Advanced Coordination Chemistry Flashcards

1
Q

Describe the solubility of crown ethers.

A

> Crown ethers show unusual solubility: they are all soluble in all common solvents (both lipophilic and hydrophilic).

> This occurs due to the flexibility of the chain which allows the lone pairs (polar bits) to be either internalised or exposed to solvent.

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2
Q

How are corand complexes formed?

A

> Crown ethers can coordinated with high affinity to alkali ions forming corand complexes.

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3
Q

How do crown ether ligands affect the solubility of alkali metal ions?

A

> The ligand lipophilises alkali metal ions e.g. in the presence of [18] crown-6, inorganic salts (KCl, KOH and KMnO4) become soluble in organic solvents such as toluene.

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4
Q

Explain the selectivity of crown ethers.

A

> Selectivity was originally thought to be due to hole size of the macrocycle i.e. each crown specifically binds to a certain alkali metal cation. THIS IS WRONG

> There is some correlation between affinities and cavity diameters but the simplest explanations is that as the number of donor groups increase (as the size of the crown ether increases), there is an increase in enthalpy due to more coordination interactions.

> ALSO: the free energy of solvation (cost before the benefit of coordination) for K+ is lower than that of Na+, Li+ and Ca2+ i.e. it is easier to remove solvated water from K+ before crown coordination. (smaller charge density = weaker solvation) .

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5
Q

What are podands?

A

> Podands are acyclic polyether analogues of crowns.

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6
Q

Describe the binding affinity behaviour of podands.

A

> Although they have multiple donor groups, podands bind to metals ions with much lower affinities than the crowns.

> This is due to the MACROCYCLIC EFFECT (different to the chelate effect which is entropy driven).

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7
Q

What is the chelate effect?

A

> Polydentate chelate ligands form more stable complexes (have a higher affinity for metal ions) compared to their monodentate analogues.

> Macrocyclic complexes are stabilised by chelate effects.

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8
Q

What is preorganisation?

A

> In addition to the chelate effect, macrocycles undergo less conformational change on binding to a metal ion than a podand (the entropic cost of binding is already paid) - this phenomenon is preorganisation.

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9
Q

What 3 factors contribute to the MACROCYCLIC EFFECT?

A

(i) A macrocycle undergoes less conformation change on binding to a metal ion than a podand - PREORGANISATION.
(ii) Donor lone pairs are held together in the macrcocycle framework so the enthalpic cost of unfavourable electrostatic interactions had been paid.
(ii) Macrocycles are less solvated than analogous podands as they have less solvent accessible surfaces.

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10
Q

What are cryptands?

A

> Cryptands are 3D bicyclic crown analogues. They are more preorganised than crowns due to the 3D element.

> Complexes of cryptands are known as cryptates.

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11
Q

How can binding preferences of cryptands be adjusted?

A

> Cryptands show a “cavity diameter effect” and binding preference can be adjusted by changing 1 arm of the cryptands i.e. the rigid structure of the macrocycle cavity can be fine-tuned.

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12
Q

How can cryptand rigidity dictate selectivity?

A

> Rigidity means that cryptands cannot flex around ions that are too small for their cavity and cannot expand to accommodate ions larger than the cavity.

> Therefore, each cryptand selectively binds specific metal ions (can be used in metal detection).

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13
Q

How do cryptand binding affinities compare to that of crowns?

A

> Cryptand binding affinities for metal ions are HIGHER than those for corresponding crowns by several orders of magnitude.

> Although cryptands contain more donor groups than crowns, their increased affinities have a larger preorganisation component.

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14
Q

What are spherands?

A

> Spherands= most rigid macrocycles and more preorganised.

> They contain binding pockets where convergent donor groups are held more rigidly in the correct geometry for metal binding (claw type structure that surrounds the metal ion).

> There is NO conformation change on binding to metal ion, therefore, it binds 10^12 more effectively than the equivalent acyclic podand (NO entropic cost of binding).

