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Atomic Structure. 26 Mg. 25 Mg. 24 Mg. 12. 12. 12. Protons 12 12 12 Neutrons 12 13 14 Electrons 12 12 12. ATOMIC NUMBER (Z) = number of protons in an atom .

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slide1

Atomic Structure

26Mg

25Mg

24Mg

12

12

12

Protons 12 12 12

Neutrons 12 13 14

Electrons 12 12 12

ATOMIC NUMBER (Z)

= number of protons in an atom

ISOTOPES = atoms of same element having the same number of protons but different numbers of neutrons.

MASS NUMBER (A)

= sum of protons and neutrons

Isotopes have the same

chemical properties

as they have the same

number of electrons

N = A-z

Deflection

Calculate Ar

Relative Atomic Mass (Ar)

The average mass of an atom

compared to 1/12 of the mass

of an atom of carbon -12

(isotope mass x % Abundance)

100

Ar = (20 x 91) + (22 x 9)

100

= 20.2

Relative Isotopic Mass (Ar)

The average mass of an isotope

compared to 1/12 of the mass

of an atom of carbon -12

Acceleration

91%

Watch

out

for SF

Detection

Relative formula mass(Mr)

Sum of individual Ar of all atoms in

one formula unit e.g.H2SO4

= 2(1.0)+1(32.1) + 4(16.0) = 98.1

Ionisation

9%

m/z

20 22

slide2

Empirical Formula

n = mol

M = Mass

Mr= molar mass

aAbB

NaOH + Cl2 NaCl + NaClO + H2O

Na= 1 2 x2 on Left

2 NaOH + Cl2 NaCl + NaClO + H2O

Cl= 2 2 balanced

2 NaOH + Cl2 NaCl + NaClO + H2O

H = 2 2 balanced

2 NaOH + Cl2 NaCl + NaClO + H2O

V

O = 2 2 balanced

n

24

2 NaOH + Cl2 NaCl + NaClO + H2O

n

C

V

p

n

NA

EF = simplest whole number ratio

of atoms in one molecule

MF = Actual number of each type

of atom in one molecule

m

Use a moles triangle to

1

Mr

n

Calculate moles of A = nA

Use Balanced Equation

Mole ratio to find moles of B

A compound 34.33% Na, 17.91% C,

47.76% O, has a Mr =134 -Find EF and MF

n = mol

C = concentration

V = Volume (dm3)

2

b x nA = nB

a

Na C O

34.33 17.91 47.76

34.3317.9147.76

23.0 12.0 16.0

=1.493 = 1.493 = 2.985

1.4931.4932.985

1.493 1.493 1.493

= 1 =1 = 2

EF = NaCO2

EF mass = 23+12+32 = 67

%

Use a moles triangle

 by Ar

 by

smallest

Calculate required

quantity of B

V, C, m, p etc..

3

n = mol

V = gas Volume (dm3)

24= Volume of 1 mol

(dm3)

V is in dm3

dm3 to cm3 x by 1000

cm3 to dm3 by 1000

n = mol

p = particles

NA =6.022 x1023

EF x N = MF

N = Mr

EF mass

N = 134 = 2

67

Moles & Equations

MF = Na2C2O4

Balanced Equations

H2SO4 -Sulfuric acid,

NH3 -Ammonia

HNO3 -Nitric acid

NaOH –sodium hydroxide

Non metal elements

Are X2 e.g. O2H2 Cl2, N2

1-Balance METAL

2-Balance NON metal

3-Balance H and O last

which ever of O and H is in

fewest substances first)

slide3

IE

Ionisation Energy:

The energy needed to remove 1 mole of electrons

from 1 mole of gaseous atoms forming 1 mole of

gaseous positive ions.

Successive Ionisation Energies for Ca

First shell 1s2

Closest to nucleus

Most difficult to

remove

2

Second shell 2s2 2p6

First IE = M (g)  M+(g) + e-

Second IE = M+(g) M2+(g) + e-

IE1

8

Decreases down group

Third shell 3s2 3p6

Group 2 xs2

full s sub-shell

8

Group 3 xs2 xp1

full s sub-shell shields outer p1 further from Nucleus

“big jump” in IE

Shows when a new

Inner shell is started

Increases across period

2

Group 5 xs2 xp3

half Full p sub-shell

no spin pairing

successive removal of e-

Outer shell 4s2

Furthest from nucleus

Most shielding

Easiest to remove

Each successive IE is greater

as there are more protons to

hold fewer electrons.

