UNIT 3

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215 Terms

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Oxidation and reduction

Oxidation is the process where electrons are lost

reduction is the process where electrons are gained

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Oxidising agent

A substance that takes electrons from another substance and so it is reduced

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Reducing agent

A substance that gives electrons to another substance and so it is oxidized

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Oxidation number

  • of an element is 0

  • in a neutral compound add to 0

  • in a charged compound adds to total charge eg. +1

  • Hydrogen is +1 (except metal hydrides = -1)

  • oxygen is -2 ( except peroxides and F2O = -1)

  • Halogens are -1

  • Group 1 are +1

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Electrochemical cell diagram

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Purpose of the wire in the half cell

O-R

Allows electrons to flow from the half cell where oxidation occurs to the half cell we're reduction occurs i

ncluding a high-resistant voltmeter if we are measuring the potential difference in the cell

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Purpose of the salt bridge in the half cell

completes

Complete the circuit and allows ions to flow without the Solutions mixing

typical salt bridge is made of a gel soaked in a solution of potassium nitrate

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The half cell of a metal/ metal ions

we would have a piece of metal as the electrode with a solution containing 1mol dm-3 solution of metal ions

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A half cell with a gas in contact with a solution of non-metal ions with an inert metal electrode

Non-metals are not conductors so we must use an inert platinum electrode to allow electrons to flow in or out of the half cell

typically used for a hydrogen electrode or oxygen half cells

the gas is bubbled over the inert electrode which is dipping into a solution of ions the changes do not cause any apparent colour change

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Half cell with a solution containing ions of a metal into different oxidation states, using an inert metal electrode

No conductor in the system so an inert platinum electrode is used to allow electrons to flow in or out of the half cell. Typically used for transition metals where the metal may have several oxidations States eg. Fe2+/ Fe3+ and Mn2+/MnO4-

both of these half cells cause color changes when oxidation or reduction occurs

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Representing half cells

Mg(s) | Mg2+ (aq) - this is the metal/ metal ion half cell

Pt (s) | H2(g) | H+ (aq) - gas/solution half cell containing in it platinum electrode

Pt(s)| Mn2+(aq), MnO4- (aq) - mixed solution half cell containing inert platinum electrode

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Standard electrode potential

The ability of a half cell to gain or lose electrons is measured using standard electrod potential E©

measured in volts

species that are easier to reduce have a positive e-value

species that are easier to oxidize have a negative e-value

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Standard hydrogen electrode

Consists of a platinum electrode dipped in 1.0 mol dm^-3 solution of H+, used as hydrochloric acid and hydrogen gas is slowly passed over the electrod at a pressure of 1 atm and a temperature of 298K

the standard electrode potential for this half cell is taken as 0.00 V

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Why is platinum foil used

Used as an electrode hydrogen isn't an electrical conductor so platinum is inert and provides electrical conduction

used when the half cell doesn't contain a metal

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To measure the standard electoral potential

writing

We must set up a half cell under standard conditions and connect it to a standard hydrogen electrode

if two half cells are combined we must use a double line between them to represent a salt Bridge and place the metals at both ends

a high resistant voltmeter is connected between the electrical conductors of both electrodes

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Using the standard electrod potentials- negative value

Reading on voltmeter gives E value

a negative signifies that the electrode potential is more negative than the potential of these standard hydrogen electrode, means electrons must be flowing along the wire from the half cell to the standard hydrogen electrode and the hydrogen electrode becomes the positive electrode

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Using the Standard electrode potential- positive value

The positive signifies that the standard hydrogen electrode has more negative potential

electrons therefore flow along the wire from the standard hydrogen electrode to the half cell and the hydrogen electrode becomes the negative electrode

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Electoral chemical series

the most reactive metals

non metals

The most reactive metals have the most negative E value the most reactive non-metals have the most positive E value

the strongest oxidizing agents are those that have the most positive standard electrode potentials the

strongest reducing agents are present in the half equations with the most negative standard electrode potentials

