron transitions between such ‘orbits’. The calculated values for the frequencies of light did NOT match with the data in the emission spectra of the elements.  The electron orbits in atoms may NOT be simple circular path.

4.3 The Wave-mechanical model of the atom (p.95) Wave nature of electrons A beam of electrons shows diffraction phenomenon • Electrons possess wave properties (as well as particle properties).

4.3 The Wave-mechanical model of the atom (p.95) Wave nature of electrons Schrödinger used complex differentialequations/wave fucntions to describe the wave nature of the electrons inside atoms (wave mechanic model). The solutions to the differential equations describes the orbitals of the electrons inside the concerned atom. An orbital is a region of space having a high probability of finding the electron.

4.3 The Wave-mechanical model of the atom (p.95) Quantum numbers Electrons in orbitals are specified with a set of numbers called Quantum Numbers: 1. Principal quantum number (n) n = 1, 2, 3, 4, …... 2. Subsidiary quantum number (l) l = 0, 1, 2, 3…, n-1s p d f 3. Magnetic quantum number (m) m = -l, …, 0, …l 4. Spin quantum number (s) s= +½, -½ The solutions of the wave functions are the orbitals -- which are themselves equations describing the electrons.

4.3 The Wave-mechanical model of the atom (p.96) 8 18 32

4.3 The Wave-mechanical model of the atom (p.97) Check Point 4-3 3d 4s 3p 3s 2p 2s Each orbital can accommodate 2 electrons with opposite spin. 1s

4.4 Atomic Orbitals

4.4 Atomic orbitals (p.98) s Orbitals Graph of probability of finding an electron against distance from nucleus

4.4 Atomic orbitals (p.98) s Orbitals

4.4 Atomic orbitals (p.100) p Orbitals The shapes and orientations of 2px, 2py and 2pz orbitals

Check Point 4-4 4.4 Atomic orbitals (p.101) d Orbitals The shapes and orientations of 3dxy, 3dyz, 3dx2-y2 and 3dz2 orbitals

The END

4.1 The electromagnetic spectrum (SB p.82) Let's Think 1 Some insects, such as bees, can see light of shorter wavelengths than humans can. What kind of radiation do you think a bee sees? Answer Ultraviolet radiation Back

4.1 The electromagnetic spectrum (SB p.87) Let's Think 2 What does the convergence limit in the Balmer series correspond to? Answer The convergence limit in the Balmer series corresponds to the energy required for the transition of an electron from n =2 to n = . Back

4.1 The electromagnetic spectrum (SB p.88) Example 4-1A Given the frequency of the convergence limit of the Lyman series of hydrogen, find the ionization enthalpy of hydrogen. Frequency of the convergence limit = 3.29  1015 Hz Planck constant = 6.626  10-34 J s Avogadro constant = 6.02  1023 mol-1 Answer

4.1 The electromagnetic spectrum (SB p.88) Back Example 4-1A For one hydrogen atom, E = h = 6.626  10-34 J s  3.29  1015 s-1 = 2.18  10-18 J For one mole of hydrogen atoms, E = 2.18  10-18 J  6.02  1023 mol-1 = 1312360 J mol-1 = 1312 kJ mol-1 The ionization enthalpy of hydrogen is 1312 kJ mol-1.

4.1 The electromagnetic spectrum (SB p.88) Example 4-1B The emission spectrum of atomic sodium is studied. The wavelength of the convergence limit corresponding to the ionization of a sodium atom is found. Based on this wavelength, find the ionization enthalpy of sodium. Wavelength of the convergence limit = 242 nm Planck constant = 6.626  10-34 J s Avogadro constant = 6.02  1023 mol-1 Speed of light = 3  108 m s-1 1 nm = 10-9 m Answer

For one mole of sodium atoms, E = hL = = = 494486 J mol-1 = 494 kJ mol-1 The ionization enthalpy of sodium of 494 kJ mol-1. 4.1 The electromagnetic spectrum (SB p.88) Back Example 4-1B

4.1 The electromagnetic spectrum (SB p.90) Check Point 4-1 • The first line of the Balmer series of the emission spectrum of atomic hydrogen corresponds to the energy emitted in the transition of an electron from the third energy level to the second energy level. It has a wavelength of 656.3 nm. What is the energy difference between the second and the third energy levels? • (Planck constant = 6.626  10-34 Js, Avogadro constant = 6.02  1023 mol-1) Answer

E = hv = • E = 6.626  10-34 J s  • = 3.03  10-19 J (for one electron) • For 1 mole of electrons, • E = 3.03  10-19 J  6.02  1023 mol-1 • = 182406 J mol-1 • = 182 kJ mol-1 4.1 The electromagnetic spectrum (SB p.90) Check Point 4-1

4.1 The electromagnetic spectrum (SB p.90) Check Point 4-1 (b) Given that the frequency of the convergence limit corresponding to the ionization of helium is 5.29  1015 Hz, calculate the ionization enthalpy of helium. (Planck constant = 6.626  10-34 Js, Avogadro constant = 6.02  1023 mol-1) Answer • For 1 mole of helium atoms, • I.E. = hvL • = 6.626  10-34 J s  5.29  1015 s-1  6.02  1023 mol-1 • = 2.11  106 J mol-1 • = 2110 kJ mol-1

E = • = • = 4.42  10-19 J 4.1 The electromagnetic spectrum (SB p.90) Check Point 4-1 (c) The blue colour in fireworks is often achieved by heating copper(I) chloride (CuCl) to about 1200 oC. The compound then emits blue light with a wavelength of 450 nm. What is the energy released per copper(I) ion at the specified condition? Answer

4.1 The electromagnetic spectrum (SB p.90) Check Point 4-1 • Name the element present in the sample when the following flame colours are observed in flame tests. • (i) Golden yellow • (ii) Lilac • (iii) Brick-red • (iv) Bluish green (d) (i) Sodium (ii) Potassium (iii) Calcium (iv) Copper Answer Back

4.2 Deduction of electronic structure from ionization enthalpies (p.94) Check Point 4-2 • Given the successive ionization enthalpies of boron, plot a graph of the logarithm of successive ionization enthalpies of boron against the number of electrons removed. Comment on the graph obtained. • Successive I.E. (in kJ mol-1): 800, 2400, 3700, 25000, 32800 Answer

4.2 Deduction of electronic structure from ionization enthalpies (p.94) (a) The first three electrons of boron are easier to be removed because they are in the outermost shell of the atom. As the fourth and fifth electrons are in the inner shell, a larger amount of energy is required to remove them. Check Point 4-2

4.2 Deduction of electronic structure from ionization enthalpies (p.94) Check Point 4-2 (b) Give a rough a sketch of the logarithm of successive ionization enthalpies of potassium. Explain your sketch. Answer

The Electronic Structure of Atoms