Transcript ionic bonding - Old Saybrook Public Schools

IONIC BONDING
A. CHEMICAL BONDS:
Attractive forces between atoms in a compound
or molecule
1. Formed by VALENCE ELECTRONS
B. IONIC BONDS
1. Formed by TRANSFER OF ELECTRONS
FROM ONE ATOM TO ANOTHER
2. Formed between
METALS with low ionization energies and
NONMETALS of high ionization energies
3. Best examples of ionic bonding:
between Groups 1A & 2A with Groups 6A & 7A
4. Electronegativity difference:
1.7 or greater
Ex. KF
4.0 - .8
= 3.2
Na2O
3.5 - .9 =
2.6
Which compound has greater ionic character?
KF
C. Energy Changes in the formation of an ionic
compound
1. Elements converted to the gaseous state
Na(s) + ½ Cl2(g)  Na(g) + Cl(g)
ΔH = +55 kcal
2. Electron transfer
Ionization of Na
Na  Na +1 + 1eΔH = 119 kcal
Formation of Chloride Ion
Cl + 1e  Cl -1
ΔH = - 83 kcal
3. Ions combine to form solid (Bond Formation)
Na+1 + Cl -1  Na+1 Cl -1
ΔH = - 184 kcal
LATTICE ENERGY
TOTAL ENERGY: 55+119-83-184 =
-93 Kcal
LATTICE ENERGY
Energy released when ions arrange themselves in a
crystal lattice
FORMATION OF IONIC BONDS (Key steps)
1.
2.
Electron transfer
Attraction of positive and negative ions
(Electrostatic attraction)
WHY DO ATOMS REACT?
Elements tend to react to acquire a stable electron
configuration, usually a noble gas configuration.
(OCTET RULE)
Charges on the ions formed depend upon the # of
electrons lost or gained to acquire the noble gas
configuration
IONS
t
1. Positive ions (metallic ions): CA IONS
smaller than their corresponding atoms
are _______
Why?
a. Outer energy level is removed
Na