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Calculating the EMF of an electrochemical cell

Given by the difference between the standard electrode potentials of two half cells. Value is positive

EMF= (Rhs) - (Lhs)

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Feasibility of reactions

In general if the overall cell potential is positive the reaction taking place is spontaneous and more favorable

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Anticlockwise rule

  1. Write the two half equations with the most negative value on top

  2. by drawing anti-clockwise arrows means that in the feasible reaction the top reaction goes in reverse and the bottom goes in the direction is written

  3. balance the electrons and equation

  4. write out the balanced cell equation

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Calculating the EMF of SEP reaction

= reduction- oxidation

the stronger oxidizing agent will have a more positive value

if the value is negative the reaction is not feasible

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Le chateliers principal

State the position of equilibrium of a system changes to minimize the effect of any change in conditions

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Temperature and cell potential

If the cell reaction as written is exothermic then as the temperature increases the position of equilibrium shifts to the left, meaning the cell potential becomes more negative

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Concentration and cell potential

If the concentration of the reactions are increased then the position of equilibrium shifts to the right and the cell potential becomes more positive

if the concentration of one of the products increases an opposite effect is observed and the potential becomes more negative

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Electrode potential and pressure

A change in pressure only affects those reactions that involve the gas

the effect is the same as concentration

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Fuel cells

electrochemical

An electrochemical method of releasing energy which avoids the need to burn the fuel followed by using this heat to cause an expansion which is used to move a motor

energy is lost at each stage of the traditional process with much of the energy released from the fuel being lost as heat out of the exhaust

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Fuel cell system

passes fuel

protons diffuse thru

electrons removed form a

The fuel cell system passes the fuel over platinum metal which acts as a catalyst but also as an electrode for the electrochemical system.

Electrons are removed from hydrogen atoms at one electrode.

  • H2 → 2H+ + 2e-

the protons diffuse through a semi permeable membrane to the other electrode where they receive electrons and oxygen molecules to form water molecules

  • O2 + 4H+ + 4e- → 2H2O

overall reaction is ( 2H2 + O2 → 2H2O) and voltage produced is 1.23V

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Benefits of fuel cells

  • They are a convenient way of storing and releasing energy

  • energy efficiency is much higher than standard fuel systems

  • emissions from fuel cells are less damaging than the carbon dioxide from traditional engines

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Drawbacks of fuel cells

conversion process

  • Hydrogen fuel must be generated Elsewhere and this is likely to use fossil fuel energy sources which will cause their own carbon dioxide emissions

  • also an energy loss here as the conversion process is not 100% efficient

  • the gas is needed are difficult to store compared to liquid fuels

  • the fuel cells operate at lower temperatures so need very efficient catalysts which use expensive metals

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ion/ half equations

Elements where the oxidation state of a single atom is changing only requires electrons to balance

eg. Mg→ Mg2+ + 2e-, Fe3+ +e-→ Fe2+

compound ions require oxygen, water, hydrogen and electrons to balance

eg. MnO4- + 8H+ +5e-→ Mn2+ + 4H2O

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Combining half equations

  • If needed multiply equations so that the electrons are balanced

  • cancel out electrons on both sides add elements if match

  • can simplify if you can

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Redox titrations

involves 25cm3

Involves 25cm3 of solution measured using a volume metric pipette and placed in a conical flask. Second solution is added a little at a time from burette swirling the mixture during addition. Continued until desired color change is seen.

Many redox titrations do not need an indicator as colours of reactants frequently allow the endpoint to be seen

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Acidified Manganate (VII) ions with iron (II) ions

purple sol of

burette

Purple solution of oxidizing agent is added from the burette. When it reacts with species to be oxidized it forms Mn2+ (colourless). The Ependpoint is reached when solution goes pale pink, this is because a very small amount of purple manganate remains appearing pink when dilute.

