Quick click to Introduction * Sc * Ti * V * Cr * Mn * Fe * Co * Ni * Cu * Zn * Ag/Pt etc.

10.1. Introduction - section 1 - 'description criteria'  READ 1st

• The elements scandium to zinc (Z = 21 to 30) are known as the 3d block of elements or 3d-block of metals because here the first of the possible d sub-shells is progressively filled (3d-block - first row of the d-blocks). (See rules on how to work out electron configurations)

• The transition elements are group of industrially important metals mainly due to their strong inter-atomic metallic bonding giving them generally high melting/boiling points and high tensile strength.

• These-called 'transition metal characteristics' arise from an incomplete d sub-shell energy level but scandium and zinc are not true transition metals i.e. Ti to Cu are the real transition elements (reasoning later).

• Note that physically, zinc is low melting and a lower tensile strength compared to the others in the 3d block.

• Although scandium is physically typical of a transition metal e.g. high melting point and high tensile strength, chemically, scandium only forms a single and colourless triple charged ion (Sc³⁺). Therefore like zinc (only Zn²⁺), shows non of the typical characteristics of transition metal chemistry e.g. variable oxidation state, coloured complex ions, catalytic properties of the metal or ion. This is all explained in detail later.

• Therefore probably the best definition of a transition metal is an element which forms at least one ion with an incomplete d sub-shell containing at least one electron. How this relates to variable oxidation state and coloured complex ions is elaborated further in section 10.2 and the subsequent sections on the individual metals (links below) and some of the appendices. Zinc (Zn²⁺, [Ar]3d¹⁰) and scandium (Sc³⁺, [Ar]3d⁰) cannot meet this criteria.

• The presence of the partially-filled d sub-shells of electrons gives transition elements properties which are not in general possessed by the main group elements, namely Groups 1-7 and 0, BUT, there are similarities with other metals, particularly in Groups 2, 3 and  4.

• PLEASE NOTE the following about these Transition Elements notes:

  • All the reactions are described with visual observations and full ionic equations whether redox reactions or not.

  • I have made extended use of standard electrode potentials to indicate not only the relative oxidising/reducing power of a half-cell reaction, but also to argue for the thermodynamic feasibility of a reaction.

    • However, if you have done little on using electrode potentials you should still be able to follow the reaction descriptions and the accompanying equations.

    • Electrode potentials (half-cell potentials) and how to work out E^Ø_reaction (which must be a +ve voltage for the reaction to be feasible) are all explained in Equilibria Part 7 Redox Equilibria and Electrode Potentials. How to use half-cell reactions to work out full redox equations is explained on Redox Reactions Part II section 6 Constructing redox equations.

  • In the latest Periodic Table convention, the 3d-block elements are considered the 'head elements' of Groups 3-12.

    • Groups 1-2 remain unchanged but Groups 3-7 and 0 become Groups 13-18. I tend to retain the Groups 3-7 and 0 convention for the moment!

    • Group number	3	4	5	6	7	8	9	10	11	12
      Period 4	21, Sc	22, Ti	23, V	24, Cr	25, Mn	26, Fe	27, Co	28, Ni	29, Cu	30, Zn
      Period 5	39, Y	40, Zr	41, Nb	42, Mo	43, Tc	44, Ru	45, Rh	46, Pd	47, Ag	48, Cd
      Period 6	57, La	72, Hf	73, Ta	74, W	75, Re	76, Os	77, Ir	78, Pt	79, Au	80, Hg

    • There are actually many 'vertical' chemical similarities in a 'classic' periodic table way of thinking to justify this latest 'numbering' of the Periodic Table. e.g.

      • 'Modern Group 3': Scandium and yttrium have very similar and relatively simple M³⁺ ion chemistries.

      • 'Modern Group 10': Nickel, palladium and platinum are good hydrogenation catalysts. They all tend to form more square planar complexes than other transition elements.

      • 'Group 11': Copper, silver and gold are relatively unreactive metals in terms of corrosion. They form linear complexes like the cationic, [Ag(NH₃)₂]²⁺ or the anionic [CuCl₂]^- and [Au(CN)₂]^-. All three are extremely good conductors of heat and electricity.

      • 'Modern Group 12': Zinc and cadmium chemistry is mainly about the M²⁺ ion.

      • From 'Group 3 to 7' the maximum known oxidation state known (albeit in some pretty unstable compounds at times) is equal to the 'new' group number i.e. Sc/Y/La (+3) to Mn/Tc/Re (+7).

      • The discontinuity of atomic/proton number from lanthanum to hafnium on period 6 is due to the insertion of the 4f-block elements ₅₈Ce to ₇₁Lu.

• Comparison of certain properties of the 3d block of metals and other elements for Z = 1 to 38 particularly the preceding Group 1 metal potassium and the Group 2 metal calcium.

  • Periodicity plots for elements 1 to 38 Look for Z = 21 (Sc) to 30 (Zn)

    •  Note: There is no direct link back to here, so used <== 'back' on browser bar.

  • Melting/boiling points: Generally higher than other elements in period 4.

  • 1st ionisation energy: The 3d block 1st ionisation energies tend to increase from left to right and fit in with the general pattern for period 4.

  • Pauling electronegativity: The 3d-block values range from a relatively low 1.3 to 1.9 and fit in with the general pattern of increasing value across period 4.