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15
Q

What is the effect of changing donor atoms?

A

> Macrocycles based on hard O donors bind to hard metal ions e.g. alkali earth metals.

> Incorporation of softer donor atoms (e.g. S) has a radical effect on the metal ion binding selectivities (LogKb).

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16
Q

How do crowns and cryptands bind to non-metallic cations?

A

> Crowns and cryptands can complex cations such a [NH4}+ via hydrogen bonding.

> Crowns interact via O——H-N interactions to give what is known as 2perching2 complex, with the [NH4]+ sitting above the crown and making usually 3 H-bond contacts.

> Cryptands can show higher selectivity through preorganisation → the large ball shaped cryptand selectively bind to [NH4]+ over K+ (x500) form 4 tetrahedrally arranged N—H-N interactions.

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17
Q

What 2 methods are used to synthesise macrocyclic ligands?

A

(i) Template Synthesis
(ii) High Dilution Techniques

Note: The methods are not mutually exclusive (sometimes syntheses require both methods).

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18
Q

Outline the synthesis of crown ethers.

A

> Most common method: 2 presynthesised fragments are broight together in the final macrocyclisation step. (OTs is a good LG for this in THF using Base (KOH)).

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19
Q

Why are these reactions so high yielding?

A

> There are no oligomeric/polymeric material produced.

> If KOH is replaced with NBu4OH, a mixture of polymeric ether products result → NO CROWN. This is a example of the TEMPLATE EFFECT.

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20
Q

What is the TEMPLATE EFFECT?

A

> Template effect: A template organises an assembly of atoms in a specific geometry in order to achieve a particular linking of atoms.

> Template process: where a metal ion (or centre) with definite stereochemistry and electronic properties serves as mould for forming a product that would normally be difficult or impossible to obtain.

> In this specific case, the intramolecular macrocyclisation step has to compete against the unwanted intermolecular polymerisation step - the metal template facilitates the intramoleular reaction.

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21
Q

What is the mechanism of the template effect?

A

> The mechanism of the template effect can be either kinetic or thermodynamic (or a mixture of the 2).

> Kinetic effect: when the metal ion provides a site for the reactants to assemble themselves in the right arrangement to produce the target product structure.

> Thermodynamic effect: coordination to a metal ion shifts a reaction equilibrium towards a desired product (stabilises intermediate to create a product not usually accessible).

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22
Q

Outline the synthesis of cryptands.

A

> Cryptand synthesis usually has several steps where A≠B≠C. Therefore unsymmetrical cryptands are easily accessed (high control).

(i) similar to crown ether - template synthesis
(ii) Classical high dilution (tiny amounts of reagent in ~1L solvent - always very low concentration).

amine + acid chloride → (High dilution) → amide → (LiAlH4) → amine

> 2nd order intermolecular reaction is favoured by high concentration (low dilution) → gives polymeric products X

> 1st order intramolecular reaction is favoured by low concentration (high dilution) → gives desired cryptand :)

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23
Q

Why does high dilution work?

A

rcycl/rpol α 1/[A-B]

> inverse proportionality means that rcycl (rate of cyclisation) increases as [A-B] (reactant concentration) decreases.

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24
Q

What role does [18] C-6 play in oxidations using KMnO4?

A

> Addition of catalytic amounts of [18]C-6 (~3-5 crystals) allows oxidations using KMnO4 to be completed at room temperature in quantitative yields.

Reactions: Alkene → Carboxylic acids, Alcohol → Ketone (using KMnO4 in Toluene → “purple toluene”)

> [18]C-6 binds to KMnO4 and forces it to dissolve in the toluene where is performs the reaction.

> [18]C-6 then precipitates out and can be reused (reason why only a small amount is needed).

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25
Q

What role do crowns and cryptands play in activating inorganic bases?

A

> Crowns and cryptands are phase transfer catalysts → they are able to transport species from one phase into another e.g. from aqueous solvent into non-aqueous solvent ( can therefore facilitate reaction of hydrophilic inorganic salts with lipophillic organic compounds).