  1. MnO4- + 8H+ +5e- → Mn2+ + 4H2O

the colour change of solution Fe2+ (pale green) is oxidised to Fe3+ (yellow) is usually difficult to see a solutions are dilute colour change associated with manganate is used for endpoint. iron ions are oxidized:

  1. Fe2+ → Fe3+ + e-

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Acidified dichromate (VI) ions with iron (II) ions

in acid

Potassium dichromate in acid solution oxidizes Fe2+ to Fe3+ with a color change from Dark orange Cr2O72- to green solution of Cr3+

  1. Cr2O72- + 14H+ +6e- → 2Cr3+ + 7H2O

iron is again oxidized

  1. Fe2+ → Fe3+ + e-

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Acid in the dichromate titration

if the ph of solutions

COLOUR change

Acid must be present for this reaction to occur usually sulfuric acid

if the pH of a solution Rises too high then dichromate ion is broken up into two chromate ions:

Cr2O72- + H2O = 2CrO42- + 2H+

dark orange. Yellow

chromium always has an oxidation state of plus six so no redox reactionchromium always has an oxidation state of +6 so no redox reaction

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Equilibrium if acid is added

Equilibrium will shift to the left and to the right if base is added

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Redox titration for copper (II) ions

Addition of colourless solution containing iodide ions to a blue solution of copper (II) ions lead to the formation of a cloudy brown solution. Cu2+ ions in solution react with iodide ions generating a brown solution of iodine and are reduced to Copper (I) in the precipitate of CuI.

  1. 2Cu2+ +4I- →2CuI + I2

iodine is titrated with Sodium thiosulfate to work out how much copper was present originally. Sodium thiosulfate is a common reducing agent.

  1. 2S2O32- → S4O62- + 2e-

  2. I2 +2e- = 2I-

Brown. Colourless

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Experiment details

adding excess iodide ions

titrate bbrown

adding a start indicator

  • Adding excess iodide ions to Cu2+ ensures all Cu2+ reacts to produce I2. Mixture formed is a cloudy Brown solution

  • titrate the brown I2 against sodium thiosulfate until mixture is straw coloured

  • adding a start indicator goes blue black and continue until colour vanishes

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Finding the molar mass of a redox titration

  • Find out the number of moles

  • workout in either 25cm3 or 250cm3

  • ratio between compounds

  • work out the other moles

  • workout the concentration then the molar mass

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Oxidation states

Elements in the P block typically have two oxidation states

the maximum oxidation state which equals the group number and a lower oxidation state which is two less

as you go down the group the trend decreases

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How does carbon display a + 4 oxidation state

An electron is promoted and excited from 2s orbitals to 2p

2s orbital is now unpaid and four orbitals are now hybridized in the same energy level

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Inert pair effect

amount of energy returned

The tendency of S2 pair of electrons in an atom to stay paid leading to a lower oxidation state

As you go down the group the amount of energy “returned” when bonds are formed decreases. This means insufficient energy is available for the promotion of S electrons and so they stay paired

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As you go down the grou

  • Atomic radius increases

  • electronegativity decreases

  • less covalent bonding

  • increase metallic character

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As you go across the group

  • Atomic radius decreases

  • electronegativity increases

  • more covalent bondingl

  • less metallic character

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As you go diagonally across the group

  • Atomic radius increases

  • electronegativity increases

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Tin (II)

Lead (IV)

  • Tin wants to be +4 oxidation state, reducing agent supplies the electrons and therefore acts as a reducing agent so it oxidized itself

  • Lead wants to be plus two oxidation state so receives/ gains electrons as an oxidizing agent so reduces itself

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Octet expansion

The ability of some atoms to use d-orbitals to have more than eight electrons in their valence shells

limits the numbers of bonds that can be formed in the first row elements

boron conform 3 covalent bonds and is electron deficient

carbon can form 4 covalant bonds

nitrogen can form 3 covalent bonds and 1 lone pair

oxygen can form 2 covalent bonds and 2 lone pairs

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Other members of the group ( 3rd period and below)

they have acess to

s orbitals promote an electron

They have access to d-orbitals that are not present in the second shell allowing them to expand their octet and every electron in the outer shell can be used to form covalent bonds

no longer a limit of eight electrons in the outer shell

the s orbitals promote an electron to the d orbitals in which singular orbitals are formed

