Chem 101 chapter 08 LEC

8/13/2019 Chem 101 chapter 08 LEC

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Chapter 8

Periodic

Properties of

the Elements

2007, Prentice Hall

Chemistry: A Molecular Approach, 1stEd.Nivaldo Tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA


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Tro, Chemistry: A Molecular Approach 2

Mendeleev

order elements by atomic mass saw a repeating pattern of properties Periodic LawWhen the elements are arranged in

order of increasing atomic mass, certain sets of

properties recur periodically put elements with similar properties in the same

column used pattern to predict properties of undiscovered

elements where atomic mass order did not fit other properties,he re-ordered by other propertiesTe & I


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Periodic Pattern

H

nm H2Oa/b

1 H2

Li

m Li2O

b

7 LiH

Na

m Na2O

b

23 NaH

Be

m/nm BeO

a/b

9 BeH2

m MgO

b

24 MgH2

Mg

nm B2O3

a

11 ( BH3)n

B

m Al2O3

a/b

27 (AlH3)Al

nm CO2

a

12 CH4

C

nm/mSiO2

a

28 SiH4

Si

nm N2O5

a

14 NH3

N

nm P4O10

a

31 PH3

P

nm O2

16 H2O

O

nm SO3

a

32 H2SS

nm Cl2O7

a

35.5 HClCl

nm

19 HF

F

a = acidic oxide, b = basic oxide, a/b = amphoteric oxide

m = metal, nm = nonmetal, m/nm = metalloid


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Mendeleev's Predictions


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What vs. Why

Mendeleevs Periodic Law allows us to predictwhat the properties of an element will be based

on its position on the table it doesnt explain why the pattern exists

Quantum Mechanics is a theory that explains

why the periodic trends in the properties exist


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Electron Spin

experiments by Stern and Gerlach showed a beamof silver atoms is split in two by a magnetic field the experiment reveals that the electrons spin on

their axis

as they spin, they generate a magnetic fieldspinning charged particles generate a magnetic field

if there is an even number of electrons, about halfthe atoms will have a net magnetic field pointingNorth and the other half will have a netmagnetic field pointing South


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Electron Spin Experiment


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Spin Quantum Number, ms

spin quantum number describes how theelectron spins on its axisclockwise or counterclockwise

spin upor spin down

spins must cancel in an orbitalpaired

mscan have values of


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Pauli Exclusion Principle no two electrons in an atom may have the same set of

4 quantum numbers

therefore no orbital may have more than 2 electrons,and they must have with opposite spins

knowing the number orbitals in a sublevel allows us todetermine the maximum number of electrons in thesublevelssublevel has 1 orbital, therefore it can hold 2 electrons

psublevel has 3 orbitals, therefore it can hold 6 electrons

dsublevel has 5 orbitals, therefore it can hold 10 electrons

fsublevel has 7 orbitals, therefore it can hold 14 electrons


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Allowed Quantum Numbers

QuantumNumber

Values Number ofValues

Significance

Principal, n 1, 2, 3, ... - distance from

nucleusAzimuthal, l 0, 1, 2, ..., n-1 n shape of

orbital

Magnetic, ml -l,...,0,...+l 2l+ 1 orientation of

orbital

Spin, ms -, + 2 direction of

electron spin


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Quantum Numbers ofHeliums Electrons

helium has two electrons

both electrons are in the first energy level

both electrons are in thesorbital of the first energy level

since they are in the same orbital, they must have opposite spins

n l ml ms

first

electron 1 0 0 +

second

electron1 0 0 -


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Electron Configurations

the ground stateof the electron is the lowestenergy orbital it can occupy

the distribution of electrons into the various

orbitals in an atom in its ground state is called itselectron configuration

the number designates the principal energy level

the letter designates the sublevel and type of orbital

the superscript designates the number of electronsin that sublevel

He = 1s2
http://wps.prenhall.com/wps/media/objects/166/170213/ElectronConfigurationsmov.htmlhttp://wps.prenhall.com/wps/media/objects/166/170213/ElectronConfigurationsmov.html