0

1

2

3

4

5

6

7

0

1

Group 6 xs2 xp4

spin pairing causes repulsion

between electrons –lowers IE

Group Number

Increases across period:

Increased nuclear attraction

Electron shielding –says the same

Atomic radius decreases

Decreases down group:

Increased nuclear attraction

outweighed by electron shielding

Atomic radius increases

Electron

Structure

Energy/distance from Nucleus

Big jump after 5th IE shows where electron

Is removed from a full inner shell,so elements in group 5)

Electrons only pair up

If they can not

occupy orbital singly

4p

Electron Pair

with

Opposite

spin

Each orbital

can hold two e-

3d

4s

4s filled

before 3d

4s emptied

before 3d

s

p

3p

3s

2p

s

p

26Fe atom

26 electrons

2s

d

= 1s22s22p63s23p64s23d6

Order of Filling

Loss of outer

shell 4s2 first

1s22s22p63s23p64s23d104p6

Fe ions

Highest energy Occupied orbital

Identifies the block

1s

Fe2+ ion = 1s22s22p63s23p6 3d6

Fe3+ ion = 1s22s22p63s23p63d5

Other electrons then lost

from inner 3d

slide4

Acids, Bases, and Salts

Acid + Carbonate Salt + water + CO2

2HCl + MgCO3  MgCl2 + H2O + CO2

Effervescence

Solid dissolves

Acid = Proton Donor

Base = Proton Acceptor

Acid + MetalSalt + Hydrogen

HCl + Mg MgCl2 + H2 (g)

Effervescence

Metal dissolves

Redox

Acids

H2SO4 = Sulfuric acid

HCl = Hydrochloric acid

HNO3 = Nitric acid

a Salt

e.g. MgCl2

Acid + Base Salt + water

HCl + MgO  MgCl2+ H2O

Salt =When the H+ of

an acid is replaced by

a metal or ammonium

ion NH4+

Acid + Alkali Salt + water

2HCl + Mg(OH)2  MgCl2 + 2H2O

Bases

NaOH = sodium hydroxide

KOH = potassium hydroxide

NH3 = Ammonia

e.g. H2SO4 + 2NH3 (NH4)2SO4

e.g 2HNO3 +Ca(OH)2 Ca(NO3)2+ 2H2O

A salt consists of a Positively charged Cation–from a metal, or Ammonium ion

And a Negatively charged Anion –which is the remains of the acid

When 4.76g of hydrated cobalt chloride

were heated, they gave 2.60g

of anhydrous cobalt chloride (CoCl2).

Cation

Salt

Anion

Acid

Mg

Mg SO4

H2 SO4

Mg2+ + SO42-

Magnesium sulfate

Sulfuric acid

Magnesium ion Sulfate (VI) ion

NH3

1) Mass of water lost = 4.76-2.60 = 2.16 g

NH4 NO3

HNO3

NH4 + + NO3 -

CoCl2 .XH2O (s)  CoCl2 (s) + X H2O (g)

Ammonium nitrate

Nitric acid

Ammonium ion nitrate(V) ion

4.76 g 2.60g 2.16 g

An anhydroussalt

MSO4

Has lost its water of

Crystallisation

2) Moles of substances (= m/Mr)

CuSO4.5H2O (s) CuSO4 (s) + 5H2O (g)

2.60

129.9

2.16

18

Hydrated salt

–blue crystals

Anhydrous salt

-white powder

= 0.02 = 0.12

A Hydrated salt

MSO4.X H2O

Has water of

crystallisation

3) Divide by smallest gives whole number ratio

X H2O = Water of Crystallisation

A number of Water molecules bonded

in the crystal Shown as a dot formula

CuSO4 .5H2O

MgSO4 .7H2O

Na2CO3. 10 H2O

= 0.02 = 0.12

0.02 0.02

= 1 = 6

CuSH10O9gives a dot formula of CuSO4. 5H2O

CoCl2 .6H2O

slide5

0 +1-2 +1-2+1 0

O.N. change

Na + H2O  NaOH + H2

Oxidation

+

0

-

Na: 0 to +1 = Oxidised

H : +1 to 0 = Reduced

0 +1-2 +1-1 +1+1-2

increases

Cl2 + H2O  HCl + HClO

Cl 0 to +1 = Oxidation

Cl 0 to –1 = Reduction

decreases

Disproportionation

= Oxidation and

reduction of the

same species

Reduction

OIL

RIG

Oxidation = Loss of electrons

Reduction = Gain of electrons

Redox

Oxidation number(state)

