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Balancing the oxidative decomposition of FeS2
Doing some practice for the GRE next year.
Please could you look at my working for this and tell me why I'm wrong?
$$\ce{? FeS2 + ? O2 + ? H2O = ? Fe(OH)3 + ? H2SO4}$$
My working so far is:

There must be the same number of of $\ce{FeS2}$ as $\ce{Fe(OH)3}$
There must be twice as many $\ce{H2SO4}$ as $\ce{FeS2}$
If there is one mole of $\ce{FeS2}$ for example, there must be 1 mole of $\ce{Fe(OH)3}$ and two moles of $\ce{H2SO4}$

Tot up the $\ce{H}$ on the right: $1\times 3 + 2\times 2 = 7$. $7$ on the right so $\ce{3.5 H2O}$ on the left. ($7/2 = 3.5$)
$\ce{O}$ on the right = $1\times 3 + 2\times 4 = 11$.
This mean $11$ on the left, too. I already have $\ce{3.5 H2O}$, so we must have $11-3.5 = 7.5$ $\ce{O}$'s from the $\ce{O2}$. So $\ce{3.75 O2}$ because $7.5/2 = 3.75$.

Following the rules I've written down I arrive at:
$$\ce{1.75 FeS2 + 3.75 O2 + 3.5 H2O -> 1.75 Fe(OH)3 + 3.5 H2SO4}$$
I've multiplied these by $4$ to give integers, because that's what the question asks of me, so $7, 15, 14, 7$ and $14$.
Why am I wrong?

Your equation (if this is it) is not balanced in H and O. Besides, having an acid and a base at the same time among the products does not feel right at all.
$$\ce{7 FeS2 + 15 O2 + 14 H2O = 7 Fe(OH)3 + 14 H2SO4}$$
Anyway, welcome to Chem.SE, and pay attention to the formatting.

I've now balanced this equation and the answers are 4, 15, 14 = 4, 8 For anyone who would like to know!

I found a solution which works,
$$\ce{4FeS_2 + 15O_2 +14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$

This is how I found it. Do you know linear algebra ? Because that is very useful. Anyway without to know this is linear algebra you'll be able to understand.
So we are looking about the stoichiometric coefficients $\{\ce{a, b, c, d, e}\}$ of this reaction :
$$\ce{aFeS_2 + bO_2 + cH_2O + d Fe(OH)_3 + eH_2SO_4 \rightarrow 0}$$
Now what you need to do is first to find the number of different elements which are involved in your reaction, here they're four : $\{\ce{\color{\green}{H}, \color{\red}{O}, \color{\purple}{S}, \color{\orange}{Fe}}\}$. I'm not that good in the use of $\LaTeX$ but now the idea is to make column vectors under each of the reagents of your reactions and in these vectors you put in the same order for each the number of each elements there are in and if there is not just put a zero. It will gives you this :
$$a\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{0} \\ \color{\purple}2 \\ \color{\orange}{1} \end{pmatrix}+b\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{2} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+c\cdot\begin{pmatrix} \color{\green}2 \\ \color{\red}{1} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+d\cdot\begin{pmatrix} \color{\green}3 \\ \color{\red}{3} \\ \color{\purple}{0} \\ \color{\orange}1 \end{pmatrix}+e\cdot\begin{pmatrix} \color{\green}2 \\  \color{\red}4 \\ \color{\purple}1 \\ \color{\orange}0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$
It can be easily transform into a system like this :
$$\begin{cases}
2c+3d+2e=0 \\
2b+c+3d+4e=0 \\
2a+e=0 \\
a+d=0
\end{cases}$$
So here we are in a good way because we have more unknowns than equation so if the system converge it has an infinity of solutions. And we can multiply by the number we want each stoichiometric coefficients and still have the same reaction.
So here we will determine four unknowns in terms of an other, I chose $e$. And I got :
$$$$\begin{cases}
a=-e/2 \\
b=-15e/8 \\
c=-7e/4 \\
d=e/2 \\
e=e
\end{cases}$$$$
I multiply then by $8$ to have only integers which give me with $e=-1$ :
$$\ce{4FeS_2 + 15O_2 + 14H_2O - 4Fe(OH)_3 - 8H_2SO_4 \rightarrow 0}$$
Then $$\ce{4FeS_2 + 15O_2 + 14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$

I will take pretty much the same approach that 9-BBN does except I will do it a little bit differently. I have never heard of setting the reactants to zero and I'm not very good at making matrices, so I'll just jump right into the systems of equations.
First of all, define your coefficients $a,b,c,d,e$ for each of the chemicals involved, being the reactants that you start with and the products that you get from the reaction.
$$\ce{a FeS2 + b O2 + c H2O ->  d Fe(OH)3 + e H2SO4}$$
Now we keep a tally of the number of each atom in molecule. If I put a zero, it means that there are none in that compound and this is for clarity; an equality sign replaces the reaction arrow separating products and reactants.

Iron (Fe): $\displaystyle 1a + 0b + 0c = 1d + 0e \Longrightarrow a = d$
Sulfur (S): $\displaystyle 2a + 0b + 0c = 0d + 1e \Longrightarrow 2a = e$
Oxygen (O): $\displaystyle 0a + 2b + 1c = 3d + 4e \Longrightarrow 2b + c = 3d + 4e$
Hydrogen (H): $\displaystyle 0a + 0b + 2c = 3d + 2e \Longrightarrow 2c = 3d + 2e$

Now then, we have four equations and 5 variables $a,b,c,d,e$ … so it shouldn't be solvable, except that we only want one solution which is the lowest integer coefficients for each molecule. So we have to make one initial guess for any of the variables. We choose the one that is the simplest and that will be $a$. We set $a = 1$ for initial guess. Note that $a < 0$ & $a = 0$ are forbidden.
In iron (Fe), $a = d$, as $a = 1$, then $1 = d$
In sulfur (S), $2a = e$, as $a = 1$, then $2 = e$
What we know: $a = 1, b =\ ?, c =\ ?, d = 1, e = 2$
With what we know, we can only solve for $c$ in the hydrogen equation next.
$c = [(3d + 2e) / 2]$, as $d = 1, e = 2$, then: $c = 7 / 2$
Solving the last equation for b:
$b = [3d + 4e -c] / 2$, given known variables, then: $b = 15 / 4$
Solutions: $a = 1, b = (15 / 4), c = (7 / 2), d = 1, e = 2$
This will balance the equation according to conservation of mass. However, it makes more sense if you multiply $a$ through $e$ by $4$ so as to get the lowest whole number solutions. This is like scaling a recipe. We are synthesizing the same thing, though.
Therefore, $a = 4, b = 15, c = 14, d = 4, e = 8$
Balanced Chemical Equation:
$$\ce{4 FeS2 + 15 O2 + 14 H2O ->  4 Fe(OH)3 + 8 H2SO4}$$

Now that your equation is balanced I will show you a little something that I've derived. So the formula for the commplete combustion of every hydrocarbon alkane ($\ce{C_nH_{2n+2}}$) such as methane, ethane, propane, butane, pentane, etc … is this:
$$\ce{C_nH_{2n+2} + $(3n + 1) / 2$ O2 -> n CO2 + (n + 1) H2O}$$
So if we are completely combusting (full airflow of oxygen, no major flickering of the flame producing a mixture of $\ce{CO}$ and $\ce{CO2}$) propane which has the molecular formula $ce{C3H8}$, where $n = 3$ then it's combustion is the following:
$$\ce{C3H8 (g) + 5 O2 (g) -> 3 CO2 (g) + 4 H2O (g)}$$
I know the general complete combustion equation is true for every alkane because all alkanes have the formula $\ce{C_nH_{2n+2}}$, $\ce{O2}$ is always involved in combustion, and $\ce{CO2}$ & $\ce{H2O}$ are always the products of the complete combustion of alkanes. I think this is really cool given that most popular general chemistry equations are combustion equations!
Additional information: I have a lot of ideas running through my head now so once you've determined the balanced chemical equation, you may be able to calculate the spontaneity of the reaction from the change in Gibbs Free Energy of reaction which is useful for knowing if the reaction is a waste of energy, that is that it consumes more energy than produces or not!
You can determine the percent yield after experimentation, how useful your reaction is from the stoichiometric amount of expected product produced. But even with ideally $100~\%$ yield, are all the elements that you find in your desired product going on to form your product or are they being wasted producing something else?
The 2nd Principle of Green Chemistry: Atom Economy, states that

Synthetic methods should be designed to maximize incorporation of all materials used in the process into the final product. (The American Chemical Society)

So what's the point of your reaction? To make $\ce{Fe(OH)3}$? What percent of the atoms $\ce{Fe, O}$, and $\ce{H}$ are being incorporated as $\ce{Fe(OH)3}$ and how much is going off to form sulfuric acid instead? We can calculate that as the percent atom economy.
$$\%\ \text{Atom Economy} = \frac{\text{mass of atoms in desired product}}{\text{mass of atoms in all reactants}} \times 100$$
where units are in grams per mole, but they are eliminated in the ratio.
$$\begin{align}\%\ \text{Atom Economy} &= \frac{4 \times 106.866}{4 \times 119.965 + 15 \times 31.998 + 14 \times 18.015} \times 100\\
&= \frac{427.464}{1212.04} \times 100\\
&= 35.27~\%\end{align}$$
So if there were multiple ways to make $\ce{Fe(OH)3}$ $\%\ \text{Atom economy}$ may be a useful factor to consider when conducting a synthesis. I think this is more important, though, in choosing a synthetic reaction where toxic byproducts are involved and you want to minimize the toxic byproducts produced for the health of the earth and of the customer or patient and also to reduce on having to spend money to deactivate and/or filter out toxic byproducts in an industrial synthesis. Therefore, you want the synthesis with the highest $\%\ \text{Atom Economy}$.
A link on Atom Economy: https://www.acs.org/content/acs/en/greenchemistry/what-is-green-chemistry/principles/gc-principle-of-the-month-2.html

I think that this chemical equation solver can solve your difficult equation in flash. It probably uses either matrices or systems of equations. You can use this if you like: http://www.webqc.org/balance.php ... anyway, it's for speed of calculation. It doesn't not explain the methodology like I did, though.

I don't believe that I said that your mathematics is unnecessary, just that I would do it a little bit differently.

You cannot comment everywhere because that is a priviledge associated with 50 reputation. Rather, since this was a simple mathematical error, you could have proposed an edit instead. I will now read on to review the remainder of your post to determine whether it answers the question or not (only up to the third paragraph).

Okay, more than $50~%$ of your post ramble on about things that are not relevant to the question. Also, I learnt a different definition of atom economy, but that is not important here.

What is the definition of atom economy that you learned?

One based on number, not on mass. (As a side note: if you include @ followed by a username e.g. Jan in your comment, I will be pinged. If more than one person answers in a thread, automatic pinging will not work. Post owners are always pinged.)

I still don't get why I got a dislike. I know that you think that some of the stuff is unrelated. But another answer got flagged and remove for providing the answer, but not a detailed long answer which is what the moderators wanted. I did the opposite and my two likes turned into one.

