What is formed when you leave iron(II) sulfate in plain air? What is formed when you leave iron(II) sulfate in plain air?

Knowing that iron(II) is easily oxidised to iron(III), and assuming that the reactive component of air is oxygen, I solved it this way:

$$\ce{FeSO4 + O2 -> Fe2O3 + SO2}$$

But in my textbook it's given that iron(III) sulfate, $\ce{Fe2(SO4)3}$, is formed. Is this because $\ce{Fe2O3 + SO2}$ react to form $\ce{Fe2(SO4)3}$?

Usually, the original iron(II) sulfate is present as green $\ce{FeSO4.7H2O}$.

Oxygen can oxidize $\ce{Fe(II)}$ salts to $\ce{Fe(III)}$ salts; for example, at $\mathrm{pH}=0$:

$$\begin{alignat}{2}
\ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}\\
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad &&E^\circ = +1.229\ \mathrm{V}
\end{alignat}$$

$\ce{Fe(II)}$ is even easier oxidized under alkaline conditions; for example, at $\mathrm{pH}=14$:

$$\begin{alignat}{2}
\ce{FeO(OH) + H2O + e- \;&<=> Fe(OH)2 + OH-}\quad &&E^\circ = -0.69\ \mathrm{V}\\
\ce{O2 + 2H2O + 4e- \;&<=> 4OH-}\quad &&E^\circ = +0.401\ \mathrm{V}
\end{alignat}$$

However, you cannot simply oxidize iron(II) sulfate $\left(\ce{FeSO4}\right)$ to iron(III) sulfate $\left(\ce{Fe2(SO4)3}\right)$ in dry air since you would need additional sulfate to balance the equation:

$$\ce{2FeSO4 + SO4^2- -> Fe2(SO4)3 + 2e-}$$

Nevertheless, $\ce{Fe(II)}$ can be oxidized to $\ce{Fe(III)}$.

The resulting $\ce{Fe(III)}$ is subject to hydrolysis. By way of comparison, the ion $\ce{[Fe(H2O)6]^3+}$ is only stable under strong acidic conditions. Already at $\mathrm{pH}=0{-}2$, it turns into yellow $\ce{[Fe(OH)(H2O)5]^2+}$ and begins to form $\ce{[Fe(OH)2(H2O)4]+}$:

$$\begin{align}
\ce{[Fe(H2O)6]^3+ \;&<=> [Fe(OH)(H2O)5]^2+ + H+}\\
\ce{[Fe(OH)(H2O)5]^2+ \;&<=> [Fe(OH)2(H2O)5]+ + H+}\\
\end{align}$$

Further addition of base causes precipitation of amorphous iron(III) hydroxide.

Accordingly, the likely product when iron(II) sulfate is oxidized is basic iron(III) sulfate, i.e. approximately:

$$\ce{4FeSO4 + O2 + 2H2O -> 4Fe(OH)SO4}$$

However, the real weathering and aging of iron(II) sulfate in dry air actually yields a mixture of various compounds, including iron(III) sulfate and iron(III) oxide-hydroxide.

You may observe this reaction in some iron fertilizers for lawns. The fresh product typically contains green iron(II) sulfate, which gradually becomes yellow.


How much lactose is there in milk (mechanism)? Iodometric titration is sometimes used to find out how much lactose there is in milk. The overall oxidation reaction is written as follows.

$$\ce{C12H22O11 + I2 + 3NaOH ->[\ce{CuSO4}]C12H21O12Na + 2NaI + H2O}$$

How does this reaction proceed? What is the role of $\ce{CuSO4}$?

The solution is then acidified with $\ce{HCl}$. Excess $\ce{I2}$ is determined via

$$\ce{2Na2S2O3 + I2 ->2NaI + Na2S4O6}.$$

What I think might be going on (with references)

Using basics of inorganic chemistry, iodine reacts with cold sodium hydroxide to give sodium hypoiodate.

$$\ce{I2 + 2NaOH <=>NaIO + NaI + H2O}$$

$\ce{NaIO}$ is the main oxidiser of lactose.$^{[1]\ [a]}$

$$\ce{C12H22O11 +NaIO + NaOH ->C12H21O12Na + NaI + H2O }$$

I have no clue as to what the complete mechanism might be here.

Alternatively, $\ce{Cu^{2}^+}$can be reduced. The following picture$^{[2]}$ shows the general transitions for (semi)linear lactose.

Sadly, neither reference mentions the other option, nor are there arrow-pushing mechanisms which is what I am after.

Sodium iodate is prone to decomposition even under room temperature, but the reaction is vigorous from $80 \ ^\circ \text{C}\ ^{[3]}$:

$$\ce{3NaIO->NaIO3 +2NaI}.$$

This is partly why the acid is added. To get an accurate reading of "unreacted" iodine we must somehow re-release $\ce{I2}$ from the ions $\ce{IO^-}$ and $\ce{IO_3^-}$.$^{[1]}$

$$\ce{NaIO + NaI + 2HCl -> I2 + 2NaCl + H2O\\ NaIO3 + 5NaI + 6HCl -> 3I2 + 6NaCl + 3H2O}$$

Thiosulphate is a widespread reagent that is subject to inaccuracies$^{[4]\ [5]}$ due to

$$\ce{Na2S2O3 + 4I2 + 10NaOH -> 2Na2SO4 + 8NaI + 5H2O}.$$

This reaction is less favoured in acidic conditions.$^{[1]}$

Clarification

I am asking for

arrow-pushing mechanisms that show the role of $\ce{NaIO}$ and/or $\ce{CuSO4}$ with cyclic lactose (if possible).

$^{[1]}$ A Study of the Method for Titrating Aldose Sugars with Standard Iodine and Alkali.  G. M. Kline, S. F. Acree. Bureau of Standards Journal of Research. Link [Comment: the study actually deals with aldose sugars. The equation for cyclic lactose might thus be a stretch.]

$^{[a]}$ Agu-Tõnis Talvik. Organic chemistry. (1996). page 490 (as far as I know, not available in English)

$^{[2]}$ Dairy Chemistry and Biochemistry. Second Edition. P. F. Fox, T. Uniacke-Lowe, P. L. H. McSweeney, J. A. O'Malhony. page 62. Link

$^{[3]}$ Иодат натрия. Russian Wikipedia. Link

$^{[4]}$ (G. Topf). Zeit. anal. Chem., 26, p. 137; 1887

$^{[5]}$ (Bougault). Compt. rend., 164, p.; 949; 1917

Using the basics of organic chemistry, solutions of iodine and sodium hydroxide retain enough free $\ce{I2}$ to do other things. I would disocunt $\ce{NaIO}$ as a significant component of this process.

What if instead, we had exactly what your picture shows (except the last few steps are prevented - we'll see how in a few lines). You are worried abut the cyclic form, but the cyclic form and the acyclic form are in equilibrium in aqeous solution, especially if acid or base is present. Note that in the reactions equations below, I am using the notation $\ce{R}$ to refer to parts of the molecule that are not involved.

  Isomerization to the ene-diol

$$\ce{R-CH(OH)-CHO <=>[\ce{OH-}] R-C(OH)=CHOH}$$

  Oxidation by copper(II)

$$\ce{R-C(OH)=CHOH + 2Cu^2+ +3OH- -> R-CH(OH)-CO2^- + 2Cu+ + 2H2O}$$

Now, if that copper(I) cations stumbles into a hydroxide anion, then it's all over since $\ce{Cu2O}$ is not soluble. However, it's iodine to the rescue! Iodine oxidizes $\ce{Cu+}$ back to $\ce{Cu^2+}$.

$$\ce{2Cu+ + I2 -> 2Cu^2+ + 2I-}$$

These last two equations add to give the one you quote (only I left out the sodium spectator cations and simplified the carbohydrate).


Different reactions of clean and rusty iron nails in vinegar (acetic acid) I have submerged a iron nail in vinegar twice recently. What is puzzling me is the different results each time. The first time I submerged it, it was covered in rust. When I put it in the solution of vinegar, I got a red precipitate dissolved in the solution that I believe was Iron(II) and Iron(III) acetate. However, when I put the newly cleaned nail in ANOTHER bath of the same concentration of vinegar, I did get some bubbles but no red precipitate that was dissolved in the solution. It's only been about 2 days, though. However, there was a black precipitate on the surface of the nail, but I do not believe it is iron acetate because of its color.

I am confused as to the results because Iron is higher than hydrogen on the activity series, so it SHOULD be reacting with the vinegar to give me that red precipitate and hydrogen gas. I added some hydrogen peroxide to the solution to oxidize the iron, and the solution is getting a reddish tint to it.

Why do you think I am not getting that reddish precipitate when I put in the cleaned iron nail in the new vinegar solution? Thank you!

In your first experiment, the rust ($\ce{Fe2O3. xH2O}$) probably reacted with vinegar to form Iron (III) acetate which makes the solution reddish in color.

In your second experiment, the iron on the clean nail surface would have reacted with vinegar to form Iron (II) acetate. There are two main reasons why the solution might appear to have no change:

Usually, Iron (II) acetate is prepared using concentrated acetic acid and scrap iron (or ferrous oxide/hydroxide). Since vinegar is extremely dilute (around 5-8% acetic acid solution in water), the reaction would be very slow and the conc. of Iron (II) acetate formed will be very low.
Iron (II) acetate happens to be very soluble in water and has a very faint light green color in solution which is not easily detectable in lower concentrations.

Then again, when you added $\ce{H2O2}$ to the second solution, you mentioned that a reddish color appeared. This shows that the dissolved $\ce{Fe^2+}$ ions are oxidized to $\ce{Fe^3+}$. The $\ce{Fe^3+}$ formed will react with the acetate ions to form red Iron (III) acetate.


Identifying Reaction Types I need help on identifying precipitation, acid-base, and redox reactions. I know what each one is in theory:

Precipitation: Creating a solid
Acid-Base: neutralization
Redox: exchange of electrons

But given a chemical reaction like:
HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)

I need to understand how to classify them. Are there any tell-tale signs to look out for when determining. For example, does a precipitation reaction ALWAYS produce a (s) as a product and therefore any reaction with a (s) on the product side is a precipitation reaction?

It's true that a precipitation reaction will always yield at least one solid product, but not every reaction that yields a solid is, strictly speaking, properly categorized as a precipitation reaction. Precipitation reactions are ones in which at least one of the reactants is in the aqueous phase (i.e., dissolved in water), and a solid forms on the product side which was not present on the reactant side. If all the reactants are solid, then a solid product forming should not be called a precipitate. To be called a precipitate, an insoluble product must form from within solution, either from a solid and solute interacting, or from an interaction strictly between solutes.

Examples of precipitation reactions:

$\ce{AgNO3_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3_{(aq)}}$

$\ce{CuCl2_{(aq)} + Zn_{(s)} -> ZnCl2_{(aq)} + Cu_{(s)}}$

Notice that at least one of the reactants was in the aqueous phase, and a new solid chemical species is produced on the product side. $\ce{AgCl_{(s)}}$ and $\ce{Cu_{(s)}}$, respectively, are the precipitates. Incidentally, the second reaction is also an example of a redox reaction, since it involved a change in the oxidation states of the reactants (here, zinc is oxidized while copper is reduced).

Identifying acid-base and redox reactions is a lengthier topic, so I suggest you consult a textbook or any of the numerous online resources.


