Electron Configuration and Chemical Periodicity

Transcription

1 Electron Configuration and Chemical Periodicity Orbital approximation: the e density of an isolated many-electron atom is approximated by the sum of the e densities of each of the individual e taken separately. 1. Every e in an atom has a set of four quantum numbers (n, l, m l, m s ) that describe its spatial distribution and spin state. 2. Every e dwells in an atomic orbital with a characteristic size, shape and energy and has a spin (up or down). 3. In many-electron atoms the energies of the orbitals do not depend solely on n as in the one-electron atoms, but they depend on both n and l, but not on m l ) 2p orbital has higher energy than a 2s 3d orbital has higher energy than a 3p A group of orbitals with exactly equal energies comprise a subshell Subshells having similar energies make up a shell of orbitals Electrons with parallel spins ( ) tend to stay apart better than those with paired spins ( ), When e - e interactions become large the orbital approximation tends to break down. Pauli Exclusion Principle : No two e in an atom may have the same set of the four quantum numbers (n, l, m l, m s ). Electron Configuration: list of the occupied orbitals and the number of e in each. Ground-State Configuration: electronic configuration of lowest energy. Factors Affecting Atomic Orbital Energies 1. The Effect of Nuclear Charge (Z effective ) Higher nuclear charge lowers orbital energy (stabilizes the system) by increasing nucleus-electron attractions 2. The Effect of Electron Repulsions (Shielding) Additional electron in the same orbital (raises the orbital energy through electronelectron repulsions.) 1

2 Additional electrons in inner orbitals (shield outer electrons more effectively than do electrons in the same sublevel. Shielding by inner electrons greatly lowers the Zeff felt by outer electrons.) Penetration and orbital energy Order for filling energy sublevels with electrons. Aufbau Principle (aufbau means build-up in German) We use it when we are placing electrons into orbitals in the construction of polyelectronic atoms This principle states that in addition to adding protons and neutrons to the nucleus, one simply adds electrons to the hydrogen-like atomic orbitals Once more: Pauli exclusion principle: No two electrons mayhave the same quantum numbers. Therefore, only two electrons can reside in an orbital (with opposing spin) p subshells have three orbitals with the same energy d subshells have five orbitals with the same energy f subshells have seven orbitals with the same energy Each of these orbitals may accommodate a maximum of two electrons. 2

3 Filling order of the Periodic Table Orbitals are filled starting from the lowest energy. Example: Hydrogen 1s 1 1s 2s 2p Example: Helium (Z = 2) 1s 2 1s 2s 2p 3

4 Lithium (Z = 3) 1s 2 2s 1 1s 2s 2p Berillium (Z = 4) 1s 2 2s 2 1s 2s 2p Boron (Z = 5) 1s 2 2s 2 2p 1 1s 2s 2p Carbon (Z = 6) 1s 2 2s 2 2p 2 1s 2s 2p Hund s Rule: Lowest energy configuration is the one in which the maximum number of unpaired electrons are distributed amongst a set of degenerate orbitals. Nitrogen (Z = 7) 1s 2 2s 2 2p 3 1s 2s 2p 4

5 A vertical orbital diagram for the Li ground state Determining Quantum Numbers from Orbital Diagrams PROBLEM: Write a set of quantum numbers for the third electron and a set for the eighth electron of the F atom. PLAN: 9F Use the orbital diagram to find the third and eighth electrons. 1s 2s 2p SOLUTION: The third electron is in the 2s orbital. Its quantum numbers are: n = 2 l = 0 m l = 0 m s = + or - The eighth electron is in a 2p orbital. Its quantum numbers are: 1 2 n = 2 l = 1 m l = -1, 0, or +1 m s = + or Oxygen (Z = 8) Fluorine (Z = 9) 1s 2s 2p 1s 2 2s 2 2p 4 1s 2 2s 2 2p 5 Neon (Z = 10) 1s 2s 2p 1s 2s 2p 1s 2 2s 2 2p 6 full 5

6 Sodium (Z = 11) Ne 1s 2 2s 2 2p 6 3s 1 [Ne]3s 1 3s Argon (Z = 18) Ne [Ne] 3s 2 3p 6 3s 3p 6

7 Condensed ground-state electron configurations in the first three periods 7

8 A periodic table of partial ground-state electron configurations The relation between orbital filling and the periodic table 8

