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Doc Brown's Advanced Physical–Theoretical Inorganic Chemistry

REDOX REACTIONS CHEMISTRY Revision Notes Part 1

Sections 1 to 4 Definitions and describing redox reactions

The definitions of oxidation and reduction are given and explained in terms of electron loss or gain, and then the ideas applied to lots of examples from inorganic chemistry reactions. The concept of oxidation state (oxidation number) is then introduced. Simple ions are dealt with first, then, via the concept of electronegativity, how to work out oxidation state for more awkward examples is explained! The rules concerning oxidation number are given and should be learned as soon as possible! Examples of how to name inorganic compounds that require Roman numeral oxidation states in the name are also given on this page.

REDOX section–index: 1. Basic redox definitions * 2. Introducing oxidation state (with sub–index) * 3. Oxidation state rules–guidelines & inorganic examples * 4. Naming inorganic compounds * 5. Using oxidation states to describe redox changes in a given inorganic reaction equation (with sub–index) * 6. Constructing full inorganic redox equations from half–equations (with sub–index) * 7. Redox titrations * 8. Organic synthesis reductions (with summary table) * 9. Organic synthesis oxidations (with summary table) * 10. Other Organic Redox Reactions (with sub–index) * 11. Carbon's ox. state in selected organic compounds and functional group level * You are advised to study sections 1. to 6. in strict order and covers the requirements of AS–A2 * See also Equilibria Part 7 Redox Reactions for Half cell equilibria, electrode potential, standard hydrogen electrode, Simple cells and notation, Electrochemical Series, EØcell for reaction feasibility, 'batteries' and fuel cell systems etc.

 

Advanced Inorganic Chemistry Notes

Redox reactions in Inorganic Chemistry sections 1–4

The very important concepts of oxidation, reduction, oxidation state/number and their application to redox reactions are explained using a variety of examples in sections 1–4. How to combine half–cell equations to write a full ionic–redox equations is further explained in Part 2 sections 5–6. Application of redox concepts to cells will be described in Equilibria Part 7 "Redox Equilibria".

1. Oxidation and reduction definitions

The basic, but inadequate descriptions of oxidation and reduction in terms of oxygen are first described, followed by some simple examples of the much more important electron transfer definitions. It is advisable to study section 1. before proceeding to section 2. on oxidation state/number. Please note that in the first parts of section 1. the equations are not meant to be balanced, but just focus on the particular oxidation or reduction change and later the full balanced redox equations are given.

The examples are not always meant to be balanced or complete, but they do show the essential oxidation or reduction change.

OXIDATION

REDUCTION

Gain/addition of oxygen by an atom, molecule or ion e.g.

  1. S ==> SO2
    • Burning sulphur – oxidised.
  2. CH4 ==> CO2 + H2O
    • Burning methane to water and carbon dioxide, C and H both gain O.
  3. NO ==> NO2
    • Nitrogen monoxide oxidised to nitrogen dioxide, rapid in air with NO from car exhaust fumes.
  4. SO32– ==> SO42–
    • Oxidising the sulphite ion to the  sulphate ion e.g. by chlorine/bromine oxidising agents.

Loss/removal of oxygen from a compound or ion  e.g. 

  1. CuO ==> Cu
    • Loss of oxygen from copper(II) oxide to form copper atoms in metal extraction with or displacement by more reactive metal.
  2. Fe2O3 ==> Fe
    • Iron(III) oxide reduced to iron in blast furnace using carbon or carbon monoxide.
  3. NO ==> N2
    • Nitrogen(II) oxide reduced to nitrogen.
  4. SO3 ==> SO2
    • Sulphur trioxide reduced to sulphur dioxide.
Loss/removal of electrons from atom, ion or molecule e.g.
  1. Fe ==> Fe2+ + 2e
    • An iron atom loses 2 electrons to form the iron(II) ion e.g. in the initial chemistry of iron rusting or in an iron–acid reaction.
  2. Fe2+ ==> Fe3+ + e
    • The iron(II) ion loses 1 electron to form the iron(III) ion, e.g. via chlorine or manganate(VII) oxidising agents.
  3. 2Cl ==> Cl2 + 2e
    • The loss of electrons by chloride ions to form chlorine molecules e.g. the +ve anode reaction in electrolysis of chlorides or the action of a very strong oxidising agent.
Gain/addition of electrons by an atom, ion, or molecule e.g.
  1. Cu2+ + 2e ==> Cu
    • The copper(II) ion gains 2 electrons to form neutral copper atoms e.g. in electrolysis at the –ve cathode or when copper is displaced from its salt by a more reactive metal.
  2. Fe3+ + e==> Fe2+
    • The iron(III) ion gains an electron and is reduced to the iron(II) ion e.g. by adding zinc to acidified iron(III) salt solution.
  3. 2H+ + 2e ==> H2
    • Hydrogen ions gain electrons to form neutral hydrogen molecules e.g. in the –ve cathode reaction in the electrolysis of acids or in a metal–acid reaction.

An oxidising agent is the species that gives the oxygen or removes the electrons

A reducing agent is the species that removes the oxygen or acts as the electron donor

Advanced Inorganic Chemistry Page Index and LinksREDOX REACTIONS – in an overall reaction, oxidation and reduction must go together  e.g. in terms of oxygen and these are complete and balanced equations

  1. copper(II) oxide + hydrogen ==> copper + water
    • CuO(s) + H2(g) ==> Cu(s) + H2O(g)
    • Heated copper oxide is reduced to copper (O loss) when hydrogen is passed over it,
    • hydrogen is oxidised to water (O gain),
    • hydrogen is the reducing agent (removes O from CuO),
    • and copper oxide is the oxidising agent (donates O to hydrogen).
  2. iron(III) oxide + carbon monoxide ==> iron + carbon dioxide
    • Fe2O3(s) + 3CO(g) ==> 2Fe(l) + 3CO2(g)
    • In the blast furnace the iron(III) oxide is reduced to liquid iron (O loss),
    • the carbon monoxide is oxidised to carbon dioxide (O gain),
    • CO is the reducing agent (O remover/acceptor from Fe2O3),
    • and Fe2O3 is the oxidising agent (O donator to CO)]
  3. nitrogen(II) oxide + carbon monoxide ==> nitrogen + carbon dioxide
    • 2NO(g) + 2CO(g) ==> N2(g) + 2CO2(g) (e.g. in car exhaust catalytic converter)
    • Nitrogen(II) oxide is reduced to nitrogen (O loss),
    • carbon monoxide is oxidised to carbon dioxide (O gain),
    • and CO is the reducing agent (O remover/acceptor)
    • and NO is the oxidising agent (O donor).
  4. iron(III) oxide + aluminium ==> aluminium oxide + iron
    • Fe2O3(s) + 2Al(s) ==> Al2O3(s) + 2Fe(s)  (the Thermit reaction)
    • Iron(III) oxide is reduced and is the oxidising agent (its the oxygen loser i.e. oxygen donor),
    • and the aluminium is oxidised and is the reducing agent (its the oxygen gainer/acceptor/remover).