Na+
1s22s22p63s1
1s22s22p6
b. Greater positive than negative charge draws
remaining energy levels closer to nucleus
Ions
2. Negative (nonmetallic) ions: ANIONS
are larger
______ than their corresponding atoms
Why?
Extra repulsion in the electron cloud weakens
pull of nucleus on e- cloud
F
F-1
1s22s22p5
1s22s22p6
F
F-
3. While metal atoms are usually larger than
nonmetal atoms in the same period,
POSITIVE ions are usually SMALLER than
negative ions (of the same period).
4. Periodic trends in ion size
Compare the sizes of the ions of Mg , Al , Na
Na+  Mg2+  Al3+
Of the ions of P , S , and Cl
P3-  S2-  Cl 1increases
b. Within a group, ion size ____________
moving down.
CHARGES OF IONS WITH NOBLE GAS
CONFIGURATIONS
Br:
Charge: -1
S:
Charge: -2
Sr:
Charge: +2
Cs
Y
Al
N
Attains e- configuration of Kr
Attains e- configuration of Ar
Attains e- configuration of Kr
NAMING IONIC COMPOUNDS
1. Monatomic positive ions have the name of their
metal.
2. In a binary compound, the negative ion is
named by dropping the element’s ending and
adding IDE.
NaCl = Sodium chloride
BaF2 = Barium fluoride
K3N = Potassium nitride
FORMULAS OF IONIC COMPOUNDS
COMPOUNDS ARE ELECTRICALLY NEUTRAL!
Magnesium & oxygen:
Mg2+O2- = MgO
Scandium & hydroxide:
Sc3+(OH)1- = Sc(OH)3
Cobalt(II) & bromine:
Co2+Br1- = CoBr2
Cobalt(III) & bromine:
Co3+Br1- = CoBr3
Many transition elements form ions with more
than one positive charge (oxidation state)
Except: Ag and Zn
Ex. Iron(II): Fe2+ Iron(III): Fe3+
3. Certain metals form positive ions with more
than one positive charge. Determine the
charge on the positive ion and write it in
Roman numerals, in parentheses, after the
name of the positive ion.
Iron(III) chloride
FeCl3
Iron(II) chloride
FeCl2
Cr(NO3)2 Chromium(II) nitrate
Cr(NO3)3 Chromium(III) nitrate
Polyatomic Ions
Charged species containing more than one atom
(Ex. NH4+, SO42-)
4. Compounds containing three or more
elements contain polyatomic ions. Name the
positive ion and then the polyatomic ion.
Al(NO3)3
Aluminum nitrate
NH4ClO3
Ammonium chlorate
Na2SO4
Sodium sulfate
Calcium & sulfate
Ca 2+(SO4)2- = CaSO4
Aluminum & sulfite
Al3+(SO3)2- = Al2(SO3)3
Gallium & hydrogen carbonate
Ga3+(HCO3)-1= Ga(HCO3)3
PRACTICE!
BaI2
KNO2
NiS
KClO4
CuBr
Au(ClO2)3
CoI3
Al2(CO3)2
AgNO3
Sn(SO4)2
GaCl3
(NH4)2O
Sr3(PO4)2
Ca(HCO3)2
HYDRATES
1. Ionic compounds which retain water
in their crystal lattice
Ex. CuSO4* 5 H2O
CoCl2 * 6 H2O
BaCl2 * 2 H2O
2. When heated, they decompose,
releasing water.
CuSO4* 5 H2O(s)  CuSO4(s) + 5 H2O(l)
CuSO4* 5 H2O(s)  CuSO4(s) + 5 H2O(l)
Hydrate:
CuSO4 * 5 H2O
Water of hydration:
H2O
Anhydrous salt:
CuSO4
EFFLORESCENCE
Loss of water by a hydrate
a. Endothermic process
Calculate the percent of water in BaCl2*2H2O.
Write the equation for the efflorescence of
BaCl2*2H2O.
Label: Water of hydration, hydrate, anhydrous
compound
WRITING FORMULAS FROM NAMES
Pay attention to ENDINGS!
Potassium bromide
K+1Br-1 = KBr
Aluminum nitrite
Al+3(NO2)-1 = Al(NO2)3
Mercury(II) sulfite
Hg2+(SO3)2- = HgSO3
Iron(III) hydroxide
Fe3+(OH)1- = Fe(OH)3
WRITING EQUATIONS FOR THE
FORMATION OF IONIC COMPOUNDS
Write the balanced equation for the formation of
copper(II) oxide from its elements.
1. Write the correct formula for the compound.
Cu2+O2-  CuO
2. Write the EQUATION for the formation of the
compound. Then balance.
Cu(s) + O2(g) → CuO(s)
2 Cu(s) + O2(g) → 2 CuO(s)
Write the balanced equation for the formation of
iron(III)chloride from its elements.
1. Fe3+Cl1- = FeCl3
2. Fe(s) + Cl2(g) → FeCl3(s)
3. 2 Fe(s) + 3 Cl2(g) → 2 FeCl3(s)
Oxidation Reduction Chemistry: Redox Chemistry
Oxidation and Reduction reactions always take
place simultaneously.
Loss of electrons – oxidation (Increase in
Oxidation Number)
Ex: Na ------> Na+1 + e-1
Gain of electrons - reduction ( Decrease in
Oxidation Number)
EX: Cl2 + 2 e-1 ------> 2 Cl-1
Oxidation occurs when a molecule
does any of the following:
Loses electrons
Loses hydrogen
Gains oxygen
If a molecule undergoes oxidation, it has
been oxidized and it is the reducing agent
Reduction occurs when a molecule does any of the
following:
Gains electrons
Gains hydrogen
Loses oxygen
If a molecule undergoes
reduction, it has been reduced
and it is the oxidizing agent.
Leo the Lion!
 LEO the lion says GER
 Loss of Electrons is Oxidation, Gain of Electrons
is Reduction
Example
Redox reactions involve electron transfer:
Lose e - =Oxidation
Cu (s) + 2 Ag (aq)
Cu2+ (aq) + 2 Ag(s)
Gain e - =Reduction
Oxidizing and Reducing Agents



Now the confusing part…
CuO + H2  Cu + H2O
Cu goes from +2 to 0


Cu is reduced, therefore it is called an oxidizing
agent because it causes some other substance to
be oxidized
H goes from 0 to +1

H is oxidized, therefore it is called a reducing agent
because it causes some other substance to be
reduced.
Identifying Agents in an Equation
Reduction: CuO is the
oxidizing agent

CuO + H2  Cu + H2O
Oxidation: H2 is the
reducing agent
Oxidation Numbers



A count of the electrons transferred or shared
in the formation or breaking of chemical
bonds
You must assign each element in the reaction
an oxidation number
Follow a set of rules…
Oxidation Number Rules
1.
The total of the oxidation numbers of all the
atoms in a neutral molecule, an isolated
atom, or a formula unit is 0
2.
In their compounds, the Group 1A metals all
have an oxidation number of +1, and the
Group 2A metals have an oxidation number
of 2+.
Rules Continued
3.
In its compounds, hydrogen has an oxidation number of
+1 (except in metal hydrides such as NaH, where it is
-1)
4.
In its compounds, oxygen has an oxidation number of -2
(except in peroxides such as H2O2, where it is -1)
5.
In their binary compounds with metals, Group7A
elements have an oxidation number of -1. Group 6A
elements have an oxidation number of -2, and Groups
5A elements have an oxidation number of -3.
Examples

Is the reactant oxidized or reduced?