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Orbital Diagrams we often represent an orbital as a square and the

electrons in that orbital as arrows the direction of the arrow represents the spin of the

electron

orbital with1 electron

unoccupiedorbital

orbital with2 electrons


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Sublevel Splitting in

Multielectron Atoms the sublevels in each principal energy level of

Hydrogen all have the same energywe call orbitalswith the same energy degenerate

or other single electron systems for multielectron atoms, the energies of the sublevels

are split caused by electron-electron repulsion

the lower the value of the lquantum number, the lessenergy the sublevel hass(l= 0)


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Penetrating and Shielding the radial distribution function shows that

the 2sorbital penetrates more deeply intothe 1sorbital than does the 2p the weaker penetration of the 2psublevel

means that electrons in the 2psublevelexperience more repulsive force, they aremore shielded from the attractive force ofthe nucleus

the deeper penetration of the 2selectronsmeans electrons in the 2ssublevelexperience a greater attractive force to thenucleus and are not shielded aseffectively

the result is that the electrons in the 2ssublevel are lower in energy than theelectrons in the 2p


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Penetration & Shielding


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Energy

1s

7s

2s

2p

3s

3p3d

6s6p

6d

4s

4p4d

4f5s

5p

5d5f

Notice the following:1. because of penetration, sublevels within

an energy level are not degenerate2. penetration of the 4thand higher energy

levels is so strong that theirs sublevel islower in energy than the dsublevel of the

previous energy level3. the energy difference between levels

becomes smaller for higher energy levels


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Order of Subshell Fillingin Ground State Electron Configurations

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d

7s

start by drawing a diagramputting each energy shell ona row and listing the subshells,

(s, p, d, f), for that shell inorder of energy, (left-to-right)

next, draw arrows through

the diagonals, looping backto the next diagonaleach time


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Filling the Orbitals with Electrons

energy shells fill from lowest energy to high subshells fill from lowest energy to highsp dfAufbau Principle

orbitals that are in the same subshell have the sameenergy

no more than 2 electrons per orbitalPauli Exclusion Principle

when filling orbitals that have the same energy, placeone electron in each before completing pairsHunds Rule


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2. Draw 9 boxes to represent the first 3 energy

levels sandporbitalsa) since there are only 12 electrons, 9 should be

plenty

1s 2s 2p 3s 3p

Example 8.1Write the Ground StateElectron Configuration and Orbital

Diagram and of Magnesium.


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3. Add one electron to each box in a set, thenpair the electrons before going to the next setuntil you use all the electrons When pair, put in opposite arrows

1s 2s 2p 3s 3p




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4. Use the diagram to write the electronconfiguration Write the number of electrons in each set as a

superscript next to the name of the orbital set

1s22s22p63s2 = [Ne]3s2

1s 2s 2p 3s 3p


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Valence Electrons the electrons in all the subshells with the

highest principal energy shell are called thevalence electrons

electrons in lower energy shells are calledcore electrons

chemists have observed that one of the most

important factors in the way an atombehaves, both chemically and physically, isthe number of valence electrons


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Electron Configuration of Atomsin their Ground State

Kr = 36 electrons1s22s22p63s23p64s23d104p6

there are 28 core electrons and 8 valence electrons

Rb = 37 electrons

1s22s22p63s23p64s23d104p65s1

[Kr]5s1

for the 5s1electron in Rb the set of quantum numbers isn= 5, l= 0, m

l= 0, m

s= +

for an electron in the 2p sublevel, the set of quantumnumbers is n= 2, l= 1, ml= -1 or (0,+1), and ms= - or (+)


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Electron Configurations


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Electron Configuration & thePeriodic Table

the Group number corresponds to the number ofvalence electrons

the length of each block is the maximumnumber of electrons the sublevel can hold

the Period number corresponds to the principalenergy level of the valence electrons


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s1

s2

d1 d2 d3 d4 d5 d6 d7 d8 d9d10

s2

p1

p2

p3

p4

p5

p6

f2

f3

f4

f5

f6

f7

f8

f9

f10

f11

f12

f13

f14

f14

d1

1234

567


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Electron Configuration fromthe Periodic Table

P = [Ne]3s23p3

P has 5 valence electrons

3p3

P

Ne123

4567

1A2A 3A 4A 5A 6A 7A

8A

3s2


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31

Transition Elements for the dblock metals, the principal energy level is one less than

valence shell one less than the Period number sometimesselectron promoted to dsublevel

