8-1 chapter 8 electron configuration and chemical periodicity
TRANSCRIPT
8-1
Chapter 8
Electron Configuration and Chemical Periodicity
8-2
Electron Configuration and Chemical Periodicity
8.1 Development of the Periodic Table
8.2 Characteristics of Many-Electron Atoms
8.3 The Quantum-Mechanical Model and the Periodic Table
8.4 Trends in Some Key Periodic Atomic Properties
8.5 The Connection Between Atomic Structure and ChemicalReactivity
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Mendeleev’s Predicted Properties of Germanium (“eka Silicon”) and Its Actual Properties
Table 8.1
PropertyPredicted Properties of eka Silicon(E)
Actual Properties of Germanium (Ge)
atomic massappearancedensitymolar volumespecific heat capacityoxide formulaoxide densitysulfide formula and solubility
chloride formula (boiling point)
chloride densityelement preparation
72 amugray metal
5.5 g/cm3
13 cm3/mol0.31 J/g.KEO2
4.7 g/cm3
ES2; insoluble in H2O;
soluble in aqueous (NH4)2SECl4; (< 100 oC)
1.9 g/cm3
reduction of K2EF6 with sodium
72.61 amugray metal
5.32 g/cm3
13.65 cm3/mol0.32 J/g.KGeO2
4.23 g/cm3
GeS2; insoluble in H2O;
soluble in aqueous (NH4)2SGeCl4; (84 oC)
1.844 g/cm3
reduction of K2GeF6 with sodium
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Figure 8.1
Observing the Effect of Electron Spin
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Table 8.2 Summary of Quantum Numbers of Electrons in Atoms
Name Symbol Allowed Values Property
principal n positive integers (1, 2, 3,…)
orbital energy (size)
angular momentum
l integers from 0 to n-1 orbital shape (l values of 0, 1, 2 and 3 correspond to s, p, d and f orbitals, respectively.)
magnetic mlintegers from -l to 0 to +l orbital orientation
spin ms+1/2 or -1/2 direction of e- spin
Each electron in an atom has its own unique set of four (4) quantum numbers.
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The Pauli Exclusion Principle
No two electrons in the same atom can have the same four quantum numbers
An atomic orbital can hold a maximum of two electronsand they must have opposite spins (paired spins)
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Figure 8.2
Spectral evidence of energy-level splitting in many-electron systems
Leads to the splitting of energy levels into sublevels of differing energies:the energy of an orbital depends mostly on its n value (size) and somewhaton its l value (shape)
Many-electron atoms: have nucleus-electron and electron-electroninteractions
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Factors Affecting Atomic Orbital Energies
1. Additional electron in the same orbital
An additional electron raises the orbital energy through electron-electron repulsions.
2. Additional electrons in inner orbitals
Inner electrons shield outer electrons more effectively than do electrons in the same sublevel.
Higher nuclear charge lowers orbital energy (stabilizes the system) by increasing nucleus-electron attractions.
Effect of nuclear charge (Zeffective)
Effect of electron repulsions (shielding)
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The effect of nuclearcharge
Figure 8.3
Greater nuclear chargelowers orbital energy
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Figure 8.4
The effect of another electron in
the same orbital
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are needed to see this picture.Each electron shieldsthe other from the fullnuclear charge, thusraising orbital energy
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Figure 8.5
The effect of other electrons in inner
orbitals
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Inner electrons shieldouter electrons very welland raise orbital energygreatly
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Figure 8.6The effect of orbital shape
2s electron farther from nucleusthan 2p electron, but penetratesnear nucleus; increased attractionresults in lower orbital energy
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General Rule for Predicting Relative Sublevel Energies
For a given n value, the lower the l value, the lower the sublevelenergy; thus….
s < p < d < f
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Illustrating Orbital Occupancies
A. The electron configuration
n l # of electrons in the sublevel
as s, p, d or f
B. The orbital diagram (box or circle)
Figure 8.7Order for filling energy sublevels with
electrons
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dark = filled, spin-paired
light = half-filled
no color = empty
Figure 8.8
A vertical orbital
diagram for the Li ground
state
Sublevel energyincreases frombottom to top
1s22s1
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Hund’s Rule
When orbitals of equal energy are available, the electronconfiguration of lowest energy has the maximum numberof unpaired electrons with parallel spins.
