Chapter 7: Structure of Atoms and periodic trends.

Transcription

1 Chapter 7: Structure of Atoms and periodic trends. End of chapter exercises: 1, 5, 11, 15, 19, 21, 27, 29, 31, 35, 37, 43, 58, Pauli Exclusion Principle No two electrons in an atom can have the same set of 4 quantum numbers (n, l, m l, m s ) governs arrangement of electrons in atoms. For a given orbital, i.e. 1s or 2p z n, l and m l are fixed, therefore m s must have different values. There are only 2 m s values orbital can only hold 2 electrons e -s must have opposite spin e.g. 3 p-orbitals each hold 2 e -s 6 e -s in p- subshell 5 d-orbitals 10 electrons n-subshells in the nth electron shell n 2 orbitals in a shell maximum 2n 2 electrons in any shell. Electron Configuration = the way electrons are distributed among the various orbitals of an atom. Ground state most stable state electrons in lowest possible energy states. Orbitals are filled in order of increasing energy Electrons assigned to shells (defined by n) of increasing energy Within a shell, electrons assigned to subshells (defined by l) of successively higher energy. Electrons assigned to ensure total energy of atom is always as low as possible Bohr model the energy of H-atom depends on n.(e = - Rhc/n 2 ) Atoms with more than 1 electron, sub-shell energies depend on both n and l. Subshell energy order 2 general rules Electrons assigned to sub-shells of increasing n + l value For 2 sub-shells with the same n + l value, electrons assigned first to sub-shell of lower n. e.g. e- assigned to 2s (n + l = = 2) before 2p (n + l = = 3) 3s (n + l = = 3) before 3p (n + l = = 4) before 3d (n + l = = 5) Atomic Electron Configurations. Hydrogen (H): Z = 1 (1 p + & 1 e - ) 1 e - in 1s orbital can use orbital notation or spdf notation 1s 1 Lithium (Li) and Group 1 A elements. Li (1st element period 2) Z = 3 (p + = 3, e - = 3) 1st 2 e-s in 1s orbital, 3rd e- in 2s: Abbreviated form: [He]2s 1 i.e. previous noble gas (which includes core e-s) with valence electrons Core electrons (i.e. e in previous E shell) can be ignored when considering the chemistry of atom. Remaining electrons i.e. 2s 1 = valence electrons electrons that determine the properties of the element. Position of Li on the periodic gives the configuration all group 1A elements have 1 valence electron in an s-orbital of the nth shell, where n = number of period in which element is found. e.g. Potassium (K), 4th period, 1st group: [Ar] 4s 1 1

2 Beryllium (Be) and group 2A Be: Z = 4 (4 p + and 4 e - ) 2 e -s in 1s and 2 e -s in 2s: [He]2s 2 All elements in group 2A: [preceding noble gas] ns 2 n = period where element is to be found Group 1A and Group 2A = s-block elements Boron (B) and group 3A B: Z = 5 (5 p + and 5 e - ) 1st element in block on RHS 1s and 2s orbitals are filled, 5th e - in 2p orbital 3 valence e-s Elements from group 3A 8A = p-block elements General configurations ns 2 np x, where x varies from 1 6 (group number 2) Carbon (C) and group 4A C: Z = 6 (6 p + and 6 e-) 1s 2 2s 2 2p 2 [He]2s 2 2p 2 2nd p-electron must be assigned to either of the remaining p-orbitals and will have the same spin as the 1st p-electron. Hund s Rule All elements of group 4A have similar outer shell configuration: [noble gas]ns 2 np 2 Nitrogen (N) and group 5A N: Z = 7 (7 p + and 7 e - ) [He]2s 2 2p 3 All p-electrons in different p-orbitals with same spin. All elements outer shell configuration [noble gas]ns 2 np 3, n = period in which element is located. Oxygen (O) and group 6A O: Z = 8 (8 p + and 8 e - ) 1s 2 2s 2 2p 4.[He]2s 2 2p 4 4th 2p-electron must pair up with one already present must have opposite spin to electron already there. Outer shell: ns 2 np 4 Fluorine (F) and group 7A (Halogens) F: Z = 9 (9 p + and 9 e - ) [He]2s 2 2p 5 Outer shell: ns 2 np 5 Neon (Ne) and group 8A (Noble gases) Group 8 (except He) 8 e-s in shell of highest value ns 2 np 6 i.e. all noble gases have filled ns and np subshells this explains their nearly complete chemical inertness. Ne: Z = 10 (10 p + and 10 e - ) 1s 2 2s 2 2p 6 [He]2s 2 2p 6 Period 3 Similar electron configuration to 2nd period preceding noble gas neon and valence shell n = 3 e.g. Si (group 4 A) compare with C 1s 2 2s 2 2p 6 3s 2 3p 2 or [Ne]3s 2 3p 2 Period 3 ends with Argon 1s 2 2s 2 2p 6 3s 2 3p 6 Q1. Give a set of quantum numbers for each of the valence electrons of aluminium: Period 4: Next element, K (Z = 19), does the electron go into 4s or 3d? 4s (n + l = = 4), 3d (n + l = = 5) n + l rule tell us 4s, also Chemical properties of K indicate it is a member of the alkali metal group implies outermost electron occupies an s-orbital and not 3d-orbital. Elements of 4th 7th periods use d- and f- subshells in addition to s- and p-subshells. 2