Advanced Inorganic Chemistry Page Index and LinksRedox reaction analysis based on the more important and advanced electron definitions

and note that the electron gain must equal the electron loss in the complete and fully balanced equation e.g.

  1. magnesium + iron(II) sulphate ==> magnesium sulphate + iron
    • Mg(s) + FeSO4(aq) ==> MgSO4(aq) + Fe(s)
    • This is the 'ordinary molecular' equation for a typical metal displacement reaction, but does not really show what really happens in terms of atoms, ions and electrons, so we use ionic equations like the one shown below.
    • The sulphate ion SO42–(aq) is called a spectator ion, because it doesn't change in the reaction and can be omitted from the ionic equation (below).
    • Also, no electrons should show up in the 'real' balanced equation because the number of electrons lost by some reactant species equals the electrons gained by other reactant species.
    • Mg(s) + Fe2+(aq) ==> Mg2+(aq) + Fe(s)
    • The magnesium atom, Mg, loses 2 electrons (oxidation) to form the magnesium ion, Mg2+,
    • the iron(II) ion, Fe2+, gains 2 electrons (reduced) to form iron atoms, Fe,
    • Mg is the reducing agent i.e. the electron donor to the Fe2+ ion,
    • and the Fe2+ ion is the oxidising agent i.e. the electron remover or acceptor from the Mg atom.
  2. zinc + hydrochloric acid ==> zinc chloride + hydrogen
    • Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g)
    • The chloride ion Cl is the spectator ion so the ionic–redox equation is ...
    • Zn(s) + 2H+(aq) ==> Zn2+(aq) + H2(g)
    • Zinc atoms are oxidised to zinc ions by losing two electrons,
    • and zinc is the reducing agent i.e. the electron donor to the hydrogen ion, H+.
    • Hydrogen ions are the oxidising agent i.e. two hydrogen ions gain/accept one electron each from a Zn atom,
    • and so are reduced to form hydrogen molecules.
  3. copper + silver nitrate ==> silver + copper(II) nitrate
    • Cu(s) + 2AgNO3(aq) ==> 2Ag(s) + Cu(NO3)2(aq)
    • The nitrate ion NO3 is the spectator ion so the ionic–redox equation is
    • Cu(s) + 2Ag+(aq) ==> 2Ag(s) + Cu2+(aq)
    • in which copper atoms are oxidised by the silver ions by a two electron loss,
    • these electrons are transferred from the copper atoms to the silver ions,
    • so they are reduced by one electron gain each to silver atoms.
    • The silver ions are the oxidising agent (e acceptor),
    •  and the copper atoms are the reducing agent (e donor).

Advanced Inorganic Chemistry Page Index and Links

  • One other, and very limited definition, is in terms of hydrogen, but such redox definitions have little use these days except in the reduction (H gain) reactions of unsaturated organic compounds such as alkenes, aldehydes, ketones, carboxylic acids and the reduction nitro–aromatic compounds to amines. However, that is not to say they are unimportant reactions, in fact all four reactions below are all of considerable industrial significance.

    • reduction is hydrogen gain: e.g.

      • (alkene) ethene + hydrogen ==> ethane (alkane)

        • C2H4 + H2 ==> C2H6 (reduction of an alkene)

      • nitrogen + hydrogen ==> ammonia

        • N2 + 3H2 ==> 2NH3 (reduction of nitrogen)

    • oxidation is hydrogen loss: e.g.

      • ammonia + oxygen ==> nitrogen(II) oxide + water

        • 4NH3 + 5O2 ==> 4NO + 6H2O (oxidation of ammonia)

      • methane + oxygen ==> carbon dioxide + water

        • CH4 + 2O2 ==> CO2 + 2H2O (oxidation of methane)

  • Electron definitions of oxidation or reduction are the much more useful and important for advanced level chemistry as is the very important associated concept of oxidation state/number which is explained in section 2.

  • Definitions and analysis of redox reactions in terms of oxygen or hydrogen may not get you the marks at advanced level! So, LEARN and USE the following as soon as possible in your advanced redox analysis.

    1. Oxidation is electron loss by a species or increase in the oxidation state/number of an element in the species involved which can be the element itself, or a covalent compound or ion containing the element (ditto for 2., 3. and 4.).

    2. Reduction is electron gain by a species or decrease in the oxidation state/number of an element in the species involved.

    3. An electron acceptor is an oxidising agent and therefore gets reduced in its action, so there is a decrease in the oxidation number/state of one of the atoms of the oxidising agent.

    4. An electron donor is a reducing agent and therefore gets oxidised in the process, so there is an increase in the oxidation number/state of one of the atoms of the reducing agent.

  • You also need to conceive of 'redox' reactions as a combination of  two half–cell reactions/equations (see Equilibria Part 7 for more on half–cell potentials and redox equilibria.

  • One half–reaction will be an oxidation and the other 'half' will be a reduction.

  • How to put the two together to gives the full ionic–redox equation to completely summarise a redox reaction in simple cases is given in section 2. below, and see sections 5. and 6. for lots of more complex examples).

Advanced Inorganic Chemistry Page Index and Links


2. Introducing, explaining and defining oxidation state and examples

  • In simple cases, particularly simple ions, the oxidation state can be defined as the number of electrons which must be added or removed to leave the element in its electrically neutral elemental state which is then considered to be in a zero oxidation state e.g.

    • chlorine, Cl or Cl2, oxidation state zero (0), the element itself in an electrically neutral state.

    • chloride ion, Cl, oxidation state minus one ( –1), because one electron must be removed to leave the electrically element itself.

    • sulfur, S or S8, oxidation state zero (0), the element itself in an electrically neutral state.

    • sulfide ion, S2–, oxidation state minus two ( –2), because two electrons must be removed to leave the electrically element itself.

    • sodium, Na, oxidation state zero (0), the element itself in an electrically neutral state.

    • sodium ion, Na+, oxidation state plus one (+1), because one electron must be added to give the electrically atom of the element sodium.

    • iron, Fe, oxidation state zero (0), the element itself in an electrically neutral state.