4s 3d

ZnZ = 30, Period 4, Group 2B

[Ar]4s23d10

for thefblock metals, the principal energy level is twoless than valence shell two less than the Period number they really belong to

sometimes delectron in configurationEuZ = 63, Period 6[Xe]6s24f 7 6s 4f


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Electron Configuration fromthe Periodic Table

As = [Ar]4s23d104p3

As has 5 valence electrons

As

123

4567

1A2A 3A 4A 5A 6A 7A

8A

4s2

Ar3d10

4p3


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PracticeUse the Periodic Table to write the shortelectron configuration and orbital diagram for each of the

following Na (at. no. 11)

Te (at. no. 52)

Tc (at. no. 43)


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PracticeUse the Periodic Table to write the shortelectron configuration and orbital diagram for each of the

following Na (at. no. 11) [Ne]3s1

Te (at. no. 52) [Kr]5s24d105p4

Tc (at. no. 43) [Kr]5s24d5

3s

5s 5p4d

5s 4d


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Properties & Electron Configuration

elements in the samecolumn have similar

chemical and physicalproperties because they

have the same number of

valence electrons in thesame kinds of orbitals


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Electron Configuration &Element Properties

the number of valence electrons largely determines the behaviorof an element chemical and some physical

since the number of valence electrons follows a Periodic pattern,

the properties of the elements should also be periodic quantum mechanical calculations show that 8 valence electronsshould result in a very unreactive atom, an atom that is verystableand the noble gases, that have 8 valence electrons are allvery stable and unreactive

conversely, elements that have either one more or one lesselectron should be very reactiveand the halogens are the mostreactive nonmetals and alkali metals the most reactive metals as a group


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Electron Configuration &

Ion Charge we have seen that many metals and nonmetals

form one ion, and that the charge on that ion is

predictable based on its position on the PeriodicTableGroup 1A = +1, Group 2A = +2, Group 7A = -1,

Group 6A = -2, etc.

these atoms form ions that will result in anelectron configuration that is the same as thenearest noble gas


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Electron Configuration of Anions

in their Ground State anions are formed when atoms gain enoughelectrons to have 8 valence electronsfilling thesandpsublevels of the valence shell

the sulfur atom has 6 valence electronsS atom = 1s22s22p63s23p4

in order to have 8 valence electrons, it must gain2 moreS2-anion = 1s22s22p63s23p6


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Electron Configuration of

Cations in their Ground State cations are formed when an atom loses all itsvalence electronsresulting in a new lower energy level valence shell

however the process is always endothermic

the magnesium atom has 2 valence electronsMg atom = 1s22s22p63s2

when it forms a cation, it loses its valence electronsMg2+cation = 1s22s22p6


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Trend in Atomic RadiusMain Group Different methods for measuring the radius of an

atom, and they give slightly different trends van der Waals radius = nonbonding

covalent radius = bonding radius

atomic radius is an average radius of an atom based onmeasuring large numbers of elements and compounds

Atomic Radius Increases down group valence shell farther from nucleus

effective nuclear charge fairly close

Atomic Radius Decreases across period (left to right) adding electrons to same valence shell

effective nuclear charge increases

valence shell held closer


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Effective Nuclear Charge in a multi-electron system, electrons are simultaneously

attracted to the nucleus and repelled by each other outer electrons are shieldedfrom full strength of nucleus

screening effect

effective nuclear chargeis net positive charge that isattracting a particular electron

Zis nuclear charge, Sis electrons in lower energy levelselectrons in same energy level contribute to screening, but

very little

effective nuclear charge on sublevels trend,s > p > d > f

Zeffective= Z - S


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Screening & Effective Nuclear Charge


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Trends in Atomic Radius

Transition Metals increase in size down the Group

atomic radii of transition metals roughly the

same size across the dblockmust less difference than across main group

elements

valence shell ns2

, not the delectronseffective nuclear charge on the ns2electrons

approximately the same


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l h h


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Example 8.5Choose theLarger Atom in Each Pair

1) Nor F,N is further left1) Nor F2) C or Ge

3) N or Al

4) Al or Ge? opposing trends

1) Nor F2) C or Ge, Ge is further down1) Nor F2) C or Ge

3) N or Al, Al is further down & left


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Electron Configuration ofCations in their Ground State

cations form when the atom loses electronsfrom the valence shell

for transition metals electrons, may be removed

from the sublevel closest to the valence shellAl atom = 1s22s22p63s23p1Al+3ion = 1s22s22p6

