Review -Atomic Theory, Periodic Table, Bonding, Solution
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 4-2 = 2 n
Ion: charged: #p ( #e
A( 16
O2- oxide ion
Z( 8 8p, 8n, 10e(more e than protons)
Isotope: Element with the same #protons, same #electrons
and different #neutrons
5) Relative Atomic Mass:
Hydrogen has 3 main isotopes; 99% 1H, 0.8% 2H and 0.2% 3H. What is its relative atomic mass? (a.m.u.: atomic mass unit)
Solved by using the average abundance of each element’s isotopes multiplied by its mass number
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 carbon-11. What is the average mass?
|Isotopes |Atomic Mass |Abundance (in decimals) |
|Carbon-11 |11.011 5 sf |0.9200 |
|Carbon-14 |13.999 5 sf |0.0800 |
(0.0800 X 13.999) + (0.9200 X 11.011)=
‘1.11’992 + ‘10.13’ = “11.25”004 = 11.25 amu or g/mol
3 sf /2 dp 4 sf/2 dp 2 dp /4sf
6) Atomic mass of oxygen: 16.002 a.m.u. (atomic mass unit)
If 99% is 16O, what the abundance of its other isotope 17O?
(0.99 X 16) + (x • 17) = 16.002
17x = 16.002 – 15.84
17x = 0.162
x = 0.0095 or 0.95%
7) Ions:
Cation: positively charged, lost e- Ex: Na+ 11p, 10e
Anion: negatively charged, gained e- Ex: Cl- 17p, 18e
What is the substance formed with aluminium and sulphite?
Al+3 SO32- (charges cross over)
Al2(SO3)3 (overall charge: zero; electrically neutral)
8a) Atomic Models and Its Scientist
[pic]
b) Quantum Mechanics Model (Electron Cloud Model)
[pic]
9a) Periodic Table-Main Families
Discovered by Russian Scientist: Demitri Mendeleev
Write the following families on the periodic table below:
Alkali, Alkaline-Earth, Halogen, Noble Gas
b) Periodic Table and Orbitals (Outer Electrons)
[pic]
10a) Energy Level Diagram and Orbitals for All Elements
[pic]
b) Number of Electrons per Orbital
|Orbital |Maximum # e- |
|s |2 |
|p |6 |
|d |10 |
|f |14 |
11) Electron Configuration Using Hund’s Rule
a) Hund’s Diagram: Complete this diagonal Sequence for electron configuration
Why? We are using this rule to find an easier way to remember the elements’ energy level diagram
[pic]
b) Sequence for filling in electrons in orbitals:
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p6
12) Electron Configuration
Quantum Mechanical Model of the Atom-Revised
1) An e- is both a particle and a wave.
2) This result is an e- cloud called orbital.
3) Each orbital can hold a maximum of 2 e-s.
4) An orbital is an e- cloud where the probability of finding an e-,
is the greatest.
5) There are 4 different kinds of orbitals: (refer to page 3 for their shapes)
Some people don’t fail
s p d f
1 type 3 types 5 types 7 types
sphere double complex complex
lobes lobes lobes
6) e- fill in orbitals from the lowest energy to the highest energy
7) Hund’s Rule(Diagonal Lines).
8) We use the atomic number to write the electron configuration
Example-1: Chlorine atom Cl: 17; 17e- to fill in
1s22s22p63s23p5
Example -2: Chloride ion Cl-: 17p, 18e-
1s22s22p63s23p6
This is the same electron configuration as Ar: 18 p, 18 e-
1s22s22p63s23p6
Cl- is isoelectronic to Ar
13) Electron Configuration: Full Vs Core Configuration
Core configuration: using the previous Noble Gas’ configuration
|Element/Atomic Number |Full Configuration |Core Configuration |
|Cl:atom17 =17p and 17e |1s22s22p63s23p5 |[Ne]3s23p5 |
|Cl-:ion= 17 p and 18e |1s22s22p63s23p6 |[Ne]3s23p6 or [Ar] |
14) Valence of an element: all the electrons in an atom except:
• Core (inner e-s)
• Filled d or f orbitals (d10 or f14)
• Substances that are isoelectronic to Noble Gases: 0
Complete these charts:
|Families |Alkalis |Alkaline-Earths |Halogens |Noble Gases |
|Group |I |II |VII |VIII |
|Valence e- |1 |2 |7 |8 |
( For Groups I to VIII, the valence electrons are the outermost electrons.