    • iron(III) ion, Fe3+, oxidation state plus three (+3), because three electrons must be added to give the electrically atoms of the element iron.

    • Note:

      • Important sign conventions:

        • The sign (+/–) for an oxidation state is quoted first e.g.

          • –3, –2, –1, 0, +1, +2, +3 etc. (obviously no sign for zero)

        • In the symbols/formulae of ions, the numerical value of the ion charge is quoted first e.g.

          • PO43–, S2–, I, NH4+, Mg2+, Al3+ etc. (no number if charge is +/– one)

      • In an electrically neutral compound or molecule, the sum of the oxidation states must add up to zero.

      • If it doesn't, then you have got something wrong in your analysis.

      • BUT,  the 'zero sum rule' provides a way of working out another unknown oxidation state of an atom, if you know the oxidation states of the other atoms in the species.

      • In the case of ions, the sum of the oxidation states equals the overall charge on the ion.

  • When dealing with atoms and simple ions, redox analysis is usually quite straight forward e.g. the three 'electron analysed' examples in section 1. above and the examples below where the concept of oxidation state/number is now further introduced via detailed examples for simple ions or molecules. It is best to study them in the order given.

  • After this, the more awkward situations are dealt with.

  • Section 2 reaction sub–index: 2.1 metal + acid * 2. reactive metal + non–metal * 2.3 * halogen–halogen displacement * 2.4 metal–metal displacement * 2.5 what about more complex molecule/ion situations?

Advanced Inorganic Chemistry Page Index and Links


Ex 2.1 Reacting a metal with an acid

  • magnesium + sulphuric acid ==> magnesium sulphate + hydrogen

    • Mg(s) + H2SO4(aq) ==> MgSO4(aq) + H2(g)

  • Since the chloride ion SO42 is a spectator ion i.e. it doesn't change chemically or physically in state, so the ionic–redox equation is

    • Mg(s) + 2H+(aq) ==> Mg2+(aq) + H2(g)

  • Magnesium atoms are oxidised to magnesium ions by the loss of two electrons.

  • Hydrogen ions are reduced by electron gain to form hydrogen molecules.

  • The oxidation states of the elements are shown in () for the species involved in the actual chemical change and would be quoted as follows:

    • Mg (0) + 2 x H(+1) ==> Mg(+2) + 2 x H (0) and the half–cell reactions are ...

      • the oxidation: Mg(s) ==> Mg2+(aq) + 2e (ox. state increases by 2 units per Mg atom)

      • the reduction: 2H+(aq) + 2e ==> H2(aq) (ox. state decreases by 1 unit per H+ ion)

    • The oxidation state for magnesium is increased by two units per atom, is balanced by the oxidation state of two hydrogens (as H+ ion) decreasing by one unit each.

  • Having now quoted some oxidation states, now is the time to explain the concept further and connect it with electron transfers in simple ionic–redox reactions like the one above.

  • Important points and ideas to take forward into section 2.2 onwards ...

    • The oxidation state defines the theoretical electron change needed to convert the oxidised or reduced 'state' of the element in an ion (or in a molecule later), back to its 'elemental state' where the oxidation state is considered as zero (0).

      • Conventions reminders to be strictly adhered to ...

      • In quoting oxidation states, the sign is placed before the number (e.g. ... –2, –1, 0, +1, +2 ... etc.), and there is always a number after the sign.

      • For ions, the charge sign comes after the number (e.g. ... 3–, 2–, –, +, 2+, 3+ ... etc.), but the case of a single +/– charge no number is required, the '1' is assumed.

    • The 'free element' is considered to have an oxidation state of zero (0) i.e. not combined with another element in an ionic or covalent compound. It can of course be combined with itself e.g. H2, but its still in a zero ox. state.

    • So, in terms of the magnesium–acid reaction ....

    • The magnesium (Mg atom) is the reducing agent or electron donor, so its oxidation state is increased.

      • Magnesium is in the +2 oxidation state in its ion (Mg2+) and theoretically needs to gain two electrons to return to elemental magnesium atoms (Mg).

    • The hydrogen ion (H+) is the oxidising agent or electron acceptor, so its oxidation state is decreased.

      • Hydrogen is in the +1 oxidation state in its ion (H+) and theoretically needs to gain one electron per ion to return to elemental hydrogen molecules (H2).

    • The total/net electron transfers or oxidation state changes must be zero i.e. for example above

      • For every magnesium atom losing two electrons, two hydrogen ions gain an electron each, i.e.

      • magnesium's ox. state increases by two units balanced by two hydrogens (ions) each decreasing their oxidation state by one unit.

      • Oxidation (e loss) = increase in ox. state and reduction (e gain) = decrease in ox. state.

      • So, numerically, electrons lost = electrons gained

      • and increase in oxidation states = decrease in oxidation states.

Advanced Inorganic Chemistry Page Index and Links


Ex 2.2 Heating reactive metals with non–metals

  • When reactive metals are heated with reactive non–metals, ionic compounds are readily formed exothermically in a combustion reaction. MgSLi2O , Na2O2, the major product of burning sodium in excess air/oxygen, and  Ca3P2 respectively.

  • Note that strict ionic formulae are also given to make the redox analysis clearer.

  • In each case the metal is the reducing agent (electron donor/loser, oxidised in process),

  • and the non–metal is the oxidising agent (electron acceptor/gainer, reduced in process).

  • Ex 2.2.1: 2Al(s) + 3Cl2(g) ==> 2Al3+(Cl)3(s) (or 2AlCl3, aluminium chloride, it is an ionic lattice)

    • Oxidation state changes:

    • Aluminium atom ==> aluminium ion: oxidation, e loss, oxidation state change ..

    • 2 x Al increasing, each from (0) to (+3), losing 3 electrons per atom,

    • chlorine molecule ==> chloride ion, reduction, e gain, ox. state change is ..

    • 6 x Cl decreasing, each from (0) to (–1), gaining 1 electron per atom (or 2 e per molecule),

    • and the half–reactions* are written as:

    • (i) oxidation: Al ==> Al3+ + 3e and (ii) reduction: Cl2 + 2e ==> 2Cl

    • adding 2 x (i) + 3 x (ii) gives the full equation (iii) 2Al + 3Cl2 ==> 2AlCl3

    • In terms of the full balanced equation, 6 electrons lost i.e. a rise of 6 ox. state 'units', which is balanced by 6 electrons gained i.e. a decrease of 6 ox. state 'units'.

    • *Note these also called half–cell potential equations, and is a way of representing the 'separate' oxidation or reduction in equation form.