It is not a ‘dislike’; likewise upvotes are not ‘likes’; using those terms would overload the voting concept of Stack Exchange in a non-intended way. Upvotes are intended to mean ‘this answer is good, factually correct, helpful.’ Downvotes are intended to mean ‘This answer is factually wrong, badly researched or otherwise not helpful.’ I feel that your answer is not as helpful because of the long paragraphs on alkane combustion and atom economy which are not part of the question. Finally, brevity is not bad in itself; see this answer of mine.

Okay, well, no worries. I can learn to accept criticism, I suppose.

This is a redox equation. You can approach it in a standard redox manner, however, you will need to use a small, helpful trick.
If we set up redox pairs, we will notice that there are two compounds that get oxidised but only one that gets reduced:
$$\begin{align}\ce{Fe^2+} &/ \ce{Fe^3+}\tag{Ox1}\\
\ce{(S^{-I})2^2-} &/ \ce{(S^{+VI})O4^2-}\tag{Ox2}\\
\ce{(O^{\pm 0})2} &/ \ce{(O^{-II})H-}\tag{Red}\end{align}$$
However, we can alleviate this problem by realising that $\ce{FeS2}$ is likely a mineral and thus we need to oxidise one iron per $\ce{S2^2-}$ unit. Also, rather than considering water it seems the question wants you to generate hydroxide, hence why $\ce{OH-}$ is my reduced partner of elemental oxygen. That lets us set up the half equations. The reduction is first because it is a lot simpler.
$$\begin{align}\ce{O2 + 4e- + 2 H+ &-> 2 OH-}\tag{Red}\\
\strut\\
\ce{FeS2 \phantom{\ce{+ 8 H2O}} &-> Fe^3+ + 2 SO4^2- + 15 e-}\tag{Ox1}\\
\ce{FeS2 \phantom{\ce{+ 8 H2O}} &-> Fe^3+ + 2 SO4^2- + 15 e- + 16 H+}\tag{Ox2}\\
\ce{FeS2 + 8 H2O &-> Fe^3+ + 2 SO4^2- + 15 e- + 16 H+}\tag{Ox3}\end{align}$$
The smallest common multiple of $4$ and $15$ happens to be $60$, thus $\text{(Ox3)}$ must be multiplied with $4$ and $\text{(Red)}$ must be multiplied with $15$ leading us to $\text{(RO1)}$ and the following:
$$\begin{align}\ce{4 FeS2 + 32 H2O + 15 O2 + 30 H+ &->\\ 4 Fe^3+ &+ 8 SO4^2- + 64 H+ + 30 OH-}\tag{RO1}\\
\ce{4 FeS2 + 32 H2O + 15 O2 &->\\ 4 Fe^3+ &+ 8 SO4^2- + 34 H+ + 30 OH-}\tag{RO2}\\
\ce{4 FeS2 + 32 H2O + 15 O2 &->\\ 4 Fe^3+ &+ 8 SO4^2- + 30 H2O + 4 H+}\tag{RO3}\\
\ce{4 FeS2 + 2 H2O + 15 O2 &->\\ &4 Fe^3+ + 8 SO4^2- + 4 H+}\tag{RO4}\end{align}$$
Everything henceforth is just adding further water molecules or regrouping at your desire. For example, to me, the equation suggests combining the free protons with sulfate ions and even the iron cations in the following way $\text{(RO5)}$:
$$\ce{4 FeS2 + 2 H2O + 15O2 -> 4 Fe(SO4)(HSO4)}\tag{RO5}$$
On the other hand, if you really had to include the generation of sulfuric acid and iron hydroxyide, note that you are lacking twelve hydroxide ions ($4 \times 3$) and 12 protons ($8 \times 2 - 4$) — meaning that twelve water molecules should be added to the left-hand side.
$$\ce{4 FeS2 + 14 H2O + 15 O2 -> 4 Fe(OH)3 + 8 H2SO4}\tag{RO6}$$
Although I must agree with Ivan’s comment: the co-presence of $\ce{H2SO4}$ and $\ce{Fe(OH)3}$ is extremely unlikely; the following will happen:
$$\ce{2Fe(OH)3 (s) + 3H2SO4 (aq) -> 2 Fe^3+ (aq) + 6 H2O (l) + 3 SO4^2- (aq)}$$

I used to be pretty good at balancing chemical equations, but now days I'm pretty rusty. Never-the-less, as far as I can see the equation you've been given must be wrong (or you mis-wrote it above). FeS2 would be assumed by a beginning student to have a ferric (+IV) ion (formally, at least) combined with two sulfide ions S=. So your reaction reduces the Iron (to Fe(+III)) and oxidizes the sulfurs (to S(+VI)). The oxidation of the sulfurs in this scenario gives you all the electrons you need to reduce the oxygen to the oxide (O=) for the formation of the sulfate anion. In other words you'd be left with an "extra" electron (from the iron) - which as long as this isn't a half reaction isn't going to happen (at least as far as elementary chemical homework is concerned). A more sophisticated student might learn that Sulfur commonly forms more exotic ions, such as the disulfide anion S-S= and that the two common forms of FeS2 both have Iron in the +2 not the +4 valence state. This means iron goes from +2 to +3 in your given reaction. Which is pretty much the same problem (in reverse, actually): the sulfur oxidation gives you 14 electrons,  the Fe oxidation 1 more and you need 16 to reduce the oxygen a discrepancy of one electron. There's no place for that electron to go! You're stuck, again.


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Distance between two atoms that lie along a body diagonal
It is given that two atoms lie along the body diagonal of a unit cell (crystalline solid). The length of such cubic cell is $a=\mathrm{0.336\,nm}$. My book says that we should take the coordinates of the two atoms to be $(0, 0, 0)$ and $(1, 1, 1)$. By my intuition is that if we place the cube at the origin then one atom should be at $(0, 0, 0)$ but the other atom should be at $(0.336, 0.336, 0.336)$. After this we can simply find out the required distance using $$l = a[(x_2-x_1)^2 + (y_2-y_1)^2 +(z_2-z_1)^2]$$
My doubt is that what should be the second coordinate? Should it be $(0.336, 0.336, 0.336)$ or $(1, 1, 1)$? If $(1, 1, 1)$ then why?

Both answers are right. It's just that they use different units.

@IvanNeretin Can you please elaborate? I mean how both of them are correct? $(0.336, 0.336, 0.336)$ is using nanometre. What unit is $(1, 1, 1)$ using?

Side of the unit cell, of course. It is pretty natural when working with crystals.

It might be worth noting for the future  that in may texts and papers on x-ray crystallography  the coordinates are often given as fractions of unit cell length i.e.  $ x/b, y/b, z/c$   and not in regular cartesian coordinates because a unit cell need not be cubic or rectangular. The values of $a, b, c $ are then given separately  together with the angles between them.

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sort: votes


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Judging whether a compound is planar or not
How can one determine whether an organic compound is planar or not?
My idea was that for a compound to be planar the hybridization of all the atoms must be the same. But I came across the compound cyclooctatetraene, which has the same hybridization of all its carbon atoms but is not planar. How do I explain above exception?

Conjugated aromatic systems and certain small molecules are planar, the rest are not.

Frankly your notion is just wrong. Pyridine for example is planar.

This is rather a broad question and there are not may ways to approach this a priori. However, there are a few guidelines that can help you make a call:

π bonds must include up to six atoms in one plane for the π overlap to be maximum. Thus, you can conclude that $\ce{C2H4}$ must be planar.
Three atoms denote a plane and thus must always be planar. From this and the previous bullet point, you can conclude that $\ce{C2H2O}$ (oxirine) must be planar.
If you have a non-cyclic contiguous π system — that usually means alternating single and double bonds but can include an atom carrying a lone pair in-between — the system has a desire to be planar to create a better overlap and thus reduction in energy. This refers to all systems which do not form a contiguous π circle, whether the circular part is on its own or not. Thus, conclude that acrolein $\ce{C3H4O}$ is planar.
If you have a cyclic π system that consists of a single circular arrangement, refer to the Hückel rules: cyclic systems with $4n + 2$ π electrons are aromatic and thus often planar. Cyclic systems with $4n$ π electrons would be antiaromatic. With the exception of butadiene (see the first and second bullet points) these distort to a non-planar state so that the double bonds may be regarded as isolated rather than conjugated. Of course in larger systems ($\ce{C12H12}$ and above), it would be entirely conceivable for parts of the π system to remain planar but there would always be at least two breaking points.
If you have a polycyclic π system or cyclic with extensions, these simple rules don’t work anymore. In most cases, they will be planar and display some kind of semi-charge separation to keep with Hückel’s rules.
Finally, there are cases which stericly just cannot be planar. The helicenes jump to mind: If they were planar, the hydrogens on one end of the helix would be inside the other end. Thus, these systems must distort to exist. Similar points can be made for 2,6-disubstituted styrenes, bisphenyl and its derivatives and more.
Honorary mention: Any molecule that contains a tetrasubstituted atom or a trisubstituted atom that is not in the thirteenth or fourteenth group or nitrogen cannot be entirely planar. Thus, dimethyl sulfoxide is not planar, nor is $\ce{PH3}$ or methane. Nitrogen compounds such as ammonia are a difficult call: typically, they would also be trigonal-pyramidal but they can undergo nitrogen inversion which comes with a planar transition.

Your last point is slightly incorrect. One counterexample is SO3. It is trisubstituted and contains a central atom not from Group 13 or 14 yet it is planar.

@TanYongBoon Argh, forgot about high oxidation states. Well, it’s two years old, I’ll leave it here as is …

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Mechanism of hydroboration oxidation reaction
I know that the hydroboration reaction is used to add $\ce{H2O}$ to the double bond in an anti-Markovnikov fashion.
I went to the Wikipedia page and found the following mechanism:

But I could not understand as to why the addition is syn.

In its current form, 's not clear to me why this got downvoted.

The addition is concerted, i.e., everything happens at the same time; there aren't multiple steps to the addition. You can expect the boron to go on one face of the double bond while the hydrogen goes on the other face. There's just no way for the hydrogen to reach over to the other side for the anti-attack. It has to be syn.

It develops a cyclic transition state between C double bond:
  CH3     H
    \    /
     C--C
 H/       \H
  |         |
  |         |
  H---  ---BH2

Then replaces BH2 by OH

Please draw an image and upload that; it is impossible to understand what you wrote down there.

I did the best I could with it, but I would heed @Jan's sage advice here.


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Copper layer doesn't stick after Copper electroplating
I am playing around with copper plating using Copper Sulfate solution with electricity.
I am using 2-2.5 volts as power source and around 100-500 mA.
It is actually a very small container around 100 ml of the solution.
I have used copper sulfate anhydrous to make the solution and I have mixed around 20 mg with around 200ml of tap water in normal room temp.
I have tried several materials (iron, copper , steel, graphite) but I end up with a layer of copper that can be easily removed, even I leave it for fair amount of time(up to 15 min), the result layer is actually is not thin at all but still can be easily removed!!
What is the thing that I am missing here, I have seen people doing this on the internet for just like 5 min. and the layer that they create seems very strong!!

