Atomic Structure and Periodicity Chapter 7 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

The speed (u) of the wave = l x n Properties of Waves Wavelength (l) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = l x n

Speed of light (c) in vacuum = 3.00 x 108 m/s Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 108 m/s All electromagnetic radiation l x n = c

l x n = c l = c/n l = 3.00 x 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m A photon has a frequency of 6.0 x 104 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? l n l x n = c l = c/n l = 3.00 x 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m l = 5.0 x 1012 nm

Mystery #1, “Heated Solids Problem” Solved by Planck in 1900 When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths. Radiant energy emitted by an object at a certain temperature depends on its wavelength. Energy (light) is emitted or absorbed in discrete units (quantum). E = h x n Planck’s constant (h) h = 6.63 x 10-34 J•s

Photon is a “particle” of light Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 hn Light has both: wave nature particle nature KE e- Photon is a “particle” of light hn = KE + W KE = hn - W where W is the work function and depends how strongly electrons are held in the metal

E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m) When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm. E = h x n E = h x c / l E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m) E = 1.29 x 10 -15 J

Line Emission Spectrum of Hydrogen Atoms

n (principal quantum number) = 1,2,3,… Bohr’s Model of the Atom (1913) e- can only have specific (quantized) energy values light is emitted as e- moves from one energy level to a lower energy level En = -RH ( ) 1 n2 n (principal quantum number) = 1,2,3,… RH (Rydberg constant) = 2.18 x 10-18J

E = hn E = hn

( ) ( ) ( ) Ephoton = DE = Ef - Ei 1 Ef = -RH n2 1 Ei = -RH n2 1 nf = 2 ni = 3 nf = 1 ni = 3 Ef = -RH ( ) 1 n2 f Ei = -RH ( ) 1 n2 i nf = 1 ni = 2 i f DE = RH ( ) 1 n2

Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. i f DE = RH ( ) 1 n2 Ephoton = Ephoton = 2.18 x 10-18 J x (1/25 - 1/9) Ephoton = DE = -1.55 x 10-19 J Ephoton = h x c / l l = h x c / Ephoton l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J l = 1280 nm

Why is e- energy quantized? De Broglie (1924) reasoned that e- is both particle and wave. 2pr = nl l = h mu u = velocity of e- m = mass of e-

What is the de Broglie wavelength (in nm) associated with a 2 What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? l = h/mu h in J•s m in kg u in (m/s) l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6) l = 1.7 x 10-32 m = 1.7 x 10-23 nm

Chemistry in Action: Laser – The Splendid Light Laser light is (1) intense, (2) monoenergetic, and (3) coherent

Chemistry in Action: Electron Microscopy le = 0.004 nm STM image of iron atoms on copper surface Electron micrograph of a normal red blood cell and a sickled red blood cell from the same person

Schrodinger Wave Equation In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e- Wave function (y) describes: . energy of e- with a given y . probability of finding e- in a volume of space Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems.

Schrodinger Wave Equation y is a function of four numbers called quantum numbers (n, l, ml, ms) principal quantum number n n = 1, 2, 3, 4, …. distance of e- from the nucleus n=1 n=2 n=3

Where 90% of the e- density is found for the 1s orbital Radial probability max. at 52.9 pm from nucleus, equal to radius of innermost orbit

Schrodinger Wave Equation quantum numbers: (n, l, ml, ms) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, … n-1 l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 Shape of the “volume” of space that the e- occupies

l = 0 (s orbitals) l = 1 (p orbitals)

l = 2 (d orbitals)

Schrodinger Wave Equation quantum numbers: (n, l, ml, ms) magnetic quantum number ml for a given value of l ml = -l, …., 0, …. +l if l = 1 (p orbital), ml = -1, 0, or 1 if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2 orientation of the orbital in space

ml = -1, 0, or 1 3 orientations is space

ml = -2, -1, 0, 1, or 2 5 orientations is space

Schrodinger Wave Equation (n, l, ml, ms) spin quantum number ms ms = +½ or -½ ms = +½ ms = -½

Schrodinger Wave Equation quantum numbers: (n, l, ml, ms) Existence (and energy) of electron in atom is described by its unique wave function y. Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time

Schrodinger Wave Equation quantum numbers: (n, l, ml, ms) Shell – electrons with the same value of n Subshell – electrons with the same values of n and l Orbital – electrons with the same values of n, l, and ml How many electrons can an orbital hold? If n, l, and ml are fixed, then ms = ½ or - ½ y = (n, l, ml, ½) or y = (n, l, ml, -½) An orbital can hold 2 electrons