  • Ex 2.2.2: Mg(s) + S(s) ==> Mg2+S2–(s)  (or MgS, magnesium sulphide)

    • Oxidation state changes:

    • Magnesium atom ==> magnesium ion: oxidation, e loss, oxidation state change ...

    • 1 x Mg increasing from (0) to (+2), loses two electrons per Mg atom,

    • which is 'electronically' balanced by the ...

    • sulphur atom ==> sulphide ion: reduction, e gain, ox. state change ...

    • 1 x S decreasing from (0) to (–2), gaining two electrons per S atom,

    • and the half–reactions are written as:

    • (i) oxidation: Mg ==> Mg2+ + 2e and (ii) reduction: S + 2e ==> S2–

    • adding (i) + (ii) gives the full equation (iii) Mg + S ==> MgS

    • In terms of the full balanced equation, 2 electrons lost i.e. a rise of 2 ox. state 'units', which is balanced by 2 electrons gained i.e. a decrease of 2 ox. state 'units'.

  • Advanced Inorganic Chemistry Page Index and LinksEx 2.2.3: 4Li(s) + O2(g) ==> 2(Li+)2O2–(s)  (or 2Li2O, lithium oxide)

    • Oxidation state changes:

    • Lithium atom ==> lithium ion: oxidation, e loss, ox. state change for lithium is from 0 to +1

    • 4 x Li increasing, each from (0) to (+1), losing one electron per Li atom, 4 electrons change in terms of the balanced equation

    • which is 'electronically' balanced by the ...

    • two oxygen molecules ==> two oxide ions: reduction, e gain, ox. state change, ox. state change of 0 to –2

    • 2 x O decreases, each from (0) to (–2), gaining two electrons per O atom (or four electrons per O2 molecule),

    • and the half–reactions are:

      • (i) oxidation: Li ==> Li+ + 2e (formation of the lithium ion)

      • (ii) reduction: O2 + 4e ==> 2O2– (formation of the oxide ion)

    • adding and balancing gives 4 x (i) + (ii) giving the full equation

      • (iii) 4Li + O2 ==> 2Li2O

    • In terms of the full balanced equation, 4 electrons lost i.e. a rise of 4 ox. state 'units' by the lithium atoms, which is balanced by 4 electrons gained i.e. a decrease of 4 ox. state 'units' by the oxygen atoms.

  • Ex 2.2.4: 2Na(s) + O2(s) ==> (Na+)2O22–(s)  (or Na2O2, sodium peroxide is the major product in excess air/oxygen)

    • Oxidation state changes: This is very similar to example 2.2.3

    • Sodium atom ==> sodium ion: oxidation, e loss, ox. state change is from 0 to +1

    • 2 x Na increases, each from (0) to (+1), losing one electron per Na atom,

    • which is 'electronically' balanced by the ...

    • oxygen molecule ==> peroxide ion: reduction, e gain, ox. state change ...

    • 2 x O decreases, each from (0) to (–1), gaining two electrons per O atom (or two electrons per O2 molecule),

    • and the half–reactions are:

      • (i) oxidation: Na ==> Na+ + e (formation of sodium ion)

      • (ii) reduction: O2 + 2e ==> O22– (formation of the peroxide ion)

    • adding and balancing 2 x (i) + (ii) gives the full equation

      • (iii) 2Na + O2 ==> Na2O2

    • In terms of the full balanced equation, 2 electrons lost i.e. a rise of 2 ox. state 'units' for the sodium atoms, which is balanced by 2 electrons gained i.e. a decrease of 2 ox. state 'units' of the oxygen atoms of the oxygen molecule.

  • Ex 2.2.5: 6Ca(s) + P4(s) ==> 2(Ca2+)3(P3–)2(s)  (or 2Ca3P2, calcium phosphide)

    • Oxidation state changes:

    • Calcium atom ==> calcium ion: oxidation, e loss, ox. state change ...

    • 6 x Ca increasing, each from (0) to (+2), losing two electrons per Ca atom,

    • which is 'electronically' balanced by the ...

    • phosphorus molecule ==> phosphide ion: reduction, e gain, ox. state change ...

    • 4 x P decreasing, each from (0) to (–3), gaining three electrons per P atom (or twelve electrons per P4 molecule),

    • and the half–reactions are:

    • (i) oxidation: Ca ==> Ca2+ + 2e and (ii) reduction: P4 + 12e ==> 4P3–

    • adding and balancing 6 x (i) + (ii) gives the full equation (iii) 6Ca + P4 ==> 2Ca3P2

    • In terms of the full balanced equation, 12 electrons lost i.e. a rise of 12 ox. state 'units', which is balanced by 12 electrons gained i.e. a decrease of 12 ox. state 'units'.

    • You can also work out this example based on (iv) P + 3e ==> P3–,

    • in which case the balanced equation is based on 3 x (i) + 2 x (iv) ==> Ca3P2.

Advanced Inorganic Chemistry Page Index and Links


Ex 2.3 A halogen displacement reaction

  • A 'more reactive'** halogen displacing a less reactive halogen (X > Y).

  • halogen X + Y halide salt  ==> X halide salt + halogen Y

  • X2(aq) + 2KY(aq) ==> 2KX(aq) + Y2(aq)

  • since the potassium ion (K+) is a spectator, the ionic–redox equation is

  • X2(aq) + 2Y(aq) ==> 2X(aq) + Y2(aq)

  • where halogen X is more reactive than halogen Y (F > Cl > Br > I).

  • X2 is the oxidising agent or electron acceptor, X has an ox. state of (0) in the halogen molecule and accepts an electron from Y, and so X gets reduced to ox. state(–1) in the X ion.

  • Y is the reducing agent or electron donor, Y has an ox. state of (–1) in the halide ion and donates an electron to X2, and so Y is oxidised to ox. state (0) in the Y2 molecules.

  • so in terms of oxidation states: 2 x X(0) + 2 x Y(–1) ==> 2 x X (–1) + 2 x Y (0)

  • and the half–cell reactions are ...

  • (i) reduction*: X2(aq) + 2e ==> 2X(aq) and (ii) oxidation*: 2Y(aq) ==> Y2(aq) + 2e

  • adding* (i) + (ii) gives (iii) X2(aq) + 2Y(aq) ==> 2X(aq) + Y2(aq)

  • *Note that the equations can be written on the basis of half–mole of the halogen molecule,

  • *i.e. (iv) 1/2X2(aq) + e ==> X(aq), (v) Y(aq) ==> 1/2Y(aq) + e

  • and  (iv) + (v) gives the equally valid balanced equation (vi) 1/2X2(aq) + Y(aq) ==> X(aq) + 1/2Y2(aq)

  • **More reactive here means the half–cell potential for X2/X is more positive than for Y2/Y,

  • i.e. X2 is a stronger oxidising agent than Y2. (see Equilibria Part 7).