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Sample Problem 8.1 Determining Quantum Numbers from Orbital Diagrams
PLAN:
SOLUTION:
Use the orbital diagram to find the third and eighth electrons.
PROBLEM: Write a set of quantum numbers for the third electron and a set for the eighth electron of the fluorine (F) atom.
9F
1s 2s 2p
The third electron is in the 2s orbital. Its quantum numbers are:
n = l = ml = ms= 2 0 0 +1/2
The eighth electron is in a 2p orbital. Its quantum numbers are:
n = l = ml = ms= 2 -1 -1/21
Up arrow = +1/2Down arrow = -1/2
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Figure 8.9
Orbital occupancy for the first 10 elements, H through Ne
He and Ne have filled outer shells: confers chemical inertness
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Figure 8.10
Condensed ground-state electron configurations in the first three periods
Similar outer electron configurations correlate with similarchemical behavior.
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Similar Reactivities within A Group
Orbitals are filled in order of increasingenergy, which leads to outer electronconfigurations that recur periodically,which leads to chemical properties thatrecur periodically.
Figure 8.11
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Cr and Cu: Half-filled and filled sublevels are unexpectedly stable!
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Figure 8.12
A periodic table of partial ground-state electron configurations
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Figure 8.13
The relation between orbital filling and the Periodic Table
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Common tool used topredict the filling orderOf sublevels
n values are constant horizontally
l values are constant vertically
combined values of n+1 are constant diagonally
p. 302
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Categories of Electrons
Inner (core) electrons: fill all the lower energy levels of an atom
Outer electrons: those electrons in the highest energy level(highest n value) of an atom
Valence electrons: those involved in forming compounds; the bonding electrons; among the main-group elements, the valence electrons are the outer electrons
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General Observations about the Periodic Table
A. The group number equals the number of outer electrons (those with the highest value of n) (main-group elements only)
B. The period number is the n value of the highest energy level.
C. The n value squared (n2) gives the total number of orbitals in thatenergy level; 2n2 gives the maximum number of electronsin the energy level.
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SAMPLE PROBLEM 8.2 Determining Electron Configuration
PLAN:
SOLUTION:
PROBLEM: Using the periodic table, give the full and condensed electron configurations, partial orbital diagrams showing valence electrons, and number of inner electrons for the following elements:
(a) potassium (K: Z = 19) (b) molybdenum (Mo: Z = 42) (c) lead (Pb: Z = 82)
Use the atomic number for the number of electrons and the periodic table for the order of filling of the electron orbitals. Condensed configurations consist of the preceding noble gas plus the outer electrons.
(a) for K (Z = 19)
1s22s22p63s23p64s1
[Ar] 4s1
4s1
condensed configuration:
partial orbital diagram:
full configuration:
K has 18 inner electrons.
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SAMPLE PROBLEM 8.2: (continued)
(b) for Mo (Z = 42)
1s22s22p63s23p64s23d104p65s14d5
[Kr] 5s14d5
(c) for Pb (Z = 82)
[Xe] 6s24f145d106p2
condensed configuration:partial orbital diagram:
full configuration:
5s1 4d5
condensed configuration:
partial orbital diagram:
full configuration: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2
Mo has 36 inner electrons and 6 valence electrons.
6s2 6p2
Pb has 78 inner electrons and 4 valence electrons.
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All physical and chemical properties of the elements are based on the electronic configurations of their atoms.
