Units 8&9:Atomic Theory, Periodic Table, Bonding, Solutions
Atomic Theory
1) Atomic Particles-Location?
2) Size and Mass of Atomic Particles
3) Charge of Atomic Particles
Electron
Proton
Neutron
Mass
1/1837
1
1
Charge
-1
+1
0
Location
around the nucleus nucleus
nucleus
4) Isotopic Formula for Atom and Ion:
Including #protons, #neutrons, #electrons
=
Atom: no charge, #p
#e
A 4
where A: mass number (protons + neutrons)
He
Z: atomic number (protons)
Z 2
2p, 2n, 2e
A – Z: number of neutrons
Ion: charged: #p
A 16
 #e
O2Z 8
8p, 8n, 10e(more e than protons)
5) Relative Atomic Mass:
Using the average abundance of main isotopes of an element
isotope 1H
+
isotope 2H
+
isotope 3H
(abundance X mass number) + (abundance X mass number) +(abundance X mass number)
(99.% X 1H)
+ ( 0.80 % X 2H)
+ ( 0.2 % X 3H)
(0.99 X 1)
+ (0.0080 X 2)
+ (0.002 X 3)
0.99
+ 0.016
+
0.006
2 dp
3 dp
3 dp
= 1.01 amu = 1.01 g/mol
2 dp/3 sf
a) Find the average atomic mass of:
95 Pd = 3.00%
100Pd = 50.0%
110Pd = 40.0%
115Pd = 7.00%
(95 X 0.0300) + (100 X 0.500) + (110 X 0.400) + (115 X 0.0700)
2.85
+
50.0
+
44.0
+
8.05 = 104.9
3sf/2dp
3 sf/1dp
3 sf/1dp
3sf/2dp 1dp/4sf
b) In a sample of carbon, there are 8.00% of carbon-14 and 92.00 % of carbon11. What is the average mass?
Isotopes
Atomic Mass
Carbon-11
11.011 5 sf
Carbon-14
13.999 5 sf
(0.0800 X 13.999) + (0.9200 X 11.011)=
‘1.11’992
+ ‘10.13’0
= 11.25004 = 11.25
3 sf /2 dp
4 sf/2 dp
2 dp /4sf
6) Ions:
Cation: positively charged, lost e-
Ex: Na+ 11p, 10e
Anion: negatively charged, gained eEx: Cl
17p, 18e
Ex: Substance with aluminium and sulphite
+3
2-
Al
SO3
Al2(SO3)3
7) Atomic mass of oxygen: 16.002 a.m.u.
Most abundant isotope is 16O: 99%
8) Atomic models-associate each model with its scientist
Model
Scientist
9) Periodic Table-Main Families
Families Alkalis Alkaline-Earths
Group
I or 1
II or 2
Groups I to VIII: valence electrons
Halogens
VII or 17
10) Electron Configuration
Full Vs Core Configuration (Hund’s Rule)
Element/Atomic Number
Full Configuration
Cl: 17
1s22s22p63s23p5
1s22s22p63s23p6
Cl :18
p.3
Complete this chart:
Substance Number of
electrons
318
P
P
15
Full
Configuration
1s22s22p63s23p6
1s22s22p63s23p3
****Ground Vs Excited States ***
Ground State
Ar (18 e)
1s22s22p63s23p6
Noble Gases
VIII or 18
Core Configuration
[Ne]3s23p5
[Ne]3s23p6
Core
Configuration
[Ne] 3s23p6
[Ne] 3s23p3
Valence
e0
5
Excited State (Glowing)
1s22s22p63s13p64s03d1
N2 ( 7 e)
1s22s22p3
Who am I?
1s22s22p53s23p64s1: Ar in an excited state
1s12s22p33s03p04s1
11) Comparing radius, ionization energy, metallic character, electronegativity
.Nuclear charge (across a period)
.Shielding effect (down a group)
Properties
Across a Period
Down a Group
(left to right)
Atomic Radii
Becomes smaller
Becomes larger
Increase in nuclear
Increase in shielding
charge (more protons
effect, inner electrons
are attracting more
are blocking the outer
electrons more tightly)
electrons. Outer e- are
loosely held.
Ionization Energy
Larger due to increase in Smaller due to shielding
(Removing e-)
nuclear charge
effect
Metallic Character
Smaller
Larger
(Give e-)
Electronegativity
Larger to increase in
Smaller due to shielding
(Attracting e-)
nuclear charge
effect
a) Which ion has a larger size, Mg2+ or Na+? Why?
Na+ since it has a smaller nuclear charge. Electrons are held less tightly than
the electrons from Mg2+.
b) Which substance has the lowest IE1(First Ionization Energy)?