Advanced Inorganic Chemistry Page Index and Links


Ex 2.4 A metal displacing a less reactive metal

  • A 'more reactive'* metal displacing a less reactive metal.

  • copper + silver nitrate ==> silver + copper(II) nitrate

  • Cu(s) + 2Ag+(aq) ==> 2Ag(s) + Cu2+(aq)

  • One copper is oxidised by losing two electrons, so its ox. number changes and increases from (0) to (+2).

  • Two silver ions are reduced by gaining one electron each, so their oxidation state decreases from (+1) to (0).

  • and the half–cell reactions are ...

  • (i) reduction: Ag(aq) + e ==> 2X(aq) and (ii) oxidation: Cu(s) ==> Cu2+(aq) + 2e

  • adding 2 x (i) + (ii) gives the full balanced equation.

  • *More reactive here means the half–cell potential for the more reactive metal is the less positive or more negative potential, or in other words, the Ag/Ag+ is more positive than for Cu/Cu2+,  i.e. Ag+ is a stronger oxidising agent than Cu2+. (see Equilibria Part 7).

Advanced Inorganic Chemistry Page Index and Links


Ex 2.5 What about oxidation state and the 'not so simple' situations?

  • How in redox terms do we deal with the structure and reactions of electrically neutral covalent molecules like H2O or more complex ions like PO43– which are held together with covalent bonds but carry an overall electrical charge?

  • These situations will be redox analysed by extending the concept of oxidation number or oxidation state introduced in section 2.1, but first a recap of the simple ion situation.

    • As pointed out already, for simple ions of one atom, you can think of it as the electron change required to return the atom to an uncombined electrically neutral atom or molecule, because this relates numerically to the relative electron loss or gain involved in forming the ion from the free element e.g.

    • Magnesium is in the +2 oxidation state in ionic compounds containing the magnesium ion Mg2+

      • Mg2+ + 2e ==> Mg, e.g. in magnesium chloride, MgCl2 or Mg2+(Cl)2

    • Phosphorus is in the –3 oxidation state in ionic compounds containing the phosphide ion P3–

      • P3– – 3e ==> P, e.g. in sodium phosphide, Na3P. or (Na+)3P3–

    • Oxygen is in the –2 oxidation state e.g. in ionic compounds containing the oxide ion O2–

      • O2– – 2e ==> 1/2O2

    • Important conventions, reminders and trends before proceeding to the 'not so simple' situations:

      • In writing oxidation states, the + or – sign is written before the number and there is always a sign except for zero (0) for the uncombined element itself.

      • There is always a sign for the electrical charge on an ion (charged particle) symbol/formula, since it can't be zero by definition, but no number is written for a single +/– charge, but must have 2, 3 etc. for double or triple charges etc., and the number comes before the electric charge sign on an ion.

      • These two conventions must be strictly adhered to e.g. ion formula Al3+, ox. state is +3.

      • Sometimes () are needed in ionic formulae, e.g. the correct ionic formula (Na+)3P3– is NOT the same as the incorrect Na+3P3–, because there is no such ion as Na+3 which implies three sodium atoms combined carrying an overall charge of single +, rather than the correct three separate correct Na+ ions, each with a single plus charge.

      • Now, very importantly conceptually for the more awkward cases ...

      • the more electropositive elements tend to have a positive oxidation state in a compound because most metals more readily lose electrons than non–metals,

      • and the more electronegative elements tend to have a negative oxidation state in a compound because most non–metals more readily gain electrons compared to metals.

  • This idea, based on the relative electronegativity of an element (table below), is exploited in a way in the next section 3. to derive the oxidation state of any element in any molecule or ion consisting of at least two atoms.

    • BUT beware, initially you may have to conceive of a molecule/molecular ion as if it is composed of a combination of simple one atom ions, even if that's not the case in reality it does help to work through the abstract nature of this topic!

    • Its a useful conceptualisation of the 'redox' situation i.e. a method of problem solving at the start of dealing with the 'not so simple' chemical species. So, read on in section 3 where, by the end, hopefully you will be able to analyse species without thinking 'falsely' in terms of combining simple ions that don't really exist in the situation. You may need to refer to the tabulated values of electronegativity below.

  • Electronegativity is the power of an atom to attract electron charge from another atom it is covalently bonded to. Some Pauling values of electronegativity are quoted below.

  • element Na Mg Al Mn Fe H Si P C S I Br Cl N O F
    electronegativity 0.9 1.2 1.5 1.5 1.8 2.1 1.8 2.1 2.5 2.5 2.5 2.8 3.0 3.0 3.5 4.0
  • Generally speaking electronegativity increases from left to right across a period of the periodic table and decreases down a group of the periodic table.

Advanced Inorganic Chemistry Page Index and Links


3. Oxidation state rules and guidelines and application to inorganic molecules/ions

  • Its really important that you have studied at least section 2.5  before reading this section, and have some idea on the relative electronegativity of elements commonly encountered in your inorganic chemistry. It will greatly help in understanding the assignment of oxidation states of elements in the quoted compounds on these three web pages.

  • Further more, as much as possible of the following guidelines and examples, 3.1 to 3.4 should be learned as quickly as possible and they will help you throughout your chemistry studies.

  • Oxidation states in naming inorganic compounds with respect to oxidation state is explained in section 4. but they are unavoidably quoted in here in section 3.

  • 3.1: Atoms in elements, i.e. not covalently or ionically combined with atoms of another elements, are considered to have an oxidation state of zero or 0 irrespective of the physical, molecular or allotropic* state of the element,

    • e.g. in Xe(g), Fe(s), H2(g), Cl2(g), Br2(l), O2(g), O3(g), P4(s), S8(s), C60(a fullerene), Cn(graphite) etc.

    • *Allotropes are different physical forms of the same element in the same physical state e.g. solid carbon (graphite, diamond, fullerenes), gaseous oxygen (dioxygen, trioxygen–ozone). The different forms arise from differences in the bonding BUT they are all in ox. state of zero)

  • 3.2: Irrespective of being in an ionic or covalent compound or the physical state of the compound, for neutral molecules or simple/complex molecular ions of selected elements the following 'rules' 3.2.1 to 3.2.4 apply ...

