[Year 2] DM Flashcards

Question

Why are complex ions coloured?

Answer 1

- Complex ions are made up of ligands and transition metals. - Ligands split the d-orbital of the transition element, as shown... |↑ |↑ | (this energy gap is ΔE) |↑↓|↑↓|↑ | - This is the ground state. - When electrons absorb light energy some electrons from the ground state are excited to a higher energy level (an excited state), as shown... |↑↓|↑ | |↑↓|↑ |↑ | - This light energy has energy = ΔE - and so has a specific frequency. - The frequencies that are not absorbed are reflected or transmitted. - And so we see the complementary colour to what was absorbed.

Answer 2

- The central metal ion and its oxidation state. - The type of ligand. - The coordination number.

Answer 3

[Cu(H₂O)₆]²⁺₍ₐᵩ₎ + 2OH⁻₍ₐᵩ₎ → [Cu(OH)₂(H₂O)₄]₍ₛ₎ + 2H₂O₍ₗ₎ This forms a pale blue precipitate. - It precipitates out because the product is not charged.

Answer 4

- insoluble in excess NaOH.

Answer 5

Cu(H₂O)₆]²⁺₍ₐᵩ₎ + 2NH₃₍ₐᵩ₎ → [Cu(OH)₂(H₂O)₄]₍ₛ₎ + 2NH₄⁺₍ₐᵩ₎ This forms a pale blue precipitate. - It precipitates out because the product is not charged.

Answer 6

[Cu(H₂O)₆]²⁺₍ₐᵩ₎ + 4NH₃₍ₐᵩ₎ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺₍ₐᵩ₎ + 4H₂O₍ₗ₎ This forms a Dark blue solution - and is known as a part ligand substitution.

Answer 7

[Fe(H₂O)₆]²⁺₍ₐᵩ₎ + 2OH⁻₍ₐᵩ₎ → [Fe(OH)₂(H₂O)₄]₍ₛ₎ + 2H₂O₍ₗ₎ This forms a darker green precipitate. - It precipitates out because the product is not charged.

Answer 8

- insoluble in excess NaOH.

Answer 9

[Fe(H₂O)₆]²⁺₍ₐᵩ₎ + 2NH₃₍ₐᵩ₎ → [Fe(OH)₂(H₂O)₄]₍ₛ₎ + 2NH₄⁺₍ₐᵩ₎ This forms a darker green precipitate. - It precipitates out because the product is not charged.

Answer 10

- insoluble in excess NH₃.

Answer 11

[Fe(H₂O)₆]³⁺₍ₐᵩ₎ + 3OH⁻₍ₐᵩ₎ → [Fe(OH)₃(H₂O)₃]₍ₛ₎ + 3H₂O₍ₗ₎ This forms an orange precipitate. - It precipitates out because the product is not charged.

Answer 12

- insoluble in excess NaOH.

Answer 13

[Fe(H₂O)₆]³⁺₍ₐᵩ₎ + 3NH₃₍ₐᵩ₎ → [Fe(OH)₃(H₂O)₃]₍ₛ₎ + 3NH₄⁺₍ₐᵩ₎ This forms an orange precipitate. - It precipitates out because the product is not charged.

Answer 14

- insoluble in excess NH₃.

Answer 15

- Change in ΔE, as there is a different ligand. | - Change in shape if the ligand is larger/smaller.

Answer 16

- As they have variable oxidation states. - they are able t receive and lose electrons in their d-orbital. - This can lead to reactions speeding up

Answer 17

- Substances adsorb to the surface - This weakens bonds and breaks them to form radicals. - Radicals react with each other to form new substances. - The new substances are then desorbed.

Answer 18

- They can attract and hold reactants long enough for them to react. - They do not attract products too strongly or desorption won't occur. - If they are ground up, they have a larger surface area, so more substance can attach to its surface.

Answer 19

- Iron catalysts (used in Harber process) - Copper sulfate (used to catalyse acid reactions with zinc) - Manganese dioxide (used to catalyse the decomposition of hydrogen peroxide)

Answer 20

- Long term exposure to copper can lead to liver damage. | - Long term exposure to manganese can cause psychiatric issues and physical tremors.

Answer 21

Normal reaction is slow... S₂O₈²⁻₍ₐᵩ₎ + 2I⁻₍ₐᵩ₎ → I₂₍ₐᵩ₎ + 2SO₄²⁻₍ₐᵩ₎ ... as like charges repel each other. So an Iron catalyst is used (Fe²⁺)... S₂O₈²⁻₍ₐᵩ₎+ 2Fe²⁺₍ₐᵩ₎ → 2Fe³⁺ ₍ₐᵩ₎+ 2SO₄²⁻₍ₐᵩ₎ Fe²⁺ must be reformed so... 2Fe³⁺ ₍ₐᵩ + 2I⁻₍ₐᵩ₎ → 2Fe²⁺ + I₂₍ₐᵩ₎

Answer 22

- A molecule that loses electrons and is oxidised itself.