  • Atomic radius: 3d-block elements have similar values and significantly less than for potassium and calcium.

  • Electrical/thermal conductivity: The 3d-block are quite good conductors of electricity/heat and very good in the case of copper (ditto silver Ag below Cu).

  • Density: 3d-block range from 3.0 to 8.9g/cm³ and significantly more than for potassium (0.86) and calcium (1.5).

  • Periodicity plots for elements 1 to 96 if you want to look for the 4d and 5d blocks!

• Other comparison points of the elements titanium to copper (true transition metals) with nearby metals.

  • Potassium (+1), calcium (+2) and scandium (+3) only have one oxidation state in compounds, whereas Ti to Cu have compounds in at least at least three oxidation states, even if some are not very stable!

Quick click to Introduction * Sc * Ti * V * Cr * Mn * Fe * Co * Ni * Cu * Zn * Ag/Pt etc.

10.2. Introduction - section 2 - information and general characteristics of the 3d block of Metals Sc-Zn READ 2nd and then 'whatever'!

Data Table 1 - summary of selected properties

CLICK for a more detailed data table 2 summary

10.2a. General Physical Characteristics

• The transition metals are the most important structural metals for industry due to their strength arising from the strong inter-atomic forces (see metal bonding and alloy structure).

• The strong bonding is due to small ionic radii and at least 3 delocalised 3d or 4s electrons contributing to the bonding which accounts for their high tensile strength, malleability (can be readily beaten into shape) and ductility (can be drawn into wire).

• They are silvery-grey solids apart from the dark orange of copper.

• They generally have high melting/boiling points and densities and readily mix with themselves or other elements to give a huge variety of alloys with a wide range of uses based on varied hardness, strength, malleability and anti-corrosion properties. 

• There is a general, but small, contraction of the atomic/ionic radii across the series as the atomic/proton number rises, i.e. an increasing positive attractive force on the outer electrons of the same sub-shells (3d and 4s).

10.2b. General Chemical Characteristics and electron configurations

21 Scandium, Sc	1s22s22p63s23p63d14s2	[Ar]3d4s
22 Titanium, Ti	1s22s22p63s23p63d24s2	[Ar]3d4s
23 Vanadium, V	1s22s22p63s23p63d34s2	[Ar]3d4s
24 Chromium, Cr	1s22s22p63s23p63d54s1	[Ar]3d4s
25 Manganese, Mn	1s22s22p63s23p63d54s2	[Ar]3d4s
26 Iron, Fe	1s22s22p63s23p63d64s2	[Ar]3d4s
27 Cobalt, Co	1s22s22p63s23p63d74s2	[Ar]3d4s
28 Nickel, Ni	1s22s22p63s23p63d84s2	[Ar]3d4s
29 Copper, Cu	1s22s22p63s23p63d104s1	[Ar]3d4s
30 Zinc, Zn	1s22s22p63s23p63d104s2	[Ar]3d4s

The chemistry is dominated by the behaviour of the 3d electrons. The 3d block corresponds to the filling of the 3d sub-shell of electrons, best appreciated by the 'box diagrams' of their electron structure.

Each half-arrow is an electron, which tend to singly occupy the sub-orbitals as much as possible to minimise repulsion (Hund's Rule of maximum multiplicity). 

The outer electrons of the element are either in the 3d or 4s sub-shell. The 4s sub-shell is initially filled by potassium [Ar]4s¹ and calcium [Ar]4s².

The electron arrangement for each element from Sc to Zn is also given at the start of each individual metal section in terms of s, p and d notation.

All 10 elements, Sc to Zn are 3d block elements (the filling of the 3d sub-shell) BUT a true transition element is one in which there is an incomplete d sub-shell holding at least one electron in one or more chemically stable ions (Ti to Cu). For 3d block metals this means at least one stable ion with the configuration within the range [Ar]3d¹ e.g. Ti³⁺ to [Ar]3d⁹ e.g. Cu²⁺ and so excludes scandium and zinc. Zinc only forms Zn²⁺, [Ar]3d¹⁰ and scandium only forms Sc³⁺, [Ar]3d⁰, so neither can meet this criteria for a true transition metal. See theory of colour in transition metal complexes.

There are two apparent anomalies in the electron configuration sequence from left to right as the 3d sub-shell energy level is filled:

Cr is not 3d⁴4s² and Cu is not 3d⁹4s²

because inner half-filled or fully-filled filled 3d sub-shells seem to a little lower in energy.

The total number of outer 3d/4s electrons is equal to the maximum oxidation state from Sc(+3) to Mn(+7) and there are many stable compounds exhibiting these maximum oxidation states.  After Mn there is significantly less stability of species with the metal in oxidation states above +3 for Fe and Co, and above +2 for Ni, Cu and Zn.

The four 'classic' chemical characteristics (but NOT unique to transition metals) are ...

(1) Complex formation: Appendix 2  offers an introduction as well as numerous examples 'en route' particularly from Ti to Cu.

(2) Formation of coloured ions: Appendix 4  offers an introduction to the origin of the colour in transition metal complex ions as well as examples 'en route' from colourless 'non-transition' Sc³⁺ complexes, coloured Ti^{II, III, IV} to Cu^II 'true transition' complexes and finally colourless 'non-transition' Zn²⁺ complexes at the end of the 3d-block.