9 Keep in mind: Elements in a group have similar chemical properties because they have similar outer electron configurations. Categories of electrons 1. Inner (core) electrons are those seen in the previous noble gas and any completed transition series. They fill all the lower energy levels of an atom. 2. Outer electrons are those in the highest energy level (hi ghest n value). They spend more of their time farthest from the nucleus. 3. Valence electrons are those involved in forming compounds. Among the maingroup elements, the valence electrons are the outer electrons. For the transition elements, all the (n -1)d electrons are counted among the valence electrons also, even though the elements Fe (Z = 26 through Zn (Z = 30) use only a few of them in bonding as we will see later towards the last weeks of the course. Key information is embedded in the periodic table 1. Among the main-group elements (A groups), the group number equals the number of outer electrons (those with the highest n). Chlorine (Group 7A) has 7 outer electrons; Tellurium (Group 6A) has 6 outer electrons. 2. The period number is the n value of the highest energy level. 3. The n value squared (n 2 ) gives the total number of orbitals and 2n 2 gives the maximum number of electrons in the energy level. Unusual Configurations: Transition and Inner Transition Elements Periods 4,5, 6 and 7 incorporate the d block transition elements. The general trend is to fill the (n -1)d orbitals between the ns and np orbitals. Period 5 follows the same general pattern as Period 4. In Period 6, the 6s sublevel is filled in Cs and Ba and then La (Z = 57), the first member of the 5d transition series, occurs. (Inner transistion elements). For these elements filling of the f orbitals intervenes. f orbitals: l = 3 m l = -3,-2,-1,0,+1,+2,+3 (7 orbitals). The Period 6 inner transition series fills the 4f orbitals and consists of the lanthanides (or rare earths) because they occur after and are similar to La. The other inner transition series holds the actinides which fill the 5f orbitals that appear in Period 7. In both series the (n 2)f orbitals are filled after which the (n -1)d orbitals proceeds. Period 6 ends proceeds with filling the 6p orbitals but Period 7 is incomplete because only four elements with 7p electrons have been synthesized so far, Anomalies Cr: 4s 1 3d 5 Mo: 5s 1 4d 2 Cu: 4s 1 3d 10 Ag: 5s 1 3d 10 Au: 6s 1 3d 10 9

10 Determining Electron Configuration PROBLEM: Using the periodic table on the inside cover of the text give the full and condensed electron configurations, partial orbital diagrams showing valence electrons, and number of inner electrons for the following elements: (a) Potassium (K; Z = 19) (b) Molybdenum (Mo; Z = 42) (c) Lead (Pb; Z = 82) PLAN: Use the atomic number for the number of electrons and the periodic table for the order of filling for electron orbitals. Condensed configurations consist of the preceding noble gas and outer electrons. SOLUTION: (a) for K (Z = 19) full configuration condensed configuration partial orbital diagram 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 [Ar] 4s 1 There are 18 inner electrons.and 1 valence electron (b) for Mo (Z = 42) full condensed configuration partial orbital diagram 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 5 [Kr] 5s 1 4d 5 There are 36 inner electrons and 6 valence electrons. (c) for Pb (Z = 82) full 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 2 condensed configuration partial orbital diagram [Xe] 6s 2 4f 14 5d 10 6p 2 There are 78 inner electrons and 4 valence electrons. 10

11 Trends in three key atomic properties All physical and chemical behavior of the elements is based ultimately on the electron configurations of their atoms. Atomic size: how closely one atoms lies next to another. Defining covalent and metallic radii The metallic radius is ½ the distance between nuclei of adjacent atoms in a crystal of the element. The covalent radius is ½ the distance between bonded nuclei in a molecule of the element. C; in a covalent compound, the bond length and known covalent radii are used to determine other radii. Bond length of C Cl : 177 pm Covalent radius of Cl: 100 pm covalent radius of C: = 77 pm Trends among main-group elements 1. Changes in n: as n increases the probability that the outer electrons spend more time farther from the nucleus increases and the atoms are larger. 2. Changes in Z eff : as the Z eff - the positive charge felt by an e - increases, outer e are pulled closer to the nucleus, the atoms are smaller. a) Down a group, n dominates. Atomic radius generally increases in a group from top to bottom. b) Across a period Z eff dominates. Atomic radius generally decreases in a period from left to right. Trends among the transition metals As we move from left to right, size shrinks through the first two or three transition elements because of the increasing Z eff. But from then on, the size remains relatively constant because shielding by the inner d electrons counteracts the increasing Z eff. More on this when we will discuss the transition metal compounds. 11

