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2012/2013 I Ashraf A. Mohamed, Prof. Dr. 1

Chapter 7: Quantum Theory and the Electronic Structure of Atoms

Properties of Waves

Wavelength (): is the distance between 2 successive crests (tops) ortroughs (bottoms).Amplitude is the vertical distance from the midline of a wave to the crest ortrough.Frequency () is the number of waves that pass through a particular pointin 1 second (Hz = 1 cycle/s).Wavenumber () is the number of waves per unit distance (cycle/cm).


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 x = c

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?

= c = c/3.00 x 108 m/s) / (6.0 x 104 Hz) = 5.0x103 m = 5.0x1012 nm Radio waves


Mystery #1, “Black Body radiation” Solved by Planck in 1900Energy (light) is emitted or absorbed in discrete units called quanta (quantum). E = h = h c/ λ= h c = c/ λ = cwhere h is Planck’s constant h = 6.63 x 10-34 J•s

Mystery #2, “Photoelectric Effect” solved by Einstein in 1905Light has both: wave nature & particle naturePhoton is a “particle” of lighth = KE + BE KE = h – BEKE: kinetic energy; BE: binding energy

When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wave-length of the X rays is 0.154 nm.

E = h = h c/ λ= 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 spectra are finger prints for a given element. See next slide


1. e- can only have specific (quantized) energy values

2. light is emitted as e-

moves from one energy level to a lower energy level

Bohr’s Model of the Atom (1913)

En = -RH ( )1n2

n (principal quantum number) = 1,2,3,…

RH (Rydberg constant) = 2.18 x 10-18J


Ephoton = E = Ef - Ei

Ef = -RH ( )1n2

f

Ei = -RH ( )1n2

i

i fE = RH( )

1n2

1n2

nf = 1

nf = 2

nf = 3


Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)

Ephoton = E = -1.55 x 10-19 J

= 1.280X10-6 m = 1280 nm

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.

Ephoton = h x c / = h x c / Ephoton

i fE = RH( )

1n2

1n2Ephoton =

= 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J


Why is energy quantized?

De Broglie (1924) reasoned that e- is both particle and wave (Dual nature of electrons)

2r = n

E = mc2 Particle behaviorE = h Wave behavior

What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?

= h/m = 6.63 x 10-34 / (2.5 x 10-3 x 15.6) = 1.7 x 10-32 m = 1.7 x 10-23 nm

Dual nature of matter expressed by replacing the speed of light with the speed of the particle.

E = h = hc/λ , E = mc2

hc/λ = mc2 λ = h/mc = h/mv (dual nature)


Laser: Light Amplification by Stimulated Emission of RadiationLaser light is :(1) intense, monoenergetic &

Coherent emission(2) involves atoms or molecules(3) operate in the gas, liquid, and

solid state systems(4) of a very precise (in the IR,

visible or UV region)

Ruby laser, the first known laser since 1960. Ruby is a deep-red mineral containing corundum, Al2O3 , in whichsome of the Al3+ ions have been replaced by Cr3+ ions. A flash lamp is used to excite the chromium atoms to ahigher energy level. The excited atoms are unstable, so at a given instant some of them will return to the groundstate by emitting a photon in the red region of the spectrum. The photon bounces back and forth many timesbetween mirrors at opposite ends of the laser tube. This photon can stimulate the emission of photons havingexactly the same wavelength from other excited chromium atoms; these photons in turn can stimulate theemission of more photons, and so on. Because the light waves are in phase —that is, their maxima and minimacoincide—the photons enhance one another, increasing their power with each passage between the mirrors.One of the mirrors is only partially reflecting, so that when the light reaches a certain intensity it emerges fromthe mirror as a laser beam. Depending on the mode of operation, the laser light may be emitted in pulses (as inthe ruby laser case) or in continuous waves.Lasers are used in eye surgery, drilling holes in metals and welding, carrying out nuclear fusions,telecommunications, compact disc players, supermarket scanners, … etc.


Schrodinger Wave EquationIn 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e-

Wave function (n,l,m) describes:

1. energy of e- with a given

2. 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.

