transition elements Flashcards
What is a transition element?
A d-block element with at least one ion with an incomplete d-orbital. Eg, Cu has [Ar] 3d104s1 where Cu2+ has [Ar] 3d94s0.
Which transition metals have anomalous e- structures and why
Copper ([Ar] 3d104s1) and chromium ([Ar] 3d54s1).
Having half-built/full-built 3d subshells is lower in energy than a full-built 4s subshell.
Why aren’t zinc and scandium transition elements
Zinc can only form 2+ ion which has a complete d-subshell. ○ Zn has [Ar] 3d104s2 whereas Zn2+ has [Ar] 3d104s0.
Scandium can only form 3+ ions which has an empty d- subshell.
○ Sc has [Ar] 3d14s2 whereas Sc3+ has [Ar] 3d04s0.
What are the characteristics of all transition elements?
Formation of coloured ions. Variable oxidation state.
Eg, Fe (II) and Fe (III) since their 4s and 3d energy levels are so close, they can easily lose another e-.
● Catalytic activity.
● Complex ion formation in solution.
Why do transition elements make good catalysts?
They can change oxidation states by gaining/losing e-’s ⇒ can transfer e-’s to speed up reactions.
How are (aq) complex ions written
● When they contain ligands that aren’t water or hydroxide ions, you need to include all water ligands.
● Eg, [Cu(NH3)4(H2O)2]2+ cannot be rewritten whereas [Cu(OH)2(H2O)4]2+ can be written as Cu(OH)2.
What is a complex ion?
A central metal ion surrounded by ligands.
What is a ligand?
An atom, ion, or molecule that can forms a coordinate bond to a central metal ion.
What is coordination number
The number of coordinate bonds formed to a central metal ions.
What is a monodentate ligand?
● Ligands that form 1 dative/coordinate bond. ● Examples include: H2O, NH3 and Cl-
What is a bidentate ligand?
● Ligands that form 2 dative/coordinate bonds.
● Examples include: C2O42- (ethanedioate), ethene-1,2-diamine
/ en.
What is a multidentate ligand? (with an example)
● Ligands that form several coordinate bonds.
● Examples include: EDTA4- (forming 6 coordinate bonds).
What complex ions show cis- trans isomerism?
Square planar and octahedral complex ions with at least two pairs of different ligands
Why is cisplatin used as an anticancer drug? What are its drawbacks?
● Prevents replication of of cancer cells by binding to the DNA. ● Can prevent replication of healthy cells (∴ suppresses immune
system).
Describe the reaction of copper (II) with hydroxide which colours
● [Cu(H2O)6]2+ (aq) + 2HO- (aq) → [Cu(H2O)4(OH)2] (s) + 2H2O (l) ○ The complex loses its 2+ charge since HO- has joined
to it.
○ Goes from pale blue to blue ppt (since no charge)
Describe the reaction of copper (II) with ammonia and excess ammonia (with colours)
● [Cu(H2O)6]2+ (aq) + 2NH3 (aq) → [Cu(H2O)4(OH)2] (s) + 2NH4+ (aq)
○ The ammonia acts as a base accepting a H+.
○ Goes from pale blue to blue ppt.
● In excess NH3, [Cu(H2O)2(NH3)4]2+ (aq) is formed which is dark
blue.
Which ligands are similar in size and which are different?
What does this mean?
The ligands H2O, NH3 and CN- are SIMILAR in size so coordinate number doesn’t change under substitution.
The ligands H2O and Cl- are DIFFERENT (Cl- is larger) in size so coordination number and shape changes under substitution.
Describe how copper (II) react with chlorine ions (with colours
[Cu(H2O)6]2+ (aq) + 4Cl- (aq) ⇌ [CuCl4]2- (aq) + 6H2O (l) ○ Goes from blue to yellow.
○ Reversible reaction so greenish colour
Describe how iron (II) react with hydroxide ions and ammonia (with colours)
[Fe(H2O)6]2+ (aq) + 2HO- (aq) → [Fe(OH)2(H2O)4] (s) + 2H2O (l) [Fe(H2O)6]2+ (aq) + 2NH3 (aq) → [Fe(OH)2(H2O)4] (s) + 2NH4+ (aq)
From pale green solution to green ppt.
Describe how iron (III) react with hydroxide ions and ammonia (with colours)
[Fe(H2O)6]3+ (aq) + 3HO- (aq) → [Fe(OH)3(H2O)3] (s) + 3H2O (l) [Fe(H2O)6]3+ (aq) + 3NH3 (aq) → [Fe(OH)3(H2O)3] (s) + 3NH4+ (aq)
From pale yellow solution to orange brown ppt.
