8/29/2011 1 Lecture 14 Professor Hicks General Chemistry (CHE131) Acids • H + + anion H+ - anion • Ionic compounds must separate into ions to dissolve • Acids are molecular compound because they can dissolve without dissociating into ions • Weak acids have a small percentage of molecules separated into H + and an anion, the rest stay together as one particle Strong acids HCl HNO 3 H 2 SO 4 H + Cl - H + NO 3 - H + SO 4 2- hydrochloric acid nitric acid sulfuric acid Weak acids HC 2 H 3 O 2 HF H + C 2 H 3 O 2 - H + F - acetic acid hydrofluoric acid HF ~ 95% H + and F - 5% Strong acids separate 100% into H+ and anion in water HCl ~ 0 % H + and Cl - ~ 100% molecular compounds all other molecular compounds dissolve Dissolved molecules ionic compounds dissolve cations (+ ions) anions (- ions) separated ions H + + F - ~5% (separated ions) HCl → H + + Cl - ~ 100% HF ~ 95% (molecules) acids dissolve Acids are molecular compounds
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8/29/2011
1
Lecture 14Professor Hicks
General Chemistry (CHE131)
Acids• H+ + anion
H+ -anion
• Ionic compounds must separate into ions to dissolve
• Acids are molecular compound because they can
dissolve without dissociating into ions
• Weak acids have a small percentage of molecules
separated into H+ and an anion, the rest stay together
as one particle
Strong acids
HCl HNO3 H2SO4
H+ Cl- H+ NO3- H+ SO4
2-
hydrochloric acid nitric acid sulfuric acid
Weak acids
HC2H3O2 HF
H+ C2H3O2- H+ F-
acetic acid hydrofluoric acid
HF ~ 95% H+ and F- 5%
Strong acids separate 100% into H+ and anion in waterHCl ~ 0 % H+ and Cl- ~ 100%
4.29 What factors qualify a compound as a salt? Specify which of the following compounds are salts: CH4, NaF, NaOH, CaO, BaSO4, HNO3, NH3, KBr?
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4.33 Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate):
HBr(aq) + NH3(aq) →
4.33 Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate):
Ba(OH)2(aq) + H3PO4(aq) →
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4.33 Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate):
HClO4(aq) + Mg(OH)2(s) →
Solubility
• Max amount of a substance that will dissolve
• Not increased with increased stirring shaking etc.
• Educated guesses of solubility can be made based on the “like dissolves like” rule for all kinds of compounds
- many exceptions to “like dissolves like”
• Detailed rules exist to predict solubility for ionic compounds based on the identity of the ions
I will give you the solubility rules for ionic compounds for the exam
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Precipitation reactions• Only some ionic compounds dissolve in water
• If solutions are mixed that contain ions that can form an insoluble compound the compound will form and then precipitate as a solid
AgClinsoluble
AgNO3
soluble
Ag+ (aq)NO3
- (aq)
NaClsoluble
Na+ (aq)Cl- (aq)
mix
Na+ (aq)NO3
- (aq)
AgCl (s)
NaNO3 issolubleAgCl solidprecipitates
initiallycloudy
Precipitation of an ionic compound is a metathesis reaction
AgNO3 + NaCl → AgCl + NaNO3
white solid formsa precipitate
metathesis reaction = ions change partners
AgNO3 (aq) + NaCl (aq) → AgCl (s)+ NaNO3 (aq)
the solubility rules tell us if a substances state is(aq) if it is soluble (s) if it will precipitate as a solid (is insoluble)
4.19 Characterize the following compounds as soluble or insoluble in water: (a) Ca3(PO4)2, (b) Mn(OH)2, (c) AgClO3, (d) K2S
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4.21 Write ionic and net ionic equations for the following reactions:
AgNO3(aq) + Na2SO4(aq) →
4.21 Write ionic and net ionic equations for the following reactions:
BaCl2(aq) + ZnSO4(aq) →
4.21 Write ionic and net ionic equations for the following reactions:
(NH4)2CO3(aq) + CaCl2(aq) →
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4.23 Which of the following processes will likely result in a precipitation reaction?
(a) Mixing a NaNO3 solution with a CuSO4 solution.