Calculating equilibrium constant given Ka value? $\ce{NH_3(aq) + H_3O^+(aq)  <-> NH_4^+(aq) + H_2O(l)}$
$\ce{K_a}$ for $\ce{NH_4^+}$ is $5.6\times10^{-10}$.

I am completely lost and I have no idea how to approach this question.

Never mind, I figured it out. All I had to do is reciprocate the value since Ka is the reciprocal of the equilibrium constant for that reaction.


Why is this atom sp² hybridised? I'm doing exercises on hybridisation, and I was given this molecule:

I'm wondering about this (electron deficient) oxygen atom. My intuition says it should be ${sp^2}$ like the answer says, but honestly I only see that it's $sp$ hybridised. Is it because you ask what it would look like if it was not electron deficient?

This would become much clearer if you show the lone pair on the oxygen atom.

One $p$ orbital of the Oxygen atom forms the $\pi$ bond with Carbon, while the other three (remaining one $s$ and two $p$) orbitals are $sp^2$ hybridized.

Two $sp^2$ orbitals are used to form the $\sigma$ bonds while the last $sp^2$ orbital is occupied by the lone pair.


What is the role of NaCl in the functionalization of gold nanoparticles? In the functionalization of gold nanoparticles (GNPs) through the sulfur-metal bond, this article (among others) states that $\ce{NaCl}$ is slowly added in the GNPs solution after addition of thiol compound.

What is the role of $\ce{NaCl}$ in the functionalization step?

$\ce{NaCl}$ acts as a charge suppressor so that the negatively charged gold nanoparticle and a negatively charged ligand (with thiol) are not longer charge repelled and can experience a chemical bond. $\ce{NaCl}$ is easily broken into $\ce{Na+}$ and $\ce{Cl-}$ so that the $\ce{Na+}$ can pacify the negative charge on the surface of the gold nanoparticle to become less negative and approach zero charge. Other higher charged salts such as $\ce{Mg^2+}$ impart even greater pacification.
Here are some good references:

J Liu, Y Lu. Preparation of aptamer-linked gold nanoparticle purple aggregates for colorimetric sensing of analytes. Nature protocols, 2006. nature.com

B Du, Z Li, Y Cheng. Homogeneous immunoassay based on aggregation of antibody-functionalized gold nanoparticles coupled with light scattering detection. Talanta, 2008. Elsevier. From this last ref:

in $\pu{10 mM}$ phosphate-buffered solution (pH 7.0) containing $\pu{1.0 mmol L−1} ~\ce{NaCl}$ at room …


What does it mean if I am unable to write RS configuration for a compound? We had to write all possible optically active monochloro derivatives of 2-methylbutane.
I was unable to write RS configurations for 1 and 3, since they have carbon atoms with more than one same groups attached to them (they are achiral). Does that mean they are optically inactive?
P.S. I’m still a beginner at this...

I think I get what your question is and will try explain the things to you.

  I was unable to write RS configurations for 1 and 3, since they have carbon atoms with more than one same groups attached to them (they are achiral). Does that mean they are optically inactive?

The crux of your question lies apparently in two problems:

you assume that molecules 1 and 3 are optically inactive, but are unable to visualize this with a center/plane of symmetry
you seem to assume that molecules without chiral carbons are optically inactive

So, let's tackle this:

  Correct me if I am wrong, these compounds don’t seem to have planes or centres of symmetry.

Well, you're wrong. Note that in simple saturated hydrocarbons (with or without halogen substituents), it is usually safe to assume that if no chiral carbons are present, then the molecule will be optically inactive (without the need to validate this assumption by manually drawing out their planes/centers of symmetry). Here are two examples:

 (source)

notice how the plane of symmetry cuts-through through the blue and green atoms, using the remaining two grey identical atoms to obtain a plane of symmetry, to prove the molecule is optically inactive. This was expected, since the molecule had no chiral carbon.

it is possible to visualize the plane of symmetry in 2-chloro-2-methylbutane from its eclipsed Newmann projection, as seen above. The plane of symmetry cuts-through the chlorine atom and the methyl radical in front of it, rendering the molecule optically inactive, as expected.

  Are all molecules without any chiral carbon optically inactive?

No, this is wrong. Common examples like ortho-disubstituted-biphenyls (tl;dr: they are propeller shaped rings), allenes (tl;dr: their terminal groups are in perpendicular planes), etc. have no chiral carbons, yet are optically active. I won't detail these as you're still a "beginner" in this, but you'll definitely come across these examples in the very end of your course on optical activity.

In regards to your edited question, it is unreasonable to draw Fischer projection for achiral carbon atoms. For example, when drawing the Fischer projection of carbohydates, the terminal achiral carbon is written as a $\ce{-CH2OH}$ substituent to the penultimate chiral carbon, and not expanded out into its own Fisher projection.

I hope it helps!


Is there a household method to decolorize potentially contaminated water? I am living in my dorms at the moment and apparently it is an accepted thing that the water in the building is brown and yellow. According to my RA, this is a characteristic of the town I live in.

The only "clean" water we get is a water filter at the zeroth floor of our building where transparent water comes out.

Other than that the water that comes out of the tap, the kitchen, the shower, and the toilet bowl is light yellowish-brown water. I haven't done enough research as the cause of this problem.

Boiling water doesn't seem to change the colour either (is this even possible?). I for one cannot live like this, so I am asking the community here if they know a household method to remove the discoloration from the water.

A simple test you can do is to boil a large amount of the water and see if a sediment is left in the vessel. This will tell you certain things about you contaminant.

Is it particulate? We expect so because a filter is removing the contaminant.
Is it's boiling point below that of water?

If you get a residue what is it's state? I'm expecting a solid, I dont anticipate a coloured liquid that is miscible in water with a boiling point higher than water.

Does it burn? How does it burn? Colour of flame? Further residue?

I'd also try adding a colourless surfactant to the water, does it remain coloured?


Why can Wurtz reaction be not used for preparation of alkanes with odd number of carbon atoms? My book says that Wurtz reaction can not be used for preparation of alkanes with odd number of carbon atoms. Why is it so?

It's always good to think of what would probably go on in the reaction vessel. Here's a mechanism of the Wurtz reaction:

First, one electron from the metal is transferred to the halogen to produce a metal halide and an alkyl radical.

$\ce{R–X + M → R. + M+X−}$

The alkyl radical then accepts an electron from another metal atom to form an alkyl anion and the metal becomes cationic. This intermediate has been isolated in several cases.

$\ce{R. + M → R^-M+}$

The last step is a $\ce{S_N2}$ reaction; the nucleophilic carbon of the alkyl anion displaces the halide, forming a new carbon-carbon covalent bond

$\ce{R−M+ + R–X → R–R + M+X−}$

Simple. Now, let's take a look at an example.

If we start with a single alkyl halide, in this case a methyl halide,
$\ce{2 CH3-X + Na ->[][-2NaX] CH3-CH3 }$

Now, let's consider two dissimilar alkyl halides, ethyl and methyl halide:
3 possible reactions take place, giving a statistical distribution of products.

$\ce{2CH3-X + Na ->[][-2NaX] X-CH3}$

$\ce{2CH3-CH2-X + Na ->[][-2NaX] CH3-CH2-CH2-CH3}$

$\ce{CH3-X + X-CH2-CH3 + Na ->[][-2NaX] CH3-CH2-CH3}$

Sure, you've produced an alkane with an odd number of carbon atoms, but at the same time you have a mixture of alkanes (this is not really a selective reaction), that you will have a hard time separating.


Relationship between heat and standard enthalpy for a reversible reaction The enthalpy change for a reaction is equal to the heat absorbed at a constant pressure.

My notes say that the standard enthalpy change is the heat absorbed when the reactants (in their standard states) are completely converted to products (in their standard states). It uses this as an explanation as to why it is only possible to use heat measurements to determine  the standard enthalpy of reaction for a reaction that goes to completion.

I am not sure why it is the case that the reaction must go to completion for heat flow to be equal to the standard enthalpy change.

In this description, the reactants in the initial state and the products in the final state are all pure.  Now, if you only let the system go partially to completion, and then separated all the remaining reactants and products into pure substances at the standard conditions (as part of your process), you could determine the heat of reaction from that also.


Why do esters have to be distilled during purification? I am currently designing a lab experiment for a school project in which I am to determine the effect of catalyst concentration ($\ce{H2SO4}$) on the rate of reaction between methanoic (formic) acid and ethanol to create ethyl methanoate. I plan to do this by varying the concentration of sulfuric acid at the beginning, and comparing the end yield of ester.
$$\ce{HCOOH +C2H5OH <=>[varied concentrations of H2SO4] HCOOC2H5 + H2O}$$

I am aware that distillation is a crucial step following reflux in pretty much every esterification for purification, but why? Can't one simply wash the crude ester with deionised water to catch all remaining ethanol, use CaCO$_3$ to neutralise the acids, and filter off via a separatory funnel?

I am asking this because my school project has a set timeframe for all experiments of officially 8 hours over two weeks, in which it is recommended that I conduct at least '5 measurements at 5 intervals'. I know I won't reach this, even if I reflux each of my esters for 10 minutes because of the distillation.

According to Wikipedia ethyl formate is 9% soluble in water, that figure may be increased by any excess ethanol present, so you will be losing a significant portion in the proposed work-up. Is this enough error to invalidate the experiment?


What is the correct temperature probe for a wet etch solution? What type of probe or probe coating should one use to measure the temperature of a roughly 100c aqueous solution of cupric chloride?

A stainless steel probe seems incorrect. Is a Teflon coated probe sufficient or is something else more appropriate?

Is there a standard way of protecting metal probes in corrosive liquids?

Teflon is resistant against corrosive chemicals and stable up to temperatures of about 200 °C, so it is suitable for use in hot aqueous salt solutions. Using a coating or casing made of plastic which is unreactive and resistant against heat should be a good way to protect temperature probes.


What is Fjord Region? I'm studying molecular motors for a project of mine and I'm trying to understand what a Fjord region is.

It's sort of labeled in the image below but I'm not entirely sure which part.

I would like a suggestion of a book or article that may help me understand what it is?

Fjords and bays

Fjord is a term originally developed by chemists studying polycyclic aromatic hydrocarbons, along with the term bay.

The descriptions aren't rigidly defined in the same way as stereochemical descriptors, but were designed to be 'self-explanatory', following the real world uses of the words1

Bay: "a broad inlet of the sea where the land curves inwards."
Fjord: "a long, narrow, deep inlet of the sea between high cliffs, as in Norway, typically formed by submergence of a glaciated valley."

In an attempt unify the usage of these terms, IUPAC issued a set of guidelines for their usage in 2014.2

Some historical context

The earliest publication of these terms was given in the references above by Harry Heaney, Keith Bartle, Denny Jones, and Peter Lees who were working at the now non-existent Institute of Technology (Bradford, UK). They were studying triphenylene, and trying to find descriptions for the various environments present.

