text:
For covalent compounds, what is the microscopic difference between a solid and a liquid?
I've got a question regarding the microscopic difference between liquids and solids. I know that there exists a precise phase transition between solids and liquids, so some discrete parameter must change in the transition. I'm letting the molecules under examination be covalent, so we can't assume that a crystal lattice is formed in the solid case.
(1) Are different forces dominant in between solid and liquid molecules? By "forces" I mean ionic bonds, dipole-dipole interactions, London forces, etcetera.
(2) What is the difference in the geometric structure of the molecules' interactions? Solids are fixed and gases are mobile, but what precisely are liquids? Solid structures with enough degrees of freedom to pivot fluidly? Do liquids generally have the same structure (simple cubic, hexagonal, etcetera) as their solid counterparts when they do interact?
I guess my primary confusion is that the distance between molecules in a material (solid or liquid) should intuitively be a continuous parameter dependent on the density of particles and their temperature, not a discrete parameter taking the values present in the "solid" and "liquid" states. Why the discrete nature?

Unfortunately it seems your question is too broad, or more precisely there are various questions in your post which are broad by themselves. Also I have no idea how you can know the types of crystal lattices, but don't know that liquids can't have one.

Let's break this down, and keeping it to simple covalent compounds...

Are different forces dominant in between solid and liquid molecules? By "forces" I mean ionic bonds, dipole-dipole interactions, London forces, etcetera.

All those forces (if applicable) will still exert their expected effect and there is no 'dominant force' which increases in strength when temperature rises. Just that when we raise the temperature, the molecule may have enough kinetic energy to overcome London Dispersion but not ionic forces, etc.

What is the difference in the geometric structure of the molecules' interactions? Solids are fixed and gases are mobile, but what precisely are liquids? Solid structures with enough degrees of freedom to pivot fluidly? Do liquids generally have the same structure (simple cubic, hexagonal, etcetera) as their solid counterparts when they do interact?

So let's take a look at a typical phase transition diagram, that of water:

At the flat stages, water is changing discreetly and exists either as solid (rigid lattice) or liquid (completely formless and structure-less). And if you look at ice melting, that is exactly what you see and experience.
But the source of this question is probably the fact that you've also seen certain solids soften under heat without melting completely. These covalent materials are called thermoplastic and have a property called the "Glass Transition Temperature". The Stiffness-vs Temperature graph looks like this:

In this state, the melting solid is able to overcome the forces that hold it to a lattice, but is not able to overcome the forces that hold the molecules together and so the molecules cannot 'flow'. The reason this exists for some compounds and not others is largely a property of how their molecules hold each other together.


text:
Acid-molybdate spray of Hanes & Isherwood
I am reading a paper where they develop the phosphate spots on a paper chromatogram using what they call " Acid-molybdate spray of Hanes & Isherwood  (1949)".
Any idea what this reagent is?
I tried looking up the original reference but unfortunately I cannot fine a full text version. http://www.ncbi.nlm.nih.gov/pubmed/15398090?dopt=Abstract&holding=npg
Sigma has "Ammonium molybdate tetrahydrate" reagent available which is described as a component of the "Hanes and Isherwood reagent " but not sure what else is supposed to go in there.
http://www.sigmaaldrich.com/catalog/product/fluka/09880?lang=en&region=IN
Any ideas?

I managed to find a re-print of the paper in question– sadly it seems like Nature are mothballing many old papers (dead/missing links).
There are many acidic-cerium/molybdenum type stains that are used nowadays, many of which are commercially available off the shelf, so you might wish to try these before going after something from 50 years ago. (Cf. CAM stain, Hanlessian's stain)
I think this paper may also have pre-dated silica backed TLC plates, which will undoubtedly give you better separation than paper chromatography.

Taken from Nature (Nature Publishing Group)

source: Stack Exchange

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sort: votes


text:
Molecule analog similar and commercially available
I wanted to carry out some experiments on the following molecule:

Unfortunately this molecule (santalene; either isomer will do) seems neither commercially available nor easily synthesizable.
Hence I am looking for analogs that I can start out with. Any idea which commercially available molecules might be similar to this molecule?
I did fine one analog (curcumene) but even that does not seem to be available with Sigma or any of the usual sources for even fairly esoteric molecules.

Any thoughts / tips? Alternatively, a synthesis for my target molecule might also help, in case I am missing any obvious routes.
PS. Any recommendations for tools / sites that will do a good search for analog molecules?

There's an old Corey paper on the synthesis of santalene DOI

Santalene is the primary constituent of sandalwood oil, which is commercially available.  You could isolate pure material by fractional distillation of the oil. http://www.sigmaaldrich.com/catalog/product/aldrich/w300500?lang=en&region=US

@CurtF. Thanks. But to the best of my knowledge it is the Santalols which are the major constituents of the oil. Santalenes are 1-2% max. Separating enough from the oil seems difficult. But I could try that.

Good point @curious_cat.  You are right.  This info might be helpful: http://www.ncbi.nlm.nih.gov/pubmed/22900258

Additionally, I found very old references in the literature that dehydration of $\alpha$-santolol leads (unfortunately) to "$\alpha$-iso-santalene" and similarly for the $\beta$ isomer.  I'm not 100% sure what the structure of  "iso-santalene" is but presumably it is not what you are interested in?

@CurtF. Any thoughts on what to use as the dehydrating agent?

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text:
Is there an abbreviation for "something bulky"?
I want to draw a scheme that indicates the difference between the products formed in a reaction with a molecule that will form one thing when it has methyl or hydrogen substitution, but another thing when it has any substituent more bulky than isopropyl.
Is there any abbreviation (like "R" or "X") for "bulky"?

In discussion about Felkin-Anh etc. $\mathrm{R_L}$ or just $\mathrm{L}$ are sometimes used.

I've never come across one, but mostly those I come across will have "R" or "X", but somewhere in the reaction scheme specify what R is, like:
R = Me, Et, Pr, iPr...
So in your case, you could specify something like:
R = iPr or bulkier

I thought the series was Methyl, Ethyl, Butyl, Futyl :)

Nice one~ @porphyrin

I've seen numerous textbooks utilizing $\ce{R^*}$ ("R-star" notation) for a bulky group, and also even $\ce{R^{**}}$ ("R-star-star" notation) for "extremely"/"mega" bulky groups. From my understanding, star "*" symbol is supposed to resemble branched structure of the substituent.
Illustration on usage from [1, pp. 85--86]:

In place of a bulky substituent, we selected the chemically very inert tri-tert-butylsilyl group $\ce{Si^tBu3}$, called supersilyl and symbolized by $\ce{R^*}$. [...] In addition, the even more sterically crowded disupersilylsilyl group $\ce{R'} = \ce{SiHR^*2}$ and the extremely bulky
methyldisupersilyl group $\ce{R^{**}} = \ce{SiMeR^*2}$, which we jokingly call “megasilyl”.

Note: An asterisk in the right superscript position next to the chemical element symbol also denotes electrical or nuclear excitation.
References

Silicon Chemistry: From the Atom to Extended Systems; Jutzi, P., Schubert, U., Eds.; Wiley-VCH: Weinheim; Cambridge, 2003. ISBN 978-3-527-30647-3.


text:
Evolution of gas during chemical reaction
Why is it that whenever hydrogen gas is produced in a reaction, it comes out in the form of tiny bubbles, but whenever Carbon dioxide is produced it comes out in the form of a brisk effervescence like lather?

Biological, thermal, electrolytic, or simple chemical reactions can all produce these gasses.  Can you specify which reactions?

Likely, any difference you've observed is due to the high solubility of $\ce{CO2}$ in water... it allows accumulation of enough dissolved gas to nucleate and form large bubbles, as in seltzer. BTW, $\ce{Na}$ or $\ce{K}$ etc. do not make tiny bubbles when put in water.


text:
Why does the mole-fraction equilibrium constant depend on pressure and volume?
The ratio of rate constants for the forward and backward reactions gives us the equilibrium constant, $K_c$. For a sample reaction:
$$
\ce{A \underset{k_{-1}}{\overset{k_1}{<=>}}  P + Q}\quad \Longrightarrow \quad K_c = {k_1\over k_{-1}} = {C_P C_Q\over C_A}
$$
Per kinetic theory, rate constants depend only on temperature and the presence of any catalysts:
$$
k_i = A_i e^{-E_{a,i}\over RT}
$$
Thus, their ratio (i.e., the equilibrium constant, $K_c$) also must depend only on temperature:
So: Why does $K_x$, the mole-fraction equilibrium constant, depend on pressure and volume?

TLDR; Because only the concentration equilibrium constant is directly related to the rate constants themselves.

Given that pressure has a minimal effect on equilibria in solution, I'm going to answer assuming you're interested in gas-phase equilibria.  The first step in showing why pressure/volume matter to $K_x$ is to move to the partial-pressure equilibrium constant, $K_p$. For a system composed entirely of ideal gases, the partial pressure of each species is related to its concentration as:
$$
{P_i\over RT} = {n_i\over V} = C_i
$$
Substituting this into the expression for $K_c$ for your example reaction gives:
$$
K_c = {C_P C_Q\over C_A} = {1\over RT}{P_P P_Q\over P_A} = {1\over RT}K_p
$$
Thus, for this reaction:
$$
K_p = RT\,K_c \tag{1}\label{Kp}
$$
For other reactions, the power of $RT$ on the RHS of Eq. $\eqref{Kp}$ can differ, depending on the change in the sum of the stoichiometric coefficients between the reactants and the products.
Since $K_c = f(T)$  and $K_p = (RT)^nK_c$, then $K_p$ is also dependent only on temperature (with ideal gases only!), not pressure or volume.

Where pressure/volume enter the picture is when one goes from $K_p$ to $K_x$.  Again assuming ideal behavior, the mole fraction of a species is defined as:
$$
P_i = x_i P
$$
Substituting this into the $K_P$ expression:
$$
K_p = {P_P P_Q\over P_A} = P {x_P x_Q\over x_A} = PK_x
$$
Thus, for this reaction:
$$
K_x = {1\over P}K_p \tag{2}\label{Kx}
$$
Similar to the above, the power of $1\over P$ in Eq. $\eqref{Kx}$ can vary, depending on the change in the sums of the stoichiometric coefficients between the reactants and the products.
In any event, this is where the dependence on $P$ enters the expression for $K_x$.

This dependence on pressure implies a related dependence on volume, since by the ideal gas law:
$$
V = {nRT \over P}
$$
and, equivalently:
$$
P = {nRT \over V}\tag{3}\label{igl-P}
$$
If one is working with a system where the volume and temperature are the control variables but the pressure is not, then in general the pressure will vary implicitly according to Eq. $\eqref{igl-P}$. If the temperature and/or volume change in any way other than proportionally $($i.e., ${T\over V} = \text{constant})$, then the pressure dependence of Eq. $\eqref{Kx}$ will manifest itself in a fashion that looks like a volume dependence.