Fe atom = 1s22s22p63s23p64s23d6Fe+2ion = 1s22s22p63s23p63d6

Fe+3ion = 1s22s22p63s23p63d5Cu atom = 1s22s22p63s23p64s13d10

Cu+1ion = 1s22s22p63s23p63d10

M ti P ti f


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Magnetic Properties ofTransition Metal Atoms & Ions

electron configurations that result in unpaired electronsmean that the atom or ion will have a net magnetic fieldthis is called paramagnetismwill be attracted to a magnetic field

electron configurations that result in all paired electronsmean that the atom or ion will have no magnetic fieldthis is called diamagnetism slightly repelled by a magnetic field

both Zn atoms and Zn2+ions are diamagnetic, showingthat the two 4selectrons are lost before the 3dZn atoms [Ar]4s23d10

Zn2+ions [Ar]4s03d10

E l 8 6 W it th El t C fi ti


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Example 8.6Write the Electron Configurationand Determine whether the Fe atom and Fe3+ion

are Paramagnetic or Diamagnetic

Fe Z = 26

previous noble gas = Ar

18 electrons

4s 3d

Fe atom = [Ar]4s23d6

unpaired electrons

paramagnetic

4s 3d

Fe3+ion = [Ar]4s03d5

unpaired electrons

paramagnetic

Trends in Ionic Radius


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Trends in Ionic Radius Ions in same group have same charge

Ion size increases down the grouphigher valence shell, larger

Cations smaller than neutral atom; Anions biggerthan neutral atom

Cations smaller than anions except Rb+1& Cs+1bigger or same size as F-1and O-2

Larger positive charge = smaller cation for isoelectronic species

isoelectronic = same electron configuration

Larger negative charge = larger anion for isoelectronic series


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Periodic Pattern - Ionic Radius ()

Li

+1

Na

+1

K

+1

Rb

+1

Cs

+1

Be

+2

Mg

+2

Ca

+2

B

Tl

+3

+1

In

+3

+1

Ga

+3

+1

Al

+3

Sn

+4

+2

Ge

-4

Si-4

N

-3

Bi

Sb

As

-3

Te

-2

O

-2

P-3

S

-2

Se

-2

Br

-1

-1

F

I

-1

Cl

-1

H

+1-1

2A 3A 4A 5A 6A 7A

Sr

+2

Ba

+2

C

-4

Pb

+4

+2

0.68

0.97

1.33

1.47

0.31

0.66

0.99

1.13

1.40

1.84

1.98

2.21

1.33

1.81

1.96

2.20

Al

+3

+3

0.23

0.51

0.62

0.81

1A

1.69 1.35

1.71

2.12

2.22

0.71

0.840.95


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Ionization Energy

minimum energy needed to remove an electronfrom an atomgas state

endothermic processvalence electron easiest to remove

M(g) + IE1M1+(g) + 1 e-

M+1

(g) + IE2M2+

(g) + 1 e-

first ionization energy = energy to remove electron from

neutral atom; 2nd IE = energy to remove from +1 ion; etc.
http://wps.prenhall.com/wps/media/objects/166/170446/IonizationEnergy.htmlhttp://wps.prenhall.com/wps/media/objects/166/170446/IonizationEnergy.html


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General Trends in 1stIonization Energy

larger the effective nuclear charge on theelectron, the more energy it takes to remove it

the farther the most probable distance the

electron is from the nucleus, the less energy ittakes to remove it

1st IE decreasesdown the group

valence electron farther from nucleus 1st IE generally increasesacross the periodeffective nuclear charge increases


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E ample 8 8 Choose the Atom in Each


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Example 8.8Choose the Atom in EachPair with the Higher First Ionization Energy

1) Al or S, Al is further left1) Al or S2) Asor Sb, Sb is further down1) Al or S2) Asor Sb

3) Nor Si, Si is further down & left

1) Al or S2) Asor Sb

3) Nor Si

4) O or Cl? opposing trends

I l iti i th T d


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Irregularities in the Trend Ionization Energy generally increases from left

to right across a Period except from 2A to 3A, 5A to 6A

Be 1s 2s 2p

B

1s 2s 2p

N

1s 2s 2p

O

1s 2s 2p

Which is easier to remove an electron from B or Be? Why?Which is easier to remove an electron from N or O? Why?