Roughen the deposition surface, perhaps by buffing with very coarse sandpaper or an abrasive scrub pad or similar.
If the surface isn't rough enough, the deposited coating has nothing to mechanically "grab onto" and it can be peeled off just as you've observed.  With a sufficiently rough surface, the deposition penetrates into the nooks and crannies of the surface texture and adheres very strongly.

I was doing exactly the opposite of what you are  saying,I was smoothing the surface as possible by very fine sand paper, I will try it and see the results!

@MuhammadNour Oh, yeah, no wonder it's not sticking. Plating on shiny, smooth surfaces is a great way to make copper foil, just as you've observed. :-)

Will the concentration of the copper sulfate in the solution  make a difference ?

Not a huge difference. If it's too low, you'll know because the deposit will look rough and darker in color. It should still deposit appreciably, though.

Yes you are right, that's actually what is happening now and I am out of copper sulfate :(

You can also dip the parts to electroplate into a slightly acidic solution, the acid will attack it and increase its roughness.

There are a couple factors here:
What you are using for an anode is probably the most important. If you’re not using copper, you’re depleting your copper sulfate electrolyte of copper ions. Use a copper anode and you’ll be continuously replenishing your electrolyte with copper ions.
Concentration of your electrolyte is another factor. If the electrolyte is too weak, you won’t have enough copper ions to continue and you will start electrolyzing water into H+ and and OH-. Since you already have plated out the copper from the copper sulfate, you’re leaving behind the sulfate ions to combine with the OH- ions leaving you with weak sulfuric acid. If it’s too strong when you start, you’ll plate out too much copper too fast (at the beginning) and have poor adherence to your cathode. Alternatively, if your electrolyte doesn’t have enough of a source of sulfate ions, you’ll have too many copper ions which will plate out too quickly and leave you with poor adherence.
Anode to Cathode area ratio is another factor. If the anode is significantly smaller than your cathode, there will be a higher current density at the anode oxidizing faster than the cathode can reduce. This can cause anode polarization due to the uneven oxidation on the surface of the anode. A certain amount of oxidation is necessary for the process, but if it builds up too much, you get too much of your current focused on breaking through the built up oxides it decreases current efficiency and you don’t have as much current for electrodeposition.
If your voltage is too high, you can also start electrolyzing water and if you have hydrogen bubbles being generated at your cathode, the bubbles can prevent a good plate.
If you’re using tap water instead of distilled water for your electrolyte, you can end up with both organic and/or other ions in your electrolyte that can affect the electrochemistry. Some
Chloride ions assist in depolarization of the cathode, but too many can cause polarization of the anode.
I actually think one of the most important factors along with your anode material is your cathode material and cleanliness. If your cathode has oils, waxes, soaps or anything else on it, it cam decrease the conductivity of that area, changing your current density and potentially preventing good adherence. If your cathode is made out of a material that is significantly more reactive (iron, zinc, etc) in the reactivity series, you’ll get an immersion “plate” which isn’t actually plating, but cementing copper out of solution which isn’t actually adhering to the surface.
My understanding of is without any formal eduction in chemistry and from my own research and [failed] experiments over the last few months, so please forgive (and correct!) my errors.

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Why is the more electronegative atom lower in energy in an MO diagram?
Let's take $\ce{NO}$ for example. Why is $\ce{O}$ lower in energy? I asked my teacher and he said that because it is more electronegative and this lowers the bonding energy of the MO. But why?

More strictly speaking it should be effective nuclear charge and not electronegativity, although there is a strong correlation between these two quantities. Very simply put, a larger effective nuclear charge means that the nucleus "pulls" on the electrons harder and therefore their energies are lower.

Nucleus pulls on the electrons harder so why the energy is lower?

Do you study physics? It's the same as electric potential energy: $V = kq_1q_2/r$. In this case we could say $q_1$ is simply proportional to the effective nuclear charge and $q_2$ is the charge of an electron. They're opposite charges so $V$ is negative. Obviously for an atom classical electrostatics won't quite cut it but the physical idea is the same: the energy of an orbital is lower if $Z_\mathrm{eff}$ is larger.

Ok now i get it. thanks a lot for your answer .

Most strictly speaking it is the ionisation enthalpy that one can use to determine the energy of the outermost orbital (CC @ortho).

"Heteronuclear AOs will be slightly different because the more electronegative atom will be placed lower on the diagram. This is due to lone pairs of electrons being more stable on more electronegative elements leading them to be lower in energy." - Chem.libretexts.org

@orthocresol Maybe sum everything up and make it an answer (and hopefully prevent other non-answers).

I am summarizing from the excellent comments while adding some glue:
The convention is that lower electron energy means electrons that are more tightly bound. An energy of zero would mean a free electron (not bound in the electrostatic field of a nucleus).
As you go across from left to right in a period of the periodic table, the number of protons in the nucleus increases and the number of electrons bound (in an atom) increases likewise. In the simplest explanation of electron energies, you conceptually separate the electrons into inner electrons and valence (or outer shell) electrons, and treat the inner electrons and the nucleus together as an entity with some charge (the effective nuclear charge) that the outer electrons "see".
Comparing oxygen to nitrogen, the effective nuclear charge increases by one (same number of inner electrons, one more proton in the nucleus), so valence electrons of oxygen are more tightly bound than valence electrons of nitrogen. Thus, they appear lower in an MO diagram.
Not the OP's question, but for completeness: If you go from one period to the next (compare F with Na, say), the effective nuclear charge decreases (because we now also count the n = 2 electrons as inner electrons), and valence electrons of Na are less tightly bound than those of F. Most atomic properties change in a periodic pattern when you plot them against atomic number for the same reason (e.g. ionization energies, atomic radii)


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Predicting a route to use chloride as a reducing agent for titanium dioxide under UV
In my lab we are deadlocked on a seemingly simple problem.  We would like to find some reaction where $\ce{Cl-}$ can be used as a reducing agent on $\ce{TiO2}$ in a UV photosynthesis.  We ideally want a reaction like the following, but without using a metal
$\ce{Cl- + reactants -> products + e-}$
Then a hole in the TiO2 combines with the electron...
$\ce{TiO2+ + e- -> TiO2}$
As a secondary restriction, we need any extra reactants added to be unable to reduce $\ce{H2PtCl6}$.
What methods can be used to approach this problem, we are fairly certain something should work as its possible to do using ethylene glycol, but this can actually directly reduce the H2PtCl6.  This seems like an issue very similar to metal half reactions, but we are unable to use any metals.


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Reaction rate in terms of temperature change
For exothermic reaction where concentration changed in hard to measure, can I interpret the rate of reaction in terms of temperature change?
For example, if the experiment is iron rusting reaction with salt as a catalyst in water and I measure the temperature change and plot a graph of temperature vs time, can I say that the steeper trendline has a faster reaction rate?

Welcome to Chemistry SE. In principle I guess that it could work but only under very well defined conditions,  same conc'n of reactants same volume liquid etc. But you will have the problem of different reactions generating different amounts of heat. You could perhaps try to normalise to a fixed final temperature change. Looks messy, but you may have no choice if you can't measure anything else.

@7_G.S.N So, basically, you're suggesting to measure $dT$, and then plot $\frac{dT}{dt}$, according to $\frac{Q}{t}=mS\frac{dT}{dt}$, and $\frac{dQ}{dt}=-\Delta H\frac{d(concentration)}{dt}$. One problem I see with this is that the reaction you're measuring the rate for should be a slow reaction, or $T$ will change very quickly. Generally, S is not constant with temperature, so you'll have to further consider $S(T)$. If the reaction is gas phase, then there will be pressure changes too. So, there are a lot of restrictions for choice of reaction by this method to determine the rate.

@ I initiate conduct the experiment with solid phase of iron and initiate it by contacting it with controlled amount of water. But I don't really get how your first equation is changed into the second one. Is 'H' in the second equation means theoretical enthalpy value? Also, what does 'S' mean in the first equation?

@7_G.S.N The first reaction will help us determine $\frac{Q}{t}$, and the second will, in turn, help us to determine the rate of change in concentration , which will be related to the heat output. They're two different equations - not deriving from each other. $S$ is specific heat capacity.


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Diastereoselective epoxidation of acyclic alkenes with VO(acac)2
I've been reading about diastereoselective epoxidation of acyclic alkenes with VO(acac)2. This is taken from Carey, Sundberg Advanced Organic Chemistry Part B:

The OH and epoxide are syn.
Then, while reading Smith's Organic Synthesis I've found this (the ratio of products is 5.6:1 in favor of the anti):

which is in contrary with the results from Carey, Sundberg. In an Evans lecture the syn stereoselectivity is explained while in this lecture the anti selectivity is explained. Apparently different sources give opposite selectivities. What is the true explanation here? Is the syn or anti selectivity predominant?

Have you looked at the classic Evans, Fu paper on directed reactions? I want to say it's in JOC.

Seems like Carey and Sundberg were wrong.


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How to avoid the appearance of extra hydrogens from lattice packing of PDB file?
I'm struggling with a pdb file of a unit cell that has a $\ce{CH3}$ which becomes a $\ce{CH6}$ when some software performs replication transformations. I've tried pack it with Avogadro and Mercury.
There is a way to avoid this by only changing something in the PDB file?

It's impossible to know without having the PDB file. Clearly there's a bug in handling the transforms (i.e., there's probably disorder), but we'd need the actual file. Please send it to the Avogadro list and I can comment further.

This is a bug, which is clear.
My guess is that you don't get $\ce{CH6}$ but that there's some disorder, so the transformations put two near-identical $\ce{CH3}$ groups on top of each other. Thus, one C becomes two, and each gains hydrogens.
There's usually a "slop" factor to detect near-equivalent atoms, but it's clearly failing.


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Does lowering temperature of a gas make it easier to compress?
My teacher said that for ideal gases,

the lower the temperature, the lower the kinetic energy of a gas will be, and it will be easier to compress the gas.

Using the ideal gas equation $PV=nRT$, if the volume is constant, a decrease in temperature will cause a decrease in the pressure of the gas. That will make it easier to compress. Will the volume remain constant, or will it change?

Why gas molecules will come closer on lowering the temperature??

The kinetic energy decreases and they cannot move freely anymore. It is similar to the way water molecules come closer and form ice when temperature is lowered

Since you are saying that water molecules come closer to form ice then why don't volume of water decreases (i don't see any decrease)

Sorry I actually gave a wrong example since "Water is the only known non-metallic substance that expands when it freezes; its density decreases and it expands approximately 9% by volume". Source:http://www.lpi.usra.edu/education/explore/ice/activities/ice_action/expanding_ice/

So,give me a suitable answer of "Why gas molecules will come closer on lowering the temperature" so that i can write my answer

I don't know. I though the lowering of kinetic energy caused them to come closer. I guess I am wrong.

YEs this is the case,i think you should edit your question or add a suitable explaination of your argument,otherwise people are eager of downvote questions here

Come to chat if you are still confused.

You first had to define what you mean with "easier to compress" the gas. I guess that for any reasonable definition you will find that easiness of compression will rather depend on pressure than on temperature.

