Topic 17: Transition Metals and their Chemistry A2

For a neutral Period 4 d-block atom, place electrons after [Ar] into 4s and 3d, remembering the accepted Cr and Cu arrangements. For a positive ion, remove electrons from 4s before 3d even though 4s filled first.

Count the electrons implied by atomic number minus positive charge. For example, Fe2+ must contain 24 electrons: 18 in [Ar] and six in 3d.

Vanadium is [Ar] 3d3 4s2. It has five electrons beyond [Ar], so oxidation states from +2 through +5 are accessible; +5 corresponds to removal or bonding involvement of all five 3d and 4s electrons.

The relative stability of particular states also depends on electronic arrangement: Mn2+ has a stable half-filled 3d5 subshell, while Fe3+ is 3d5.

To decide whether a species can act as a ligand, locate an available lone pair and a donor atom able to approach the metal. Ethane has no suitable lone pair, whereas an amine does.

Show the bond initially with an arrow from the ligand donor atom toward the metal ion: donor → metal. Once formed, it is a covalent bond; the arrow records the source of the electron pair.

A ligand with several suitable donor atoms can make several coordinate bonds to the same metal. The donor atoms, not unrelated lone pairs elsewhere in the ligand, determine its denticity.

In [Fe(H2O)5SCN]2+, five water ligands contribute zero charge and SCN- contributes -1, so iron is +3. Square brackets enclose the whole coordination entity and the overall charge is written outside.

Aqueous transition-metal ions commonly exist as aqua complexes and are often coloured. Examples include green [Cr(H2O)6]3+, pale green Fe2+, green Ni2+ and blue Cu2+ complexes.

Colour can support identification or show that oxidation state or ligand environment has changed, but observations should be linked to a named or formula species rather than colour alone.

When ligands approach a transition-metal ion, their interactions split the five d orbitals into groups with different energies. An electron can absorb a photon whose energy matches this gap and move from a lower to a higher d level.

The absorbed wavelength is removed from the incident visible light. The colour seen is produced by the wavelengths transmitted or reflected, so it is complementary to the light absorbed.

Cu2+ is 3d9, so a d electron can be promoted between split levels and its aqua complex is coloured. Cu+ is 3d10, so that transition is unavailable and its comparable complexes are colourless.

A colour change is evidence that the electronic environment changed, but identify the chemistry from reagents and equations. Air oxidises pale-green Fe2+ to brown Fe3+; chloride can replace water and change both colour and shape without changing metal oxidation number.

Monodentate means that one ligand particle attaches through one donor atom and forms one coordinate bond to a metal ion. Water uses one oxygen lone pair; ammonia uses one nitrogen lone pair.

A coordination number of six commonly gives an octahedral complex: six donor atoms point toward the central metal along three perpendicular axes. Adjacent ligand–metal–ligand angles are 90°, and opposite positions are 180°.

Small monodentate ligands such as H2O, OH- and NH3 can pack six donor atoms around the centre, as in [Cu(H2O)6]2+ or [Co(NH3)6]2+. Each metal–ligand link is coordinate covalent.

In a 3D drawing, show four bonds in one plane plus one bond projecting forward and one backward, with all six donor atoms connected to the metal.

Chloride ions are larger than water or ammonia ligands. Around some transition-metal ions, only four chloride donor atoms can fit without excessive crowding, so the coordination number falls from six to four and a tetrahedral complex forms.

[CuCl4]2- and [CoCl4]2- are four-coordinate tetrahedral complexes. Their ideal bond angles are about 109.5°, unlike the 90° adjacent angles of an octahedral aqua complex.

Cis-platin, [Pt(NH3)2Cl2], is a square-planar platinum(II) complex: the four donor atoms lie in one plane with adjacent angles of 90°. In the cis isomer the two chloride ligands occupy adjacent positions; in the trans isomer they are opposite.

Cis-platin is used in cancer treatment and must be supplied as the single cis isomer. Replacement of its two nearby chloride ligands enables binding at two sites on DNA, disrupting replication; the trans arrangement cannot make the same effective two-point link.

Identify separate donor atoms with available lone pairs that can reach the same metal centre. A bidentate ligand makes a chelate ring with two coordinate bonds; EDTA4- can wrap around a metal and occupy six coordination sites.