    • BUT please note that ...

      • the total oxidation numbers of all atoms in a compound is zero (more details guideline 3.3),

      • the total oxidation numbers of all atoms in an ion equals the overall charge on the ion (see guideline 3.4) and watch the sign conventions for ion charge and oxidation state/number,

      • the most electronegative element in the compound or ion carries the negative oxidation state (see above for Pauling values of electronegativity),

      • there are important exceptions to the usual oxidation state of an element mentioned, in terms of the advanced chemistry you will encounter – so watch out and away we go!

    • Ex 3.2.1: fluorine: F is always (–1) in all compounds,

      • because it has the highest electronegativity of any element and only one electron short of a stable noble gas structure electronically e.g. formation of fluoride ion F [2.7] + e ==> F [2.8]

      • The (–1) state relates to either gaining an electron to form the fluoride ion, F, e.g. in ionic sodium fluoride, NaF (Na+ Cl ionic bond, Na is +1),

      • or share with one other electron to form a covalent bond e.g. in the molecules hydrogen fluoride, HF (δ+H–Fδ– polar bond, H is +1), or tetrafluoromethane (carbon tetrafluoride), CF4 (δ+C–Fδ– polar bond, C is +4).

    • Ex 3.2.2: hydrogen H is (+1) in most compounds.

      • This relates to the one outer electron either being lost in the formation of a hydrogen ion, H+, as in acids, or being shared with another electron to form a covalent bond pair, when bonding with a more electronegative atom.

      • e.g. HCl (δ+H–Clδ– polar bond, Cl is –1), H2O (δ+H–Oδ– polar bond, O is –2), NH3 (δ+H–Nδ– polar bond, N is –3).

      • One exception is in the formation of hydrides with metals where hydrogen's ox. state is (–1)

        •  e.g. in ionic sodium hydride, Na+H, Na is +1) or the covalent beryllium hydride, BeH2 (δ+Be–Hδ– polar bond, Be is +2) because most metals are more electronegative than hydrogen.

        • This situation happens if the other atom of the bond has a very low electronegativity, i.e. with very electropositive metals of low ionisation energy such as Group 1 (Li, K, Rb, Cs), Group 2 (Ca, Sr) and also in many covalent metal hydrides of less electropositive metals than Gps1/2, but still less electronegative than hydrogen e.g. beryllium BeH2 and aluminium AlH3.(δ+Al–Hδ– polar bond, Al is +3)

    • Advanced Inorganic Chemistry Page Index and LinksEx 3.2.3: oxygen: O is (–2) in most compounds, since it is the 2nd most electronegative element.

      • This readily relates to the formation of the oxide ion, O2– with electropositive metals (very low electronegativity) e.g. MgO (Mg2+O2– ionic bond, Mg is +2), Al2O3 (Al3+ O2– ionic bond, Al is +3) etc.,

        • where the two electron gain by oxygen O [2.6] giving the stable noble gas configuration of [2.8] in the O2– oxide ion,

      • or by sharing two electrons from other less electronegative atoms to form two single bonds or one double bond e.g. water, H2O, δ+H–Oδ2––Hδ+ (H is +1), or carbon dioxide, CO2, δ2–O=Cδ4+=Oδ2– (C is +4) molecules represented in terms of partial charges of the polar bonds due to the electronegativity difference.

      • BUT there are two important oxidation state exceptions for O

      • (i) In peroxides H2O2 (covalent, H is +1)) or Na2O2 (ionic, Na is +1)/O22–,where it is (–1)

        • Here two oxygen atoms are linked together e.g. H–O–O–H,

        • which changes the situation compared to H–O–H.

      • and (ii) O is (+2) in F2O

        • because the electronegativity of fluorine is higher at 4.0 than oxygen at 3.5

        • so you get a δ–F–Oδ2+–Fδ– polar bond as F is always –1.

      • (There is also the curious situation regarding the 'super–oxide ion' O2 as in KO2 (K is +1), where each oxygen has an average oxidation state of 0.5, since the ion is the equivalent of an dioxygen molecule plus one electron.)

    • Ex 3.2.4: chlorine Cl is –1 in many simple compounds combined with a less electronegative.

      • This is related electronically to the formation of the chloride ion by chlorine gaining one electron.

        • e.g. Cl [2.8.7] + e ==> Cl [2.8.8]) e.g. in sodium chloride NaCl (Na+ Cl bond, Na is +1).

      • or sharing with one electron from a less electronegative element in covalently bonding

        • e.g. HCl (δ+H–Clδ– polar bond, H is +1), PCl5 (δ+P–Clδ– polar bond, P is +5)

      • BUT when combined with the more electronegative O or F, Cl has a (+) oxidation state

      • e.g. Cl is (+1) in chlorine(I) oxide, Cl2O (δ+Cl–Oδ polar bond, O is –2)

      • or (+3) in chlorine(III) fluoride, ClF3 (δ+Cl–Fδ– polar bond, F is –1)

        • There dozens of similar so–called 'interhalogen compounds where the most electronegative halogen carries the negative oxidation state and the positive oxidation state of the less electronegative halogen is shown with (Roman numerals) and it retains its 'elemental' name.

      • There are also a whole series of 'chlorate ions' and 'oxides', where the more electronegative oxygen results in chlorine having oxidation states of (+1), (+3), (+4), (+5) and the maximum possible (+7). The latter is the maximum ox. state for any halogen, and is dictated by using all 7 outer electrons in bonding e.g.

      • chlorate(I), ClO, chlorate(III), ClO2, chlorate(V), ClO3, chlorate(VII), ClO4,

        • These sort of ion examples are explained in more detail in section 3.4

        • Note that the name of a non–metallic oxy–anion changes from ...ide to ...ate when the non–metal is combined with oxygen to form the polyatomic anion and the oxidation state is denoted after the name by a (Roman numerals number) in brackets.

      • chlorine(IV) oxide, ClO2 (chlorine dioxide), bromine(IV) oxide, BrO2,

        • These sort of molecule examples are explained in more detail in section 3.3

      • (for more details and examples see 3.4 charge on ions and 3.5 maximum oxidation state patterns)

      • These guidelines for chlorine also apply to bromine and iodine.

  • 3.3: Since compounds have no net overall electrical charge, the sum of the oxidation states of all the atoms in a compound is also zero, again, this is easy to follow in simple ionic compounds e.g.

    • Ex 3.3.1: Magnesium oxide, MgO, Mg2+O2–,

      • One Mg2+ balances one O2–, or ...

      • one Mg in the +2 ox. state balances one O in the –2 ox. state.

