Topic 15: Transition Metals

Question 1

What are transition metals

Answer 1

Transition metals= d block elements that form one or more stable ions with incompletely filled d orbitals.

They are hard solids w high mpts, can act as catalysts, from coloured ions, and form ions with different ox.numbers.

Question 2

Why are scandium and zinc not transition metals?

Answer 2

Scandium loses both of its 4s e- and its only 3d e- to form Sc3+. Zn loses both of its 4s e- and none of its 3d e- to form Zn2+. Bc these elements form 1 ion each w no incompletely filled d orbitals they are not transition metals.

Question 3

Describe the relationship between transition metals and oxidation number

Answer 3

Each transition metal can lose a variable number of e-, so forms ions with diff ox.numbers.

Transition metal ions w high ox.numbers usually contain an electroneg element like oxygen.

From Fe to Cu across the period hay increased nuclear charge. e-s are attracted more strongly so less likely to bond. Therefore ions w higher ox.numbers are less common across a period.

Question 4

what are Ligands?

Answer 4

Ligands are the electron rich species that form dative covalent bonds with the central metal ion.

Transition metals have smaller radii, so they attract electron rich species more strongly, including aq water molecules.

Question 5

What is a complex ion? Why do transition metals regularly form complex ions?

Answer 5

A complex ion has ligands attached to a central metal cation by a dative covalent bond.

Cations which form complex ions must have: high charge density to attract e- from ligands, and empty low energy orbitals to accept the lone e- pair from the ligands.

Metal d-block cations are small, have a high charge density and have empty low energy 3d and 4s orbitals. So they form complex ions readily.

Question 6

What is coordination number?

Answer 6

The number of lone e- pairs a cation can accept/ total no of dative bonds around the metal ion.

Coordination number is not the number of Ligands! eg EDTA forms six coordinate bonds so the coordination number is six, but it is one ligand

Question 7

How many dative bonds can 1 ligand form?

Answer 7

Ethanedioate (C2O42-) and 1,2-diaminoethane donate 2 lone pairs per ligand and are bidentate.

Ligands such as H2O, Cl- and CN- form only one dative covalent bond per ligand (monodentate).

EDTA4-, is HEXAdentate.

Question 8

Draw the following: [Fe(H2O)6]2+ [Cr(NH3)6]3+ [Cu(NH3)4(H2O)2]2+

Answer 8

Question 9

Draw the following multi dentate ligand complexes:

[Fe(C2O4)3]3- [Cr(H2NCH2CH2NH2)3]3+ [Cu(edta)]2

Answer 9

Question 10

How do you determine the shape of complex ions?

Answer 10

2-coordinate complexes are generally linear, and are formed w Ag+ ions.

Pt ions form square planar complexes. Coordination number=4, bond angle 90°

Cl- and other large ions form tetrahedral complexes. Coordination number=4, bond angle 109.5°

Most other transition metal complexes are octahedral w coordination number 6, bond angles 90°

Question 11

Draw [CoCl4]2- and explain its shape

Answer 11

Question 12

What must occur for an ion to be coloured?

Answer 12

If the ion is to be coloured:

there must be splitting of the d-orbitals. This only happens in the presence of ligands so only complex ions are coloured. Anhydrous ions don’t have split d-orbitals, so cannot aborb light in the visible spectrum so are white.

d-orbitals must be partially filled. If empty, then no hay e- which can be excited into the higher energy d-orbitals, ions will be colourless. If the d-orbitals are full then no hay empty orbitals where the e-s can be excited, ions will be colourless.

Question 13

Describe and explain why the copper 2+ ion is coloured

Answer 13

The copper ion has an incompletely filled d orbital, which splits into different energy levels by ligands.

The electron from the lower level absorbs light from the visible spectrum. The e- is excited and jumps from the lower energy to the higher energy level

In Cu2+, the SMALL energy level difference means LOW Hz radiation is absorbed from the red end of the spectrum. So blue light is transmitted.

Question 14

Explain change in oxidation number of iron after exposure to air.

Answer 14

After exposure to air iron goes from pale green in aq solution to brown. Ox.number increases from +2 to +3.

The type and number of ligands dont change.