The number of electrons that

have to be gained or lost for

an atom to acquire noble gas

configuration.

e.g. Cl =(Ne) 3s23p5 has to:

gain 1 (-1) or lose 7 (+7)

Other states are possible

in between the extremes

  • Oxidation Number of S in H2SO4
  • element, Ion,or Compound =compound so Sum of O.N = 0
  • H+H+S+O+O+O+O =0
  • Using rules +1+1+S-2-2-2-2 = 0
  • As a sum +2+S-8 = 0
  • S = +6

Oxidising Agents

gain electrons and

get Reduced

e.g. Cl2 , H2SO4

O2 MnO4-, HNO3

Reducing Agents

lose electrons and

get oxidised

e.g. Na, H2, Fe2+

CO, C,

0 +1 –1 +2-1 0

Mg + 2HCl  MgCl2 + H2

Mg: 0 to +2 = Oxidation

H : +1 to 0 = Reduction

Oxidation number in names shown as a Roman Numeral

Iron(III) = Fe+3, Nitrate(V) = NO3-,

Chlorate(I) =ClO- sulphur(IV) oxide = SO2

slide6

Ionic

Intermolecular Forces

Covalent bonds only

:

:

H – O:

H – O:

H

H

+

-

.

.

.

.

.

.

.

x

x

x

x

x

x

x

+

.

x

H

.

.

.

x

x

H

N

.

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

2+

H

Group 1 or 2

Dot and cross

Diagrams

Show

Outer electron

and charges

brackets

Electrostatic attraction

between oppositely

charged ions

Hydrogen Bonds

Van-der Waals forces

+

-

-

+

:

2+

2-

Conduct electricity

Whenmelted or

Dissolved as ions

free to move

Strongest IMF

A permanent dipole force

found in compounds that have

F-H, O-H, or N-H bonds

Mg

O

Ions form by transfer of

electrons from metal to

Non metal

:

e.g. NaCl, MgO

  • Uneven electron distribution
  • causes a temporary dipole.
  • This induces a dipole on
  • adjacent molecule
  • Temporary dipoles attract

+

-

Covalent

A Shared pair

of electrons

Dot and cross

diagrams

show outer

electrons only

:

H-bond

..

..

..

Lone Pair

:

2 molecules

Lone Pairs

Dipoles

Labelled bond

  • The only IMF in non- polar
  • molecules e.g. I2, Cl2, F2, CH4
  • In compounds with
  • C-C, C-H, Cl-Cl, F-F, I-I bonds
  • stronger in molecules with more
  • electrons

O C O

H O

Between non-metals

electrons contribute

to outer shell of

both atoms

x

..

..

+

.

.

Bond Pair

x

Dative Covalent (coordinate)

H

Attraction between + H of one

molecule and a lone pair from

a neighbouring electronegative

atom, O, or N or F

+

Molecules

Are made from

Atoms covalently

Bonded together

Ammonium ion

NH3 +H+  NH4+

Permanent Dipole Forces

H

Covalent bonds have permanent dipoles due to differences in Electronegativity

Both bond electrons

from one atom (N)

Bp

Bonding

Actual Bp

x

x

H2O

C-Cl, C-N, C-O,H-Cl, H-Br, C=O

Metallic

H2Te

+

-

+

-

x

H2Se

Bond strength

increases

As electron density

increases

H – Cl

H – Cl

x

Positive ions attracted to sea of

delocalised

electrons

H2S

x

The attraction between the + dipole of one bond to the - dipole of another

Expected Bp

Hydrides

Of Group 6

Bond strength decreases

As electron density decreases

Anomalous properties of water

due to H-Bonding

O=C=O

In Symmetrical

moleculesdipoles

cancel so overall molecule is

Non-polar e.g. CO2, BF3, CCl4

Conduct electricity

due to mobile

delocalised electrons

1) Water has a higher Bp than

expected

2) Ice is less dense than water

H-bonds are longer so

water molecules further apart

-

High Mp/Bp

Strong bonds

Malleable and Ductile

O

In unsymmetricalmolecules

dipoles do not cancel

e.g. H2O, NCl3, CHCl3

+

+

H

H

slide7

Shapes of Molecules & Ions

N

H

104.5o

N

xx

xx

H

  • Shape is determined by the number of electron pairs
  • or groups around the centre atom of the molecule
  • Electron pairs repel each other as far away as possible
  • to minimise repulsion between them.