How many 2p orbitals are there in an atom? If l = 1, then ml = -1, 0, or +1 2p 3 orbitals l = 1 How many electrons can be placed in the 3d subshell? n=3 If l = 2, then ml = -2, -1, 0, +1, or +2 3d 5 orbitals which can hold a total of 10 e- l = 2

( ) Energy of orbitals in a single electron atom 1 En = -RH n2 Energy only depends on principal quantum number n n=3 n=2 En = -RH ( ) 1 n2 n=1

Energy of orbitals in a multi-electron atom Energy depends on n and l n=3 l = 2 n=3 l = 1 n=3 l = 0 n=2 l = 1 n=2 l = 0 n=1 l = 0

“Fill up” electrons in lowest energy orbitals (Aufbau principle) C 6 electrons ? ? B 5 electrons B 1s22s22p1 Be 4 electrons Be 1s22s2 Li 1s22s1 Li 3 electrons He 1s2 He 2 electrons H 1 electron H 1s1

The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). Ne 10 electrons Ne 1s22s22p6 F 9 electrons F 1s22s22p5 O 1s22s22p4 O 8 electrons N 7 electrons N 1s22s22p3 C 6 electrons C 1s22s22p2

Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

in the orbital or subshell Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. number of electrons in the orbital or subshell 1s1 principal quantum number n angular momentum quantum number l Orbital diagram 1s1 H

What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½

Outermost subshell being filled with electrons

Paramagnetic Diamagnetic unpaired electrons all electrons paired 2p 2p

When the Elements Were Discovered

Ground State Electron Configurations of the Elements ns2np6 Ground State Electron Configurations of the Elements ns1 ns2np1 ns2np2 ns2np3 ns2np4 ns2np5 ns2 d10 d1 d5 4f 5f

Classification of the Elements

Electron Configurations of Cations and Anions Of Representative Elements Na [Ne]3s1 Na+ [Ne] Atoms lose electrons so that cation has a noble-gas outer electron configuration. Ca [Ar]4s2 Ca2+ [Ar] Al [Ne]3s23p1 Al3+ [Ne] H 1s1 H- 1s2 or [He] Atoms gain electrons so that anion has a noble-gas outer electron configuration. F 1s22s22p5 F- 1s22s22p6 or [Ne] O 1s22s22p4 O2- 1s22s22p6 or [Ne] N 1s22s22p3 N3- 1s22s22p6 or [Ne]

Cations and Anions Of Representative Elements +1 +2 +3 -3 -2 -1

Isoelectronic: have the same number of electrons, and hence the same ground-state electron configuration Na+: [Ne] Al3+: [Ne] F-: 1s22s22p6 or [Ne] O2-: 1s22s22p6 or [Ne] N3-: 1s22s22p6 or [Ne] Na+, Al3+, F-, O2-, and N3- are all isoelectronic with Ne What neutral atom is isoelectronic with H- ? H-: 1s2 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]4s23d6 Mn: [Ar]4s23d5 Fe2+: [Ar]4s03d6 or [Ar]3d6 Mn2+: [Ar]4s03d5 or [Ar]3d5 Fe3+: [Ar]4s03d5 or [Ar]3d5

Effective nuclear charge (Zeff) is the “positive charge” felt by an electron. Zeff = Z - s 0 < s < Z (s = shielding constant) Zeff  Z – number of inner or core electrons Zeff Core Z Radius (pm) Na Mg Al Si 11 12 13 14 10 1 2 3 4 186 160 143 132

Effective Nuclear Charge (Zeff) increasing Zeff increasing Zeff

Atomic Radii covalent radius metallic radius

Trends in 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.

The Radii (in pm) of Ions of Familiar Elements

Chemistry in Action: The 3rd Liquid Element? 117 elements, 2 are liquids at 250C – Br2 and Hg 223Fr, t1/2 = 21 minutes Liquid?

Ionization energy is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state. I1 + X (g) X+(g) + e- I1 first ionization energy I2 + X+(g) X2+(g) + e- I2 second ionization energy I3 + X2+(g) X3+(g) + e- I3 third ionization energy I1 < I2 < I3

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

General Trends in First Ionization Energies Increasing First Ionization Energy Increasing First Ionization Energy

Electron affinity is the negative of the energy change that occurs when an electron is accepted by an atom in the gaseous state to form an anion. X (g) + e- X-(g) F (g) + e- X-(g) DH = -328 kJ/mol EA = +328 kJ/mol O (g) + e- O-(g) DH = -141 kJ/mol EA = +141 kJ/mol

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