  Circumnavigating triphenylene, the British crew sighted two essentially different kinds of H's; and Derry Jones spontaneously called one type "bay" and the other "peninsular." They published these descriptive "self-explanatory" names,40 but not without some apprehension. In fact, Professor Jones confessed to having dreamed that referees rejected their manuscript on account of its "bay" and "peninsular" nomencla ture

Reference:

1 Spectrochim. Acta 1966,22,pp941-951; Trans. Farad.  Soc. 1967,63,pp2868-287

2 J. Polycyclic Aromatic Hydrocarbons, 2015, 35, pp 161-176


How can we define optical isomerism?What is the example of non enatiomeric optical isomers? My book has very vaguely explained these concepts,I can't understand anything from there.
Through a little bit of research on the internet I came to following conclusion:

that optical isomers are either dextrorotatory or levorotatory
all enantiomers are either dextro or levo so they are optical isomers.
Some diastereomers may also be related as d- or l- (though I feel the angle of rotation will be different) and thus can be optical isomers.

I feel I am missing something or have the wrong concept entirely.
These are pics from my book.
In the heading of optical isomers there are 4 isomers of tartaric acid mentioned,why is that?So how many types of optical isomers are there?

There are, in the sense you are asking the question, three types of optical isomers: +, -, and optically inactive. This is detailed in pages of the textbook you have uploaded.
You should think about optical isomers in how they are determined. Imagine you have a machine which we will call and "optical isometer". To use this machine, you put in any compound in solution, and it will tell you one of three outputs:

"This compound polarises light in the clockwise direction"
"This compound polarises light in the anti-clockwise direction"
"This compound does not polarise light"

Now, our optical isometer doesn't know any chemistry, but it does know how to tell if light is being polarised by our compound in solution. If you fundamentally think of isomers as identical in their molecular formula, but differing in their spatial arrangement of atoms, you need only imagine as optical isomers being different in how the spatial arrangement of atoms polarises light.
Let's imagine we have a solution of a tartaric acid, but we don't know what isomeric form it is in. The options are:

meso-tartaric acid
(+)tartaric acid
(-)tartaric acid
A mixture of (+) and (-)tartaric acid.

Where I think you have been confused by the textbook is that option number 4 doesn't actually describe different spatial arrangements of tartaric acid, rather it describes a mixture of options 2 and 3. So when it says there are four isomeric forms, it isn't strictly correct about that.
Now, how will we determine what isomer of tartaric acid we have in solution? One thing that we can do to find this out would be to use our optical isometer. We put our solution in and it outputs: "This compound polarises light in the clockwise direction". We know that of our four possible options, only (+)tartaric acid polarises light in the clockwise direction. Likewise, we would know it to be (-)tartaric acid if the light were polarised in the anti-clockwise direction.
What if our optical isometer output, "This compound does not polarise light"? Then this means that our compound is optically inactive. We know that both the meso- form and the racemic (mixture of (+) and (-)) don't polarise light in either the clockwise or anti-clockwise direction, and so we would need further analysis to find out for sure!
There are other types of light polarisation, but not in the sense this book is talking about optical isomers. This textbook seems to be a little confusing, but these ideas are actually really well explained on places like wikipedia or khan academy, so next time you get stuck, don't worry too much about the textbook but give other resources a shot!
Hope this helped,
Edan


Is methyl orange acidic or basic? It has been saying that if adding too much indicator in solution, the ph value of solution would be changed. From this, it can be saying that indicators have a pH value. Methyl orange is the indicator I used in the experiment. "My question is whether methyl orange is basic or acidic so that it affects the pH value of solutions."

In an acidic solution, the protonation methyl orange takes place, as shown below

The deprotanted form depicted on the left is basic, and is yellow in colour. The red form is acidic and is depicted on the right.

In general, a lot of acid base-indicators rely on the following mechanism of action

$$\ce{In^-}+  \ce{<=>[\ce{H^+}][\ce{OH^-}]} \ce{HIn}$$

At low pH, i.e high $\ce{H^+}$ concentration the $\ce{HIn}$ form is dominant since the equilibrium lies to the right, however, at higher pH $\ce{HIn}$ (which is weakly acidic) dissasociates to give $\ce{In^-}$ (its conjugate base).

Herein, $\ce{HIn}$ (the acid), and $\ce{In^-}$ (its corresponding conjugate base) have different colours. The colour changes are due to changes in absorption, due to change in the length of the $\pi$ conjugated system (as can be seen in the case of methyl orange pictured above).


Theoretically, Which d orbital participates in sp3d and sp3d2? This question is regarding the old way (inaccurate) $\ce{PCl5}$ and $\ce{SF6}$ are taught in elementary chemistry.

In the crystal field splitting diagram for triagonal bypyramidal geometry, $\mathrm{d}_{z^2}$ has the highest energy. Now, since $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$ of same shell hybridise, then the $\mathrm{d}$ orbital closest in energy to $\mathrm{s}$ and $\mathrm{p}$ should take part, which should be $\mathrm{d}_{xz}$ or $\mathrm{d}_{yz}$. Instead, the hybridisation is said to be $\mathrm{sp^3d}_{z^2}$.

A similar problem arises for molecules with an octahedral shape. It has $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ which are higher in energy than the rest and have higher energy difference with $\mathrm{s}$ and $\mathrm{p}$ orbitals of the same shell. There is no such problem for $\mathrm{d^2sp^3}$, $\mathrm{d^3s}$, $\mathrm{dsp^2}$.

EDIT: Let me change the question to ask the same thing. In outer shell octahedral complexes ($\mathrm{s,p,d}$ of same principal quantum numbers take part) why are the $\mathrm{e_g}$ orbitals taking part in sigma bond formation when they are furthest away from $n\mathrm{s}$ and $n\mathrm{p}$ orbitals?

There is no such problem for inner shell complexes.

The answer to this lies in the symmetry of the orbitals involved.

Naturally, an octahedral complex will have the point group $O_\mathrm{h}$ — the octahedral point group. In molecular orbital theory, we will first need to work out how the ligand group orbitals transform; then we can find out which metal orbitals transform identically, i.e. belong to the same irreducible representation; only these will form bonds with each other. To arrive at the group orbitals’ irreducible representations, we need to see how they transform under the given symmetry operations of the group; thankfully, a copy of the character table can be found on meta (Attention: MathJax heavy link! Give the page about 30 seconds time to load!) Since your diagram does not include π interactions, I will ignore them, too.

\begin{array}{lcccccccccc}\hline
\text{Irrep} & E & 8 C_3 & 6 C_2 & 6 C_4 & 3 C_4^2 & i & 6 S_4 & 8 S_6 & 3 \sigma_\mathrm{h} & 6 \sigma_\mathrm{d} \\ \hline
6 \sigma_{\ce{L\bond{->}M}} & 6 & 0 & 0 & 2 & 2 & 0 & 0 & 0 & 4 & 2 \\ \hline\end{array}

We now need to use the reduction formula to determine how many times each irreducible representation is part of this reducible representation.

$$n = \frac 1h \sum_{\chi} N \chi_\mathrm{r} \chi_\mathrm{i}\tag{1}$$

Wherein $n$ is the number of times an irreducible representation is included, $N$ is the coefficient of a symmetry operation group, $h$ is the sum of all coefficients ($48$ in $O_\mathrm{h}$) and $\chi$ is the character of the irreducible or reducible representation, respectively. Thankfully, a large number of reducible group characters end up as $0$, easing our calculations. This should be the result:

$$\begin{align}n(\mathrm{a_{1g}}) &= \frac 1{48} (6 + 2 \times 6 \times 1 + 2 \times 3 \times 1 + 4 \times 3 \times 1 + 2 \times 6 \times 1) &&= 1\\
n(\mathrm{a_{2g}}) &= \frac 1{48} (6 + 2 \times 6 \times (-1) + 2 \times 3 \times 1 + 4 \times 3 \times 1 + 2 \times 6 \times (-1)) &&= 0\\
n(\mathrm{e_g}) &= \frac 1{48} (12 + 2 \times 6 \times 0 + 2 \times 3 \times 2 + 4 \times 3 \times 2 + 2 \times 6 \times 0) &&= 1\\
n(\mathrm{t_{1g}}) &= \frac 1{48} (18 + 2 \times 6 \times 1 + 2 \times 3 \times (-1) + 4 \times 3 \times (-1) + 2 \times 6 \times (-1)) &&= 0\\
n(\mathrm{t_{2g}}) &= \frac 1{48} (18 + 2 \times 6 \times (-1) + 2 \times 3 \times (-1) + 4 \times 3 \times (-1) + 2 \times 6 \times 1) &&= 0\\
n(\mathrm{a_{1u}}) &= \frac 1{48} (6 + 2 \times 6 \times 1 + 2 \times 3 \times 1 + 4 \times 3 \times (-1) + 2 \times 6 \times (-1)) &&= 0\\
n(\mathrm{a_{2u}}) &= \frac 1{48} (6 + 2 \times 6 \times (-1) + 2 \times 3 \times 1 + 4 \times 3 \times (-1) + 2 \times 6 \times 1) &&= 0\\
n(\mathrm{e_u}) &= \frac 1{48} (12 + 2 \times 6 \times 0 + 2 \times 3 \times 2 + 4 \times 3 \times (-2) + 2 \times 6 \times 0) &&= 0\\
n(\mathrm{t_{1u}}) &= \frac 1{48} (18 + 2 \times 6 \times 1 + 2 \times 3 \times (-1) + 4 \times 3 \times 1 + 2 \times 6 \times 1) &&= 1\\
n(\mathrm{t_{2u}}) &= \frac 1{48} (18 + 2 \times 6 \times (-1) + 2 \times 3 \times (-1) + 4 \times 3 \times 1 + 2 \times 6 \times (-1)) &&= 0\end{align}$$

We see, that the six σ-symmetric ligand orbitals transform as $\mathrm{a_{1g} + e_g + t_{1u}}$ — this is also shown in your picture. Now that we have this settled, we need to ask ourselves which metal orbitals these irreducibles correspond to. $\mathrm{a_{1g}}$, the totally symmetric irreducible is easy; it must be the $\mathrm{s}$ orbital. Furthermore, the character table tells us that $(x,y,z)$ transforms as $\mathrm{t_{1u}}$ which means that the $\mathrm{p}$ orbitals must transform as that. (You can verify this by performing the same exercise on the three central $\mathrm{p}$ orbitals.)

This leaves the $\mathrm{d}$ orbitals which — for symmetry reasons as there is no pentadegenerate irreducible — have to split up into at least a group of three and two ($\mathrm{t}_x$ and $\mathrm{e}_x$). Again, glancing at the character table tells us that $\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ transform as $\mathrm{t_{2g}}$ and $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$ as $\mathrm{e_g}$. The first can easily be verified by the same exercise as for the $\mathrm{p}$ orbitals; the second not so easily due to the ‘unnatural’ shape of the $\mathrm{d}_{z^2}$ orbital. However, do trust the calculations that they are equivalent.

Therefore, it is only $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ that can interact with the ligand group orbitals and which therefore form the σ bonds as your scheme shows. And this also makes sense from a purely qualitative point of view: The $\mathrm{d}_{z^2}$ orbital is pointing towards the $z$-axis ligands (and the ring is somewhat interacting with the $x,y$-axis ligands) while each of the four lobes of $\mathrm{d}_{x^2-y^2}$ is pointing towards a ligand on the $x$ or $y$ axis. The $\mathrm{t_{2g}}$ orbitals are pointing in between the coordinate axis directions and therefore between the ligands and are thus nonbonding with respect to σ bonds.


Supermarket HDPE plastic bags safety Supermarket plastic bags are made of high density polyethylene, and I am interested in the decomposition products at high temperature: what type of compounds are released and how dangerous are they? I am interested in a situation where one has placed some kind of heated food inside one of these plastic bags.