In multiple bonds

the shape is

determined by the

number of groups

of electrons that are

bonded to the

Centre atom

CO2 = 2 –Linear

NO3- = 3 –Trigonal plane

MnO4- = 4 -Tetrahedral

Effect of Lone Pairs:

Lone Pairs repel Bond Pairs more strongly

and so bond angles are decreased by

2.5o for each Lone Pair

Delete two bonds

and replace with

Lone Pairs

In ions with a positive charge e.g. PCl4+ , NH4+ subtract one

Electron from the total before calculating number of

Electron Pairs

slide9

Uses of Group 2

compounds

Thermal decomposition

of carbonate

Mg

Mg

Group Properties

(Ne) 3s2

Atomic Radius

Melting

Point

Heat

Medical:

Mg(OH)2 “milk of Magnesia”

Ant acid indigestion remedy

-neutralises stomach acidity

MCO3 MO + CO2

Ca

Ca

Stability of carbonate increases

down group

Reactivity

(Ar) 4s2

Ionisation

energy

Sr

Group 2 Oxide + Acid

CaSO4 –Gypsum,

“Plaster of Paris”

Sr

MO + 2 HNO3  M(NO3)2 + H2O

(Kr) 5s2

Ba

BaSO4 “Barium meal”

Before X-Ray

Observation: Solid Dissolves

–No effervescence

Metallic bonds are

weaker as sea of

delocalised electrons

more diffuse so

less energy needed

to melt metal

Increased nuclear

attraction outweighed

by more electron shielding.

Outer electrons are further away and easier to loose

Ba

(Xe) 6s2

Acidbase reaction forms a SALT

MO is a base (proton acceptor)

HNO3 is a proton donor (acid)

Agricultural:

Ca(OH)2 “slaked Lime”

Use:

Neutralise soil acidity

Excessive use makes soil

Too alkaline

Group 2

  • s-block
  • metals
  • Have s2 outer
  • shell electrons
  • All form M2+
  • ions
  • Oxidation state
  • +2
  • in compounds
  • Alkaline earth
  • metals

Lime Chemistry

Metal + acid

Metal + H2O

Limestone

M + 2 HCl  MCl2 + H2 (g)

CaCO3

Heat

M + H2O  M(OH)2 + H2

Lime

CaCO3  CaO + CO2

Observations

CaO

Observations

H2O

  • Effervescenceof H2(g)
  • Metal dissolves
  • Effervescence of H2(g)
  • Metal dissolves

Slaked lime

CaO + H2O Ca(OH)2

Ca(OH)2

Mg only reacts with steam

2Mg + 2H2O 2MgO + H2

Metal + O2

CO2

Metal + O2

“milky”

Test for CO2

Ca(OH)2 + CO2 CaCO3 + H2O

2 M + O2 2 MO

Burns with

Bright light

CaCO3

M(OH)2more soluble

Down the group

pH increases

as concentration of OH-

increases

XS CO2

Oxide is soluble in water

(more down group)

CaCO3 + CO2 +H2O  Ca(HCO3)2

Ca(HCO3)2

MO + H2O  M(OH)2

pH of solution =12

Calcium hydrogencarbonate

slide10

Reactions and Uses of Cl2

Physical Properties Trends

-1

+1

0

Cl2 + H2O  HCl + HClO

Cl2 used in drinking water– it kills germsbut is toxic

F2

(g)

Electronegativity

Mp /Bp

Disproportionation

Cl2

Oxidising power

Atomic Radius

Cl is oxidised from 0 to +1 and reduced from 0 to -1

(g)

Reactivity

Cl2 reacts with cold dilute NaOH

Disproportionation

Br2

0

-1

+1

(l)

Cl2 + 2 NaOH  NaCl + NaClO + H2O

Decreases

Down Group

Increases

Down Group

I2

NaClO= Sodium chlorate(I) =Bleach

(s)

Cl is oxidised from 0 to +1 and reduced from 0 to -1

This solution will turn blue litmusred (HCl) then bleaches white (NaClO)

Bigger atoms with more

electrons have more electron shielding.

stronger Van der Waals

forces so more energy needed to break bonds

Smallest atom with least

electron shielding and

most effective nuclear

charge attracts

electrons most strongly

Cl2 reacts with hot concentrated NaOH

Disproportionation

0

-1

+5

Cl2 + 6 NaOH  5NaCl + NaClO3 + 3H2O

Group 7

Cl is oxidised from 0 to +5 and reduced from 0 to -1

Displacement Reactions

Tests for Halide ions with Ag+

A higher halogen (more oxidising) will

oxidise a lower halide ion. The Oxidising

Halogen gets reduced.