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Metallic character

elements at the top of each group are non metals eg, B, C, N

elements at the bottom of each group are metals

change in properties lead to the charactistic zigzag line between metals and non metals

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Amphoteric

materials that react with both acids and bases

typically show the material reacting with an acid such as hydrochloric or nitric acid and sodium hydroxide

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Reactions with aluminium oxide

HCL and NaOH

Al2O3 + 6HCl → 2AlCl3 + 3H2O

Al2O3 + 2NaOH + 3H2O → 2Na[Al(OH)4]

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Reactions with Aluminium hydroxide

H+ and OH-

Al(OH)3 + 3H+ → Al3+ + 3H2O

Al(OH)3 + OH- → [Al(OH)4] -

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Reactions with lead oxide

HNO3 and NaOH

PbO + 2HNO3 → Pb(NO3)2 + H2O

PbO + 2NaOH + H2O → Na2[Pb(OH)4]

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Reactions with lead Hydroxide

H+ and OH-

Pb(OH)2 + 2H+ → Pb2+ + 2H2O

Pb(OH)2 + 2OH →[Pb(OH)4]2-

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Reactions with aluminium

OH-

ppt,

Solutions containing amphoteric metal compound form precipitates when sodium hydroxide is added. These precipitates are metal hydroxides. Since the hydroxide can react with more sodium hydroxide these precipitates will then redisolve

Al3+ + 3OH-(aq) → Al(OH)3(s)

Then: Al(OH)3 (s) + OH-(aq) → [Al(OH)4] -

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Reaction with lead- ppt

OH-

Pb2+(aq) + 2OH-(aq) → Pb(OH)2 (s)

then: Pb(OH)2(s) + 2OH- (aq) → [Pb(OH)4]2- (aq)

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Electron deficiency

An electron deficient atom is one that doesn't have a full outer shell, when elements in group 3 form compounds they commonly form three covalent bonds such as BF3, BCl3, AlCl3

To fill their outer shell, these atoms will often form coordinate bonds to gain extra electron pairs (they are electron acceptors)

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AlCl3 → Al2Cl6

in this case each electron deficient aluminium atom uses a lone pair on a chlorine atom to form a coordinate bond

the aluminium chloride dimer no longer has any electron deficient atoms as each aluminium atom has eight electrons in its outer shell

other compounds that can form coordinate bonds to remove their electron deficiency our classified as doner- acceptor compounds where one molecule donates a lone pair and the other accepts it eg. BF3 and NH3

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Boron Nitride

B-N bond Is similar to C-C bond

there are a total of 12 electrons on the two atoms the atomic radii of all the atoms are similar with carbon being almost exactly the average of the radii of boron and nitrogen, sharing a similar relationship in their electronegativities

boron nitride can have several forms which are similar to the several different forms of carbon

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Hexagonal boron nitride- graphite structure

  • Can form hexagonal sheets similar to those found in graphite

  • in this case the atoms in different layers lie directly above one another with each boron having a nitrogen atom directly above and below it

  • differs from graphite as the layers in graphite are arranged so that atoms on adjoining layers are not directly above one another

  • forces between layers are weak so boron nitride shares ability for layers to slip over one another with graphite - used as lubricant

  • each carbon is covalenty bonded to 3 others

  • IS ISOELECTRIC

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Electrical properties of boron nitride

  • No delocalized electrons present, with electrons localized as lone pairs on N atoms

  • the B-N bonds are polar due to the different electronegativities of the two atoms making BN an insulator and leads to its use in electronics as a substrate for semiconductors, microwave transparent windows.