The question is, are harmful compounds from the plastic bag passed into the food (food is hot, like baked bread) contained inside the plastic bag?

Polyethylene is generally considered suitable for food handling and even for warming food. Unlike some plastics such as polycarbonate, it does not need an additional plasticizer that could be toxic.

That said, the melting point of polyethylene is ~1150C, and I've found that it tends to shrink-wrap around microwaved food. For that reason, polypropylene, m.p. ~1300C, is often used for food containers intended for microwave oven use and for automatic dish-washing.

BTW, both polyethylene and polypropylene absorb some of the natural lycopene coloring of tomatoes or other carotenoid pigments and slowly develop a pale brown discoloration that cannot be removed, but that is of no significance.


Reaction involving markovnikov's rule Question:

Attempt:
The first reaction is addition of hydrogen halide. The reaction follows markovnikov's rule. The two possible products of the first reaction are:

For each product, the carbocations formed are:

After the formation of carbocation, $\ce{Cl-}$ attacks the carbocation forming a dihalide.
The final product after second reaction depends on the above intermediates.
The second carbocation faces electron withdrawing inductive effect and electron donating resonance effect from chlorine. But for halogens, inductive effect is more than resonance effect. So the carbocation is destabilized. So I chose option (A) as the answer since the first carbocation is formed.
But the answer given is (C) which means the second carbocation is formed.

Indeed the second carbocation is formed.  It's favored by the chlorine atom being able to delocalize the charge through resonance, draw resonance hybrids for the second cation and you should see it.  In a sense you might say that the carbon bonded to chlorine is more effectively substituted for stabilizing the cation and so the Markovnikov rule is "bent" accordingly.

The usual version of the Markovnikov rule is for hydrocarbon substrates.


Is it possible that diamond can be grown from aqueous solutions? Assume a cubic diamond crystal with hydrogen-terminated surface bonds. The terminating hydrogen atoms will form pairs that are geometrically close together in alternating diagonal pairs (think of the rectangles on corrugated walking steel).

Next, picture a formaldehyde molecule being pushed oxygen-first into one of these terminating hydrogen pairs. If the oxygen atom can be made to react with the hydrogen pair, a water molecule would be expelled and the remaining $\ce{CH2}$ group would presumably bond to the two bonds of the diamond lattice.

If you repeat this process over the entire cubic face, the result will be a new layer of diamond that is terminated with the same paired hydrogen atoms as the original layer. The entire layer-growth process thus in principle could be repeated indefinitely, resulting in a diamond grown at or near room temperature from an aqueous or other low-temperature solution.

Question: Can anyone think of a fundamental reason why the above process, or some variant of it, could not be used to grow diamond crystals at room temperature? (A "variant" might start with carbohydrate units larger than $\ce{OCH2}$, for example.)

One or more geometrically sophisticated catalyst molecules would be required. Each catalyst molecule would snag, carry, and orient a formaldehyde molecules onto some part of the textured surfaces provided by the cubic diamond face, such as the face itself or along a growth edge.

I did this a long time ago in high school, but my recollection is that the reaction should be mildly exothermic. But even if it is not, energy-source molecules (think of the role of ATP in living systems) could provide energy to enable an endothermic reaction.

So... thoughts, anyone? Is there any deep reason why this approach to room-temperature or near-room-temperature diamond synthesis is clearly not possible?

The reaction you propose would be formally equivalent to the protection of aldehydes with 1,3-propanedithiol to form the correspondent dithiane. There are many organic reactions in which aldehydes participate this way, but all of them have something in common: the other species(lets call it donor) must have a free pair of electrons to attack the carbon atom in the aldehyde. In case this pair is not present, but the donor has an acidic hydrogen, we could use a strong base to remove it and form in situ the pair of electrons. (See aldol reactions)

Here's the problem: diamonds have no free pairs of electrons. And terminal hydrogens in a diamond lattice are by no means acidic, hence can't be removed by any base to the best of my knowledge. So there's no way a diamond will react with aldehydes.


Formation of a stable hydrate Question: Which of the following does not form a stable hydrate?

I think they mean nucleophilic addition of water. For compound (C), on reaction with water, the product formed is

The product has lesser bond angle strain than the reactant. Hence, the compound formed is stable. This is because $\ce{sp^2}$ hybridised carbonyl carbon requires $120^{\circ}$ but only gets $60^{\circ}$ in cyclopropanone. But converting it into a diol makes the angle $109^{\circ}$ which reduces strain to a large extent. In fact, cyclobutanone is also reactive to water.

For (B) however, there is not much change in the bond angle strain. The initial angle is already close to the required angle.

What can I say about (A) and (D)?

What can I say about (A) and (D)?

When a carbonyl compound and water are present together, an equilibrium is set up between the carbonyl compound and the corresponding gem-diol. Like any equilibrium, this equilibrium can be shifted to one side or the other by factors that stabilize (or destabilize) one side or the other.

If water is present, poly-vicinal carbonyl compounds usually exist as hydrates. Drawing the resonance structure of a carbonyl compound we see the the $\ce{C=O}$ bond is highly polarized with the carbon terminus being the positive end.  If we have several carbonyls adjacent to one another, then each carbonyl has this same polarization and an electrostatic destabilization results. This is particularly true for the central carbon since there is a partial positive charge on each of the carbons attached to it.

In order to avoid this electrostatic destabilization, such compounds will form a hydrate at the central (the most electrostatically destabilized) carbon. In the hydrated form there is no repulsive partial positive charge on adjacent carbons.

You've already figured out why cyclopropanone exists as a hydrate.  So the correct answer is B, since there is no driving force that would provide any special stabilization of its hydrate.


Can a molecule be neither gerade nor ungerade? Or can any molecular orbit always be written a linear combination of gerade and ungerade basis states?

g/u is not a property of a molecule; I assume you meant molecular orbital.
All functions can be written as a linear combination of even + odd functions: there is a short explanation on Wikipedia. Basically, you have a function $f(x)$; now define

$$\begin{align}
f_\mathrm e(x) &= \frac{1}{2}[f(x) + f(-x)] \\
f_\mathrm o(x) &= \frac{1}{2}[f(x) - f(-x)] \\
\end{align}$$

It is clear from the definition of parity that $f_\mathrm e$ is even (since $f_\mathrm e(x) = f_\mathrm e(-x)$) and likewise that $f_\mathrm o$ is odd. Now since $f(x) = f_\mathrm e(x) + f_\mathrm o(x)$, we have shown that any function $f(x)$ can be expressed as a linear combination of even and odd functions.

In three dimensions where you have $\psi(x,y,z)$ you simply need to define

$$\begin{align}
\psi_\mathrm g(x) &= \frac{1}{2}[\psi(x,y,z) + \psi(-x,-y,-z)] \\
\psi_\mathrm u(x) &= \frac{1}{2}[\psi(x,y,z) - \psi(-x,-y,-z)] \\
\end{align}$$

and analogously to above one can see that every function $\psi(x,y,z)$ can be expressed as the sum of a gerade component $\psi_\mathrm g$ and an ungerade component $\psi_\mathrm u$. These components don't have any physical meaning (unless $\psi$ itself is either g or u, in which case one component is simply $\psi$ and the other is zero), but that's somewhat beside the point.

If a molecule does not possess a centre of inversion then its (canonical) molecular orbitals will not possess g/u symmetry.

So the answer is yes, and yes. There is no "or".


Avogadro's gas Law doesn't make sense? (it does actually) It says equal volumes of all gasses under same conditions of temp. and pressure contains the same number of atoms/molecules.

Now suppose you have 1L of Chlorine in a box and 1L Hydrogen in another box , now the comparitively huge size of the Cl atom means less of it will fit in the box , whereas lots of hygrogen will. ( breaking the law)

But, well this should have an effect in pressure , how does all that boil down to avogadro being right ?

Oh it definitely makes sense since it allows us to describe a lot of systems with very simple formulas.

The problem is that this law assumes "ideal gases" (just like the ideal gas law) where atoms/molecules are "points" (no volume) and they don't interact with each other. This works well at rather low temperatures, low pressure and for gases without strong intermolecular interactions.

But you are correct, there are problems and the law doesn't work in a lot of cases. For example if you go to high pressures the volume of the indiviual atoms/molecules does play a role, since if they are bigger the available space will be smaller and the pressure will be higher than predicted by the ideal gas law.


What kind of plastic is (really) safe to store vegetable oil? Storing colza or olive oil in a plastic bottle might be risky since the oil can leach out phthalates and BPA (and others molecules?).

Some manufacturers sell "phthalates and BPA free bottles", but are they safe to store oil?

What kind of plastic would you use to store oil?

If you are concerned, I would suggest that high-density poly ethylene (HDPE) or poly propylene (PP) to be better options. HDPE is a heavy duty plastic resistant to most chemicals and physical deformation- it is often used for household products including bleach, detergent etc, but is also commonly used for milk bottles and the like.


Using the electrical potential created by the electron transport chain to power electrical systems My friend and I were discussing the extremely low energy density of modern batteries and read that food has a much higher energy density. Thus, we were considering the possibility of using biological processes to extract energy from food and use that to create an electric potential or do work in some other way.

The electron transport chain creates an electrochemical gradient and thus and electric potential using protons. If some sort of system could be devised to mimic/perform glycolysis and the citric acid cycle and also mimic the ETC membrane and necessary proteins, is it feasible/is there already research on using glucose and cellular respiration (or a similar biological process) to power electrical systems? How would this relate to fossils fuels and what scale could this be useful at?

There are fuel cells that run on alcohols such as ethanol or methanol and on glucose.

That said, the energy density in current Li-ion cells is quite high - a 18650 cell holds ~30,000 joules, about 9 kCal. By comparison, a  "biochemically" powered shotgun shell (well, the nitrate and carbon of gunpowder can come from bio sources) packs ~2,000 J.


Determining the torsion energy profile of butadiene by molecular dynamics Disclaimer: this question is related to my previous question, although it is much more detailed

Consider the following rotation of butadiene:

I want to determine the relative energy of each dihedral angle by molecular dynamics (MD).

My initial idea was to simulate butadiene in the NVT ensemble and use the relative frequencies of each dihedral angle to determine the relative energy. In the NVT ensemble, the probability of each microstate $P_i$ is:

$$P_i = \frac{\mathrm e^{-E/RT}}{Z}$$

Thus, I performed MD simulations in the NVT ensemble and recorded the dihedral angle periodically. With $N$ total dihedral angles, I made a histogram of the dihedral angles (e.g. a bin every 5 degrees). If we define the number of angles contained in a given bin as $n_i$, I am using the following equation:

$$n_i = \frac{\mathrm e^{-E/RT}}{N}$$

This equation can be rearranged to isolate the approximate energy associated with that dihedral angle:

$$-RT \ln(n_i N)= E$$

However, the curve obtained does not match the distribution I obtained by a simple relaxed scan of the dihedral:

Where is the problem in my method?

I really only want to make a comment, but it is too long...

Your equation $n_i = \exp^{-E_i/RT}/N$ does not make sense to me.

I agree with
\begin{equation}
P_i = \frac{\exp ^{-E_i/RT}}{Z}
\end{equation}

where $Z = \sum \exp ^{-E_i/RT}$ is the configurational partition function.