Pb  PbO3


SnO2  SnO
KClO3 KCl
Oxidized
Reduced
Reduced
Identifying Redox Reactions





0 +3 -2
0 +3 -2
2 Al + Fe2O3  2 Fe + Al2O3
Al increases from 0 to +3, it is ______
Oxidized!
Fe decreases from +3 to 0, it is _______
Reduced!
Problems



What is the oxidation number of each
element?
I
0
Why? Its alone!
I2
Cr2O3
Cr +3 O -2
Al +3
Cl -1

AlCl3

Na2SO4
Na +1 S+6

CaH2
Ca +2 H -1
O4-2
Oxidation Numbers with Charges
(PO
PO43)31P
4O
@ 5+
3- =
5+
3@ 2- = 8113-
?

Fluorine and oxygen are highly electronegative and will attract
electrons very strongly meaning they will have a negative
oxidation number. Generally, phosphorus will be 3- oxidation
state: however, when combining with oxygen, phosphorus will
lose five electrons and take on a 5+ oxidation charge.
SYNTHESIS (COMBINATION)
One or more elements or compounds combine
to from a single product (compound).
A
B
AB
One Product
Examples:
2 Cu +
CO2 +
O2
H2O
2 CuO
H2CO3
DECOMPOSITION
A single compound is broken down into 2 or
more simpler substances
AB
A
B
One Reactant
Examples:
2 H2O2(aq)
2 KClO3(s)
2 H2O(l) + O2(g)
2 KCl(s) +3 O2(g)
SINGLE REPLACEMENT
A more reactive element replaces another
from a compound
A
BC
Examples:
Zn(s) + 2 HCl (aq)
Cl2 (g)+ 2 KI(aq)
AC
B
ZnCl2 (aq) + H2(g)
2 KCl(aq) + I2(s)
DOUBLE REPLACEMENT
Ions in a solution combine to form a product that
leaves the scene of the reaction
AB
CD
AD
CB
Example:
AgNO3 (aq) + NaCl (aq)
AgCl (s) + NaNO3 (aq)
ONLY occurs if one product is removed
from the solution (precipitate (s), gas (g)
or water forms.)
AgNO3 (aq) + NaCl (aq)
AgCl (s) + NaNO3 (aq)
Actual reaction:
Ag+1 + Cl-1
Spectator Ions:
Na+1 & NO3-1
AgCl
COMBUSTION
Element or compound reacts
with oxygen, producing light
and heat
CxHy+ O2
1
CO2 + H2O
C2H2+ 2.5 O2
* 2 C2H2 + 5 O2
2

CO2 + 1 H2O
4 CO2
+
2 H 2O
Nuclear Reactions

The composition of the nucleus is changed.
Stable Nuclei



Stable nuclei are NOT
radioactive
Stable nuclei are
elements #1-83
(#84are radioactive)
Strong nuclear forces
= attraction between
particles in nucleus
that hold it together

VERY STRONG!
P
P
• #1-20 equal number
of protons & neutrons
for stable nuclei
• #21-83 nuclei need more
& more neutrons to be
stable
• #84 radioactive (all
isotopes
Types of Radioactive Decay



1. Alpha particles
2. Beta particles
3. Gamma particles
Alpha Particles




=
α
2
4
4
2
Consists of 2 protons
& 2 neutrons
Has a +2 charge
Identical to a He-4
nucleus
Stopped by paper
226
88
Ra  α + Rn
4
2
222
86
He
Alpha decay problems
Write the nuclear equation for the alpha
decay of 231 Pa
91
231
Pa

91
4
2
α
227
89
+ Ac
Write the nuclear equation for the alpha
decay of 244 Pu
244
94
Pu
94
 α
4
2
+
240
92
U
Beta particles




0
-1
0
-1
β= e
High speed electron is
emitted out from atom
-1 charge
Stopped by heavy
clothing
Neutron changes into
a proton & an electron
I  e + Xe
131
53
0
-1
131
54
Beta Decay problems
Write a nuclear equation for the beta decay of
223
Fr
87
223
87
Fr 
0
-1
223
e + Ra
88
Write a nuclear equation for the beta decay of
50
Ti
22
50
22
Ti 
0
-1
50
23
e+ V
Gamma Radiation
0
0




High energy
Radiant energy
0 charge, 0 mass
Most penetrating

Stopped by lead or
concrete
Recap
Other Nuclear Reactions


Fission is splitting of the nucleus (atomic bomb)
Fusion is joining of nuclei (H-bomb)
Fission
Chain Reaction
Fusion