Unit 5: Transition Metals and Organic Nitrogen Chemistry A2

Assign oxidation numbers before and after, name the element whose value changes, and state both the direction and values. In Hg + nitrate to Hg(II) + NO, Hg changes 0 to +2 and is oxidised, while N changes +5 to +2 and is reduced.

The same rules apply to s-, p- and d-block species. Oxidation and reduction occur together because electrons lost by one species are gained by another.

A single half-cell potential cannot be measured in isolation: a voltmeter measures a potential difference. Assigning the standard hydrogen electrode E° = 0.00 V supplies the common reference.

A more positive E° means the written reduction has a greater thermodynamic tendency relative to H+/H2; a more negative value means a weaker tendency to be reduced under standard conditions.

These fixed conditions make tabulated E° values comparable. Every ion participating in a half-equation must have the stated concentration; stoichiometry may require choosing solution concentrations carefully.

Connect it to the test half-cell through a salt bridge and a high-resistance voltmeter. The sign and magnitude of the measured emf relative to the 0.00 V reference define the other half-cell's E°.

Use a conducting solid that participates when the redox couple contains a metal, but use inert platinum when every redox species is gaseous or aqueous. Include all species and standard conditions.

Clean electrodes, prepare stated concentrations, connect the salt bridge so it touches both solutions, attach the voltmeter, record polarity and a stable voltage, and repeat while controlling temperature. For a Zn/Cu cell, use Zn in Zn2+ and Cu in Cu2+.

A practical bridge may be filter paper soaked in saturated KNO3. Choose ions that do not react with either half-cell.

Keep both tabulated half-equations written as reductions. The more positive couple operates as reduction; reverse the other half-equation for oxidation, balance electrons, then subtract its tabulated reduction potential.

For Ag+/Ag, +0.80 V, paired with Ti3+/Ti2+, -0.37 V, E°cell = +0.80 - (-0.37) = +1.17 V.

Write the oxidation half-cell on the left and reduction half-cell on the right, placing each electrode at an outer end. Include state symbols and every redox species.

E depends on temperature, gas pressure and the activities approximated by solution concentrations. Changing them shifts the half-cell equilibrium and changes the potential, so an E value cannot automatically be compared with E°.

For M^z+ + ze- ⇌ M, diluting M^z+ favours the left side and makes the reduction potential more negative. Apply the same equilibrium reasoning to every species in the actual half-equation.

Record actual conditions and allow the cell to equilibrate. A concentration change in either half-cell can alter Ecell and may even change a borderline prediction.

Write the proposed oxidation and reduction half-reactions, use their tabulated reduction potentials, calculate E°cell, and accept the proposed direction as thermodynamically feasible under standard conditions when E°cell is positive.

Test sequential reductions separately. A reagent may give positive E°cell for conversion to one oxidation state but negative E°cell for further reduction, explaining selective products.

n is electrons transferred in the balanced overall equation, F = 96500 C mol^-1, R = 8.31 J mol^-1 K^-1 and T is kelvin. A positive E°cell gives positive total entropy change and K greater than 1.

Balance the redox equation before choosing n, use volts as J C^-1, and exponentiate ln K only after evaluating the full expression.

A small positive or negative E°cell is especially sensitive to concentration changes. Concentrated reactants can shift both relevant half-cell equilibria enough to reverse the standard prediction.

Select the two couples that share the intermediate species, reverse the oxidation branch, balance and combine them, then calculate E°cell = E°(reduction branch) - E°(oxidation branch). A positive value predicts feasibility.

Verify that the shared starting species appears on the reactant side of both branches after directions are chosen; otherwise the calculation is not for disproportionation.

For a burette titre formed from two readings, include the uncertainty of both readings. Add percentage uncertainties for quantities combined by multiplication or division to estimate the total percentage uncertainty.

Compare the total percentage uncertainty with the reported precision and with differences between results. An answer quoted to digits far smaller than the experimental uncertainty implies unjustified precision.

Rinse apparatus appropriately, pipette the analyte, acidify permanganate work with sulfuric acid, titrate while swirling, add reagent dropwise near the endpoint, and obtain concordant titres.