  • Structural material for seals, electrodes

  • catalyst carriers in Fuel Cells and batteries

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Cubic boron nitride- diamond structure

  • Like Diamond cubic boron nitride is extremely hard with a high melting point as covalent bonds must be broken to break or melt the solid

  • leads to its use as a resistant coating or an industrial abrasive

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Group 4 chemistry

Has a range of metals and non-metals

as you go down the group The oxidation states go from +4 to +2 due to inert pair effect

carbon is a non-metal which forms huge range of covalent compounds whereas lead is a metal which forms ionic compounds, being the most stable

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Carbon

Co2 is the most stable oxide of carbon. Carbon monoxide is the only stable compound to contain carbon in the +2 oxidation state

CO will act as a reducing agent as it easily becomes oxidized from +2 to +4. Carbon monoxide is used as a reducing agent especially for extracting metals from the oxides

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Carbon monoxide with iron oxide and copper oxide

Fe2O3 (s) + 3CO (g) → 2Fe (s) + 3CO2 (g)

CuO (s) + CO(g) → Cu (s) + CO2 (g)

method can only be used for the oxides of the less reactive muscles the oxides of the more reactive metals are too stable and will not react eg. Anything above zinc

processes used to extract metals above sync is electrolysis

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Lead

Lead (II) oxide, PbO, is the most stable oxide

Lead (IV) oxide, PbO2, will act as an oxidizing agent as it easily becomes reduced from +4 to +2 all lead compounds are oxidizing agents

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PbO2 with HCl

PbO2 (s) + 4HCl (conc) → PbCl2 (s) + Cl2 (g) + 2H2O (l)

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Oxides of carbon- acid base properties

Carbon dioxide is a colourless gas made up of small covalent molecules

this is an acidic oxide as the oxide is soluble

in water to give the very weak acid carbonic acid

Co2 (g) + H2O = H+ (aq) + HCO3- (aq)

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Carbon dioxide reacted with alkalis

Forms a salt. All salts produced in this way are carbonates or hydrogen carbonates

CO2 (g) + 2NaOH (aq) → Na2CO3 (aq) + H2O (l)

CO2 (g) + NaOH (aq) →NaHCO3 (aq)

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Oxides with lead: acid-based properties

Lead (II) oxide is an orange solid which contains bonding which is mainly ionic

Lead (II) oxide is amphoteric

acting as a base:

PbO (s) + 2HNO3 (aq) → Pb(NO3)2 (aq) + H2O (I)

acting as an acid:

PbO(s) + 2NaOH(aq) + H2O(I) → Na2[Pb(OH)4] (aq)

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Carbon and silicon

liquids

Stable chlorides of carbon and silicon are tetrachlorides CCl4 and SiCl4

both colourless liquids containing individual covalent molecules

the molecules found in both are tetrahedral due to the 8 electrons in the valence cells

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Reactions w Water CCl4

doesnt react

doesnt combine readily

CCl4 doesn't react with water as it forms a separate layer under the water

the carbon atom cannot combine easily with water molecules. This lack of reactivity is due to the absence of available d-orbitals in the valence shell, meaning that the octet cannot be expanded to allow the water molecules to be combined with the carbon atom

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Reaction with water SiCl4

Reacts very quickly with water in a hydrolysis reaction this reaction produces fumes of hydrogen chloride gas and silicon dioxide SiO2 as a solid precipitate

the reaction becomes more vigorous down the group as the Bonds in the compound become weaker

SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl (g)

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Increase reactivity in silicon tetrachloride

complex molecule

Silicone possesses available 3d orbitals in addition to the 3s and 3p orbitals involved in bonding to the chlorine atoms

the lone pairs of the water can form co-ordinate bonds with these empty d- orbitals giving a complex molecule that can then eliminate two HCl molecules

this molecule can then eliminate two more molecules of HCL leaving silicon dioxide

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Reactions with water lead chloride

white

Lead (II) chloride is the most stable chloride of Lead. It is a white ionic solid made up of Pb2+ and Cl- ions as it is an ionic compound it doesn't react with water, but neither does it dissolve in cold water although it can be dissolved in hot water

