Ch-8, Periodic Relationships Among the Elements

Development of the Periodic Table 1.Periodic Table based on atomic masses Law of Octaves: Every eighth element had similar properies (Newlands), inadequate beyond Ca 2. Mendeleev’s Periodic Table: Based on regular, periodic recurrence of properties and atomic mass – Still had some discrepancies. 3. Mosely refined Mendelev’s periodic Table based on atomic number, which represents number of protons in the element and the number of electrons when the elements are at ground state.

When the Elements Were Discovered

ns 1 ns 2 ns 2 np 1 ns 2 np 2 ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 d1d1 d5d5 d 10 4f 5f Ground State Electron Configurations of the Elements

Classification of the Elements

Electron Arrangement and the Periodic Table Electron configuration - describes the arrangement of electrons in atoms. The electron arrangement is the primary factor in understanding how atoms join together to form compounds. Valance electrons - the outermost electrons in the outermost shell. – These are the electrons involved in chemical bonding. 3

For the representative elements: –The number of valance electrons is the group number. –The period number gives the energy level (n) of the valance shell. For an atom of fluorine, how many valance electrons does it have and what is the energy level of these electrons? Fluorine has 7 electrons in the n=2 level Valance Electrons

Let’s look at fluorine more closely. What is the total number of electrons in fluorine? –The atomic number is 9. It therefore has 9 protons and 9 electrons. If there are 7 electrons in the valance shell, (with n = 2 energy level) where are the other two electrons? –In the n = 1 energy level. This level holds two and only two electrons. –n=2, 2s 2, 2p 5, 2 p can accept one moe e -, F - –Then F will get Ne configuration.

Isoelectronic - they have the same electron configuration (same number of electrons) Nonmetallic elements tend to form negatively charged ions called anions. Nonmetals tend to gain electrons so they become isoelectronic with its nearest noble gas neighbor. O [He]2s 2 2p 4 + 2e - O 2- [He]2s 2 2p 6 or [Ne]

Electron Configurations of Cations and Anions Na [Ne]3s 1 Na + [Ne] Ca [Ar]4s 2 Ca 2+ [Ar] Al [Ne]3s 2 3p 1 Al 3+ [Ne] Atoms lose electrons so that cation has a noble-gas outer electron configuration. H 1s 1 H - 1s 2 or [He] F 1s 2 2s 2 2p 5 F - 1s 2 2s 2 2p 6 or [Ne] O 1s 2 2s 2 2p 4 O 2- 1s 2 2s 2 2p 6 or [Ne] N 1s 2 2s 2 2p 3 N 3- 1s 2 2s 2 2p 6 or [Ne] Atoms gain electrons so that anion has a noble-gas outer electron configuration. Of Representative Elements

Cations and Anions Of Representative Elements

Na + : [Ne]Al 3+ : [Ne]F - : 1s 2 2s 2 2p 6 or [Ne] O 2- : 1s 2 2s 2 2p 6 or [Ne]N 3- : 1s 2 2s 2 2p 6 or [Ne] Na +, Al 3+, F -, O 2-, and N 3- are all isoelectronic with Ne What neutral atom is isoelectronic with H - ? H - : 1s 2 same electron configuration as He

Electron Configurations of Cations of Transition Metals When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals. Fe: [Ar]4s 2 3d 6 Fe 2+ : [Ar]4s 0 3d 6 or [Ar]3d 6 Fe 3+ : [Ar]4s 0 3d 5 or [Ar]3d 5 Mn: [Ar]4s 2 3d 5 Mn 2+ : [Ar]4s 0 3d 5 or [Ar]3d 5

Trends in the Periodic Table We will look at the following trends – effective nuclear charge – atomic size – ion size – ionization energy – electron affinity

Periodic Properties Two factors determine the size of an atom. One factor is the principal quantum number, n. The larger is “n”, the larger the size of the orbital. The other factor is the effective nuclear charge, which is the positive charge an electron experiences from the nucleus minus any “shielding effects” from intervening electrons.