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

@Martin-マーチン Aww, nuts, I did it again... old habits die hard. Better?

Do you say the total kilograms, or would you say the total mass? What would be the problem with 'the total amount of substance'?

@Martin-マーチン Because 'total amount of substance' implies an overall mass balance to my ear, not the overall molecularity of the reaction.  (And "molecularity" isn't even the right word here, if this is correct.) The 'total amount of substance' never changes in any reaction -- it's just reordered.

I don't understand your point, does the total mass change in any reaction? The added amounts of the substances of the components can be referred to as the total amount of substance of the components, if you really mean that. If you want to say the power of RT depends on the molecularity of the reaction, then say that. Or you could say the power of RT depends on the difference of the amounts of substances of the reactants and the amounts of substances of the products. In any case, when you say 'moles', you are referring to a quantity by its unit. It only makes an inaccurate statement worse.

To chat.

To circumvent the argument about amount of substance/number of moles, I would probably define it using the difference between stoichiometric coefficients. It is more abstract but also technically more accurate. The reason being that $K_p$ and $K_c$ (and by extension $K_p/K_c$) depend not on the actual amount of substance reacting, but on the stoichiometric coefficients...

... i.e. the reactions $\ce{2NO2(g) <=> N2O4(g)}$ and $\ce{4NO2(g) <=> 2N2O4(g)}$ have different equilibrium constants, despite the fact that both equations could describe the same reaction perfectly well.

@orthocresol How's this?

Awkward wording (not that I could word it any better, really), but it's correct, and Martin will be happy too!

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text:
Open Source SDF (Chemical Table File) Parser in any language
I'm looking to parse a large .sdf file and put it into a .csv file.
The only software I could find that parses .sdf files is OpenBabel's Pybel, but it won't properly install for my El Capitan, for either python2.7 or 3.+. (It's an open issue, and might hopefully be fixed with the next release). Are there any other languages/OSs that can easily parse this file?  If there aren't any, is there a recommended operating system I could spin up with VirtualBox that will easily run OpenBabel/Pybel?

Instead of using python why not use Notepad++ or Atom (both free) to extract the data you want (using column select tool) or use their syntax to convert to csv. Alternatively you could write a small python prog just to read data you have. If you look at some .sdf files you will soon see that the way data is organised is very clear.

afaik, chemsketch has scripting capabilities. But it isn't opensource, just freeware.

I'm not sure what "open issue" you indicate. I run Open Babel and Pybel all the time on El Capitan. Please contact me (or the Babel mailing list). As indicated below, there are many tools for parsing SD files.

The free, open-source RDKit chemoinformatics software can read .sdf files.  This software is most often used via it's Python API.
You can learn more about RDKit at the project's home page.
I recommend installing it via conda.  (If you use python regularly but don't use conda for controlling environments and package distribution, you probably should.)  The installation command is conda install -c https://conda.anaconda.org/rdkit rdkit
.
The python code to parse an sdf file and write to a csv file is very simple.
import pandas as pd
from rdkit import Chem
from rdkit.Chem import PandasTools

my_sdf_file = '/Users/curt/Desktop/sdf-isothiocyanates.sdf'

frame = PandasTools.LoadSDF(my_sdf_file,
                            smilesName='SMILES',
                            molColName='Molecule',
                            includeFingerprints=False)

frame.to_csv('/Users/curt/Desktop/excel_isothiocyanates.csv')

The .png below has some stand alone python that will read .sdf files and produce comma separated output. Its come out rather small but should still be readable. You can run the code from Atom on an apple mac or notepad++ on a pc.

You can also have a look at Bioclipse, which mixes a graphical interface with scripting (JavaScript, Python, and Groovy):

I blogged about an example script for Bioclipse before, which looks like:
hmdbIndex =
  molTable.createSDFIndex(
    "/WikiPathways/hmdb.sdf"
);
props = new java.util.HashSet();
props.add("HMDB_ID");
molTable.parseProperties(hmdbIndex, props);

idIndex = new java.util.HashMap();
molCount = hmdbIndex.getNumberOfMolecules();
for (i=0; i<molCount; i++) {
  hmdbID = hmdbIndex.getPropertyFor(i, "HMDB_ID")
  idIndex.put(hmdbID, i);
}

Bioclipse is developed by a team predominantly working on OS/X.


text:
Why does the Lewis definition of acids specify a PAIR of electrons and not a single electron?
The Lewis definition of an acid is: "a compound or ionic species which can accept an electron pair from a donor compound."
Why does it specify a pair of electrons and not just a single electron?
I initially thought maybe covalent bonds require a pair of electrons but it says that is not the case here : https://en.wikipedia.org/wiki/Covalent_bond#One-_and_three-electron_bonds
__
For reference:
Question from textbook :
Which of the following species can act as a Lewis acid?
$\ce{NH3}$
$\ce{F-}$
$\ce{H2 O}$
$\ce{NH4+}$
$\ce{BF3}$   (answer)
I thought the answer was $\ce{NH4+}$  as it could accept an electron, but the Lewis definition specifies accepting a pair of electrons.
Thanks.

If only electrons are transferred you have redox not acid-base reaction.

You know, could you delete the bad homework second part? Otherwise it's a nice question.

Thanks Mithoron. In the past I have been asked to provide an example. I also find sometimes there are other parts to the problem I am working on which I have not thought of in my posted question (leading me to the wrong answer) and posting the textbook question below, people often point it out.

It partially has something to do with the history of acid-base definitions. We started with the Arrhenius definition which was based on the generation of $\ce{H+}$ or $\ce{OH-}$ in aqueous solution, and then to the Bronsted-Lowry definition which moved towards acceptance or donation of $\ce{H+}$. Finally came the Lewis Definition which removed $\ce{H+}$ and left us only with electron pairs, to be consistent with the past two definitions. We couldn't just say electrons because that would have included a large class of reactions called redox reactions which are clearly not the same as acid-base.
Now, with respect to your answer to the multiple-choice question:

I thought the answer was $\ce{NH4+}$ as it could accept an electron, but the Lewis definition specifies accepting a pair of electrons.

There are two points that I would like to draw your attention to:

You're right in that $\ce{NH4+}$ is an acid (more specifically we call it a conjugate acid), but by the two older definitions.
However, $\ce{NH4+}$ cannot accept even one electron in this case. If you draw the Lewis structure, you'll find that although there is a net positive charge, all hydrogens have 2 electrons and the central Nitrogen has eight, fulfilling the octet rule.


text:
Why is coulomb integral is called local while the corresponding exchange integral is called non-local
In the book "A chemist's guide to DFT" page 27-29 they said that coulomb integral is local interaction term while exchange integral is non-local. I can visualize, to some extent, that repulsion between two orbital/wave function at two different position (say x1 and x2) is called coulomb integral and that is why it's local. But I am having problem with the exchange term. It is also local, I feel, just the difference is two orbital can interchange their position between them. Can anyone please shed light on these concepts?

The definitions of local and non-local operators are given at p. 12 of the book. Basically, an operator $\hat{A}$ is called local if the value of $\hat{A} f(x)$ at some point $x=x'$ depends only on the value of $f(x)$ at the very same point $x=x'$ and not on the values of $f(x)$ at any other points. Otherwise an operator is called non-local.
For instance, any $j$-th Coulomb operator in Hartree-Fock theory is local since the value of $\hat{J}_{j} \psi_{i}(q)$ at $q=q'$ depends only on the value of $\psi_{i}(q)$ at $q=q'$, while any $j$-th exchange operator is non-local since the value of $\hat{K}_{j} \psi_{i}(q)$ at $q=q'$ depends on the values of $ψ_{i}(q)$ at all points $q$.


text:
Methanol adducts with CaCl2
Methanol forms adducts with $\ce{CaCl2}$ but I am not finding much information about the nature of these adducts & their stability. Are these solid adducts? Crystalline?  Is the reaction substantially exothermic etc.?
In this link, it is written that:-

At temperatures below 55°C, $\ce{CaCl2}$ forms $\ce{CaCl2·4CH3OH}$ with methanol and below 55°C, $\ce{CaCl2·3CH3OH}$ is formed.

Obviously, they seem to have gotten their Temperatures wrong. Any more authoritative information about methanol adducts?

Yes, they are solid, crystalline, and moderately exothermic.


text:
Wood burning and carbon dioxide or monoxide?
I am building greenhouse and i want to operate greenhouse at winter time. For heating i ll use wood. And also i am thinking to give back  carbon monoxide which ll come from burning process of wood. There is a question, how can avoid to get carbon monoxide from burning wood, do i must give more air flow (or oxygen flow) to burning process? Or when you burn wood there is just one gas output that is carbon dioxide?

related http://chemistry.stackexchange.com/questions/35484/what-causes-incomplete-combustion

A good reference to read is Laboratory and field investigations of particulate and carbon monoxide emissions from traditional and improved cookstoves Atmospheric Environment February 2009, Pages 1170–1181.
Unforntunately, there is a significant amount of carbon monoxide, from 29 to 118 grams of CO per kg of wood.
They find that the drier the wood, the less CO, but don't expect to completely eliminate CO.


text:
Diastereoselectivity in synthesis problem (4 stereocenters)
Why is the particular diastereomer the major product of the sequence starting from racemic enone? I know that hydroboration is syn, therefore the arrangement of ring junction H atom and OH group. Why is one of the methyl groups anti and one syn with respect to the H atom and OH group?

This is a part of a synthesis from Warren's Strategy and Control. Sadly, he didn't explain anything.

Draw the middle intermediate in its 3D conformation and it becomes a little clearer (or post if you cant and i'll upload a picture).        [Also, do you have a page ref for the reaction? I still have a copy from when I was an undergrad covered in my scribbles, would be interested to see what I thought of it then]

I'm not sure how to draw the half chair at the ring junction part. Please add a picture.

You don't use half chairs to draw those fused bicyclic systems - it's more of a convex/concave face thing as described in the answer. http://i.imgur.com/YhkrFEH.png

@orthocresol... there will still be a 'chair' and a 'half-chair', they're just a bit fiddlier to draw than straight decalins. Unless you draw chair + half chair you'd struggle to nice the concave/convex faces

@NotNicolaou I know they still have defined conformations - I certainly should have been clearer, I meant more like the way to approach the question is not to focus on the cyclohexene ring & its half-chair conformation

It is page 441 of the workbook.

If we get rid of the two methyl groups and just consider what the 3D conformation of the molecule is likely to be (I don't have a pen on me right now, but the model below was minimised using very low level molecular mechanics).

