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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