3 Elements filling d-subshells transition elements f-subshells lanthanides (4f) and actinides (5f) Transition elements always preceded by 2 s- block elements, i.e. fill ns orbital and then (n 1)d Period 4: Scandium (Sc) 1st transition element Sc: [Ar]3d 1 4s 2 or [Ar] 4s 2 3d 1 Ti: [Ar] 3d 2 4s 2 or [Ar] 4s 2 3d 2 Use Hund s rule singly into each d-orbital until all 5 have one electron each. All orbitals of given subshell i.e. 5 x 3d have the same energy degenerate Exceptions: Chromium (Cr) [Ar]3d 5 4s 1 not [Ar]3d 4 4s 2 3d and 4s orbitals have approximately the same energy in Cr each of the 6 valence electrons are assigned to one of the orbitals, same spin. Copper (Cu): [Ar] 3d 10 4s 1, more stable when all d orbitals have paired electrons. Q2. Give a set of quantum numbers for each of the valence electrons for Arsenic. Q3. a) Give the orbital energy diagram as well as the spdf electron configuration for the nickel atom in the ground state. b) Give a full set of quantum numbers for an electron in the highest subshell energy. c) If one of these e -s is excited by absorbing a photon, and 10 line spectra are observed, to which subshell was the electron excited? Lanthanides and Actinides 6th period includes the lanthanide series, starting with Lanthanum (La):[Xe] 5d 1 6s 2 (or [Xe] 6s 2 5d 1 ) La 1st element in d-block Next: Cerium in separate row 7 degenerate f-orbitals corresponding to 7 allowed m l values from +3 to -3. Elements in this row, Ce Lu are first to have electrons assigned to f-orbitals. Ce: [Xe] 4f 1 5d 1 6s 2 Energies of 4f and 5d orbitals are very similar occasional variations in occupation of 5d and 4f orbitals followed by 6p orbitals. Ends with Radon (Rn) the heaviest of known nobel gases. 7th period first fill the 7s orbitals also includes elements using f-orbitals, the actinides, which begin with Actinium (Ac). Ac: [Rn] 6d 1 7s 2 Next Thorium (Th): [Rn] 5f 1 6d 1 7s 2 first using 5f orbitals. Actinide elements are radioactive, most not found in nature. Q4: Using the periodic table, write condensed electron configurations for: (a) phosphorus, (b) cobalt, (c) tellurium, (d) bismuth. How many valence electrons? Paramagnetic or diamagnetic? Q5. In the ground state of mercury, Hg a) How many electrons occupy orbitals with n =3? b) How many electrons occupy d atomic orbitals? c) How many electrons occupy p z atomic orbitals? d) How many electrons have spin up (m s = +1/2) 3

