Transition Metals (AQA A Level Chemistry): Flashcards

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  • What is the definition of a transition metal?

    A Brønsted–Lowry acid is a proton donor — a species that donates an H+ ion to a base in an acid–base reaction.

  • Cr: [Ar] .......... (not [Ar] 3d4 4s2)

    Cu: [Ar] .......... (not [Ar] 3d9 4s2)

    Cr: [Ar] 3d5 4s1

    Cu: [Ar] 3d10 4s1

    These configurations are more energetically stable due to the extra stability of a half-full or full 3d subshell.

  • True or False?

    Scandium and zinc are classified as transition metals because they are in the d-block.

    False.

    Sc only forms Sc3+ with a 3d0 configuration and Zn only forms Zn2+ with a 3d10 configuration. Neither ion has a partially filled d-subshell, so they do not meet the definition of a transition metal.

  • What is a ligand in a transition metal complex?

    A buffer solution is a solution that resists large changes in pH when small amounts of acid or alkali are added to it. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable concentrations.

  • Ligand

    Type

    Dative bonds formed

    H2O

    ..........

    ..........

    C2O42-

    ..........

    ..........

    EDTA4-

    ..........

    ..........

    Ligand

    Type

    Dative bonds formed

    H2O

    Monodentate

    1

    C2O42-

    Bidentate

    2

    EDTA4-

    Hexadentate (multidentate)

    6

  • Why do transition metals make good catalysts?

    Transition metals have variable oxidation states, allowing them to gain and lose electrons readily during a reaction. They can also adsorb reactants onto their surface. Both properties help provide alternative lower-energy pathways.

  • True or False?

    Chloride ligands give a coordination number of 4 while water and ammonia ligands typically give a coordination number of 6.

    True.

    Cl- is a large ligand, so only 4 fit around the central metal ion (coordination number 4, tetrahedral). Water and ammonia are small, so 6 can fit (coordination number 6, octahedral).

  • What is ligand exchange (ligand substitution)?

    A Brønsted–Lowry base is a proton acceptor — a species that accepts an H+ ion from an acid in an acid–base reaction.

  • [Co(H2O)6]2+ is .......... in colour.

    Addition of excess ammonia gives [Co(NH3)6]2+, which is .......... in colour.

    In excess concentrated ammonia, oxidation to Co(III) gives [Co(NH3)6]3+, which is .......... .

    [Co(H2O)6]2+ is pink in colour.

    Addition of excess ammonia gives [Co(NH3)6]2+, which is pale yellow/straw in colour.

    In excess concentrated ammonia, oxidation to Co(III) gives [Co(NH3)6]3+, which is brown.

  • True or False?

    When [Cu(H2O)6]2+ reacts with concentrated HCl, the coordination number changes from 6 to 4 because chloride ions are larger than water ligands.

    True.

    [Cu(H2O)6]2+ + 4Cl- (aq) → [CuCl4]2- (aq) + 6H2O (l). The large Cl- ligands mean only 4 fit around Cu2+, changing the geometry from octahedral to tetrahedral and the colour from blue to yellow.

  • Why is carbon monoxide toxic in terms of haem chemistry?

    CO is a stronger ligand than O2 and binds irreversibly to the Fe(II) centre in haemoglobin, forming carboxyhaemoglobin. This prevents oxygen from binding and being transported to cells.

  • What drives the chelate effect and why is it energetically favourable?

    The chelate effect is driven by a large positive entropy change (ΔS). Replacing monodentate ligands with bidentate or multidentate ligands produces a net increase in the number of particles, increasing entropy. The resulting negative ΔG makes chelation spontaneous.

  • In the EDTA chelation of aqueous cobalt(II):

    [Co(H2O)6]2+ (aq) + EDTA4- (aq) → .......... + ..........

    The number of particles goes from .......... to .......... , giving a .......... entropy change.

    [Co(H2O)6]2+ (aq) + EDTA4- (aq) → [CoEDTA]2- (aq) + 6H2O (l)

    The number of particles goes from 2 to 7, giving a positive entropy change.

  • True or False?

    Adding water to a solution of [CuCl4]2- (yellow) converts it back to [Cu(H2O)6]2+ (blue) because water displaces the chloride ligands.

    True.

