21

Yes, you can use a common stove to test for cations. But a stove is designed to minimize the risk of incomplete combustion (which could lead to production of carbon monoxide), hence its flame always appears as an intense blue flame. Such color contamination could be problematic when testing for cations. In contrast, the combustion (and the color of the flame)...


15

This is not in general true Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms. Both the symmetry number and order of the group are 2. Matrix representations All rotations ...


11

The figure below shows the situation between configuration for a $p^2$ configuration, terms, levels and states. The word 'state' tends to be used colloquially to mean any of Term, Level or State. The Configuration such as $(1s)^2$ or ...$(2p)^2$ etc. tell us which orbitals are occupied. These are split with electrostatic (Coulomb) coupling to form Terms. ...


8

Why not send in white light (with all the wavelengths at once)? This is certainly and routinely done today. The main thing is how much price are we willing to pay? In a rigorous sense, colorimetry refers to the fact that we use optical filters to isolate wavelengths rather than monochromators. When the latter are used, you call them spectrophotometers. ...


7

To give some numbers to MaxW's comment, you can see on this webpage the x-ray absorption energies of basically all the elements. For example, the K-edge of potassium is $3.61~\mathrm{keV}$. The L-II edge of silver is $3.52~\mathrm{keV}$ and the L-1 edge of silver is $3.81~\mathrm{keV}$. So, it seems like you are somehow automatically assigning the peaks and ...


7

To simplify slightly both Raman and IR spectroscopy show the vibrational modes of a molecule (though the techniques used to reveal these are very different). IR spectroscopy relies on coupling between the electromagnetic field of light passing through a sample and the electric dipole of the molecule. But that absorption is only possible is the vibration in ...


7

Contributing to the first half of the answer: The MPI-Mainz UV/VIS Spectral Atlas of Gaseous Molecules of Atmospheric Interest (public access here) allows you to search by chemical name, sum formula, CAS registry number, InChi key or author. The output is a plot, annotated with the literature reference, e.g. (reference) It does not cover IR data to cover ...


7

As Ivan commented, there are actually an infinite number of possible series of this type. So your question is really why there are only six named series. The reason is part of the culture of science. Typically, a result is named if it is, to use Mithoron's term, sufficiently notable. Thus we have the Diels-Alder reaction, the Hartree–Fock method, the ...


6

Since I do not know which compound it is, my guess is that you have two different $\ce{H^1}$ that have almost the same chemical shift at room temperature. The chemical shifts change with temperature as for instance in sucrose: Here are represented mainly the $\ce{H^1}$ of the $\ce{-OH}$ but the $\ce{C-H}$ (the peak of 1) shifts also. In your case, ...


6

To put the other answer in a graphical context, I took the X-ray emission spectra of K and Ag from some place online*, and plotted them together, crudely resized to fit the same scale. Some of the Ag L-lines (I think it's Lβ 1 and/or 2) are at exactly the same position as the K Kɑ line. So, the software sees a certain amount of counts at K, and identifies it....


6

To my understanding, it is simply single atom versus many number of atoms. For example, suppose one atom with an electron at energy level 7 ($n_2=7$). That electron can "de-excite" from $n_2=7$ to $n_1=6, 5, 4, 3, 2,$ or $1$. All those transitions give one spectral line for each. Thus, total of $1 \times 6 = n_1(n_2-n_1)$ (foot note 1) spectral lines would ...


6

The previous answers by @Buck Thorn and @M. Farooq are very good, but not quite complete. The main advantages of FT UV-Vis spectroscopy would be 1) accuracy of wavelength (or wavenumber) determination and 2) better spectral resolution of bands, relative to conventional UV Vis spectroscopy. But it is very hard to do well, and very expensive. The disadvantage ...


6

Could you also mention the correction method? Your auto-fluorescence background is really huge and your signal to noise ratio is low. One of the most cited methods in automated background correction is indeed based on polynomial fitting, albeit little but more sophisticated than just least square fitting. This article is worth reading. Automated Method for ...


6

Since I was reviewing this in some detail recently, I will try to supplement Prof. Hutchison's already good answer with some more detail as to why one would choose to work with normal coordinates, rather than the already quite simple Cartesian coordinates. The short answer is that, under the assumption of small displacements, normal coordinates decouple the ...


5

You can find plots and further explanations online (as I easily did). This is a homework-type question, yet I do feel that the information request for diatomic molecular spectroscopic constants warrants at least this partial answer. See this writeup for the "Vibronic Absorption Spectrum of Molecular Iodine" for more details and illustrations. From NIST: T$...