4 Electron Configuration and Periodic Table Electron configuration of elements related to its position on the periodic table elements with same pattern of outer shell (valence) electrons arranged in groups (columns). e.g. Group 3A: B: [He] 2s 2 2p 1 Al: [Ne] 3s 2 3p 1 Ga: [Ar] 3d 10 4s 2 4p 1 In: [Kr] 4d 10 5s 2 5p 1 Tl: [Xe] 4f 14 5d 10 6s 2 6p 1 Electron configuration of Ions To form a cation from a neutral atom, one or more valence electrons are removed. electrons are always removed from electronshell of highest n. if several subshells are present within the nth shell, electron/s of maximum l are removed. e.g. Na ion remove 3s 1 electron from Na atom. Na: [Ne] 3s 1 or 1s 2 2s 2 2p 6 3s 1 Na+: 1s 2 2s 2 2p e - Electron configuration of Ions. Germanium: remove 2 x 4p electrons Ge: [Ar] 3d 10 4s 2 4p 2 Ge 2+ : [Ar] 3d 10 4s 2 + 2e - Ge 2+ : [Ar] 3d 10 4s 2 Ge 2+ : [Ar] 3d e - Same general rule applies to transition metals ns electrons always lost before (n-1)d electrons. e.g. Fe: [Ar] 3d 6 4s 2 Fe 2+ : [Ar] 3d 6 + 2e - Fe 2+ [Ar] 3d 6 Fe 3+ : [Ar] 3d 5 + e - cations formed have the general electron configurations [noble gas] (n-1)dx Electron configuration of Ions magnetic properties of transition metals determined by the number of unpaired electrons in d-orbitals Fe 3+ - ions are paramagnetic 5 unpaired electrons. e.g. Cu, Cu+ and Cu2+ Cu: [Ar] 3d 10 4s 1 Cu+: [Ar] 3d 10 + e - all e-s paired, diamagnetic Cu 2+ : [Ar] 3d 9 + 2e - one unpaired e-, paramagnetic Effective Nuclear Charge (Z*) i.e. the nuclear charge experienced by a particular electron in a multi-electron atom, modified by the presence of other electrons. In H: only 1 e -, 2s and 2p have the same Energy. Li: 3 e -s, presence of 1s e -s alters the energy of 2s and 2p subshells. Li: 3 protons in nucleus (Z = 3), nuclear charge = +3. If third e - (2s) is very far from nucleus experiences +1 nuclear charge (screened from nucleus by 2 x 1s e -s ). Due to the penetration of the 2s orbital into 1s orbital region, 2s e - experiences charge < +3 but > +1 Z* is greater for s-electrons than p-electrons in same energy shell s-electrons have lower E than p- electrons difference becomes larger as n becomes larger, i.e. down in a group. Penetrating power of sub-shells: s > p > d > f Effective nuclear charge experienced by orbitals: ns > np> nd > nf within the same e - - shell Z increases from left to right across the period increased number of protons value of Z* also increases across the period. Z* = Z S where S = screening constant, how much the inner (core) electrons shield or screen the outer electrons from the nucleus. 4

5 Atomic Properties and Periodic Trends Similarities in properties of elements result from similar valence shell electron configuration. Atomic electron configuration related to some of the physical and chemical properties of elements and why they change. Atomic Size Orbital has no sharp boundary Distance between atoms in element used to determine the radius. e.g. Cl Cl bond distance 198 pm radius of Cl = 99 pm Main Group Elements: atomic radii generally increase going down a group in the Period Table, and decrease across a period. Size of atom determined by From top to bottom outermost electrons assigned to orbitals with higher value of n (principle quantum number) Core electrons require space In a given period, n of valence electrons is the same. effective nuclear charge, Z*, increases slightly from left to right attraction between nucleus and electrons increases and radius decreases. Transition Metal Atoms Slight decrease initially from left to right, then slight increase. Electrons added to d-orbitals, increase in Z* causes the decrease only slight due to e - e - repulsion. Slight increase due to increasing e - - e - repulsion when d-electrons are pairing up. Ionisation Energy i.e. the energy required to remove an electron from an atom in the gas phase. Atom in ground state (g) Atom + (g) + e - E ionisation energy, IE To separate the electron from atom, energy must be supplied to overcome the attraction of the nuclear charge, IE positive. Atoms other than H have a series of IE s, more than 1 electron can be removed. Removing each subsequent electron requires more energy due to electron being removed from increasingly positive ion. e.g. 1st ionisation Mg (g) Mg + (g) + e - IE 1 = 738 kj/mol 1s 2 2s 2 2p 6 3s 2 1s 2 2s 2 2p 6 3s 1 2nd ionisation Mg + (g) Mg 2+ (g) + e - IE 2 = 1451 kj/mol 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 3s 0 3rd ionisation Mg 2+ (g) Mg 3+ (g) + e - IE 3 = 7733 kj/mol* 1s 2 2s 2 2p 6 1s 2 2s 2 2p 5 *Removal of core electrons requires much more energy than removal of valence electrons. Core electrons are not lost in chemical reactions. Main Groups (s- and p- block) First ionisation energies generally increase across a period and decrease down in a group. Across period: increase in effective nuclear charge, Z*, with increasing atomic number attracts the electrons more strongly atomic radius decreases energy to remove an electron increases. Variations due to e- e- repulsions i.e. group 5 6, 4th e - in 2p makes it easier to remove. Down in group: electron being removed is further from nucleus less nuclear electron attractive force less energy required to remove electron. 5