    The ligand exchange is reversible. Water molecules displace the larger chloride ligands when excess water is added, restoring the blue hexaaqua complex.

  • Complete the table showing the coordination numbers and bond angles for each complex ion shape.

    Shape

    Coordination number

    Bond angle

    Linear

    ..........

    ..........

    Tetrahedral

    ..........

    ..........

    Square planar

    ..........

    ..........

    Octahedral

    ..........

    ..........

    Shape

    Coordination number

    Bond angle

    Linear

    2

    180°

    Tetrahedral

    4

    109.5°

    Square planar

    4

    90°

    Octahedral

    6

    90°

  • True or False?

    Both the cis and trans isomers of [Pt(NH3)2Cl2] can cross-link DNA and are used in cancer treatment.

    False.

    Only cisplatin (the cis isomer) can cross-link DNA in cancer cells because of its ligand arrangement. The trans isomer (transplatin) cannot bind to DNA in the same way and has no therapeutic use.

  • What conditions are required for a complex ion to show optical isomerism?

    The complex must have no plane of symmetry — the structure and its mirror image must be non-superimposable.

    This most commonly occurs in octahedral complexes with bidentate ligands (e.g., [Co(en)3]3+), but can also arise with an appropriate mix of monodentate ligands.

    The two optical isomers rotate the plane of polarised light in opposite directions.

  • Why do octahedral complexes and square planar complexes both show cis-trans isomerism?

    Both geometries are rigid — the positions of ligands around the central metal ion are fixed and cannot interchange without breaking bonds.

    This means when two or more different ligands are present, the arrangement is locked: the same ligands can sit either adjacent (cis, 90°) or opposite (trans, 180°), giving two distinct, non-interconvertible isomers.

  • 2 coordinate bonds: shape = ..........

    4 coordinate bonds (large ligands): shape = ..........

    6 coordinate bonds: shape = ..........

    2 coordinate bonds: shape = linear (example: [Ag(NH3)2]+)

    4 coordinate bonds (large ligands): shape = tetrahedral (example: [CoCl4]2-)

    6 coordinate bonds: shape = octahedral (example: [Fe(H2O)6]2+)

  • True or False?

    [Ag(NH3)2]+ is a linear complex and is the active species in Tollens' reagent.

    True.

    The diamminesilver(I) ion [Ag(NH3)2]+ is a linear complex with a coordination number of 2 and bond angle 180°. It is the complex present in Tollens' reagent used to test for aldehydes.

  • Define coordination number.

    The coordination number of a complex ion is the total number of coordinate bonds formed between the central metal ion and its ligands. Common coordination numbers are 2 (linear), 4 (tetrahedral or square planar), and 6 (octahedral).

  • Why is cisplatin used in cancer treatment when the trans isomer is not?

    Cisplatin's cis arrangement places the two Cl ligands adjacent (90° apart), allowing it to cross-link guanine bases on the same DNA strand.

    The trans isomer places the Cl ligands opposite (180° apart), which prevents this cross-linking — so it cannot bind to DNA in the same way and has no therapeutic use.

  • Why do transition metal complexes appear coloured?

    Ligands split the 3d orbitals into two sets of different energy. An electron absorbs light of frequency ν = ΔE/h and is promoted to the higher set. The colour observed is the complementary colour of the wavelength absorbed.

  • In an isolated transition metal ion, the five 3d orbitals are .......... (equal in energy).

    When ligands bond to the metal, the orbitals split into .......... sets.

    The energy difference is called .......... .

    An electron is promoted when it absorbs energy equal to .......... .

    In an isolated transition metal ion, the five 3d orbitals are degenerate (equal in energy).

    When ligands bond to the metal, the orbitals split into two sets.

    The energy difference is called ΔE.

    An electron is promoted when it absorbs energy equal to ΔE (= hν).

  • True or False?

    Changing the ligand on a transition metal complex can change its colour even when the oxidation state of the metal remains the same.

    True.

    Different ligands cause different degrees of d-orbital splitting (ΔE), so the wavelength of light absorbed changes. For example, [Cu(H2O)6]2+ is light blue but [Cu(NH3)4(H2O)2]2+ is deep blue, both with Cu in the +2 state.

  • How is a colorimeter used to determine the concentration of a coloured transition metal solution?