Compressing a gas by applying an external force means to work against the force the gas exerts on the walls of its container, i.e. its pressure.
The definition of compressibility
$$\beta = -\frac{1}{V}\frac{\partial V}{\partial p} \tag{1}$$
for an ideal gas with the well-known equation of state
$$p V = n R T \tag{2}$$
gives
$$\beta = -\frac{1}{V}\frac{\partial V}{\partial p} = \frac{1}{V} \frac{n R T}{p^2} = \frac{1}{p} \tag{3}$$
which is independent of the temperature $T$.
For non-ideal gases there will be higher order terms that also depend on temperature.
It is also worth noting that at the critical point the compressibility of a real gas is infinite. So if you are below the critical temperature you will be able to compress the gas more easily at a state (the critical point) with higher temperature.


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How to quantify caffeine content in coffee?
How do I titrate caffeine from coffee to estimate the percentage of caffeine in coffee?
I have tried using iodine solution, sulfuric acid, and sodium thiosulfate to no avail.

Coffee is quite a complex mixture that contains much more than just the caffeine. There's chlorogenic acid, trigonelline, polysaccharides, proteines and lignines too. Your titrations are a dead end.

Beckers, J. L. The Determination of Caffeine in Coffee: Sense or Nonsense? J. Chem. Educ. 2004, 81, 90-93; doi 10.1021/ed081p90.

According to a paper published named: "Chemical Test for Caffeine" for the "Hong Kong Chemistry Olympiad for Secondary Schools (2013-14)" (Hyperlink provided below). You can titrate caffeine, with the help of iodine and potassium iodide, and sulfuric acid as the acid. You will have to add an excess of iodine, which means it will be an iodometric backtitration. The iodine will react in two times the number of moles compared to your caffeine.
The only problem I have with this method, is that I cannot found any elaboration on how this reaction is possible on the internet (I haven't tested it myself). If there is someone who knows more about how this reaction if possible, please provide me with more information.

If someone knows how or why the reaction can take place, please keep me notified about it. I will provide the reaction below:
Reaction:
Reactants --- Caffeine + 2 Iodine + Potassiumiodide + Sulfuric Acid
Products ---- Potassium Bisulflate + Caffeine (in a peculiar complex with the remaining substances, notated like a hydrate. No water present.)
$$\begin{align}\ce{C8H10N4O2 + 2 I2 + KI + H2SO4 &-> C8H10N4O2 . HI . I4 + KHSO4}\\
\ce{\color{red}{I2} + 2 Na2S2O3 &-> 2 NaI + Na2S4O6}\end{align}$$
EDIT 1: This URL should lead to the paper I mentioned (if it works).
Chemical Test for Caffeine
EDIT 2: Another URL provided, this is the source used in the actual paper. The site is in Chinese (if I'm not mistaken), but can be translated via google. For anyone interested: Source used in mentioned paper, from "www.med66.com"

You might want to consider cutting out the first half of your second paragraph. If someone's going to tell you something (about the said reaction), they will do it in the comments section (and you will be automatically notified).


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Gel filtration separation of cellulose, starch, oxytocin and palmitic acid
I am studying biochemistry and I have been asked by my teacher to find a procedure for separating cellulose, starch, the peptide oxytocin (9 a.a) and palmitic acid ($\ce{C16}$) using gel filtration chromatography. I have some ideas about using Fehling's test to tell the cellulose from the starch and the peptide since they all react with the $\ce{Cu^{2+}}$ in the solution. And I'm thinking that the palmitic acid will not dissolve in the solution and can therefore be separated from the others.
However, I have a problem with how the gel filtration would work. Obviously cellulose and starch are much bigger than palmitic acid and oxytocin but, how could I possibly separate cellulose from starch using gel filtration??
Gel filtration: https://en.wikipedia.org/wiki/Size-exclusion_chromatography

Are you sure you have to use gel filtration for this? Cellulose might be the right size for gel filtration, but starch is normally not soluble and the two smaller molecules, oxytocin and palmitic acid will probably elute in the void volume.

I think by using a gradient mobile phase that changes by time. At the beginning, the mobile phase would have higher affinity to cellulose and lower to starch and then the gradient changes where it would have higher affinity to starch than cellulose.
Specifically, which mobile phase you would use and what gradient would take some time and research to find the answer but it think you get the idea

Using a gradient on a gel filtration column doesn't really work.


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Computational chemistry: is k a quantum number
I am trying to use the natural band offset method to calculate band offsets of junctions. In order to do so, I need to find the core electron eigenstates. I tried using FP-LAPW software Elk. But in the output of the core electron eigenstates there is a "k" value.
The value is listed among "n" and "l", so my thought is that it may a quantum number. But I am not sure. I was able to find that it may be the "projection of the total spin I on the nuclear symmetry axis". Could someone help me identify what k means?


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Why can phosphorus have 5 bonds while nitrogen cannot?
Phosphorus can form 5 bonds like in the case of phosphate. While nitrogen is known to make a maximum of 4 bonds (3 covalent, 1 dative covalent). But why is this so?
Is it because nitrogen doesn't have available d orbitals while phosphorus  has available d orbitals? Or does the presence of d orbitals have nothing to do with this and it's the sizes of both atoms that allows them to make more bonds?

This is a complex subject, see https://en.wikipedia.org/wiki/Hypervalent_molecule and http://pubs.acs.org/doi/abs/10.1021/ja00178a014

They (and all the other elements) have "available" d (and higher) orbitals - whether they're occupied is another matter.

related http://chemistry.stackexchange.com/questions/42750/how-can-the-size-of-atom-be-the-cause-of-octet-expansion

Does this answer your question? Why can't NH5 form?


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What makes heat resistant proteins heat resistant?
We were doing the chapter Biomolecules in class the other day, and a doubt popped up once we reached the section on proteins.
We were taught that the stuff that keeps proteins together are: Ionic bonds (not always), Covalent bonds (peptide bonds and di-sulfide bridges), van der Waals' forces and hydrophobic interactions.
Now I've heard that certain bacteria are known to survive extremely hot conditions (Inside thermal vents, volcanoes, the like), like Thermus aquaticus, the source of taq polymerase. I'm told they're able to survive in those conditions because their proteins (enzymes and other cell components) show a very high level of thermal tolerance.
I tried reading up more on that, but Wikipedia adopts a very vague, generalized approach to the topic and doesn't specifically mention what imparts this high degree of stability to these proteins. https://en.wikipedia.org/wiki/Thermus_aquaticus
Frankly, if the earlier stated interactions are the only interactions seen in proteins, I don't see how some proteins can possess such high resilience to heat, especially since most of those interactions are fairly weak (and ionic interactions aren't very common).
Sure, maybe an incredibly extensive network of those interactions (say, hydrogen-bonding) would impart a good deal of stability, but I suppose such a protein would be incredibly compact, and twisted in all sorts of wacky shapes to generate the required stability.
I even tried comparing the structures of taq polymerase (First image; which denatures at about $100\ \mathrm{^\circ C}$) and salivary amylase (Second image; which denatures at a little under $40\ \mathrm{^\circ C}$), but to my (untrained) eye, there doesn't seem to be some absolutely fantastic and unique character to taq polymerase going by the structure that is:

So what exactly is it that imparts such high thermal stability to some proteins? Is it just a vastly intricate network of the earlier mentioned interactions, or is it because of something else?

I recommend reading this journal http://link.springer.com/article/10.1007/PL00000935

related http://biology.stackexchange.com/questions/23246/how-does-protein-denaturation-work

Szilágyi and Závodszky published an article in the journal Structure which analyses a number of different structural parameters of proteins of moderately thermophilic ($45~\mathrm{^\circ C} < \vartheta_\mathrm{opt} < 80~\mathrm{^\circ C}$) and hyperthermophilic ($\vartheta_\mathrm{opt} \approx 100~\mathrm{^\circ C}$) organisms compared to homologous mesophilic organisms.[1] I shall summarise their results in this answer.
They analysed the parameters number of cavities, volume of cavities, surface area of cavities (all relating to cavities on the surface of a given protein), number of hydrogen bonds, number of unsatisfied potential hydrogen-bonding candidates, number of ionic pairs (the ionic pairs were analysed for distances of $4~\mathrm{Å}, 6~\mathrm{Å}$ and $8~\mathrm{Å}$) number of α helices, number of β strands, number of unordered loops and the ratio of polar to apolar amino acids (both of exposed and of buried residues). These were all assigned a $P$-value; this defining the probability of the observed effect occuring if the null hypothesis ($H_0$, ‘this feature does not contribute to heat stability’) were true.
The single most important influence on the stability of both moderately thermophilic and hyperthermophilic proteins proved to be additional ions pairs as shown in table 1 below.
$$\textbf{Table 1: }\text{$P$-values of ion pair distances in both}\\\text{moderately thermophilic and hyperthermophilic proteins}\\
\begin{array}{lcc}\hline
d/\mathrm{Å} & P_{45{-}80} & P_{100}\\ \hline
4 & 0.178\phantom{0} & 0.026\\
6 & 0.024\phantom{0} & 0.016\\
8 & 0.0048 & 0.014\\ \hline \end{array}$$
While close-contact ion pairing is not as important for moderately thermophilic proteins ($P_{45{-}80}$), this quickly changes for less-close contact ion pairings; in hyperthermophiles, all types of ion pairing are significantly important. Other influences were:

polarity of buried areas; while the difference was not as significant for only moderately thermophilic ($P_{45{-}80} = 0.156$) or hyperthermophilic proteins ($P_{100} = 0.101$), over the full data set there was a greater proportion of polar amino acids buried (i.e. not superficial) in thermophilic proteins when compared to mesophilic proteins ($P_\mathrm{full} = 0.047$).
polarity of exposed ares; this mainly stabilised moderately thermophilic proteins ($P_{45{-}80} = 0.024$).
number of cavities; this was mainly a feature of hyperthermophilic proteins ($P_{100} = 0.033$).

Furthermore, the authors point out that the choice of amino acid is somewhat influenced; however, the effects are not as strong. Moderately thermophilic proteins tend to favour arginine over lysine and use less lysine than mesophiles; hyperthermophiles incorporate both more lysine and arginine residues. Tryptophan, glutamate and aspartate are significantly increased in hyperthermophiles; moderate thermophiles incorporate less tryptophan and slightly more aspartate and glutamate than mesophiles. Also, hyperthermophiles exhibit a preference for large unpolar amino acids (leucine and isoleucine) and and have a reduced methionine and asparagine content. Since both latter amino acids decompose more readily at elevated temperatures, their reduction was expected. Cysteine content is either similar or reduced, showing that increased disulphide bridging does not contribute to thermic stability.
Previous studies and theoretical models[2] have shown that ion pairing is expected to have a marked effect on stability. At room temperature, the formation of a salt bridge leads to an unfavourable reduction of solvation which overcompensates any enthalpy gained. At higher temperatures however, this desolvation penalty is strongly reduced and the stability gain from salt bridges thus increased.
The authors do point out one major caveat concerning hyperthermophilic proteins they studied:[1]

It should be noted, however, that the number of known structures of proteins from organisms with $T_\mathrm{opt} \approx 100~\mathrm{^\circ C}$ is very small (there are only five such structures in our data set). This means that our conclusions pertaining to proteins from extremely thermophilic organisms might be of limited validity; in fact, it is possible that much of the observed deviations are because of ‘sample bias’ (i.e., the peculiarities of the available protein structures).