Adding starch only near the iodine endpoint avoids a strongly bound iodine-starch complex that can make the endpoint slow or unclear.

Hydrogen, methanol and other hydrogen-rich fuels can be used. Hydrogen-oxygen cells form water at point of use; methanol is easier to store but its oxidation forms carbon dioxide.

Hydrogen is oxidised at the negative electrode and oxygen is reduced at the positive electrode in both electrolytes. Multiply the hydrogen half-equation by two before adding; H+ or OH- then cancels.

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.

The evidence is consistent with six π electrons delocalised around a planar six-carbon ring. Benzene may be drawn as a hexagon with a circle or as a Kekulé hexagon when equations and curly-arrow mechanisms require explicit electron movement.

Each carbon in benzene is trigonal planar and uses three orbitals to make σ bonds: two C-C bonds in the ring and one C-H bond. This leaves one unhybridised p orbital perpendicular to the ring plane on every carbon.

The six parallel p orbitals overlap sideways with both neighbours. Their electron density joins into one continuous delocalised π system above and below the carbon ring, containing six π electrons rather than three isolated electron pairs.

Because the π electrons are shared across all six carbon atoms, every C-C bond has the same order, length and strength, intermediate between a localised single and double bond.

Benzene can react with an electrophile, but formation of the non-aromatic intermediate has a larger activation-energy barrier. Substitution then restores the delocalised ring; addition would destroy its stabilisation in the product.

1. A curly arrow starts from the benzene π system and ends at E+, forming a C-E bond and a positively charged non-aromatic intermediate. 2. A base removes H+ from that carbon; the curly arrow from the C-H bond returns into the ring and restores delocalisation. The net result is replacement of H by E.

The catalyst is regenerated: FeBr4- + H+ → HBr + FeBr3, or AlCl4- + H+ → HCl + AlCl3. In nitration, HSO4- accepts H+ to regenerate H2SO4.

One lone pair on the phenol oxygen overlaps with the ring π system and donates electron density into it. The ring is therefore more electron-rich, especially at the 2, 4 and 6 positions, so it polarises bromine and undergoes electrophilic substitution much more readily than benzene.

For a structural formula, show the atom connectivity in groups; for a displayed formula, show every bond; for a skeletal formula, omit carbon labels and carbon-bound H atoms but show N, O and their attached H atoms. Classify an amine as primary, secondary or tertiary by the number of carbon groups bonded directly to nitrogen.

Butylamine represents a primary aliphatic amine and phenylamine a primary aromatic amine. In both, the lone pair can bond to $\ce{H+}$, attack an electron-poor carbon, or donate to $\ce{Cu^{2+}}$; their different basicities change how readily protonation occurs.

A small amine mixes with water because its polar C-N/N-H region can form hydrogen bonds with water. The nitrogen lone pair accepts a hydrogen bond from water, and an N-H bond can donate one. These favourable amine-water attractions replace the attractions disrupted on mixing.

To make butylamine by reduction, begin with butanenitrile: $\ce{CH3CH2CH2CN}$ becomes $\ce{CH3CH2CH2CH2NH2}$. Count the nitrile carbon as part of the product chain.

Heat the aromatic nitro compound under reflux with tin and concentrated hydrochloric acid. Six reducing equivalents replace the two nitro oxygens by hydrogen, converting $\ce{-NO2}$ into $\ce{-NH2}$. In the acidic mixture the amine is initially present as an arylammonium salt; adding alkali liberates the free aromatic amine.

Thus nitrobenzene gives phenylamine. Preserve every other substituent on the ring when applying the route to a substituted nitroarene, and show the same carbon skeleton before and after reduction.

Alkali converts phenol into the more electron-rich phenoxide ion. Its activated ring couples mainly at the para position when that position is available, giving a conjugated azo compound; extended delocalisation absorbs visible light, so the product is coloured.

Ammonia attacks the electron-deficient carbonyl carbon of an acyl chloride; chloride is displaced and an amide forms. A second ammonia molecule neutralises the hydrogen chloride, so excess ammonia is used.