(3) Variable oxidation state - variable valency:

• From Sc to Mn the maximum oxidation state is determined by the total maximum number of 3d and 4s electrons. After that, things get very complicated but the maximum tends to fall down to +2 for zinc after +3 for Fe and Co (there are some higher oxidation state species, but not that common and not that stable in aqueous media). See Table 10d-2 Examples of compounds/complexes for different oxidation states of the 3d-block elements.

• The relative ease of oxidation state change for Ti to Cu AND the maximum oxidation state formed by Sc to Mn, is partly explained by considering the ionisation energies involved and a comparison with Group 1, 2 and 3 metals helps too.

  • In the sequences below the atoms and ionised species are all in the gaseous state as is the convention for ionization energy data.

  • The energies (kJmol^-1) required to remove the next most loosely bond electron to give the next more highly charged ion (the next higher oxidation state) are shown as a sequence.

  • Only for the first example, potassium, are the full formal equations shown.

  • The successive ionisation energy sequences for Group 1 (potassium), Group 2 (calcium), the 3d-block (e.g. titanium) and Group 3 (gallium) are now considered for period 4.

  • Gp1: K_(g) == +418 ==> K⁺_(g) == +3070 ==> K²⁺_(g) 

    • they would be formally written as:

      • for the 1st ionisation energy: K_(g) - e^- ==> K⁺_(g)  

      • and for the 2nd ionisation energy: K⁺_(g) - e^- ==> K²⁺_(g)  

  • Gp2: Ca_(g) == +590 ==> Ca⁺_(g) == +1150 ==> Ca²⁺_(g) == +4940 ==> Ca³⁺_(g) 

  • 3d-block: e.g. Ti_(g) == +661 ==> Ti⁺ == +1310 ==> Ti²⁺ == +2720 ==> Ti³⁺ == +4170 ==> Ti⁴⁺ == +9620 ==> Ti⁵⁺

  • Gp3: Ga_(g) == +577 ==> Ga⁺ == +1980 ==> Ga²⁺ == +2960 ==> Ga³⁺ == +6190 ==> Ga⁴⁺ 

• So, for Groups 1, 2 and 3, the ionisation energy dramatically rises after the outer shell of s or p electrons are removed, i.e. a very stable electronic noble gas structure ([Ar], 1s²2s²2p⁶3s²3p⁶) is left. This gives a maximum positive stable oxidation state equal to the group number. The energy required (very endothermic) to make Na²⁺, Ca³⁺ and Ga⁴⁺ is too high to be compensated by exothermic bond formation with other elements like oxygen or chlorine etc.

  • Also note that intermediate lower oxidation states Ca⁺ and Ga²⁺ (and  Ga⁺?)are not very stable either.

  • I'm afraid ionisation energies and electron arrangements are not the only factors to be considered, you also need to study the Born Haber Cycle in some detail to prove this, but not here and not usually on a pre-university course!

• For the transition metals, at first, the successive ionisation energies rise relatively gradually, due to the 3d/4s electron levels being of similar energy. When all the outer s and d electrons are removed to leave an [Ar] core, there is, as with Groups 1-3 etc., a dramatic rise as an electron must be removed from the inner very stable noble gas (argon) core.

  • Therefore Ti has a maximum oxidation state of +4, but +2 and +3 species are also formed, but NOT +5 compounds.

  • This does mean however, across the 3d-block, there is the potential for very high oxidation states if there are enough 3s and 3d electrons that can be energetically favourably removed or become involved in stable bonding e.g. Mn has a maximum oxidation state of +7 by 'removing'(*) or 'sharing' the outer 3d⁵4s² electrons. (see data table).

    • Similarly you can argue that the maximum oxidation states for vanadium would be +5 and chromium +6, as is indeed is the case!

    • After manganese, things get complicated and there is a general decrease from Mn (+7) to Zn (+2) in the maximum possible higher oxidation states, and many higher oxidation state compounds of Fe, Co, Ni and Cu are unstable and uncommon.

  • (*) Of course e.g. in manganese (VII) compounds, 7 electrons are not removed to give an Mn⁷⁺ ion, but all 7 outer electrons are involved in the bonding and, unlike calcium and gallium, true transition metals form many stable compounds of the 'intermediate' oxidation states e.g. manganese forms +2, +3, +4, +6, +7 oxidation sate compounds.

  •  This is due to closeness of the energies of the 3d sub-shell electrons and the stabilising influence of ligand molecules like water or ammonia and ligand ions like chloride or cyanide. Vacant 3d orbitals (and 4s/4p orbitals too) can accept pairs of electrons to for stable dative covalent bonds.

    • I'm afraid arguments for the characteristic variable oxidation states of transition metals based on ionisation energies and the similar energies of the 3d orbitals is a bit limited, but better than nothing!

    • I have done detailed notes on oxidation state/oxidation number and redox reactions - a lot of which come up in transition metal chemistry.

(4) Catalytic activity by the elements and their compounds:

• Many examples are quoted in the two pages of detailed notes and the theory of heterogeneous and homogeneous catalysis is outlined in Appendix 6.

Quick click to Introduction * Sc * Ti * V * Cr * Mn * Fe * Co * Ni * Cu * Zn * Ag/Pt etc.