References:
[1]: A. Szilágyi, P. Závodszky, Structure 2000, 8, 493. DOI: 10.1016/S0969-2126(00)00133-7.
[2]: A. H. Elcock, J. Mol. Biol. 1998, 284, 489. DOI: 10.1006/jmbi.1998.2159. Reference [42] in [1].

To be frank, thermostable proteins are stable because they as you state contain a huge amount of stabilizing interactions. [notice my dry humor]
You state that the presence of large and complex hydrogen bonding networks would cause the protein to adopt a compact conformation, however you also show structures with alpha-helixes which certainly have a lot of stabilizing h-bonds.
Notice how the thermostable taq polymerase have nearly all of its amino acid residues in stable secondary structure, while the amylase have a lot of unordered loops, which certainly wont help with thermostability.
I can tell you from personal experience that protein structures can easily have hundreds or even thousands of small stabilizing interactions. If you want to see some examples, download a free version of pymol and start browsing the structures on (RCSB.org). I suggest reading a couple of articles about protein engineering on enzymes with the focus of improving thermostability (for example https://www.ncbi.nlm.nih.gov/pubmed/27899916).


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Why are alkyl halides immiscible with water?
Why are alkyl halides, though polar, immiscible with water?
I have tried to solve this question by myself and I believe that it's due to the large energy is needed to break the hydrogen bonds, which is insufficient to overcome the forces present in water. But I'm not sure if I'm right. Can anyone help me out?

Think about the energy considerations for a substance to get dissolved.


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Reasons for weakness of London dispersion force in graphite?
The layers of graphite are held together via London dispersion force. This force is weak. I know carbon having a smaller atomic radius and few electrons is one factor for this force's weakness.
But, the free electron is weakly bound with the adjacent three atoms. Does this constrains the free electron's ability to disperse, making the London force weaker?
P.S. This might be a duplicate question. Inform me if you find similar one(I haven't found any)

Welcome to Chemistry.Stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about how it works.

I think I don't get your point. What do you mean by constrained by attraction?

Hope the new edit will make things more clear.

London force is generally weak; graphite is not special at all in this regard.

But the bonding with 3 adjacent atoms still has some effect, no?

Typically when electron is "free", it has larger polarizability and the dispersion interaction is somehow enhanced.

We need to look at the orbitals, where are the electrons localized?
because the hybridization of the carbons in grpahite are sp2 we have one free p orbital with one electron, which is denoted as the pz orbital.
The free pz electron then forms the resonance bond with the adjacent carbons. Therefore there are no non bonding valence electrons to contribute much for the London dispersion interactions.


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Ketal formation of ethyl acetoacetate
Why does the ketone carbonyl group react in preference to the ester carbonyl?
Is it something to do with the resonance occurring into the ester carbonyl from the oxygen? but how does this make the ketone carbonyl more likely to form the ketal when reacted with a diol and non aqueous acid?

That resonance makes the ester less electrophilic.

Well, if ester carbonyl reacted this way the result would be orthoester not ketal. (nitpicking here :)


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Rate of substitution and elimination reactions at high temperatures
From High School, I've been reading that elimination reactions are favorable at high temperatures because their entropy change is positive, so at higher temperature the Gibbs free energy will be more negative so the equilibrium will be more towards product's side.
But I've encountered a different statement in Clayden's Organic Chemistry:

This explanation is simplified because what matters is the rate of the reaction, not the stability of the products. A detailed discussion is beyond the scope of the book, but the general argument still holds.

So what is the reason behind this? Why are substitution reactions not that much faster at high temperature than elimination reactions?

What makes you think they aren't?

When you are discussing the kinetics of a reaction, what is central to the discussion $\Delta G$ or $\Delta G^ \dagger$ ?

As Far as I know-
$\Delta G^ \dagger$  determines the rate of reaction . And It is used to be positive otherwise the transition state  will be more stable  .
More It is Positive the more reaction gets slows down .

With temperature, the rates of substitution reactions and elimination reactions both increase.
For the purpose of answering this question, let us consider $S_N2$ and $E_2$ type reactions.
For $S_N2$ type reactions, the HOMO of the nucleophile has to be oriented with the LUMO of the electrophile. Here, a linear orientation is required. So, the entropic requirement is that the nucleophile should approach the electrophile in a certain way.
For $E2$ type reactions, the base attacks the $\alpha$-proton, which finally leads to the elimination. So, the HOMO of the base interacts with the $\sigma^*$ orbital of the $C-H$ bond, and then the $\sigma$ orbital interacts with the $\sigma^*$ orbital of the leaving group. So here the entropic requirement is the linear orientation of the base, the proton and the carbon it is attached to, and the leaving group should be antiperiplanar with respect to the proton.
With an increase in temperature, the "rate of bond rotation" increases and hence, the probability that a collision with a proper orientation will occur in a given time frame increases. This increase is more prominent in the case of the elimination reaction because it is quite dependent on the orientation of bonds (2 constraints vs. 1 constraint) compared to the substitution reaction.
So, a rise in the temperature will not only affect the thermodynamics of the equation, but it will also directly affect the rate of reaction.


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Sodium Naphthalenide
The sodium fusion test done in the lab for the systematic identification of a given organic compound is basically fusing sodium metal in a fusion tube, and then adding the compound in a small quantity, then gently heating the compound and then after the completion of initial reaction,  strongly heating to red heat and then plunging it into water.
I found that Sodium naphthalenide can be produced by the direct reaction of sodium naphthalenide with THF as the solvent. It is used for the solubilisation of sodium metal and is used as a reagent in place of just sodium metal as it is more effective.
So, the question is whether sodium naphthalenide will form in the fusion tube? If yes, why? If no, why not?
I asked a professor who said it is very unlikely to form due to the presence of oxygen (open to the atmosphere) and without the solvent. I did not get it exactly.
If the reaction does take place, then I think naphthalenide anion will probably undergo oxidation due to the oxygen, but no evidence of oxygen is specifically found. So, the reaction probably doesn't take place. Another evidence pointing towards the "no reaction takes place" side is that the contents of the fusion tube remain colourless (except for the piece of sodium metal) till we start heating it strongly, and the excess naphthalene in the tube vaporises and catches fire, blackening the tube from the outside.
Also, if it is possible, in the comments section, or in an answer, please include the energy difference between naphthalene and naphthalenide radical anion.
So, why does the reaction not take place in the absence of solvent?

Under the conditions for the sodium fusion test naphthalene should either evaporate or burn. I suppose it is possible that some sodium naphthalenide  is transiently formed sometime during the process. However, even if it did survive until the end of the testing procedure, it would be returned to the naphthalene form upon the water quenching step.

@BiggChemT That is sort of what I was thinking, that naphthalene (fused) serves as the solvent for the transient sodium naphthalenide (if formed). However, the naphthalenide radical anion would undergo disproportionation on quenching with water.


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Why does cotton dissolve in a dilute solution (1/2) of fuming sulfuric and nitric acids?
A nitration experiment carried out showed that a solution of half diluted sulfuric and nitric acids would turn the submerged cottons into a gooey substance. The remnants of the cotton balls did not flash as nitrocellulose would have.
On the other hand, cotton balls submerged in a mixture of fuming nitric acid and fuming sulfuric acid remained intact and were nitrated, producing nitrocellulose.
What I'm really curious and confused about is how the dilution not only interferes with the nitration itself, but also affects its dissolution.
Any guidance is appreciated.

Could  it be hydrolysis?

It happened to me today. Turns out my sulfuric acid was wet (titrated 14M), so the water content was enough to hydrolize the cellulose. Switched to another bottle of H2SO4, now works fine.
Using 6mL HNO3 70% + 14mL H2SO4 98% + 1g cotton

To address you second question first, in neither case did the cotton dissolve.  In the diluted acid mixture the cotton was just wetted, possibly also having enough oxidation of the cotton by the nitric acid to give it a different texture.
Regarding the effect of dilution on the nitration of the cotton, the reaction was basically just too dilute to proceed.  Also, according to this Wikipedia article, the sulfuric acid is present as a catalyst, producing the nitronium ion, $\ce{NO2+}$. The net reaction is given as:
$$\ce{3HNO3 + C6H10O5 → C6H7(NO2)3O5 + 3H2O}$$
Since water is a product of the reaction, by adding water to the reaction via dillution we further inhibit the formation of the products.


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White creamy substance (not calcium) observed when boiling water is filtered through coffee paper filter
We have hard water from our well. It is filtered twice before boiling it: by the water filter and with a manual on-the-counter filter. When boiled in a clean kettle a white substance floats in the water, and it isn't calcium which is hard and sinks at the bottom. Then we filter that hot water through a coffee paper filter and a white creamy substance remains. I called water related companies but no one knows. Anyone have a answer to this one?

You could always separate the white creamy substance and test it...

It may be an organic matter - proteins that is. Try drying it and burning it - if it turns black then it is organic.

Thank you. I'll do the drying and burning part for certain.

did you try to burn the precipitate and if so what was the result? I have observed something similar with a white colloidal suspension forming after boiling water which eventually precipitates to the bottom of the kettle. This is totally different in appearance and behaviour to normal kettle scale that I've seen previously in the same kettle. I’ve tested the precipitate with indicator, it’s around pH10 BUT a flame test does not give any red/orange flame expected of calcium ions. We have a scale inhibitor cartridge treating our water supply so wonder if this has stopped working and/or is now le

I assure you it is calcium carbonate.
The particles are so small they do not just sink to the bottom like a stone. Instead, they aggregate at the nearest phase boundary. All small impurities do that, unless they have a very low interfacial tension with water. It's energetically much more favourable than sinking.
Over time, this creamy substance aggregates further, up to a point where the aggregates are large and dense enough to dissociate from the surface, and sink to the bottom.

Thank you Karl. This week I'll be installing a bigger water filter (adding salt to it or either potassium I'm told), hoping that it will get rid of that "calcium" as you say. If it is calcium carbonate in very small particles, then such calcium is unharmful to the body I presume. Small amount of natural calcium (126 mg/l from the analysis) can only be healthy. I really appreciate you feed back. Thank you.

"adding salt to it or either potassium I'm told" What? Do you mean you have an ion exchanger like in a washing machine? That's not a filter, and I've never heard of using a Ca-Na exchanger for drinking water.

This is what they call a water softener. (Autotrol)  And yes they add salt... unfortunately.

