3.2.5 Transition metals Flashcards

Question

What are monodentate, bidentate and multidentate ligands?

Answer 1

**Monodentate**: One lone electron pair for bonding so forms 1 coordinate bond (e.g. H₂O, NH₃, Cl⁻) **Bidentate**: Two lone electron pairs for bonding so forms 2 coordinate bonds (e.g. ethanedioate, en) **Multidentate**: Forms more than 2 coordinate bonds (e.g. EDTA)

Answer 2

Neutral: H₂O, NH₃, en (1,2-diaminoethane) Negative: Cl⁻, OH⁻, CN⁻

Answer 3

**Stereoisomerism** – complex ions can have ligands in **different spatial arrangments** around the central metal ion, giving different isomers with the **same structural formula**.

Answer 4

Occurs in **octahedral** complexes with **3 bidentate** ligands. They form two non-superimposable mirror images, known as enatiomers.

Answer 5

1. Octahedral complexes with **4 ligands of one type** and **2 of another** Example: [Ni(NH₃)₄Cl₂] Cis: Cl⁻ ligands **adjacent** (next to each other) Trans: Cl⁻ ligands **opposite** (sides) 2. Square planar complexes with **2 pairs of monodentate** ligands Example: [Pt(NH₃)₂Cl₂] Cis: Cl⁻ ligands **adjacent** Trans: Cl⁻ ligands **opposite**

Answer 6

The amount of energy needed to make an **electron jump** from a **lower-energy d orbital** to a **higher-energy d orbital** in a transition metal ion. This energy often comes from visible light - the colour of the light absorbed depends on the size of the energy gap (the colour we see is the light that isn't absorbed)

Answer 7

* Identity of the central metal ion * Oxidation state of the metal ion * Identity of the ligands * Coordination number Changing these factors alters the energy gap and therefore the colour of the complex

Answer 8

Degenerate orbitals

Answer 9

Orbitals that have **different enegy levels**.

Answer 10

When ligands bond to transition metal ions, they **split** the energies of the **3d orbitals** into **two energy levels:** The lower energy (**ground state**) and a higher energy (**excited state**). The degenerate orbitals are now non-degenerate orbitals. This splitting creates an **energy gap (ΔE)** between the ground and excited states.

Answer 11

There needs to be an energy input **equal** to the energy gap.

Answer 12

1. When wavelengths of visible light hit the complex ion, electrons can jump from **ground to excited** state by **absorbing** specific wavelengths of light with energy **equal to ΔE** 2. The remaining wavelengths of light are **transmitted** or **reflected** by the complex. 3. These transmitted or reflected wavelengths **combine** to form the colour we see of the complex ion. 4. The observed colour is the **complementary colour** of the wavelength that was absorbed.

Answer 13

**ΔE= hc ÷ λ** ΔE = energy gap (J) h = Planck’s constant (6.63 × 10⁻³⁴ J / s) c = speed of light (3.00 × 10⁸ m/s) λ = wavelength of absorbed light (m)

Answer 14

ΔE = h x f (f is the fequency and measured in Hz or s-1)

Answer 15

* Smaller ΔE → longer wavelength absorbed (e.g., red) * Larger ΔE → shorter wavelength absorbed (e.g., blue or violet) This changes which wavelengths are transmitted and affects the complex’s observed colour.

Answer 16

**ΔE= hc ÷ λ** λ = 7.96 x10-7 m λ = 796 nm

Answer 17

1. Measure absorbance of standard solutions at a specific wavelength. 2. Plot a calibration curve (concentration vs absorbance). 3. Measure absorbance of unknown solution. 4. Use the curve to find the unknown concentration.

Answer 18

Because the 3d and 4s electrons have **similar energies**, allowing multiple stable oxidation states to exist.

Answer 19

+5: VO₂⁺ - Yellow +4: VO²⁺ - Blue +3: V³⁺ - Green +2: V²⁺ - Violet

Answer 20

1. Yellow VO₂⁺ solution changes to blue VO²⁺. The reduction half-equation is: VO₂⁺ + H⁺ + e- ➔ VO²⁺ + H2O 2. Blue VO²⁺ solution changes to green V³⁺. The reduction half-equation is: VO²⁺ + 2H⁺ + e- ➔ V³⁺ + H2O 3. Green V³⁺ solution changes to violet V²⁺. The reduction half-equation is: V³⁺ + e- ➔ V²⁺

Answer 21

It indicates how **easily** an ion is **reduced** (gain electrons) Higher, more positive E⦵ = easier to reduce to lower oxidation states.

Answer 22

* **Ligands** - ligands that form **stronger bonds** than water with the metal ion can raise or lower the potential by **stabilising** oxidation state. * **pH** - acidic conditions provide excess H⁺ ions needed for reduction of some metal ions.

Answer 23

MnO₄⁻ acts as an **oxidising agent**, turning from **purple** to **colourless/pale pink** (Mn²⁺) as it accepts electrons. It is a self-indicator

Answer 24

1. Use a pipette to transfer a measure volume of **reducing agent** (e.g. Fe²⁺) into a conical flask 2. Add dilute **sulfuric acid** to the conical flask to provide excess acidic conditions. 3. Fill a burette with **MnO₄⁻** (aq) ions and slowly add it to the conical flask while swirling continuously. The **purple** MnO₄⁻ ions react to produce a colourless solution of Mn²⁺ ions. 4. Record the volume of MnO₄⁻ (aq) needed to reach the end point. The end point is indicated by the sudden appearance of a **permanent pale pink/purple** solution, indicating an **excess of MnO₄⁻ ions**. 5. Repeat the titrations until concordant titres are obtained (within 0.10 cm3 of each other).

