Principles of Chemistry I
Principles of Chemistry I CHEM 1307
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The Structure of Atoms and Periodic Trends Chapter 6 Reminders Each orbital can hold no more than two electrons A set of 4 quantum numbers n 6 me ms describes the energy state and orbitals available to the electrons in an atom Name Symbol Permitted Property values Principle n 1 2 3 Orbital size Angular 6 0 1 2 n 1 Orbital shape momentum 0 s 1 p 2 d 3 f Magnetic me i 6 Orbital orientation in space Spin ms 12 or 12 Direction of e39 spin To describe the electron in an H atom n 1e0me0ms12 Therefore the electron in an H atom is found in the ls orbital Pauli Exclusion Principle No two electrons in an atom can have the same set of four quantum numbers n a me ms No atomic orbital can be assigned more than two electrons so each electron must have a different value of ms To describe the electrons in a He atom n1 0me0ms 12 n 1 0me0ms12 The two electrons fill the ls orbital We can represent this pictorially using an orbital box diagram 15 Two electrons in 15 orbital This electron has n 1 C 0 My 0 m5 12 This electron has n 1 6 O m 0 m5 12 EraaksCol Canaan Learmrn Number of Electrons in a ShellSubshell mam Orbital m leztm s mu Suhshells Available Possible in Slllnllell Possible or m 71 Available 2 1 22 1 Shell 211 l S 1 Z 2 2 5 1 2 E p a s 3 S l 2 18 p 3 a d 5 l0 4 S 1 Z 32 p 3 s d 5 ll f 7 14 s x 1 z 50 p 3 6 d 5 10 f 7 14 g 9 n 5 5 1 2 72 3 6 d 5 m r 7 14 g39 9 121 h 1 1 22 Them orhmls are no occuplcd in me ground ml or any known cicmcm n1 0ml0 15 orbital contains 2 e n2 0ml0 25 orbital contains 2 e n 2 1ml10 1 2p orbitals contain 6 e n3 0ml0 35 orbital contains 2 e n 3 1ml10 1 3p orbitals contain 6 e n 3 2 n1l 2 1 0 1 2 3d orbitals contain 10 e Assigning Electrons to Subshells The Aufbau Principle states that electrons occupy orbitals with the lowest possible energy This causes the total energy of the atom to be as low as possible The Rules 1 Electrons are assigned to subshells in order of increasing n e 2 If two subshells with the same value of n e are possible the electrons are assigned first to the subshell with a lower n For Example Electrons are assigned to 2s n e 2 before 2p n e 3 Electrons are assigned to 2p n e 3 before 3s n e 3 because 2p has a lower n value Electrons are assigned to 45 n e 4 before 3d n e 5 because the n Evalue is lower Assigning Electrons to Subshells 6 value 1 n value 0 2 3 so u 8 l n m M ch ox Ln 6P m w v 1 5d 5d 5f ngg V 4df 4f W 4 n 6 n 7 a w w v 3P n 4 39n 5 2 p n 3 z 1 A 1 N 4 H m 2 Effective Nuclear Charge 2 The net attractive charge experienced by a particular electron Let s consider the electrons in a Li atom 152251 25 CED 7 g 15 25 2p 13 This 9 IS easier loremove m The nucleus holds 3 protons 3 Li 25 At large distances from the nucleus the electron in the 2s orbital will only feel a net 1 charge due to shielding provided by the 2 electrons in the 1s orbital Inner electrons shield well for electrons farther from the nucleus ie a 2s e39 shields well for a 2p e39 but not the other 2s e39 Element Effective Nuclear Charge 2 Let s see how 2 changes as we move across a row on the periodic table p shieding 2 O numberpmtons 3 1397 13 shielding 4 205 195 quotFF 15 5 24 2 6 E u 6 275 325 7 31 39 8 345 455 9 38 52 3 4 5 6 7 s 9 10 10 415 585 Ammic Number Shielding increases by 035 with each added proton allowing 2 to increase by 065 Effective Nuclear Charge 2 Let s see how 2 changes as we move from a 2p orbital to a 3s orbital 16A 14 7 Onumberpmtons I shielding 12 A A Zquot Element p shielding 2 u 10 I I I Ne 10 415 585 g 8 Na 11 88 22 6 A Mg 12 915 285 4 a I l A A 2 i ol ll lll39ll 3 45 67 891011121314 Atomic Number Moving to a new energy level leads to an increase in shielding and a decrease in Z This is because the electrons in the 2s and 2p orbitals shield well for the electrons in the 3s orbital Effective Nuclear Charge 2 2 increases across a row of the periodic table As a result elements on the right on the periodic table hold their electrons more tightly It is very difficult to remove an electron from F and much