4.23 Which of the following processes will likely result in a precipitation reaction?
(b) Mixing a BaCl2 solution with a K2SO4 solution. Write a net ionic equation for the precipitation reaction.
4.24 With reference to Table 4.2, suggest one method by which you might separate (a) K+ from Ag+, (b) Ba2+ from Pb2+, (c) from Ca2+, (d) Ba2+ from Cu2+. All cations are assumed to be in aqueous solution, and the common anion is the nitrate ion.
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Lecture 15Professor Hicks
General Chemistry (CHE131)
Molarity (M)
• Unit of concentration
molarity =moles
liter
• Conversion factor from volume of
a homogenous solution to moles
molarity
moles
literx liters
x volume
= moles
= # moles in that volume
molarity
divide
stoichiometric
number
mass
compound
moles of
formula units,
molecules,
atoms,
or ions
number
formula units,
molecules,
atoms,
or ions
multiply
molar
mass
divide
molar
mass
multiply
molar
mass
divide
molar
mass
divide
Avogadro's
number
multiply
Avogadro's
number
divide
Avogadro's
number
multiply
Avogadro's
number
equivalents
multiply by
volume litersM x V = # moles
2
Example: What is the molarity of glucose (C6H12O6) in a solution made by dissolving 100.0 grams of glucose in enough water to make 250.0 ml of solution
1) molarity has units of liters
convert mL liters 250.0 mL x10-3 liter
1.0 mL= 0.2500 L
2) Convert mass C6H12O6 to moles
100.0 grams x 1 mole
180.16 grams= 0.5551 moles
Molarity = # moles
#liters=
0.5551 mol
0.2500 L= 2.220 M glucose
Example: How many moles of NaCl are in
50.5 ml of a 0.250 M NaCl solution?
1) molarity has units of liters
convert mL liters
0.250 M NaCl x 0.0505 liters
50.5 mL x10-3 liter
1.0 mL= 0.0505 L
= 0.0126 moles NaCl
x 0.0505 liters solution = 0.0126 moles NaCl0.250 moles NaCl
liter solution
or if you write out the base units of molarity
3
How many moles of CaCl2 are present in 160.0 mL of
0.36 M MgCl2 solution?
How many grams of NaOH are present in 36.0 mL of a
2.50 M solution?
4
Calculate the volume in mL of a solution required to
provide the following:
(a) 2.14 g of sodium chloride from a 0.270 M solution
(b) 4.30 g of ethanol from a 1.50 M solution
(c) 0.85 g of acetic acid (CH3COOH) from a 0.30 M
solution.
Dilution Formula MconcVconc = MdilVdil
concentrated
stock solution
Mconc = 12.0 M HCl
1) add 10 ml
HCl stock
Vconc = 0.010 lit
water
2) add water to
500 ml markMconc x Vconc = # moles HCl
12 M x 0.010 lit
= 0.12 moles HCl
added
500 ml flask
# moles HCl = # moles HCl
taken from stock after adding water
used to prepare a dilute solution
from a concentrated stock solution
Mdil =MconcVconc
Vdil
12 x 0.010
0.50= M = 0.24 M adding water did not change # moles
HCl but it did change molarity of HCl
Vdil = 0.50 lit
Mdil = 0.24 M
added
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Serial dilution
• Homogenous solutions uniform throughout
1.00 gram
of blue compound
in 1000 mL of water
1 mL
+ 999 mL
water
1 mL
+ 999 mL
water
1 mL
contains
0.000000000100 grams
(1.00 nanogram)
of blue compound
0.00100 grams
of blue compound
in 1000 mL
compared to a scale
that is limited to
0.001 g ( 1 mg)
0.00000100 grams
of blue compound
in 1000 mL
6
Water is added to 125.0 mL of a 0.66 M NaNO2 solution
until the volume of the solution is exactly 250 mL. What
is the concentration of the final solution?
How would you prepare 50.0 mL of 0.100 M HCl from a
stock solution of 3.60 M HCl?
4.67 A 35.2-mL, 1.66 M KMnO4 solution is mixed with
16.7 mL of 0.892 M KMnO4 solution. Calculate the
concentration of the final solution.