12 Atomic radii of the main-group and transition elements Periodicity of atomic radius Trends in Ionization Energy The ionization energy (IE) is the energy in kj required for the complete removal of 1 mol of electrons from 1 mol of gaseous atoms or ions. Many-electron atoms can lose more than one electron. The first ionization energy (IE 1 ) removes an outermost electron (highest sub-level) from the gaseous atom: Atom (g) ion + (g) + e E = IE 1 The second ionization energy (IE 2 ) removes a second electron: Ion + (g) ion 2+ (g) + e E = IE 2 (IE 2 > IE 1 ) 12

13 Periodicity of first ionization energy (IE 1 ) The lowest values occur for the alkali metals and the highest for the noble gases. First ionization energies of the main-group elements In general, IE a) Decreases down a group b) Increases across a period (since Z eff increases and atomic size decreases) However there are several small dips as shown in the adjacent figure For Be: 13

14 EXAMPLE Identifying an Element from Successive Ionization Energies PROBLEM: Name the Period 3 element with the following ionization energies (in kj/mol) and write its electron configuration: IE 1 IE 2 IE 3 IE 4 IE 5 IE ,230 PLAN: Look for a large increase in energy which indicates that all of the valence electrons have been removed. SOLUTION: The largest increase occurs after IE 5, that is, after the 5th valence electron has been removed. Five electrons would mean that the valence configuration is 3s 2 3p 3 and the element must be phosphorous, P (Z = 15). The complete electron configuration is 1s 2 2s 2 2p 6 3s 2 3p 3. Trends in Electron Affinity The electron affinity (EA) is the energy change in kj accompanying the addition of 1 mol of electrons to 1 mol of gaseous atoms or ions. The first EA refers to the formation of 1 mol of monovalent (1 ) gaseous anions: Atom (g) + e ion - (g) E = EA 1 In most cases energy is released when an e is added because it is attracted to the atom s nuclear charge. The second EA 2 must always be positive because energy must be absorbed to overcome electrostatic repulsions and add an e to a negative ion. 14

15 Electron affinities of the main-group elements Despite the irregularities a) elements in group 6A and especially in 7A have high IE and high EA. They lose e with difficulty but attract e strongly. In their ionic compounds they form negative ions. b) Elements in groups 1A and 2A have low IE and slightly negative EA. They lose e readily but attract e very weakly. In their ionic compounds they form positive ions. c) Noble gases (Group 8A) have high IE and slightly positive EA. They tend not to lose or gain e. Trends in three atomic properties Trends in metallic behavior 15

16 Example Ranking Elements by First Ionization Energy PROBLEM: Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE 1 : (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs PLAN: IE decreases as you proceed down in a group; IE increases as you go across a period. SOLUTION: (a) He > Ar > Kr (b) Te > Sb > Sn (c) Ca > K > Rb (d) Xe > I > Cs These three elements are all in Group 8A(18), IE decreases down a group. These are all in Period 5, IE increases across a period. Ca is to the right of K; Rb is below K. I is to the left of Xe; Cs is furtther to the left and down one period. Acid-base behavior of the element oxides Most main-group metals transfer e to oxygen, so their oxides are ionic. In water the oxides act as bases producing OH ions from O 2 and reacting with acids. Non-metals share e with oxygen, so non-metal oxides are covalent. In water they act as acids producing H + ions and reacting with bases. Some metals and many metalloids form oxides that can act as acids or as bases in water. They are called amphoteric (Al 2 O 3 ) The trend in acid-base behavior of element oxides As the elements become more metallic down a group, the oxides become more basic (blue) As the elements become less metallic across a period, their oxides become more acidic (red) Sb 2 O 5 : weakly basic, SiO 2, As 2 O 3 : weakly acidic 16