    • Ex 3.3.2: Aluminium sulphide, Al2S3, ionically (Al3+)2(S2–)3,

      • two 3+ ions electrically balance three 2– ions, or

      • two Al's in the +3 ox. state balance three S's in the –2 ox. state.

    • But it is not so easy at first in covalent molecules because you need to know how to assign the relevant oxidation states. This is again is readily done by using the difference in electronegativity guideline, and, particularly at the start, thinking in a sort of 'ionic' conceptual way, even though the atoms in such an ion are covalently bonded together and the 'ionic' nature is really due to the overall net electrical charge carried by the particle.

    • In covalent molecules the most electronegative elements carry the negative oxidation states e.g. using the guidelines 3.2.1 to 3.2.4 ...

      • Advanced Inorganic Chemistry Page Index and LinksEx 3.3.3: Water, H2O

        • Two hydrogens in the +1 state balance one oxygen in the –2 state because oxygen's electronegativity is higher than hydrogen's.

        • In other words, the assignment of oxidation states will coincide with bond polarity δ+H–Oδ2––Hδ+ based on electronegativity differences and you can crudely think of water made up of two H+ ions and one O2– ion, even though in reality water is neutral covalent molecule!

      • Ex 3.3.4: Phosphorus oxychloride, POCl3.
        • The two most electronegative elements are oxygen and chlorine and they will carry the negative oxidation states of –2 and –1 respectively.
        • You can think of this covalent compound as if it is made up of one P at (+5), one O at (–2) and three Cl at (–1), so (+5) + (–2) + (3 x –1) = 0.
        • Note that by using the guidelines you can deduce that P is in the +5 ox. state even though there is no rule specified to phosphorus quoted here.
      • Ex 3.3.5: Pure liquid sulphuric acid, H2SO4
        • From the guidelines hydrogen is +1, oxygen is –2, so sulphur will be sulphur in the +6 ox. state.
        • so, (2 x +1) + (+6) + (4 x –2) = 0
        • Note again,  that by following the guidelines you can deduce that S is in the +6 ox. state even though there is no rule specified for sulphur.
  • 3.4: In simple ions of one atom the oxidation state is numerically the same as the charge on the ion.

    • Ex 3.4.1: Sodium as Na+ has an ox. state of (+1), oxygen as the oxide ion O2– has an ox. state of (–2), for aluminium in the ion Al3+ it is (+3), chlorine as the chloride ion Cl (–1), sulfur as the sulfide ion S2– (–2) etc.

    • For ions of at least two atoms, the charge on an ion is numerically equal to the sum of the oxidation numbers of all the atoms in the ion.

    • Therefore you can now deal with non–electrically neutral molecular ions like PO43– in the same way outlined for neutral covalent molecules in 3.3, BUT you must take into account the overall charge on the ion.

    • Advanced Inorganic Chemistry Page Index and LinksEx 3.4.2: For ions like phosphate(V), PO43–

      • Oxygen is more electronegative than phosphorus, so ..

      • P is +5, O is –2, and therefore +5 + (4 x –2) = –3 in total, = 3– for overall charge on the ion.

      • You can conceive of it made up of one fictitious P5+ combined with four fictitious  O2– ions, even if they don't exist here, but (5+) + (4 x 2–) = 3–.

      • Note again that the Roman numeral in brackets denotes the positive oxidation state of phosphorus,

      • the –2 of oxygen is correctly assumed (more on naming ions/compounds later).

    • Ex 3.4.3: In the 'inorganic' carbonate ion, CO32–, again oxygen is the most electronegative element.

    • Ex 3.4.4: The compound potassium nitrate, KNO3
      • In the nitrate ion, NO3, to give an overall charge of –1 on the ion, nitrogen must be +5 and oxygen at –2.
      • So, overall in the compound: one of K at (+1) and one of N at (+5) is balanced by three of O at (–2).
    • Ex 3.4.5: The chlorate(V) ion, ClO3
      • Oxygen is in the –2 oxidation state, so with 3 O's at (–2) each, the chlorine must be in the (+5) state to give the ion an overall charge of a single minus.
    • Ex 3.4.6: The dichromate(VI) ion, Cr2O72–
      • With 7 O's at (–2), total 14–, the 2 Cr's must equal a total of +12 to give the 2– surplus charge on the ion, so the chromium is in the (+6) ox. state.
    • Ex 3.4.7: Transition metal complex ions e.g. [CrCl2(H2O)4]+ ,
      • H2O is neutral, 2H (+1) and O (–2), so you can focus on the Cl and Cr.
      • The two chloride ligands are Cl ions (ox. state –1), contributing a total charge of a 2–,
      • therefore the Cr must be in the +3 state to give an overall charge of a single + on the complex ion.
      • This complex might be called the tetraaquadichlorochromium(III) ion.
      • Other examples of transition metal complexes
        • [TiCl6]2–, titanium is in a +4 ox. state (Cl is –1), (6 x –1) + (+4) = 2– overall.
        • [VO]2+(aq), or [VO(H2O)5]2+(aq), vanadium is in +4 ox. state (H is +1, O is –2),
  • 3.5: The maximum oxidation state possible is often the number of outer electrons in the valency shell.

    • For Groups 1 to 7 and even the Noble Gases (group 0/8) there is, not surprisingly, a nice simple pattern related to the Group number of the Periodic Table e.g.

    • Groups 1–4: Na is +1 e.g. in NaCl, Ca is +2 e.g. in CaO, Al is +3 e.g. in AlCl3, C is +4 e.g. in CO2,

    • Groups 5–8: P is +5 e.g. in H3PO4, S is +6 e.g. in SO42–, Cl is +7 e.g. in ClO4, and Xe is +8 e.g. in XeO4.

    • After K and Ca on Period 4 of the Periodic Table come the 3d block of metals, where things are not so clear cut.

    • Up to manganese, the maximum ox. state is still dictated by the number of valency electrons e.g.

      • Sc +3 in ScCl3, Ti +4 in TiO2, V +5 in VO43–, Cr +6 in K2Cr2O7, Mn +7 in MnO4,

        • (x–ref outer electron configuration in terms of 3dx4s2),

      • but for Fe, Co, Ni, Cu and Zn the pattern is complex, and generally one of decreasing max. ox. state,

      • e.g. effectively a max. of +3 for Fe since only a few unstable compounds with over +3 ox. state exist (e.g. FeO42– [ferrate(VI) oxyanion], Co max. +3 and usually in complex ions, for Cu there are a few unstable complex ions showing the +3 state, but effectively the max. is +2 in e.g. CuSO4 etc. and Zn only exhibits the +2 ox. state in its compounds e.g. ZnO, ZnCl2 etc.