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Titrations• Burette has solution of known molarity = titrant
• Flask has unknown solution = analyte
• Titrant added until the endpoint is reached
• At endpoint the titrant has been added in slight
excess
• Excess added reacts with an indicator that
changes color
• Calculations use the approximation
endpoint = equivalence point
• Equivalence point is the point where the
number of moles of added titrant exactly,
completely, reacts with all moles of analyte
titrant
analyte
In a titration
# moles = # moles
• For titrations of strong acids and bases
# moles H+ = # moles OH-
at the equivalence point
• This leads to
M1V1= M2V2
• Molarities of H+ and OH- must be used
• M2 is often the unknown
M2 =M1V1
V2
4.79 Calculate the volume in mL of a 1.420 M NaOH
solution required to titrate the following solutions:
1.25.00 mL of a 2.430 M HCl solution
2.25.00 mL of a 4.500 M H2SO4 solution
3.25.00 mL of a 1.500 M H3PO4 solution
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4.94 Acetic acid (CH3COOH) is an important ingredient
of vinegar. A sample of 50.0 mL of a commercial vinegar
is titrated against a 1.00 M NaOH solution. What is the
concentration (in M) of acetic acid present in the vinegar
if 5.75 mL of the base were required for the titration?
1
Lecture 16Professor Hicks
General Chemistry (CHE131)
What is energy?
• Ability to do work
• Work means moving something against a force
• Energy thought of as an imaginary liquid that gets moved from one container to another when processes occur
Types of energy• Kinetic energy- energy of motion.
KE = ½ mv2
• PE (opposite charges) = increases with separation (like gravity)
-+
• Potential energy- associated with position.
different types
PE gravity = mass x gravity x height
• Heat
2
Units of energy
James Joule
• SI unit Joule
2) 1 dietary calorie = 1 kilocalorie
(energy to warm 1 kg H2O 1 C)dietary and scientific calorie are not the same
dietary calorie written uppercase Calorie
1 Joule = 1 kg m2
s2
Other common units
1) 1 calorie = 4.18 Joules
First law of thermodynamics
kinetic
energypotential
energy
heat
Rudolf Clausius
applied when energy changes forms
energy like a liquid
energy cannot be created or destroyed
When things go downhill
A 10 kg bowling ball at a height of 100 m has potential energy
due to gravity
Potential energy = mgh = 10 kg x 9.8 m/s2 x10 m = 9800 J
As it falls it speeds up converting
potential energy into kinetic energy
Right before it hits ground all the potential
energy has become kinetic energy = 9800 J
After it hits the ground it is not moving so kinetic energy
reaction products are trapped in the lowest energy
state after heat is released, like the bowling ball
Second law:
Energy and matter spread out
Fuels
heat
CO2 (g)
Heat and matter spread out
prevents products from turning back into reactants!!!
System and Surroundings
• System – region of interest
• Surroundings – everything else
system surroundings
heat = q work = w
energy
Energy can be either…
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Internal Energy (U)
(ALL the energy that is in there)
system surroundings
q heat released
– (negative in sign)
w work done by system
- (negative in sign)
q heat absorbed
+ (positive in sign)work done on system
+ (positive in sign)
Change in internal energy (U)
U = q + w
Work• Movement performed against a force
(resistance) requires energy
• Amount of energy=Work=force × distance
Examples of work
• Lifting an object against gravity
• Stretching a spring
• Gas forming and blowing up a balloon (PV
work)
P × V =
×V = ×Area × Length
work
Force
Area
Force
Area
Enthalpy (H)
• Heat released under constant pressure = Enthalpy change (H)
• Constant pressure – PV work can be done
- Conditions for reactions of life
- Conditions of most chemical reaction unless they are run in rigid containers
• Only reflects part of the internal energy change since it is only the heat-not the work
• The alternative to constant pressure is constant volume like the aging of wine in a glass bottle
6
system
surroundings
C6H12O6 (g) + 6O2 (g)
work heat
Enthalpy is heat released
under constant pressure
6CO2 (g) + 6H2O (g)
change in
enthalpy
H = q
change in
internal energy
E = q + w
Thermochemical equations
• Chemical equation with H or E included
6 CO2 + 6 H2O C6H12O6 + 6 O2H = 2830 kJ
• Endothermic = heat as a reactant = positive
• Exothermic = heat as a product = negative
• The heat is associated with the whole reaction
• Heat + moles of any substance
conversion factors
2830 kJ
6 moles CO2
2830 kJ
6 moles O2
1 mole C6H12O6
2830 kJetc.