What we can hopefully see is that there is very much a convex face and a concave face of this molecule.
In the absence of all other factors, the hydroboration should be faster from the convex face (the lower face as drawn).
This would be simple facial selectivity, as is often observed on ring systems.
However, we have additional chiral centres that are going to influence the reaction (diastereoselective reactions can be matched/mismatched such that the methyl groups present might help us or hinder us.
The problem in this question is that it isn't specified whether the methyl groups are up or down etc. So presumable what we're actually seeing is a resolution (i.e. we're only forming a small amount of product, but its optically pure because certain combinations just don't react at a sufficient rate).    [I did try and find the paper on Reaxys but had no luck, there is a chance he just made the question up].
So if we now consider the methyl group next to the acetal functionality, it would help us if it were pointing up (as I've drawn it, you could flip the rings to get it pointing down as shown in your product above) to leave the convex face free.
If it were pointing down (as I've drawn it..) then the approach of H-B would be hindered, making attack from convex/concave both unfavourable relative to the approach in which the methyl group is orientated into the concave face.

Using molecular mechanics is cheating :D

I can delete if you'd prefer ;)       If i get time later i'll upload some hand drawn images too- I do acknowledge that they'd be more useful to the community.

And do add the methyl groups for clarity.


text:
When naming alcohols based on alkanes, when do you keep the "e" at the end of the alkane's name?
For example, you would write pentane-1,2-diol but pentan-2-ol, correct? Is the difference simply the number of hydroxyl groups?

The names pentane-1,2-diol and pentan-2-ol are correct.
Systematic elision of the terminal letter ‘e’ or other vowels appears on several occasions in the IUPAC nomenclature.
A brief rule for simple alcohols is mentioned early in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-15.1.7.2.1 Alcohols are named by attaching the suffix ‘ol’ to the name of the parent hydride, with elision of the final letter ‘e’ in the parent hydride, if present. When alone in the structure, the characteristic group(s) must receive the lowest locant(s) possible, which is (are) cited immediately in front of the suffix (see P-14.3.2).

For example: pentane → pentan-2-ol
Extensive general rules for the elision of vowels can be found in Subsection P-16.7:

P-16.7 ELISION OF VOWELS
P-16.7.1 Vowels are systematically elided as follows:
(a) the terminal letter ‘e’ in names of parent hydrides or endings ‘ene’ and ‘yne’ when followed by a suffix or ‘en’ ending beginning with ‘a’, ‘e’, ‘i’, ‘o’, ‘u’, or ‘y’;
(…)

Clearly, this rule also applies to the suffix ‘ol’ for alcohols.
A similar rule is mentioned later in Chapter P-6 concerning the applications of general principles to specific classes of compounds:

P-60.2 PRESENTATION OF NAMES.
(…)
(a) names formed substitutively using suffixes follow the general method described in P-15.1. Substitutive names are formed by adding a suffix such as ‘al’, ‘ol’, ‘yl’, ‘carbaldehyde’, ‘carboxylic acid’, etc., to the name of a parent hydride, with elision of the final letter ‘e’ of the parent hydride, if any, before ‘a’, ‘i’, ‘o’, ‘u’, and ‘y’;

Clarification concerning multiplying prefixes and numerals can be found in Section P-59 concerning name construction:

P-59.1.9 Characteristic groups
(…)
If, and only if, the complete suffix (that is, the suffix plus its multiplying prefixes, if any, as described in Appendix 1) begins with a vowel, a terminal letter ‘e’ (if any) of the parent hydride name is elided. For example, ethanol (not ethaneol). Elision or retention of the terminal letter ‘e’ is independent of the presence of numerals between it and the following letter, for example, propan-2-ol (not propane-2-ol).
(…)

Since the complete suffix (the suffix plus its multiplying prefixes) of pentane-1,2-diol is ‘diol’ and thus begins with ‘d’ and not with a vowel, the terminal letter ‘e’ of the parent hydride pentane name is not elided.


text:
Energy of hydrogen bonds vs. kinetic energy of water molecules
I watched an extremely engaging video "Why does ice float in water?" by George Zaidan and Charles Morton, https://www.youtube.com/watch?v=UukRgqzk-KE. In particular, video says "Below 4oC kinetic energy of water molecules falls below the energy of hydrogen bonds. So hydrogen bonds form much more frequently then they break". I decided to verify this statement, and on my way to do so I found energy of hydrogen bonds: "6-30 kJ/mol" according to this source http://book.bionumbers.org/what-is-the-energy-of-a-hydrogen-bond/. However I was unable to find a way to calculate kinetic energy of water molecules. Your help on this matter would be highly appreciated.

It looks like they calculated the average thermal energy of water molecules, which by classical equipartition theorem is $3RT/2$ kJ mol$^{-1}$. This works out at approx 4 kJ mol$^{-1}$ at 40 C. This amount of energy includes translational, vibrational and rotational energy. The vibrational quanta are large in water and almost all molecules will have only zero point vibrational energy. Rotational quanta are smaller and many will be excited at these temperatures. The smallest rotational quantum value for water is approx 0.2 kJ mol$^{-1}$ the next 6 times larger and the next 12 times (increasing as $J(J+1)$ for quantum number J) so a few rotational levels will be populated. Energy will flow between translation and rotations as random collisions occur between molecules of water. A full calculation involves using the partition functions. Calculating these and their contribution to heat capacity is treated in most undergraduate phys. chem. textbooks.


text:
The Reaction Between Nickel and Hydrochloric Acid
Having conducted an investigation involving multiple metals, I have found that nickel refuses to react with concentrated hydrochloric acid. I have even left the reaction overnight and still have seen no mass change in the solid piece of nickel. Is the reaction simply extremely slow or does it not react despite being above hydrogen in the reactivity series?

Define "concentrated".

I think that the nickel passivation is the problem. Not sure if the nickel is reacting with water to form oxides or HCl (to form chlorides), but I would bet that nickel which was ground to powder in an inert atomphere, would react with dry HCl (gas) quickly, and at lower temperature.

What is the color of HCl now? Does it remain colorless? Solutions of NiCl2 are yellow or green, so you don't have to weight Ni to see if reaction is going.

Aye the HCL remains the same colour.

Try 30% and heat it to 80C. All of the chloride should be exhausted after a day or less. You can then evaporate the solution to generate crystals of nickel chloride. Any work with nickel compounds should always be done in a well-ventilated hood.

Thank you for this. I was using 30% hydrochloric acid I believe and I have done the same with zinc, magnesium, tin, and aluminium. All had immediate noticeable reactions at even 10 degrees celsius. My question is why, despite having a greater reactivity, does it react so slowly? Does it have something to do with a difference in its oxide form?

One last question: Would a similar result be expected of all other acids or would it react faster with them?

The acid should be in a well-ventilated hood too.

The reaction is given here.
$$\ce{Ni + 2HCl → NiCl2 + H2}$$

Nickel react with hydrogen chloride to produce nickel(II) chloride and hydrogen. Hydrogen chloride - diluted solution. This reaction takes place slowly.

In the wikipedia page of nickel, the reaction is explained elaborately:-

Nickel(II) chloride is produced by dissolving nickel or its oxide in hydrochloric acid. It is usually encountered as the green hexahydrate, the formula of which is usually written $\ce{NiCl2•6H2O}$. When dissolved in water, this salt forms the metal aquo complex $\ce{[Ni(H2O)6]^2+}$. Dehydration of $\ce{NiCl2•6H2O}$ gives the yellow anhydrous $\ce{NiCl2}$.

This is called passivation of metal surface. For a similar reason aluminum doesn't dissolve in water, but Al/Hg amalgam reacts with water. In case of Al Al2O3 film prevents the reaction. In case of Ni you probably form a thin layer of NiCl2 that isn't very soluble in concentrated HCl. Try diluting HCl by adding 2-3 volumes of water. Stirring should help. Fine powder will react faster than big pieces of metal.

Passivation is indeed important. Nickel alloys are used where corrosion resistance is important. I upvoted your answer. One other item not mentioned in the answers is the cathode (allowing evolution of H2)? Connect the Ni to a piece of Cu...Not all metals are good at evolving H2 - consider Hg applied to Zn to reduce reactivity (evolution of H2) by increasing its overvoltage.

In my PM refining lab I frequently use straight HCl 32% to remove surface nickel from my metallic products (post-first melt) containing predominately Au/Ir. The nickel easily bubbles away into a green solution that eventually turns yellow. I then pour it off into my nickel-saver beaker. If the nickel is extremely susceptible to HCl, the nickel might be in the form of an oxide.


text:
Can orbitals in atoms be compressed?
Consider an atom (carbon) with a ground state valence electron configuration of:
$$ \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \; \; ]} $$
In some molecules such as carbon tetrafluoride, carbon can "expand" to a more energetic electron configuration similar to:
$$ \underset{3s}{[\uparrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \uparrow ]} $$
Although these orbitals would usually be mixed together and also paired together with the opposite atoms they are attached to.
Are there any cases where the reverse can happen?
For example could phosphorous "contract" from
$$ \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \uparrow ] }$$
to
$$ \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow  \downarrow \vert \uparrow \vert \; \; ] }$$
Looking at the geometry of p orbitals and taking into account molecular orbital theory I think this is always of non benefit unless possibly the materials are put under external strains such as high pressure or heat.
Also, in the cases of atoms like chromium which are "expanded" by default I'd imagine contraction to the typical state could happen.

No. The electrons in carbon do not 'expand' when carbon forms bonds. The electrons reside in molecular orbitals which are wholly different to the atomic orbitals.

Carbon electrons are not in $3s$ and $3p$ orbitals. They occupy $2s$ and $2p$ orbitals.

@bon is right in that you are talking about molecular orbital theory. Speaking broadly, the redistribution of electrons the way you mention happens if the orbitals actually become energetically non-equivalent. This can happen if an atom is surrounded by ligands in certain way. Because orbitals have certain orientations, certain complexes (f.e, tetrahedral) will energetically be more favorable than others (f.e, octahedral).
A simple example is the Jahn-Teller effect: have a look here and here. Essentially, for low-spin d$^7$ configurations you will have something resembling your example, although those are d orbitals; I actually do not know if this happens with p orbitals that you use in your example.
This Wikipedia article on crystal field theory also provides a nice overview of the different ways the orbital energy levels can distort in. However, for better understanding, I suggest a good textbook (I can give a few suggestions if you're interested).

There are a few things worth clarifying here.
Hybrid orbitals do not exist in physical reality. I think I have said this enough recently for it to be clear. Hybridisation is a mathematical tool to produce nicer 2-centre-2-electron descriptions of bonding. When carbon (or any other atom) forms bonds it does not actually promote electrons into hybrid orbitals. The electrons in a molecule occupy molecular orbitals which are delocalised over the whole molecule (although in some cases there may be very little contribution from some atoms). These molecular orbitals are filled by electrons so as to produce the lowest energy state, in just the same way as atomic orbitals are filled to produce the lowest energy state.
Species (atoms, molecules, ions or whatever) always adopt the lowest energy state unless excited. For carbon the lowest energy state of an atom is:
$$\underset{2s}{[\uparrow \downarrow]} \underset{2p}{[\uparrow \vert \uparrow \vert \; \; ]}$$
For chromium it is (see here for an explanation): $$\underset{4s}{[\uparrow]} \underset{3d}{[\uparrow \vert \uparrow \vert \uparrow \vert \uparrow \vert \uparrow ]}$$
Unpairing electrons in carbon or pairing them in chromium requires energy and so these atoms will not exhibit this electron configuration under normal circumstances. It might be possible to generate these configurations by exciting the atom with light of the right frequency or by producing excited states through some decay process. Singlet oxygen (an excited state of $\ce{O2}$) can be produced in both of these ways.

