Lecture outline: Chapter 7 Periodic properties

Lecture outline: Chapter 7 Periodic properties 1. Electrostatic effects 2. Atomic size 3. Ionization energy 4. Electron affinity it 5. Summarize some periodic properties 1

Some important terms Electron configuration: the order of filling of orbitals; tells us which orbitals are filled or partially filled Core electrons: those in filled shells inner shell electrons Valence electrons: those beyond the filled shells outer shell electrons Effective nuclear charge Z eff = Z - S 2

Electron configurations of the elements, color coded by subshell type 3

Ra adial electr ron density 1s e - 2s & 2p e - 1s e - How far are the electrons from the nucleus for three Noble gases (He, Ne, Ar)? He: 1s 2 Ne: 1s 2 2s 2 2p 6 Ar: 1s 2 2s 2 2p 6 3s 2 3p 6. 2s &2 2p e - 3s & 3p e - 1s e - Distance from nucleus ( ) 1 Å 4

Electrostatic interactions are responsible for the differences in orbital sizes for different atoms particles with like charges repel one another (electrostatic repulsion) particles with opposite charges are attracted to each other (electrostatic attraction) The strength of an electrostatic attraction is directly related to the relative magnitudes of the charges on the particles The strength of an electrostatic attraction is inversely related to the distance between the interacting particles 5

The potential energy of two interacting charged particles Q Q E k 1 2 d 1 d = Q 1 Q 2 Q Q E 1 2 d 6

Electrostatic interactions are responsible for the differences in orbital sizes for different atoms Q1Q d E 2 n = 4 Z eff = Z -S eff Effective nuclear charge n = 3 Nuclear charge (atomic #) n = 2 # of shielding electrons Zp+ n = 1 A simplified model for electronic structure: treat all electrons of a given n value as if they were identical (they re really not, of course) Assume that electrons in the same shell do not shield each other effectively 7

Electrostatic interactions are responsible for the differences in orbital sizes for different atoms He: 1s 2 Z eff = Z -S n = 3 eff Effective nuclear charge n=4 Nuclear charge (atomic #) # of shielding electrons n = 2 2 p+ n = 1 Treat all electrons of a given shell (n value) as identical Assume that electrons in the same shell do not shield each other effectively Q1Q d E 2 8

Electrostatic interactions are responsible for the differences in orbital sizes for different atoms Ne: 1s 2 2s 2 2pp 6 Z eff = Z -S n = 3 eff Effective nuclear charge n=4 Nuclear charge (atomic #) # of shielding electrons 10 p+ n = 1 n = 2 Treat all electrons of a given shell (n value) as identical Assume that electrons in the same shell do not shield each other effectively Q1Q d E 2 9

Electrostatic interactions are responsible for the differences in orbital sizes for different atoms Ar: 1s 2 2s 2 2pp 6 3s 2 3pp 6 Z = Z -S eff n = 3 Effective nuclear charge n=4 Nuclear charge (atomic #) # of shielding electrons 18 p+ n = 1 n = 2 Treat all electrons of a given shell (n value) as identical Assume that electrons in the same shell do not shield each other effectively Q1Q d E 2 10

Ra adial electr ron density 1s e - : Z eff = +18 2s & 2p e - 1s e - : Z eff = +10 How far are the electrons from the nucleus for three Noble gases (He, Ne, Ar)? He: 1s 2 Ne: 1s 2 2s 2 2p 6 Ar: 1s 2 2s 2 2pp 6 3s 2 3pp 6. 2s &2 2p e - 3s & 3p e - 1s e - : Z eff = +2 Distance from nucleus ( ) 11

The concepts of nuclear charge, inner electron shielding and electrostatic t ti attraction/repulsion are important in predicting three important chemical properties p of atoms: Atomic size Ionization energy Electron affinity 12

Atomic size (radius) 18 p+ n = 1 n = 2 n 1 n=3 Ar: 1s 2 2s 2 2p 6 3s 2 3p 6 13

Atomic size (radius) Ar: 1s 2 2s 2 2pp 6 3s 2 3pp 6 n = 3 n=2 18 p+ n = 1 is really: 14