> e.g. In organic solutions, CO32- also behaves as a very strong base (usually a very mild base) i.e. it is activated in non-aqueous solvent. In water, CO32- is a weak base due to solvation sphere. It is taken across into non aqueous solvent where it is a much stronger base (bare carbonate).

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26
Q

Why does a solvation sphere inhibit the action of inorganic ions?

A

> When ions are solvated by water molecules, these water molecules need to be removed so that a reaction can occur (part of the reaction activation energy).

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27
Q

How can non aqueous solvents affect ion action?

A

> In non-aqueous solvents, ions can be strongly ion paired as electrostatic interactions in low polarity solvents are high → also leads to activation energy.

28
Q

Describe how Monencin (natural antibiotic) and other crown ether derivatives act as switchable binding sensors.

A

> Monencin works by binding to Na+ and transporting it across membranes (ionosphere). It has 2 structures, a LINEAR NON-BINDING MODE → deprotonation of the terminal acid group results in CYCLIC Na+ BINDING MODE.

> can therefore use pH to “switch” this molecule on and off.

> Crown ether derivatives with similar properties can behave as IONOPHORES and carry metal ions from basic aqueous solutions across membranes into acidic aqueous conditions.

29
Q

Describe a sulphur based redox switch.

A

> A sulphur based system can be switch using common oxidising and reducing agents to reversibly form a DISULPHIDE BRIDGE.

> In the absence of metal ions, oxidation of the podand produces mostly polymeric material.

> In the presence of Cs+, the main product is the macrocycle (kinetic template effect).

30
Q

Describe ferrocene based redox switching.

A

> The ferrocene unit shows a reversible oxidation to form the ferrocenium ion.

> The structure = cryptand where one arm is replaced by the redox active ferrocene unit (reduces the no of donor groups which is compensated for by an increased degree of preorganisation.)

31
Q

Describe an example of photo-switchable binding.

A

> Azobenzene photo isomerises between cis and trans forms.

> The cis form is less stable mostly due to steric reasons.

> The trans isomer binds to metal ions (K+) with almost identical affinities as [18]C-6. On photoisomerisation to the cis form, the binding site of the macrocycle is blocked by intramolecular complexation of the ammonium ion.

> High energy light causes the trans → cis transition. ( need to keep the cis form cool and away from blue light to prevent it switching back).

32
Q

What are butterfly crowns?

A

> Butterfly crowns can switch between 2 binding modes.

> The trans mode binds to alkali metals (with similar preferences as [18]C-6. The trans isomer can potentially bind to metal ions in each of the crown moieties.

> The cis form preferentially binds to larger alkali metal ions (e.g. K+, Rb+, Cs+). This isomer binds metal ions in an intramolecular 1:2 sandwich structure (selectivity).

33
Q

Desribe how [18]C-6 and anthracene act as alkali metal sensors.

A

> The anthracene part of the molecule is luminescent → this is QUENCHED until K+ complexation which cause the luminescence to switch on.

34
Q

Why is emission quenched?

A

> In the unbound sensor, the lone pairs (non-bonding of the crown ether) are available to quench the excited state of the anthracene through PHOTO-INDUCED ELECTRON TRANSFER (PET).

> When K+ is bound, the lone pairs on the crown are NO LONGER AVAILABLE to quench the excited state through PET.

35
Q

What is magnetic susceptibility (χ)?

A

> Magnetic susceptibility is a quantitative measure of the response of a material - The intensity of the magnetic dipole acquired by a material - its magnetisation (M) when a magnetic field (H) is applied to the material.

> χ=M/H

36
Q

What is a more useful measure?

A

> Mass susceptibility (К) and the molar susceptibility (χmol) are more useful.

> χmol = К . Mr/10^3 where К= χ/ρ

> A high value of χ means the material is easy to magnetise (applies to materials with unpaired electrons).