easier to remove an electron from Na quot24 2P Elll Electron Configuration Arrangement of electrons in the orbitals of an atom We can connect the electron configuration of an element with its position on the periodic table A hydrogen atom has 1 electron n1e0me0ms 12 The electron is in the 1s orbital number of electrons assigned to designated 151 orbital 15 E orbitaltype 4 Orbital Box Notation electron shell n spdf Notation Electron Configuration A lithium atom has 3 electrons n1e0me0ms 12 n1e0me0mS 2 n2e0me0ms 12 The electrons ll the ls orbital and partially fill the 2s orbital 152251 spa f notation 15 25 one 5 two two 5 onequot BOX notation valencee Noble gas notation He251 Valence electrons are the outermost electrons in an element These electrons determine the chemical reactivity of an element Electron Configuration Boron 15 25 2p Valence e39 3 Carbon I 15 25 20 Valence e39 4 Nitrogen 15 25 2P Valence e39 5 1522522p1 He2522p1 1522522p2 He 2522p2 1522522p3 He 2522p3 Hund39s Rule when filling orbitals with equal energy fill each orbital with one electron of the same spin before pairing the spins maximize of unpaired e39 Electron Configuration Oxygen 1522522p4 15 25 2p He2522p4 Valence e39 6 Fluorine 1522522p5 15 25 2p He2522p5 Valence e39 7 Neon 1522522p6 15 25 2p He2522p6 Valence e39 8 Noble gases have full sp shells and are chemically inactive Electron Configuration Practice Identify the element with the electron configuration ls2 Zs2 2p6 3s2 3p4 Write the noble gas notation and box notation Give the quantum numbers for the valence electrons 16 e39 S Ne3523p4 Ne 35 3p n3e0me0ms 2 n3e1me0ms 2 n3e0me0mS 2 n3e1me1ms 2 n3e1me1ms12 n3e1me1mS 2 Period Electron Configurations Within Groups Elements in the same group have similar electron configurations Group 11A 1 valence electron Become highly reactive by losing 1 e39 ie Na ns1 Group 17 7A 7 valence electrons Become highly reactive by gaining 1 e39 ie F39 nsznp5 1A 1 1 1 H 131 2A 2 3 2 Li He 2s1 Be He 232 11 3 Na Ne SS1 12 Ne 352 Elements lose or gain electrons to obtain noble gas con gurations Electron Configuration Summary Electrons fill lower energy orbitals first Use the n 6 rule Pauli Exclusion Principle No two electrons can have the same four quantum numbers Each orbital holds two electrons with opposing spin Hund s Rule Electrons fill orbitals of equal energy with one electron each in order to maximize the number of unpaired electrons After each orbital is half full electron pairing can commence What about transition metals Electron Con guration of Transition Metals Fill the 4s orbital in the period then begin lling the 3d orbitals Electron Configuration of Transition Metals 8 of the transition metals have typical electron configurations 5c mumsZ I Ti Ar3d 42 III Chromium is an exception One electron moves from the 4s orbital into the 3d orbital to half fill the 3d orbitals making six unpaired electrons crquot Ar 3d545 Copper is also an exception One electron moves from the 4s orbital into the 3d orbital to completely fill the 3d orbitals Cu Ar3d 4s Half and completely filled orbitals are more stablelower in energy Lanthanides and Actinides The sixth period includes the lanthanides Lanthanum has the electron configuration Xe 5d1 6s2 Cerium is the first element with e39 that occupy f orbitals Cerium has the electron configuration Xe 4f1 5d1 6s2 At the end of the lanthanide series is Lu lutetium with an electron configuration Xe 4f14 5d1 6s2 so 62 so as as E Tm 6524 6524 55 Ce szA Sd w Th 751 7a 7o 1 Nd Pm Sm En ca Tb Dy Ha Yb Lu 652w mp 65145 552w 55247 552473 6524 6mm 65W 6524 Ssh sen BI 92 53 34 95 9s 97 93 w 102 ms Pa u Np I u Am Cm Bk cs Es Fm Md No Lr 75152m17925w7515m 751m 7315 725pm 7515p 7515pquot 7523 752311 752513 7525 malw Electron Configuration Practice Time Give the electron configuration of technetium Tc Tc has 43 electrons 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d5 Kr 5s2 4d5 Give the electron configuration of polonium P0 P0 has 84 electrons 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p4 Xe 6s2 4f14 5d10 6p4 What element has the electron configuration Kr 5s2 