Haemoglobin contains an Fe2+ complex held by a polydentate ligand. An additional coordination site can bind O2 reversibly so oxygen can be transported in the blood.

Carbon monoxide acts as a ligand and binds more strongly to the Fe2+ centre than oxygen. It replaces bound O2 by ligand exchange, occupying the site and reducing haemoglobin's ability to carry oxygen.

Stepwise reduction therefore gives yellow → blue → green → purple. Link every observation to an oxidation state or species; an intermediate mixture of yellow and blue can also appear green without being pure V3+.

For each step, calculate E°cell = E°(vanadium reduction) - E°(reducing-agent reduction couple). A positive result predicts that step is feasible; repeat independently for the next oxidation state.

Iron metal with E°(Fe2+/Fe) = -0.44 V can reduce V3+ to V2+ because -0.26 - (-0.44) = +0.18 V. Tin with E°(Sn2+/Sn) = -0.14 V cannot: -0.26 - (-0.14) = -0.12 V.

Use the relevant reduction potentials to show that Zn gives positive Ecell values for the stated reductions. Conditions matter: peroxide oxidises Cr3+ to chromate in alkaline solution, and acidification then forms dichromate.

Record the initial solution, precipitate colour, whether it changes on standing, and whether it dissolves in excess. Write an equation for the actual process rather than reporting colour alone.

Cr(OH)3 is also amphoteric and dissolves in excess OH- to form a green hydroxo complex. Amphoteric means reacting with both acid and base; it is not merely 'soluble in excess'.

Replacing ligands changes the d-orbital splitting and therefore colour. Replacing six small water ligands by four larger chloride ions also lowers coordination number from six to four and changes shape.

Replacing several monodentate ligands by one bidentate or hexadentate ligand often releases several small ligand molecules into solution. The number of independently moving particles increases, so ΔSsystem is positive and the chelated complex is thermodynamically more stable.

The left side has two solute species, while the right contains one complex plus six liberated water molecules. For another equation, count the particles shown rather than assuming that all polydentate ligands give the same numerical change.

Both provide an alternative pathway with lower activation energy and are regenerated overall. Transition metals are suited to catalytic redox cycles because they can change oxidation state, while metal surfaces can adsorb reactants.

Finely divided catalyst exposes more active sites, so more reactant particles can be adsorbed at once and the reaction rate can increase.

Vanadium(V) oxide catalyses SO2 oxidation through two redox steps. SO2 first reduces vanadium from +5 to +4 while becoming SO3; oxygen then oxidises vanadium(IV) back to +5, regenerating V2O5.

Adding the two steps cancels V2O5/V2O4 and gives SO2 + 1/2 O2 → SO3. The catalyst participates but has no net consumption.

Add all mechanism steps and cancel species that appear on both sides. A catalyst appears as a reactant early and a product later; an intermediate appears as a product early and a reactant later.

The sequence replaces a slow direct reaction with faster steps having lower activation barriers. Because the catalyst is regenerated, it is absent from the overall equation.

The direct reaction between I- and S2O8^2- is slow because both ions are negative and repel. Fe2+/Fe3+ ions provide two favourable electron-transfer encounters.

Adding the steps gives S2O8^2- + 2I- → 2SO4^2- + I2. Fe2+ is consumed then regenerated; Fe3+ is the intermediate oxidation state.

In acidic permanganate–ethanedioate reaction, Mn2+ is both a product and a catalyst. Little Mn2+ is present initially, so the reaction is slow; as Mn2+ accumulates the catalytic route speeds up. Later the rate falls as reactants are depleted.

Mn3+ is the intermediate and Mn2+ is regenerated. Combining suitable multiples gives the uncatalysed overall stoichiometry while exposing the faster pathway.

Write the balanced ligand-substitution or complex-formation equation with correct brackets and overall charges. For a substitution step, show displaced ligands or counter-ions explicitly rather than treating the complex as an uncharged formula.

Calculate percentage yield from the limiting reagent. Discuss losses during transfer, incomplete crystallisation and product remaining dissolved; observations such as colour support formation but do not alone prove purity.