Inorganic Chemistry

Front 1

Why is there is a small drop between Mg + Al in ionisation energy?

Back 1

Mg has its outer electrons in the 3s sub shell, whereas Al is starting to fill the 3p subshell. Al’s electron is slightly easier to remove because the 3p electrons are higher in energy.

Front 2

Why is there is a small drop between phosphorous and sulphur in ionisation energy?

Back 2

Sulphur’s outer electron is being paired up with an another electron in the same 3p orbital.

When the second electron is added to an orbital there is a slight repulsion between the two negatively charged electrons which makes the second electron easier to remove.

Front 3

Magnesium burned in air (oxygen)

Back 3

Mg burns with a bright white flame. 2Mg + O2  2MgO

MgO is a white solid with a high melting point due to its ionic bonding.

Front 4

Steam and Magnesium

Back 4

Magnesium reacts in steam to produce magnesium oxide and hydrogen. The Mg would burn with a bright white flame.

Mg (s) + H2O (g) MgO (s) + H2 (g)

Front 5

The other group 2 metals will react with cold water equations (not magnesium)

Back 5

Ca + 2 H2O (l) Ca(OH)2 (aq) + H2 (g)

Sr + 2 H2O (l) Sr(OH)2 (aq) + H2 (g)

Ba + 2 H2O (l) Ba(OH)2 (aq) + H2 (g)

Front 6

Group 2 reactions with cold water

Back 6

One would observe:

fizzing, (more vigorous down group)

the metal dissolving, (faster down group) the solution heating up (more down group)

with calcium a white precipitate appearing (less precipitate forms down group with other metals)

Hint 6

Observation

Front 7

Steps in extracting titanium

Back 7

TiO2 +2Cl2 +2CTiCl4 +2CO

Titanium is extracted by reaction with a more reactive metal (e.g. Mg).1. 2.

3.

TiO2 (solid) is converted to TiCl4 (liquid) at 900C:

The TiCl4 is purified by fractional distillation in an argon atmosphere.

The Ti is extracted by Mg in an argon atmosphere at 500C

Front 8

Group 2 hydroxides solubility

Back 8

Group II hydroxides become more soluble down the group.

All Group II hydroxides when not soluble appear as white precipitates.

Front 9

Group II sulfates

Back 9

Group II sulfates become less soluble down the group. BaSO4 is the least soluble.

Front 10

Filtration step up

Back 10

Back (Definition)

Front 11

Reaction with Flouride/Chloride and concentrated sulfuric Acid

-observations and Equations

Back 11

NaF (s) + H2SO4(l)NaHSO4(s) + HF(g)

Observations: White steamy fumes of HF are evolved.

NaCl(s) + H2SO4(l)NaHSO4(s) + HCl(g)

Observations: White steamy fumes of HCl are evolved.

Hint 11

Equation and observation

Front 12

Iodide with concentrated sulphuric acid

Back 12

Observations:

White steamy fumes of HI are evolved. Black solid and purple fumes of Iodine are also evolved

A colourless, acidic gas SO2

A yellow solid of sulfur

H2S (Hydrogen sulfide), a gas with a bad egg smell,

NaI(s) + H2SO4(l)NaHSO4(s) + HI(g)

2HI+H2SO4I2(s) + SO2(g) + 2H2O(l)

6HI+H2SO4 3I2 +S(s)+4H2O(l)

8 HI + H2SO44 I2(s) + H2S(g) + 4 H2O(l)

Hint 12

Equation and observation

Front 13

Chloride with water

Back 13

Cl (g) + HO(l)⇌ HClO(aq) + HCl(aq

Hint 13

Equation

Front 14

Chloride Reaction with water in sunlight

Back 14

2Cl2 + 2H2O4H+ + 4Cl- + O2

The same reaction occurs to an equilibrium mixture of chlorine water when standing in sunlight. The greenish colour of chlorine water fades as the Cl2 reacts and a colourless gas (O2) is produced.

Hint 14

equation

Front 15

Reaction of chlorine with cold dilute NaOH solution

Back 15

Cl2,(and Br2, I2) in aqueous solutions will react with cold sodium hydroxide. The colour of the halogen solution will fade to colourless.