Answer 25

HCl contains Cl⁻ ions, and these can be **oxidised by MnO₄⁻.** Cl⁻ competes with the intended reducing agent, interfering with results. Toxic chlorine gas (Cl₂) is released → it's harmful and unpleasant to work with.

Answer 26

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Answer 27

RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag + 4NH₃ + 2H₂O

Answer 28

MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq) 41.9%

Answer 29

Transition metals have variable stable oxidation states. Their vacant d orbitals can form coordinate bonds with ligands, facilitating electron transfer. They can donate or accept electrons, lowering activation energy and increasing reaction rates.

Answer 30

Heterogenous catalysts exist in a **different phase** from the reactants. They catalyse reactions by **adsorbing** reactants onto **active sites**, on the surface of the catalyst. **Increasing surface area to volume** ratio increases the number of exposed active sites which leads to a **lower activation energy**.

Answer 31

Provide a **high surface area** for the catalyst particles to adhere to. **Maximises** the number of **active sites** available for the reactants to adsorb and react on. This minimises the costs as less catalyst is required.

Answer 32

Vanadium(V) oxide in the Contact process to produce sulfuric acid. Iron in the Haber process to produce ammonia.

Answer 33

Vanadium(V) oxide **catalyses the oxidation** of SO₂ to SO₃. Step 1: SO₂ is oxidised to SO₃: **SO₂ + V₂O₅ → SO₃ + V₂O₄** Step 2: V₂O₄ is re-oxidised back to V₂O₅: **V₂O₄ + ½O₂ → V₂O₅** This regeneration of V₂O₅ completes the cycle, maintaining the catalyst.

Answer 34

Homongenous catalysts exist in the **same phase** as reactants They catalyse reactions by forming **intermediate** compounds with reactants, which decompose to products. Two-step mechanism

Answer 35

**Fe²⁺** in the reaction between S₂O₈²⁻ and I⁻ ions. **Mn²⁺** in the reaction between MnO₄⁻ and C₂O₄²⁻ ions (Mn²⁺ is also an autocatalyst)

Answer 36

Impurities block the active sites on the catalyst’s surface. - Reduces available catalyst SA - Lowers reaction rate - Increases operating costs - less product output - May need to replace catalyst

Answer 37

Sulfur poisons the iron catalyst in the Haber process. The sulfure adsorb onto the iron catalyst surface as iron sulfide.

Answer 38

Fe²⁺ is oxidised to Fe³⁺ by S₂O₈²⁻: **S₂O₈²⁻ + 2Fe²⁺ → 2Fe³⁺ + 2SO₄²⁻** Fe³⁺ then oxidises I⁻ to I₂, regenerating Fe²⁺: **2Fe³⁺ + 2I⁻ → I₂ + 2Fe²⁺** I⁻ and S₂O₈²⁻ are both negative ions, so ions repel each other and few collisions happen = Very slow reaction. But with Fe²⁺, each step of the reaction involves a positive ion, there is **no repulsion** to slow the reaction.

Answer 39

Autocatalysis occurs when a **product** of the reaction **acts as a catalyst**, speeding up the reaction as the product accumulates.

Answer 40

**5C₂O₄²⁻(aq) + 2MnO₄⁻(aq) + 16H⁺(aq) → 10CO₂(g) + 2Mn²⁺(aq) + 8H₂O(l)** **Mn²⁺ acts as an autocatalyst** by speeding up the reaction as it accumulates. Step 1: Mn²⁺ reacts with MnO₄⁻ to form Mn³⁺: **MnO₄⁻ + 4Mn²⁺ + 8H⁺ → 5Mn³⁺ + 4H₂O** Step 2: Mn³⁺ reacts with C₂O₄²⁻ to regenerate Mn²⁺: **2Mn³⁺ + C₂O₄²⁻ → 2Mn²⁺ + 2CO₂** As Mn²⁺ builds up, the reaction speeds up.

Answer 41

Ligand substitution occurs when ligands bonded to a central metal ion in a complex are **exchanged** with other ligands in solution. It is **generally** reversible Often leads to a **colour change** in the solution.

Answer 42

If the incoming and outgoing ligands are **similar in size**, the coordination number and shape **do not change**. Example: [Cr(H₂O)₆]³⁺ + 6NH₃ → [Cr(NH₃)₆]³⁺ + 6H₂O Colour changes from pale purple to dark purple

Answer 43

The coordination number and shape **can change**. For example: [Cu(H₂O)₆]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 6H₂O Colour changes from pale blue to yellow

Answer 44

**Some** of the ligands are replaced, but the coordination number and geometry can remain the **same**. For example: [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O Colour changes from pale blue to dark blue

Answer 45

1. Ligands that form **stronger bonds** with the central metal ion. 2. The replacement of monodentate ligands with multidentate ligands, which leads to more stable complexes due to the **chelate effect**.

Answer 46

The chelate effect refers to the **increased stability** of complexes when monodentate ligands are replaced by multidentate ligands. This is because the number of seperate particles in solution increases Which **increases the entropy** of the system, making the reaction **more thermodynamically favorable**. (Reversing the reaction would decrease the entropy)

Answer 47

Cl⁻ ions are larger than H₂O ligands [Co(H₂O)₆]²⁺ + 4Cl⁻ → [CoCl₄]²⁻ + 6H₂O Tetrahedral