chapter 8: electron configurations and the periodic...

Chem 6A Michael J. Sailor, UC San Diego

Chapter 8:Electron Configurations and the

Periodic TableChem 6A, Section D Oct 25, 2011


The Periodic Table of the Elements

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Electron ConfigurationsAufbau ="Building up"

As each proton is added to the nucleus, you add an electron to the hydrogen-like orbitals.

Add to s, then d, then p orbitals of the same principle quantum number


Quantum numbersQuantum Number

Called Describes

n Principle quantum number SIZE and ENERGY

l Angular momentum (Azimuthal) quantum number

SHAPE

mlMagnetic quantum number ORIENTATION

msElectron spin quantum number

INTRINSIC ANGULAR MOMENTUM OF THE ELECTRON

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1s2s 2p3s 3p4s 3d 4p5s 4d 5p6s 5d 6p7s

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Electron ConfigurationsHow to fill up orbitals with electrons?• Electrons go in lowest energy orbital first,

they spread over all the empty orbitals with the same spin, then they pair up.

• Pauli Exclusion Principle: No more than 2 electrons per orbital, must be of opposite spin.

• Hund's rule: When there is more than one orbital with the same energy, fill up empty orbitals first, keeping the spins the same.


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from T. Moeller “Inorganic Chemistry” Wiley 1952

Similar figure in your textbook

The Building-up Principle(aufbau)

Pauli exclusion principle: no more than 2 electrons in each orbital. Pairs of electrons in the same orbital must have opposite spins.

Hund’s rule: When there is more than one orbital with the same energy (degenerate), fill up empty orbitals with one electron before pairing the electrons, keep the spins the same.

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Stern-Gerlach Experiment:

OVENContaining Ag

ScreenAg atoms

Magnet

The atoms split into two paths in a magnetic field

This experiment tells us that each individual electron has a magnetic moment; there must be a 4th quantum number: Electron

spin, or ms


Electron ConfigurationsExample Electron configurations:Potassium 1s22s22p63s23p64s1

shorthand: [Ar]4s1

Vanadium [Ar]4s23d3

Selenium [Ar]4s23d104p4

Note: (n+1)s orbitals always fill up before nd.4f and 5d orbitals have similar energies, Lanthanides fill up 4f and 5d orbitals in somewhat random order. Don’t worry about lanthanides for now.

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Electron Configurations


Electron Configurations-ExceptionsExceptions: everything fills up normally, with

a few exceptions:Cr: [Ar]4s13d5

Why? half-filled shells are unusually stable. This one you can't predict, just memorize.

Cu: [Ar]4s13d10

Why? fully-filled shells are stable.Memorize these

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Ordering of s, p, d, and f orbitalsRelative ENERGIES:

s < p < d < fFill ns, then np, then nd orbital. Example (V):

Vanadium: 1s22s22p63s23p64s23d3

You always fill up (n+1)s before nd. Why?


Ordering of s, p, d, and f orbitalsYou always fill up (n+1)s before nd. Why?Main factors determining relative energy ordering for

orbitals with the same value of n:– Nuclear charge– Electron-electron repulsion (shielding)

17www.rsc.org/chemsoc/visualelements/orbital/

3 px orbital3 s orbital 3 dxz orbital

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0 1 2 3 4 5 6

Radial Probability Functions

Ψ2 (p

roba

bbilit

y)

r, Bohr radii

1s

2s2p

Ordering of s, p, d, and f orbitalsORBITAL SHAPE: Which orbital allows the

electrons to get closer to the nucleus?

The electron likes to get

closer to the nucleus if it can,

so it goes into the s orbital before the p

orbital

2s has greater probability close to the nucleus than 2p“Electron Penetration”


Electron PenetrationThe lower the value of l, the greater the

penetration. Relative ENERGIES: s < p < d < f

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Periodic TrendsATOMIC RADII Relative size of atomsIONIC RADII Relative size of ionsIONIZATION ENERGYELETRON AFFINITY

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Periodic Trends:ATOMIC RADII

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TREND:• small on top, big on

bottom• shrink going L to R,

because nuclear charge is increasing w/o increasing n

Fig 8.8


Periodic Trends: ATOMIC RADIIFig 8.9

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Problem: Atomic RadiusRank the following set of main group elements in order of decreasing atomic size: Br, Rb, Kr, Ca, Sr


Solution: Atomic RadiusRank in order of decreasing atomic size: Br, Rb, Kr, Ca, Sr

Elements with n = 4: Br, Kr, CaSize increases going L->R, so big to small is Ca > Br > Kr

Elements with n = 5: Rb, SrSize increases going L->R, so big to small isRb > Sr

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Solution: Atomic RadiusRank in order of decreasing atomic size: Br, Rb, Kr, Ca, Sr

Is Sr > Ca or is Ca > Sr?

Sr > Ca, so the final ranking is

Rb > Sr > Ca > Br > Kr


Periodic Trends: IONIC RADIIFig 8.20

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Periodic Trends: IONIC RADIIFig 8.21


Periodic Trends: IONIZATION ENERGY and ELECTRON AFFINITY

• IONIZATION ENERGY: A → A+ + e-

• ELECTRON AFFINITY: A + e- → A-

They are not the reverse of each other!

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Periodic Trends: IONIZATION ENERGY

Fig 8.10



Fig 8.11It is really hard to pull an electron from He

It is really easy to pull an electron from Cs

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Periodic Trends: ELECTRON AFFINITY

Fig 8.13


Periodic Trends: ELECTRON AFFINITY

ELECTRON AFFINITY:

Can be either exothermic or endothermic

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Periodic Trends: SUMMARY

Fig 8.14



Fig 8.12

1st, 2nd, and 3rd ionization energies

for Beryllium (Be)

Why is 3rd IE so much larger than 1st or 2nd IE?