Standard Conditions
• To be able to compare reactions Standard
Conditions are defined
• All substances at concentration of 1.0 M
gases at a partial pressure of 1.0 atm
• Heat released under constant P and
standard conditions is the Standard
Enthalpy Change (Ho)
7
First law and thermochemical equations
• If a process is reversed an equal amount of
energy must flow in the opposite direction
H and E for reverse process will just have the
sign reversed
• H and E depend upon the amount of material
reacted so doubling, tripling reaction doubles,
triples the H and E
• Hess’ Law states that if a series of chemical
reactions occurs the overall H or E will just be
the sum of each step’s H and E values
First law and thermochemical equations
• Example: Determine the Ho for
2CO2 (g) 2C (s) + 2O2 (g)
given that
C (s) + O2 (g) CO2 (g) Ho = - 393 kJ
Ho = 2 x 393 kJ
= 786 kJ
Hess’ law• If a process happens in steps the Ho is the
sum of the Ho for the steps
A + B C Ho = +100 kJ
C + D E Ho = -150 kJ
__________________________________
A + B + C + D C + E Ho = +100 – 150 kJ
A + B + D E Ho = – 50 kJ
overall
8
6.15 A sample of nitrogen gas expands in
volume from 1.6 L to 5.4 L at constant
temperature. Calculate the work done in joules if
the gas expands
(a) against a vacuum
(b) against a constant pressure of 0.80 atm
(c) against a constant pressure of 3.7 atm.
6.17 A gas expands and does P-V work on the
surroundings equal to 325 J. At the same time, it
absorbs 127 J of heat from the surroundings.
Calculate the change in energy of the gas.
6.19 Calculate the work done when 50.0 g of tin
dissolves in excess acid at 1.00 atm and 25°C:
Assume ideal gas behavior.
9
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Lecture 17Professor Hicks
General Chemistry (CHE131)
Heat Capacity and Specific Heat
• Heat capacity (C) - extensive property -units are Joule
C
• Properties that describe how much temperaturechanges when heat is absorbed
• Specific heat capacity (s) - intensive property-units are Joule
g*CSpecific heat capacity is sometimes
referred to as just specific heat
Heat capacity (C)
• Imaginary container that holds
the imaginary liquid heatcopper
water
heat capacitywater
Temperature
heat capacitycopper
same massof both
add 100 J add 100 J
100 J
T
T 100 J
water has a larger HC than copper so its temperature changes less when same amount of heat is added
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T x C = q (heat)(C) x (Joule) = (Joule)
(C)
Heat capacity (C)
• Conversion factor
change temperature heat
answers the question “if I have a certain change in temperature
for an object how much heat must have flowed to cause it”
changes in quantities are given symbol always calculated final – initialT = T final – T initial = Tf - Ti
old unit
conversion factor
new unit
Specific heat capacity (s)
Conversion factor
mass heat capacity
mass x specific heat = heat capacity
m x s = C
multiplying a mass by the specific heat capacity answers the question
“if I have a certain mass of a substance what is its heat capacity?”
specific heat capacity is heat capacity per gram
old unit
conversion factor new unit
(grams) x (Joule) = (Joule)(gramC) (C)
How much heat would be required to raise the temperature of 50.0 grams of water from 12.0 C to 22.0 C ? (s (water) = 4.18 J/g*C)
Given Information
Ti = 12.0 Tf = 22.0 calculate T = 22.0 – 12.0 = 10.0 C
mass = 50.0 grams water
s (water) = 4.18 J/g*C
s x mass = HC
4.18 J/g*C x 50.00 grams = 209 J/C (heat capacity)
HC x T = q
209 J/C x 10.0 C = 2.09 x103 J
(use this equation to calculate amount of heat to raise temperature of 50 g water 10 C )
(use this equation to calculate heat capacity of 50 g water)
these two equations can be combined:q = mass x specific heat x change temperature
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Calorimetrymeasuring heat flow
surroundingsand
system
qno heat
can escapeperfectly insulated
coffee cup calorimeter
calorimetry – measure T calculate heat
add room temperature water
10 C
add hot metal sample at about 100 C
heat
20 C
First law - energy cannot be created or destroyed
heat lost by metal was gained by the water
(and the cup which we neglect here)
qwater = - qmetal
qwater = mass water x s (water) x (Tf-Ti)
qmetal = mass metal x s (metal) x (Tf-Ti)?
