brightsky's Chem Thread

[brightsky]

Individual bands correspond to energy differences between pairs of orbitals in different atoms. By themselves, they may happen to coincide for different elements; after all, a 2s to 2p jump for one element may well be a 4s to 5s jump for another.

Using the Rydberg equation for the hydrogen atom, we can see that the higher energy levels get closer and closer in energy; energy differences decrease. Therefore, it is entirely possible that two elements absorb/emit light of a similar frequency, if one allows the two similar frequencies to be due to different electronic transitions.

Thanks nliu! But I thought the energy absorbed needs to correspond EXACTLY to energy difference between two energy levels in order for an electron in the lower energy level to be excited to the higher energy level. For example, if Na atoms absorb at 589.0 nm, they won't be interested in absorbing waves of length 589.1 nm, because close is not good enough.

Also I've added a new question to my original post. 

[lzxnl]

If you superimpose an emission and an absorption spectrum for the same atom on top of each other, you'll get a smooth, uninterrupted spectrum.

And you're right, the electronic energy difference must correspond exactly to the energy of the photon emitted. I'm personally not entirely sure about the chances two atoms will have exactly the same energy difference between two orbitals.

[Mao]

Thanks nliu! But I thought the energy absorbed needs to correspond EXACTLY to energy difference between two energy levels in order for an electron in the lower energy level to be excited to the higher energy level. For example, if Na atoms absorb at 589.0 nm, they won't be interested in absorbing waves of length 589.1 nm, because close is not good enough.

In theory, yes, at 0K in an infinitely diffuse gas.
In practice, neighbouring atoms will have small effects on other atoms, temperature will also have small effects. These mean spectral lines are not infinitely sharp nor infinitely thin, and overlap is possible.

[brightsky]

Thanks Mao!

Also, this might seem a dumb question, but say we wanted to determine the sodium content of a sample via some spectroscopic technique. Does it matter whether the sodium is present in the form of Na(g) or Na+(aq)? Will there be slight differences in wavelengths at which the two absorb?

[Mao]

Also, this might seem a dumb question, but say we wanted to determine the sodium content of a sample via some spectroscopic technique. Does it matter whether the sodium is present in the form of Na(g) or Na+(aq)? Will there be slight differences in wavelengths at which the two absorb?

Yes, it does matter. Transition energies depend on electron occupancy as well as core charge. The dependence on electron occupancy is much smaller than the dependence on core charge, but it is affected nonetheless.

Also, you wouldn't want to use those kind of spectroscopic techniques on Na+(aq), because the water would scatter most of the radiation. You would probably want to use Na+(g).

[brightsky]

Thanks Mao! So the Na (g) would absorb at a slightly different wavelength than Na+(g)? And also, surely the scattering can be compensated by using a blank solution?

A few more:

1. What is the purpose of the 'slit' in a spectrophotometer? Does it function simply to focus the light on the monochromator?

2. What is the difference between a colorimeter and a spectrophotometer? Is there a difference? As far as I'm aware, spectrophotometers are used in colorimetry as well.

3. Why are group 1 and 2 metals white in solid state and colourless in aqueous state, while transition metals are colourful in both states? Is it because group 1 and 2 metals do not absorb in the visible region of the electromagnetic spectrum, while transition metals do?

[Mao]

1. What is the purpose of the 'slit' in a spectrophotometer? Does it function simply to focus the light on the monochromator?

The simplest assembly of a monochromator is a prism + slit. This is the way we 'select' a particular color/wavelength.

2. What is the difference between a colorimeter and a spectrophotometer? Is there a difference? As far as I'm aware, spectrophotometers are used in colorimetry as well.

I don't really want to get into the semantics of which of colorimetry or spectroscopy is the broader term. Colorimeter and spectrophotometer generally have very similar instrumental setups, but colorimeters tend to use much broader wavelengths (i.e. terrible monochromator). Where spectrophotometers tend to have spectral purity of <1 nm (high spectral purity),  colorimeters can have the incident spectrum spanning 10s or even ~100nm if using a cheap color filter. Colorimeters tend to be a low accuracy, low cost companion to spectrophotometers. These are good for portable use and preliminary analysis.