      • Apart from Sc (+3) and Zn (+2), all the rest of the 3d block are true transition metals exhibiting variable oxidation states, a fact that is important in the catalytic activity of there compounds e.g. the use of V2O5 in the Contact Process for making sulphur trioxide in the manufacture of sulphuric acid or the iron(III) ion catalysis of iodide oxidation by peroxodisulphate..

  • 3.6: Other Periodic Table 'patterns', e.g. for those elements encountered at UK advanced AS–A2 and IB academic level.

    • For Group 1 (+1) and group 2 (+2) elements there is only one stable oxidation state,

    • Group 3 mainly +1 and +3,

    • Group 4 mainly +2 and +4,

    • Group 5 mainly –3, +3 and +5, (though see the variation for nitrogen in section 3.7 below)

    • Group 6 mainly –2, +4 and +6,

    • Group 7 can be very variable from –1 all the way to +7 (see 3.2.4 for Cl and Br examples).

    • 3d Block – Transition metals show values from +1 to +7 depending on the element.

      •  Scandium (+3) and zinc (+2) have only one stable oxidation state in compounds and are not true transition metal elements.

      • From titanium to copper, all of the elements exhibit at least two stable oxidation states in simple compounds or complex ions and are considered true transition metals.

      • Lastly a 'curiosity', nickel(0) carbonyl is Ni(CO)4, a sort of neutral complex formed on heating nickel in carbon monoxide (C +2, O –2) so Ni is in a zero ox. state!

  • 3.7: Oxidation state can be quite variable even for a single element e.g. nitrogen's oxidation state in various compounds or ions:

    • for ammonia NH3, ammonium ion NH4+, ammonium chloride NH4Cl,

      • nitride ion N3– and magnesium nitride Mg3N2 it is (–3)

    • for hydrazine N2H4 it is (–2),

    • for azo compounds R–N=N–R it is (–1), R is usually aromatic,

    • for nitrogen N2 it is (0),

    • for nitrogen(I) oxide N2O (+1),

    • for nitrogen(II) oxide NO (+2),

    • for sodium nitrite NaNO2, nitrite/nitrate(III) ion NO2 and nitrogen (III) oxide N2O3 it is (+3)

    • for nitrogen(IV) oxide NO2 (+4)

    • for nitrogen(V) oxide N2O5, nitronium cation NO2+, nitrate ion NO3, NaNO3 and HNO3 it is (+5)

    • and just to complicate matters further in ammonium nitrate NH4NO3,

      • nitrogen's oxidation state is –3 and +5.

    • This sequence illustrates why the concept of oxidation number/state is so important in redox chemistry e.g. the nitrogen cycle, which is all about electron transfer via enzymes!

Advanced Inorganic Chemistry Page Index and Links


4. Oxidation states and naming inorganic compounds

  • 4.1: Different compounds or ions can be formed from the same two (or more) different elements, so at least one of the elements is in a different oxidation state in each compound or ion.

  • Therefore there is a need to indicate this when writing the name of the compound or ion.

    • (Roman numerals) are used to indicate the value of the oxidation state but with no preceding + sign.

    • Some examples have already been dealt with in section 3.4, so section 4. acts as a collective recap and extension of using Roman numeral notation depicting different oxidation states of the same element in two or more compounds.

    • A positive oxidation state is assumed, and if you have worked through the guidelines in section 3. you should have no trouble in analysing the examples below and connecting the (II) etc. in the name with an oxidation state analysis of the compound or ion.

    • The Roman numerals number is placed in () after the name, but with no space between it.

      • Sometimes, especially in transition metal complexes, you might even see the Roman numerals in a formula,

      • e.g. the octahedral complex [CrIIICl2(NH3)4]+ (can you see why Cr is in the +3 state?)

      • and this is called the tetraamminedichlorochromium(III) ion,

      • but this 'superscript' notation is rarely encountered until undergraduate level studies,

      • so we will start with some simpler naming examples!

  • 4.2: For metallic or non–metallic elements the name of the element is used if NOT in an anion

    • Some of the old names are still in common use, but try to use the correct systematic name e.g.

    • copper(I) oxide Cu2O and copper(II) oxide CuO

      • (once called cuprous oxide and cupric oxide)

    • iron(II) chloride FeCl2 and iron(III) chloride FeCl3

      • (once called ferrous chloride and ferric chloride)

    • iron(II) oxide FeO, iron(III) oxide Fe2O3 and diiron(II) iron(III) oxide, Fe3O4

      • (once called ferrous oxide and ferric oxide and tri–iron tetroxide)

      • Historic note: ...ous was the lower oxidation state, ...ic the higher.

    • vanadium(II) sulphate for VSO4 or V2+SO42– and vanadium(III) sulphate V2(SO4)3

    • sulphur(IV) oxide SO2 (sulphur dioxide) and sulphur(VI) oxide, SO3 (sulphur trioxide)

    • nitrogen(I) oxide N2O (dinitrogen oxide) nitrogen(II) oxide NO (nitrogen monoxide), nitrogen(IV) oxide NO2 (nitrogen dioxide) and nitrogen(V) oxide N2O5 (nitrogen pentoxide).

    • transition metal complex cations e.g.

      • diaquatetraamminecopper(II) ion, [Cu(H2O)2(NH3)4]2+

        • (water and ammonia are electrically neutral ligands attached to the central Cu2+ ion)

  • 4.3: For elements (metal or non–metal) combined with oxygen or other more electronegative element, giving an anion, the ion name ends in ...ate with the prefix derived from the elements name. In such cases the oxygen carries the negative oxidation state of (–2) or chlorine (–1) e.g.

    • vanadate(V) ion, VO43–,

    • manganate(VI) ion, MnO42–, manganate(VII) ion, MnO4, (was called the permanganate ion)

    • sulphate(IV) ion, SO32– (sulphite) and sulphate(VI) ion,SO42– (sulphate)

    • nitrate(III), NO2 (nitrite) and nitrate(V), NO3 (nitrate)

      • Historic note: ...ite was the lower oxidation state, ...ate the higher.

    • chlorate(I), ClO, chlorate(VII), ClO4 etc. oxygen is more electronegative than chlorine.

      • (once called the hypochlorite ion and the perchlorate ion respectively)

    • transition metal complex anions e.g.

      • tetrachlorocuprate(II) ion, [CuCl4]2– (oxidation states Cu +2, Cl –1)


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