Hard water often contains magnesium as well as calcium. A ppt of Mg(OH)2 or Mg2CO3(OH)2 would give a pH ~10 (Tee's result).
The ppt could be a function of the pH of the water; temporary hard water contains bicarbonates which are more soluble than the carbonates. When the water is heated, it drives off CO2, which could drive nano particles to the surface until they grow and fall.
Try dissolving the creamy ppt with acid (vinegar or HCl). That would give some indication of metallic nature. To be complete, you could try adding household ammonia (no soap in it) to see if you get dissolution (clarification of the liquid).
You could also take the twice-filtered, but cool, water and add sodium bicarbonate to see if you get a ppt. Then take the supernatant and boil it - there should be no more ppting. And likewise, to be complete, try adding a little vinegar to get a pH ~ 4-5; this would likely keep Ca and Mg in solution.
Magnesium gives no color in a flame test. https://en.wikipedia.org/wiki/Flame_test


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Chemical bonding and dipole moment
We know that ammonia undergoes amine inversion. Why doesn't the dipole moment decrease in ammonia, since the direction of the dipole changes to the opposite direction every time there's inversion?
Shouldn't the effective dipole moment become zero?

Come to think of it, the effective dipole moment of anything is essentially zero, because the molecules rotate all the time, so the direction of dipole changes to the opposite every now and then.

Probably due to the preservation of symmetry - not a true "inversion". See this Wikipedia article on molecular symmetry: "Similarly, ammonia (NH3) has two equivalent pyramidal (C3v) conformations which are interconverted by the process known as nitrogen inversion. This is not an inversion in the sense used for symmetry operations of rigid molecules, since NH3 has no inversion center. Rather it is a reflection of all atoms about the centre of mass (close to the nitrogen), which happens to be energetically feasible for this molecule.

The 'inversion' in ammonia is also called an 'umbrella' motion as seen when one is opened the wrong way on a windy day;  C3v changes to D3h to C3v to pointing the other way and back again. Inversion in the symmetry sense is replacing x, y, z  by -x, -y, -z.  The 'effective' or long time  average dipole should really be zero as you suggest because during the vibration, the H atoms more from one side of the N atom to the other, so the dipole goes to zero at halfway in the motion and has the opposite sense every half cycle. A dipole assumes the molecule has its canonical structure.

Ivan Neretin is correct in stating that no particle in a (nondegenerate) state can have a permanent electric dipole moment in the laboratory frame. In fact, a permanent dipole moment would require a violation of time (T) and parity (P) symmetry. This is because any such dipole should be directed along the angular momentum vector of the molecule and time-reversal and parity operators have a different effect on the rotational angular momentum and direction of the dipole.
What chemists call a permanent dipole moment refers to the dipole moment in the molecular point group to which the molecule belongs. Only when a polar molecule is subjected to an electric field (e.g. an EM wave or DC field), this field will mix rotational states of opposite parity in such a way that the molecule "orients" itself in the field. You can only speak of the inversion of ammonia in the presence of an electric field (the eigenstates of the Hamiltonian are time independent in the absence of a field or other perturbing force).
Although degenerate states could in principle have a permanent dipole moment, it was shown by Klemperer et al. (J. Phys. Chem. 1993,97, 2413) that the degeneracy of these states is removed by higher-order terms in the Hamiltonian.

Of course a molecule can have a dipole, e.g. $\ce{HCl}$ but in the gas phase (or solution) thermal motion in the ensemble of molecules has them pointing in all possible directions so the overall experimentally measured dipole is zero. Applying an external field breaks the symmetry and the average dipole can be measured and from this the true one.  In small molecules in gas phase quantum effects are more obvious & restrict orientation. You can quite easily measure the inversion frequency of ammonia by observing  its spectrum and its about $0.7 \pu{cm^{-1}}$

A molecule only has a permanent dipole in the molecule fixed-frame, not in the laboratory fixed frame. It is more fundamental than an effect of averaging over the motion of the molecules (and it took me quite some time to understand). The concept of a molecular fixed frame is a consequence of the Born-Oppenheimer approximation where the nuclei are considered to move in the effective potential of the electrons.

If you consider beyond BO wavefunctions, you'll see that this concept indeed breaks down. Of course you can measure the inversion of ammonia, but you'll need a EM field, otherwise you cannot speak of inversion at all.

The BO approx does allow us to imagine potential energy surfaces, and its generally applicable. However, experiment (x-ray diffraction) does show molecules have real shapes: atoms don't just wander around on the surface, not at normal energies anyway. So BO approx is good. As I mentioned dipole are only measurable in the presence of a field (e.g. directed force) to break spherical symmetry. This is how we get to molecule frame from lab frame. Your last comment seems to imply that an em field causes the inversion instead of measuring the effect. The tunnelling (inversion) occurs anyway.

It makes no sense to speak about the inversion in the absence of a field. If you solve the SE for the double well potential of the ammonia molecule you find eigenstates that do not depend on time, i.e., there is no transition from one state to the other. Only when you introduce a field and couple the different parity states it makes sense to speak of the inversion. See for instance the treatment of Townes and Schawlow.

Well, I'm not convinced by your arguments.  Solving the SE produces eigenstates, single or double well makes no odds. The energy of these obviously does not depend on time if time is not introduced anywhere. When a photon is absorbed, say, you can if you wish think of the levels moving together under the influence of the em field and so a transition occurs and the molecule now has more energy.  What you imply, however, is that a molecule, say in a harmonic well does not vibrate if a field is not present, i.e. the photon induces vibrations.

Since the parity operator commutes with the Hamiltonian because of the symmetry of the potential, the eigenfunctions of the SE should have even or odd parity and are completely delocalized over the system. By adding or subtracting the odd and even functions you can create "localized" states in the form of a wavepacket that moves between the left and the right well. The only way to mix the different parity states is by the interaction with a field.

And yes, an isolated harmonic oscillator does not vibrate, such a localized motion of the nuclei is a "classical" picture and requires the formation of a wavepacket which you can do by disturbing the (isolated) system.

Yes I agree  you can interpret this as no motion: but using the classical idea is convenient and not altogether incorrect  as you can make a wavepacket. For example with  a laser pulse of shorter duration that the vibrational frequency then several eigenstates are simultaneously excited and so are no longer stationary states and so evolve in time.   ... continued

continued.. The motion can be observed by a second probe photon. The frequency of the wavepacket (particle) motion is the same as that obtained by steady state absorption spectrum obtained with narrow band excitation. Like many things in QM there are two opposing but equivalent viewpoints.

That was exactly my point; you need a field (such as a short fs laser pulse) to be able to speak of localized states. But I agree that it certainly makes sense in many cases to ignore the rigorous physics and treat the system semi-classical even in the absence of a perturbing force. (Especially since we never measure an unperturbed system anyway).


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Which contributes more strongly to ionic conductivity, charge or mobility?
Conductivity depends on the magnitude of charge as well as mobility. Which will be the dominating factor in the case of opposite trends in these two properties?


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E/Z configuration of the central double bond in a highly branched poly-ene
What would be the configuration of the marked double bond in the attached molecule?

From Cahn Ingold Prelog rules, it can not be assigned E/Z configuration.

The substituents on the right side of the marked double bond are symmetrical. Thus, the marked double bond doesn’t give rise to cis/trans isomerism.

Maybe Cahn-Ingold-Prelog (CIP) rules cannot describe such a system but IUPAC rules can. In IUPAC rules, everything should have a priority and, according to what I obtained on the latest ChemDraw 16, E has priority over Z (just like R has priority over S in atom configuration. See Pseudo-asymmetry). In your example, the central double bond should have geometry Z according to ChemDraw, but I do not have the IUPAC rule for that. I will go on searching.

For simplicity, I shall assign numbers to the double bonds so that it is easier to reference them. This numbering is chosen solely after their position in space and does not correspond to the number of carbons in a PIN or otherwise systematic name.
1                 10
  > 5         8 <
2    \       /    11
       > 7 <
3    /       \    12
  > 6         9 <
4                 13

The terminal bonds, 1–4 and 10–13, are easy since they are all disubstituted: (1Z,2Z,3E,4E,10E,11Z,12Z,13E)
1Z                 10E
   > 5         8 <
2Z    \       /    11Z
        > 7 <
3E    /       \    12Z
   > 6         9 <
4E                 13E

Next up are the intermediate double bonds. If we simply follow the atoms, neither side will ever have priority, thus the somewhat less common rule that Z has priority over E takes effect. This allows us to label the double bonds (8Z) and (9Z).
1Z                  10E
   > 5         8Z <
2Z    \       /     11Z
        > 7 <
3E    /       \     12Z
   > 6         9Z <
4E                  13E

However, we cannot determine a stereodescriptor for bonds 5 and 6: both 1 and 2 have the same geometry (Z) as do both 3 and 4 (E). Thus, neither of these double bonds is asymmetric as they only feature three different substituents with the two identical ones being on the same side.
1Z                   10E
   > 5Ø         8Z <
2Z     \       /     11Z
         > 7 <
3E     /       \     12Z
   > 6Ø         9Z <
4E                   13E

Moving on to the final double bond 7, we face the same problem. The substituents 8 and 9 are identical, both feature an E double bond attache to E and Z. Thus again, we are unable to determine a stereodescriptor since the double bond is not asymmetric.
1Z                    10E
   > 5Ø          8Z <
2Z     \        /     11Z
         > 7Ø <
3E     /        \     12Z
   > 6Ø          9Z <
4E                    13E

Note that we can assign priorities to the left-hand side of double bond 7: the bottom half (3,4,6) has a lower priority than the top half (1,2,5) since the latter features Z double bonds where the former features E. But assigning a descriptor to 5, 6 and 7 is impossible due to their symmetric other side.

So what is the answer to the question?

@ketbra Last sentence.


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Structure of Sodium Naphthalenide
This question is a sort of follow-up to my earlier question.
I know about a method of solubilising potassium where the potassium atoms are "trapped" between layers of graphite, but I think that sodium naphthalenide exists only in solution (if a contradiction is found, please inform).
In the solution phase, say in THF, I would expect some THF molecules co-ordinated to the sodium, and an $\eta^4$ complex with naphthalene, involving one ring. An $\eta^{10}$ complex seems possible, involving both the rings, as the negative charge is distributed further, but I haven't read about any $\eta^{10}$ complex. An $\eta^1$ complex might be possible,  but the distribution of negative charge will be lesser.
I expect sometehing similar to the tetrahedral form of butyllithium.
What is the structure of sodium naphthalenide?

Wikipedia says "It has not been isolated as a solid", which can be understood probably since the anion is a radical, so at the moment there is no single crystal structure (it actually should be doubted in such a case that something like a compound with name sodium naphthalenide really deserves to be called existing). Structures in solution are very difficult to determine, one of the many reasons is that complicated equilibria between different species can occur. Anyway the question is surely interesting, maybe one should look for computational studies (I could'nt find anything, yet).

That being said, there are quite a few known crystal structures containing naphtalene radical anion (or cation, for that matter).

@Eashaan Godbole Not about sodium naphthalenide, but as partially related compound: the complex of THF, anthracene, and Mg -- useable, for example, to trigger Grignard reactions (though there are other methods as well); Acc. Chem. Res., 1988, 21, 261–267 (DOI: 10.1021/ar00151a002) by Bogdanovic himself.

TL;DR: Naphthalene anion radical won't tolerate a presence of unshielded alkali metal cation nor in solution, neither in crystal. Reduction of naphthalene is possible strictly under air- and moisture free conditions by alkali-metals in the presence of coordinating solvents (cryptands). In this case solvent-separated cations don't form coordination bonds with naphthalenide radical.
On the other hand, further reduced (dianion) naphthalene readily forms various compounds not only with alkali metals, but also with lanthanides, typically $\eta^4$-complexes.