Complete this chart:
|Substance |Number of electrons |Full |Core |Valence e- |
| | |Configuration |Configuration | |
|P3-(15) |18 |1s22s22p63s23p6 |[Ne] 3s23p6 |0 |
|P (15) |15 |1s22s22p63s23p3 |[Ne] 3s23p3 |5 |
15) Electron Configuration: Ground Versus Excited States
| |Ground State |Excited State (Glowing) |
|Ar (18 e) |1s22s22p63s23p6 |1s22s22p63s13p64s03d1 |
|N2 ( 7 e) |1s22s22p3 |1s12s22p33s03p04s1 |
Who am I?
1s22s22p53s23p64s1: Ar in an excited state
16) Periodic Trends
In the periodic table, a family/group is: a vertical column (up and down)
a) Elements of the same family have: same properties
Why is that so?
3 main reasons explain such similarities:
• Same # outer e-s
• Same # valence e-s
• Same outer electron configuration
b) What is special about these outer electrons?
n: period, energy level, orbit
s, p, d: orbital, shape of electron cloud (refer to page 3)
|Element |Full e- Configuration |General e- Configuration |
|Li (3) |1s22s1 | ns1 |
|Na (11) |1s22s22p63s1 |Group: I |
|K (19) |1s22s22p63s23p64s1 |Charge:+1 |
|F (9) |1s22s22p5 | ns2np5 |
|Cl (17) |1s22s22p63s23p5 |Group: VII |
|Br (35) |1s22s22p63s23p64s23d104p5 |Charge: -1 |
( These outer e-s provide a general e- configuration and the group number
c) What is the general e- configuration for:
i) Alkaline-Earth Family: ns2
ii) Noble Gases Family: ns2np6
d) What is the element with 4p5? Br
4: period number p: “p block” 5: 5 electrons in p block
e) Comparing radius (size) and metallic character
.Nuclear charge (across a period) horizontal
.Shielding effect (down a group) vertical
|Properties |Across a Period ** |Down a Group |
| |(left to right) | |
|Atomic Radii |Becomes smaller |Becomes larger |
|(atomic size) |Increase in nuclear charge,more protons are|Increase in shielding effect, inner |
| |tightly pulling e-s |electrons are blocking the outer electrons.|
| |inward. |Outer e- are loosely held. |
|Metallic Character |Smaller |Larger |
|(Giving away e-) | | |
|Electronegativity |Larger |Smaller |
|(Attracting e-) | | |
i) Which atom has a larger size, Cl or Mg? Why?
Mg since it has a smaller nuclear charge. Electrons are held less tightly
Mg: 12 p, 12 e Cl: 17 p, 17 e (17 p: more inward pull than for 12 p)
( size Mg > Cl
ii) Which ion has a larger size, Mg2+ or Na+? Why?
Na+ since it has a smaller nuclear charge. Electrons are held less tightly.
Na+:11p, 10e Mg2+:12p, 10e (12 p: more inward pull than for 11 p)
( size Na+ > Mg2+
iii) Which substance is the most electronegative: F2, Ne, I2 or Al?
Answer: F2
Explain: It is the smallest substance with a strong nuclear pull and
it needs to receive an e- to be like Ne.
F2: 9 p, 9e Ne: 10 p, 10 e I2: 53 p, 53e Al: 13 p, 13 e
Most Least None (Electronegative)
F2 >>>>> I2 >>> Al > Ne
Metallic character: tendency to give electrons (like any metals)
Electronegativity: tendency to take electrons (like any non metals)
iv) Which substance has the most metallic character; Ca, V, S, F2?