Based on computational work by Martin Rahm, Roberto Cammi, N. W. Ashcroft, and Roald Hoffmann [1], as well as review of both experimental and computational work by John Tse [2], atoms do indeed change their electron congfigurations in response to a change in pressure.
This is not surprising:  Changing the pressure changes an atom's environment, and it's broadly known that changing an atom's environment can affect the relative energies of its orbitals, and in turn alter its ground state electron configuration.  This is simply one more example of that.
[1] Martin Rahm, Roberto Cammi, N. W. Ashcroft, and Roald Hoffmann. Squeezing All Elements in the Periodic Table: Electron Configuration and Electronegativity of the Atoms under Compression. Journal of the American Chemical Society 2019 141 (26), 10253-10271. DOI: 10.1021/jacs.9b02634
https://pubs.acs.org/doi/10.1021/jacs.9b02634
["Popular press" summary of above available at: https://phys.org/news/2019-08-rewriting-periodic-table-high-pressure.html]
[2] John S Tse. A chemical perspective on high pressure crystal structures and properties. Natl Sci Rev. 2020; 7(1):149-169.  DOI:10.1093/nsr/nwz144
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC8289026/

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How to name this benzene alcohol?
The following is the molecule I am confused about.

Earlier in my textbook it gives an example of how to name benzenes with two or more hydroxyl groups:

My textbook (Nelson 12 Chemistry) is infamous for its crappiness but my school is too lazy to change it, so I've been asking a few of these questions in recent weeks.

What is the proper, IUPAC name of the first molecule depicted in this question?
Are there any other commonly accepted names for this molecule and under what convention are they named?

Personally, I would name the first molecule benzene-1,2,4-triol, but the textbook gives the name 1,2,4-trihydroxybenzene, using the hydroxyl groups as substituents.

The compounds given in the question contain the characteristic group $(\ce{-OH})$. Since there is only one characteristic group, the seniority order of classes is not relevant in this case; thus, the $\ce{-OH}$ substituent corresponds to the principal characteristic group that is expressed as a suffix (‘ol’).
Therefore, the preferred IUPAC name (PIN) for compound (e) is benzene-1,2,4-triol.

Various traditional names exist for hydroxy compounds. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), only the name ‘phenol’ is retained as a PIN.

P-63.1.1.1 Only one name is retained, phenol, for $\ce{C6H5-OH}$, both as a preferred name and for general nomenclature. The structure is substitutable at any position. Locants 2, 3, and 4 are recommended, not o, m, and p.

Therefore, the PIN for compound (c) is indeed phenol.

Furthermore, according to Subsection P-63.1.1.2, the names

pyrocatechol (benzene-1,2-diol),
resorcinol (benzene-1,3-diol), and
hydroquinone (benzene-1,4-diol)

are retained but only for general nomenclature and only when unsubstituted.
Therefore, the PIN for compound (d) is the systematic name benzene-1,2-diol (the traditional name ‘pyrocatechol’ may be used in general nomenclature).

Loong gives an excellent overview of IUPAC approved names for your molecule.  (He always does.)
However you should also be aware that the common name for this compound is hydroxyquinol, which is still in common use.  The other two isomers of benzenetriol also have non-IUPAC-approved common names which are also in wide use.  They are phloroglucinol for benzene-1,3,5-triol and pyrogallol for benzene-1,2,3-triol.
A look at the Google N-grams viewer for these terms shows that pyrogallol and phloroglucinol are used far, far more often than the word "benzenetriol", while hydroxyquinol is used at about the same frequency.


text:
3 component (ternary) system in equilibrium
If the components of a system were Cl2, Br2, and I2, how could I describe this system under equilibrium conditions when the number of phases is at a maximum?
If there are 3 components, using the formula, F=M+2-P and solving for P under equilibrium conditions there would be a total of 5 phases.
Since these above components are all gases, and we know there are 5 phases, can I assume that 3 of the 5 phases will be Cl2 (gas), Br2 (gas), and I2 (gas).
For the other two phases, can I assume they consist of liquid iodine as it's commonly in the liquid phase, and cl2 (solid) as it's commonly in the liquid phase.
I'm not sure if this is correct.

No, you cannot assume much of anything. The Gibbs Phase Rule only tells you how many phases can coexist, not what they are. For example, here you have left out things like ClBr as a species, or dissociated Cl (which would be stable at high temperatures). What species will be present will depend on the temperature and pressure of the system, and the Gibbs free energies of the various possible components.

I know that it is not part of your question, but you also need to consider that $$\ce{Cl2 + I2 <=> 2ICl}$$

This makes perfect sense. Thank you for clarification.

If there are 3 components, using the formula, F=M+2-P and solving for P under equilibrium conditions there would be a total of 5 phases.

I understand from that you set $F=0$. Equilibrium does not necessarily mean that you have no degrees of freedom. In this case, you actually have two degrees of freedom: the molar fractions of two components, for example $X(\rm{Cl}_2)$ and $X(\rm{I}_2)$.
$X(\rm{F}_2) = 1 - X(\rm{Cl}_2) -X(\rm{I}_2)$ and cannot vary independently, so it is not a degree of freedom. Once you consider $F=2$, you only need three phases.
What about liquids and solid? Let's say that you vary temperature, and you reach the melting/condensation/etc curve in $T$ space. On the curve itself you will have four phases, three gases and (for instance) liquid chlorine. This causes you to lose a degree of freedom, which in this case is temperature. These four phases can exist only in that specific temperature.
EDIT
Following the comments on the original question, I would like to note that I was addressing a simplified case where the only phases possible are the ones listed in the original question. Once you have intermediate compounds (ICl for example) it's a whole different story.


text:
Does vibrating an NMR sample produce the same benefits and artifacts as spinning an NMR sample?
Does vibrating an NMR sample produce the same benefits and artifacts as spinning an NMR sample?  Are there "vibration sidebands" above the C13 satellites and is there a line thinning effect on the sample spectrum?  Please list any references that you may have read or seen.
Thank You

When you spin a sample you make sure that orientation effects are averaged faster than the relaxation time. This holds true if you spin a sample at 8 s^-1 and H1 relaxes at 1 s^-1. Now, wobbling (aka shaking) will keep orientation of the molecules for the most part (unless it creates spinning motion in the solution). It will change several %, but the general direction will be the same. So, no, shaking will not do the molecule averaging like spinning does.
Btw, why do we spin samples. I mean, if the tube is half decent and you spend 5 minutes on shimming you get about as good of a result. All spinning does is hiding poor XY shims by averaging them.

Possibly worth noting that with a good modern NMR machine we rarely spin samples these days. The combination of better probes and better NMR tubes makes the practice often unnecessary

This is true but none of our spectrometers except a 600 cold probed Agilent vnmrs are modern.  We never spin that one.  I only asked the question because when calibrating (FID shimming) using a D2O standard there appeared a harmonic oscillation in the upper barrel of one of our Mercury NMR's and the lineshape instantly became much thinner than I have ever seen before and the FID decayed like it was an CDCl3/acetone standard sample.  I never was able to force the vibrations to occur again and it has haunted me since.

Also when I observed the line thinning during the vibration event I was fairly certain that it was not a wobbling vibration but rather and up-down oscillation in the z-axis (vertical for NMR magnets) because of the tolerances of the spinners. One might expect homogenization effects from poor Z shims also.

sixtytrees;  =)> Yes many researchers here at UF also do not spin to protect the probe and sample which if it's a natural products distillation could be worth many-many $K.  This event that I observed with my instruments and my own two eyes still haunts me.  Care to try it on one of your machines whereby you induce a vibration in the z-axis ONLY and observe the results?  Anyone???

PS: I am Robert Harker - Engineer -   rharker@ufl.edu AKA 3dalliance here and other places.

https://nmr.chem.ufl.edu/


text:
Why do we need three equations to find the pH of NaCN, given Ka(HCN)?
To find the pH of a 1.0 M solution of $\ce{NaCN}$, given
$\ce{Ka(HCN) =4.9E(-10)}$
The solution (refer end of question) uses three equations:
Equation 1 : $\ce{HCN + H2O <-> CN- + H+}$
Equation 2: $\ce{NaCN <-> CN- + Na+}$
Equation 3: $\ce{CN- + H2O <-> HCN + OH-}$
I am confused as to:
1) the logic behind why we need these three equations (up to this point in the course, all the similar questions have been of the form e.g. find pH of NH3 given Kb(NH3), and so we only used the NH3 acid base reaction equation.
I assume equation 1 is necessary because the Ka is given for HCN and Equation 2 is necessary because we are asked to find the pH of NaCN.  Where does Equation 3 come from? We did talk about spectator ions in class so does it have something to do with Na being a spectator ion?
2) why the ICE (intial, change, at equillibrium) table to find x is written in terms of Equation 3 and not, for example, Equation 2.
The solution from the textbook is given below:

Thank you.

So strictly speaking you only really need equation 3 to solve this problem, but the first two equations help you figure out how to solve it if you're not too familiar. What do I mean:

Equation 1: $\ce{HCN + H2O <=> H3O+ + CN-}$

As you correctly note, this is to remind you of the definition for $K_a$

Equation 2: $\ce{NaCN -> Na+ + CN-}$

Note that I've changed $\ce{<=>}$ to $\ce{->}$. This is because most sodium salts fully dissociate in aqueous solution and so this "equilibrium" doesn't exist and you can then consider all $\ce{NaCN}$ to be converted to $\ce{CN-}$ for your "initial" stage in the ICE table. Hence there's no need to write an ICE Table with respect to Equation 2.
Value of writing this out:
In the case that you were given something sparingly soluble or did not fully dissociate, you would have to consider another equilibrium and this equation 2 would become significant. But in this case, $\ce{Na+}$ is a spectator ion which you can ignore.

Where does Equation 3 come from?