5

Remote sensing of the simple (few atoms) greenhouse gases works by comparing the measurement spectra with calculated spectra. These calculations are very precise and probably the reason why you do not find so many measured reference spectra. The reason for calculation is that the atmosphere has pressure, concentration and temperature gradients which even ...


5

If we consider N atoms in a non-linear molecule, then each atom can move independently on the X, Y, and Z axes. That's $3D \times N = 3N$ degrees of freedom. But if I move the whole molecule 3 Å along the x-axis, that doesn't constitute a vibration - it's just a translation. So we have to remove 3 degrees of freedom for translations. Similarly, if I ...


5

Long time ago, I wrote mass spectroscopy, by mistake, in an undergraduate exam and the instructor told me "You were the only one in my class who wrote mass spectroscopy." The explanation was that MS is mass spectrometry because it does not deal with light or electromagnetic radiation and that's what the textbook said. At that time internet was not that ...


5

You have assessed the problem correctly. Physicists and chemists use just the opposite directions. Keep in mind that both are arbitrary choices, however the physicist's convention is consistent with electrostatics. All you have to do is to be aware of this direction in the chemistry literature. I have rarely seen any student ever thinking about it which is ...


5

The product $xyz$ is an odd function and as such should produce the same selection rules as 1 photon (electric dipole) transitions. Thus CO has a 1 photon transition at 146.7 nm in the vacuum uv that can be more easily reached with a 3-photon transition with a pulsed laser at 440 nm. ( Three photon transitions are also used in bio-imaging where a ir photon ...


5

This is an interesting question. However, disadvantage is having only $\ce{^1H}$-$\mathrm{NMR}$ to deal with. Yet, it has pretty good resolution (probably using $\pu{400 MHz}$ machine) so we can resolve the splitting pattern very easily and predict the structure. First, OP has correctly assigned two degree of unsaturation. For molecular formula, $\ce{...


5

Yes, of course the Raman scattering can suffer from additional scattering on its way through the sample, just like it will generally be attenuated on its way to the detector by any element in its path (mirrors, lenses, glass surfaces, etc.). If scattering is a problem or not very much depends on your sample. The ideal sample will not scatter your ...


4

If current is the flow of charge why doesn't it stop once all the atoms have been ionized? The short answer is that it is impossible to ionize all the gas at once. It just requires huge amount of energy to ionize the whole atoms at once. Just for the sake of discussion, even at 10000 °C, still a very small fraction of atoms is ionized! This temperature is ...


4

The first step you can take to make the spectral bands narrow is to reduce the slit width. It is slightly wide at this moment. We are not sure where you were looking for wavelengths? In the compiled tables in handbooks or schematic diagrams on the web, the emission wavelengths are listed for the excited atoms using very high temperature flames using air ...


4

If you rotate the molecule by $45^\circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$. Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. ...


4

Consider also water in addition to the solvents you mention: water-d2 (or, $\ce{D2O}$) is an extensively H-bonding polar solvent (dielectric constant $\epsilon = 79$) $\ce{DMSO-d6}$) is polar ($\epsilon= 47$) but aprotic (no-H-bond donation ability) chloroform-d is an apolar aprotic solvent ($\epsilon=4.81$) Now consider the state of a trace amount of $\...


4

An explanation is provided in this abstract (1): Fourier transform spectrometry in the UV-Vis region (FT/UV-Vis), because it is source shot-noise limited, has a signal-to-noise ratio (S/N) disadvantage in comparison to dispersive spectrometry, especially with dense spectra. At the expense of poorer S/N, FT/UV-Vis can be satisfactory for high-...


4

Absorption bands are very small: smaller than the band which a typical monochromator is capable of isolating. If a continuum light source was employed, the absorbed band would be much smaller than the band which would be isolated by the monochromator and which would be read by the detector. This implies a low SNR (signal-to-noise ratio), and a very poor ...


4

Normal modes are technically orthogonal so no energy could flow from one to another, but of course this is just a mathematical construct to make life simple for us, and in reality 'anharmonicity' (physical and electronic) will allow the energy to move about. This is to say that these modes are able to couple to one another. Fermi golden rule is the usual ...


4

Firstly, regarding the extra question: Atkins' Molecular Quantum Mechanics (5th ed.) uses "term" for $^2\!S$, and "level" for $^2\!S_{1/2}$. Back to the main question. It's been a long time since I did term symbols, so I am happy to be corrected, but if I am not wrong, your $^4\!P$, $^4\!D$, and $^2\!S$ terms are not allowed because of the Pauli exclusion ...


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