Formula for both: [Fe(H20)6]2+ and [Fe(H20)6]3+

Question 15

Describe the reaction between sodium hydroxide and copper sulphate solution.

Answer 15

When aq Na0H is added to aq CuS04, a pale blue solution forms a blue ppt:

[Cu(H20)6]2+ + 2OH- –> [Cu(H20)4(0H)2] + 2H20

This is not ligand exchange, but an acid-base reaction; the 2 0H- ions removed H+ from 2 water ligands and converted them to water molecules. The 2 water ligands that lost H+ are now 0H- ligands.

This is amphoteric bc it’s reversible– when acid is added to the blue ppt, H+ from the acid reacts with the OH- ligands and converts them to water molecules.

Question 16

Describe the reaction between copper sulphate solution and ammonia

Answer 16

Copper sulphate solution+ ammonia= a pale blue solution forms a blue ppt. 2 of the water ligands transfer a H+ to the ammonia:

[Cu(H20)6]2+ + 2NH3- –> [Cu(H20)4(0H)2] + 2NH4+

When aq ammonia is added to the blue ppt, a deep blue solution forms:

[Cu(H20)4(0H)2] + 4NH3—> [Cu(NH3)4(H20)2]2+ +2H20 +20H-

This IS a ligand exchange – 4 ammonia molecules replace 2 water molecules and 20H- ions.

Question 17

Describe change in coordination number, using HCl and copper sulfate

Answer 17

When conc HCl is gradually added to aq CuS04, colour changes from blue to green to yellow:

[Cu(H20)6]2+ + 4Cl- —> [CuCl4]2- + 6H20

All 6 water ligands are substituted by 4 Cl- ions. This is also a change in coordination n0 from 6 to 4. No change in ox.number.

[Cu(H20)6]2+ = pale blue. [CuCl4]2- = yellow. Green= a mix of both.

Question 18

Describe the reaction between Hexaaquacobalt(II) and NaOH

Answer 18

Hexaaquacobalt(II) + NaOH: a pink solution forms a blue ppt:

[Co(H20)6]2+ + 2OH- –> [Co(H20)4(0H)2] + 2H20

This reaction is acid base. The 2 0H- ions removed H+ from 2 water ligands and converted them to water molecules. The 2 water ligands that lost H+ are now 0H- ligands.

Upon standing, ppt gradually changes to pink.

Question 19

Describe the reaction between hexaaquacobalt and ammonia.

Answer 19

Hexaaquacobalt(II) + NH3: a pink solution forms a blue ppt:

[Co(H20)6]2+ + 2NH3- –> [Co(H20)4(0H)2] + 2NH4+

When EXCESS ammonia is added the ppt dissolves to a brown solution. This is ligand exchange:

[Co(H20)4(0H)2] + 6NH3—> [Co(NH3)6]2+ +4H20 +20H-

Upon standing, the brown solution turns yellow. Ox.number increases from +2 to +3 bc of 02 in the atmosphere.

Question 20

Describe the reaction between hexaaquacobalt(II) and HCl

Answer 20

A pink solution gradually turns blue:

[Co(H20)6]2+ + 4Cl- –> [CoCl4]2- + 6H20

This is ligand exchange and there is a change in coordination number from 6 to 4.

Question 21

Describe the reaction between Hexaaquairon(II) and NaOH and ammonia

Answer 21

A pale green solution forms a green ppt: [Fe(H20)6]2+ + 2OH- –> [Fe(H20)4(0H)2] + 2H20

like Cu and Co, this is an acid base reaction.

SAME observations w adding aq ammonia: [Fe(H20)6]2+ + 2NH3 –> [Fe(H20)4(0H)2] + 2NH4+

Upon standing the green ppt is oxidised to brown from the air. Triaquatrihydroxoiron(III), or [Fe(H20)3(0H)3]3+ is formed

Question 22

Describe the reaction between Hexaaquairon(III) and NaOH and ammonia

Answer 22

A yellow-brown solution forms a brown ppt:

[Fe(H20)6]3+ + 3OH- –> [Fe(H20)3(0H)3 + 3H20. This is an acid base reaction.