I can also accept that using a histogram, you could say $P_i = n_i/N$ where $n_i$ is number of samples in that bin, and N is total samples.

I do not see however how

\begin{equation}
n_i N = \exp ^{-E_i/RT}
\end{equation}

This does not make sense to me. Perhaps you could swap $P_i$ for $n_i/N$ so that you have
\begin{equation}
P_i =\frac{n_i}{N} = \frac{\exp ^{-E_i/RT}}{Z}
\end{equation}

which could lead to
\begin{equation}
n_i = \frac{N}{Z}\exp ^{-E_i/RT}
\end{equation}

Which could be rearranged to get
\begin{equation}
E_i =-RT \ln \left(\frac{n_i }{N}Z\right)
\end{equation}

If you are comparing to a reference energy then you could get

\begin{eqnarray}
E_i - E_{\rm ref} = \Delta E_{\rm i,ref} &=&-RT \ln \left(\frac{n_i }{N}Z\right) +RT \ln \left(\frac{n_{\rm ref} }{N}Z\right) \\
&=& -RT\ln \left(\frac{n_i }{n_{\rm ref}}\right)
\end{eqnarray}

It would be interesting to see if this works? I don't have the raw data, so I cannot try it (I also don't have the time, only the interest).


Why are isotopically pure diamonds 50% more thermally conductive than other diamonds? A laboratory grown type-IIa diamond (no nitrogen defects) has a lambda of $\pu{1800-2200 W/mK}$, but an isotopically pure diamond of $\ce{^{12}C}$ can have up to $\pu{3320 W/mK}$.

Why are $\ce{^{12}C}$ diamonds so much more thermally conductive than diamonds with $\approx \pu{1.1\%}$ $\ce{^{13}C}$ atoms?

Sourcing for my question:
https://www.sciencedirect.com/science/article/pii/092596359290197V?via%3Dihub
https://adsabs.harvard.edu/abs/1990PhRvB..42.1104A

From THERMAL  CONDUCTIVITY  OF  NATURAL  AND  ISOTOPICALLY  ENRICHED  DIAMOND  - EFFECT OF NEUTRON IRRADIATION, D.P. White, Department of Physics, Merrimack College, N. Andover, MA 01845:

  The thermal conductivity was calculated using the Callaway method.  This method takes into account  the  effects  of  intrinsic  three-phonon  normal  (N)  processes.    These  processes  do  not create thermal resistance directly but affect the thermal resistance by transferring phonons from frequencies where they are not scattered to frequencies where scattering is more efficient.

[...]

  In order to  explain  the  large  increase  (originally  much larger  than  expected)  in  terms  of  isotope scattering it was necessary to include the effects of N processes.  Subsequent analyses of the thermal  conductivity  of  enriched  diamond  have  included  N  processes.    Wei et.al.  used  the Callaway  expression  to  fit  experimental  thermal  conductivity  data  on  diamond  with  the  natural isotope  concentration  and  isotopically  enriched  (0.1% $\ce{^{13}C}$)  diamond  over  a  wide  temperature range (100-1000K).


What are the disadvantages of using the InChI format? What are the disadvantages (if any) in producing and using the InChI format of chemical compound representation in chemoinformatics? Are there any particular problem one can face when using InChI?

You have to evaluate shortcomings as compared to something. You can make a list and say "but InChI doesn't represent X" where X is ranging around

mixtures
Markush groups
polymers from repeat units
reactions
toasters

The point is: so what? All representations have some limitations. But limitations aren't shortcomings unless you have a different representation that doesn't have those limitations and that is not traded off by other limitations or bloat or other disadvantages.

What do you have to compare InChI with? In my world: SMILES and with MOLFILEs. InChI is most closely related to SMILEs. SMILES is easier to just type, but it isn't canonicalized and it doesn't have the layers of specificity that allow you to find approximate equivalences so easily as with InChI. InChI connection layer is very similar to SMILEs in principle, but it is a lot harder to read because you need to replace the numbers with the atoms from the preceding formula layer. But it is easy to convert InChI to SMILES, except where InChI is better. For example, parenthesized groups, so you represent the carboxyl group -C(=O)OH as -C(O,O)H, which isn't a thing in SMILES.

Where InChI shines is that many isomerisms are abstracted or pushed into more detailed layers. There is no other format that does this as well or at all as InChI, which is why it is good. But InChI is conceived as an "identifier" and some of the details are not at all represented, most specifically the atom positions. That's where MOLFILEs are so easy to work with, if you want to just display the molecule. But in that regard, InChI is no worse than SMILES.

So, you should be more specific in terms of what your goals are that you are asking for "disadvantages" and you need to stay very critical of text that berates InChI or any other standard with a bunch of rationalizations for why not to choose standard A (because the author is committed to standard B). I know this sort of superficial argument is being made a lot in any informatics sector, people like their baby and berate the other's baby. Especially be weary of those who say "don't use InChI, InChI is bad" and then "you really should follow the work on XYZ, because XYZ will be able to represent everything including the toaster."

InChI has it's firm place when the purpose is to represent, refer to, identify basic molecular structures. That's it. If you want mixtures, reactions, polymer repeat units, consider using InChI as the core building block and then build higher order structures around it. For example, obviously there is no reason to push reactions into the InChI representation, just make a simple structure that lists the reaction participants on both sides, their stoichiometry and the enthalpy, activation energies, etc. etc., these can be done in higher order data structures much better than shoving that information into a single InChI++ string. Likewise for polymers, represent each repeat unit as an InChI and then combine them. If your goal is such higher order molecules, you would profitably use InChI as the representation of those core building blocks.

Now that said, I give you my personal pet peeve about InChI, i.e., that is in scope of current InChI where I think it has, let's say, pitfalls, not necessarily that it made the wrong choice. And that is, ionic bonds and coordination bonds. Especially ionic bonds are represented as two separate structures, recognizable with the period . in the formula separating the multiple structures. This is actually a rudimentary representation of mixtures. So in my view, it should not even have introduced this dot notation but simply forced people to represent the parts of these ionic mixes as separate InChIs. I come from the biochemical / physiology / pharmaceutical / medical world where such ionic bonds usually do not even really exist in nature, as the ions are all dissociated in solution. But all that can be disputed. Same for metal coordination or chelate bonds, you want to say where the metal ion associates with, and I think that's not in InChI. So in my world it would be better not to have any of this mixing and coordination inside InChI and we would use higher order structures for it instead, but always use InChI as the basic building blocks.


Is butanoic acid more soluble in water than ethyl acetate? I was asked which of the two following isomeric compounds, a carboxylic acid and an ester, was more soluble in water.

I believe both molecules should be somewhat soluble in water as they have polar C=O and C–O bonds, which should allow them to form dipole–dipole interactions with water. However, I'm not sure how the difference in the structures would affect the solubility.

Try to solve the question by analogy. Imagine that you perform a couple of experiments on different glasses of water:

Add vinegar to one glass with water. Does it mix? Yes, it does!
Add olive oil to another glass with water? Does it mix? No, it does not.

Vinegar is a carboxylic acid, whereas olive oil is made of esters. This should suggest to you that carboxylic acids in general are more soluble than esters.

Carboxylic acids, $\ce{R-COOH}$, are acids, that partly dissociate in water

$$\ce{R-COOH + H2O -> H3O+ + R-COO-}$$

to form ions. Supposing that the alkyl chain isn't extremely long, even the undissociated part of the acid added is solvated in water, a polar protic solvent, by hydrogen bonding interactions. In other words, the carboxylic acid is highly soluble in water.

The solubility of ethyl acetate in water is 8.3 g/100 ml (Wikipedia), whereas butanoic acid is completely miscible in water. That is to say, butanoic acid completely dissolves in water, regardless of the proportion of acid to water.


How does one export a PDB structure to SVG? This question is one that I'm wondering whether it belongs here or on Graphic Design SE. All I want is some software that can export a PDB structure of a protein (displayed in ribbons style) or DNA double helix, or a combination of the two, to a high-quality SVG file. I know that several pieces of software for 3D molecular modelling can create raster images of PDB files quite easily (e.g., Accelrys DS Visualizer, Jmol, PyMOL, etc.). But SVG files have several advantages over PNGs, now I don't mind if the colours are all bland (i.e., each shape having one colour that is not subject to any shading or other gradients; I am saying this is OK as I realize that vector images often have problems with intricate colour gradients) just so long as the end result is a decent-looking SVG file. I have tried using Avogadro for this purpose but the resulting SVG was 264 MB in size -- a little too big for my taste. If you'd like an acceptable size range it'd be <5 MB.

If you haven't found answer till now, you can use VESTA (Visualization for Electronic and STructural Analysis) package. It is a freeware and can be downloaded from
http://jp-minerals.org/vesta/en/

It is available for both Windows and Linux. It can be used to import any many structural files (pdb included) and get an output in any supported format(which includes SVG). From their website:

  VESTA is a 3D visualization program for structural models, volumetric data such as electron/nuclear densities, and crystal morphologies.

However, if you do not strictly require SVG format them my personal favorite is "QuteMol". It gives way too many options to adjust visual representation
It is an open source (citeware, i.e. you have to cite their paper for any academic publishing) molecular rendering software available from:

http://qutemol.sourceforge.net/

It can generate PNG, JPEG anf GIF(animated) images of maximum dimensions up to 8192x8192 pixels so you can use them for high resolution printing/posters etc. too. One small flaw is that it is rather ill maintained and considerable amount of bugs are there.For e.g. using default options is fine but in case you want to use customized colour scheme then their example ART(atomic rendering table, just a table of atomic symbols, their radii, and colour in RGB codes) file gives following example:

  X    0       0.853  204    0    0

  H    1    0.930  199  199  199

  He    2  1.085  107  115  140

  Li    3  3.100    0  128  128

However in reality you will require 2 colums of radii for programme to recognize it (core and vander waal radii). Similarly atomic symbols with single alphabet are not rendered properly!!! But all these problems have work around and results are worth it (specially for large protein molecules).
sample this:

(source: sourceforge.net)


How do I find Rf values & length size of paper in chromatography? A student is designing an experiment using paper chromatography. From previous results, she knows that the Rf values for two dyes that she wishes to investigate are 0.72 and 0.80.
It is proposed that the dyes be run on the same piece of paper and, for various reasons, that the sports should have a separation of a at least 1cm at the end of the procedure.
What is the minimum height of the chromatography paper that she should use in this experiment?

Can someone correct my working out. I don't know what the correct answer is.

distance travelled by dye x = $x$
distance travelled by dye y = $y$
distance of solvent front = $p$
$x/p=0.72$ therefore $x = p(0.72)$
$y/p=0.8$, therefore $y = p(0.8)$
$y-x=1$, therefore $y=1+x$
new equation: $1+x=p(0.8)$ where $x=0(0.72)$, if we solve for $p$ we get $12.5$

so the piece of paper should be at-least 12.5 cm long.

$$\Delta R_f = 0.08$$
Let the separation distance be $\Delta r$ = 1 cm and let $l$ be the length of the paper in cm.

$$\Delta R_f = \frac{\Delta r\, \mathrm{cm}}{l\, \mathrm{cm}} $$

Solve for $l$, obtain $l$ = 12.5 cm.


What organizational methods pre-date the Periodic Table? Dmitri Mendeleev noticed patterns in elements which allowed him to design the periodic table which ultimately led to the modern periodic table.

How were elements organized before this? Was there any method that served as a crude standard?

