Presentation on theme: "Periodic Table – Ionization Energies Energy is required to remove an electron from an isolated gaseous atom. This energy, the ionization energy, can be."— Presentation transcript:
Periodic Table – Ionization Energies Energy is required to remove an electron from an isolated gaseous atom. This energy, the ionization energy, can be determined very accurately. The energy is often supplied in the form of light. High frequency/short wavelength light is needed to ionize atoms. The energy needed to ionize one gaseous atom can be obtained from E photon = hν Photon.
Ionization Energies – cont’d: The H atom is unique in that H has only one electron. After this electron is removed no further ionization processes are possible. For all other atoms several ionization steps are possible. Increasing amounts amounts of energy are required to remove successive electrons. Why? There are clear periodic trends for ionization energies. Ionization energies will be used eventually in discussions of chemical bonding.
Ionization Energies – Group Trends For chemical families (Groups) ionization energies drop as atomic radii increase. We need to consider both atomic radii and effective nuclear charge. In alkali metals the single valence electron (ns 1 ) is removed first. As n increases the valence e - is located on average further from the nucleus. The effect of growing nuclear charge is again offset by inner or core electron screening.
Ionization Energies – Group Trends (I 1 Values in kJ∙mol -1 ) Alkali Metals & HAlkaline Earth Metals The HalogensNoble Gases H 1312 He 2372 Li 520 Be 899 F 1681 Ne 2080 Na 496 Mg 738 Cl 1251 Ar 1520 K 419 Ca 590 Br 1140 Kr 1351 Rb 403 Sr 549 I 1008 Xe 1170 Cs 376 Ba 503 AtRn 1037
Ionization Energies – Period Trends As we move from left to right across a period values of first ionization energies increase and, at the same time, atomic radii decrease. Moving across the 2 nd and 3 rd periods the atomic number, the nuclear charge and the total number of valence electrons increases steadily. The number of core electrons remains constant and causes screening to produce an effective nuclear charge for the outermost electrons. Z effective ≈ Z – S (Very roughly!)
Third Period – Effective Charges Trends in Atomic Radii (pm)and Ionization Energies Na Mg Al Si P S Cl Ar Z Effective 184.108.40.206.220.127.116.11.8 Atomic Radius(pm) 18616014311711010499 I 1 (kJ∙mol -1 )4967385787861012100012511520
Successive Ionization Energies - Trends Ionization energies increase as successively more electrons are removed from an atom. The relative values of the ionization energies show surprising jumps or “discontinuities”. This is illustrated by data for Mg where the first second and third ionization energies are: I 1 = 738 kJ∙mol -1, I 2 = 1451 kJ∙mol -1 and I 3 = 7733 kJ∙mol -1 respectively. Let’s see if electron configurations help us understand these data.
Magnesium – Successive Ionizations Mg(g) → Mg + (g) + e - I 1 = 738 kJ 1s 2 2s 2 2p 6 3s 2 1s 2 2s 2 2p 6 3s 1 Mg + (g) → Mg 2+ (g) + e - I 2 = 1451 kJ 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 Mg 2+ (g) → Mg 3+ (g) + e - I 3 = 7733 kJ 1s 2 2s 2 2p 6 1s 2 2s 2 2p 5 The first large jump in ionization energy corresponds to the removal of the first non- valence electron. Why is this reasonable?
Electron Affinities In the gas phase most nonmetal atoms will pick up an electron (or two!) to form a negatively charged monatomic ion. The process is usually exothermic and the term electron affinity tells us the size of the energy change associated with this process. Neglecting the Noble Gases the EA values are generally most exothermic for the non-metals with the smallest atomic radii.
Electron Affinities – cont’d: Surprisingly, most gaseous metal atoms can also pick up an electron (usually an exothermic process)! (Discuss this again when consider chemical bonding.) There are surprises with nonmetals! E.g., the second EA value for oxygen is +ve (an endothermic process). The O 2- ion is stable in binary ionic compounds, though, due to the lattice energy of compounds such as MgO(s).
Magnetic Properties For both atoms and monatomic ions we can readily identify paramagnetic species using orbital diagrams for the atoms and monatomic ions. Often one writes electron configurations first. With some practice one can jump to considering the valence electrons and write partial orbital diagrams for the valence electron subshells.
“Periodic Physical Properties” The next slide shows pictures of three molecular halogens – chlorine, bromine and iodine. “Obviously”, from the pictures alone!, one can see that chlorine has the lowest melting and boiling point. Why? Let’s estimate values for the missing melting point and boiling point of bromine and compare the estimates to the experimental values.
Class Examples: 1. Which of the following atoms and ions are paramagnetic (i.e. have unpaired electrons). Note: An even number of electrons does not indicate that all electrons are paired. (a) He atom, (b) F atom, (c) As atom, (d) F - ion (e) Al 3+ ion and (f) Fe atom. 2. Arrange the following in order of increasing atomic radius: (a) Mg, Ba, Be, Sr (b) Rb +, Se 2-, Br - and Sr 2+ (c] Ca, Rb, F, S (d) Fe, Fe 3+, Fe 2+.
Class Examples – cont’d: 3. Write balanced chemical equations to represent the reactions of the following oxides with water: (a) SO 3 (g), (b) P 4 O 10 (s), (c) BaO(s) and (d) Li 2 O(s). 4. Arrange the following atoms in order of increasing first ionization energy: (a) Fr, He, K, Br (b) P, As, N, Sb and (c) Sr, F, Si, Cl. 5. Why are transition metal atoms and ions so often paramagnetic?
Class Examples 6. What information does the term “degenerate orbitals” convey? 7. How do a ground state and an excited state electron configuration differ? 8. How many electrons are described using the notation 4p 6 ? How many orbitals does this notation include? What is the shape of the orbitals described using the 4p 6 notation?
Class Examples 9. Are all (neutral) atoms having an odd atomic number paramagnetic? Are all atoms having an even atomic number necessarily diamagnetic? Explain. 10. The following electron configurations do not correspond to the ground electronic state of any atom. Why? (a) 1s 2 2s 2 2p 6 4s 1, (b) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 (c) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 8 4s 2 4p 2.