Effective nuclear charge (Z eff ) is the “positive charge” felt by an electron. Na Mg Al Si Z eff Core Z Radius (pm) Z eff = Z -  0 <  < Z (  = shielding constant) Z eff  Z – number of inner or core electrons

17 Effective Nuclear Charge (Z eff ) increasing Z eff Attr.force between nucleus and an electron directly proportional to eff.nuclear charge, inversely proportional to the square of the distance of sepn.

The radius of the atom is defined as one-half the distance between the centers of the atoms in a molecule (diatomic) The radius of the atom is defined as one-half the distance between the centers of two adjacent atoms(metals)

Trends

Atomic Size 1.The size of the atoms increases from top to bottom. This is due to the valance shell being higher in energy and farther from the nucleus. 2.The size of the atoms decreases from left to right. This is due to the increase in magnitude of positive charge in nucleus. The nuclear charge pulls the electrons closer to the nucleus. 7

Atomic Radii

Comparison of Atomic Radii with Ionic Radii

Cation is always smaller than atom from which it is formed. Anion is always larger than atom from which it is formed.

Ion Size 7 Cations are always smaller than their parent atom. –This is due to more protons than electrons. The extra protons pulls the remaining electrons closer. –Which would be smaller, Fe 2+ or Fe 3+ ? –Fe 3+ –This size trend is also due to the fact that it is the outer shell that is lost.

Anions are always larger than their parent atom. –This is due to the fact that anions have more electrons than protons.

The Radii (in pm) of Ions of Familiar Elements

Ionization energy is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state (kJ of energy to remove one mole of electrons from one mole of gaseous atoms). I 1 + X (g) X + (g) + e - I 2 + X + (g) X 2 + (g) + e - I 3 + X 2+ (g) X 3 + (g) + e - I 1 first ionization energy I 2 second ionization energy I 3 third ionization energy I 1 < I 2 < I 3 ionization energy + Na  Na + + e -

Filled n=1 shell Filled n=2 shell Filled n=3 shell Filled n=4 shell Filled n=5 shell Variation of the First Ionization Energy with Atomic Number

Ionization decreases down a group because the outermost electrons are farther from the nucleus. Ionization increases across a period because the outermost electrons are more tightly held. Why do you think that the noble gases would be so unreactive?

General Trend in First Ionization Energies Increasing First Ionization Energy

Application of Ionization energy and Emission Spectrum Inductively Coupled Plasma- Atomic Emission Spectroscopy (ICP-AES)

The principle of the ICP-AES is quite simple. The sample is exposed to the extremely high temperature of an argon plasma (up to K) that breaks the sample into atoms, ionizes these atoms, and electronically excites the resulting ions. When the excited electrons in these ions fall back to lower energy levels, they emit light. The wavelengths of light emitted by a particular element serve as a “fingerprint” for that element. Therefore by measuring the wavelengths of light emitted by our sample, we can identify the elements in the sample; and by measuring the amount of light emitted by a particular element in our sample, we can determine the concentration of that element.

8.4 (a)Which atom should have a smaller first ionization energy: oxygen or sulfur? (b)Which atom should have a higher second ionization energy: lithium or beryllium?

8.4 Strategy (a)First ionization energy decreases as we go down a group because the outermost electron is farther away from the nucleus and feels less attraction. (b)Removal of the outermost electron requires less energy if it is shielded by a filled inner shell. Solution (a) Oxygen and sulfur are members of Group 6A. They have the same valence electron configuration (ns 2 np 4 ), but the 3p electron in sulfur is farther from the nucleus and experiences less nuclear attraction than the 2p electron in oxygen. Thus, we predict that sulfur should have a smaller first ionization energy.