KEY PRINCIPLE
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Figure 8.14
Defining metallic and covalent radii
A. Metallic radius: 1/2 the distance between adjacent nuclei in a crystal
B. Covalent radius: 1/2 the distance between bonded nuclei in a molecule
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Figure 8.15
Atomic radii of main-group and transition elements
Opposing forces: Changes in n andchanges in Zeff
Overall Trends
(A) n dominates within a group; atomicradius generally increases in a group
from top to bottom
(B) Zeff dominates within a period; atomic
radius generally decreases in a periodfrom left to right
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Figure 8.16
Periodicity of atomic radius
Large size shifts whenmoving from oneperiod to the next
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SAMPLE PROBLEM 8.3 Ranking Elements by Atomic Size
PLAN:
SOLUTION:
PROBLEM: Using only the periodic table, rank each set of main group elements in order of decreasing atomic size.
(a) Ca, Mg, Sr (b) K, Ga, Ca (c) Br, Rb, Kr (d) Sr, Ca, Rb
Size increases down a group; size decreases across a period.
(a) Sr > Ca > Mg These elements are in Group 2A.
(b) K > Ca > Ga These elements are in Period 4.
(c) Rb > Br > Kr Rb has a higher energy level and is far to the left. Br is to the left of Kr.
(d) Rb > Sr > Ca Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr.
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Ionization Energy
The amount of energy required for the complete removal of 1 molof electrons from 1 mol of gaseous atoms or ions; an energy-requiring process; value is positive in sign
IE1 = first ionization energy: removes an outermost electron from
the gaseous atom: atom(g) ion+(g) + e- ∆E = IE1 > 0
IE2 = second ionization energy: removes a second electron from the
gaseous ion: ion+(g) ion+2(g) + e- ∆E = IE2 > IE1
Atoms with a low IE1 tend to form cations during reactions, whereas those
with a high IE1 (except noble gases) often form anions.
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Ionization Energies: Correlations with Atomic Size
1. As size decreases, it take more energy to remove an electron.
2. Ionization energy generally decreases down a group.
3. Ionization generally increases across a period.
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Figure 8.17
Periodicity of first ionization energy (IE1)
Lowest values for alkali metals; highest values for noble gases
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Figure 8.18
First ionization energies of the
main-group elements
Increase within aperiod and decrease
within a group
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SAMPLE PROBLEM 8.4 Ranking Elements by First Ionization Energy
PLAN:
SOLUTION:
PROBLEM: Using the periodic table, rank the elements in each of the following sets in order of decreasing IE1:
(a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs
IE decreases down in a group; IE increases across a period.
(a) He > Ar > Kr
(b) Te > Sb > Sn
(c) Ca > K > Rb
(d) Xe > I > Cs
Group 8A elements- IE decreases down a group.
Period 5 elements - IE increases across a period.
Ca is to the right of K; Rb is below K.
I is to the left of Xe; Cs is further to the left and down one period.
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Figure 8.19
The first three ionization energies of beryllium (in MJ/mol)
Successive IEs increase, but a largeincrease is observed to remove the firstcore electron (for Be, IE3)
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SAMPLE PROBLEM 8.5 Identifying an Element from Successive Ionization Energies
PLAN:
SOLUTION:
PROBLEM: Name the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration:
IE1 IE2 IE3 IE4 IE5 IE6
1012 1903 2910 4956 6278 22,230
Look for a large increase in energy that indicates that all of the valence electrons have been removed.
The largest increase occurs at IE6, that is, after the 5th valence electron has been removed. The element must have five valence electrons with a valence configuration of 3s23p3, The element must be phosphorus. P (Z = 15).
The complete electronic configuration is: 1s22s22p63s23p3.
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Electron Affinity (EA)
The energy change accompanying the addition of 1 mol of electronsto 1 mol of gaseous atoms or ions.
atom(g) + e- ion-(g) ∆E = EA1 (usually negative)
EA2 is always positive (adding negative charge to negativelycharged ion).