H = E

Heisenberg uncertainty principle: It’s impossible to determine simultaneously the position and speed of electron with absolute precision or (position uncertainty)(speed uncertainty) = constant

mhvx 4


Principal quantum number (n): determines the energy of electron & its distance (radius) from the nucleus

Secondary (Azimuthal or angular momentum) quantum number (l) : determines the shapes of orbitals l = 0, 1, 2, 3, 4, …, (n-1)

Magnetic quantum number (ml ): determines the orientation of orbitals in spacem = - l, …., 0, ….., + l

Spin quantum number (s): determines the spin of orbitals clockwise or anticlockwise = (+1/2) , (-1/2)

Quantum numbers

(n-1)….….3210l

….….….fdpssymbol

n = 1 2 3 4 5 6 7 ….

symbol K L M N O P Q ….


f-type atomic orbitals - complex shapes


1s2s

3s

Radial Distribution Wave Functions


number of orbitals in a subshell

e-capacity of subshell

e-capacity of shell

n l m s (2 l + 1) (4 l + 2) 2 n2

1 0 (1s) 0 +1/2, -1/2 1 2 2

2 0 (2s)1 (2p)

0-1, 0, +1

+1/2, -1/2 ½ for each value of s

13

26

8

3 0 (3s)1(3p)2(3d)

0-1, 0, +1

-2, -1, 0, +1, +2

+1/2, -1/2½ for each value of s ½ for each value of s

135

2610

18

4 0 (4s)1(4p)2(4d)3 (4f)

0-1, 0, +1

-2, -1, 0, +1, +2-3, -2, -1, 0, +1, +2, +3

+1/2, -1/2½ for each value of s ½ for each value of s ½ for each value of s

1357

261014

32

5 0 (5s)1(5p)2(5d)3 (5f)

4 (5g) !!

0-1, 0, +1

-2, -1, 0, +1, +2-3, -2, -1, 0, +1, +2, +3

+1/2, -1/2½ for each value of s ½ for each value of s ½ for each value of s

1357

261014

32


Shell (Level): electrons with the same value of nSubshell (Sublevel): electrons with the same values of n and lOrbital – electrons with the same values of n, l, and ml Pauli exclusion principle : no two electrons in an atom can have the same four quantum numbers.Hund’s rule: The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins. (in a subshell, degenerate orbitals are occupied singly first).

Order of orbitals filling in multielectron atom (Aufbau (Building up) principle)1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

1- orbitals with lower (n+l) values are filled 1st

2- when two orbitals have the same (n+l), the orbital with lower (n) is filled 1st. Therefore, 4s is filled before 3d, for example.

How many electrons can an orbital hold? 2How many 2p orbitals are there in an atom? 3How many electrons can be placed in the 3d subshell? 10What is the electron configuration of 12Mg? 1s2, 2s2, 2p6, 3s2

Abbreviated as [Ne]3s2; where [Ne] 1s2,2s2,2p6

What are the possible quantum numbers for the last (outermost) electron in 17Cl?


1- what are the possible (l) values when (n = 5)2- what are the possible (m) values for an electron in the (4f) orbital?3- what are the possible (s) values for an electron in the (5dxy) orbital?4- how many electrons are there in a shell with (n = 4)?5- how many electrons are there in a (3p) subshell?6- how many electrons are there in the (4dxz) orbital?7- what are the possible quantum numbers for an electron in the 4p sublevel?8- which of the following sets of quantum numbers is incorrect:(n=4, l=4), (n=2, l=0, m=0), (n=3, l=2, m=1), (n=5, l=3, m=4), (4,3,3,+1/2), (3,4,2,-1/2), (5,4,0,-1), (5,4,0,-1/2)9- is there a 2d subshell? Why?10- write the electronic configuration of the following species:

7N, 13Al, 12Mg, 11Na+, 19K, 29Cu2+, 26Fe2+

11- Why 24Cr and 29Cu have unusual electronic configurations

42Mo 47Ag79Au


The electron configuration of chromium (24Cr) is [Ar]4s1,3d5 and not[Ar]4s2,3d4, as we might expect. A similar break in the pattern is observedfor copper, whose electron configuration is [Ar]4s1,3d10 rather than[Ar]4s2,3d9. The reason for these irregularities is that a slightly greaterstability is associated with the half-filled (3d5) and completely filled (3d10)subshells. Electrons in the same subshells (in this case, the d orbitals) haveequal energy but different spatial distributions. Consequently, their shieldingof one another is relatively small, and the electrons are more stronglyattracted by the nucleus when they have the 3d5 (3d10) configuration.According to Hund’s rule, the orbital diagram for Cr & Cu are:

