Discussion

PART A:

How do bulk metals interact with light?

Again, I have no idea what the question want. In general, bulk metals are very dense which prevent light from deeply penetrating into the metal, thus most of these interactions occurs on the surface of metals. Firstly, bulk metals can absorb light (theoretically, any surface can do this, but amount and type absorbed varies). The absorption mechanism is that electrons within the metal lattice (including surface electrons) all tend to vibrate at certain frequency (natural frequency). When the light of same natural frequency impinges upon the atom, then the electrons of that atom will be set into continuous vibrational motion (resonance). Thus light energy is converted into kinetic energy which is later converted into thermal energy via various interactions between these vibrating electrons and the metal lattice, therefore, light is absorbed and never released. This interaction allows bulk metal to have an observable colour, e.g. silicon is black since it can effectively absorb all visible spectrum. Secondly, the shininess of bulk metals is the result of reflecting some of the incident light into our eyes. Bulk metals are much more denser than air, thus when light "enters" the metal, it experience a huge refraction index which will cause the light to refract at an angle much greater than the critical angle, hence light can "bounce off" the metal surface. Thirdly, there is photoelectric effect, this effect states that electrons of an atom will selectively absorb a photon (particle model of light) to promote it self into an unstable, high energy state (transition between valence band and conduction band). Then, since this state is unstable, these electron will rapidly fall back into their ground state and a corresponding photon is released to conserve the energy, thus light is re-emitted by the metal surface. In the case of transparent material (just in case if there is some transparent metals which I don't know), this energy is passed onto the neighbouring electron until been re-emitted on the other side of the material.

What causes different sizes of silver nanoprisms to appear as different colours in solution? Explain your answer.

When come to nanoparticles and nanoprisms, they don't behave as mentioned above, the colour of the solution is actually proportional to the particle/prism size. Instead of having a constant band gap (minimum amount of energy required to promote a valence electron into a conduction band - previously unoccupied orbital) which allows the metal to have a consistent appearance. Size of band gaps of nanoparticles/nanoprisms is actually size depended. For example the band gaps of CdSe nanoparticles are actually decreasing with increasing size (Pic_7). Thus the amount of energy released when a electron fall from conduction band is different, a larger band gap means more energy must been released as the electron fall (conservation of energy). Subsequently electromagnetic waves of shorter wavelength must been emitted (the colloid appears darker). 

Which wavelength of light (photon) has higher energy: red or blue?

Pic_8 a)

Derive the equality A = 2 – [log(%T)] from A = - [log(T)].

Pic_8 b)

What colours were observed for each concentration of KBr (potassium bromide) added?

Pic_3 (in the "Methods and Observation" page)

From left to right: 59 μL, 0 μL, 29 μL and 9 μL

Corresponding colours: light yellow, gray - blue, dark yellow, and dark blue.

These are all mine, waiting for other group members to share their results and/or demonstrators to send me the class result.

Does nanoprism size increase or decrease with addition of KBr?

According to the Lab Manual ​(link provided in the "Methods and Observation" section; P3), potassium bromide (KBr) can be added at substoichiomteric amounts to the reaction to alter the size at which the particle stops growing. At higher concentrations, potassium bromide limits the growth of the nanoparticle to a larger degree, leading to smaller sizes of nanoprisms being produced (It regulates the size of silver nanoprisms by introducing Br^- ions which strongly bond to silver to prevent further growth).

Just again out of interest, since this is provided in the Lab Manual, I'll assume is reliable and valid. After linking this known fact with the experimental observation (general trend across all samples prepared in the class), I can say that the band gap of silver nanoprisms is decreasing as their size decrease. This is exactly opposite to the CdSe nanoparticle (Pic_7) which is in our lecture 14 (again assumed reliable and valid). Is this differenced in behaviour got to do with the differences in element? OR The geometrical differences: Ag nanoprisms are prisms and CdSe nanoparticle are "irregular" particles? OR The technique used to synthesise them? OR One of the source is unreliable which provided wrong knowledge.

Also a significant inconsistency was observed, since  a quite number of class members prepared red solutions with addition of 20-30μL of KBr. The colour red indicates these silver nanoprisms are smaller than all my colloids, even the one with 59 μL of KBr since that one is still yellow (red is less energetic than yellow photons). This just contradicted my answer to this question...

What is the reduction reaction in the formation of silver nanoprisms?

Ag⁺_(aq) + e^- ----> Ag_(s)

Above is all I had before viewing Hsin's ELN (technically the question only asked for the reduction half of this redox reaction) . However, I think the oxidation is necessary to explain how electrons are provided, especially after seen Hsin's ELN; but his equation is unbalanced. Also, Our Lab Manual didn't mention anything about the spontaneous combustion of B₂H₆ gas at room temperature (a risk?) and no sweet odour was smelt by me during my reaction, hence instead of his B₂H_6 (g), I suggest 2BH_{3 (g)}.

2NaBH_{4 (aq)} ----> 2Na⁺_(aq) + 2BH_3 (g) + H_2 (g) + 2e^-  (I don't even know if such reaction exist since all information on the internet is about reduction of ketones and aldehydes, but this is the best I can offer)

Overall reaction: 2Ag⁺_{(aq) +} 2NaBH_{4 (aq)} ----> 2Ag_(s) + 2Na⁺_(aq) + 2BH_3 (aq) + H_2 (g)

What is the role of sodium borohydride?

Again according to the Lab Manual (P2), sodium borohydride is the reductant that reduces Ag⁺ ions to neutral Ag atoms, which also explains why this reagent is added last since immediately after the addition of this reactant, silver nanoprisms will start to grow.

How many silver atoms are in a single silver nanoprism with the shape of an equilateral triangle, with an edge length of 40 nm and thickness of 5 nm? The density of bulk silver is 10.5 g/cm3.

Pic_9