One attempt to order chemical elements was Döbereiner's system of triades, published in Annalen der Physik und Chemie, back in 1829 (doi 10.1002/andp.18290910217 with Wiley); or (open access with Gallica).

Although an actual view into his paper permits the speculation he was not using the atom masses we know today; rather than using specific weight and molecular weight as known / defined at his time of oxides of Ca, Ba, Sr, for example:

(The dots on top of Ca, Ba, Sr seem to adhere a nomenclature, as on top of S, Se, Te three of them; Cl, Br, I are topped by five.)


How to define a new solvent in Gaussian09 to run SCRF? I want to run computation of my system in sulphuric acid with Gaussian 09; however sulphuric acid is not among the built-in solvents in G09. I know I have to give some parameters for the new solvent I define (including static and dynamic dielectric constants). Here's an example of my input files where I try to redefine a solvent water2.

#p b3lyp/6-31G(2df,p) int(grid=ultrafine)  SCRF(Solvent=Water2,Read)

Comment

0 1

...

stoichiometry=H2O1
solventname=Water2
eps=78.3553
epsinf=1.77556

G09 keeps telling me that "SCRF(Solvent=Water2,Read)" is not correct. I want to know what the correct input will be.

The correct input should be more like (replacing your Water2 with Generic)

#p b3lyp/6-31G(2df,p) int(grid=ultrafine) SCRF(Solvent=Generic,Read)

Comment

0 1
...

stoichiometry=H2O1
solventname=Water2
eps=78.3553
epsinf=1.77556

as it is stated in the Documentation under Additional Input for PCM Calculations.

Note - this only gets your calculation running. Your shown current input section might be insufficient to properly define sulphuric acid.

As Geoff noted, the Gaussian Documentary states an important fact short before showing some $\varepsilon$-values of common solvents that should be taken into account.

  We list the ε values here for convenience, but be aware it is only one of many internal parameters used to define solvents. Thus, simply changing the ε value will not define a new solvent properly.


Moment of inertia for a triatomic molecule What does the moment of inertia for a molecule mean? How is it useful? Most importantly, how would I go about solving it for a triatomic molecule?

The moment of inertia is defined for a molecule as it would be for any other solid: you consider the nuclei as points with mass, ignore the mass of the electrons, and you can then calculate either the scalar moment of inertia or the tensor of the same name.

The moment of inertia of a molecule is involved in the quantification of its rotational movements. Of the top of my head, the two areas of chemistry where this plays a big role are:

rotational spectroscopy: the rotation of a molecule is quantized, and the value that the angular momentum can take are related to the moment of inertia of the molecule.
the rotational entropy of a molecule is linked through statistical mechanics to its inertia tensor.


Electroplating an alloy from separate metal solutions Similarly to a question asked here, would mixing two solutions containing metallic ions in different proportions allow to electroplate something with the alloy of the metals? For example if I take a $\ce{CuSO4}$ solution and mix it with a $\ce{ZnSO4}$ solution, will I be able to electroplate brass? If I want to make 70/30 brass, would mixing a 0.7 mol/L and 0.3 mol/L solution work?

I know a "brass" solution can be made by reverse electroplating (is that the term?) a brass anode in an acid solution so mixing two solutions should have the same result, copper and zinc ions in a solution.

In short, yes, it is possible to electroplate alloys, but it isn't as simple as just mixing the ions you want in the ratio you want.
In the simplest approximation, the alloy deposition would behave as two simultaneous, separate depositions at the same voltage and the composition could be controlled by changing the voltage, assuming each component has a different current-voltage response.
In the real world, unfortunately, there are numerous interactions between the ions in solution, the metals on the electrode, etc. and the outcome becomes quite complex to predict.
Here is a decent open-access review of the chemistry and applications of this type of process.


Why water vapors exerts pressure rather than diffusing in air before reaching the boiling point of water? I am discussing only two types of vaporization in my question that are, evaporation and boiling. We know that evaporation is the phenomena which describes vaporization of liquid under a certain temperature. And boiling point is the temperature at which vapors of liquid exceeds their pressure by $760\ \mathrm{Torr}$ and starts escaping. The liquid doesn’t absorb more heat and utilize all of the heat to change its state and is said to boil.

There are more details about these two kinds but there is no need to discuss them right now so we should go back to the question that asks that why vapors exerts pressure rather than diffusing in atmosphere.

I be mistaken, but are you referring to vapor pressure?

Vapor pressure  or equilibrium vapour pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The equilibrium vapor pressure is an indication of a liquid's evaporation rate. It relates to the tendency of particles to escape from the liquid (or a solid). A substance with a high vapor pressure at normal temperatures is often referred to as volatile. (http://en.wikipedia.org/wiki/Vapor_pressure)

What is a closed system?

A system in which no mass is transferred.

Thus, vapor pressure would not be exerted if the container is open and the liquid is constantly diffusing. It would only occur in a closed system.

Therefore, liquids DO diffuse (evaporate) in air, rather than exert vapor pressure.

You can leave a bottle of water, one with a lid and one without. The one without the lid will soon be without water, but the one with a lid would remain with water.

*Vapor pressure does relate to evaporation rate, which occurs in the open systems we normally observe in life


Why do not metals gain electrons and non-metals donate electrons? My teacher said that metals lose electrons and non-metals gain electrons to complete the noble gas configuration because of stability as it consumes less energy but why are systems having high energy less stable as compared to systems having less energy.
Why do metals and non-metals lose and gain electrons respectively instead of gaining and losing them respectively?

First, it is a high-school rubbish that was kept only because of lack of desire to revise basic courses.

Can metal gain electrons? Certainly, YES. Sodium has several compounds with negative oxidation state, many transition metals have extensive chemistry where the metal is in formal negative oxidation state and so forth.

Can nonmetal lose electron? Certainly, as in nonmetal-nonmetal compounds it is unavoidable.

So, what is it really about? What is the difference between metals and non-metals?

Before we continue, we have to revisit definition of what metal is.

Metallic solid (sometimes shorthanded as metal) is a solid with metallic conductivity. It arises from having continuous half-filled orbitals allowing electrons to move freely. Metallic solid perfectly can be a compound (say, $\ce{Ag2F}$). Furthermore, many non-metals under heavy pressure undergo transition into metallic phase.

However, when we are talking from chemical PoV, metal is an element. But what kind of element? The problem is, that the definition of metal in chemistry has a history, and the term was introduced long before many of metals known today were discovered. As such, only common metals were considered and no exotic compounds was known, and certainly before extremely high pressures became available.

Consequently, a metal was characterized by its ability to lose electron (but many elements are capable of doing so) and having metallic conductivity when in form of simple compound. The border cases were either discarded or counted as 'metalloids' - a special kind of non-metal.

For example, tin has metallic and non-metallic allotropes at ambient pressure. Oups, is it a metal or non-metal ? Well, legally it is considered as a metal. However, antimony, with pretty much the same case, is typically considered as metalloid.

There is however a strict difference between tin and antimony in than tin is capable of forming normal salts and having a basic oxide, while antimony has only slightly acidic oxides. The reason for discrimination DOES exist. On the other hand, some heavy transition metals, such as rhenium and tungsten do not have true basic oxides.

So, it would be best to define metals by enumerating them. It is, however, easier to enumerate non-metals: commonly recognized non-metals are boron, silicon, arsenic, tellurium, iodine, everything to right and up of that elements and hydrogen. Optionally, germanium and antimony may be included. (Polonium, astatine and some others are questionable cases, since their chemistry is virtually unexplored thanks to their high radioactivity. But for that reason they can be safely ignored)

Said elements have some things in common: they have relatively high electronegativity, form covalently bound or molecular solids, form acidic oxides (if form at all) and do not form simple cations stable in water. All this comes from them having relatively high number of electrons in valence shell and tight binding of said electrons.

By no mean you should assume that sodium, for example, is eager to lose an electron. Nope, this process results in energy consumption. Only subsequent stabilization of by electron affinity of its partners and ionic packing results in net energy gain. On the other hand, addition of electron to an atom often results in slight energy release.

TL; DR. Non-metals typically has compact electron shells tightly tied to their nucleus and as such are uneager to loose them.


Nucleophilic attack on cyclic intermediates in dibromination and oxymercuration reactions of alkenes Why does the nucleophile attack the most substituted carbon after the formation of a cyclic intermediate (three membered ring) in the cases of halogenation and oxymercuration of alkenes, as if there were a real positive charge. Shouldn't the most electrophilic carbon be attacked by the nucleophile?

Just to clear your doubts these are electrophilic addition reactions

Case 1: oxymercuation-demercuration

The reaction is initiated by electrophilic addition of $\ce{Hg^2+}$(mercuric) ion to the alkene to give an intermediate mercurinium ion, whose structure resembles that of a bromonium ion. Nucleophilic addition of water as in halohydrin formation, followed by loss of a proton, then yields a stable organomercury product.
The final step, demercuration of the organomercury compound by reaction with sodium borohydride, is complex and involves radicals.

The regiochemistry of the reaction corresponds to Markovnikov
addition of water; that is, the $\ce{-OH}$ group attaches to the more highly substituted carbon atom, and the  $\ce{-H}$ attaches to the less highly substituted carbon.

  When the π electrons of the alkene attack the mercury cation, a
  three-membered intermediate forms in which both carbons share a single
  bond with the mercury and a partial positive charge. It turns out that
  rearrangements are unlikely to occur from this intermediate (as
  opposed to the discreet carbocation intermediate in the acidic
  hydration reaction).

A water molecule then attacks the more substituted carbon, which bears a greater share of the positive charge. Cleavage of the carbon-Hg bond is accomplished with a reducing agent called sodium borohydride $\ce{NaBH4}$.

See images:

So where is the carbocation present?

The end result of an oxymercuration-reduction reaction is the Markovnikov addition of water to an alkene without rearrangement

The regioselectivity on this reaction is partly based on inductive effects resulting from the shifting of electrons in a s bond in response to the electronegativity of nearby atoms. I believe the other reason is based on hydride shift which converts the carbonation into a tertiary one and this is thermodynamically favourable. (Not entirely sure on second reason though)

Case 2: Dibromination

Halogenation of Alkenes: Addition of $\ce{Br2}$

Follows the reaction sequence (proposed by George Kimball and Irving Roberts)

  that the reaction intermediate is not a carbocation but is instead a
  bromonium ion, $\ce{R2Br+}$, formed by electrophilic addition of $\ce{Br+}$ to the
  alkene.

The bromonium ion is formed in a single step by interaction of the alkene with $\ce{Br2}$ and simultaneous
loss of $\ce{Br-}$.

See image:

I have updated to include an even more detailed mechanism:


Work done by a gas in thermodynamics A given mass of gas expands from state A to state B by three paths 1, 2, 3 as shown in the figure below.

If $W_1,$ $W_2$ and $W_3$ are respectively, be the work done by the gas along three paths, then
(A) $W_1 > W_2 > W_3$
(B) $W_1 < W_2 < W_3$
(C) $W_1 = W_2 = W_3$
(D) $W_1 < W_2 : W_1 < W_3$

The work done is $W = -P \,\mathrm dV$ that is the negative of the area under the graph; since Area 3 > Area 2 > Area 1, thus, $W_1 > W_2 > W_3.$
But the answer given is option (B). Why is this so?