    A colorimeter passes light of a selected wavelength (the complementary colour to the solution, chosen with a filter) through the sample. Absorbance is measured and compared to a calibration curve of absorbance vs. known concentration.

  • Why is a red filter used when measuring the absorbance of a blue solution in a colorimeter?

    A blue solution absorbs light from the red region of the spectrum. The red filter selects the wavelength most strongly absorbed, giving maximum absorbance and the most accurate concentration measurement.

  • Stronger-field ligand: ΔE .......... , wavelength absorbed .......... , colour ..........

    Higher oxidation state: ΔE .......... , colour changes to absorb .......... energy light

    Stronger-field ligand: ΔE increases, wavelength absorbed decreases (shorter), colour changes

    Higher oxidation state: ΔE increases, colour changes to absorb higher energy light

  • True or False?

    The Beer-Lambert law states that at low concentrations, absorbance is directly proportional to concentration, producing a straight-line calibration graph.

    True.

    At low concentrations, the Beer-Lambert law holds and a plot of absorbance vs. concentration is linear. This straight-line graph is used to determine the concentration of unknown solutions.

  • d-d transition

    A d-d transition occurs when an electron in a transition metal complex absorbs a photon of visible light and is promoted from a lower-energy 3d orbital to a higher-energy 3d orbital. The energy absorbed equals ΔE (the splitting energy), and the complementary colour of the absorbed wavelength is observed.

  • Why do transition metals have variable oxidation states?

    The successive ionisation energies of transition metals are relatively close together for the first few electrons. This means removing additional electrons beyond the 4s requires little extra energy, allowing multiple stable oxidation states.

  • VO2+ (+5): ..........

    VO2+ (+4): ..........

    V3+ (+3): ..........

    V2+ (+2): ..........

    VO2+ (+5): yellow

    VO2+ (+4): blue

    V3+ (+3): green

    V2+ (+2): violet/purple

  • True or False?

    Transition metal ions always lose their 3d electrons before their 4s electrons when forming ions.

    False.

    When transition metals form ions, they lose 4s electrons first, then 3d electrons. This is because in the presence of other electrons, the 4s orbital is at higher energy than the 3d, giving a common +2 oxidation state.

  • How does changing the pH affect the reduction of MnO4-?

    In acidic conditions MnO4- is reduced to Mn2+ (+2 state, colourless). In neutral conditions, the product is MnO2 (+4 state, brown solid) because fewer H+ ions are available. The E value also changes with pH.

  • Tollens' reagent contains .......... . An aldehyde reduces Ag+ to .......... , producing a .......... . The equation is:

    [Ag(NH3)2]+ + e-.......... + ..........

    Tollens' reagent contains [Ag(NH3)2]+ (diamminesilver(I)). An aldehyde reduces Ag+ to Ag (s), producing a silver mirror. The equation is:

    [Ag(NH3)2]+ + e-Ag (s) + 2NH3 (aq)

  • True or False?

    Changing the ligand from water to ammonia makes nickel(II) harder to reduce, as shown by a more negative E value.

    True.

    Different ligands cause different degrees of d-orbital splitting. Ammonia binds more strongly to Ni2+ than water, stabilising the Ni2+ ion and making it harder to reduce. This is reflected in a more negative standard electrode potential for the ammonia complex compared to the aqua complex.

  • What is autocatalysis and how does it occur in the MnO4- / C2O42- titration?

    Autocatalysis is when a product of a reaction catalyses the same reaction. In the MnO4- / C2O42- titration, Mn2+ ions formed as a product catalyse the further reduction of MnO4-, causing the reaction to initially speed up.

  • Autocatalysis

    Autocatalysis is when a product of a reaction acts as a catalyst for that same reaction, causing the reaction rate to increase as the product accumulates. A classic example is the MnO4- / C2O42- titration, where Mn2+ ions produced in the reaction catalyse the further reduction of manganate(VII).

  • What is a redox titration and why are indicators not always needed?

    A redox titration involves titrating an oxidising agent against a reducing agent. Many transition metal ions used in redox titrations (e.g. MnO4-) are self-indicating: they are intensely coloured in one oxidation state and colourless or a different colour in another.

  • MnO4- (aq) + .......... H+ (aq) + .......... e- → Mn2+ (aq) + .......... H2O (l)

    Colour change: .......... to ..........