10.3. Chemistry of Scandium Sc, Z=21, 1s²2s²2p⁶3s²3p⁶3d¹4s² 

• Sc data table 1 summary * extended scandium data table 2 * Scandium & electrode potential chart of 3d-block

• Summary of some complexes-compounds & oxidation state of scandium compared to other 3d-block elements

• Scandium is not a significant metal.

• Scandium's chemistry is entirely based on the +3 oxidation state (Sc³⁺), the result of losing the outer 3d and 4s electrons.

• So it forms a typical series of binary compounds with non-metals e.g. Sc₂O₃,  ScCl₃ etc.

• The Sc³⁺ ion has an empty sub-shell, 3d⁰, which does not allow the electronic transitions which account for the colour in transition metal compounds (see Appendix 4. complex ion colour theory).

• The aqueous octahedral hexa-aqua ion, [Sc(H₂O)₆]³⁺ is therefore colourless and with no other oxidation state possible i.e. no ion with an incomplete 3d sub-shell with at least one electron, although a member of the 3d-block, scandium is NOT a true transition metal.

  • Apart from Sc³⁺ and Zn²⁺ all the other M²⁺ or M³⁺ hexaaquaions are coloured.

• The aqueous ion Sc³⁺ forms a white scandium hydroxide ppt. with alkali, Sc(OH)₃, which is a basic oxide and not amphoteric i.e. it does not dissolve in excess alkali, but it readily dissolves in acids to form salts e.g. to form scandium chloride, scandium nitrate or scandium sulphate ...

  • Sc(OH)_3(s) + 3HCl_(aq) ==> ScCl_3(aq) + 3H₂O_(l)

  • Sc(OH)_3(s) + 3HNO_3(aq) ==> Sc(NO₃)_3(aq) + 3H₂O_(l)

  • 2Sc(OH)_3(s) + 3H₂SO_4(aq) ==> Sc₂(SO₄)_3(aq) + 6H₂O_(l)

  • The chemistry of scandium is not very colourful or exciting!

  • The equations are similar to those for aluminium hydroxide i.e. you can substitute Al for Sc.

Quick click to Introduction * Sc * Ti * V * Cr * Mn * Fe * Co * Ni * Cu * Zn * Ag/Pt etc.

10.4. Chemistry of Titanium Ti, Z=22, 1s²2s²2p⁶3s²3p⁶3d²4s² 

• Ti data table 1 summary * extended titanium data table 2 * Titanium & electrode potential chart of 3d-block

• Summary of some complexes-compounds & oxidation sates of titanium compared to other 3d-block elements

• Titanium is a very important metal for various specialised uses.

• Titanium alloys are amongst the strongest of metal alloys. There is a GCSE note about the bonding and structure of alloys on another page.

  • It is used in aeroplanes, in nuclear reactor alloys and for replacement hip joints.

  • With a lighter density of 4.4 g/cm³ compared to steel (~7.9 g/cm³) its just as strong as steel and with the added advantage of being unreactive towards oxygen and water at room temperature so does not suffer the rusting of iron corrosion.

  • Titanium(IV) oxide, TiO₂, is an important white pigment used in the paints industry.

    • Note that Ti4+ has a [Ar]3d⁰ structure, hence, with no 3d electrons it is colourless (see colour theory).

• Titanium extraction and Ti(IV) CHEMISTRY

• It is more difficult  to extract from its ore than other more common metals so is not cheap!

  • Titanium is extracted from the raw material rutile ore which contains titanium dioxide. This is a high melting ionic compound Ti⁴⁺(O^2-)₂.

  • Carbon reduction of the oxide to the metal is not that practical due to titanium carbide formation so the titanium(IV) oxide is initially converted to titanium(IV) chloride which is then reduced to the metal with a more reactive metal in a displacement reaction.

    • Tungsten (W), another transition metal, cannot be obtained from reduction of its oxide for the same reason.

  • The rutile titanium oxide ore is heated with carbon and chlorine to make titanium(IV) chloride

    • TiO₂ + 2Cl₂ + C ==> TiCl₄ + CO₂

  • After the oxide is converted into TiCl₄ which is then reacted with sodium or magnesium to form titanium metal and sodium chloride or magnesium Chloride. The sodium and magnesium act as the reducing agent in this batch process.

    • This reaction is carried out in an atmosphere of inert argon gas so non of the metals involved becomes oxidised by atmospheric oxygen.

      • TiCl₄ + 2Mg ==> Ti + 2MgCl₂  or  TiCl₄ + 4Na ==> Ti + 4NaCl

    • These are examples of metal displacement reactions e.g. the less reactive titanium is displaced by the more reactive sodium or magnesium.

    • Overall the titanium oxide ore is reduced to titanium metal (overall O loss from ox. state +4, oxide => metal with ox. state 0)

    • TiCl₄ is covalent liquid which (i) hydrolyses back to the oxide in water and (ii) dissolves in conc. hydrochloric acid to form the hexachlorotitanate(IV) complex ion.

      • (i) TiCl_4(l) + 2H₂O_(l) ==> TiO_2(s) + 4HCl_(aq/g) (fumes in air!)

      • (ii) TiCl_4(l) + 2Cl^-_(aq) ==> [TiCl₆)]^2-_(aq) (typical complex anion)

  • When titanium(IV) compounds are dissolved in water of acid the oxo-cation [TiO]²⁺ is formed.