Sign conventions can make 'work' confusing, but the question here asks for the work done by the gas. When a gas expands, it's doing work (it's losing energy), so it's actually $W = + P \, \mathrm dV$, inverting all the inequality signs. So option B) is correct.


Why do we use molality on describing colligative property? I'm satisfied with the explanation about Raoult's law that the chance of vaporizing will depend on the surface exposed to the atmosphere and solute will decrease such chance by taking portion of the surface at rate of $X_{solute}$.
However, incase of elevation and depression of boiling point, I can't find any explanation why it should depends on the molality. There's a molarity, molality, mole fraction, etc.. why molality among them?
I know that since it's microscopic, we can't be sure about anything since we cannot observe. But still I'm looking for some satisfactory explanation.
And below is the example of boiling-point-elevation equation that I made by looking at the graph of tripoint and considering the fact that boiling point elevation occurs due to the vapor pressure change. Why can't we use such method? it seems accurate to me though..
Let $$P(t):T \to P, when$$ $$t_{tri}<t, t \in T (temperature), t_{tri} \in T (temperature), p\in P(pressure)$$,
$$P^{-1}(p): P \to T $$ $$p_{tri}<p, t \in T (temperature), p_{tri} \in P (pressure), p\in P(pressure)$$
$p_{tri}, t_{tri}$ are pressure and temprature at tripoint $P, P^{-1}$ are function for solvent before dissolve.
According to the Raoult's law, we can calculate vapor pressure of solution's solvent $P_{2}(t)$ as such
$$P_{2}(t)=P(t)*X_{solvent}, X_{solvent}=mol_{solvent}/(mol_{solvent}+mol_{solute})$$
Let's say that we want to know the elevation of boiling point at pressure $P(t_0)$ when $t_0$ is former boiling point. According to the Raoult's law stated above, we can say that boiling point elevation $\Delta t=P^{-1}(\frac {P(t_{0})}{X_{solvent}})-t_0$.
(I'd tried to write down the specific process. But I couldn't do it well on words. Still you can see it by looking at the graph of any tri point.)

It is much easier to work with molality than molarity when measuring freezing point depression. Imagine. You weigh $10$ g, or $20$ g ice (or any other amount); you add $1$ g salt, $2$ g salt, or any other amount of any sort of solute. You drop a thermometer in the mixture, and you measure the fusion temperature. It is done quickly, and the scientist does not mind about the molarity of the obtained solution.


Is my analysis of quantum numbers describing the ground state of At correct? Which series of quantum numbers describes the highest (energy) occupied orbital in a ground state of At atom?

a) n = 6, l = 0
b) n = 6, l = 2
c) n = 5, l = 2
d) n = 4, l = 3
e) n = 6, l = 1

According to my teacher, the answer is e.

My analysis:

The electron configuration of At atom:

$$\begin{aligned}1s^22s^22p^63s^23p^6\ce{3d^10}4s^24p^64\ce{d^10}5s^25p^64\ce{f^14}\ce{5d^10}6s^26p^5
\end{aligned}$$

Knowing for levels 0,1,2,3 correspond to s,p,d,f respectively we can say that:

a) $$\begin{aligned}6s^2\end{aligned}$$
b) $$\begin{aligned}Nothing \end{aligned}$$
c) $$\begin{aligned}\ce{5d^10}\end{aligned}$$
d) $$\begin{aligned}\ce{4f^14}\end{aligned}$$
e) $$\begin{aligned}\ce{6p^5}\end{aligned}$$

What I thought here first is that d is the answer since 4f14 has the largest number of electrons, and thus will occupy the highest energy around.

Then I saw on another question that the orbitals are filled so that the ones of lowest energy are filled first.

$$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s ...$$

In this case, e is the answer since 6p5 has the highest energy regardless of the number of electrons it is holding?

Yes , there is another name for it (n+l) rule or Aufbau principle

http://en.wikipedia.org/wiki/Aufbau_principle

Even though it is not correct to check this rule at such higher values of Z where the trend sometimes changes due to very less differnce between higher values of shell and orbitals .But if you are a beginner you should not heed to this statement.


What is the hybridization of diatomic nitrogen, diatomic anything like oyxgen, bromine; hydrogen and chlorine in HCl? I'm guessing sp hybridization for nitrogen in diatomic nitrogen based on the lone pair on nitrogen and the triple bond but there is no mixing of s and p orbitals so I want to say that there is no hybridization. Are there any chemicals without hybridization? What are some examples of chemicals without hybridization

Why are lone pairs counted in hybridization if they do not contribute to the mixing of the orbitals, a.k.a. hybridization?

Hybridisation is something you typically deduce from the geometry of the compound you are observing. It is primarily a mathematical concept.

That said, to form the diatomic molecules you asked about, no hybridisation is needed. In fact, you get much better results from assuming unhybridised atoms and then combining their atomic orbitals to molecular ones. Except for hydrogen (which only has 1σ and 2σ), the molecular orbitals of all these diatomic compounds ($\ce{N2, O2, F2, Cl2, Br2, I2}$) are basically identical save the energy levels of the MOs and their populations. Thus, these would all count as unhybridised compounds.

Also, there is no nead to invoke hybridisation to explain the bonding situation in $\ce{HCl}$ either. Instead agan consider a σ bond formed between hydrogen’s 1s orbital and chlorine’s 3p. All of chlorine’s other orbitals will sit there happily filled and unhybridised.


Can an excessive reactant amount result in a longer reaction time? I am performing an experiment with DTZA and Ferrate. After a point of increasing the DTZA concentration too much the Ferrate takes longer (lower reaction rate) to react with the DTZA. Is this something reasonable or is it due to error?

Let me first say that I don't know this specific reaction, so I am not sure whether my answer holds for it, but let me answer the general question: can a high concentration slow down the reaction rate?

The answer is, yes it can. The mechanism by which this happens is called inhibition and it is often (although not necessarily) connected to reactions on surfaces. I will explain inhibition in terms of a surface reaction here, but keep in mind that it can also happen in solution due to 'crowding', in which there is so much of 1 reactant that it basically 'cannot find' the other reactant.

For surface reactions the story is as follows. You will get a competition between adsorption on the surface and reaction on the surface.

Qualitatively, if you have a large amount of one of the two reactants than many of the surface sites will have this reactant adsorbed to it meaning that some of the reactant on the surface has neighbors that it cannot react with (because it is the same molecule), this can slow down the reaction rate.

Quantitatively, the reaction order will be concentration dependent. A famous example of this is Langmuir-Hinshelwood kinetics
$$r =k C_s^2 \frac{k_1 k_2 C_A C_B}{(1+k_1 C_A+k_2 C_B)^2} $$
where $C_s$ is the concentration of available adsorption sites and $C_A$ and $C_B$ are the concentrations of your reactants.

There are many possible ways in which this reaction rate equation can behave, depending on the values of the rate constants $k$, $k_1$ and $k_2$, but let me single out two cases that will explain a shift from increasing to decreasing reaction rate with increasing concentration:

a very high concentration of $C_A$ will result in $k_1C_A>>>1, k_2C_B$. This allows us to neglect the $1$ and $k_2C_B$ terms in the denominator, turning the rate equation into this:
$$r =k C_s^2 \frac{k_2 C_B}{k_1 C_A} $$
which indeed decreases with increasing concentration of $C_A$.
a very low concentration of $C_A$ will result in $k_1C_A<<<1, k_2C_B$. Then the rate equation becomes:
$$r =k C_s^2 \frac{k_1 k_2 C_A C_B}{(1+k_2 C_B)^2} $$
which has an increasing rate with increasing concentration of $C_A$

So what you see here is that a change of $C_A$ from low to high concentration can change the reaction order in $C_A$ and therefore indeed result in an initial increase of the reaction rate with increasing $C_A$, but later a decrease if you increase it even further. When this crossover occurs depends on the values of the rate constants ($k$, $k_1$ and $k_2$) and the concentration of the other reactant.


Using electrolysis of water as a proton generator I'm working on a project right now, where the challenge is to try to use electrolysis as a way of controlling the $\ce{pH}$ of water.

I've set up an experiment where I have two separate chambers or volumes of water.
Both chambers are connected to the same electrolytic unit. The flow of water is from a respective chamber, into the unit and then back into the same chamber again, in an attempt to separate the two half reactions. For this reason I call the chamber that flows over the anode for the anodic chamber, and the chamber that flows over the cathode for the cathodic chamber. The bodies of water are separated in the electrolysis cell by a proton-exchange-membrane (PEM) that should, in theory, only allow positively charged ions to cross.

Apologies for the Danish text on my setup, but the figures should help you understand what I'm trying to do.

The anodic chamber (the water that goes to the anode half-reaction of the electrolysis cell) contains tap water while the cathode chamber contains Milli-Q water with added $\ce{NaCl}$ for conductivity purposes.

The $\ce{pH}$ of the water is alkaline in both chambers when the experiment starts, but I've also tried to run the experiment where the anodic chamber is acidic.

Anyway, here is what I hoped would happen in alkaline water:

Anode (oxidation):

$$\ce{4OH- (aq) -> 4e- + O2(g) + 2 H2O(l)}$$

Cathode (reduction):

$$\ce{2H2O (l) +2e- -> H2(g) + 2OH-(aq)}$$

In theory, I was expecting the $\ce{pH}$ to drop in the anodic chamber, since removing $\ce{OH-}$ would necessitate a flow of positively charged ions across the PEM membrane, instead of the $\ce{OH-}$ produced in the cathodic chamber in order to satisfy electronegativity. This would also mean that the $\ce{pH}$ should rise in the cathodic chamber.

However, what I see is invariably that the $\ce{pH}$ rises very rapidly, and continuously, in the cathodic chamber, whereas the $\ce{pH}$ in the anodic chamber remains either stagnant or rises very slowly.

Now, part of the reason for the rapid rise in the cathodic chamber is the complete lack of alkalinity, that much is clear to me. However, I don't understand why only the reduction half cell reaction seems to occur.

The only explanation I have been able to come up with is that something else on the anode half-reaction is donating electrons to the reaction. I tried running the experiment at different voltages in the hopes of finding a "sweet spot" that would prevent this "something" from reacting but as of now I haven't been able to do that. I'm worried it may be that my anodic electrode is donating the electrons.
My electrodes are nickel based.

A cursory glance at Wikipedia shows the standard electrode potential of nickel as

$$\ce{Ni^2+ + 2 e− <=> Ni(s)} \;\;\; \pu{E^{\circ} = −0.25 V}$$

And for my hydroxides

$$\ce{O2(g) + 2H2O + 4 e− <=> 4 OH−(aq)} \;\;\; \pu{E^{\circ} = +0.401 V}$$

When I run the reaction, I also see a yellowish/greenish precipitate forming in the anodic chamber, which according to some lab technicians at my location may be the nickel reacting.

Anyway, the question(s) are then - what do you think is happening  in my system, given this information?

Am I on the right track? should I try to change my electrode?

Have I made some obvious, basic mistakes?

What suggestions do you have, so that I can get this to work?

I hope that you have the time to advice me on this, and I wish you all a pleasant day.