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Figure 8.20
Electron affinities of the main-group elements
Negative values =energy is released whenthe ion forms
Positive values =energy is absorbedto form the anion
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General Trends Involving IEs and EAs
Reactive non-metals: Groups 6A and 7A; in their ionic compoundsthey form negative ions (have high IEs andvery (-) EAs)
Reactive metals: Group 1A; in their ionic compounds, they formpositive ions (have low IEs and slightly (-) EAs)
Noble gases: Group 8A; they do not lose or gain electrons (have veryhigh IEs and slightly (+) EAs)
8-47Figure 8.21
Trends in three atomic properties
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Metallic Behavior
Metals: shiny solids; tend to lose electrons in reactions withnon-metals (left and lower 3/4 of periodic table)
Non-metals: tend to gain electrons in reactions with metals;upper right-hand quarter of periodic table
Metalloids: have intermediate properties; located between themetals and non-metals in the periodic table
Metallic behavior decreases left to right and increases top to bottom in the periodic table
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Figure 8.22
Trends in metallic behavior
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Moving down a GROUP: elements at the top tend to formanions and those at the bottom tend to form cations
Moving across a PERIOD: elements at the left tend to formcations and those at the right lend to form anions
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Figure 8.23
The change in metallic behavior in Group 5A(15) and Period 3
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Acid-Base Properties
Main-group metals: transfer electrons to oxygen; their oxidesare ionic; in water these oxides act as bases (produce OH-)
Nonmetals: share electrons with oxygen; their oxides arecovalent; in water these oxides act as acids (produce H+)
Some metals and many metalloids form oxides that areamphoteric (can act as an acid or a base in water).
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The trend in acid-base behavior of element oxides
Figure 8.24
red = oxides are acidic blue = oxides are basic
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N2O5(s) + H2O(l) 2HNO3(aq)
P4O10(s) + 6H2O(l) 4H3PO4(aq)
Bi2O3(s) + 6HNO3(aq) 2Bi(NO3)3(aq) + 3H2O(l)
Al2O3(s) + 6HCl(aq) 2AlCl3(aq) + 3H2O(l)
Al2O3(s) + 2NaOH(aq) + 3H2O(l) 2NaAl(OH)4(aq)
Examples
Amphoteric Behavior
Na2O(s) + H2O(l) 2NaOH(aq)
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Monatomic Ions
Elements in Groups 1A, 2A, 6A and 7A that readily form ions either loseor gain electrons to attain a filled outer level and thus a noble gas configuration.Their ions are said to be isoelectronic with the nearest noble gas.
Elements in Groups 3A, 4A and 5A form cations via a different process; they attain pseudo-noble gas configurations.
Sn ([Kr]5s24d105p2) Sn+4 ([Kr]4d10) + 4e-
Sn+2 ([Kr]5s24d10) + 2e-Sn ([Kr]5s24d105p2)
Main Group
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Figure 8.25
Main-group ions and the noble gas configurations
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SAMPLE PROBLEM 8.6 Writing Electron Configurations of Main-Group Ions
PLAN:
SOLUTION:
PROBLEM: Using condensed electron configurations, write reactions for the formation of the common ions of the following elements:
(a) iodine (Z = 53) (b) potassium (Z = 19) (c) indium (Z = 49)
Ions of elements in Groups 1A, 2A, 6A and 7A are usually isoelectronic with the nearest noble gas.
Metals in Groups 3A to 5A can lose the np, or ns and np, electrons.
(a) Iodine (Z = 53) is in Group 7A and will gain one e- to be isoelectronic with Xe:I([Kr]5s24d105p5) + e- I- ([Kr]5s24d105p6)
(b) Potassium (Z = 19) is in Group 1A and will lose one e- to be isoelectronic with Ar: K ([Ar]4s1) K+ ([Ar]) + e-
(c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three electrons: In ([Kr]5s24d105p1) In+ ([Kr]5s24d10) + e-
In ([Kr]5s24d105p1) In3+([Kr]4d10) + 3e-
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Monatomic Ions
Transition Metal Ions
Rarely attain a noble gas configuration
Form more than one cation by losing all of their ns andsome of their (n-1)d electrons
For Period 4 transition metals, the 4s orbital is morestable that the 3d orbitals; thus the rule “first in, first out” applies.