• TITANIUM(III) CHEMISTRY

• Titanium(III) compounds can be obtained from Ti(IV) salts by using a zinc/dil. sulphuric acid reduction agent.

  • colourless Ti(IV), [TiO]²⁺ ==> Ti(III) in acid solution giving purple [Ti(H₂O)₆)]³⁺_(aq),

  • but it is readily oxidised back to Ti(IV) by dissolved oxygen from the atmosphere (see electrode potential chart TiO²⁺/Ti³⁺ +0.10V is less positive than O₂+H⁺/H₂O +1.23V in acid solution).

  • TiCl₃ is a violet solid.

• TITANIUM(II) CHEMISTRY

• TiCl₂ is a black solid.

• The violet? [Ti(H₂O)₆)]²⁺ ion can be formed by reducing Ti(IV) or Ti(III) with a metal/acid mixture but it is very unstable in redox terms.

• Ti²⁺ will reduce water to hydrogen and it is rapidly oxidised by air - dissolved oxygen and cannot exist in aqueous solution (see electrode potential chart Ti³⁺/Ti²⁺ -0.37V is less positive than O₂+H⁺/H₂O +1.23V in acid solution).

• Comparison with a Group 4 metal e.g. tin

  • Tin only exhibits oxidation states of +2 and +4, there is no intermediate +3 compounds.

  • The compounds are usually colourless.

  • Tin is a much weaker metal physically with much lower melting/boiling point.

Quick click to Introduction * Sc * Ti * V * Cr * Mn * Fe * Co * Ni * Cu * Zn * Ag/Pt etc.

10.5. Chemistry of Vanadium V, Z=23, 1s²2s²2p⁶3s²3p⁶3d³4s² 

• V data table 1 summary * extended vanadium data table 2 * Vanadium & electrode potential chart of 3d-block

• Summary of some complexes-compounds & oxidation states of vanadium compared to other 3d-block elements

• Vanadium is one of many transition metals alloyed with iron to make specialist steels.

• Vanadium(V) oxide, V₂O₅, is used as a catalyst in the 'Contact Process' in the production of sulphur trioxide for the manufacture of sulphuric acid.

  • The catalysing of the conversion of sulphur dioxide into sulphur trioxide is explained via change in oxidation state changes i.e. some classic transition metal chemistry.

  • 2SO_2(g) + O_2(g) ==> 2SO_3(g) 

  • The mechanism, somewhat simplified, goes via the catalytic cycle ...

    • (i) SO₂ + V₂O₅ ==> SO₃ + V₂O₄, then (ii) V₂O₄ + ¹/₂O₂ ==> V₂O₅ 

    • The vanadium changes oxidation state from +5 to +4 and back to +5 in the catalytic cycle, a classic combination of two characteristics of transition metals - variable oxidation state and catalytic properties.

    • This is an example of heterogeneous catalysis - reactants (g) and catalyst (s) in different phases.

• Vanadium shows a 'classic' display of variable oxidation states of varying colours when a solution of e.g. ammonium vanadate(V), is reduced by a zinc/dilute sulphuric acid mixture. do picture

  • Acidification changes the vanadate(V) ion into the pale yellow oxo-cation VO₂⁺ (oxovanadium(V) ion)

  • VO₄^3-_(aq) + 4H⁺_(aq) VO₂⁺_(aq) + 2H₂O_(l) [an acid-base reaction, NOT a redox change]

    • Note: Highly charged cations >3+ rarely exist as the simple 'hydrated' tetra or hexa-aqua ion.

    • The theoretical polarising power of the 'central metal ion' is so strong that they form oxocations (see above) or oxyanions e.g.

    • orange dichromate(VI) Cr₂O₇^2-, yellow chromate(VI) CrO₄^2-, purple manganate(VII) MnO₄^- etc.

    • For transition metals they may be coloured even if electronically the theoretical 'central metal ion'  has a noble gas structure e.g. [Ar] in its maximum oxidation state like V(V), Cr(VI) and Mn(VII).

    • These oxyanions are called charge transfer complexes and the theory is beyond pre-university chemistry.

  • Three successive reduction steps then follow to eventually give V²⁺ ions, shown as half-cell equations:

  • (i) V(V, +5) ==> V(IV, +4): VO₂⁺_(aq) + 2H⁺_(aq) + e^- VO²⁺_(aq) + H₂O_(l)

    • E^Ø_{half-cell potential} = +1.00V, pale yellow to the blue oxovanadium(IV) ion

  • (ii) V(IV, +4) ==> V(III, +3): VO²⁺_(aq) + 2H⁺_(aq) + e^- V³⁺_(aq) + H₂O_(l)

    • E^Ø_{half-cell potential} = +0.34V, blue to the green vanadium(III) ion

    • V³⁺ is actually the green hexaaquavanadium(III) ion, [V(H₂O)₆]³⁺

    • Both V(IV0 and V(III) species slowly oxidised by dissolved oxygen back to the V(V) compound in acid solution (see electrode potential comments later).

  • (iii) V(III, +3) ==> V(II, +2): V³⁺_(aq) + e^- V²⁺_(aq)

    • E^Ø_{half-cell potential} = -0.26V, green to the purple-violet vanadium(II) ion.

    • V²⁺_(aq) is powerful reducing agent and is unstable in the presence of air. Any dissolved oxygen will oxidise V²⁺_(aq) back to the vanadium(III) cation.