Cl2 (aq) + 2 NaOH (aq)  NaCl (aq) + NaClO (aq) + H2O (l)

The mixture of NaCl and NaClO is used as bleach and to disinfect/ kill bacteria.

Hint 15

Equation and uses of products

Front 16

Observation of period 3 reactions with oxygen

Back 16

The elements all react with oxygen to form oxides.

Sodium burns with a yellow flame to produce a white solid

Mg, Al, Si and P burn with a white flame to give white solid smoke

S burns with a blue flame to form an acidic choking gas.

Front 17

Amphoteric Oxides

Aluminium acting as acid and base

Back 17

Aluminium oxide can act as both an acid and an alkali and is therefore called amphoteric Aluminum oxide acting as a base

Al2O3 (s)+ 3H2SO4 (aq)Al2(SO4)3 (aq) + 3H2O (l)

Al2O3 + 6HCl2AlCl3 + 3H2O

Aluminum oxide acting as a acid

Al2O3 (s)+ 2NaOH (aq) + 3H2O (l)  2NaAl(OH)4 (aq)

Front 18

Si02 acting as acid

And why only acts as acid

Back 18

SiO2 has a giant covalent structure with very strong bonds. This stops SiO2 dissolving or reacting with water and weak solutions of alkali.

It will, however, react with very concentrated NaOH 2NaOH (l) + SiO2 (s) Na2SiO3 (aq) + H2O

Front 19

What is an Transition Metal

Back 19

incomplete d sub-level in atoms or ions

Front 20

Transition Metal Properties

Back 20

complex formation, •

formation of coloured ions,

•variable oxidation state

•catalytic activity

Front 21

How colours arise- Transition Metals

Back 21

Colour arises from electronic transitions from the ground state to excited states: between different d orbitals.

A portion of visible light is absorbed to promote d electrons to higher energy levels. The light that is not absorbed is transmitted to give the substance colour.

Front 22

Changing colours- Transition Metal

Back 22

Changing a ligand or changing the coordination number will alter the energy split between the d- orbitals, changing E and hence change the frequency of light absorbed.

Front 23

Spectrophotometry Method

( Colourimtery)

Back 23

Add an appropriate ligand to intensify colour

Make up solutions of known concentration

Measure absorption or transmission

Plot graph of absorption vs concentration

Measure absorption of unknown and compare

Front 24

Vanadium colours

Back 24

VO2+ Oxidation state +5 ( a yellow solution)

VO2+ Oxidation state + 4 (a blue solution)

V3+ Oxidation state + 3 (a green solution)

V2+ Oxidation state + 2 (a violet solution)

Front 25

Manganate Redox Titration

Back 25

MnO4-(aq) + 8H+ (aq) + 5Fe2+ (aq)Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)

Purple to colourless

Hint 25

Equation and observation

Front 26

Vanadium colours

(You better get vanadium)

Back 26

VO2+ Oxidation state +5 ( a yellow solution)

VO2+ Oxidation state + 4 (a blue solution)

V3+ Oxidation state + 3 (a green solution)

V2+ Oxidation state + 2 (a violet solution)

Front 27

How does Catalyst Poisoning occur?

Back 27

Catalysts can become poisoned by impurities and consequently have reduced efficiency.

Front 28

2+ ion summary

Back 28

Back (Definition)

CuCl4 is green

FeCl4 is yellow

Hint 28

Observations and Equations

Cu and Fe

Front 29

3+ ion summary

Back 29

Back (Definition)

Hint 29

Fe and Al

Observations and Equations

Front 30

Equation for Al2O3 and sodium hydroxide

Back 30

Back (Definition)

Front 31

poisoning reduces effectiveness of a heterogenous catalyst because the...

Back 31

the poison attaches to the surface and blocks active site/ surface

Front 32

reasons why spectrometer can be appropriate i

Back 32

- Rapid determination of concentrations

-Does not use up any of the reagent/does not interfere with the reaction

-easy to get many readings

-possible to measure very low concentrations

Front 33

What changes ∆E?

Back 33

-(Identity of the) metal

-Charge (on the metal) / oxidation state / charge on complex

-(Identity of the) ligands

-Co-ordination number / number of ligands

-Shape