37
Q

How is χmol related to the effective magnetic moment (μeff)?

A

μeff = 2.84 √χmolT

μeff = √n(n+1)

38
Q

What is diamagnetism?

A

> Any material with paired electrons is diamagnetic (most materials).

> When a field H is applied to a material, M (its magnetisation) can be more or less than H. In diamagnetism, H>M, χ is negative (the material is repelled by a magnet).

39
Q

What is paramagnetism?

A

> Paramagnetism is when HOnly materials with unpaired electrons can be paramagnetic.

Note: a material that is paramagnetic also has diamagnetism is there are pair electrons but the effect is very small and swamped by the larger paramagnetic effect.

40
Q

How is χ measured?

A

> The mass of the sample is measured in a magnetic field and outside of it.

> The difference in weight (due to susceptibility effect) is the force (F) the field exerts on the sample.

> If F of the sample is compared to a sample with known χmol then χmol of the unknown sample can be calculated.

> Can also use SQUID (quantum effect) to directly measure the magnetic susceptibility.

41
Q

What effect does temperature have on χ?

A

> Diamagnetism = unaffected by temperature

> Paramagnetic materials = affected by temperature and follow CURIE LAW.

42
Q

What is Curies Law?

A

χ α 1/T = C/T

where C = a constant for that material

43
Q

What is a magnetically dilute material?

A

> A substance that displays simple paramagnetism (the magnetic centres do not interact with each other and act as isolate dipoles).

44
Q

What is a magnetically non-dilute material?

A

> A substance in which the magnetic dipoles do interact i.e. the show cooperative behaviour (long range order in materials allows spins to “see” each other and interact) - they do not obey Curie’s law.

?Anything that can be thought of as “a magnet”.

45
Q

What is antiferromagnetism?

A

↑↓↑↓↑↓↑↓
↓↑↓↑↓↑↓↑
↑↓↑↓↑↓↑↓
↓↑↓↑↓↑↓↑

> A manifestation of ordered magnetism that only exists at low temperatures (below the Neel temperature).

> Spin pairing (antiferromagnetism) occurs at lower temperatures because it is possible to overcome the thermal energy.

46
Q

What is ferromagnetism?

A

> The basic mechanism by which certain materials form permanent magnets.

47
Q

What is ferrimagnetism?

A

> Many transition metal compounds sow a cooperative behaviour - a combination of ferromagnetic and antiferromagnetic behaviour to give an overall ferromagnetic effect.

> This interaction involves 2 magnetic centres distributed in a lattice with unequal rations (one can produce more magnetism than the other).

> Magnetic ordering results in an effect that seems like ferromagnetism.

> This occurs because the interaction between neighbouring metal ions in different lattice sites is more intense than interaction between metal ions in the same site.

48
Q

How do magnetically dilute species begin to interact?

A

> Through suitable bridging ligands providing a superexchange pathway.

49
Q

Give a real example of tuning magnetic interactions and explain how it works.

A

> Hydroxyl Bridged Cu(I) Dimers:

Cu(II) → d9 → S=1/2 (1 unpaired electron)

> Depending on the N-donor ligands, the Cu-O-Cu angle could be used to tune the interactions from antiferromagnetic to ferromagnetic.

> If L=bulky, Cu-O-Cu angle > 97.5 degrees → antiferromagnetic

> If L=small, Cu-O-Cu angle < 97.5 degrees → ferromagnetic.

> For each Cu(II) centre, the single unpaired electron is in the dx2-y2 orbital. L supplies 2 electrons (1 for each bond) → total electrons = 4.

> When the ligands produce an “open” Cu-O-Cu angle (>97.5), the superexchange is mediated by a singlet oxygen p-orbital. The mixing of 3 atomic orbitals produces MO’s yielding a SINGLET GROUND STATE.

> When the Cu-O-Cu angle ~90 degrees, hen superexchange is mediated 2 orthogonal oxygen p orbitals. This produces a pair of both bonding and antibonding orbitals and leads to a TRIPLET GROUND STATE.