Lab experiment
on specific heat
capacity of a metal
qmetal = -qwater
qmetal = mass metal × s (metal) × (Tf-Ti)
qwater = mass water × s (water) × (Tf-Ti)
mass metal × s (metal) × (Tf-Ti) = - mass water × s (water) × (Tf-Ti)
s (metal)
mass metal × (Tf-Ti)
= - mass water × s (water) × (Tf-Ti)
metal and water have same final temperature
metal and water have different initial temperatures
A sample of a pure substance with a mass 125 g has a heat capacity of 307.5 J/°C. What is the specific heat of this substance?
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A 36 kg piece of silver metal is heated from 0.0°C to 273°C. Calculate the heat absorbed (in kJ) by the silver.
Calculate the amount of heat liberated (in kJ) from 366 g of mercury when it cools from 100.0°C to 1.0°C.
A sheet of gold weighing 5.0 g and at a temperature of 99.9°C is placed flat on a sheet of iron weighing 25 g and at a temperature of 16.5°C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings. (Hint: The heat gained by the gold must be equal to the heat lost by the iron. The specific heats of the metals are given in Table 6.2.)
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A 44.0-g sample of an unknown metal at 199.0°C
was placed in a constant-pressure calorimeter
containing 155 g of water at 37.0°C. The final
temperature of the system was found to be 48.4°C.
Calculate the specific heat of the metal. (The heat
capacity of the calorimeter is 12.4 J/°C.)
1
Lecture 18Professor Hicks
General Chemistry (CHE131)
Wavelength ()
wavelength distance between high points
take a snapshot of any wave
this waves wavelength
is ½ the wavelength of
the wave above
Light waves
• Many types of “light” we cannot see
• All light is called electromagnetic radiation
• Speed of all “light” is c = 2.99 x 108 m/s
aka visible spectrum
2
Why do we have to call it
electromagnetic radiation?
• water waves amplitude? height
• sound waves amplitude? pressure
• electromagnetic waves amplitude?
strength of force felt by charged object
and/or a magnet
+
--
+
e-
e-
like charges repelled
opposite charges
attracted
Diffraction
opening close to size of
wavelength of the wave
When waves collide with objects
or try to pass through openings
close to the size of their
wavelength they undergo…
Diffraction
Diffraction!
if the wave happens to be light then diffraction
spreads it out magnifying the image
for visible light a pinhole
will cause some diffraction
wave
opening wave bumps into
as it tries to pass through
d’Oh!!!
3
Class demo on pinhole diffraction of light
1. take a pin and make the smallest hole you can in a
piece of aluminum foil
2. hold the foil up to the light and try to view the tip of
the pin through the hole
3. after you see the pin move the foil out of your line of
sight and try to continue viewing the pin
4. move the pinhole back and forth
5. you will observe the pin is magnified when viewed
through the pinhole
this is the result of light undergoing diffraction at the pinhole
Electrons have wave properties
• Electrons also undergo diffraction!
Water waves Electron waves
Particle/Wave Nature of Matter
• Louis de Broglie theorized that if light can have wave and particle properties all matter should exhibit wave properties.
• He demonstrated that the relationship between mass and wavelength was
=h
mv
applies to electrons, protons, atoms, bowling balls, people, etc.
• Electrons in the highest energy levels are called valence electrons
• Exact definition is different for main group elements and the rest of the periodic table
1) For main group elements all electrons in the highest n value orbitals are valence electrons
2) For transition metals all electrons in highest n value and those in the n-1 value d orbitals are valence electrons
6
Ca = 1s22s22p63s23p64s2
Valence electrons
Br = 1s22s22p63s23p64s23d104p5
5 valenceelectrons
V = 1s22s22p63s23p64s23d3
2 valenceelectrons
7 valenceelectrons
n = 4 is highest n valueBr is in main group
n = 4 is highest n valueCa is in main group
n = 4 is highest n value V is transition metalso n-1 level 3d electrons are also included
7
Electron configurations and the Periodic Table
elements in
same group
similar electron
configurations
chemical properties are determinedby the number of valence electrons!