8.4 (b) The electron configurations of Li and Be are 1s 2 2s 1 and 1s 2 2s 2, respectively. The second ionization energy is the minimum energy required to remove an electron from a gaseous unipositive ion in its ground state. For the second ionization process, we write Because 1s electrons shield 2s electrons much more effectively than they shield each other, we predict that it should be easier to remove a 2s electron from Be + than to remove a 1s electron from Li +.

8.4 Check Compare your result with the data shown in Table 8.2. In (a), is your prediction consistent with the fact that the metallic character of the elements increases as we move down a periodic group? In (b), does your prediction account for the fact that alkali metals form +1 ions while alkaline earth metals form +2 ions?

Electron Affinity Electron Affinity - The energy change when a single electron is added to an isolated atom. Br + e -  Br - + energy Electron affinity gives information about the ease of anion formation. –Large electron affinity indicates an atom becomes more stable as it forms an anion. 7 8

Variation of Electron Affinity With Atomic Number (H – Ba)

E.A. generally decreases down a group. E.A. generally increases across a period.

8.5 Why are the electron affinities of the alkaline earth metals, shown in Table 8.3, either negative or small positive values?

8.5 Strategy What are the electron configurations of alkaline earth metals? Would the added electron to such an atom be held strongly by the nucleus? Solution The valence electron configuration of the alkaline earth metals is ns 2, where n is the highest principal quantum number. For the process where M denotes a member of the Group 2A family, the extra electron must enter the np subshell, which is effectively shielded by the two ns electrons (the ns electrons are more penetrating than the np electrons) and the inner electrons. Consequently, alkaline earth metals have little tendency to pick up an extra electron.

Mainly, the first three elements in the second period show similarities.

General Observations Several general observations can be made about the main-group elements. The metallic characteristics of these elements generally decrease across a period from left to right in the periodic table. Metallic characteristics of the main-group elements become more pronounced going down any column (group).

46 Group 1A Elements (ns 1, n  2) M M e - 2M (s) + 2H 2 O (l) 2MOH (aq) + H 2(g) 4M (s) + O 2(g) 2M 2 O (s) Increasing reactivity

47 Group 2A Elements (ns 2, n  2) M M e - Be (s) + 2H 2 O (l) No Reaction Increasing reactivity Mg (s) + 2H 2 O (g) Mg(OH) 2(aq) + H 2(g) M (s) + 2H 2 O (l) M(OH) 2(aq) + H 2(g) M = Ca, Sr, or Ba

48 Group 3A Elements (ns 2 np 1, n  2) 4Al (s) + 3O 2(g) 2Al 2 O 3(s) 2Al (s) + 6H + (aq) 2Al 3+ (aq) + 3H 2(g)

49 Group 4A Elements (ns 2 np 2, n  2) Sn (s) + 2H + (aq) Sn 2+ (aq) + H 2 (g) Pb (s) + 2H + (aq) Pb 2+ (aq) + H 2 (g)

50 Group 5A Elements (ns 2 np 3, n  2) N 2 O 5(s) + H 2 O (l) 2HNO 3(aq) P 4 O 10(s) + 6H 2 O (l) 4H 3 PO 4(aq)

51 Group 6A Elements (ns 2 np 4, n  2) SO 3(g) + H 2 O (l) H 2 SO 4(aq)

52 Group 7A Elements (ns 2 np 5, n  2) X + 1e - X - 1 X 2(g) + H 2(g) 2HX (g) Increasing reactivity

53 Group 8A Elements (ns 2 np 6, n  2) Completely filled ns and np subshells. Highest ionization energy of all elements. No tendency to accept extra electrons.

54 The metals in these two groups have similar outer electron configurations, with one electron in the outermost s orbital. Chemical properties are quite different due to difference in the ionization energy. Comparison of Group 1A and 1B Lower I 1, more reactive

55 Properties of Oxides Across a Period basicacidic

Homework- Ch-8 Example problems, practice problems, review of concepts 8.2,8.7,8.20,8.37,8.38,8.43,8.45,8.55,8.63, 8.64,8.72,8.74,8.78,8.84,8.107,8.120