    • V²⁺ is actually the purple-violet hexaaquavanadium(II) ion, [V(H₂O)₆]²⁺

  • Note

    1. The standard electrode potential E^Ø_{Zn(s)/Zn2+(aq)} is -0.76V, so the reducing power of zinc is sufficient to effect any of the three vanadium oxidation state reduction changes described above.

       • See electrode potential chart VO₃⁺/VO²⁺ (+1.00V), VO²⁺/V³⁺ (+0.34V) and V³⁺/V²⁺ (-0.26V) are less positive than O₂+H⁺/H₂O +1.23V in acid solution so the vanadium(II/III/IV) cations can theoretically be oxidised back to vanadium(V) i.e. VO₂⁺.

       • Electrode potentials (half-cell potentials) and how to work out E^Ø_reaction, which must be a +ve voltage for the reaction to be feasible, are all explained in Equilibria Part 7 Redox Equilibria and Electrode Potentials. How to use half-cell reactions to work out full redox equations is explained on Redox Reactions Part II section 6 Constructing redox equations.

    2. The reduction occurs on the surface of the zinc metal i.e. the site of electron transfer and you can write the above reductions as a fully balanced complete equations ...

       • (i) 2VO₂⁺_(aq) + 4H⁺_(aq) + Zn_(s) ==> 2VO²⁺_(aq) + 2H₂O_(l) + Zn²⁺_(aq)

         • E^Ø_reaction = E^Ø_reduction - E^Ø_oxidation = +1.00 - (-0.76) = +1.76V

         • The half-cell reaction of the reduction will have the most +ve E^Ø_potential.

       • (ii) 2VO²⁺_(aq) + 4H⁺_(aq) + Zn_(s) ==> 2V³⁺_(aq) + 2H₂O_(l) + Zn²⁺_(aq)

         • E^Ø_reaction = +0.34 -(-0.76) = +1.10V

       • (iii) 2V³⁺_(aq) + Zn_(s ==> 2V²⁺_(aq) + Zn²⁺_(aq)

         • E^Ø_reaction = -0.26 - (-0.76) = +0.50V

         • BUT the vanadium(II) cation is unstable in the presence of dissolve oxygen in air.

         • ¹/₂O_2(g) + 2H⁺_(aq) + 2e^-   H₂O_(l) has a standard electrode potential of +1.23V,

         • so, for the vanadium(II) oxidation reaction ...

         • ¹/₂O_2(g) + 2H⁺_(aq) + 2V²⁺(aq) ==> 2V³⁺_(aq) + H₂O_(l)

         • E^Ø_reaction = E^Ø_reduction - E^Ø_oxidation = +1.23 - (-0.26) = +1.49V

         • hence the if left standing open to air, the violet V²⁺_(aq) solution will gradually change to a green V³⁺_(aq) solution and in turn V³⁺_(aq) will revert back to VO²⁺_(aq) in the presence of air because of oxidation by dissolve oxygen unless protected by an inert atmosphere. (see Redox Electrode Potential Chart, V²⁺/V³⁺ and V³⁺/VO²⁺ potentials are below that for O₂/H₂O/H⁺ potentials).

    3. You will see hydrogen formed simultaneously from the unavoidable metal-acid reaction.

       • Zn_(s) + 2H⁺_(aq) ==> Zn²⁺_(aq) + H_2(g)

• Does vanadium chemistry show an example of disproportionation?

  • This is just a little academic exercise using standard electrode potential data.

  • A disproportionation reaction is where a species in one oxidation state spontaneously and simultaneously changes into two species of different oxidation states - one higher and one lower in oxidation number.

  • Examples: disproportionation in manganese(VI) chemistry and disproportionation in copper(I) chemistry

  • Question: In terms of aqueous ions, is the disproportionation of vanadium(III) into vanadium(II) and vanadium (IV) feasible?

    • (i) VO²⁺_(aq) + 2H⁺_(aq) + 2e^- V³⁺_(aq) + H₂O_(l)   (E^Ø_VO2+/V3+ = +0.34V)

    • (ii) V³⁺_(aq) + e^- V²⁺_(aq)   (E^Ø_V3+/V2+ = -0.26V)

    • The disproportionation equation would be (iii) 2V³⁺_(aq) + H₂O_(l) V²⁺_(aq) + VO²⁺_(aq) + 2H⁺_(aq)

    • For equation (iii), (ii) will be the reduction half-cell equation and (i) reversed will be the oxidation half-cell reaction.

    • E^Ø_reaction = E^Ø_reduction - E^Ø_oxidation = = E^Ø_V3+/V2+ - E^Ø_VO2+/V3+ = (-0.26) - (+0.34) = -0.60V

    • showing the disproportionation is thermodynamically NOT feasible i.e. E^Ø_reaction is less than zero.

    • In fact what can actually happen is if you mix salt solutions of vanadium(IV) and vanadium(II) on an equimolar basis, you end up with a solution of vanadium(III) salts, a sort of 'anti-disproportionation' reaction!

Quick click to Introduction * Sc * Ti * V * Cr * Mn * Fe * Co * Ni * Cu * Zn * Ag/Pt etc.