3. Why are group 1 and 2 metals white in solid state and colourless in aqueous state, while transition metals are colourful in both states? Is it because group 1 and 2 metals do not absorb in the visible region of the electromagnetic spectrum, while transition metals do?

As far as I know, only Cu and Au are coloured in their pure, elemental metal phase. This is because Cu and Au have electron configurations that absorb red and green/blue light more than yellow. Other metals are all 'silver'-like (though silver is a bit of an odd case, as it will appear pale yellow if you place two silver mirrors parallel to each other, I won't go into the details here).

As for salts and aqueous solutions, the answer lies in coordination chemistry. In short, transition metal ions use their d-orbitals to form weaker forms of covalent bonds with other molecules (sometimes neutral, sometimes charged), this modifies the orbital energies, and can affect the color of the compounds. This area is called inorganic chemistry, it is not part of VCE Chemistry due to its complexity, but is extensively studied throughout university (it is one of the main branches of chemistry)

[brightsky]

Thanks Mao!

For example, a solution of Na+ is colourless, while a solution of Cr2O72- is orange. Why is this?

And why is a solution of Na+ colourless when Na(g) absorbs wavelengths in the visible range (as evidenced by the intense yellow it imparts to a non-luminous Bunsen flame)?

[Mao]

Thanks Mao!

For example, a solution of Na+ is colourless, while a solution of Cr2O72- is orange. Why is this?

And why is a solution of Na+ colourless when Na(g) absorbs wavelengths in the visible range (as evidenced by the intense yellow it imparts to a non-luminous Bunsen flame)?

It's to do with accessible electron transitions in the atom/molecule. There is no easy way to tell if a compound will be coloured or not, but if you have a computer powerful enough to solve the Schrodinger equation for these systems, then you can estimate what the absorption bands are.

[jgoudie]

Another thing you might be confusing emission and absorption.  The intense yellow colour is the emission of light from an electron falling to a lower wavelength.  The absorption is what causes the electron to be promoted in the first place, could it be that the energy required to promote the electron is not in the visible spectrum but the energy released is, a similar theory behind fluorescence.

Thanks Mao!

For example, a solution of Na+ is colourless, while a solution of Cr2O72- is orange. Why is this?

And why is a solution of Na+ colourless when Na(g) absorbs wavelengths in the visible range (as evidenced by the intense yellow it imparts to a non-luminous Bunsen flame)?

[brightsky]

Ah okay thanks! A few more:

1. What do we call an alcohol or a carboxylic acid with a double bond/triple bond in it?
2. What is the name of the molecule CH2=CHCH2OH? Is it '2-propen-1-ol' or 'prop-2-en-1-ol'?
3. What are the IUPAC rules with regard to naming aldehydes and ketones? I know that if the formyl group or carbonyl group is acting as a primary functional group, you tack on the suffix -al, or -one. But what happens if they are acting instead as secondary functional groups (for example, if there is a carboxyl group at one end of the molecule, or if formyl and carbonyl groups are both present in the molecule)? What are the prefixes used?

[lzxnl]

1. Carboxylic acids and alcohols have higher priority than double/triple bonds. We could still have prop-2-ynoic acid and prop-2-en-1-ol.
2. Answered above; it's preferable to have the number next to the functional group it refers to.
3. Oxo for both of them. If we had propanal, and then added a carboxy group on carbon 3, that would be called 3-oxypropanoic acid. As aldehydes and ketones are very similar, they have the same prefix.

[brightsky]

Is HPLC a type of partition chromatography or adsorption chromatography?

[Mao]

Is HPLC a type of partition chromatography or adsorption chromatography?

Adsorption. Partition chromatography usually requires the analyte to go "into" the stationary phase, which isn't possible in HPLC given the silica column is solid.

[brightsky]

The proton NMR spectrum of an alcohol C5H12O has peaks at chemical shift 1.52, 1.39, 1.21, and 0.93. The measured integration trace gives the ratio 0.90:045:2.70:1.35. The first peak is a quartet, the second a singlet, the third a singlet and the fourth a triplet. The peaks at delta 1.52 and 0.93 arise from the presence of a single alkyl group. What is this group?

Might be a dumb question, but can't see my way clear. Any help appreciated!