Various studies on stabilization of radical anions, especially in solid state, show that complexation (preferably with polar aprotic solvent or polydentate ligands) of counter-cations is essential:
$$\ce{C_nH_m + [M^0]_{\infty} + $x$ Solv + $n$ L <=> [M+(Solv_x|L_n)](C_nH_m^{.-})},$$
where $\ce{[M^0]_{\infty}}$ - any metal in oxidation state $0$ as a bulk reactant, $\ce{Solv}$ and $\ce{L}$ - solvent and ligand, correspondingly.
If cation is optimally solvated (complexated/sterically protected), radical counter anions $\ce{C_nH_m^{.-}}$ with extended $\pi$ systems are preferred because they lack extensively charged centers which favor contact ion formation (1). For example, crystals of bis(diglyme-O,O',O'')-sodium naphthalenide radical
$\ce{[Na+(diglume)2](C10H8^{.-})}$ ([Na]) (1) has been isolated based on a principle of increasing enthalpy (calcd. based on crystal structures):
$$\Delta H (\ce{[Na+(THF)6]}) = \pu{-587 kJ mol^{-1}}$$
$$\Delta H (\ce{[Na+(DME)3]}) = \pu{-671 kJ mol^{-1}}$$
$$\Delta H (\ce{[Na+(diglume)2]}) = \pu{-677.3 kJ mol^{-1}}$$
Under certain conditions, the reduction of the naphthalene can proceed further to the naphthalene dianion, $\ce{C10H8^{2-}}$. For example, a complex with radical ion $\ce{[Li+(TMEDA)2](C10H8^{.-})}$ ([Li]) can reversibly be transformed into naphthalene dianion-complex $\ce{[Li+(TMEDA)]2(C10H8^{2-})}$ (2):

Of course, bigger cations require better shielding, commonly resulting in a usage of bulkier solvents or supramolecular assemblies, such as ([2,2,2]-cryptand)-potassium naphthalenide radical, $\ce{[K+(crypt-222)](C10H8^{.-})}$ ([K]) (3), in comparison to previously mentioned [Li] and [Na] naphthalenides (in the following crystal structures sphere-packing model is applied to metal atoms, and the rest is represented with sticks for simplicity):

It is also worth mentioning that naphthalene dianion forms numerous complexes with rare earth elements (4), e.g. with [Y] (see picture): $\ce{CpYC10H8(DME)}$, ($\eta^5$-Cyclopentadienyl)-(1,2-dimethoxyethane)-($2\eta^1:\eta^2(2\sigma,\pi)$-naphthalenide)-yttrium(III) with a $\ce{C6}$ ring being bent by $26.1^\circ$:

$\ce{La}$ and $\ce{Eu}$ complexes with bridged $\mu-\eta^4:\eta^4$-naphthalene are also obtained ($\ce{[LaI2(THF)3]2(C10H8)}$, $\ce{[EuI(DME)]2(C10H8)}$).
(1) Bock, H.; Arad, C.; Näther, C.; Havlas, Z. J. Chem. Soc., Chem. Commun. 1995, 23, 2393–2394. DOI: 10.1039/C39950002393
(2) Melero, C.; Guijarro, A.; Yus, M. Dalton Trans. 2009, 0 (8), 1286–1289. DOI: 10.1039/B821119C
(3) Rosokha, S. V.; Kochi, J. K. J. Org. Chem. 2006, 71 (25), 9357–9365. DOI: 10.1021/jo061695a
(4) Protchenko, A. V.; Zakharov, L. N.; Fukin, G. K.; Struchkov, V. T.; Bochkareva, M. N. Russ Chem Bull 1996, 45 (4), 950–953. DOI: 10.1007/BF01431330

Please what does the notation $\ce{[M^0]_\infty}$ mean ?

@Hexacoordinate-C Any metal in oxidation state $0$ as a bulk reactant.


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Electron configurations for d-block and p-block
Question 1: Why is it that when you get to the transition metals (or the d-block), the energy of the $\ce{3d}$ orbitals becomes slightly less than that of $\ce{4s}$, making $\ce{4s}$ have the electrons with the most energy?
Also, is it correct to write electron configurations for transition metals with the $\ce{4s}$ orbital being the outermost one? For example like this $\ce{Sc = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$ instead of this $\ce{Sc = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1}$
Question 2: If my first thought is true, then do you write electron configurations for the p-block just like the d-block, with the $\ce{4s}$ orbital being the outermost one? Or do you follow the regular pattern of letting the $\ce{3d}$ block comes after the $\ce{4d}$ block?
For example, would it be like this $\ce{Ga = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^1}$ or would it be like this $\ce{Ga = 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^1}$

According to google, if you search up "scandium electron configuration" it would say my first one). If you search up "gallium electron configuration", you would see my second one)

Related: http://chemistry.stackexchange.com/questions/8357/why-does-the-3rd-electron-shell-start-filling-up-with-scandium

I did not completely understand you question here, but I am gonna give it a try, so maybe I might answer your question.
I do not think that it matters how you write them, but its just depends on how you fill them. The order for filling orbitals is as follows:
$$\ce{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s}$$
In high school, I just used to write the $(n+1)s$ before the $nd$. Because with some exceptions, the rule says that you will always fill the $(n+1)s$ orbital before you fill the $nd$ orbital. Meaning that you fill $\ce{4s}$ before you fill $\ce{3d}$, $\ce{5s}$ before you fill $\ce{4d}$, and so on.
The reason is that the $\ce{s}$ actually has a lower energy than the $\ce{d}$ orbital. You wrote that wrong in your question. That is the only you fill the s before the d, and when you make ions, you will have to remove the electrons from s before you do from $\ce{d}$.
But like I said, you can write them like you want as long as you have filled them correctly in the right order. Likewise once you have completely filled your $ns$, only then you can move to fill the $(n-1)d$ orbital and then again proceed to fill the $\ce{np}$ orbital. Its just how we fill them.
In some books, you will find that they write the s as the outermost but some write d as the outermost. Frankly, in my book and myself have learned to write $\ce{d}$ at the end. But I do not think that writing it matters.
For example, take a look at this site. They write $s$ orbital at the end of the configuration and also note the energy diagram. But in the following Wikipedia website, they have written the $\ce{d}$ at the end instead of $\ce{s}$. So basically, the writing of the orbitals doesnt matter, but the order of filling matters.

Is it true though that when you reach the transition metals (d-block), the energy of 3d becomes slightly less than the energy of 4s?

No it doesnt. It is the opposite. The energy of the 4s is lower than that of 3d, therefore when filling the orbitals, we fill the 4s before the 3d, as per the Aufbau principle.
You can have a look at an energy diagram of orbitals on this website: http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html.

Yes but during ionization, electrons from the 4s subshell are removed first, and then the 3d, right? Doesn't this mean that 4s has a higher energy?

The 4s sublevel in an atom have a lower energy than the 3d sublevel if the atom only has 4s electrons and no 3d electrons, so 4s "fills first" as you progress across period 4 of the P.T., and group 1 and 2 metals will have the s sublevel at the top, but if you have an element in the d block, the  ELECTRONS IN the 4s orbital have higher energy than the electrons in the 3d orbitals (they flip), so the 4s electrons leave first and fill last.
http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html
So, in other words,
1) Order of filling depends on what you mean by order of filling.
a) As you progress along a period on the P.T. each successive element will (with exceptions) have electrons added to their d orbitals compared to previous elements.
b) As you fill up a given element in the d-block, the electrons of known sublevels will fill in the 3d sublevel before the 4s (4d before 5s etc.) however you can not predict the electron configuration by using the rule that the electrons will go into the sublevels of a given d-block atom, by first entering nd and then into the (n-1)p because, for example, the entire 3d sublevel is not at a lower energy than the entire 4s sublevel for Gallium, ONLY the 3d electrons that Gallium HAS are lower energy than Gallium's 4s electrons.
Google now lists the "b" answer for both Scandium and Gallium
2b) Ga = $1s^22s^22p^63s^23p^63d^{10}4s^24p^1$
1b) Sc = $1s^22s^22p^63s^23p^64s^23d^1$
https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=scandium%20electron%20configuration
xs electrons are higher than (x-1)d electrons that are present in a given atom
only the  (x-1)d electrons that are present in an atom are lower energy than the xs present in that atom.  You can't know for sure how many electrons will be in the (x-1)d and xs, but ones with 2 electrons in the xs will have the rest in the (x-1)p, its just that the xs electrons are higher.
d-orbitals still get "fuller" as you move across the P.T.


text:
Are the $+\leftrightarrow+$ and $-\leftrightarrow-$ methanol rotational transitions dipole-allowed?
I am quite confused about the origin and justification of the rotational selection rules of methanol, and I would appreciate any help or references. As far as I understand, there are three irreducible representations of the symmetry group of methanol ($A_1$, $A_2$,and $E$) and the two $A$-type are linked by dipole-allowed radiative transitions. Now, I think the literature also calls these ($A_1$ and $A_2$) $A^+$ and $A^-$ representations, I suppose because the parity of the eigenfunctions which generate the $A_1$ and $A_2$ are $+1$ and $-1$, respectively. Is this correct?
From what I have studied, because the electric dipole generates the $A_2$ representation, usually the allowed transitions follow the selection rule $A_1\leftrightarrow A_2$ (because $A_2\times A_2=A_1$), and I have seen this in many references. However, a lot of strong $A$-type methanol rotational lines are $++$ or $--$. I have noticed that all of these are associated with transitions with $|\Delta J|=1$ and all of the $+-$ and $-+$ transitions are associated with $\Delta J=0$. In fact, I have seen this statement as a selection rule in some papers (in this reference, they say that the $+-$ refers to K-type doubling ... ?), but it is not at all clear to me where does these rules come from.
I assume that the symmetry group of methanol is $C_{3v}$ (in books sometimes they call it $G_6$), and the representations above are usually the three which appear in the character tables. I think the group $C_{3v}$ represent permutations and permutation-inversions involving the protons of only the methyl group, because the other proton is too far and cannot be easily swapped with the others. The $J$ above represent the total rotational angular momentum of the molecule. Apparently, the usual approach is to express the wavefunctions as linear combinations of the product of symmetric top rotor wavefunctions and the wavefunctions  associated with the (hindered or not) rotation of the methyl group around the CO axis. I am interested in transitions at the torsional ground state ($\nu_t=0$) so I guess we can assume that the rotational potential is large. The $K$ I mention briefly above is the projection of the angular momentum in the main principal axis of the molecule, which is roughly aligned with the CO bond. Since methanol is asymmetric, there are two energy levels with the same $J$ and $K$ in the rotational spectrum, which is what I think they call the $K$ doubling.
I am in no way an expert on this, so if you want to point out inaccuracies in what I have just wrote I would be thankful.

crosspost of http://physics.stackexchange.com/questions/291290/are-the-leftrightarrow-and-leftrightarrow-methanol-rotational-transiti

It would help is you state  what you consider the point group to be, C1, Cs or something else. Are you worried about the restricted rotation of the OH group or do you ignore this, in which case the molecule is C3v. Have a look at http://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613  which may help in identifying transitions.