50
Q

How can magnetic interactions be extended?

A

> By selecting suitable bridging ligands and controlling the coordination sphere of the metal ion, the interaction can be extended.

51
Q

How can charge transfer salts be used as magnets?

A

> TCNE and TCNQ are easily reduced to radicals (electrons will be delocalised). They are relatively small and flat so produce HIGH SPIN DENSITY.

> In the right conditions, these building blocks react o form ionic charge transfer salts (2 systems with unpaired electrons).

> The crystal structure of the salts show that because both are small, they can STACK close together in the solid state (stacking interactions).

> The ions are close enough for spin interactions to occur and at 4.8 K the solid becomes a ferromagnet. As the material cools, long range magnetic ordering occurs.

> Since the material is ionic, it is soluble in polar organic solvents, however, SOLUTIONS ARE NOT MAGNETIC as long range order is broken.

Note: these are the first materials mentioned where magnetism is (partially) due to unpaired p electrons. (up til now all been metals not organic salts.

52
Q

Define and describe the phenomenon of hysteresis of charge transfer salts at low temperatures.

A

> The magnetic phase of the charge transfer salt is a hard magnet and so displays hysteresis.

> Hysteresis= any situation in which the value of one variable depends on whether the other has been increasing or decreasing i.e. the value of the first variable is path dependent.

> Hysteresis is the basis of all current memory devices based on magnetism - allows data to be “written” and “read”.

> Hard magnets: retain their magnetism after being magnetised (can disrupt by heating).

> Soft magnets: easily magnetised and demagnetised when magnetic field is applied.

53
Q

What are potential synthetic room temperature magnets?

A

> Potentially useable magnetic materials need to be room temperature magnets. e.g. V(TCNE)2(CH2Cl2)0.5 → black amorphous material , it is a soft room temperature magnet with Tc > 400 degrees (not thermally stable).

> Structure = not completely known but xray powder analyses show that it is composed of a network of 6 [TCNE].- anions coordinated to V (II) units.

> Also possible: C(CO)6 + TCNE → V(TCNE)2 via chemical vapour deposition (CVD) to produce thin magnetic layers on most surfaces which can then be plastic coated (air stable for days).

54
Q

Describe the synthesis,magnetism and optical properties of Prussian Blue.

A

> Prussian Blue: (Fe(III)4 [Fe(II) (CN)6]3) (1st synthetic pigment).

> Synthesis: 4Fe 2+ (any Fe (II) salt) + 3[Fe(CN)6]3- → Fe(III)4 [Fe(II) (CN)6]3 (mixed valence compound). - REDOX

> Photo-switched magnetism: although PB is a ferromagnet with Tc=5.6 K, only the Fe (III) sites are spin active (Fe(II) → LS d6).

> Many synthetic magnets have been made by substituting paramagnetic metal ions into every site of the PB structure e.g.
CoCl2 + K3[Fe(CN)6] → K0.4Co1.3[Fe(II)(CN)6]3 (non-stoichiometric formula). This structure contains Fe(II) → d6 -CN - Co(III) → LS d6 ∴ is not magnetic (diamagnetic).

> PB complexes are optically transparent and can be photo excited. Excitation into an absorption band at 500-700 nm (red) causes the follow charge transfer process within the solid:
Fe (II) (d6, LS)-CN-Co(III) (d6, LS) → Fe (III) (d5, LS)-CN-Co (II) (d7, HS) - CAUSE THE MATERIAL TO BECOME A HARD FERROMAGNET.

55
Q

Describe the use of the Faraday effect in these materials.

A

> Thin films of the optically transparent magnetic materials display the Faraday effect → in a magnetic field (or when magnetised) the material rotates plane polarized light by a characteristic angle. For these materials:

(i) Faraday rotation is observed at r.t. in the ferromagnetically ordered phase - but only at wavelengths associated with the charge transfer bands observed in the absorption spectrum.
(ii) The exact wavelength of the bands depends on the stoichiometry of the material, which can be varied by reaction conditions.

