chemistry 11 review group period family electronic configuration orbital long version short version...
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Chemistry 11 Review
Group
Period
Family
Electronic configuration
Orbital
Long version
Short version
Atoms
Ions: cations & anions
XAZ
Valence Atomic charge
Periodic Trends
Atomic radius vs Electronegativity vs Electron affinity
Metals
Non Metals
Metalloids
Avogadro # = 6.02 x 1023
Diagonal rule
Balancing Equations
Types of Chemical Equations (S), (D), (SD), (DD), (CC), (IC)
STP / SATP
Mole Calculations
Stoechiometry
Common Ions (+) & (-)
Lewis DiagramBonding
% composition
Nomenclature
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1 IA
2 IIA
3 IIIB
4 IVB
5 VB
6 VIB
7 VIIB
8 VIIIB
9 VIIIB
10 VIIIB
11 IB
12 IIB
13 IIIA
14 IVA
15 VA
16 VIA
17 VIIA
18 VIIIA
Group New Version Old Version
Period
1
2
3
4
5
6
7
6
7
Family
Alkali metals
Alkaline earth metals
Metals of transition
Bo
ron
Fam
ily
Carb
on
Fam
ily
Nitro
gen
Fam
ily
Oxyg
en F
amily
Halo
gen
Fam
ily
No
ble G
as Fam
ily
Orbitals
pd
f
End of the electronic configuration
ELECTRONIC CONFIGURATION
s
Diagonal rule
1s1 1s2
2s1 2s2 2p1 2p2 2p3 2p4 2p5 2p6
3s1 3s2 3p1 3p2 3p3 3p4 3p5 3p6
4s1 4s2
Lantanide series
Actinides series
3d1 3d2 3d3 3d4 3d5 3d6 3d7 3d8 3d9 3d10 4p1 4p2 4p3 4p4 4p5 4p6
5s1 5s2 4d1 4d2 4d3 4d4 4d74d5 4d6 4d8 4d9 4d10 5p1 5p2 5p3 5p4 5p5 5p6
6s1 6s2 5d1 5d2 5d3 5d4 5d75d6 5d8 5d9 5d10
6d1 6d2 6d3 6d4 6d76d6 6d8 6d9 6d10
5d5
6d57s1 7s2
6p1 6p2 6p3 6p4 6p5 6p6
7p1 7p2 7p3 7p4 7p5 7p6
4f1 4f2 4f3 4f4 4f5 4f6 4f7 4f8 4f9 4f10 4f11 4f12 4f13 4f14
5f1 5f2 5f3 5f4 5f5 5f6 5f7 5f8 5f9 5f10 5f11 5f12 5f13 5f14
Metals
Non-Metals
Metalloids
Atoms vs Ions
Atomic
number
(Z)(p+)
Mass number
(A)(p++no)
proton
p+
neutron
no
electron
e-
Mg2412
Cl3517
12 12 12
22412Mg
12 24.3u
12 24.3u 12 12 10
Cl3517
17 35.45u 17 18 17
Cat+ions (+)
Anions (-)
17 35.45u 17 18 18
XAZ
Electronic Configuration Long version vs Short version and Ions
Long Version
20Ca: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2 since it has 20 e- and 20 exp.
20Ca2+: 1s2, 2s2, 2p6, 3s2, 3p6 since it has 18 e- and not 20 e-
53I: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p5
53I-: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6
Short VersionPick the previous noble gas + …
20Ca: [Ar], 4s2
20Ca2+: [Ne], 3s2, 3p6
53I: [Kr], 5s2, 4d10, 5p5
53I-: [Kr], 5s2, 4d10, 5p6
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VALENCE ELECTRON: An electron that occupies the outermost energy level, or shell, of an atom.
ATOMIC CHARGE: The electric charge of an ion, equal to the number of electrons the atom has gained or lost.
VALENCE
CHARGE
1
1+2
2+
Most of the metal of transition have a valence of 2 which mean they would have a charge of 2+
3
3+
6
2-
5
3-
4
4+-
7
1-
2 or 8
0
Periodic TrendsAtomic radius: The distance from the centre of the nucleus to the outermost shell of an atom.
Periodic TrendsElectronegativity: A relative measure of an atom’s ability to attract the shared electrons in a chemical bond.
Periodic TrendsIonization Energy: the energy needed to remove electrons from an atom. Large atoms require low ionization energy while small atoms require high ionization energy.
This means that each human on earth has to eat 35000 eggs per second for 80 years?????? What about the # of chicken to produce that amount of eggs????
Avogadro # = 6.02 x 1023
How many eggs does each human on earth has to eat every second to add up to 1 mol in 80 years?
80 years x 365 day x 24 hours x 60 min. x 60 sec. = 2 522 880 00 s
Current earth population = 6 800 000 000 x 2 522 880 000 s = 1.72 x 1019 s
1 mole of eggs = 6.02 x 1023 eggs
6.02 x 1023 eggs = 35000 eggs / s
1.72 x 1019 s
1 year 1 day 1 hour 1 min.
Avogadro # = 6.02 x 1023
Avogadro’s Number:
1 mole of gas @ …
STP – Standard Temperature and Pressure:
V= 22.4L, 0ºC(273 K), 101.3 kPa
SATP – Standard Ambient Temperature Pressure
V= 24.8L, 25ºC(298 K), 101.3kPa
% Composition
• Percent Composition: The percentage, by mass, that each element that makes up a compound, out of 100%.
• To find the percent composition you must divide the mass of the element by the mass of the compound and multiply the result by 100.