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Figure 8.26
The Period 4 crossover in
sublevel energies
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General Rules For Ion Formation
Main group, s-block metals: remove all electrons with highest n value
Main group, p-block metals: remove np electrons before ns electrons
Transition (d-block) metals: remove ns electrons before (n-1)d electrons
Non-metals: add electrons to the p orbital of highest n value
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Magnetic Properties of Transition Metal Ions
Chemical species (atoms, ions, molecules) with one or moreunpaired electrons are affected by external magnetic fields.
Ag (Z=47) [Kr]5s14d10
Cd (Z=48) [Kr]5s24d10
Species with unpaired electrons exhibit paramagnetism (attracted byan external magnetic field).
Species with all electrons paired exhibit diamagnetism (notattracted by an external magnetic field).
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Figure 8.27
Apparatus for measuring the magnetic behavior of a sample
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Some Examples
Fe+3 exhibits greater paramagnetism than Fe.
Fe ([Ar]4s23d6) Fe+3 ([Ar]3d5) + 3e-
Zn, Zn+2 and Cu+ are diamagnetic, but Cu is paramagnetic.
Cu ([Ar]4s13d10) Cu+ ([Ar]3d10) + e-
Zn ([Ar]4s23d10) Zn+2 ([Ar]3d10) + 2e-
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SAMPLE PROBLEM 8.7 Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions
PLAN:
SOLUTION:
PROBLEM: Use condensed electron configurations, write the reaction for the formation of each transition metal ion and predict whether the ion is paramagnetic.
(a) Mn2+(Z = 25) (b) Cr3+(Z = 24) (c) Hg2+(Z = 80)
Write the electron configuration and remove electrons starting with the ns electrons to attain the ion charge. If the remaining configuration has unpaired electrons, the ion is paramagnetic.
paramagnetic(a) Mn2+(Z = 25) Mn([Ar]4s23d5) Mn2+([Ar]3d5) + 2e-
(b) Cr3+(Z = 24) Cr([Ar]4s13d5) Cr3+([Ar] 3d3) + 3e- paramagnetic
(c) Hg2+(Z = 80) Hg([Xe]6s24f145d10) Hg2+([Xe] 4f145d10) + 2e-
not paramagnetic (is diamagnetic)
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Ionic Size vs Atomic Size
Ionic radius: an estimate of the size of an ion in a crystallineionic compound
Cations are smaller than their parent atoms (decrease in electron-electronrepulsions).
Anions are larger than their parent atoms (increase in electron-electron repulsions).
General Observations
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Figure 8.28
Depicting ionic radii
8-67Figure 8.29
Ionic vs atomic radius
Ionic size increasesdown a group
Trends in periodsare complex
For atoms that form morethan one cation: thegreater the ionic charge,the smaller the ionic radius
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Summary on Ionic Size
Ionic size increases down a group.
Ionic size decreases across a period but increases from cation to anion.
Ionic size decreases with increasing (+) (or decreasing (-)) chargein an isoelectronic series
Ionic size decreases as charge increases for different cations of agiven element
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SAMPLE PROBLEM 8.8 Ranking Ions by Size
PLAN:
SOLUTION:
PROBLEM: Rank each set of ions in order of decreasing size, and explain your ranking:
(a) Ca2+, Sr2+, Mg2+ (b) K+, S2-, Cl- (c) Au+, Au3+
Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons.
(a) Sr2+ > Ca2+ > Mg2+
(b) S2- > Cl- > K+
These are members of the same Group (2A) and therefore decrease in size going up the group.
These ions are isoelectronic; S2- has the smallest Zeff and
therefore is the largest while K+ is a cation with a large Zeff and is the smallest.
(c) Au+ > Au3+ The higher the positive charge, the smaller the ion.