> Potential use: opto-magnetic memory devices using specific laser colours to read/write data.

56
Q

What is the spin crossover phenomenon?

A

> The spin crossover (SCO) phenomenon is only observed in specific transition metal complexes (d4-d7; 1st row metal ions) and is when the spin state of the complex changes due to external stimuli such as temperature.

> SCO occurs when P (pairing energy) ~ Δoct i.e. the ligand is intermediate to strong/weak field. In such systems, the LS spate is always the lowest energy state.

57
Q

What are the 4 main criteria to consider when designing SCO systems?

A
  1. Virtually all SCO systems involve octahedral Fe (II) metal ions with aromatic N-donor ligands.
  2. Placing substituents close to the the N-donors prevents the ligands closely approaching the metal ion - bond lengths are shorter in LS complexes.
  3. Changing a 6-membered ring to a 5-membered ring decreases σ donor and π acceptor character. In bi/tridentate systems this also introduces strain into chelate rings. All this lowers Δoct (desirable).
  4. Replacing some aromatic N-donors with aliphatic N donors .
58
Q

Describe how hysteresis can occur through weaker interactions.

A

> Cooperativity and hysteresis do not require direct bonding between spin active units. Weak interactions such as hydrogen bonding and aromatic stacking between Fe(II) units can also produce these effects.

> example: stacking of the pendent phenyl rings in an Fe(ii) complex creates the long range order required for cooperativity and therefore hysteresis,

59
Q

Describe how SCO complexes can be light-switched.

A

> Spin crossover is a d-d- transition. At room temperature such transitions are VERY shortlived (~2-3 ns).

> In solids at low temperatures, different behaviour is observed e.g. when [Fe(propyl-trz)3]2+ is irradiated with green laser light at 5K, the low spin state converts to high spin. This is known as LIGHT-INDUCED-EXCITED-SPIN-STATE-TRAPPING (LIESST).

At these temperatures, high spin states are stable for days, arming up to ~70 K provides enough thermal energy to deactivate the metastable high spin state.

60
Q

Describe the process of photo-excitation for metal complexes.

A

> Octahedral Complex:

(i) Ligand based: π→ π* (intense, usually in the UV).
(ii) Metal based: d-d transitions (sometimes called ligand field transition (usually weak in UV/Vis region)

> Charge Transfer Transitions:

(iii) Ligand to metal transfer, LMTC (intermediate to intense in visible).
(iv) Metal-to-ligand charge transfer, MLTC (intermediate to intense in visible).

61
Q

How can the energy of a complex displaying MLCT be tuned?

A

> The HOMO of the state can be tuned by varying the metal ion.

> The LUMO of the state can be tuned by varying the ligand.

62
Q

How can emission from MLCT states be tuned?

A

> More N in the ligands makes the metal more electron deficient lowering the LUMO →excited state is less energetic → gap between HOMO and LUMO is closed.

> Changing the central metal to one with more electrons raises the LUMO (higher energy) → smaller energy gap.

63
Q

What is Stokes shift?

A

> Stokes shift = difference in wavelength between the absorption and emission band.

GS → 1.MLCT (singlet state) → (intersystem crossing)→ 3. MLCT (Triplet state)→ (luminescence) →GS

64
Q

What are the problems associated with water splitting?

A

(i) the reductant/oxidant product pairs can undergo back electron transfer at rates comparable to the required forward process.
(ii) The excited state of the metal complex is relatively short lived.
(iii) Only a small part of the solar spectrum is being used.

65
Q

What are the advantages of Dyes sensitized solar cells (DSSC)?

A

(i) relatively high efficiency (up to 11%)
(ii) They work in lower light conditions (cloudy days)
(iii) Cheap
(iv) can easily be laid onto flexible plastic substrates.