Atomic size (radius) 18 p+ n = 1 n = 2 n = 3 r = 0.97 Å = 97 pm Golf ball nucleus 15

Atomic radius: the typical distance from the nucleus to a boundary of the surrounding electron cloud What defines the boundary? Van der Waals radius (unbonded touching) Covalent radius (calculated l from bond lengths measured for molecules) Calculated radii from theoretical models 16

Atomic radii of elements 1-57 300 250 atomic radius (p pm) 200 150 100 50 0 0 10 20 30 40 50 60 atomic number of element 17

Calculated atomic radii of s and p block elements Increasing atomic radius fr rom T to B in a per riod Increasing atomic radius from R to L in a row 18

Two opposing effects from Left Right and Top Bottom: increase n, increase orbital size Increase Z eff, decrease orbital size Increasing atomic radius from T to B in a period Increasing atomic radius from R to L in a row 17 19

Li vs. Be (L to R) Li: 1s 2 2s 1 Be: 1s 2 2s 2 n = 1 n = 3 n=2 3 p+ 4 p+ n = 3 n=2 n = 1 Q1Q d E 2 20

Li vs. Na (Top to Bottom) Li: 1s 2 2s 1 Na: 1s 2 2s 2 2p 6 3s 1 n = 1 n = 3 n=2 3 p+ 11p+ n = 3 n=2 n = 1 Q1Q d E 2 21

Chapter 7 material 1. Electrostatic effects 2. Atomic size 3. Ionization energy 4. Electron affinity 22

Ionization energy: The minimum energy required to remove an electron from the ground state of the isolated gaseous atom X (g) X + + e (g) - When reported with units of kj/mol, it is the energy required to remove one mol of electrons from one mol of gaseous atoms 23

Ionization of the hydrogen atom: H (g) H + (g) + e - Zero point E = 0 n = Increasing E n = 2 n = 3 n = 4 n = 5 n = 6 n = 8 n = 7 ΔE (+) for removing an electron from the atom s ground state orbit n = 1 Ground state E = (-) 24

Ionization of the sodium atom: Na: 1s 2 2s 2 2p 6 3s 1 Na (g) Na + (g) + e - Zero point E = 0 n = Increasing E n = 4 n = 5 n = 6 n = 7 n = 8 ΔE (+) n = 3 Ground state E = (-) n = 2 n = 1 25

Successive ionizations for Silicon Si: 1s 2 2s 2 2p 6 3s 2 3p 2 I. # Elec. Config. before and after ionization Ion formed Z eff felt by departing Energy req. (kj) to remove e- e- 14 p+ n = 4 n=3 n = 2 n = 1 I 1 [Ne]3s 2 3p 2 Si + +4 786 [Ne]3s 2 3p 1 I 2 [Ne]3s 2 3p 1 Si 2+ +4 1,577 [Ne]3s 2 I 3 [Ne]3s 2 Si 3+ +4 3,228 [Ne]3s 1 I 4 I 5 [Ne]3s 1 Si 4+ +4 4,354 [He}2s 2 2p 6 [He}2s 2 2p 6 Si 5+ +12 16,100 [He]2s 2 2p 5 Q1Q d E 2 26

Periodic trends for the first ionization energy: X (g) X + (g) + e - 2500 energy first ion nization 2000 1500 1000 500 0 0 10 20 30 40 50 60 atomic number of element 27

What determines the energy required to remove an electron from an outer shell? Z eff felt by the valence electron Distance of electron from the nucleus (reflected by atomic radius values) - Z p+ - - Q1Q d E 2 28

What happens to Z eff and atomic radius when we move from left to right in a period? Increase Z eff, increase I 1 Decrease size, increase I 1 Q1Q d E 2 29

What happens to Z eff and atomic radius when we move down in a group? Z eff stays same for a given group Increase size, decrease I 1 Q1Q d E 2 30

Some irregularities in the trend for first ionization energy 2500 He Ne 2000 energy nization first io 1500 1000 500 H Li Be B N C F O Mg Na Al P Si Cl S Ar 0 0 5 10 15 20 atomic number of element 31