Other Irregular Electron Configurations• Mo = [Kr]5s24d4 [Kr]5s14d5

• Ru = [Kr]5s24d6 [Kr]5s14d7

• Pd = [Kr]5s24d8 [Kr]5s04d10


• Result of the spin of electrons• diamagnetism - no unpaired electrons (zero magnetism)• paramagnetism - one or more unpaired electrons (small magnetism)• ferromagentism – a case of paramagnetism where the substance retains

its magnetism even after removing the applied external magnetic field (high magnetism, characteristic to Fe, Co & Ni)

Paramagnetism

Without external applied field With external applied field

Ferromagnetism

ParamagneticAt least one unpaired e-

2p2p

Magnetism (magnetic properties)Ferromagnetic

All unpaired e- are parallel to the external field &

strengthen their magnetism

Diamagneticall electrons paired & net

magnetism = zero


The Periodic TableIn 1864, Newlands noticed that when the elements were arranged in order ofatomic mass, every eighth element had similar properties “the law of octaves”.However, this “law” was inadequate for elements beyond calcium.

In 1869, Mendeleev and Meyer independently arranged the 62 knownelements based on the regular, periodic recurrence of properties. Mendeleev’sclassified elements based on their chemical properties, whereas Meyer’sclassified elements based on their physical properties. Later on, Mendeleevnoted that both chemical and physical properties of the elements varyperiodically with atomic mass. Mendeleev’s classification has the greatimprovement of predicting the properties of several elements that had not yetbeen discovered, e.g. Ga.

In 1913, Moseley arranged the elements based on their atomic numbers togive the modern periodic table.


Classification of the Elements

Lanthanides

Actinides

Halogens

Nobel gases

Alkali m

etals

Alkaline earth’s

metals

Transition metalsInner transition metals


ns1

ns2

ns2 n

p1

ns2 n

p2

ns2 n

p3

ns2 n

p4

ns2 n

p5 ns2 n

p6

ns2

(n-1

)d1

ns2

(n-1

)d5

ns2

(n-1

)d10

4f

5f

Ground State Electron Configurations of the Elements


Group IA alkali metalsGroup IIA alkaline earth metalsGroup VIIA halogensGroup VIIIA noble gasestransition metals (incomplete d-orbitals) (4 periods & 8 groups)(N.B. Group 8B contains 3 subgroups 8, 9, 10)inner transition metals (incomplete f-orbitals)• lanthanide series (rare earths, incomplete 4f-orbitals)• actinide series (radioactive, trans-uranic elements, incomplete 5f-orbitals)Types of Elements: metals, nonmetals & metalloids(semimetals)

Family NamesGroup Names

Isoelectronic Substances And Excited States

Isoelectronic species (atoms or ions) have the same number of e’s, e.g.

13Al3+, 12Mg2+, 11Na+, 10Ne, 9F-, 8O2-, 7N3- are all isoelectronic with 10 e

21Sc3+, 20Ca2+, 19K+, 18Ar, 17Cl-, 16S2-, 15P3- are all isoelectronic with 18 e

The electron configuration of an element in an excited state will have anelectron in a high-energy state, e.g. the excited-state electron configuration for

12Mg may be [Ne]3s13p129Cu may be [Ar]4s13d94p1


Electron Configurations of Cations and AnionsOf Representative Elements

11Na [Ne]3s1 Na+ : [Ne]12Mg [Ne]3s2 Mg2+: [Ne]13Al [Ne]3s2 3p1 Al3+ : [Ne]

Atoms lose (or gain) electrons so that cation (or anion) has a noble-gas outer electron configuration

9F 1s2 2s2 2p5 F- :1s22s22p6 or [Ne]8O 1s2 2s2 2p4 O2- :1s22s2 2p6 or [Ne]

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.

26Fe: [Ar]4s2 3d6

Fe2+: [Ar]4s0 3d6 or [Ar]3d6

Fe3+: [Ar]4s0 3d5 or [Ar]3d5

25Mn: [Ar]4s2 3d5

Mn2+: [Ar]4s0 3d5 or [Ar]3d5


From the electronic structure, how can you determine the group, period,block type, metallic and magnetic properties of an element?