17 Writing Electron Configurations of Main-Group Ions PROBLEM: Using condensed electron configurations, write reactions for the formation of the common ions of the following elements: (a) Iodine (Z = 53) (b) Potassium (Z =19) (c) Indium (Z = 49) PLAN: Ions of elements in Groups 1A(1), 2A(2), 6A(16), and 7A(17) are usually isoelectronic with the nearest noble gas. Metals in Groups 3A(13) to 5A(15) lose the np and ns or just the np electrons. SOLUTION: (a) Iodine (Z = 53) is in Group 7A(17) and will gain one electron to be isoelectronic with Xe: I ([Kr]5s 2 4d 10 5p 5 ) + e - I - ([Kr]5s 2 4d 10 5p 6 ) (b) Potassium (Z = 19) is in Group 1A(1) and will lose one electron to be isoelectronic with Ar: K ([Ar]4s 1 ) K + ([Ar]) + e - (c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three electrons: In ([Kr]5s 2 4d 10 5p 1 ) In + ([Kr]5s 2 4d 10 ) + e - In ([Kr]5s 2 4d 10 5p 1 ) In 3+ ([Kr] 4d 10 ) + 3e - The Period 4 crossover in sublevel energies For main-group, s-block metals remove all e with the highest n value For main-group, p-block metals remove np e before ns e For transition (d-block) metals, remove ns e before (n-1)d For non-metals, add e to the p orbitals of highest n value Magnetic Properties A species with unpaired e exhibits paramagnetism, attracted by an external magnetic field. A species with all electrons paired exhibits diamagnetism, it is not attracted and, in fact, is slightly repelled by a magnetic field. 17

18 PROBLEM: PLAN: SOLUTION: Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic. (a) Mn 2+ (Z = 25) (b) Cr 3+ (Z = 24) (c) Hg 2+ (Z = 80) Write the electron configuration and remove electrons starting with ns to match the charge on the ion. If the remaining configuration has unpaired electrons, it is paramagnetic. (a) Mn 2+ (Z = 25) Mn ([Ar] 4s 2 3d 5 ) Mn 2+ ([Ar] 3d 5 ) + 2e - paramagnetic (b) Cr 3+ (Z = 24) Cr ([Ar] 4s 1 3d 5 ) Cr 3+ ([Ar] 3d 3 ) + 3e - paramagnetic (c) Hg 2+ (Z = 80) Hg ([Xe] 6s 2 4f 14 5d 10 ) Hg 2+ ([Xe] 4f 14 5d 10 ) + 2e - not paramagnetic (diamagnetic) 18

19 Ionic radius Ionic vs. atomic radius Ionic size increases down a group Ionic size decreases across a period but increases from cations to anions Ionic size decreases with increasing positive charge in an isoelectronic species and the opposite Ionic charge decreases as charge increases for different cations of the same element Example Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca 2+, Sr 2+, Mg 2+ (b) K +, S 2-,, Cl - (c) Au +, Au 3+ (a) Sr 2+ > Ca 2+ > Mg 2+ These are members of the same Group 2A(2), and decrease in size going up the group. (b) S 2- > Cl - > K + The ions are isoelectronic; S 2- has the smallest Z eff and therefore, is the largest while K + is a cation with a large Z eff and is the smallest. (c) Au + > Au 3+ The greater the + charge, the smaller the ion 19

20 The Periodic Table Types of Chemical Bonding 1. Metal with nonmetal: electron transfer and ionic bonding 2. Nonmetal with nonmetal: electron sharing and covalent bonding 3. Metal with metal: electron pooling and metallic bonding 20

21 Lewis Electron-Dot Symbols For main group elements - The A group number gives the number of valence electrons Place one dot per valence electron on each of the four sides of the element symbol Pair the dots (electrons) until all of the valence electrons are used For Periods 2 and 3 Example Use partial orbital diagrams and Lewis symbols to depict the formation of Na + and O 2- ions from the atoms, and determine the formula of the compound the ions form. 21