A primary amine reacts by the same acylation pattern to give an N-substituted amide. For example, ethanoyl chloride plus butylamine gives N-butylethanamide; an additional amine molecule accepts the released proton.

Each amino acid contains both $\ce{-NH2}$ and $\ce{-COOH}$. Condensation between these groups forms a peptide (amide) link and eliminates water; repeating the change produces a polypeptide or protein.

For propenamide, $\ce{CH2=CHCONH2}$ gives a backbone bearing $\ce{-CONH2}$ side groups. The poly(ethenol) repeat unit has $\ce{-OH}$ side groups on an addition-polymer carbon backbone.

For an addition polymer, open the monomer C=C to a C-C single bond, keep every substituent on its original carbon, place the smallest repeating section in brackets, and draw a bond through each bracket edge.

For a polyamide, remove the small-molecule fragments at complementary functional groups and join carbonyl carbon to nitrogen. Choose bracket boundaries so repeating the unit reconstructs the uninterrupted chain.

A soluble laundry bag or liquid-detergent capsule uses a poly(ethenol)-based film that is strong enough while dry but disperses or dissolves when water penetrates. Water forms hydrogen bonds to the many O-H groups and competes with polymer-polymer attractions.

Polyalkenes of similar molar mass have only London forces between chains, so separating their chains generally needs less energy than overcoming the hydrogen-bond network in a polyamide.

Use the same concentration, path length and wavelength when comparing optical rotations. In acid the cationic form is favoured; in alkali the anionic form is favoured.

Write a decision sequence before testing. For every branch record the reagent, concentration or heating condition, observation and inference. Use fresh portions so acid, alkali, silver ions or oxidants from one test cannot create a false result in the next; include a blank or known comparison when an observation is subtle.

Work on a microscale, wear eye protection, control heating and volatile reagents, and follow separate disposal routes for heavy-metal, silver, oxidising and organic waste. Unknowns must be treated as hazardous until identified.

Work from independent constraints toward a small candidate set. Draw a candidate, predict its formula, functional tests and every spectral feature, then reject it if even one reliable observation conflicts. Equivalent atoms reduce the number of NMR environments, so signal count is not automatically atom count.

React a halogenoalkane or halogenoarene with magnesium in dry ether to form a Grignard reagent, $\ce{RMgX}$. The C-Mg bond is strongly polarised, so the carbon bonded to Mg behaves as a carbon nucleophile and attacks electron-deficient carbon.

For a carbonyl compound, nucleophilic addition first forms a magnesium alkoxide; acid hydrolysis then protonates oxygen to give the alcohol. Choose R by disconnecting the target C-C bond next to the future $\ce{OH}$ or $\ce{COOH}$ carbon.

After proposing a route, redraw every intermediate and audit each arrow: reagent, essential condition, product class, carbon count and selectivity. Check that a reagent does not also attack another group already present. If protection is not in the specification or supplied information, do not invent it.

Risk depends on both hazard and exposure. Prefer smaller scale or a less hazardous reagent when feasible; otherwise use a closed addition, condenser, cooling, fume cupboard or ignition control matched to the hazard, then appropriate eye/skin protection. State how the measure interrupts the exposure or runaway pathway.

Salicylic acid reacts with ethanoic anhydride to form aspirin (2-ethanoyloxybenzoic acid). A small amount of concentrated sulfuric or phosphoric acid catalyses acylation of the phenolic $\ce{-OH}$ group; the catalyst speeds the reaction without being consumed.

1. Mix measured reactants with the acid catalyst, controlling the initial exothermic mixing, then warm in a water bath. 2. Add crushed ice/water to destroy excess ethanoic anhydride and lower aspirin solubility. 3. Cool to crystallise, collect by suction filtration and wash with cold water. 4. Recrystallise from a minimum of hot solvent, collect and dry the purified crystals.

Use eye protection and controlled addition for corrosive acid and ethanoic anhydride; avoid inhaling vapour and heat with a water bath rather than a naked flame. Dry before weighing or measuring melting temperature.

Choose from physical properties: use extraction for unequal solubility between two liquid phases, distillation for volatility differences, steam distillation for a high-boiling steam-volatile material, and recrystallisation for temperature-dependent solid solubility.