Transition Metals (Inorganics) Flashcards
Define:
- Ligand
- Complex
- Co-ordination number
- Lewis base
- Lewis acid
- Ligand - particle with lone pair of electrons that bonds to metal by a co-ordinate bond.
- Complex - metal ion with coordinately bonded ligands.
- Co-ordination number - number of co-ordinate bonds from ligands to metal ion.
- Lewis base - lone pair donor
- Lewis acid - lone pair acceptor
Name the four shapes a complex can hold, and give:
- bond angles
- occurrence with which ions/ligands
- co-ordination number
- Linear - bond angles = 180 degrees; occurs as Ag+ complexes; co-ordination number = 2
- Square planar - bond angles = 90 degrees; occurs as Pt(2+) and Ni(2+) complexes; co-ordination number = 4
- Tetrahedral - bond angles = 109.5 degrees, occurs as complexes containing larger ligands (e.g: Cl-); co-ordination number = 4
- Octahedral - bond angles = 90 degrees, occurs as most complexes with small ligands (e.g: H20); co-ordination number = 6
Name, describe and give the examples of the three types of ligands.
Unidentate ligands: ligands which form one co-ordinate bond to a metal ion. (e.g: Cl-, OH-, CN-, H2O, NH2)
Bidentate ligands: ligands which form two co-ordinate bonds to a metal ion. (e.g: 1,2-diaminoethane (en), ethanedioate ion)
Multidentate ligands: ligands which form more than two co-ordinate bonds to a metal ion. (e.g: EDTA+, porphyrin)
Describe how and where Isomerism can occur in complexes.
Stereoisomerism:
- Cis-trans isomerism -
> special case of E-Z isomerism, can occur in octahedral and square planar complexes where there are two ligands of one type different to the other ligands.
- Optical isomerism -
> can occur in octahedral complexes with three bidentate ligands.
Define ligand substitution and explain what will happen to the co-ordination number for:
- substitution by similar sized ligands
- substitution by larger/smaller ligands
- substitution by ligands that form more co-ordinate bonds
- a reaction where one ligand is replaced by another ligand.
- Similar sized; no change in co-ordination number as the same number of ligands are swapped.
- Larger/Smaller ligands; co-ordination number may change as different sized ligands may form fewer/more co-ordinate bonds.
- Ligands that form more co-ordinate bonds; co-ordination number will decrease.
Why are ligand substitutions by ligands that form more co-ordination bonds very feasible reactions?
- Increase in entropy as more product particles made than reactant particles were used.
- Enthalpy change is negligible as the same number and type of bonds are being broken and made, so doesn’t change.
- Temperature doesn’t change.
- Therefore only entropy looked at when deciding feasibility, and as entropy is very positive, Delta G will be very negative.
Define a chelating agent.
Ligands that form more than one co-ordinate bond, especially those that form many, as they are very good at bonding to a metal ion and are very difficult to them remove, which renders the metal ion harmless as it cannot bond to anything else.
Why are transition metal compounds colourful?
- When ligands bind in transition metal compounds, the five d-orbitals don’t all have the same energy. This gap corresponds to the energy of UV/visible light, (d-orbital splitting). The electrons in the d-orbitals absorb UV/Visible light to excite the electrons into the higher energy level.
- The frequency of wavelength that doesn’t match to the electrons wavelength cannot be absorbed and will instead be reflected, which corresponds to the colour we see.
Name four factors that can affect which colour you see that’s reflected back off the electrons in transition metal compounds.
- Identity of metal - (e.g: complex 1 - [Cu(H2O)6]^(2+) = blue, : complex 2 - [Fe(H2O)6]^(2+) = green)
- Oxidation state of metal (e.g: complex 1 - [Fe(H2O)6]^(3+) = pale violet : complex 2 - [Fe(H2O)6]^(2+) = green)
- Identity of ligands (e.g: complex 1 - [Cu(H2O)6]^(2+) = blue. : complex 2 - [Cu(H2O)4(NH3)2]^(2+) = deep blue)
- Co-ordination number (e.g: complex 1 - [Cu(H2O)6]^(2+) = blue. : complex 2 - [CuCl4]^(2-) = yellow)
Describe how calorimetry can be used to find the concentration of unknown concentration solutions.
- If pale colour complex, then change ligand to intensify colour. (e.g: use SCN- instead of H20)
- Choose a complementary colour filter to your sample colour for your calorimeter. This will give you the largest range of absorbance.
- Measure absorbance for several known concentrations and plot a calibration curve.
- Measure absorbance of the unknown solution and use the calibration curve to find the concentration.
What is the Beer-Lambert Law?
A = ECL
- A = absorbance.
- E (curly E) = molar absorbance coefficient (fixed value that depends on sample)
- C = concentration
- L - cell path length (=1cm ( how much the light has to travel through)
(A and C are directly proportional)
What effect can:
- pH
- ligand type
:have on how easily transition metals can change oxidation states.
How can this be shown using electrode potentials?
- Generally it is easier to:
> oxidise a transition metal in alkaline conditions. (e.g: H2O2 oxidising agent, adding NaOH for alkaline conditions.)
> reduce a transition metal in acidic conditions. (e.g: Zinc reducing agent, adding H2SO4 for acidic conditions.) - When looking at the E cell values for the reversible redox reactions of metal ion compounds, the more positive value prefers the forward reaction (usually reduction in acidic conditions) and the more negative value prefers the backwards reaction (usually oxidation in alkaline conditions).
Describe how the Vanadium in Ammonium Vanadate can be reduced from v(+5) eventually to V(+2) in acidic conditions.
- Ammonium Vanadate added to a flask with either HCl or H2SO4 and Zinc as a reducing agent. Cotton wool is inserted as a plug to prevent air from getting in and V(+2) being formed.
V(+5) ——> V(+4) ——> V(+3) —> V(+2)
VO2(+) ——> VO(2+) ——> V(3+) ——> V(2+)
yellow ——> blue ——> green ——> violet
Name the suitable acid, and some unsuitable acids to use during redox titrations.
- Why can’t you use these?
- dilute sulphuric acid
- hydrochloric acid - the MnO4- would also oxidise Cl- to Cl2 so affect the volume of KMnO4 required in the titration.
- conc sulfuric acid or conc nitric acid - they are oxidising agents themselves so affect the volume of KMnO4 required in the titration.
- ethanol acid - it is a weak acid and would not provide enough H+ ions.
Describe how acidified KMnO4 can be used in redox titrations, and the process.
Give the equation.
- The dark purple potassium manganate is in the burette, with the sample in the conical flask being mixed with an excess of sulfuric acid.
- As you add the purple potassium manganate it reacts and forms colourless Mn2+. At the endpoint, no more of the MnO4- ions can react, so the purple colour of the MnO4- remains, allowing this to be a self-indicating titration.
MnO4- + 8H+ + 5e- —> Mn2+ + 4H2O (MnO4- = dark purple, Mn2+ = colourless)