10.6. Chemistry of Chromium Cr, Z=24, 1s²2s²2p⁶3s²3p⁶3d⁵4s¹ 

• Cr data table 1 summary * extended chromium data table 2 * Chromium & electrode potential chart of 3d-block

• Summary of some complexes-compounds & oxidation sates of chromium compared to other 3d-block elements

• Chromium is a hard white metal that is extremely resistant to chemical attack at room temperature. It is used in the production of extremely hard steel alloys e.g. ball bearings and coating steel articles for corrosion protection ('chrome plating') and chromium metal is an important component in 'stainless steel'.

• CHROMIUM(III) chemistry

• Chromium forms the stable green (greyish green almost violet sometimes?) chromium(III) ion, [Cr(H₂O)₆]³⁺_(aq).

• With aqueous ammonia (alkaline) or sodium hydroxide colour green chromium(III) hydroxide is precipitated.

  • Cr³⁺_(aq) + 3OH^-_(aq) ==> Cr(OH)_3(s) (but the structures can be quite complex)

  • or [Cr(H₂O)₆]³⁺_(aq)  + 3OH^-_(aq) ==> [Cr(OH)₃(H₂O)₃]_(s) + 3H₂O_(l) 

    • The hydroxide readily dissolves in acids to form salts,

    • Cr(OH)_3(s) + 3H⁺_(aq) ==> Cr³⁺_(aq) + 3H₂O_(l) 

      • or more elaborately: [Cr(OH)₃(H₂O)₃]_(s) + 3H₃O⁺_(aq) [Cr(H₂O)₆]³⁺_(aq)  + 3H₂O_(l)

      • or more simply Cr(OH)_(s) + 3H⁺_(aq) Cr³⁺_(aq)  + 3H₂O_(l)

      • thus showing amphoteric behaviour, since the hydroxide ppt. also dissolves in excess strong alkali to give a dark green solution and the hydroxide ppt. does not dissolve in the weak base aqueous sodium carbonate. However, it will dissolve in excess ammonia because a new green complex ion is formed. (more details on these reactions below)

  • VIEW ppts. with OH^-, NH₃ and CO₃^2-, & complexes, if any, with excess reagent.

• With aqueous sodium carbonate the hydroxide ppt. is formed (as above) and carbon dioxide because of the acidity of the hexaaquachromium(III) ion (see Appendix 1.):

  • *initially 2[Cr(H₂O)₆]³⁺_(aq) + CO₃^2-_(aq) ==> 2[Cr(H₂O)₅(OH)]²⁺_(aq) + H₂O_(l) + CO_2(g)   

  • this process of proton donation (deprotonation) continues until [Cr(OH)₃(H₂O)₃]_(s) is formed

  • No Cr₂(CO₃)₃ is formed because of the acid-base reaction above, due to the acidity of the chromium(III) ion. Note the similarly highly charged and small ions Al³⁺ and Fe³⁺ behave in the same way.

  • * the acidity of a the hexa-aquachromium(III) ion can be expressed as ...

    • [Cr(H₂O)₆]³⁺_(aq) + H₂O_(l) [Cr(H₂O)₅(OH)]²⁺_(aq) + H₃O⁺_(aq)

• With excess sodium hydroxide or ammonia, further complex ions are formed by ligand replacement reactions:

  • [Cr(H₂O)₆]³⁺_(aq) + 6OH^-_(aq) ==> [Cr(OH)₆]^3-_(aq) + 6H₂O_(l)  (from original hexa-aqua ion)

    • or [Cr(OH)₃(H₂O)₃]_(s) + 3OH^-_(aq) ==> [Cr(OH)₆]^3-_(aq) + 3H₂O_(l) (from hydroxide ppt.)

    • or more simply Cr(OH)_3(s) + 3OH^-_(aq) ==> [Cr(OH)₆]^3-_(aq)

      • showing amphoteric behaviour, since the hydroxide ppt. also dissolves in acid (above)

  • [Cr(H₂O)₆]³⁺_(aq) + 6NH_3(aq) ==> [Cr(NH₃)₆]³⁺_(aq) + 6H₂O_(l)   (from original hexa-aqua ion)

    • or [Cr(OH)₃(H₂O)₃]_(s) + 6NH_3(aq) ==> [Cr(NH₃)₆]³⁺_(aq) + 3OH^-_(aq) + 3H₂O_(l) (from hydroxide ppt.)

    • or more simply Cr(OH)_3(s) + 6NH_3(aq) ==> [Cr(NH₃)₆]³⁺_(aq) + 3OH^-_(aq)

  • The uncharged ligand molecules ammonia NH₃ and water H₂O are similar in size and ligand exchange occurs without change in co-ordination number. They all octahedral complexes with a co-ordination number of 6.

• Chromium(III) complexes are extremely numerous and varied, including many examples of isomerism. (see Appendix 2. and 3. for an introduction to complexes)

• Ionisation isomerism in chromium(III) chloride based on Cr³⁺, 3Cl^- and 6H₂O 

  • [Cr(H₂O)₆]³⁺(Cl^-)₃  (violet or grey-blue?)

  • [CrCl(H₂O)₅]²⁺(Cl^-)₂.H₂O  (pale green)

  • [CrCl₂(H₂O)₄]⁺ Cl^-.2H₂O  (dark green)

  • [CrCl₃(H₂O)₃]^0*.3H₂O ?  (brown?, this I found reference to on a Russian site, doesn't seem to be in textbooks? ^*the ⁰ to signify an overall electrically neutral complex can be omitted)

  • and this is not all, the 3rd one down with two chloride ligands can exist as cis (1) or trans (2) isomers (illustrated below, and also serve as models for representing the other complexes with cis/trans isomers).