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Reactions of Solutions of Lead (II) compounds

naoh

excess naoh

hcl

ki

  1. Solution added- NaOH, anions present- OH-

  2. initial white precipitate of Pb(OH)2 is formed

Pb2+(aq) + 2OH- (aq) → Pb(OH)2 (s)

excess NaOH, OH-

  1. white precipitate redisolves in excess hydroxide to form tetrahydroxoplumbate (II) ion

Pb(OH)2 (s) + 2OH-(aq) → [Pb(OH)4]2- (aq)

  1. HCl, Cl-, dense white precipitate of Lead chloride

Pb2+(aq) + 2Cl- (aq) → PbCl2(s)

  1. KI, I-. Dense bright yellow precipitate of Lead iodide

Pb2+ (aq) + 2I-(aq) → PbI2 ()

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Oxidizing power of halogens

value for cl and i

readily gains

Oxidising power decreases down the group

the ability to remove electrons from other species can be measured used standard electrode potentials

  • The value for chlorine is the most positive it readily gains electrons to form chlorine ions. This also shows that it is difficult to oxidize chloride ions to chlorine molecules

  • the value for iodine is the least positive. It gains electrons less readily than bromine and chlorine, also shows that it is easier to oxidise iodide ions than it is to oxidise bromide and chloride

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Reactions of concentrated sulfuric acid with sodium halides

Concentrated sulfuric acid is a strong acid and an oxidizing agent

when any sodium halide is added to concentrated sulfuric acid the hydrogen halide is formed as a steamy gas

sulfuric acid or products formed from it may oxidize the halide in the hydrogen halide to form the halogen, if the halide is relatively easy to oxidize process becomes easier lower down the group

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Sodium chloride and sulfuric acid

Produces HCL gas. Hydrochloric acid is difficult to oxidize so the sulfuric acid doesn't cause any redox reactions

NaCl(s) + H2SO4(aq) → NaHSO4(s) + HCl(g)

steamy fumes of HCL

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Sulfuric acid and sodium bromide

Produces HBr gas

NaBr(s) +H2SO4 (conc) → NaHSO4(s) + HBr(g)

sulfuric acid oxidizes some of the hbr to form Brown fumes of BR2 and SO2 gas. the Hydrobromic acid is slightly easier to oxidize and so sulfuric acid causes a redox reaction

2HBR(s) + H2SO4 (conc) → SO2(g) + Br2(g) +2H2O(l)

steamy films of HBr, orange fumes of BR2

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Sodium iodide and sulfuric acid

Initially produces HI gas

NaI(s) + H2SO4(conc)→ NaHSO4(s) + HI(g)

sulphuric acid easily oxidizes HI

2HI(s) + H2SO4 (conc) → SO2(g) + I2(s) + 2H2O (l)

steamy fumes HI, purple fumes of I2 or black/solid Brown solution, yellow solid of S

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Disproportionation reaction

One in which the same element is both oxidized and reduced forming products containing the element in two different oxidation states

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Reaction with chlorine and water

When chlorine is bubbled through water a reversible reaction

Cl2(g) + H2O(l) → HCl (aq) + HOCl(aq)

chlorine is both oxidized and reduced

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Reaction with chlorine and sodium hydroxide

Will force the equilibrium to the right

Cl2 + 2OH- → Cl- + OCl- + H2O

the ClO- ion is stable in the solution at room temperature

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Reactions with chlorine and heated concentrated sodium hydroxide

Disproponination reaction

3Cl- + 6OH- →5Cl- + ClO-+ 3H2O

the chlorate ion

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Uses of chlorine and chlorate ions

ClO-+ 2H+ + 2e- → Cl-+ H2O

ions are being reduced

chlorine is an oxidising agent: Cl2 +2e- → 2Cl

oxidising power of chlorine and of chlorate ions is used in bleachers. The oxidizing ability of CO- leads to its ability to kill bacteria as the microbes are oxidized

this is the basis of chlorination of water supplies to disinfect them

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The dblock

outer electrons

Group of elements whose outer electrons are found in the d orbitals

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Transition elements

metal that possess a partially filled d sub shell in its atom or stable ions

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Why chromium and copper exceed the rule

4s orbitals donate an electron to the 3d orbitals, to give them a filled shell. Makes more stable configurations

it is more energetically favorable to have a full 3D shell

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Why can elements form different oxidation states

energies

The energies of the 4S and 3D orbitals are very similar so the energy required to remove any of these electrons are similar. As the element form compounds, energy is released either through the formation of covalent bonds or ionic lattice. The energy needed to reach higher oxidation states and the energy released in compound formation is balanced allowing a range of oxidation states to form.