This question is asking which one is the easiest to remove an electron.
It has to be the substance that is the largest in size
i) Br or F
Answer: Br (shielding effect)
ii) Na or Mg
Answer: Na (smaller nuclear charge)
Reason:
More shielding, easiest to remove an electron.
Less nuclear charge, easiest to remove an electron
c) Which substance has the most non metallic character; F2, Ne, I2 or Al?
F2: It is the smallest substance, there is less shielding and fluorine needs
to receive an electron.
Reason:
Metallic character: tendency to give electrons
Non metallic character: tendency to take electrons
d) Which substance has the most metallic character; Ca, V, S, F2?
Answer: Ca
12) Valence of element
Number of outershell electrons
Family
I
II
III
Valence 1
2
3
electron
Name
Alkalis Alkaline
Earths
IV
4
V
5
VI
6
VII
7
VIII
0
Halogens Noble
Gases
13) Bonding
3 types of bonds: ionic, polar covalent, non polar covalent
Based on electronegativity difference between 2 atoms (chart provided)
Electronegativity
Type of bonds
Electron Sharing
Diagram using
Difference
charges or dipoles
less than 0.2
non polar
more or less
covalent
equal
H-H
- +
0.2 -1.7
polar covalent
unequal
 
O-H
3.5 -2.1 =1.4
more than 1.7
ionic
gain and loss of
Na+ Cl–
electrons
3.0 – 0.9 = 2.1
Consider the following bonds:
- +
+ + N-Br
Mg-P
C-O
Na+Cl3.0 -2.8
1.2-2.1
2.5-3.5
0.9-3.0
∆EN : 0.2
∆EN: 0.9
∆EN:1.0
∆EN: 2.1
Which of these bonds is
a) the least polar:
N-Br
(smallest ∆EN)
b) the most ionic bond:
NaCl
(largest ∆EN)
c) polar covalent:
N-Br, Mg-P, C-O
14) Drawing Lewis Structures (Electron Dot Structure)
Valence e- (total number of electrons)
Bonded eRemaining eOctet [Exceptions: H(2), B(6), Be(4): don’t have 8 e- around them]
Tidy Up
Write the following Lewis structures:
a) CO
c) SO2
b) C2H6
d) NCl3
15) Molarity Calculations: (HIA Bkt p.28)
Dilution Calculations and Ion Concentrations
Dilution Formula: C1V1 = C2V2
C1: initial concentration in mol/L
V1: initial volume in L
C2: final concentration in mol/L
V2: total volume in L
Ex: 30.0 mL of 0.25 M K2SO4 is mixed with 70.0 mL of 0.35 M MgCl2.
Find each ion concentration
i) What are the products?
Cations
Anions
Products
+
2KCl
K
SO4
MgSO4
Mg2+
Clii) Calculate the reactants’ final concentration by using the dilution formula:
C2 = C1V1
V2
[K2SO4] = 0.25 M X 30.0 mL = 0.075 M
100.0 mL Total Volume
[MgCl2] = 0.35 M X 70.0 mL = 0.245 M
100.0 mL
iii) Write the dissociation equations and each species’ concentration:
K2SO4
 2K+ + SO4 20.075 M
0.15 M
0.075 M
2+
MgCl2
 Mg
+ 2Cl 0.245 M
0.245 M
0.490 M
iv) Write each ion ‘s concentration
[K+] = 0.15 M
[SO4 2-] = 0.075 M
[Mg2+] = 0.25 M
[Cl-] = 0.49 M
16) Equations
Formula, Complete Ionic, Net Ionic
Write the formula, complete ionic and net ionic equations when:
Potassium sulphate is mixed with barium hydroxide.
What are the products?
Cations
Anions
Products
+
2KOH (aq)
K
SO4
2+
BaSO4(s)
Ba
OHsoluble: aqueous insoluble: solid
i) Balanced Formula Equation:
K2SO4(aq) + Ba(OH)2(aq)  2KOH(aq) + BaSO4(s)
ii) Complete Ionic Equation:
Any soluble substance breaks down into cation and anion.
2K+(aq)+ SO42-(aq) + Ba2+(aq)+ 2OH-(aq) 
2K+(aq)+ 2OH-(aq) + BaSO4(s)
iii) Net Ionic Equation:
Remove spectator ions (appear both as reactant and product)
SO42-(aq) + Ba2+(aq)  BaSO4(s) precipitate
Spectator Ions: K+, OH17) Predicting solubility of products (soluble or precipitate)
Using the Solubility Chart
Soluble: dissolved in water
Low solubility: forms a precipitate, insoluble, is solid
For the equations in #19, BaSO4(s): insoluble, forms a precipitate (ppt)