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The ionization energies for lead are given below. Based on this information, predict the common oxidation states for this element.

Problem: Predicting Common Oxidation States from Ionization Energy


Lead in the Washington, DC water supply

March 12, 2004 Steve Curwood radio interview with Marc Edwards on “Living on Earth:”

D.C. WATER WOESCURWOOD: “Last year, at the request of some

Washington, D.C. residents, Marc Edwards, a civil engineer and corrosion specialist with Virginia Tech, began testing the quality of drinking water being piped into their homes. Soon, he says, he found concentrations of lead in that water that he describes as being literally off the charts. Some of the levels were so high that the water could be considered hazardous waste.”

33http://www.loe.org/ETS/organizations.php3?action=printContentItem&orgid=33&typeID=18&itemID=195#feature3

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Pb2+, Pb4+ are the most common oxidation states for Pb.

Solution: Predicting Common Oxidation States from Ionization Energy

0

1000

2000

3000

4000

5000

6000

7000

1 2 3 4 5

IE, K

J/m

ol

State

Pb+ → Pb2+

Pb3+ → Pb4+


Inert PairsSn2+, Tl+, Pb2+, Bi3+ are stable ions, even though they do not have a noble gas configuration:Pb: [Xe] 6s25d106p2

Pb2+: [Xe] 6s25d10

Pb4+: [Xe] 5d10

Common oxides of lead: PbO, PbO2

The s electrons are not as easily removed in these elements--a pair of s electrons is “inert”

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Lead in the Washington, DC water supply

March 12, 2004 Steve Curwood radio interview with Marc Edwards on “Living on Earth:”

D.C. WATER WOESCURWOOD: “Last year, at the request of some

Washington, D.C. residents, Marc Edwards, a civil engineer and corrosion specialist with Virginia Tech, began testing the quality of drinking water being piped into their homes. Soon, he says, he found concentrations of lead in that water that he describes as being literally off the charts. Some of the levels were so high that the water could be considered hazardous waste.”

http://www.loe.org/ETS/organizations.php3?action=printContentItem&orgid=33&typeID=18&itemID=195#feature3


Use of monochloramine in water purification

April, 1999 Guidance Manual from EPA:“Use an alternative or supplemental

disinfectant or oxidant such as chloramines or chlorine dioxide that will produce fewer DBPs (Disinfectant Byproducts).”

“Monochloramine, and chlorine dioxide are typically used to maintain a disinfectant residual in the distribution system”

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Use of monochloramine in water purification

Chlorine: Monochloramine:

• Monochloramine kills bacteria too• Not as strong an oxidant as chlorine or hypochlorite• Produces fewer potentially toxic or carcinogenic

byproducts

Cl2 + H2O → HOCl + H+ + Cl-

NH3 + HOCl → NH2Cl + H2O


Presence of lead in simulated drinking water

Rebecca Renner, ENVIRONMENTAL SCIENCE & TECHNOLOGY JUNE 15, 2004

water disinfected with chlorine

2 months after switching to chloramine disinfectant

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Chemistry of lead oxides

Hypochlorite reaction: Monochloramine reaction:

Solubility in water: PbO: 0.017 g/L at 20 °CSolubility of PbO2 << PbO

Pb + 2HOCl → PbO2 + 2HCl

Pb + NH2Cl + H2O → PbO + NH3 + HCl

oxidation state: + 4

oxidation state: + 2


Acid-Base Behavior of OxidesSome metal oxides and most metalloid oxides are

amphoteric (react with acid or base):

acid: Al2O3 + 6H+ → 2Al3+ + 3H2O

base: Al2O3 + 2OH- + 3H2O → 2Al(OH)4-

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Acid-Base Behavior of Oxides

Metal oxides produce basic solutions in water:

MgO + H2O → Mg2+ + 2OH -

Nonmetal oxides produce acidic solutions:CO2 + H2O → HCO3

- + H+

Because metals are ionic

Because nonmetals are covalent


Acid-Base Behavior of Oxides

Metal oxides produce basic solutions in water:

PbO + H2O → Pb2+ + 2OH -

Neither is very soluble in water, but PbO >> PbO2

PbO2 + 2H2O → Pb4+ + 4OH -

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Chemistry of lead oxides“Changes in pH, drops in ORP, or both could destabilize these PbO2

films, and thus increase plumbosolvency…observations that have been made by some water systems of erratic lead release from lead service lines, and increases or decreases without clear correlations with pH, DIC, and temperature, may be caused at least in part by effects of ORP changes.” --Lytle, D.A. and Schock, M.R., U.S. Environmental Protection Agency, 2005

ORP = oxidation-reduction potentialDIC = dissolved inorganic carbon


Periodic TrendsMain Points:Quantum numbers summaryElectron configurations for d-block elements:

– First-in, first-out for transition metals: ns electrons removed before (n-1)d electrons

– Magnetic properties of transition metal ions: paramagnetism, diamagnetism

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Measurement of Magnetic Properties Fig 8.19

No unpaired electrons in compound

Compound has unpaired electrons


Possible electron configurations for the paramagnetic ion Fe2+ are given below. Which is correct?

Problem: Predicting Diamagnetism or Paramagnetism from Electron

Configurations

(a)

(b)

(c)

(d)

(e) None of the above are correct

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Electron configuration for Fe:

4s23d6

Electron configuration for Fe2+:

3d6

Solution: Predicting Diamagnetism or Paramagnetism from Electron

Configurations

Always remove (n+1)s electrons before nd electrons when you make ions.

How to place electrons in the d orbital? Follow Hund’s rule:

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So (c) is the correct answer:

Solution: Predicting Diamagnetism or Paramagnetism from Electron

Configurations

How many unpaired electrons in Fe2+?

Fe2+

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