Equation 3 comes from the question itself (or rather what is happening in the system described). Remember, you've got a 1.0M solution of $\ce{NaCN}$. Since we've already established that it fully dissociates, at the "initial" stage of the system you have only $\ce{Na+}$ and $\ce{CN-}$ ions floating around. But from Equation 1 you know that $\ce{HCN}$ is a weak acid and so free $\ce{CN-}$ in water may act as a base and take a proton from water, and that is given by the equation:
$$\ce{CN- + H2O <=> HCN + OH-}$$

Great. Thank you for explaining that IT Tsoi :)

First and Third (sic!) equations are necessary. The second one just says "when I say that 1M solution of NaCN was formed, understand it as 1M of Na+ and 1M of CN- were added to a system".
Your key equilibrium is [HA]=[H+][A-] and [HA]/[H+][A-]=Ka. You then say HA+A- = total_molarity, [HA]=X, [H+]=X, [A-] = total-molarity - X and
x/([x]*[  total-molarity - X] = Ka. You know Ka, total_molarity. You now need to find x. In case of NaCN HCN is HA, CN- is A-.
Just plug your concentrations in the formula and get the result. Quite boring, indeed.


text:
Calculating vacancies in crystal structure
I have this question on a practice exam. I'm not quite sure how to find the number of vacancies when replacing atoms.
Question:
The stable oxide for lithium is $\ce{Li2O}$. How many oxygen vacancies in $\ce{SiO2}$ will be created if 100 Si atoms are replaced by 100 Li atoms (remember, O has a -2 charge in an ionic material)?
I think that because Li is +1 charge and Si is +4 that when I replace the Si with Li it leaves 150 unbound O. So 150 vacancies? Am I totally wrong?

That's it, you have the right answer. Conservation of charge (electroneutrality) is what you use to determine the number of vacancies.

Well, Jon, I think it's the question's fault, not the answers. What can F'x's answer possibly contain more?

re more explanation; on the face of it one might think that the crystal has a charge of -300 after replacement,  so another sentence might help.


text:
Stability of Bohr Orbits
In class we had been taught that Rutherford's model was unsuccessful because it failed to show that the orbits are stable because the electrons would lose energy because of electromagnetic radiation.

Bohr had suggested that electrons would move in orbits having well defined energy levels, but my question is that what would prevent these electrons from emitting radiation and collapsing into the nucleus.

Well, there's something "special" about those orbits. Quantisation is key. https://en.m.wikipedia.org/wiki/Bohr_model

ok, but what stops them from emitting radiation?

I think you should do some reading first. I'm sure you'll be able to answer your own question. But to be honest, as far as I know, Bohr just said those orbits are stationary states and are stable. Full blown quantum theory addressed this question with complete rigour. Imo

But, quantum theory came after Bohr's theory, why did the scientists accept Bohr's theory at that time?

Because they had nothing better. Why do we accept general relativity and quantum mechanics even though we know they are incompatible? They are still both useful theories.

Yeah exactly. For lack of a better option, and even though it was replaced by quantum mechanics, the Bohr model worked well for explaining the spectra of hydrogen. It's not a bad model..but it was replaced by something better later. That's how science works.

I'm not sure I would say that Rutherford's model was unsuccessful at all. It addressed the question of where the protons were quite well - the demonstration that there is a nucleus is pretty important, and still quite useful. That did leave the question of the electrons. Given that we still do Rutherford scattering, I would say it was pretty successful.

Could any of you guys also add the quantum mechanics explanantion..

@TanmayKulkarni That is like "please tell me all quantum mechanics and stuff" - a super unspecific and broad question.

Not every model is perfect. Rutherford's model suffered from the problem of electromagnetic radiation but answered important questions about the structure of atoms by showing the existence of a positively charged nucleus.
Bohr improved on this by realising that if electrons could only have certain energies then he could explain other things such as the spectra of hydrogen. As far as I am aware he did not attempt to explain why electrons are fixed in these orbits - this was only explained later by the advent of quantum mechanics. However, this does not mean that Bohr's model was useless since it accurately predicts useful things.
We do not have to know everything about why a model works in order for it to be useful. Einstein did not know why the speed of light was constant but he assumed that it was (based on all the available experimental evidence) and derived special relativity, which is an extremely useful theory. Nowadays we have quantum mechanics and general relativity which we know are incompatible but they are both exceptionally useful theories so we do not ignore them.

Few people continue to investigate this issue, but it was an active field of research in Bohr's days. Bohr himself basically sidelined the question with a postulate of stability, and quantum mechanics avoided it altogether by postulating that the classical laws of physics do not hold on the atomic scale.
In the course the 20th century, successive generations of scientists did manage to formulate a so-called nonradiation condition based purely on the mathematical application of Maxwell's laws. They found that, although an orbiting  point-charge must radiate, this is not necessarily the case for distributions of charge. Goedecke published on this subject in 1964 and ended with the suggestion that there might be "a theory of nature in which all stable particles (or aggregates) are merely nonradiating charge-current distributions whose mechanical properties are electromagnetic in origin."
Around the turn of the millennium, Mills brought this work to its full conclusion : if a point-charge in orbit around a nucleus must radiate, then the electron is not a point-charge. Instead, it is a sphere of distributed spinning charge and mass centered on the nucleus — a physical electron is like a soap bubble that encompasses the atom's core, in the world according to Mills.
Whilst this idea may seem outlandish to the current generation, it seems to be a useful extension of the Bohr model as it allows accurate calculation of a wide range of atomic and molecular phenomena, including such things as the ionization energy of $\ce{He}$ and other atoms.

G.H. Goedecke "Classically Radiationless Motions and Possible Implications for Quantum Theory" (Physical Review, Volume 135 number 1B, July 1964)
Randell L. Mills, "Classical Quantum Mechanics" Physics Essays, Volume 16: Pages 433-498, 2003

As you might have studied Bohr's model includes various orbits. An orbit is a Defined as a particular region in space where the electron moves and has constant energy.
A particle that has constant energy will not collapse in the nucleus. He even went forward to calculate these specific distances from the nucleus he came to the conclusion equation mvr=nh/2pi where the terms stand for their general meaning. By this equation we get to know about specific distances from the nucleus where the electron can be present without losing any energy.
See the thing is Bohr's model was only created for hydrogen, what you are actually seeing is the adaptations made to the Bohr's model that help us calculate values like velocities, energies, radii, etc.
The Logic behind the discarding of the Rutherford model was that the fact that classical electromagnetism states that ACCELERATING  charges product EM waves and hence an electron would lose energy.(Circular Motion is a motion where constant centripetal acceleration is needed hence the electron actually experiences the constant acceleration). This constant energy loss would collapse the structure proposed in almost 10^(-8) Seconds as you might have studied. But this does not happen.

The same flaw applies to the Bohr model as does the Rutherford model. Bohr just ignored it because he realised the model was still a useful one.


text:
Why is O2 (Oxygen) colorless and odorless while O3 (Ozone) is a pale blue gas with a distinctively pungent smell?
Why does adding one more oxygen molecule to the bond make it smelly and have color?!

see this:-http://chemistry.stackexchange.com/questions/15035/does-o2-have-a-color-in-the-gas-phase

also this:-http://chemistry.stackexchange.com/questions/767/does-ozone-o%E2%82%83-gas-have-a-color

The real answer for why ozone smells the way it does has little to do with chemistry but perhaps a lot to do with evolution.
The specific reason why it is coloured far more intensely than oxygen is explained well in the previous answer but it is worth adding a simple point. There is very little reason to think that small changes in structure make only small changes to chemical properties. Hydrogen peroxide is only a single oxygen different from water, but is a strong and dangerous oxidising agent; chlorine gas is just two electrons away from two chloride ions but one is a poison gas and the other a large part of table salt.
As for the smell it would be very inconvenient if we smelled oxygen. The level of oxygen is essentially constant and we need it to breathe. Smelling it would be useless and would get in the way of the ability to smell other things. And the brain would filter out a constant signal anyway (as it does with plenty of other stimuli). Ozone is uncommon and harmful. We might not have evolved specifically to smell it, but we have generic smell senses for unusual things many of which are also harmful. It is useful to be able to smell unusual, rare substances as it helps either avoid them or find them depending on whether they are dangerous or not.

Wow! Thanks for elobarating on the other person's answer, lots of smart people on these forums, will defintely come back here if I got any other chemistry related questions. I hope to dig deeper into chemistry in the Fall :)

Our perception of colour and smell involve different chemistry so there is no reason why they should be related. For example, what ever the chemistry of smell is (and its not certain what this is) it clearly does not involving photons and subsequent isomerisation of a rhodopsin.
Ozone is slightly coloured as it has two low lying electronically excited states (Chappius bands 550 to 600 nm) and a second state (at approx 700 to 1000 nm) both of whose energy gaps from the ground state correspond to a wavelength in the visible spectrum. The combined absorption is rather broad, weak and peaks at approx 600 nm which in the red part of the spectrum, thus ozone appears blue. Oxygen also has two low lying electronic states at around 700 nm and 12000 nm but are very weak and the longer wavelength transition is beyond the range of our eyes to detect it (liquid oxygen appears blue as does liquid ozone). Both O$_2$ and O$_3$ have electronic states in the ultra violet at a wavelength beyond the range of our eyes.
( I have added a moderately detailed description as to why high pressure gas and liquid oxygen molecules are coloured. see Does O2 have a color in the gas phase )

Incredible answer, you really went into detail why it shows color. I'm glad the others explained why it's smells. I really learned from this one thread. This site is truly incredible!

Smell: this is going to be lie. Simple answer: ozone is way more reactive (more poisonous than Cl2 that was used in WWI), so it burns your nose, nakes reactive chemicals that you smell. Correct answer. Brain receives the map of smells. There are no receptors (in the usual understanding) in the nose. That is why people change smell perception as they age.
Color: A simple answer: ozone has a system of conjugated bonds and a resonance structure. The longer the resonance structure the more "red-shifted" the absorption will be. Ozone has its absorption tail reaching visible blue light. Dioxygen doesn't. Long answer. Well, what is the color of water? Is it colorless or blue? Why oxygen gas is transparent, but liquid oxygen is blueish? Why skies are blue? These effects are quantitative and are not derivable apriori (from first principles).

I'd like to add, is there a such thing as liquid Ozone? If there can be liquid oxygen??

@Alex It exists. Wiki chemistry has more about it. https://en.wikipedia.org/wiki/Ozone


text:
What is "gold number"?
I was reading about gold number and a book defined gold number as following:

It is the minimum mass in milligrams of a stabilizing agent which is added to the 10 mL of red gold sol to protect it against coagulation caused by 1 mL of 10% (by mass) NaCl.

Is this the definition? Because I am not able to understand above definition.
Is NaCl used as the standard?

A red gold sol is a colloidal suspension of gold nanoparticles with average size of particles less than 100 nm. The suspension is quite stable — the nanoparticles don't aggregate because of the existence of an electrical double layer on the surface of particles which causes electrostatic repulsion between the particles.
If you add a sufficient amount of an electrolyte (e.g., NaCl) to the colloid, the ions will disrupt the electric double layer which will cause aggregation (coagulation) of gold particles. A stabilizing agent (e.g., citric acid) is a compound that is able to adsorb on the surface of gold nanoparticles and protect them from aggregation (it acts like a shield), even in a high electrolyte concentration.

Why does colour change from red to violet?