4d10 5p2 Tin Sn with 50 electrons Electron Configuration Practice Time The electron configuration for sulfur is Is2 2s2 2p6 3s2 3p4 How many unpaired electrons would you expect in an atom of sulfur Sulfur has two unpaired electrons How many valence electrons does an atom of phosphorus have Phosphorus has five valance electrons Would you expect Ti to be paramagnetic or diamagnetic Why Paramagnetic because only two electrons in the cl orbitals Electron Configurations of Ions Once we know the electron configuration of neutral atoms we can determine the configuration of ions When a cation is formed valance electrons are removed from the shell with the highest n value If multiple subshells are possible the electrons are removed from the orbital with the highest 6 value Na ls2 Zs2 2p6 3s1 gt Na ls2 Zs2 2p6 e39 Ti Ar 452 3d2 gt Ti2 Ar 3d2 2 e39 Transition metals lose their s electrons before their d electrons Fe Ar 452 3d6 gt Fe2 Ar 3d6 2 e39 Fe2 has four unpaired electrons Fe Ar 4s2 3d6 gt Fe3 Ar 3d5 3 e39 Fe3 has five unpaired electrons Periodic Trends Electron configurations dictate the properties of an element Elements with the same valence shell configuration have similar properties You should be able to use the periodic table to predict the properties of the elements and their compounds 3d 4d 5d 6d 7 l 1 ll 1 7 ll l i ll 1 l l l l u i l i l l J4 i l 5 block elements l d block elements transition metaLs I pblock elements 1 f block elements lanthanides 4f and actinides 5f mmmmmmmmmmmmmmmmmmmm m1 Atomic Size Covalent Radii One way to determine atomic size is to compare the distance between atoms of the same element c i C C lt gti 151 pm 198 pm radius 99 pm radius 77 pm To test the estimates add the radii together to determine a CCI bond distance C Cl 175m radius 176 m This method only works for molecular compounds nonmetals and metalloids 1A H37 H Li 152 Ne 186 K 227 Rb 2A3 Cs 255 2A Be 113 Mg 150 Ca 197 Sr 215 Ba 217 Atomic Size Trends For main group elements atomic radii increase going AA C 77 Si 117 Ga 123 Sn 141 Pb 150 5A N 71 P 115 As 125 Sb 111 Bi 155 6A 0 56 5 1M Se 117 7A F 71 Cl 99 Br 114 T2 143 Pa 157 I 133 down a group and across the table to the left The size of an atom is determined by the outermost electrons As n increases radius increases Going right across the PT 2 increases so electrons are held more tightly Greater attraction with the nucleus leads to a decrease in the atomic radius Increase Increase Atomic Size Trends For transition metals the change in the atomic radius is less dramatic Increase Increase Atomic radii Atomic radii 250 N O O Radius pm 4 Ln 0 100 1A 2A 35 4B 53 6B 73TH 23 Transition metals The radius initially decreases moving from left to right The sizes of elements in the middle of the series change very little A small increase occurs at the end of the series Atomic Size Practice Rank the following elements in order of increasing size Se Br Cl Se and Br are in the same period therefore Br lt Se Br and Cl are in the same group therefore Cl lt Br Cl lt Br lt Se Rank the following elements in order of increasing size I Xe Ba Sr I and Xe are in the same period therefore Xe lt I I and Sr are in the same period therefore I lt Sr Sr and Ba are in the same group therefore Sr lt Ba Xe lt I lt Sr lt Ba Ionization Energy The energy required to remove an electron from an atom in the gas phase Energy must be supplied to overcome nuclear attraction so ionization energy is always a positive value First Ionization Energy Mg g gt Mg g e39 IE1 738 kJmol Second Ionization Energy Mg g gt Mg2 g e39 IE2 1451 kJmol Third Ionization Energy Mg2 g gt Mg3 g e39 IE3 7732 kJmol Removing each subsequent electron requires more energy The third ionization of Mg removes an electron from an inner orbital 2p so this requires a huge input of energy Increase 3 Increase Ionization Energy First ionization energy Increase Increase gt First ionization energy It Alor main group elements first ionization energies increase going up a group and across a period to the right 2500 N O O O 1500 1000 First ionization energy kJmol U1 0 O gtr V3 739 a E x i lit quot