Thank you for your comment. I include a paragraph with more details.


text:
I suspended iodine over steel nails. What are these droplets that formed?
I suspended about 15g of iodine crystals over a couple of nails to see how quickly they would begin to rust.
After a week, these droplets had formed. When prodded they broke apart like water drops.
What are these? My guess is that the iodine sublimated and then combined with water vapor.
The original question that prompted this experiment.

The setup (~one week prior)

Iodine vapors combine with iron to form iron(II) iodide:$$\ce{I2 + Fe->FeI2}$$ In an anhydrous form the compound is dark red-purple. But iron(II) iodide is hygroscopic and deliquescent which means that it dissolves in moisture it absorbs from the air, and forms those droplets. This complicates a bit the chemistry of the droplets - the $\ce{Fe^2+}$ ion exists most likely in a form of various aquo-, hydroxo- and iodo- complexes. Also, when an iodide anion and elementary iodine are present together they form triiodide anions:$$\ce{I- + I2->I3-}$$ Therefore, the droplets are a solution of a rather complicated mixture of iron and iodine compounds.


text:
What is the difference between hydrogen chloride and hydrochloric acid?
My teacher insists that hydrogen chloride is a misnomer, and that the proper nomenclature demands HCl to be named hydrochloric acid. However, I saw the name hydrogen chloride used in my textbook, but it was never further explained.

The two are not identical. Hydrogen chloride is the anhydrous gas. Hydrochloric acid is an aqueous solution of hydrogen chloride. // PS - Forget arguing with the teacher. It isn't worth it...

@MaxW That is a very stupid and egoistic advise, to not argue with teachers. How will they ever learn? And the classmates and future pupils no less?

@Karl It depends on your situation. In my experience, if you are not extremely friendly with a teacher, it is impractical to argue straight out with them even if you are being polite. Some people just don't want to hear it.

Hydrogen chloride is a gas. It is the molecule made up of a hydrogen and a chlorine atom bonded by a single bond. It comes in gas cylinders for use in chemistry.
Hydrochloric acid is a solution (no, not to your problem of a stubborn teacher). It is composed of water molecules and hydrogen chloride molecules. In terms of mass percentage, concentrated hydrochloric acid is often $37~\%\ \ce{HCl}$ in water. It comes in glass bottles (often brown glass for ease of bottle use; the chemical company does not have to distinguish between compounds that can live in clear bottles and those that must live in brown ones) for use in chemistry.
Hydrogen chloride can also be dissolved in other solvents such as ether. In this case too, it may be termed hydrochloric acid, although a specification of the solvent should be added.
Whether or not either of the two names is a misnomer depends on what you are addressing. Using hydrogen chloride for the liquid solution is as wrong as using hydrochloric acid for the gas.

In ether, the Sigma Aldrich choice is actually "hydrogen chloride in ether."

Does HCl gas ionize  in ether and even if it does my book says that "An acid is a compound which dissociates in aqueous solution to produce H+ ions as the only positively charged ions"

@MrAP That’s using the Arrhenius definition. By the Brønsted-Lowry one, HCl is acidic in etheric solution (the ether oxygen acts as a base).

It is true for all solvents by Lowry-Bronsted definition or it is true only for ether?

@MrAP For all solvents that are Brønsted bases.

Pure HCl(g) is a covalent bonded compound and does not break into ions and is neutral to litmus and is not acidic. In case of HCl acid it is a monobasic acid and exhibits acidic properties.
Secondly HCl gas has smooth reaction with alkali metals whereas HCl acid has explosive reaction with alkali metals.


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Which is more stable chemically: DNA or peptide
Assuming enzyme-free environment, and the DNA and peptide are lyophillized, and then stored in vials under the same conditions (temperature and humidity).
On average, which would be more chemically stable,

in terms of backbone cleavage?
in terms of loss of side chain?

You know what holds each of them together, don’t you?

Yes, so is it correct that the answer boils down to the relative size of Gibbs Free Energy of the hydrolysis of Phosphodiester bond and Peptide bond? If it is true, where can I find these information? It seems that there's not much information on solid state stability.

Why solid state?

Because the samples are freeze-dried and stored in low humidity, so I can assume the samples to be in solid state.

I would consider the liquid state, though, since they won’t be cleaved in the absence of water.

Proteins are more stable, but this is only tre for collagen type 1 specifically. Not all proteins.

Well, more than 1 year has passed, and now the field of palaeo-proteomics is born, I think the verdict is clear:
The winner is PEPTIDE.
Still found intact in tens of millions old extinct animals, where DNA had degraded for long.
Source: https://www.nature.com/articles/nature14249


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What is the difference between a mixture and a solution?
Is there an actual difference, or are they synonymous?

Related: Are all homogeneous mixtures just solutions?

Let's try some examples for mixtures and solutions. Think about the answers yourself before you uncover them ;-)

We cut carrots, celery, and green beans into pieces and throw them into boiling water to prepare a vegetable soup. Mixture or solution?

It's a mixture, because the pieces of the different vegetables can be retrieved by a physical method, such as picking them with a fork.

Throw a spoonful of sugar into a cup of water and stir until there's no sugar left on the ground of the cup. Mixture or solution?

 This is a solution. It it homogenous; even the tiniest part of the water contains the same amount of sugar and there is no way to get back the sugar (unless you boil the water off).

Take a handful of razorblades and a handful of ants. Throw everything in a bowl. Wear gloves! Shake. Mixture or solution?

 This is a mixture. You can separate the components with a magnet - or by luring the ants out of the bowl with a trace of sugar water.

Melt iron. Add cobalt. Don't mix with bare hands ;-) Let it cool down until it is a solid again. Mixture or solution?

 This is a solution. Solutions do not need to be liquid. They can be solids as well. You can cut the metal block into pieces and mill every piece down to a fine powder and the amount of iron and cobalt in every grain of powder will be the same. Again, you cannot fiddle the components apart by any physical method.

The gist is clear, but "mixture" is not so clearly defined, i think.  Even "solution" does not have a clear-cut definition, if you try to avoid discussing it in terms of Gibbs free energy.  Think of a solution of virus particles. Or polystyrene in an organic solvent. Interesting question, actually.

Mixture- two or more substances physically combined and can be separated. Solution- a homogeneous mixture that has two or more substances in it


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Cycloaddition of Silyl Dienol Ethers with Sodium Bromodifluoroacetate: What's the scope?
I have recently came across this article "Copper(I)-Catalyzed [4 + 1] Cycloaddition of Silyl Dienol Ethers with Sodium Bromodifluoroacetate: Access to
β,β‐Difluorocyclopentanone Derivatives". (Org. Lett. 2016, 18 (18), 4502–4505)
Can anyone explain what is the use of these β,β‐Difluorocyclopentanone Derivatives or where can I find more information about this cycloaddition procedure.
Thanks in advance!


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Oxidation of dihydropyridine

This is a step in a experiment I have carried out. Usually DDQ (dichlorodicyanoquinone) would be used here to oxidise the dihydropyridine. However in this case Nitric Acid is being used as the oxidising agent. I'm unsure as to how this would happen mechanistically, my initial guess that since the product is aromatic and Nitrogen can help expel the hydrogen atom from the 4 position this will happen quite fast in one step. What role will the Nitric Acid have in this reaction? Mechanistic detail would be helpful, I can't find an example of this in any textbook.

This isn't an answer to your question on the mechanism, but did you consider to perform the oxidation of your Hantzsch DHPs with nitrite in methylene chloride using acidified silica (add silica to hydrochloric acid, stirr and evaporate) as a solid support?

@KlausWarzecha No, this is a reaction I performed in an undergrad lab. Simply dissolved the DHP in glacial acetic acid and added sodium nitrite drop wise, and left to stir for an hour. It specifically says not to worry about mechanistic detail in the procedure but I was curious how this step happened.

Did you mean nitrous acid? Not that I know the mechanism. Loss of H2 via an ionic mechanism has to involve a proton transfer and a hydride transfer to somewhere. The proton transfer isn't a big deal and under acidic conditions you might even get the pyridinium cation instead. In any case it's going to be some kind of hydride transfer to HNO2. Exactly where I don't know. Edit: Actually, I have a better idea. HNO2 isn't going to hang around, it's probably going to decompose to give water and $\ce{NO+}$. Now that's a suitable electrophile for "$\ce{H-}$" to attack.

@orthocresol Yes that does explain why it wasn't done in an ice bath either, keeping it cold would prevent it decomposing as much into NO+ but since it's basically done at room temp (advised to keep under 50 degrees) it's probably to ensure the NO+ electrophile is generated, so i presume the mechanism would work just like the one with DDQ but the hydrogen pushed towards the NO+ instead. I mentioned Nitric Acid because that's what Clayden said was the oxidising agent.

A quick Google search tells me that both HNO2 and HNO3 can be used as oxidising agents (really, a lot of things work, because there is a strong driving force for the oxidation of a dihydropyridine - namely aromaticity). Hydride transfer to NO+ gives you a very weird HNO species, I'd suspect that would decompose to give H2O + N2O, but again as far as I'm concerned all bets are off.

Again, this doesn't answer your question, but Arthur Hantzsch himself described in Ann. Chem., 1882, 215, 1-82 that - even under cooling - dropwise addition of nitric acid to a dihydropyridine, which he called "Hydrocollidindicarbonsäureather", led to a vigorous reaction in which a large amount of the starting material was destroyed. With nitrous acid, on the other hand, the oxidation was smooth.


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Air ionization, positive ions
Sharp air cleaners has a technology called PlasmaCluster. They say that it produces negative ions $\ce{O-}$ and positive ions $\ce{H+}$ (the difference with classical ionization is that ionization produces only $\ce{O-}$, PlasmaCluster also produces $\ce{H+}$)
The air contains oxygen molecules so oxygen can be ionized. But the question is how H+ is created ? I know that the air is $5.5 \times 10 ^{-5}\%$ Hydrogen gas. Is this small quantity of hydrogen in the air ionized, or is water in the air is split via electrolysis, creating the $\ce{H+}$ ions?

The plasmacluster supposedly works by ionizing water molecules in the air into $\ce{H+}$ and either $\ce{O^2-}$ or $\ce{HO-}$ ions. These ions then, supposedly, actively attach themselves to and break down pollutants, neutralizing their charges by pulling them apart. After they achieve this, the ions come back together to reform water.
At least, this is what their website says. They are correct that water can be split into positive and negative ions as well as the fact that these ions can come back together to reform water. I am unsure that the ions will stay separated from each other long enough to break up pollutants, and it is not very clear how positive and negative ions break up "pollutants." I find it unlikely that these ions will kill bacteria, which is one of the claimed benefits of this type of ionizer (I am not a biologist, so I could be wrong about this). Even if these ions are fatal for bacteria, odds are the plasmacluster cannot produce concentrations sufficient to do this. Also, if it could produce such concentrations, I might be somewhat concerned about what those concentrations of ions could do to humans.