Answer: Ca (Always choose Alkali or Alkaline-Earth)
Explain: Alkaline-Earths are more reactive than transition metals
Most Least (Metallic)
Ca > V >> S > F2
v) Which substance is larger ?
a) Br > F Explain: Br has more shielding, more orbits
b) Na > Mg Explain: Na has less nuclear pull
[pic]
|Properties |Metal |Non Metal |
|Electron |Lost |Gained |
|Conductor of heat and electricity |Good |Poor |
|Ion |Cation Na+ |Anion Cl- |
|Shiny |Yes |No |
|Melting Point |High |Low |
a) Which substance has the most non metallic character: F2, Ne, I2 or Al?
b) Which substance has the most metallic character: Ca, V, S, F2?
c) Which substance is most electronegative: Cl2, Al, Cu or I2?
d) Which substance is least electronegative: Cl2, Al, Cu or I2?
18) Bonds
3 types of bonds: ionic, polar covalent, (non polar) covalent
Based on: electronegativity difference between 2 atoms (refer to chart below)
Electronegativity: ability of an atom to attract a bonded e- pair
Electronegativity Chart:
[pic]
Types of Bonds
|Electronegativity Difference |Type of bonds |Electron Sharing |Diagram using |
|((EN) | | |full or partial charges |
|less than 0.2 |(non polar) |more or less | H-H |
| |covalent |equal |2.1 – 2.1 = 0 |
| 0.2 -1.7 |polar covalent |unequal | (- (+ |
| | | |O-H |
| | | |3.5 -2.1 =1.4 |
| more than 1.7 |ionic |gain and loss of electrons | Na+ Cl– |
| | | |3.0 – 0.9 = 2.1 |
(: partial charge or dipole, a fraction of a charge
Consider the following bonded atoms: NBr, MgP, CO and NaCl
a) Find their electronegativity difference and show the full or partial charge for each atom
(- (+ (+ (- (+ (-
N-Br Mg-P C-O Na+Cl-
3.0 -2.8 1.2 2.1 2.5 3.5 0.9 3.0
2.1 -1.2 3.5 -2.5 3.0 - 0.9
∆EN: 0.2 ∆EN: 0.9 ∆EN:1.0 ∆EN: 2.1
b)Which of these bonds is/are:
i) the least polar: N-Br (smallest ∆EN=0.2)
ii) the most ionic bond: NaCl (largest ∆EN= 2.1)
iii) polar covalent: 3 substances: N-Br, Mg-P, C-O
19) Drawing Lewis Structures (Electron Dot Structures)
Rules of the Game:
(i) Valence e- (total number of electrons)
For an ion: positive charge: remove e-, negative charge: add e-
(ii) Bonded e- (single bond: 2 e-, double bond: 4 e-, triple bond: 6 e-)
(iii) Remaining e- (Valence – Bonded), divided by 2: for the # of lone e pairs
(iv) Octet (8 e- around each atom)
Exceptions: H (2e-), B (6e-), Be (4e-)
(v) Tidy Up: Check that each atom has its correct number of valence e-
Write the following Lewis structures:
[pic]
20) Molarity Calculations When Mixing Two Solutions
Dilution Calculations and Ion Concentrations
Dilution Formula: C1V1 = C2V2
C1: initial concentration in mol/L or M
V1: initial volume in L
C2: final concentration in mol/L or M
V2: total volume in L
Example-1: 30.0 mL of 0.25 M K2SO4 is mixed with 70.0 mL of 0.35 M MgCl2.
There is no reaction taking place.
i) Find each ion concentration
Calculate the reactants’ final concentration by using the dilution formula:
C2 = C1V1
V2
C1 V1
[K2SO4] = 0.25 M X 30.0 mL = 0.075 M
100.0 mL ( V2: Total Volume
[MgCl2] = 0.35 M X 70.0 mL = 0.’24’5 M
100.0 mL
ii) Write the dissociation equations(similar to a decomposition rxn) and each species’ concentration:
K2SO4 ( 2K+ + SO4 2-
1 mol 2 mol 1 mol
1 M 2 M 1 M
0.075 M 0.15 M 0.075 M (same as mole ratio)
MgCl2 ( Mg2+ + 2Cl -
1 mol 1 mol 2 mol
0.’24’5 M 0.’24’5 M 0.’49’0 M
iii) Write each ion’s concentration
[K+] = 0.15 M [SO42-] = 0.075 M
[Mg2+] = 0.25 M [Cl-] = 0.49 M
Example-2: 30.0 mL of 0.100 M calcium nitrate mixed with 45.0 mL of 0.300 M aluminium nitrate. No reaction is taking place!