Practically it means:
Dissolve different amounts of your reagent in 10 mL water. Make a solution of colloidal gold. Specifically this colloidal should be "red gold" (there are different ways to prepare colloidal gold, but "red gold" is a common standard. Add 1 mL of solution of NaCl. It will cause precipitation in a sample with no stabilizing reagent, but if the amount of stabilizing reagent is sufficient then the gold will remain in "red gold" colloidal state. The amount of required stabilizing agent (in grams) is the gold number of this reagent.
The reason they do it is because it provides an standard, easy and reproducible way to compare different stabilizing agents. Part about 10 (by mass) is unclear.


text:
Transition state and intermediate very close to each other
I've been studying a particular reaction with DFT, where a transition state and the following intermediate are very close to each other. The change in geometry is rather small and therefore the electronic energies differ only by about ~ 1-2 millihartrees. Still, the TS has the higher electronic energy (everything else would be surprising?), but adding thermal corrections to both structures leaves the intermediate about 2 kJ/mol higher in energy.
Does that make sense? Or is there something I did not consider in my calculations?

It happens relatively often. The calculations of the electronic energy are the most accurate, vibrational & rotational correction calculations are rather crude. And besides, the typical accuracy of energetics with hybrid functionals (you didn't mention, but I assume you use some hybrid functional) is $\pm 10-15 , \mathrm{kJ/mol}$ (MAD), so that a difference of few kJ/mol is not a difference at all.

Which reaction is it? I realize it is secret and stuff, but at least mention the class. Is it organic, organometallic, other?

It is organometallic, transition metal / ligand / substrate. The transition state is a hydride transfer from the metal center onto the ligand.

1 Kcal/mol is indeed quite low. This is around 1 hydrogen bond energy. In general - compute frequencies using freq command in Gaussian or whatever you are using. You should have a single negative frequency. It should be relatively low (abs value compared to lowest positive frequencies). Visualize it using Moleken or a similar tool. If movement is along the reaction coordinate then you are good.
Also, compare all bond lengths and angles to see what changed. Did bonds along the reaction coordinate change the most?
Please use Kcal/mol instead of Hartrees. I realize that Gaussian prints out
Hartrees, but it is easy enough to convert to Kcal/mol.


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Heisenberg's uncertainty principle
Can anyone explain Heisenberg's uncertainty principle and it's derivation.
Im a 12th standard student.Please make it simple enough for me to understand.

Why put it on hold? the question is valid. Very briefly. For some operations Ab = BA (sum, product, max value, etc) Other operations are not. ("Then" in cooking recipe).  Quantum mechanics has Measure_Energy (M_E) and Measure_Time. Interestingly "M_E than M_T" is not the same as "M_T then M_E". So, AB-BA is not zero. Heisenberg's uncertainty is directly related to that. Now, you can also Measure_Impulse (M_I). Interestingly "M_E then M_I" = "M_I then M_E". Thus you can simultaneously measure energy and impulse, but not energy and time of the system. You need to know matrix algebra.

@sixtytrees Quantum mechanics does not have a time operator. There is no action called "measure time" in QM. http://physics.stackexchange.com/questions/6584/time-as-a-hermitian-operator-in-qm The energy-time uncertainty principle, unlike the position-momentum one, is more subtle and nuanced. The question was put on hold neither because of the sentiment in my initial comment, nor because it was too difficult to answer. If you read the close reason, you would see that it was because there was no or little effort put into the question. Also, please do not spam comments.

related http://chemistry.stackexchange.com/questions/14848/heisenbergs-uncertainty-principle

also http://chemistry.stackexchange.com/questions/16392/heisenberg-uncertainty-principle http://chemistry.stackexchange.com/questions/37254/justify-the-value-of-the-product-of-the-uncertainty-in-position-and-momentum

There is a complete derivation in Atkins & Friedmann "Molecular Quantum Mechanics', ed 3, p25

It is a basic postulate of quantum mechanics. It states that it is impossible simultaneously to measure the position of a particle, say an electron, an atom or a molecule, and its momentum. It places limits on these measurements such that uncertainty in position $\Delta x$ and momentum $\Delta p$ is given by $\Delta x\Delta p \ge \hbar/2$ where (hbar) $\hbar =h/2\pi$ and $h$ is the Planck constant $6.626 \times 10^{-34}~\mathrm{J~s}$. In a classical world no such restriction occurs, but this is not what is observed experimentally.
The important part is simultaneous measurement. We can measure where an atom is to whatever precision an instrument will allow, (e.g. by xray crystallography) but we cannot in the same measurement (i.e. at same time) know what its momentum is. We can measure an atom's speed v as accurately as we like and hence momentum mv. We note its position in two places a and b, say 10 m apart, and divide distance by time to cover this distance.
The uncertainty principle places limits on the measurement of momentum mv and position that allows us to know both the position and momentum an atom (particle) has at any given instance. Ask again if this is not clear


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Why would permeability for krypton be higher than argon?
So, I've been looking over data on permeabilities and there's something that bothers me.  Permeability data on the noble gases is somewhat hard to find, but the Parker O-Ring Handbook has them for a number of materials:
http://www.parker.com/literature/O-Ring%20Division%20Literature/ORD%205700.pdf
(as an example material to choose in the list, say, PTFE)
What I don't get is... why is the permeability rate for krypton usually higher than that of argon for most materials? ?  Their chemical properties are basically identical, krypton is just a larger atom.  By all standards one would expect it to permeate slower.  Xenon permeates slower than both of them, as expected.  So what's up with krypton?

The tables are a little hard to compare - the temperatures aren't always the same between, say, the argon entries and the krypton entries for the same elastomer. However, for those that are directly comparable it does appear that Kr can be a little faster. I think, sadly, that one might have to see if there is a paper from one group directly comparing measurements on the same system - I could see there being large differences for measurements done in different ways on different systems. Oh, and Kr should of course be stronger since it comes from the planet Krypton just like Superman...

Could possibly be a solubility thing and not just about straight diffusion.

Related: http://chemistry.stackexchange.com/questions/54826/why-does-co2-diffuse-through-a-butyl-rubber-membrane-more-readily-than-air

@Jon: But as you note, in the cases that are directly comparable... yeah, krypton seems faster than argon, and significantly faster than xenon... the latter is expected, while the former is not.  And it's not like there's just one entry... there's a lot of them, for a variety of materials.  And given that the measured temperatures, etc are identical and not some overly generic thing (for example, "149C" for PTFE), it sure looks like they're all from the same testing run.

@KarenRei - when I'm at work tomorrow I'll try and dig up a paper or two...

@matt_black Why would krypton be an anomaly in that regard?

@aventurin - CO2 is much easier of a case, CO2 is chemically very different from N2 and O2.  It's like how polar molecules with strong interactions with polymers like NH3 and H2O tend to have fast diffision rates.  The oddity about the krypton case is that, Ar, Kr and Xe are all noble gases.  Their properties are almost identical except for those determined by size.  But krypton appears to be the "odd man out".

@KarenRei - actually Google pulled up an article by Schowalter et al in Nuclear Instruments and methods A - they have data for all the noble gases through 4 polymers. The diffusivity decreases monotonically from He to Xe.

@KarenRei I don't know for sure, but there are known anomalies in apparent diffusion rates for some gases (carbon dioxide in some rubbers, for example) that are driven by the solubility. I'm guessing that sometimes a noble gas of just the right size might have some special affinity for atom-sized voids.

@Jon Custer - Suberb, I just found the paper (Permeability of noble gases through Kapton, butyl, nylon, and "Silver Shield").  I'll consider that a better source than the Parker data and just scale my permeation curves by their figures.  Thanks!  (I'd vote yours if it was an answer).

@KarenRei - I made it an answer. In the more general picture, unless your o-ring of interest is one of those materials you might just have to replicate their experiment to sort out the effects of $D$ from $b$.

As noted in some of the comments above, a paper that directly compares noble gas permeation through several materials is S.J. Schowalter et al., Permeability of noble gases through Kapton, butyl, nylon, and "Silver Shield", in Nucl. Instr. Methods A 615 267-271 (2010) . They note that $K = Db$ where $K$ is the permeability, $D$ is the diffusivity, and $b$ is the solubility of the noble gas in the material. For the noble gases, the diffusivity monotonically decreases from He to Xe in all materials. However, the solubility has an odder behavior, with Ne being the oddball with lower solubilities than He or Ar on either side of it.
The net result is that the permeation can end up not being monotonic in nature across the noble gases. In particular, for butyl, $K$ is highest for He, drops for Ne, increases for Ar, drops a bit for Kr, then continues lower for Xe.
Lots of interesting data in the paper, a nice direct comparison of the noble gases and several materials in the same aparatus under the same conditions .

I'll assume that the data is for the same pressure and temperature and concentration, and other experimental conditions AND that the units of measure are molar rather than mass based. 300 pages exceeds my patience. OK, if your concept of atoms is that they are hard billiard (pool) balls, then it doesn't make much sense. But consider that Xe is more reactive than Kr which is more than Ar which is more than Ne. That implies that those rascally electrons are not as tightly bound (in the valence shell) for Kr than for Ar. Unfortunately, this is just hand-waving. If you were to say that Ar diffuses faster than Kr, then I'd say "Ah, of course! Because it is smaller!" So, all I can say for sure is that very few situations have only one contributing variable - in the real world. Most things are a balance between two (or more) factors. If there were just one "factor" you'd expect straight line behavior. With two factors, you'd expect a change in slope, at least, or in the most severe cases parabolic behavior would be possible. This is a post hoc (after knowing the facts) explanation. In this case, you have three factors, I think: vdW size, deformability and mass. How they interact in the real world isn't easy to predict. That's one reason why compatibility tables such as for O-rings exist. Note that 1st Ionization E is higher for Ar than Kr, supporting the "Kr is more deformable, more reactive" argument.

This is very speculative, and the question refers to hard data, with specific examples. You should perhaps support your answer with similar references.

You can just search for "krypton", for example.  Starts on page 40 in the file, "3-30" in the page numbering, and I gave an example material to choose ("PTFE"), although you can feel free to choose others.  The reactivity of Xe, Kr, and Ar are tiny, I can't imagine that it applies to this situation.  They have full outer valence shells and these are not abnormally energetic scenarios..  And Kr is the middle entry in reactivity, yet the highest diffusion rate.  And Kr is, again, the middle entry in terms of 1st ionization energy.


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Derivation of formula for Hydrolysis constant
While studying acid-base equilibrium I thought I would derive the formula for hydrolysis constant myself. But I ended up getting a weird result and I am not able to understand where I went wrong.
The equilibrium constants for reactions that tend to favour products are large. As the concentration of the reactants decrease the equilibrium constant slowly should move toward infinity. For example the ionisation of strong acids is taken to be a reaction that goes to completion. According to my understanding the equilibrium constant $K_\mathrm a=\infty$. The equilibrium constant for its reverse reaction would be zero. Similarly the equilibrium constant for the ionisation of salts should be $K_\mathrm s=\infty$.
Let us consider the reaction $\ce{HA + BOH <=> AB + H2O}$. This can be carried out in a number of steps:
$\ce{HA <=> A- + H+}$  with $K_\mathrm a$
$\ce{BOH <=> AB + H2}$ with $K_\mathrm b$
$\ce{H+ +OH- <=>H2O}$ with  $1/K_\mathrm w$
$\ce{A- +OH1+<=> AB}$ with $1/K_\mathrm s=0$
So the hydrolysis constant should be
$1/K_\mathrm h=K_\mathrm aK_\mathrm b/(K_\mathrm wK_\mathrm s)=0\cdot K_\mathrm aK_\mathrm b/K_\mathrm w=0$
hence $K_\mathrm h=\infty$
I am sure there is something I havent understood well and hence I've ended up on this result. Can somebody explain where I am wrong?
Note: In most proofs the equilibrium constant of ionisation of salts is not considered but I see no reason to not consider it.