The element belongs to• a period = the greatest principal Q No. (n).• a block = the last subshell.• a group = the total No. of electrons in the last shell

= (ns+np) electrons for the (s & p) -block elements= [ns + (n-1)d] electrons for the d-block elements

• Paramagnetic 1 or more unpaired electrons.• Diamagnetic no unpaired electrons.

11Na: 1s2, 2s2, 2p6, 3s1

Period = 3; Group = 1; s-Block; metal; single electron paramagnetic.

35Br: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p5 OR [18Ar], 4s2, 3d10, 4p5

Period = 4; Group = (2+5) = 7; p-Block; nonmetal; single electron paramagnetic.

42Mo: [36Kr], 5s2, 4d4 ? OR [36Kr], 5s1, 4d5 ?Period = 5; Group = (1+5) = 6; d-Block; metal; 6 single electrons paramagnetic.


Chapter 8Periodic relationships among the elements

Effective nuclear chargeAtomic and ionic radii

Ionization energyElectron affinity Electonegativity


Effective nuclear charge (Zeff) is the “positive charge” felt by an electron.

Zeff = Z – where is the shielding or screening constant (0 < < Z )Zeff Z – number of core electrons

Slater (Phys. Rev., 1930, 36, 57) set simple rules for determining :1. The electronic structure of the atom is written in groupings as follows:

(1s) (2s,2 p) (3s,3 p) (3d) (4s,4 p) (4d) (4f) ( 5s,5p) etc .(everything is grouped according to n)

2. Electrons in higher groups (to the right in the list above) do not shield those inlower groups.

3. For ns or np valence electrons: (a) electrons in the same ns, np group contribute0.35 (except the Is, where 0.30 works better); (b) electrons in the n-1 groupcontribute 0.85; (c) electrons in the n-2 or lower groups contribute 1.00.

4. For nd and nf valence electrons: (a) electrons in the same nd or nf groupcontribute 0.35; (b) electrons in groups to the left contribute 1.00.

5. Zeff = (Z-). Some examples are given below showing that the original Slater’srules applies well for periods but not for groups for which the Waldron’scorrection should be applied (J. chem. Edu., 78 (2001) 635).

increasing Zeff

decr

easi

ng Z

eff


8O: (1s2) (2s2, 2p4)=(5x0.35)+(2x0.85)=3.45Zeff=8-3.45=4.5528Ni: (1s2) (2s2, 2p4) (3s2, 3p6) (3d8) (4s2) 3d = (2x0)+(7x0.35)+(18x1) = 20.45Zeff = 28-20.45 = 7.55 4s =(1x0.35)+(16x0.85)+(10x1)=23.95Zeff = 28-23.95 = 4.05

3Li: (1s2) (2s1) =(2x0.85) = 1.7 Zeff=3-1.7=1.311Na: (1s2) (2s2, 2p6) (3s1) =(8x0.85)+(2x1)=8.8 Zeff=11-8.8=2.219K: (1s2) (2s2, 2p6) (3s2, 3p6) (4s1) = (10x1)+ (8x0.85)=16.8 Zeff=19-16.8=2.2

12Mg: (1s2) (2s2, 2p6) (3s2) =(0.35x1)+(8x0.85)+(2x1)=9.15 Zeff=12-9.15=2.8513Al: (1s2) (2s2, 2p6) (3s2, 3p1) =(0.35x2)+ (8x0.85)+(2x1)=9.50 Zeff=13-9.5=3.5

Atomic Radius & Size(a) In metals such as Be, the atomic radius = one-half the distance between thecenters of two adjacent atoms.(b) For elements that exist as diatomic molecules, such as Cl2, the atomic radius =one-half the distance between the centers of the atoms in the molecule.

11Na 12Mg 13Al 14SiZ 11 12 13 14

Core electrons

10 10 10 10

Zeff rough 1 2 3 4Radius 186 160 143 132


Atomic radii increase going down a group because electronsare being added to shells farther from the nucleus.

Atomic radii decrease from left to right across a period owingto increasing effective nuclear charge (a proton is added tothe nucleus and an electron is added to the particular shell).