H r cc 39 1A 2A 3A 4A 5A 6A 7A 8A Group As 2 increases the energy to remove an electron increases As the principal quantum number gets smaller the electron is located closer to the nucleus and held more strongly Ionization Energy Exceptions Moving from the s block to the p block leads to a decrease in ionization energy The 2p electrons are slightly higher in energy than the Zs electrons so the ionization energy is slightly lower for the p block Moving from ns2 np3 ie N to ns2 np4 ie 0 leads to a decrease in ionization energy In group 166A two electrons are assigned to the same p orbital The paired electrons experience greater e39e39 repulsion which makes it easier to remove the fourth electron 1314 kJmol 0 oxygen atom gt 0 oxygen cation e He He 25 2p 25 2p Ionization Energy Practice Rank the following elements in order of increasing ionization energy Na 0 P Mg Na P and Mg are in the same period Na lt Mg lt P O has the lowest principle quantum number Na lt Mg lt P lt O K Kr Se As A the elements are in the same period Suggests order of K lt As lt Se lt Kr However moving from As to Se is an exception K lt Se lt As lt Kr Electron Affinity The energy required for an atom in the gas phase to acquire an electron A g e39 gt A39 g AH EA1 The greater the affinity an atom has for an electron the lower the energy will be for the resulting ion As a result electron affinity values are always negative F g e39 gt F39 g EA1 328 kJmol B g e39 gt 839 g EA1 267 kJmol As 2 increases across a period to the right electrons are held more tightly making it more difficult to ionize the atom but easier to add an additional electron af nity for electron EA becomes more negative Electron af nity Electron af nity kJmol Increase in Increase in af nity for electron EA becomes more negative Electron Affinity Increase in affinity for electron EA becomes more negative Electron afFI m39ty crease in affinity for electron EA becomes gt lt For main group elements electron affinities become more negative going up a group and across a period to the right 350 300 250 200 150 100 Atoms with filled orbitals have no electron affinity This is true for the alkaline earth metals noble gases and nitrogen Second period elements do not follow the trend This is because B C O and F are small so there is a greater effect from e39e39 repulsion more negative Electron Affinity Practice Which has the most negative electron affinity Se Cl or Br Se and Br are in the same period Se lt Br Cl has the lowest principle quantum number least negative Se lt Br lt Cl most negative Compare C O and Si Which element has the largest atomic radius 0 lt C lt Si Which element has the largest ionization energy Si lt C lt 0 Which element has the most negative electron affinity C lt Si lt O exception Trends in Ion Size When an atom becomes ionized what happens to its radius When an electron is removed from an atom to form a cation the size of the ion shrinks considerably rcation lt ratom 139 Li 78 pm 1electron I 15 25 15 25 The remaining electrons are drawn closer to the nucleus because they feel more attraction to the positive charge Trends in Ion Size The decrease in ion size is especially large when the last electron of a particular shell is removed A large decrease in size is also expected when multiple electrons are removed Al atom radius 143 pm Al3 cation radius 57 pm 3 electrons Ne gt Nel D33 35 3p 35 3p The greater the positive charge on the cation the smaller the ion Trends in Ion Size When an atom becomes ionized what happens to its radius When an electron is added to an atom to form an anion the size of the ion increases considerably ranion gt ratom 133 pm 25 2p 25 2p When an additional electron is added there is more e39e39 repulsion in the ion which increases the size Trends in Ion Size Isoelectronic ions have the same number of electrons but a different number of protons Ion N339 0239 F39 Na Mg2 electrons 10 10 10 10 10 protons 7 8 9 11 12 Ionic radius pm 146 140 133 98 79 Rank the ions in order of increasing ionic radius F39 CI39 Br39 All the ions are in the same group F39 lt CI39 lt Br39 Crzt Cr3 Cr3 lt Cr2