Well, your reaction is wrong and there's no so thing as "infinite" pKa.

@Mithoron Equilibrium of DNA burning in F2 is practically infinite. In other words, you will not be able to detect original DNA molecule once the system reaches equilibrium.

Related: http://chemistry.stackexchange.com/questions/47262/is-every-chemical-reaction-in-equilibrium and http://chemistry.stackexchange.com/questions/47384/does-an-irreversible-reaction-have-an-equilbrium-between-reactants-and-products

Sorry, I didn't get that. first of all, which hydrolysis are you talking about? I will talk about weak acid and strong base equilibrium. You have three concentrations [HA], [H+], [A-]. For instance, if they took 1L of 1 mM of HOAc (acetic acid) and added 0.3 mmoles of NaOH (concentrated, so no dilution) you get [A-]+[HA] = 1mM Then you say HA <-> H+ + A- has Ka. Then
[H+][A-]/[HA] = Ka. Then you say  [HA] = x, so [A-] = concentration_total- x (in this case 1mM -x). Then you get an equation
[H+]*[concentration_total-x]/x=Ka. Then you solve this square equation.
Regarding your question. I think you are mixing up to things: protonation of weak acid (stuff, relevant to analytical chemistry with a lot of drama, pH swings etc) and the behavior of other salts (they just dissociate completely and we forget about them). In stories like "to a solution of NaCl and K-acetat a drop of acid was added. What happen?" you should first forget about NaCl. It has no impact on the question. You should also forget about K+. Now you have "to a solution of Acetate- a drop of acid was added. What happen to pH?" Now you can easily calculate it using a square equation.
Run away from chemistry. Carcinogens, stinky chemicals, no  jobs... It teaches you how to make poisons and drugs, but how useful is that?

Thank you for your answer but I still don't understand what went wrong in my derivation...


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Calculate ion concentrations in mixed buffers
I have to state that I'm not a chemist, nor a chemistry student, but an electrical engineer with some chemistry knowledge. However, for my work I have to deal with a problem, from which my question arose.
There is a buffer solution made by dissolving $\ce{Na2HPO4}$, citric acid and KCl for a certain measurement. I would like to calculate what the ion concenctrations are in the solution.
This is where i got.

[$\ce{K+}$], [$\ce{Na+}$] and [$\ce{Cl-}$] are quite obvious.
Denoting the hydrogen phosphate anion as $\ce{BH2-}$, I have written the equation for the phosphate: $\ce{BH^2- (aq) + H2O <=>[K_b] BH2^- (aq) + OH- (aq)}$ and [$\ce{BH^2-}$]+[$\ce{BH2^-}$]=$M_{\ce{Na2HPO4},init}$
Denoting citric acid simply as $\ce{AH3}$ I have also written the equations for its three hydrolisation steps, such as $\ce{AH3 (aq) + H2O <=>[K_{a1}] AH2- +H3O+}$ and so on two equations more, plus [$\ce{AH3}$]+[$\ce{AH2^-}$]+[$\ce{AH^2-}$]+[$\ce{A^3-}$]=$M_{citric~acid,init}$
I have written the water autohydrolysis equation $\ce{H2O <=>[K_W] 1/2 H3O+ + 1/2 OH-}$
I have so far have the 9 unknowns [$\ce{H2O}$], [$\ce{H3O+}$], [$\ce{OH-}$], [$\ce{BH2-}$], [$\ce{BH-}$] [$\ce{AH3}$], [$\ce{AH2^-}$], [$\ce{AH^2-}$], [$\ce{A^3-}$] but only 7 equations. I could do the approximation that the water concentration is constant, which would be more than enough for practical purposes, but still there are 8 unknkowns remain. I could make the approx. that the dissolved citric acid concentration is equal to the final undissociated citric acid contentration, since hardly any dissociates, leaving 7 unknowns. But this is the point where it gets interesting for me, because I don't know how I should proceed without making these approximations, since I don't see clearly the physical and chemical processes in the system. Can you help me what other equations are there which (maybe at the cost of introducing some more variables) finally result in an exactly solvable equation system?

You can ignore K+, Na+ and Cl-. You have to deal with H2O, H3O+, OH-, Ap3-, HAp2-, H2Ap-, H3Ap, Ac3-, HAc2-, H2Ac-, H3Ac. Here H3Ap = H3PO4 and H3Ac = citric acid in neutral state. You have 11 unknowns.
(+1)    [$\Delta$H+] = 3*[citric3-]+2*[Hcitric2-]+1*[H2citric-]+[(PO4)3-] - [HPO4-]-2*[H3PO4]
(+1) [H(from_water)+ + $\Delta$H+][OH(from_water)-]=10^-14 (dissociation of H2O)
(+1) [H(from_water)+]  = [OH(from_water)-]
(+3) Three dissociation constants of H3PO4 are known
(+3) Three dissociation of citric acid are known
(+1) Concentration of [Ac3- + HAc2- + H2Ac- + H3Ac] is known
(+1) Concentration of [Ap3- + HAp2- + H2Ap- + H3Ap]is known
You have 11 equations.
OH(from_water)- corresponds to total OH-
H3O+ corresponds to H(from_water)+ + $\Delta$H+]
//(+1) Water concentration is assumed 1 in solutions
Forget about (PO4)3- and H3PO4. They will have negligible impact on overall pH.
You didn't tell the ratios of Na2HPO4 and citric acid and these ratios are critical. Given pKa of citric (3.15, 4.77, 5.19) and phosphoric (2.15, 7.20, 12.35) acids you will most likely deal with either of these regions:
2.15-3.15 H3Citric + H2Phosphoric-
3.15-4.77 H3Citric+H2Citric-
4.77-5.19 H2Citric- - HCitric2-
5.19-7.20 H2PO4- - (HPO4)2-
If you know which interval you are working with (check ratio of the reagents, aka stoichiometry) then you can pick the relevant equilibrium and set all other concentrations to zero for initial evaluation. 10^-15 M of ions will have zero impact on overall pH.

I regrouped some equations and expanded the one you asked about.
Can you solve it for a general case. Yes, you can. But it is like accounting for air friction, Special and General Theories of Relativity and quantum effects. It is harder, but it does cover more cases. In this case you will be dealing with x^11+a*x^10+...=0 instead of a square equation. You can solve it using Wolframalpha. But at the end of a day you will find that in a given region the impact of majority of members is negligible. Think of it this way. You can approximate the system with a spline (5 polinomials of second degree) or one polinomial of 11th degree.

Very nice answer, thank you. But I don't understand the last equation with "Added $\ce{H+}$". Do you mean I should set $[\ce{H3O+}]=[\ce{Na2HPO4}]+[\ce{H2Ac^-}]+[\ce{HAc^2-}]+[\ce{Ac^3-}]$? Why? $\ce{HPO4^2-}$ generates $\ce{OH-}$ and $\ce{H3O+}$ too, and the ions generated will be partly vanishing because the extra hydroxide and hydroxonium ions generated will partly react to form water, right? And lastly: can you at least write the principles of what you would do if you didn't want to make this approximation of constant water concentration? This I'd like to know out of personal interest.

Be careful. AcOH is acetic acid. It only have 1 acidic hydrogen and a single pKa. Your original question was about citric acid. It has three acidic hydrogens and three corresponding pKa. Which one are you talking about?

You need to use right coefficients in the equation. If going from Na2HPO4 to H3PO4 requires 2 hydrogens then you take it as -2*[H3PO4]

Thank you again, I almost get it, but I see another problem. The way I see it, you've introduced a new variable, $\Delta\ce{H+}$ or $[\ce{H+}]_{total}=[\ce{H+}+\Delta\ce{H+}]$ (one of them is a new variable), but didn't increase the number of equations.  The problem is, that I can't do $[\Delta\ce{H+}]\approx[\ce{H+}]$, because $[\Delta\ce{H+}]$ is sometimes negative, giving unphysical results. Second, I think in the equation for $[\Delta\ce{H+}]$ you wanted to write "$-[\ce{H2PO4^2-}]$" instead of "$-[\ce{HPO4^-}]$".

@user3237992 Good catch.  I've fixed it. When working with such large systems accounting errors (forgot "-" sign and a coefficient) are quite common and difficult to spot. It is much easier to spot them in "acetic acid was added to sodium acetate". The fact that glass walls also have buffering properties should be ignored because it complicates picture and doesn't impact the system.

About ion concentration? I think dihydrogene phosphate ion instead of hydrogene phosphate ion, maybe I'm wrong.


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Carbaldehyde or formyl?
Which is correct 5-methylbenzyl-3-carbaldehyde-1-oic acid or 3-formyl-5-methylbenzoic acid, MarvinSketch recognizes them both as correct?

Where did you got the first one??

The name ‘5-methylbenzyl-3-carbaldehyde-1-oic acid’ is not correct for several reasons. In the following, I focus on the question in the title concerning the choice of ‘carbaldehyde’ or ‘formyl’.
Both, ‘carbaldehyde’ and ‘formyl’, denote the substituent group $\ce{-CHO}$ and can be used, among other ways, to systematically name some aldehydes.
The part ‘carbaldehyde’ is a suffix. It can be used when the $\ce{-CHO}$ group is attached to a ring, for example cyclohexanecarbaldehyde. However, the preferred IUPAC name (PIN) for the analogous benzene derivative is ‘benzaldehyde’, which is derived from the retained name of the corresponding carboxylic acid.
The part ‘formyl’ is a prefix. It can be used in the presence of a characteristic group having priority to be cited as a suffix or when the $\ce{-CHO}$ group is attached to a side chain.
The compound given in the question contains two characteristic groups, $\ce{-CHO}$ and $\ce{-COOH}$. The principal characteristic group must be expressed at the end of a name by means of a suffix or a class name, or implied by a trivial name. Since $\ce{-COOH}$ is the senior characteristic group in this case, the compound is named ‘3-formyl-5-methylbenzoic acid’ (not ‘3-carboxy-5-methylbenzaldehyde’). Note that the $\ce{-CHO}$ group cannot be expressed at the end of a name; it has to be expressed as a suffix (i.e. ‘formyl’).