red

0

-1

Dilute nitric acid added to stop other silver compounds from

precipitating

Silver Nitrate (AgNO3)solution is added

X2 + 2Y- 2X- + Y2

-1

0

ox

Ionic equation

observation

Non-polar solvent

e.g. hexane

Solubility

In ammonia

Ionic equation

Cl2 + 2Br- 2Cl- + Br2

Ag+ + F- AgF (aq)

F-

Colourless solution

becomes

yellow solution

Colourless solution

orange

(s)

Soluble

in dilute NH3

Cl-

Ag+ + Cl- AgCl (s)

Cl2 + 2I- 2Cl- + I2

White ppte

Colourless solution

becomes

Brown solution

(s)

Soluble in

concentrated NH3

Br-

purple

Ag+ + Br- AgBr (s)

Cream ppte

Br2 + 2I- 2Br- + I2

Colourless solution

becomes

Brown solution

insoluble in

concentrated NH3

(s)

I-

Ag+ + I- AgI (s)

purple

Yellow ppte

slide11

Periodicity

  • Patterns of physical & chemical behaviour repeat from period to period
  • Physical properties show gradual changes depending on the structure and bonding, and particularly
  • NP = Number of Protons (Nuclear attraction)
  • ES = Electron Shielding due to full inner electron shells.
  • AR = Atomic Radius or size of the atom- proximity of nucleus to outer shell.

Atomic Radius

First Ionisation Energy

Electronegativity

0.25

3.5

1600

3

1400

Melting Point

0.2

-1

1200

2.5

/ kJ mole

1000

0.15

1800

2

Atomic Radius / nm

800

Electronegativity Index

1600

I

1

1.5

0.1

1400

600

1

1200

400

0.05

1000

0.5

200

Melting Point / K

800

0

0

0

600

10

11

12

13

14

15

16

17

18

10

11

12

13

14

15

16

17

18

10

11

12

13

14

15

16

17

18

400

Atomic Number (Z)

Atomic Number (Z)

Atomic Number (Z)

200

Trend: Increases

NP: Increases

ES: Constant

AR: Decreases

Bonding electrons are attracted more and are closer

0

10

11

12

13

14

15

16

17

18

Atomic Number (Z)

Atomic Radius

0.25

Electronegativity

M Pt /K

Melting Points

0.2

2000

1.6

1.4

Atomic Radius /nm

0.15

1500

1.2

Electronegativity Index

1.0

0.8

1000

0.1

0.6

0.4

500

0.05

0.2

0

0.0

0

Be

Mg

Ca

Sr

Ba

Be

Mg

Sr

Ba

Ca

Be

Mg

Ca

Sr

Ba

Across Period

Number of Protons Increasing

Electron Shielding constant

Conductivity

1.2

1

0.8

0.6

Conductivity relative to Al

0.4

0.2

0

10

11

12

13

14

15

16

17

18

Atomic Number (Z)

Trend: Decreases

NP: Increases

ES: Constant

Electrons attracted more

strongly so atoms decrease

in size

Trend: Increases

NP: Increases

ES: Constant

AR: Decreases

Electrons attracted more

strongly so more energy

needed to remove outer e-

Trend: Increase- decrease

Giant lattice structures

Na-Al – Strong metallic bonds

Si –vStrong Covalent bonds

Simple Molecular structures

P –Ar -weak Van der Waals forces easily broken

Trend: Inc -Dec-None

Na-Alnumber of delocalised electrons per atom increases

Si –semi-conductor

P –Ar – None as no

Delocalised electrons

Down Group Electron Shielding Increasing outweighs Increasing Number of protons

Chemical properties

are similar for Group

elements as they have

a similar outer shell

electron arrangement

First Ionisation Energy

1000

800

I.E.1 /kJ mole-1

600

400

200

0

Be

Mg

Ca

Sr

Ba

Trend: Increases

NP: Increases

ES: Increases

ES outweighs NP

More full shells, Electrons further away attracted less strongly

Trend: Decreases

NP: Increases

ES: Increases

AR: Increases

Electrons further away attracted less strongly

so less energy needed -

Trend: Decreases

NP: Increases

ES: Increases

AR: Increases

Bonding electrons further away attracted less strongly

Trend: Decrease

AR: Increases

Metallic bonding is weaker

electron density in

delocalised sea decreases

so Less attraction for positive ions