i) Find the chemical formulas
Using an ion box
|Cations |Anions |Reactants |
|Ca2+ |NO3- |Ca(NO3)2 |
|Al3+ | |Al(NO3)3 |
ii) Using the dilution formula, find their final concentration(C2)
C2 = C1V1
V2
[Ca(NO3)2] = 0.100 M X 30.0 mL = 0.0400 M
75.0 mL ( V2: Total Volume
[Al(NO3)3] = 0.300 M X 45.0 mL = 0.180 M
75.0 mL
iii) Write the dissociation equations
Ca(NO3)2 ( Ca2+ + 2NO3-
1 mol 1 mol 2 mol
0.0400 M 0.0400 M 0.0800 M
Al(NO3)3 ( Al3+ + 3NO3-
1 mol 1 mol 3 mol
0.180 M 0.180 M 0.540 M
iv) Write each [ion]’s:
[Ca2+]= 0.0400 M [Al3+]=0.180 M
[NO3-]= 0.0800 M + 0.540 M = 0.620 M
21) Types of Equations
Chemical Formula, Complete Ionic, Net Ionic
Example-1: Write the formula, complete ionic and net ionic equations when:
Potassium sulphate is mixed with barium hydroxide.
i) Find the chemical formula (Set up an ion box)
|Cations |Anions |Reactants |Products |
|K+ |SO4 2- |K2SO4(aq) |KOH (aq) |
|Ba2+ |OH- |Ba(OH)2(aq) |BaSO4(s) |
Using a Solubility Chart, we need to specify each substance’s state:
aq, l, g or s
We refer to the Solubility Chart below to find out if a substance is
Soluble: aqueous(aq) or Insoluble(not soluble): solid(s)
Solubility Chart:
[pic]
ii) Balanced Chemical Formula Equation:
K2SO4(aq) + Ba(OH)2(aq) ( 2KOH(aq) + BaSO4(s)
iii) Complete Ionic Equation:
Any soluble substance(aq) breaks down (dissociates) into a cation and an anion.
2K+(aq)+ SO42-(aq) + Ba2+(aq)+ 2OH-(aq) (
2K+(aq)+ 2OH-(aq) + BaSO4(s)
iv) Net Ionic Equation:
Remove any spectator ions (soluble ions appear both as reactant and product)
Spectator Ions: K+, OH-
SO42-(aq) + Ba2+(aq) ( BaSO4(s) precipitate
Example-2: When a solution of tin (II) chloride is added to a solution of silver nitrate, we notice a white precipitate.
Complete the following:
i) Find the chemical formulas (Set up an ion box)
Using the solubility chart,
record each reactant’s and product’s solubility (aq or s)
|Cation |Anion |Reactants |Products |
|Sn2+ |Cl- |SnCl2(aq) |Sn(NO3)2(aq) |
|Ag+ |NO3- |AgNO3(aq) |AgCl(s) |
ii) Write a balanced chemical formula equation
SnCl2(aq) + 2AgNO3(aq) ( Sn(NO3)2(aq) + 2AgCl(s)
iii) Write a complete ionic equation (highlight the spectator ions)
Sn2+(aq) + 2Cl-(aq) + 2Ag+(aq) + 2NO3-(aq) (
Sn2+(aq) + 2NO3-(aq) + 2AgCl(s)
iv) Write the net ionic equation
Spectator Ions: Sn2+(aq) and NO3-(aq)
2Cl-(aq) + 2Ag+(aq) ( 2AgCl(s)
Cl-(aq) + Ag+(aq) ( AgCl(s)
22) Predicting Solubility of Products (soluble or precipitate)
Using the Solubility Chart on page 15
Soluble: dissolved in water, is aqueous (aq)
Low solubility: forms a precipitate, insoluble, is a solid (s)
What is BaSO4? insoluble(s), forms a precipitate (ppt)
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