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Any example of colloidal solution where liquid Helium is solvent?
I would like to see any information, if it exists, about colloidal solutions where Helium is "solvent".
Background of that question is, I'm wondering giving surface energy effects, how potentially small can be water ice particles at low temperatures.

Problem is that helium has very low density, so particles would tend to plummet to the bottom IMO.

@Mithoron Yes, considered it as issue, although background have another few layers of backgrounds, one of them is sedimentation coefficient and  brownian  motion. I come-up with that question after thinking about ultra centrifugal separation. So this way it's not an issue but more what I would like to have. Methods of getting fine particles also not clear in this case, so it's more about how theoretically fine they could to be in such circumstances. Not sure at all if Helium  may have some effect of preventing them to agglutinate.

A solution implies that there are solvent-solute interactions. Helium is about the most chemically inert atom there is. In the case of water ice, the hydrogen bonding will be orders of magnitude stronger than the van der Waals forces, so colloids will not be stable at 'normal' pressures. I guess if I were looking for a colloid in He(liq), I'd look at N2 or perhaps H2. We've made a lot of progress since I had colloidal (surface) chemistry class, but I doubt that we could predict size of particles in any such system. But I don't know.

as helium is diamagnetic, yes probably ice particles will to agglutinate.But which size ice particles will stop to merge. this way it's less interaction with He, but more properties of fine particles. Although  H in He, specially isotopes hm interesting. Also I will try to search later information about less extreme "solvent". Actually I did, but not so deep at the moment.

This is a question to physics department. Chemists only see liquid helium when refilling NMR instruments. Boiling point and comfort working with changes in this order: He < H2 < N2< NH3 < H2O. I didn't find any papers regarding colloidal systems in N2. So, chances are nobody looked into He yet. Try a deep search on N2 colloids first.
Currently colloidal systems are interesting as drug delivery methods (water, water soluble polymers as a solvent) or nanotechnology (organic solvents at room temperature). There isn't enough motivation to study colloids at low temperatures at the moment.

that's physical chemistry, but more physics, yes. On se.physics  looks like colloids discussed less then here. I also quick searched(gdoogdled), and also have not found, but have not expected it to be found so easy, too narrow. This way I decided to ask collective wisdom - maybe some one have saw something interesting. Can be used to refine(sort) asteroid materials as example(maybe). Selective drug transport, yes good, interesting field, useful.

@MolbOrg Which physicists? Majority of them are searching for energy sources. They see matter (including colloidal systems) as a substrate to transfer beloved energy.

Biophysicists, they are totally mad, for good). Maybe other xxx-physicists too. There are some amount of them of different kind.


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CO2 conversion to a harmless gas like oxygen
I am doing some research about indoor purifiers and I was wondering if there is any approach to purify co2 or convert it to harmless gases like oxygen just like what plants do.

You could use sodium peroxide: $\ce{2Na2O2 +2 CO2 -> 2Na2CO3 + O2}$

Plants do the job pretty well: why not use them? Besides carbon dioxide isn't harmful in the quantities normally found in the atmosphere (you produce it when you breathe). Only when there is a lot more than normal levels does it become a problem for living things which have adapted to low levels because they produce it.

Submarines currently use NaOH/Ca(OH)2 to absorb CO2. They generate O2 separately either from gas tanks or by electrolysis of water. Peroxides seem attractive at first, but they are a fire hazard and you don't want fire on a submarine. British subs are using Zeolite. The role of zeolite isn't to deal with CO2, but to remove the smell of volitile organic compounds (of dirty socks, so to say). Air is first dried and purified.
What exactly are you trying to achieve?


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Isolation of cis-3-hexanol from grass
If I wanted to isolate cis-3-hexanol (cis-3-hexen-1-ol;  IUPAC name: (Z)-Hex-3-en-1-ol ) from grass I could start the extraction by dissolving the grass in a solution of ethanol. EtOH is a common solvent for extractions, and dissolving grass in it would leave some remaining solid, as well as a crude solution containing the cis-3-hexanol. Cis-3-hexanol is a liquid at room temperature and is soluble in ethanol but not (only to a small degree) soluble in water.
After filtering out the remaining undissolved solids, I think the next best step would be to boil off the remaining ethanol and water with both of their boils points being lower than cis-3-hexanol's boiling point at around 156.5 degrees C. This would presumably leave a crude solution containing cis-3-hexanol and other organic components extracted from the grass.
What would be the best method to further purify the remaining solution for a higher purity? Please correct me if my thought process is wrong.
Cis-3-hexanol is used in perfumes and the demand for it has increased in industry.

Hexenol and hexanol are not the same thing

First, how much compound do you want to make?
Second, how much cis-3-hexanol is there in grass? It is probably around 0.01%, so be ready to process a lot of grass.
Third, it might be reactive (that is it can decompose).
Your best bet is to:
(1) Figure out theoretical yield
(2) Do steam distillation rather than extraction (bp is 156.5 °C - good for distillation).
(3) Extract collected fraction (mostly water+ some organics including hexenol) with some light organic solvent (hexane, ether). You can try ethyl acetate (aka MEK substitute, bp = 77C), but that is the highest bp you want to go for.
(4) Split organic solution in several fractions and try washing it with water solutions of 5% baking soda and then brine (saturated solution of NaCl). Hexenol should stay in organic solvent unless it decomposes.
(5) Concentrate it under reduced pressure. If you don't have access to rotavap - let it sit in open air. You can direct fan at it if you will, but you might catch fire if you are not accurate.
You will get a drop of something that is a mix of 20 compounds. How you need to do column chromatography, which is pain even with equipment.
It doesn't sound like you have access to chemistry lab. In this case for practice purposes try extracting eugenol from cloves first. Grind cloves, place in distillation apparatus, add water, distill. Oil on top is eugenol.

Steam distillation: Shred grass, add to a gallon of water and distill it off using a lab distillary apparatus.


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Product of the addition of 1 eq. bromine to 1-methylcyclohexa-1,4-diene

When we add one equivalent of $\ce{Br2}$ then which double bond will get brominated?

The double bond in the molecule with methyl group will be more stable as it will have more hyper conjugating structures. Thus the other double bond should get brominated.
But my textbook says the double bond with methyl group will get brominated. Can someone explain why?

You misunderstood what hyperconjugation is. It occurs when the electrons in a $\ce{C-H}$ $\sigma$-bond are shared with an adjacent atom's empty or partially filled $\mathrm{p}$-orbital, giving the species an extended molecular orbital and thereby stabilizing it. This means that hyperconjugation doesn't really come into play until bromine is added across one of the double bonds. Let's consider the two possible intermediates:
$\hspace{7.cm}$
$\hspace{6.5cm}$
The first one is a tertiary carbocation, and the second is a secondary carbocation. The first intermediate is more stable due to the hyperconjugative effects of the methyl group, and will require less energy to form than the second. This means the reaction will occur much faster at the more substituted double bond, and is why that is the major product.

Thanks, ringo. I thought when br2 is added to a double bond then cyclobromium ring is form and br- ions comes from the opposite and the step becomes concerted without formation of carbocation.

@nikola It is formed, but the tertiary carbocation is the most important resonance structure.

You might also add that the substituted alkene will be more nucleophilic due to the increased electron density from hyperconjugation with the $\pi ^*$ orbital.

Hyper conjugation supports transition state structures. More hyperconjugation = lower energy of TSS = faster reaction = this moiety is more prone to react.


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Why P2O5 does not show oxidising properties while N2O5 is the strongest oxidising agent among pnictogen oxides?
Stability of oxides of higher oxidation state decreases with increasing atomic number
Then why P2O5 is thermally stable while N2O5 not.
Infact N2O5 is the strongest oxidising agent among pnictogen oxides because N2O3 is most stable.
Please Offer a lucid explaination to understand this...

related http://chemistry.stackexchange.com/questions/32534/why-is-hno3-a-stronger-oxidising-agent-than-h3po4

A very easy half-answer would base this on electronegativity.
Nitrogen’s electronegativity is $\approx 3$ while phosphorus’ is $2.19$. Oxygen’s is, of course, $3.5$. So the electronegativity difference between nitrogen and oxygen is much lower, meaning that oxygen is much less able to stabilise nitrogen’s high oxidation state.
Contrarily, phosphorus’ low electronegativity means oxygen can stabilise its high oxidation state rather well and the compound is much less likely to act as an oxidating agent by itself.

For the same reason you have Na3PO4 (not NaPO3) and NaNO3 (never Na3NO4). Nitrogen doesn't have an orbital to form five bonds, while phosphorus has. This explains why PF5 exists, but NF5 doesn't. In N2O5 n=o bonds have character similar to bonds in NH4+. But in NH4+ nitrogen is surrounded by electron donating H atoms, so it is happy to donate it's lone pair of electrons. In N2O5 it is surrounded with electronwithdrawing O atoms and is not to happy to give away its lone pair. thus it is happy to transfer one oxygen atom elsewhere (in other words - to oxidize something).
tldr N is the only element in the group that doesn't have p orbital.

"NaPO3" is actually a trimer Na3P3O9

Na3NO4 actually has been created 2) "doesn't have an orbital" is terrible wording 3) there's no clear connection between tendency to lower coordination number and electronegativity about which is later part of your answer.

Link for alkali metal orthonitrates $M_3NO_4$: https://en.wikipedia.org/wiki/Orthonitrate.  They are created in molten salt conditions, but they break down upon exposure to even mild acids including water and carbon dioxide (thus air).


text:
Converting benzoic acid into meta-fluorobenzoic acid
How do I convert benzoic acid to meta-fluorobenzoic acid in not more than 3 steps? I did it but the process was a little too long.
Here is my attempt

Your approach might be "three steps" according to some people. The last two reactions in your synthesis, converting the amine to the diazo salt followed by fluoridation, are usually done in the same reaction vessel with the diazo salt never isolated.
Your three steps would be:
$$\begin{aligned}
1.\ &\ce{->[\ce{HNO3}][\ce{H2SO4}]}\\
\\
2.\ &\ce{->[\ce{Sn}][\ce{HCl}]}\\
\\
3.\ &\ce{->[1)\ \ce{NaNO2,\ HCl}][2)\ \ce{HBF4}]}
\end{aligned}$$

Two steps should be enough ;-)

Nitrate benzoic acid. In J. Chem. Educ., 1967, 44, 464
(DOI), the authors added benzoic acid to fuming nitric acid at 0 °C. 3-nitrobenzoic acid is the major isomer (~ 75%). Recrystallize from ethanol.

Your product is an electron-deficient arene and $\ce{-NO2}$ is a decent nucleofuge. Try to replace it by fluorine in a nucleophilic aromatic substitution!

In the case of 4-nitrobenzoic acid, the transformation to 4-fluorobenzoic acid has been described to work with a yield of >95% in a room temperature reaction using DMSO as a solvent and anhydrous tretrabutylammonium fluoride (TBAF) as the reagent. (DOI)

but here the NO2 group will be in resonance with the ring, therefore how can you you remove it so easily?