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8
• determine the atomic # (Z) of Mg from PT
• this is the number of electrons in the atom
Mg Z = 12, so Mg has 12 electrons
• determine the highest energy occupied orbital from its position on the PT
Example: Write the electron configuration of magnesium.
3s orbitals are highest energy orbitals for Mg
you have to remember this pattern
9
1s 2s 2p 3s
Example: Write the electron configuration of Mg.
1) 3s orbitals are highest energy orbitals for Mg so write
all orbitals up to 3s
1s22s22p63s2 or abbreviated
[Ne]3s2
2) add electrons following the Pauli Exclusion and Aufbau Principles
10
Main Group Elementselectron configuration
and ion charge
• atoms form ions that will have the same electron configuration of a noble gas
• metals lose electrons so their ions have the same electron configurations as the noble gas with a lower atomic number
• non-metals gain electrons so their ions have the same electron configurations as the noble gas with a higher atomic number
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11
Electron configurations of ions
• For positive ions valence electrons are always removedfrom the
orbital of highest n value when an ion is formed
- if more than 1 orbital has same n value e.g. 2s22p1 the electron is taken from p orbitals before s orbitals.
• for negative ions electrons are added by the Aufbau
principleExample: the magnesium atom has 2 valence electrons
Mg = 1s22s22p63s2
when it forms an ion, it loses its valence electrons from its
s orbitals since they have the highest n value
Mg2+ = 1s22s22p6
12
Electron configurations ofmain group positive ions
Na = 1s22s22p63s1
Na = [Ne]3s1
Na+ = 1s22s22p6
Na+ = [Ne]
Ne = 1s22s22p6
Ne = [Ne]
or written as the abbreviated electron configurations
Na+ has same electron configuration as the noble gas Ne
same as previous noble gas
Na forms Na+ ion
remove highest n value electron
13
Electron configurations ofmain group negative ions
S2- = 1s22s22p63s23p6
orbital diagram for S orbital diagram for S2-
same as Ar
same as nearest noble gas to the right
S = 1s22s22p63s23p4
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Isoelectronic Species
• Substances with the same number of valence electrons are chemically similar
• Especially if the electron configurations are the same or similar
Examples
• Elements on the PT
- Elements in a group have the same electron configurations with higher n values similar properties
- Stable monatomic ions and noble gases, N3-, O2-, F-, and Ne all have 1s22s2p6 electron configurations
- Unstable metals and ions
- Mg+1 and Na are both [Ne]3s1 and both have similar reactivity-both react by losing 1 electron Na+ or Mg2+ achieving [Ne]
7.80 The ground-state electron configurations listed here are incorrect. Explain what mistakes have been made in each and write the correct electron configurations.
Al: 1s22s22p43s23p3
B: 1s22s22p5
F: 1s22s22p6
7.83 Write the ground-state electron configurations for the following elements: B, V, Ni, As, I, Au.
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8.15 In the periodic table, the element hydrogen is
sometimes grouped with the alkali metals (as in this
book) and sometimes with the halogens. Explain
why hydrogen can resemble both the Group 1A and
the Group 7A elements.
8.17 Group these electron configurations in pairs
that would represent similar chemical properties of
their atoms:
1s22s22p63s2
1s22s22p3
1s22s22p63s23p64s23d104p6
1s22s2
1s22s22p6
1s22s22p63s23p3
Without referring to a periodic table, write the
electron configurations of elements with these
atomic numbers: (a) 7, (b) 19, (c) 28, (d) 35
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8.24 What do we mean when we say that two ions
or an atom and an ion are isoelectronic?
8.27 Write ground-state electron configurations for
(a) Li+
(b) H−
(c) N3−
(d) F−
(e) S2−
(f) Al3+
(g) Se2−
8.29 Write ground-state electron configurations for
these transition metal ions:
(a) Sc3+
(b) Ti4+
(c) V5+
(d) Cr3+
(e) Mn2+
(f) Fe2+
(g) Fe3+
m) Au+,
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8.30 Name the ions with +3 charges that have these
electron configurations: (a) [Ar]3d3, (b) [Ar], (c)
[Kr]4d6, (d) [Xe]4f145d6.
8.31 Which of these species are isoelectronic with