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Why can't scandium and zinc exhibit variable oxidation states

Ions must be stable to exist

zinc and scandium aren't stable enough after losing electrons to exhibit other oxidation states

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A ligand

An ion or molecule that donates a pair of electrons to a transition metal atom or ion, forming a coordination complex eg. H2O, NH3, Cl-

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Examples of complex ions

  1. [Fe(H2O)6]2+ : Pale green, octrahedral

  2. [Fe(H2O)6]3+ : yellow, octrahedral

  3. [Cu(H2O)6]2+ blue

  4. [Cr(H2O)6]3+ : dark green, octrahedral

  5. [Co(H2O)6]2+ : pink, octrahedral

  6. [CuCl4]2- : yellow or green, tetrahedral

  7. [CoCl4]2- : blue, tetrahedral

  8. [Cu(NH3)4(H2O)2]2+ : royal blue, octrahedral

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Structure of complexes: [Co(H2O)6]2+ and [Cu(H2O)6]2+

One lone pair of oxygen atoms of the water molecules are used for the bonding to the metal ion

<p>One lone pair of oxygen atoms of the water molecules are used for the bonding to the metal ion</p><img src="https://knowt-user-attachments.s3.amazonaws.com/ddd2badf-6ca3-433d-af84-cf1c5513a4e2.png" data-width="100%" data-align="center" alt=""><p></p>
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<p>Structure of [Cu(NH3)4(H2O)2]2+</p><p>Trans and cis isomers</p>

Structure of [Cu(NH3)4(H2O)2]2+

Trans and cis isomers

knowt flashcard image

addition of ammonia causes for ammonia molecules to replace water. It's contains two different ligands.

<img src="https://knowt-user-attachments.s3.amazonaws.com/d154968e-83e2-4aa7-adba-97a01c428447.jpg" data-width="100%" data-align="center" alt=""><img src="https://knowt-user-attachments.s3.amazonaws.com/e091453a-1e9e-4796-9ed3-bfb2e245bc0c.jpg" data-width="100%" data-align="center" alt="knowt flashcard image"><p>addition of ammonia causes for ammonia molecules to replace water. It's contains two different ligands.</p>
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Structure of [CuCl4]2- and [CoCl4]2-

complexes are formed when copper or cobalt ions react with concentrated hydrochloric acid which displaces the water molecules

<img src="https://knowt-user-attachments.s3.amazonaws.com/9f4bac16-64c8-4b49-9fc7-3e26a8bdef2b.jpg" data-width="100%" data-align="center" alt=""><p>complexes are formed when copper or cobalt ions react with concentrated hydrochloric acid which displaces the water molecules</p>
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Ligand exchange

when a transistion

exchanged

When a transition metal ion is exposed to a mixture of ligands they can be exchanged to form a new complex, equilibrium process.

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Ligand exchange with [Cu(H2O)6]2+

NH3

[Cu(H2O)6]2+ + 4NH3 = [Cu(NH3)4(H2O)2]2+ +4H2O

equilibrium to the right producing more [Cu(NH3)4(H2O)2]2+

addition of extra water forces the equilibrium to the left producing more [Cu(H2O)6]2+ complex.

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Ligand exchange with [Co(H2O)6]2+

Cl

[Co(H2O)6]2+ + 4Cl- = [CoCl4]2- + 6H2O

if large amounts of chloride is used eg. by adding concentrated hydrochloric acid the equilibrium shifts to the right