3s 2s 1s 3p 2p Explanation of trend irregularities: Be to B and Mg to Al Z eff = Z -S eff 3p 2p 3s n = 3 2s n = 3 10p+ n = 1 n = 2 10p+ n = 1 n = 2 Ne: 1s 2 2s 2 2p 6 Ne: 1s 2 2s 2 2p 6 Simplistic onion shell model treats s and p orbitals of a given n value as having equal energies In reality s orbitals are lower in energy than p orbitals for a given n value 32

y 2500 Explanation of trend irregularities : Be to B He first ionization energy f 2000 1500 1000 H Be B N C F O Ne Mg 500 Al Li Na P Si Cl S Ar 0 0 5 10 15 20 atomic number of element 3s 3p 2p 2s n = 3 2s 2p Be: 1s 2 2s 2 4 p+ n = 1 n = 2 1s Be 33

y 2500 Explanation of trend irregularities : Be to B He first ionization energy f 2000 1500 1000 H Be B N C F O Ne Mg 500 Al Li Na P Si Cl S Ar 0 0 5 10 15 20 atomic number of element 3s 3p 2p 2s n = 3 2s 2p B: 1s 2 2s 2 2p 1 5 p+ n = 1 n = 2 1s B A filled s orbital provides some shielding of the nuclear charge felt by an electron in a p orbital of the same n value 34

y 2500 Explanation of trend irregularities : Mg to Al He 3s 2s first ionization energy f 2000 1500 1000 3p H Be B N C F O Ne Mg 500 Al Li Na 0 0 5 10 15 20 atomic number of element P Si Cl S Ar 2p 10p+ n = 1 Mg: 1s 2 2s 2 2p 6 3s 2 2p 3p 2s 3s n = 2 n = 3 1s Mg A filled s orbital provides some shielding of the nuclear charge felt by an electron in a p orbital of the same n value 35

y 2500 Explanation of trend irregularities : Mg to Al He 3s 2s first ionization energy f 2000 1500 1000 3p H Be B N C F O Ne Mg 500 Al Li Na 0 0 5 10 15 20 atomic number of element P Si Cl S Ar 2p 10p+ n = 1 Al: 1s 2 2s 2 2p 6 3s 2 3p 1 2p 3p 2s 3s n = 2 n = 3 1s Al A filled s orbital provides some shielding of the nuclear charge felt by an electron in a p orbital of the same n value 36

Explanation of trend irregularities: N to O and P to S 2s 2p 2s 2p 1s N: 1s 2 2s 2 2p 3 1s O: 1s 2 2s 2 2p 4 The e - configuration np 3 (a spin party ) is good 37

Explanation of trend irregularities: N to O and P to S 3p 3p 3s 3s 2s 2p 2s 2p 1s P: 1s 2 2s 2 2p 6 3s 2 3p 3 1s S: 1s 2 2s 2 2p 6 3s 2 3p 4 The e - configuration np 3 (a spin party ) is good 38

Electron affinity y( (EA) The energy change associated with the addition of an electron to a gaseous atom or ion X (g) + e - X - (g) X +e - (g) X (g) X + (-) E.A. : E is released (usually, but not always the case) (+) E.A. : E is added When reported with units of kj/mol, it is the energy change associated with adding one mol of electrons to one mol of gaseous atoms or ions 39

Electron affinity values/trends More negative EA (in general) ex. n = 2, 5 and 8 No ot much ch hange in EA (in gen neral) 40

Why the trends in electron affinity values? - - Z p+ - general) Not much change in EA (in More negative EA (in general) 41

Explanation for why group 2A elements have positive or small negative electron affinities 3p 2p 2s 3s n = 3 n = 2 10p+ n = 1 Add e -.. Mg: 1s 2 2s 2 2p 6 3s 2 42

Explanation for why group 6A elements have positive or out of pattern negative electron affinities 2s 2pp e - 2s 2p 1s 1s N: 1s 2 2s 2 2pp 3 N - : 1s 2 2s 2 2pp 4 The e - configuration np 3 (a spin party ) is good 43