For transition elements, the variations arenot so regular because electrons are beingadded to an inner shell.

Hydrogen atoms are the smallest andcesium atoms are the largest naturallyoccurring atoms.

Cation is smaller than its parent atom.Anion is larger than its parent atom.


Chemistry in Action: The 3rd

Liquid Element?Among the known 117 elements, there are 2stable liquid elements at 250Cnamely Br2 & Hg.

223Fr, t1/2 = 21 minutes. Is it a liquid?

Arrange the following isoelectronic ions according to its size:21Sc3+, 20Ca2+, 19K+, 17Cl-, 16S2-, 15P3-

13Al3+12Mg2+

11Na+9F-

8O2-7N3-

Z 13 12 11 9 8 7

e’s 10 10 10 10 10 10

r 0.68 0.85 1.16 1.19 1.26 1.71

13Al3+ < 20Ca2+ < 19K+

35Br- < 34Se2- < 62Te2-

Liqu

id?


Ionization energy (Ionization potential) is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state.

*Noble gases have the highest IE’s (lowest R, highest Zeff, highest stability)*Group IA have the lowest IE’s (highest R, lowest Zeff)*2He has the highest and 55Cs has the lowest IE’s


I1 + X (g) X+(g) + e- I1 first ionization energy

I2 + X+(g) X2+

(g) + e- I2 second ionization energyI3 + X2+

(g) X3+(g) + e- I3 third ionization energy

I1 < I2 < I3

The 1st IE’s for the Group IIA elements (Be, Mg, Ca, Sr, Ba) are exceptionally higherthan those of the IIIA elements (B, Al, Ga, In, Tl) in the same periods (5B: 1S2, 2S2,2p1 & 4Be: 1S2, 2S2). Because less energy is required to remove the first p electron(to leave an empty p-subshell) of GIIIA than to remove the second s-electron from afilled s-subshell of GIIA.

Similarly, The 1st IE’s for the Group VA elements (N, P, As, Sb, Bi) are exceptionallyhigher than those of the VIA elements (O, S, Se, Te, Po) in the same periods (7N: 1S2,2S2, 2p3 & 8O: 1S2, 2S2, 2p4). Because less energy is required to remove the 4th pelectron (to leave a half filled p-subshell) of GVIA than to remove the 3rd p electronfrom a half-filled p-subshell of Group VA.


I2 (Mg) vs. I3 (Mg) I1 (Mg) vs. I1 (Al) I1 (P) vs. I1 (S)


Electron affinity (EA) is the energy change that occurs when an isolatedatom in the gaseous state accepts an electron to form an anion. The sign ofEA is opposite to the one we use for ionization energy. As we know, a +ve IEmeans that energy must be supplied to remove an electron. A +ve EA, onthe other hand, signifies that energy is liberated when an electron is addedto an atom.

X (g) + e- X-(g)

F (g) + e- F-(g) H = -328 kJ/mol (exothermic) EA = +328 kJ/mol

O (g) + e- O-(g) H = -141 kJ/mol (exothermic) EA = +141 kJ/mol

Thus, (1) The EA of an element is equal to the enthalpychange that accompanies the ionization process of itsanion, and (2) a large +ve EA means that the negative ionis very stable, i.e., its atom has a great tendency to acceptan electron.Experimentally, EA is determined by removing theadditional electron from an anion. The electron affinities ofmetals are generally lower than those of nonmetals. Thevalues vary little within a given group. The halogens (Group7A) have the highest electron affinity values.


Why are the electron affinities of the alkaline earth metals, shown in Table 8.4, eithernegative or small positive values?Answer The valence configuration of the alkaline earth metals is ns2. For theProcess (M(g) + e M-

(g)) where M denotes a member of the Group 2A family, theextra electron must enter the np subshell, which is effectively shielded by the two nselectrons and the inner electrons. Consequently, alkaline earth metals have littletendency to pick up an extra electron.F has a slightly lower EA compared to Cl. Why?


The electronegativity (EN) is the relative ability of an atom to attractelectrons to itself when it is chemically bonded with another atom.EN’s are expressed on a somewhat arbitrary scale, called the Pauling scale

Electronegativity - relative, F is highestElectron Affinity - measurable, Cl (not F) is highest X (g) + e- X-

(g)


Periodic Properties of the Elements: summary

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