SAME observations when adding aq ammonia:

[Fe(H20)6]3+ + 3NH3 –> [Fe(H20)3(0H)3 + 3NH4+. There is NO further change upon standing

Question 23

Describe the reaction between Hexaaquachromium and NaOH

Answer 23

A green or violet solution forms a green ppt in an acid base reaction:

[Cr(H20)6]3+ + 3OH- –> [Cr(H20)3(0H)3] + 3H20.

When excess NaOH is added, the green ppt dissolves to form a green solution:

[Cr(H20)3(0H)3] +OH- —> [Cr(H20)2(0H)4]- + H20.

If NaOH is more concentrated, more 0H- will replace the H20s in the acid-base reaction. Adding acid reverses these reactions so the complex is amphoteric.

Question 24

Describe the reaction between Hexaaquachromium and ammonia

Answer 24

When adding aqueous ammonia, a green or violet solution forms a green ppt:

[Cr(H20)6]3+ + 3NH3 –> [Cr(H20)3(0H)3] + 3NH4+

When XS ammonia is added to the green ppt it slowly dissolves into a violet solution:

[Cr(H20)3(0H)3] + 6NH3 —> [Cr(NH3)6]3+ + 3H20 + 3OH-

Question 25

How do you oxidise chromium complexes?

Answer 25

Add H202 in an alkaline solution. A green solution reacts to form yellow chromate(VI):

2Cr 3+ + 10OH- + 3H202 —> 2CrO4 2- + 8H20

Chromate(VI) CrO4 2- ions exist in eqm with dichromate(VI) Cr2O4 2- ions. Adding acid to yellow chromate forms orange dichromate. This is in eqm and the reaction is reversible.

Question 26

How do you reduce chromium ions?

Answer 26

Reduction: Zn + dichromate ions decreases Ox.number to +3. Colour change from orange to green:

Cr2O7 2- + 14H+ + 3Zn –> 2Cr3+ + 7H20 + 3Zn 2+.

Zinc can reduce Cr3+ further to blue Cr2+ but this is unstable and readily oxidised back to Cr3+ from oxygen in the air:

2Cr3+ + Zn —> 2Cr 2+ + Zn 2+

Question 27

Vanadium has…

Answer 27

Vanadium has many oxidation states:

+2 Vanadium(II) V 2+ = violet

+3 Vanadium(III) V 3+ = green

+4 Oxovanadium(IV) VO 2+ = blue

+5 Dioxovanadium(V) VO2 + =yellow

Question 28

How can you reduce vanadium from +5 to +2?

Answer 28

An acidic solution of ammonium vanadate, NH4VO3, contains V02 +.

When Zn is added to this solution, reduction begins y hay gradual colour change from yellow to blue to green to purple as Ox.N decreases from +5 to +2.

Question 29

Explain the feasibility of reduction of vanadium using half equations.

Answer 29

For redox reactions to be feasible, the E values from the half eqns must be positive. Reduction from +5 to +2 is feasible as all the E values are positive.

But, for reduction from +2 to 0, the E value is negative. This means Zn is not electron releasing w respect to V2+, so the reaction is not thermodynamically feasible.

Question 30

Explain the reaction of the peroxdisulphate ion

Answer 30

S208 2- = Peroxodisulfate ion. This acts as an oxidising agent with iodide ions: S208 2- + 2I- –> 2SO4 2- + I2

Reaction= slow at room temp bc the negative reactants repel each other. Fe2+/Fe3+ is not repelled so acts as an aq catalyst:

Step 1: S208 2- + 2Fe2+ —> 2SO4 2- + 2Fe3+

Step 2: Fe3+ ions formed react w I- : 2Fe3+ + 2I- –> 2Fe2+ + I2

Fe2+ ions used in step 1 regenerate in step 2 so the 2 steps continuously repeat.

Question 31

Explain autocatalysis using oxidation of ethanedioate.

Answer 31

Oxidation of Ethanedioate by KMnO4:

2MnO4- + 5C2O4 2- + 16H+ —> 2Mn2+ + 10CO2 + 8H20

Reaction= slow bc negative reactants repel each other. Mn2+ product acts as a catalyst in the reaction. This explains why rr increases as the titration proceeds. This is autocatalysis – when a reaction product acts as a catalyst for the reaction.