    MnO4- (aq) + 8 H+ (aq) + 5 e- → Mn2+ (aq) + 4 H2O (l)

    Colour change: purple to colourless (very pale pink)

  • True or False?

    In a manganate(VII) titration the end point is indicated by a permanent purple/pink colour persisting in the solution.

    True.

    KMnO4 is placed in the burette. The solution remains colourless as it reacts with the reducing agent. At the end point MnO4- is in slight excess, giving a permanent pink/purple colour.

  • A 2.25 g iron tablet was dissolved and titrated against 0.100 mol dm-3 KMnO4, requiring 26.50 cm3.

    moles KMnO4 = ..........

    moles Fe2+ = .......... (ratio MnO4- : Fe2+ = .......... )

    mass Fe = .......... g

    % Fe = .......... %

    moles KMnO4 = 0.00265

    moles Fe2+ = 0.01325 (ratio MnO4- : Fe2+ = 1:5)

    mass Fe = 0.740 g

    % Fe = 32.9%

  • Why is the MnO4- / C2O42- titration carried out at approximately 60 °C?

    The reaction between MnO4- and C2O42- is slow at room temperature. Heating to ~60 °C provides enough activation energy for the reaction to proceed at a reasonable rate (the Mn2+ produced then autocatalyses further reaction).

  • 2MnO4- + 5C2O42- + .......... H+ → 2Mn2+ + .......... CO2 + .......... H2O

    2MnO4- + 5C2O42- + 16 H+ → 2Mn2+ + 10 CO2 + 8 H2O

  • True or False?

    In the MnO4- / Fe2+ titration, Fe2+ is oxidised to Fe3+ and Mn is reduced from +7 to +2.

    True.

    The oxidation half-equation is Fe2+ → Fe3+ + e-. The reduction half-equation is MnO4- + 8H+ + 5e- → Mn2+ + 4H2O. Five Fe2+ ions are oxidised per MnO4- reduced.

  • Self-indicating

    A self-indicating reagent is one that shows a visible colour change at the end point of a titration without the need for a separate indicator. Transition metal ions such as manganate(VII) (MnO4-) are self-indicating because they have a strong colour in one oxidation state (purple) that disappears when reduced.

  • What is a heterogeneous catalyst?

    A cascade reaction is a series of consecutive reactions in which the product of one step immediately becomes the reactant for the next, allowing complex molecules to be built up efficiently.

  • True or False?

    Transition metals make good catalysts because they can form ions with more than one stable oxidation state.

    True.

    Transition metals have variable oxidation states, which allows them to gain and lose electrons readily during a reaction, providing alternative lower-energy reaction pathways.

  • In the Contact process, the catalyst .......... converts SO2 to SO3, being reduced from .......... to .......... in the process, then regenerated by reaction with O2 (g).

    In the Contact process, the catalyst V2O5 converts SO2 to SO3, being reduced from V(V) to V(IV) in the process, then regenerated by reaction with O2 (g).

  • What is a homogeneous catalyst?

    A chiral centre is a carbon atom bonded to four different groups. A molecule containing a chiral centre exists as two non-superimposable mirror-image forms called enantiomers.

  • True or False?

    In the Fe2+-catalysed reaction between I- and S2O82-, the reaction is slow without a catalyst because both ions are negatively charged.

    True.

    Repulsion between the two negative ions reduces the frequency of successful collisions. Fe2+ acts as an intermediate, reacting with each ion in turn and avoiding this repulsion.

  • Autocatalysis is when a .......... of the reaction acts as a .......... . In the reaction between MnO4- and C2O42-, the autocatalyst is .......... .

    Autocatalysis is when a product of the reaction acts as a catalyst. In the reaction between MnO4- and C2O42-, the autocatalyst is Mn2+ (aq).

  • How does the concentration–time graph for an autocatalytic reaction differ from a normal reaction?

    The gradient becomes steeper during the reaction — the rate increases over time as the autocatalyst product accumulates — rather than slowing as reactants are consumed.

  • Why are heterogeneous catalysts often spread over a honeycomb support rather than used as solid blocks?

    To maximise surface area while minimising the amount of (often expensive) catalyst used. More surface area means more active sites and a faster rate.

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