22 The Ionic Bonding Model Central idea: transfer of e from metal atoms to nonmetal atoms to form ions that come together in a solid ionic compound. Three ways to represent the formation of Li + and F through electron transfer The total number of e lost by the metal atoms equal the total number of e gained by the nonmetal atoms. Energy consideration: the importance of lattice energy Consider, Li (g) Li + (g) + e IE 1 = 520 kj F (g) + e F (g) EA = kj The two-step electron-transfer process by itself requires energy: Li (g) + F (g) Li + (g) + F (g) IE 1 + EA = 192 kj The total energy needed is even greater than this because metallic Li and diatomic fluorine must be first converted to separate gaseous atoms, which also requires energy. o The H f of LiF (s) is 617 kj/mol, which means that 617 kj of energy is released when 1 mol of of LiF (s) forms from its elements. This means that there must be some exothermic components large enough to overcome the endothermic component discussed earlier. This component arises from the strong attraction between many oppositely charged ions. Li + (g) + F (g) LiF (s) H o = 755 kj And more energy is released when the gaseous ions coalesce in the crystalline structure because each ion attracts others of opposite charge, H o = 1050kJ. The negative of this value is 1050 kj and it is called the lattice energy, the enthalpy change that occurs 22

23 when 1 mol of ionic solid separates into gaseous ions. It indicates the strength of ionic interactions that influence melting point, solubility and other properties. Coulomb s law Periodic Trends in Lattice Energy Electrostatic force charge A x charge B distance energy = force x distance therefore, cation charge x anion charge Electrostatic energy cation radius + anion radius α H 0 lattice 1. Effect of ionic size: as we move down a group in the periodic table, the ionic radius increases and the electrostatic energy between cations and anions decreases. 2. Effect of ionic charge: LiF : 1050 kj/mol, (1 x1 charge), (Li is +1, F is -1) MgO: 3923 kj/mol (2 x 2 charge) (Mg is +2, O is -2) Lattice energy of MgO is about 4 times as of LiF. 23

24 Covalent Bond (sharing of e between atoms) Covalent bond formation in H 2 Distribution of electron density of H 2 24

25 Electronegativity Linus Pauling (the greatest American chemist!) It is a measure of the ability of an atom to attract e to itself. Robert Mulliken (U of Chicago) (1934): 1 electroneg ativity(χ ) ( IE1 + EA) 2 The Pauling electronegativity (EN) scale Polar covalent bond: (dipole moment) and percent ionic character Electron density distributions in H 2, F 2, and HF dipole moment, µ µ = (eδ) R To denote polarity, the arrow should point toward the negative end. 25

26 The ionic character of chemical bonds Determining Bond Polarity from EN Values PROBLEM: (a) Use a polar arrow to indicate the polarity of each bond: N H, F N, I Cl. (b) Rank the following bonds in order of increasing polarity: H N, H O, H C. PLAN: (a) Use above figure to find EN values; the arrow should point toward the negative end. (b) Polarity increases across a period. SOLUTION: (a) The EN of N = 3.0, H = 2.1; F = 4.0; I = 2.5, Cl = 3.0 N - H F - N I - Cl (b) The order of increasing EN is C < N < O; all have an EN larger than that of H. H C < H N < H O 26

27 Oxoacids and their strength They contain the group X O H Oxoacids of the same structure show increasing acid strength as the electronegativity of the central atom increases. Their strength with a given central element increases with the oxidation state of the central atom, or equivalently, with the number of lone oxygen atoms attached to the central atom. Example Which is stronger? H 2 SO 3 or H 2 SeO 3 H 2 SO 3 because S is more electronegative than Se. HIO 3 or HIO? HIO 3 = IO 2 (OH) - 2 more O HIO =I(OH) Example Write the electron configuration of the first excited state of F -, O 2- F: 1s 2 2s 2 2p 5 F - : 1s 2 2s 2 2p 6 Excited state F - : 1s 2 2s 2 2p 5 3s 1 O: 1s 2 2s 2 2p 4 O 2- : 1s 2 2s 2 2p 6 Excited state O 2- : 1s 2 2s 2 2p 5 3s 1 27