Describe how manganese (II) react with hydroxide ions and ammonia (with colours)
[Mn(H2O)6]2+ (aq) + 2HO- (aq) → [Mn(OH)2(H2O)4] (s) + 2H2O (l) [Mn(H2O)6]2+ (aq) + 2NH3 (aq) → [Mn(OH)2(H2O)4] (s) + 2NH4+ (aq)
● From pale pink solution to pink ppt.
Describe how chromium (III) reacts with hydroxide ions and excess hydroxide ions with colours
WITHOUT EXCESS:
● [Cr(H2O)6]3+ (aq) + 3HO- (aq) → [Cr(OH)3(H2O)3] (s) + 3H2O (l)
● From violet solution to grey-green ppt.
WITH EXCESS:
● [Cr(H2O)6]3+ (aq) + 6HO- (aq) → [Cr(OH)6]3- (aq) + 6H2O (l)
● From violet solution to dark green solution.
What is ‘haem?’
An Fe (II) complex with a multidentate ligand.
What does haemoglobin do and how
Enables O2 to be transported in the blood as O2 bonds to the Fe2+ ions in haemoglobin (and is released when required).
How does CO affect haemoglobin?
It forms a coordinate bond with haemoglobin stronger than O2 (more stable) preventing O2 from attaching.
Describe how iron (II) is oxidised (with colours)
● Fe2+ → Fe3+ using acidified manganate (VII) which is H+/MnO4- .
● The reaction goes from purple (MnO4-) to colourless (Mn2+). ● Yet, Fe2+ (aq) is green and Fe3+ (aq) is brown.
Oxygen in the air can bring this change also.
The equation below is not required according to the specification, relevant information will be provided to work it out.
● Overall Equation:
○ MnO4- (aq)+ 8H+ (aq) + 5Fe2+ (aq) → Mn2+ (aq) + 4H2O
(l) + 5Fe3+ (aq) ● Formed From:
○ MnO4- (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (l)
○ Fe2+ → Fe3+ + e-
Oxygen in the air can bring this change also.
Describe how iron (III) is reduced (with colours)
● Fe3+ (aq) is brown and Fe2+ (aq) is green.
The equation below is not required according to the specification, relevant information will be provided to work it out.
● Fe3+ → Fe2+ using I- (from KI)
● Equation: 2Fe3+ (aq) + 2I- (aq) → I2 (aq) + 2Fe2+ (aq)
Describe how Cr3+ is oxidised to Cr2O72- (with colours
Step One:
● Cr3+ in [Cr(OH)6]3- is oxidised to CrO42- by warming with H2O2 in alkaline conditions.
● This goes from dark green to yellow.
Step Two:
● Add dilute sulfuric acid to the chromate (VI) solution to produce dichromate (VI) Cr2O72-.
● This goes from yellow to orange.
The equation below is not required according to the specification, relevant information will be provided to work it out.
● Step 1 full equation:
○ 3H2O2 + 2Cr(OH)63- → 2HO- + 2CrO42- + 8H2O
● Step 1 half equations:
○ H2O2 + 2e- → 2HO-
○ 2Cr(OH)63- + 4HO- → 2CrO42- + 8H2O + 6e-. ● Step 2 full equation:
○ 2CrO42- + 2H+ → Cr2O72- + H2O
Describe how Cr2O72- is reduced to Cr3+ (with colours)
● Cr2O72- is reduced to Cr3+ by acidified zinc. ● This goes from orange to green.
The equation below is not required according to the specification, relevant information will be provided to work it out.
● Full equation:
○ Cr2O72- + 14H+ + 3Zn → 2Cr3+ + 7H2O + 3Zn2+
● Half equations:
○ Zn → Zn2+ + 2e-
○ Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
Describe how Cu2+ is reduced to Cu+ (with colours)
● Cu2+ is reduced to copper (I) iodide by iodide ions, I-. ● From pale blue to off-white ppt.
The equation below is not required according to the specification, relevant information will be provided to work it out.
● Full equation:
○ 2Cu2+ (aq) + 4I- (aq) → 2CuI (s) + I2 (aq)
Describe the disproportionation of Cu+ (equation and why it happens
2Cu+ (aq) → Cu (aq) + Cu2+ (aq)
This spontaneous disproportion happens since Cu+ is unstable