  • 

  • 

• A similar case of isomerism occurs with the chromium(III) complexes with ammonia and chloride ligands shown above. All the complex ions above have a plane of symmetry and cannot exhibit optical isomerism.

  • Again, these are all octahedral complexes with a coordination number of 6.

• CHROMIUM(VI) chemistry

• When hydrogen peroxide is added to an alkaline chromium(III) solution, oxidation occurs to give the yellow chromate(VI) ion CrO₄^2- . 

  • 2Cr³⁺_(aq) + 3H₂O_2(aq) + 10OH^-_(aq) ==> 2CrO₄^2-_(aq) + 8H₂O_(l)

  • Redox changes: oxidation 2Cr(+3) ==> 2Cr(+6), reduction 6 O(-1) in 3H₂O₂ ==> 6(-2) in 6 of the 8H₂O, total of 6 'units' oxidation state change.

  • Both H₂O₂ and Cr(VI) compounds/ions are oxidising agents but in alkaline solution H₂O₂ is the stronger.

    • E^Ø = +?V details to add.

  • When the resulting solution from above is acidified with dilute sulphuric acid, the orange dichromate(VI) ion Cr₂O₇^2-  is formed.

  • The equilibrium is pH dependent. From 'Le Chatelier's Principle':

    • in more acidic solution, more H⁺, decrease pH ==> more orange (net change L to R) or in

    • more alkaline, less H⁺ (removed by OH^-), increase pH <= more yellow (net change R to L).

  • 2CrO₄^2-_(aq) + 2H⁺_(aq) Cr₂O₇^2-_(aq) + H₂O_(l) (no change in ox. state)

• The dichromate(VI) ion is reduced in two stages by a zinc/dilute sulphuric acid mixture.

  • Cr(VI, +6) ==> Cr(III, +3): Cr₂O₇^2-_(aq) + 14H⁺_(aq) + 6e^- 2Cr³⁺_(aq) + 7H₂O_(l)

    • orange (+6) ==> green (+3),  E^Ø = +1.33V 

  • Cr(III, +3) ==> Cr(II, +2): Cr³⁺_(aq) + e^- Cr²⁺_(aq)

    • green (+3) ==> blue (+2), E^Ø = -0.41V, so Cr(II) is readily oxidised by dissolved oxygen and can only be retained in an inert atmosphere.

  • Note the  E^Ø_{Zn(s)/Zn2+(aq)} is -0.76V, so the reducing power of zinc is sufficient to effect either of the two chromium oxidation state reduction changes.

    • The full redox equations for the reactions which happen on the surface of the zinc are:

    • Cr₂O₇^2-_(aq) + 3Zn_(s) + 14H⁺_(aq) 2Cr³⁺_(aq) + 3Zn²⁺_(aq) + 7H₂O_(l)

    • 2Cr³⁺_(aq) + Zn_(s) 2Cr²⁺_(aq) + Zn²⁺_(aq)

    • You will see hydrogen formed as a by-product of the zinc-acid reaction.

• Potassium dichromate(VI), K₂Cr₂O₇,  can be crystallised to high purity standard without water of crystallisation, and is a valuable 'standard' redox volumetric reagent.

  • E.g. It can used to titrate iron(II) ions in solution acidified with dilute sulphuric acid, using a redox indicator like barium diphenylamine sulphonate which is less readily oxidised than iron(II) ions, but once all the iron(II) ions are oxidised the indicator is oxidised to a blue colour. The iron(III) ions formed affect the indicator to give an inaccurate end point so phosphoric(V) acid is also added at the start to complex the Fe³⁺ ions as they form.

  • Cr₂O₇^2-_(aq) + 14H⁺_(aq) + 6Fe²⁺_(aq) ==> 2Cr³⁺_(aq) + 6Fe³⁺_(aq) + 7H₂O_(l)

  • See also fully worked examples of redox volumetric titration calculation questions.

• The dichromate(VI) ion is a strong oxidising agent - examples of oxidising action:

• See above for oxidation of iron(II) ions.

• It oxidises iodide ions to iodine.

• Cr₂O₇^2-_(aq) + 14H⁺_(aq) + 6I^-_(aq) ==> 2Cr³⁺_(aq) + 3I_2(aq) + 7H₂O_(l)

  • The released iodine can be titrated with standard sodium thiosulphate solution.

  • 2S₂O₃^2-_(aq)  +  I_2(aq)  ==>  S₄O₆^2-_(aq) + 2I^-_(aq) (black/brown ==> colourless endpoint)

  • This reaction between the released iodine and sodium thiosulfate can be used to estimate oxidising agents like dichromate(VI) ions. The iodine is titrated with standardised sodium thiosulphate (e.g. 0.10 mol dm^-3) using a few drops of starch solution as an indicator. Iodine gives a blue colour with starch, so, the end-point is very sharp change from the last hint of blue to colourless.

    • See also fully worked examples of redox volumetric titration calculation questions.

    • Soluble chromate(VI) salts give yellow solutions, but lead(II) ions give a yellow ppt. of lead(II) chromate(VI) and silver ions a dark red ppt. of silver chromate(VI).