Answer 23

- A molecule that gains electrons and is oxidised itself.

Answer 24

1. balance atoms that are not Oxygen and Hydrogen. 2. balance any oxygens with water 3. balance any hydrogens with H⁺ ions 4. Balance any charges with electrons.

Answer 25

Add excess dilute sulfuric acid to the conical flask of unknown concentration, in order to have sufficient H⁺ ions to allow the reduction of the oxidising agent.

Answer 26

1. write out balenced equation. 2. work out the number of moles of known concentration substance by doing... mol = conc x vol (in dm³) 3. Use molar ratios to work out moles of desired product/reactant. 4. With the known volme and moles now work out its concentration... conc= mol/vol (in dm³)

Answer 27

- half of an electrochemical cell. - made from metal in its aqueous ion. - or platinum with 2 aqueous ions in solution.

Answer 28

- Wear gloves for safety and to prevent contamination - obtain metals under investigation and clean with sandpaper/emery paper to ensure surface is free of impurities. - Wash surface with propanone to remove grease - place each metal electrode into a solution containing the ions of the same metal. (if you are using an oxidising agent then add acid). - make a salt bridge from filter paper in saturated KNO₃ or KCl linking the beakers. Do not touch the electrode with this though. - Connect the electrodes with wires and crocodile clips to a voltmeter. - The reading on the voltmeter is the E⁰cell.

Answer 29

Compare electrode potentials (E⁰) from an electrochemical series. - The larger E⁰ will be reduced.

Answer 30

the standard electrode potential of hydrogen at 298K 100kPa with a conc of 1 moldm³ of ions.

Answer 31

The more positive ones.

Answer 32

E⁰cell = E⁰reduced - E⁰oxidised.

Answer 33

- write out full equation. - identify half equations and their E⁰. - identify which will be reduced and which will be oxidised. - work out Ecell. - if Ecell is positive the reaction is feasible.

Answer 34

- non-standard conditions, as the electrochemical series assumes standard conditions. - Change in concentration of reactants, as the electrochemical series assumes it is 1 moldm³. - Reaction may not be kinetically favourable, so will react too slow. - Reaction may have high activation energy so it may stop before the reaction has happened.

Answer 35

Fe²⁺₍ₐᵩ₎ + 2e⁻ ⇌ Fe₍ₛ₎ O₂₍𝓰₎ + 2H₂O₍ₗ₎ + 4e⁻ ⇌ 4OH⁻₍ₐᵩ₎ 1. 2Fe₍ₛ₎ + O₂₍𝓰₎ + 2H₂O₍ₗ₎ ⇌ Fe²⁺₍ₐᵩ₎ + 4OH⁻₍ₐᵩ₎ 2. Fe²⁺₍ₐᵩ₎ + 2OH⁻₍ₐᵩ₎ → Fe(OH)₂₍ₛ₎ 3. 4Fe(OH)₂₍ₛ₎ + O₂₍𝓰₎ + 2H₂O₍ₗ₎ → 4Fe(OH)₃₍ₛ₎ 4. overtime Fe(OH)₃₍ₛ₎ becomes hydrated and becomes Fe₂O₃xH₂O

Answer 36

1. Fe²⁺₍ₐᵩ₎ + 2e⁻ ⇌ Fe₍ₛ₎ 2. O₂₍𝓰₎ + 2H₂O₍ₗ₎ + 4e⁻ ⇌ 4OH⁻₍ₐᵩ₎ - in alkaline conditions, the conc of OH⁻ ions are increased. - This shifts the equilibrium in equation 2 to the left to remove the excess OH⁻. - This means there are more e⁻ available so equilibrium in equation 1 shifts to the right using up the electrons. - This forms more iron from ion ions so it is less likely for rust to occur.

Answer 37

Barrier methods: - Oil/Grease act as a physical barrier to prevent water from accessing the Iron/ - Painting. Sacrificial methods: - galvanisation... - Zn/Zn²⁺ half-reaction has a more negative E⁰ compared to Iron. - This means it is a better reducing agent, and so oxidises before the iron does.

Answer 38

VO₂⁺ Oxidation state: +5 Colour: Yellow. VO²⁺ Oxidation state: +4 Colour: Blue. V³⁺ Oxidation state: +3 Colour: Green. V²⁺ Oxidation state: +2 Colour: Purple.