• H2CO3 H x 2 = 2.0g/mol %H = 2/62 x 100 = 3.2%C x 1 = 12.0g/mol %C = 12/62 x 100 = 19.4%O x 3 = 48.0g/mol %O = 48/62 x 100 = 77.4%
Total = 62g/mol
1+ Charge 2+ Charge 3+ Charge 4+ Charge
Li+ Lithium Be2+ Beryllium Al3+ Aluminium Sn4+ Tin (IV)
Na+ Sodium Mg2+ Magnesium Cr3+ Chromium (III) Pb4+ Lead (IV)
K+ Potassium Ca2+ Calcium Fe3+ Iron (III)
Ag+ Silver Ba2+ Barium Au3+ Gold (III)
Cu+ Copper (I) Fe2+ Iron (II) Co3+ Cobalt (III)
NH4+ Ammonium Pb2+ Lead (II) Ni3+ Nickel (III)
H+ Hydrogen Ni2+ Nickel (II)
Zn2+ Zinc
Hg2+ Mercury (I)
Sn2+ Tin (II)
Co2+ Cobalt (II)
Cu2+ Copper (II)
Common Ions (+)
Nomenclature
1. Magnesium Chlorate
2. Mercury II Chlorite
3. Lithium Sulfate
4. Zinc acetate
5. Sodium phosphide
4. Zinc acetate
5. Sodium phosphide
3. Lithium Sulfate
4. Zinc acetate
3. Lithium Sulfate
4. Zinc acetate
Mg(ClO3)2
Hg(ClO)2
Li2SO2
Zn(CH3CH2)2
Na3P
Nomenclature Game
Dimitry Vinagradov sofware
Bonding
A covalent bond is a form of chemical bonding that is characterized by the sharing of pairs of electrons between atoms, and other covalent bonds
You tube video: Covalent Bond
An ionic bond is a type of chemical bond formed through electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a metal and a nonmetal ion.
You tube video: Ionic Bond
Lewis Symbols or Diagrams Elemental properties and reactions are determined only by electrons in the outer energy levels. Electrons in completely filled energy levels are ignored when considering properties. Simplified Bohr diagrams which only consider electrons in outer energy levels are called Lewis Symbols. A Lewis Symbol consists of the element symbol surrounded by "dots" to represent the number of electrons in the outer energy level as represented by a Bohr Diagram. The number of electrons in the outer energy level is correlated by simply reading the Group number. Lewis symbols for oxygen, fluorine, and sodium are given in the diagram on the left. Lewis Symbols for the elements of the second period. Correlate the number of dots with the group number.
Balancing EquationsA chemical equation is an expression for a chemical reaction. It is a
quantitative statement indicating the number of moles of each reactant and of each product.
Reactants Products
In chemical equations, matter must be conserved. The number of atoms of each kind on the reactant side must equal those on the product side.
__Fe + __H2O → __Fe2O3 + __H22 3 3
__P4 + __Cl2 → __PCl346 DONE!
DONE!
__Fe2O3 + __CO → __CO2 + __Fe33 2 DONE!Σ R = Σ P
Types of Chemical EquationsSUMMARY
Let E= element and C = Compound
Reactants Reaction Type Products General Equation
E1 + E2 Combination (C) C R + S RS
C Decomposition (D) E1 + E2 RS R + S
E1 + C1 Single Displacement (SD)
E2 + C2 T + RS TS + R
C1 + C2 Double Displacement (DD)
C3 + C4 RS + TURU + TS
Hydrocarbon + O2
Complete Combustion (CC)
CO2 + H2O CH4+2O2 CO2+ 2H2O
Hydrocarbon + O2
Incomplete Combustion (IC)
CO + H2O 2CH4+3O2 2CO+ 4H2O
Stoichiometry
• is a branch of chemistry that deals with the quantitative relationships that exist between the reactants and products in chemical reactions.
• In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of whole numbers.
Mole Calculations
Mass – Mass: 2H2 + O2 2H2O
How many g of O2 is needed to burn with 10g H2?
10g H2 x 1 mol H2 x 1 mol O2 x 32g O2 =
2.02g H2 2 mol H2 1 mol O2
80g O2
Mole Calculations
Mass – Mol: 2KClO3 2KCl + 3O2
With 5g of KClO3, how many moles of KCl would you get?
5g KClO3 x 1mol KClO3 x 2 mol KCl =
122.45g KClO3 2 mol KClO3
0.041mol KCl
Mole Calculations
Mass – Molecules: 2H2 + O2 2H2O
How many molecules of O2 would you get if you start with 10g H2?
10g H2 x 1 mol H2 x 1 mol O2 x 6.02 x 1023 molecule O2 =
2.02g H2 2 mol H2 1 mol O2
1.5 x 1024 molecule O2
Mole Calculations
Mass – Volume: N2 + 3H2 2NH3
With 25g N2, find the volume of NH3 at STP
25g N2 x 1 mol N2 x 2 mol NH3 x 22.4L NH₃ =
28g N2 1 mol N2 1 mol NH3
40L NH3
Limited and Excessive Reactant
Limited Reactant (LR) – The substance in the formula that is used up first.
Excessive Reactant (ER) – The substance you have more of.
NH3 + HCl NH4Cl
What mass of ammonium chloride would you get if you add 1g NH₃ and 1g HCl?
1g NH3 = 0.0588 1g HCl = 0.0274
17g NH3 36.45g HCl
1.47g NH4Cl
HCl is your limited reactant since you get a smaller ratio. Therefore you get…
0.0274mol HCl x 1mol NH4Cl x 53.5g NH4Cl =
1 mol HCl 1 mol NH4Cl
This means that if you started with 1g HCl and it gave you 1.47 g of NH4Cl you can conclude that you used 0.47g of NH3 therefore since we started with 1g of NH3 we know that we have 0.53 g extra.