periodic table – ionization energies

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 Ephoton = hνPhoton.

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 Process

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Ionization EnergyMg(g) → Mg+(g) + e- I1 = 738 kJ

Mg+(g) → Mg2+(g) + e- I2 = 1451 kJ

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 9Slide 5 of 35

I = RH n2

Zeff2

Ionization energies decrease as atomic radii increase.

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 (ns1) 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(I1Values in kJ∙mol-1)

Alkali Metals & H Alkaline Earth Metals

The Halogens Noble Gases

H1312

He2372

Li520

Be899

F1681

Ne2080

Na496

Mg738

Cl1251

Ar1520

K419

Ca590

Br1140

Kr1351

Rb403

Sr549

I1008

Xe1170

Cs376

Ba503

At Rn1037

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 2nd and 3rd 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. Zeffective ≈ Z – S (Very roughly!)

Third Period – Effective ChargesTrends in Atomic Radii (pm)and

Ionization Energies

Na Mg Al Si P S Cl Ar

ZEffective 2.5 3.3 4.1 4.3 4.9 5.5 6.2 6.8

Atomic Radius(pm)

186 160 143 117 110 104 99

I1 (kJ mol∙ -1) 496 738 578 786 1012 1000 1251 1520

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First ionization energies as a function of atomic numberFIGURE 9-10


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:

• I1 = 738 kJ∙mol-1, I2 = 1451 kJ∙mol-1 and I3 = 7733 kJ∙mol-1 respectively. Let’s see if electron configurations help us understand these data.

Magnesium – Successive Ionizations

• Mg(g) → Mg+(g) + e- I1 = 738 kJ 1s22s22p63s2 1s22s22p63s1

• Mg+(g) → Mg2+(g) + e- I2 = 1451 kJ 1s22s22p63s1 1s22s22p6

• Mg2+(g) → Mg3+(g) + e- I3 = 7733 kJ 1s22s22p6 1s22s22p5

• The first large jump in ionization energy corresponds to the removal of the first non-valence electron. Why is this reasonable?

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I2 (Mg) vs. I3 (Mg)

77331451

I1 (Mg) vs. I1 (Al)

577.6

I1 (P) vs. I1 (S)

1012 999.6

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 O2- ion is stable in binary ionic compounds, though, due to the lattice energy of compounds such as MgO(s).

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Electron Affinity

F(g) + e- → F-(g) EA = -328 kJ


F(1s22s22p5) + e- → F-(1s22s22p6)

Li(g) + e- → Li-(g) EA = -59.6 kJ

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Electron affinities of main-group elements FIGURE 9-11


Values are in kilojoules per mole for the process

X(g) + e- X-(g).

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Second Electron Affinities

O(g) + e- → O-(g) EA = -141 kJ


O-(g) + e- → O2-(g) EA = +744 kJ

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Magnetic Properties• Diamagnetic atoms or ions:– All e- are paired.– Weakly repelled by a magnetic field.

• Paramagnetic atoms or ions:– Unpaired e-.– Attracted to an external magnetic field.


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.

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Paramagnetism

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Slide 21 of 35General Chemistry: Chapter 9

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Periodic Properties of the Elements


FIGURE 9-12•Atomic properties and the periodic table – a summary

“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.


332266 ??

Three halogen elementsFIGURE 9-13

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Melting points of the third-period elementsFIGURE 9-14


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Reducing Abilities of Group 1 and 2 Metals



2 K(s) + 2 H2O(l) → 2 K+ + 2 OH- + H2(g)

Ca(s) + 2 H2O(l) → Ca2+ + 2 OH- + H2(g)

I1 = 419 kJ

I1 = 590 kJ

I2 = 1145 kJ

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Oxidizing Abilities of the Halogen Elements

(Group 17)



2 Na + Cl2 → 2 NaCl

Cl2 + 2 I- → 2 Cl- + I2

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Acid-Base Nature of Element Oxides



Basic oxides or base anhydrides:Li2O(s) + H2O(l) → 2 Li+(aq) + 2 OH-(aq)

Acidic oxides or acid anhydrides:SO2 (g) + H2O(l) → H2SO3(aq)

Na2O and MgO yield basic solutions

Cl2O, SO2 and P4O10 yield acidic solutions

SiO2 dissolves in strong base, acidic oxide.

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) Al3+ ion and (f) Fe atom.

• 2. Arrange the following in order of increasing atomic radius: (a) Mg, Ba, Be, Sr (b) Rb+, Se2-, Br- and Sr2+ (c] Ca, Rb, F, S (d) Fe, Fe3+, Fe2+.

Class Examples – cont’d:

• 3. Write balanced chemical equations to represent the reactions of the following oxides with water: (a) SO3(g), (b) P4O10(s), (c) BaO(s) and (d) Li2O(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 4p6? How many orbitals does this notation include? What is the shape of the orbitals described using the 4p6 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) 1s22s22p64s1, (b) 1s22s22p63s23p63d2 (c) 1s22s22p63s23p63d84s24p2.

periodic table – ionization energies

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