In a chemical sense, "anode" of an electrolytic cell is the most powerful oxidizing reagent known, so much so that it oxidize F(-) to elemental fluorine. Don't forget that the electron balance in an electrolytic cell is always maintained so the gram-equivalents of x reduced at the cathode must equal gram equivalents of y oxidized at the anode. As a result, half cell reactions cannot occur alone. So something is being oxidized at the anode, this is your nickel electrode. Platinum is the best one but it is too expensive. Graphite electrode might be an economical substitute. Secondly, you have to add an electrolyte to the anode chamber as well because tap water resistance is pretty high. You can add a non-corrosive electrolyte such as sodium sulfate. With these changes, hopefully you will see the desired trend.


What is the analogue of "molecule" for ionic compounds? In a system, if we have $\pu{18 g}$ of $\ce{H2O}$ ($M_\mathrm r = 18$), we can say we have a mole of water molecules or $6 \times 10^{23}$ molecules. But in another system, if we have $\pu{342 g}$ of $\ce{Al2(SO4)3}$ ($M_\mathrm r = 342$), what is the best way to say that there is one mole of "aluminum sulfates" or $6 \times 10^{23}$ "aluminum sulfates"?

Since it is a ionic compound, we can't say molecules, but I can't find an analogous form for ionic compounds. I thought about saying "a mole of aluminum sulfate ionic aggregates", but it still doesn't too good, since I've never seen a definition of ionic aggregate as a unit of the ionic compound formula. It only tells the way the ions organize themselves: as an aggregate.

I know it is a silly question, but in terms of language and symbolism, it is important to me as a chemistry teacher.

The term you are looking for is formula unit, I think. Wikipedia doesn't really describe it super well, but just to give an example, you could write the sentence

  $\pu{58.44 g}$ of $\ce{NaCl}$ contains $6.022 \times 10^{23}$ formula units of sodium chloride

and it would be pretty well understood.

See also: What is the difference between a Chemical Formula and Formula Unit?


Atomic coordinates for DNA bases as they appear in a B-form or A-form duplex Where can I find atomic coordinates for guanosine / cytosine / thymine / adenine (either the ribonucleotide or deoxyribonucleotide bases) where the atoms are configured in the manner we would expect in the context of an idealized B-form or A-form nucleic acid duplex?  I'd like to calculate the angle between various bonds in these bases, and to play around with a few other things, but I'm having trouble finding proper atomic coordinate sets.

I played around with a program called GROMACS about 4 years ago and don't remember much, but I do remember doing a tutorial simulating a DNA double helix and part of it included making the DNA by giving it a sequence and running a script. It can probably make A or B form.

EDIT: This was a long time ago and I forgot. It wasn't GROMACS, it was Amber. Look here http://ambermd.org/tutorials/ Tutorial B1 looks like the same tutorial I did with the DNA.


How does formylation of a nitrogen change the hydrophobicity of a target protein? How does formylation, of the sort shown here:

(from this Wikipedia page) affect the hydrophobicity of the target protein?  Can we say anything general about adding a formyl group to a nitrogen?

Formylmethionine is used as the start amino acid for a protein chain in certain organisms (the prokaryotes.)

As the alpha nitrogen is formylated, it cannot be in the middle of the chain. This is never seen in real proteins, partly because there would be two carbonyl groups attached to the nitrogen, which would have a very strong electron withdrawing effect and be quite unstable.

http://en.wikipedia.org/wiki/N-Formylmethionine

Other organisms (the eukaryotes, including the animals and humans) use methionine as the start amino acid of the chain.

http://en.wikipedia.org/wiki/Genetic_code

As there can only ever be one N-formylated methionine in a real protein (and it is often removed after protein synthesis in a step called post-translational modification) the effect on hydrophobicity will not be very much.

If on the other hand, you were to take an existing protein and formylate the -(CH2)4-NH2 side chains on all the lysine amino acid residues, I would expect a slight increase in hydrophobicity in neutral solution. In acid solution I would expect the difference to be more marked, as the formylated NH2 groups would lose their ability to become ionised by accepting a proton from the acid.


Can nitrogens in urea be protonated? Is it possible in very acidic solutions for the nitrogens ($\ce{N}$) to get another $\ce{H+}$, making it $\ce{R-NH2+-R}$? Since it has a free electron pair, it might be possible right? If so, at which pH?

Reagents/solutions that are acidic enough will protonate the urea group, just not where you think. The most basic site is the oxygen. Perhaps this is a surprise.

The lone pairs on the nitrogen atoms are delocalized (see resonance depiction below).

The oxygen atom has the most electron density. Below is an electron density map, where ${\mathrm{\color{red}{red}}}$ represents areas of high electron density and $\mathrm{\color{blue}{blue}}$ represents areas of low electron density. You can see the area around the oxygen atom is red and yellow, while the areas around the nitrogen atoms are green.

This image is from "The effect of intermolecular hydrogen bonding on the planarity of amides" by James A. Platts, et al. published in Phys. Chem. Chem. Phys., volume 2012, issue 14, page number 11944-11952, DOI:10.1039/C2CP41716B. The image is copyright the Royal Society of Chemistry and released under a Creative Commons Attribution 3.0 Unported license.

Protonating the oxygen atom leads to a cation where the positive charge is delocalized (better, more stable).

Protonating one of the nitrogen atoms leads to a cation where the positive charge is localized (worse, less stable).


is Li ion less stable than Li atom? so our teacher introduced the concept of ionization energy... heres my doubt :  if we remove an electron from , say, Lithium , we need to supply energy first and that will be very less compared to the energy we'll have to supply to , say, oxygen. now the Li ion will have more energy than the lithium atom. but doesnt more enegry = less stability? does that mean Li ion is less stable?

what i really want to ask is,  when we remove an electron from an atom, does the atom ' gain' energy or is the gain actually the energy supplied to the electron?

What do you mean by "stable"? In a vacuum, $\ce{Li+}$ is quite stable. In water, Li ionizes readily to $\ce{Li+}$, with $\ce{OH-}$ accepting that electron. Of course, in vacuo, if an $\ce{Li+}$ ion were to be placed near an $\ce{Li-}$ ion, they'd quickly redistribute the charge.

There is also a difference between chemical (thermodynamic) stability and kinetic stability. Lifting a ball up a hill, or lifting an electron to a higher level, gives it potential energy, but if the ball rests in a cup on the hill, or the electron is in a metastable state, it will not go back on its own without a push.

Thermoluminescence is a case in point. Over time, some materials such as ceramics absorb radiation from cosmic sources and nearby radioactive minerals, putting atoms into excited states. Upon reheating, electrons hop back to lower states, emitting light. This is used to date ancient pottery, since the metastable states are preserved for millenia.

Another example of metastability is diamond. It takes considerable pressure and a specific temperature range to convert carbon, e.g. graphite, to diamond -- but once that pressure is removed, graphite is the preferred state. However, once formed, diamond is stable at room temperature for eons.

See Difference between thermodynamic and kinetic stability, answered by thomij, for a much more complete answer.


What causes Starch's lump creation when solvent (butter/water/oil) is added and how to prevent them When liquid(water/butter) is added to starch lumps of starch are formed. Which are completely dry from inside. Adding more solvent doesn't seem to help therefore it is not a case of saturated solution. So what does cause the lump formation ? and how can it be prevented ?

The trick here is that you're not actually dissolving the starch. Starch is made up of small semi-crystalline granules of amylose and amylopectin which are actually pretty much insoluble at room temperature. Some water can get in as amylose breaks up the crystallinity, but it doesn't really dissolve. When you drop it into water, it hydrates a bit on the outside, holding it all together, but water doesn't make it into the inside. If you stir or shake it, you can break the clumps up into smaller pieces, eventually getting at most individual hydrated granules.

It's not until you heat the mixture that the granules really break apart and allow water in, in a process called gelatinization. This is what makes starch useful in cooking as the starch can form a more diffuse network, thickening the liquid.


Scientific Reason for Salt Solution Gaining Volume A little background to this question: my mom placed a glass with salt and enough water to just cover the salt in a room to "absorb negative energy" out of the room at her office. It obviously "worked". So when I arrive home from university she asked me if salt water draws moisture out of the air — and explains to me what happened.

The scientist part of me is curious, because you cannot "absorb negative energy" using salt water. So I placed salt in a glass, covered it with enough water and left it overnight. To my surprise, the volume of water increased by around 20-25%, the glass is significantly more fill.

My question is this: why?

I have 2 thoughts:

Somehow the ions in the solution as the salt dissolved have charges that repel other molecules making more space in between molecules and therefore increasing the apparent volume.
Because it is a salt solution some form of osmosis occurs to actually absorb moisture out of the air.

My next idea is to redo this later today but weigh the glass and solution first to see if it gains mass. That would support 2. If no mass gain, then I assume something similar to 1 is happening.

Thoughts?

P. S. I am an applied mathematician so I have no background in the specifics of chemistry and how molecules and ions etc. etc. interact.

P. S. S.: I also couldn't find an answer to this on the internet. IN FACT salt added to water appears to reduce the volume in a lot of information I read.

The vapour tension over the saturated table salt solution { the best  with the solid salt excess) is equivalent to the relative humidity about 75 %.

If the rel. humidity is lower than 75%, the solution evaporates and the salt crystallizes.

If the rel. humidity is higher than 75%, the solution absorbs the air moisture and the salt is dissolving.

Such systems may be informally called "relative humidity buffers". They are useful for hygrometer calibration or for making a constant humidity aerial baths for long term weathering of devices or materials.

See the articles with tables of relative humidity above saturated solutions of various salts:

Researchgate.net
PDF on NIST
PDF on Omega.com


What does the molecular orbital scheme of beryllium chloride and hydride look like? Beryllium is a somewhat fascinating element since it is the only member of the second group that behaves somewhat non-metalish and, e.g. forms a somewhat covalent chloride and a covalent hydride.

In an answer to a different and unrelated question, I tried guessing what the bonding picture of $\ce{BeCl2}$ would be. Given the monomer’s linear shape, I was inclined to assume something that introduces a four-electron-three-centre bond, which has the nice side-advantage of allowing beryllium to contribute to bonding orbitals using its s orbital only. However, I would guess that the more traditional picture would include two $\mathrm{sp^2}$ type bonds to the neighbouring chlorines and then a nonzero amount of π backbonding making the MO similar to that of $\ce{CO2}$.

What does the actual, calculated MO scheme of $\ce{BeCl2}$ look like; what does $\ce{BeH2}$’s look like, how do they compare and what can we learn from them concerning the bonding properties of beryllium?

I calculated molecular orbitals for $\ce{BeCl2}$ at MP2/jun-cc-pVDZ level of theory; given below are the doubly occupied MOs at this level of theory along with their symmetry designations and energy in a.u. For some reason my Avogadro kept crashing when I tried to select an iso value I thought appropriate for the last two MOs; anyway these are completely localised $\ce{Be}$ atoms. If we look at the 8 valence orbitals, there is indeed delocalised $\pi$ bonding as you suggest, and do bear a striking resemblance to MOs for $\ce{CO2}$.

For $\ce{BeH2}$ I again ran into trouble generating nice visuals using the cube file in Avogadro. I will update this post with visuals as soon as possible (It would be great if someone can help out).

I got the following doubly occupied MOs at MP2/jun-cc-pVDZ level of theory:      -4.644660 ($\ce{a_g}$) ;     -0.504366  ($\ce{a_g}$) ;  -0.469857  ($\ce{b_{3u}}$). However, it is evident from the symmetry designations that $\pi$-type orbitals aren't involved in bonding.

P.S: If you are interested, I can provide the input, output, and cube files on request.