F332 Revision Notes

Topics: Atom, Hydrogen, Electron Pages: 7 (1428 words) Published: January 29, 2013
OCR B (Salters) AS level Chemistry
UNIT 2 – F332
Elements from the sea
Halogens and Halides (group 7 chemistry)

* Halogen is the elemental molecule, eg/ Cl₂, Br₂
* Halide is in a compound, eg/ KBr, KCl

| Fluorine| Chlorine| Bromine| Iodine|
Appearance at room temp| Pale yellow gas| Green gas| Dark red volatile liquid| Shiny black solid – sublimes to purple gas|

Halide| Colour precipitate with silver nitrate|
KCl| White|
KBr| Cream|
KI| Pale Yellow|

Redox reactions

‘Reduction is gain of electrons, oxidation is loss of electrons’


Overall reaction = 2KBr + Cl₂ 2KCl + Br₂

Halogens get more reactive going up the group, and in redox reactions the less reactive element come out of the compound/solution, and is oxidised. Oxidising agent is chlorine because it oxidises bromine, bromine is the reducing agent.

Half equations = 2Br⁻ Br₂ + 2e⁻
Cl₂ + 2e⁻ 2Cl⁻

This shows that Chlorine is reduced because it has gained electrons, and Bromine is oxidised because it has lost electrons.

Oxidation states

Element| Oxidation number|
Fluorine | -1|
Oxygen| -2 (unless with Fluorine)|
Chlorine| -1 (unless with Fluorine or Oxygen, then +1)|
Hydrogen| +1|
A spectator ion, is used in the redox reaction, but is not reduced or oxidised itself.

Working out oxidations states is really just simple maths.

Ion| Oxidation state of chlorine| Name of ion|
ClO⁻| +1| Chlorate (I)|
ClO₂⁻| +3| Chlorate (III)|
ClO₃⁻| +5| Chlorate (V)|
ClO₄⁻| +7| Chlorate (VII)|
Hydrogen becomes negative when it’s more electronegative then what it is reacting with, because it has a higher ability to draw electrons to itself.

Electrolysis is a form of redox reaction and may come up in the exam.

Example of this could be the electrolysis of potassium chloride: 2KCl Cl₂ + 2K
2K⁺ + 2e⁻ 2K
2Cl⁻ Cl₂ + 2e⁻

Atom economy

% Atom economy = (formula mass of useful products) / (formula mass of reactants) x 100

Electrolysis of water to produce hydrogen
2H₂O 2H₂ + O₂ Mr of 2H₂O = 36 Mr of 2H₂ = 4

(4/36) x 100 = 11.1%

Permanent dipole – Permanent dipole (when an element is more electronegative) ‘When two unlike atoms are convalently bonded, the shared electrons will be more strongly attracted to the atom of greater electronegativity. Such a bond is said to be polar. A polar bond results in the unequal sharing of the electrons in the bond’

Types of intermolecular bonds
* Instantaneous - Induced dipole (IDID) < 10KJ/mol
* Permanent dipole - Permanent dipole (PDPD) 10-20KJ/mol * Hydrogen bonding 10-40KJ/mol
Instantaneous – Induced dipoles (can occur in ALL molecules) * Occurs when more of the electrons happen to be at one end than the other, so one end is slightly (delta) negative and the other slightly (delta) positive. * Can cause induced dipoles if near other molecules.

* Permanent dipoles can also induce dipoles if near other molecules. * Larger molecules have more electrons so strength of IDID is greater. * Straight chained molecules have more ‘contact points’ so have more opportunities for IDID than branched molecules. * Weakest of intermolecular forces.

Amount (g)
Moles Mr
Percentage yield

% yield = (actual yield) / (theoretical yield) x 100


100g of AgNO₃ (Mr = 169.9) reacts with something to produce Ag (Ar = 107.9). In the actual reaction 50.8g of Ag was produced. Calculate the percentage yield.

First we need to work out the theoretical yield.

Moles = 100/169.9 = 0.588 moles

Because 1 mole of AgNO₃ reacts to make 1 mole of Ag,
0.588 moles of Ag should be formed which =
(0.588 x 107.8) = 63.5g (theoretically)

Because only 50.8g was produced the % yield =
(50.8/63.5) x 100 = 80%

* Needs to be covalently...
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