5.3 Flashcards
Transition metal definition
A transition metal is an element that has an incomplete d-sub-shell as a stable ion.
What two metals are not transition elements in the d-block, and what are their ions’ electronic configurations?
Sc³+: 1s², 2s², 2p⁶, 3s², 3p⁶
Zn²+: 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰
How come transition metals have variable oxidation states?
Because the 3d and 4d energy levels are so close in energy, the 3d electrons can be lost when an atom forms are stable 2+ ion
When are transition metals’ highest oxidation states found?
In strong oxygen agent compounds
For example, potassium manganate (VII) and potassium dichromate (VI)
Molecular formulae of:
Potassium manganate (VII)
Potassium dichromate (VI)
Sodium Thiosulphate
Chromate
KMnO4 (purple)
K2Cr2O7 (orange)
S2O3²-
CrO4²- (yellow)
How do we see colour?
The colour that we observe is a mix of the wavelengths of light not absorbed.
What are the 2 methods of transition metals as catalysts?
A) adsorption and desorption
B) t.m. changes oxidation state then bind to reactants forming intermediates with lower activation energy
Pros and cons of transition metal catalysts
Pros
Reduce time and energy needed for reactions
Cons
Can be toxic, must be handled with care, and disposal must be in such a way as not to cause pollution.
Complex ion definition
A complex ion is a transition metal bonded to 1 or more ligands by coordinate bonds.
Ligand definition
A ligand is a molecule or ion that can donate a pair of electrons to a transition metal to form a coordinate bond.
Conditions for cis-trans isomerism in complex ions
Octahedral - 4 monodentate and 2 different monodentate ligands OR 2 bidentate and 2 monodentate ligands
Square planar - 2 monodentate and 2 different monodentate
Conditions for optical isomerism in complex ions
Octahedral - 3 bidentate OR 2 bidentate and 2 monodenate OR 1 hexadentate
5 reaction names and their transition metal catalysts
Contact process:
Vanadium (V) oxide, V2O5
Alkenes hydrogenation:
Nickel
Haber process:
Iron
Hydrogen production:
Copper (II)
Decomposition of H2O2:
Manganese (IV) oxide, MnO2
Name a hexadentate ligand
EDTA
What is EDTA used for?
Decreases concentration of metal ions in a solution by binding with them
Treats mercury or lead poisoning
What is the name of the cancer treatment drug?
Cis-platin
Describe cis-platin
PtCl2(NH3)2
Prevents cancer cell division
Tetrahedral
109.5°
Chloride exception
Tetrahedral as ions are too big and only 4 can fit
Describe haemoglobin
FeN4 in octahedral shape with O2 bonded temporarly above and globin below
O2 is replaced with H2O when delivered
CO permanently bonds so can displace O2
General precipitation ligand substitution reactions
[M(H₂O)₆]²⁺ (aq) + 2OH⁻ (aq) –> [M(H₂O)₄(OH)₂]²⁺ (s) + 2H₂O (l)
If NH₃ (aq), forms same precipitate but NH₄⁺
Precipitation reactions colours
Fe²⁺: pale green → dark green
Fe³⁺: pale yellow → rusty brown
Mn²⁺: pale pink → brown
Cu²⁺: pale blue → blue
Cr³⁺: violet → green
Cu²⁺ other reactions (vague)
+ Excess NH₃ (aq)
+ Hydrochloric acid (aq)
Cu²⁺ and excess NH₃ (aq) reaction and colours
[Cu(H₂O)₆]²⁺ (aq) + 4NH₃ (aq) ⇄ [Cu(NH₃)₄(H₂O)₂]²⁺ (aq) + 4H₂O (l)
Pale blue sol. ⇄ Blue sol.
Cu²⁺ and Hydrochloric acid reaction and colours
[Cu(H₂O)₆]²⁺ (aq) + 4Cl⁻ (aq) ⇄ [CuCl₄]²⁻ (aq) + 6H₂O (l)
Pale blue sol. ⇄ Yellow sol.
Cr³⁺ reactions (vague)
+ Excess NH₃ (aq)
+ Excess NaOH (aq)
Oxidation to Cr⁶⁺
Cr³⁺ and excess NH₃ (aq) reaction and colours
[Cr(H₂O)₆]³⁺ (aq) + 6NH₃ (aq) ⇄ [Cr(NH₃)₆]³⁺ (aq) + 6H₂O (l)
Violet sol. ⇄ Purple sol.
Cr³⁺ and excess NaOH (aq) reaction and colours
[Cr(H₂O)₆]³⁺ (aq) + 6OH⁻ (aq) → [Cr(OH)₆]³⁻ (aq) + 6H₂O (l)
Violet sol. → Green sol.
Oxidation of Cr³⁺ to Cr⁶⁺ reaction and colours
Alkaline conditions
2[Cr(OH)₆]³⁻ (aq) + 3H₂O₂ (aq) HEAT→ 2CrO₄²⁻ (aq) + 2OH⁻ (aq) + 8H₂O (l)
Green sol. → Yellow sol.
Chromate (VI) to Potassium dichromate (VI)
2CrO₄²⁻ (aq) ⇄ Cr₂O₇²⁻ (aq)
Yellow sol ⇄ Orange sol.
–> +H⁺
<– +OH⁻
Which is more stable - Fe²⁺ or Fe³⁺?
Fe³⁺ due to the half filled d orbital (while Fe²⁺ is partially filled)
How can the concentration of oxidising agents in a solution be found?
React unknown conc. oxidising agent (e.g. Cu²⁺) with potassium iodide.
2Cu²⁺ + 4I⁻ → 2CuI + I₂
Then titrate this with Sodium Thiosulphate (S₂O₃²⁻) to find [I₂] and calculate [Cu²⁺]
2S₂O₃²⁻ + I₂ → 2I⁻ + S₄O₆²⁻
Brown sol. → Pale yellow.
The end-point is hard to see so we add starch solution. This turns the solution blue and changes the colour change to Dark blue sol. → Pale yellow sol.