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Irregularities in the

First Ionization Energy Trends

Be

1s 2s 2p

B

1s 2s 2p

Be+

1s 2s 2pTo ionize Be you must break up a full sublevel, cost extra energy

B+

1s 2s 2pWhen you ionize B you get a full sublevel, costs less energy


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Irregularities in the

First Ionization Energy Trends

To ionize N you must break up a half-full sublevel, cost extra energy

N+

1s 2s 2p

O

1s 2s 2p

N

1s 2s 2p

O+

1s 2s 2p

When you ionize O you get a half-full sublevel, costs less energy

T d i S i


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Trends in SuccessiveIonization Energies

removal of each successiveelectron costs more energy shrinkage in size due to having more

protons than electrons

outer electrons closer to the nucleus,therefore harder to remove

regular increase in energy for each

successive valence electron large increase in energy when startremoving core electrons


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d i l ffi i


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Trends in Electron Affinity energy released when an neutral atom gains an electron

gas stateM(g) + 1e-M-1(g) + EA

defined as exothermic (-), but may actually beendothermic (+)

alkali earth metals & noble gases endothermic, WHY? more energy released (more -); the larger the EA generally increases across periodbecomes more negative from left to right

not absolute

lowest EA in period = alkali earth metal or noble gas

highest EA in period = halogen


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Metallic Character


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Metallic Character Metalsmalleable & ductile

shiny, lusterous, reflect light conduct heat and electricitymost oxides basic and ionic form cations in solution lose electrons in reactions - oxidized

Nonmetalsbrittle in solid statedull electrical and thermal insulatorsmost oxides are acidic and molecular

form anions and polyatomic anionsgain electrons in reactions - reduced

metallic character increases left metallic character increase down


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Example 8 9 Choose the


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Example 8.9Choose theMore Metallic Element in Each Pair

1) Snor Te2) P or Sb

3) Ge or In, In is further down & left

1) Snor Te2) P or Sb

3) Ge or In

4) S or Br? opposing trends

1) Snor Te, Sn is further left1) Snor Te2) P or Sb, Sb is further down

Trends in the Alkali Metals


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Trends in the Alkali Metals atomic radius increases down the column ionization energy decreases down the column

very low ionization energiesgood reducing agents, easy to oxidizevery reactive, not found uncombined in nature react with nonmetals to form salts

compounds generally soluble in water found in seawater electron affinity decreases down the column melting point decreases down the column all very low MP for metals

density increases down the column except K in general, the increase in mass is greater than the increase

in volume


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2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)

Trends in the Halogens


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Trends in the Halogens atomic radius increases down the column ionization energy decreases down the column very high electron affinitiesgood oxidizing agents, easy to reducevery reactive, not found uncombined in nature react with metals to form salts

compounds generally soluble in water found in seawater reactivity increases down the column react with hydrogen to form HX, acids melting point and boiling point increases down the

column density increases down the column in general, the increase in mass is greater than the increase in

volume


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Example 8 10 Write a balanced chemical


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Example 8.10 Write a balanced chemicalreaction for the following.

reaction between potassium metal and brominegas

K(s) + Br2(g)

K(s) + Br2(g)K+Br2 K(s) + Br2(g)2 KBr(s)

(ionic compounds are all solids at room temperature)



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reaction between rubidium metal and liquidwater

Rb(s) + H2O(l)

Rb(s) + H2O(l) Rb+(aq) + OH(aq)+ H2(g)2 Rb(s) + 2 H2O(l) 2 Rb

+(aq) + 2 OH(aq)+ H2(g)

(alkali metal ionic compounds are soluble in water)



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reaction between chlorine gas and solid iodineCl2(g) + I2(s)

Cl2(g) + I2(s) IClwrite the halogen lower in the column first

assume 1:1 ratio, though others also exist

2 Cl2(g) + I2(s)2 ICl(g)(molecular compounds found in all states at room

temperature)

Trends in the Noble Gases


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Trends in the Noble Gases atomic radius increases down the column ionization energy decreases down the columnvery high IE

very unreactiveonly found uncombined in natureused as inert atmosphere when reactions with other gases

would be undersirable melting point and boiling point increases down the

column all gases at room temperaturevery low boiling points

density increases down the column in general, the increase in mass is greater than the increase in

volume


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Chem 101 chapter 08 LEC