6 Electron Affinity (Electron Attachment Enthalpy EA H) Affinity = liking (for electrons), i.e. atoms acquire one or more electrons to form a negative ion. EA : Energy change for a process in which an electron is acquired by an atom in the gas phase. A (g) + e - A (g) U = EA H The greater the affinity an atom has for an electron, the more E is released the more negative the value of EA H. e.g. Fluorine: F (g) + e F (g) EA H = 328 kj/mol large value, reaction is exothermic and product favoured. Boron much lower affinity for an electron, EA H = 26.7 kj/mol EA H and IE represent energy involved in the gain or loss of electrons by an atom the formation of negative and positive ions. Increase in Z* across a period more difficult to ionise the atom (lose e - ) increased attraction for additional electrons element with high IE usually has a more negative EA H. Values for EA H generally become more negative moving L to R across period. Trend not smooth group 2 lower adding e- would start p-subshell, more difficult group 5 lower adding e- would start pairing p electrons. EA generally becomes less negative going down in a group more difficult to form neg. ion. Electrons added progressively further from nucleus decrease in attractive force between nucleus and electrons. Does not apply to the 2nd period smaller volume of n = 2 shell causes significant electron electron repulsions. Few elements, e.g. N, noble gases have no affinity for electrons, EA H = 0 e.g. for chlorine: Ionisation energy: Cl (g) Cl + (g) + e - E = 1251 kj/mol [Ne]3s 2 3p 5 [Ne]3s 2 3p 4 Positive E value E must be added to remove the e- Electron Affinity: Cl (g) + e - Cl - (g) E = -349 kj/mol [Ne]3s 2 3p 5 [Ne]3s 2 3p 6 noble gas structure Negative E value energy released Cl - more stable. Trends in Ion sizes down in a group same as neutral atom - positive and negative ions increase in size. Compare atom and ion sizes: electrons removed to form positive ions nucleus has stronger hold on remaining electrons i.e. radius of cation smaller than neutral atom, e.g. Li (152 pm); Li + (78 pm). large decrease also expected when 2 or more electrons are removed. e.g. Al (143 pm); Al 3+ (57 pm) loses n = 3 shell anions are larger than atoms from which they are formed When electron removed in Li, attractive force of 3 p + exerted on only 2 electrons 6

7 Compare isoelectric elements (same number of electrons) N 3- O 2- F - Na + Mg 2+ No. of electrons No. of protons Ionic radius (pm) electron proton attraction increases radius decreases electron electron repulsion increases Group 1 metals form compounds containing 1+ ions, i.e. Na +, K + 2 Na (s) + Cl 2 (g) 2 NaCl (s), i.e. Na + and Cl - ions 2 K (s) + 2 H 2 O (l) 2 KOH (aq) + H 2 (g) i.e. K + and OH - ions Not NaCl 2 or K(OH) 2 as the 2+ ion is not favourable C - does not have a favourable IE or EA H - does not easily form a cation or an anion don t find many ionic compounds containing C, C usually shares electrons with other elements, CO 2, CCl 4 Q6. Compare C, O and Si place in order of increasing atomic radius. Which has the largest ionisation energy? Q7. Compare B, Al and C for atomic radius and ionisation energy. Q8. Why is fluorine a smaller atom than carbon, but also smaller than chlorine? Use spdf electron configurations of all elements to explain. Q9. Give the trend in the sizes of the ions K +, S 2- and Cl -. Explain. Also Textbook No. 34, 41, 42, 48, 50 and 52. Q10. Using your knowledge of the trends in element sizes, explain why the density of the elements increases from K through V 7