is it possible to store formic acid longer if frozen? I would like to know if formic acid (concentrated to 85%) is stable (no chemical reaction) when frozen.
Is it a good solution to store it for many years?

Formic acid is thermodynamically unstable to decomposition to water and carbon monoxide: HCOOH ==> CO + H2O. I assume this is what you are referring to.

A common use is acidification of silage to prevent smells. The fact that it decomposes makes it more environmentally friendly than alternatives. I was not aware that it was used in bee keeping to control varroa parasitic mites.

Lowering the temperature would reduce the decomposition rate from both a thermodynamic and kinetic point of view.

The MSDS I have looked at indicate that the bottle should be vented periodically to release the pressure built up from decomposition. However this one also says that formic acid should not be stored below its freezing point of 8C. http://ionics.ca/wp-content/uploads/MSDS-Formic-Acid.pdf Whether this is just for convenience of use or for reasons or safety I do not know.

To me, the build up of pressure seems far more of an issue than any loss of strength in the acid. In order to lose 50% of its strength, a 1L bottle of formic acid (1200g/46=26mol) would have to lose 13mol (about 312L) of carbon monoxide. That's a lot of venting! I would think a bottle should have a shelf life of 2 or 3 years.

I did my university chemical engineering project on a formic acid plant. It is made as follows:

1. CO + Methanol (20 bar, 80C, potassium methoxide catalyst) == Methyl Formate
2. Methyl Formate + H2O (acidic autocatylisis)== Formic acid + Methanol

From this you might conclude that formic acid is stable to decomposition at 20 bar, but that is not the case, because excess reactants and careful use of distillation are used to push both reactions in the desired direction.

That was a long time ago, but I will see if I can find any relevant data in the decomposition reaction. However you would probably be best off just following the manufacturer's recommendations.

Oh, and keep it in the original glass bottle. Many metals catalyse the decomposition.


Why is the solubility of calcium sulfate so low even though it is a salt? What is the reason the solubility of calcium sulfate in water is so low? Compared to other salts like potassium sulfate (110 g/L) or calcium nitrate (1470 g/L) the solubility of calcium sulfate seems to be very low (2.4 g/L).

Solubility of calcium sulphate is quite high, compared to many much less soluble salts, like barium sulphate, calcium phosphate or calcium fluoride.

Most of insoluble minerals are salts.

Insoluble/limited solubility salts have as crystals lower Gibbs energy than dissolved, leading to solution being thermodynamically  unfavourable.

That is related to mutual relation of energy of crystal lattice and of hydration of ions, and also to dissolution entropy change.


How to count free rotatable bonds I am having difficulties to count the number of free rotating bonds for the Lipinski and Veber rules. What are the rules in doing so?

In Vemurafenib (shown above), which bonds are considered rotatable? Do the sulfonamide N-S bond, or the single bond linking two aryl groups, rotate?

Many thanks

The idea of a 'rotatable bond' is somewhat arbitrary, given that under the right conditions basically any bond can be rotated. When thinking about concepts in medicinal chemistry like rotatable bonds, its therefore important to make sure you're working with the definition used when defining the rule.

The original Veber papers (J. Med. Chem. 2002, 45, 2615) define what they consider to be a rotatable bond:

  Rotatable bonds were defined as any single bond, not in a ring, bound to a nonterminal heavy (i.e., non-hydrogen) atom. Excluded from the count were amide C-N bonds because of their high rotational energy barrier.

Based on this rule, the Ar-Ar single bond would count as a rotatable bond, as would the Ar-N bond. The sulfonamide is generally excluded for the same reason as amide C-N bonds (although I can't find a quote by Veber that actually discusses this).

In reality, these properties are usually calculated automatically, for instance using ChEMBL.


To which state of matter does the flame belong to? I had this question from the day (9 years old, now 16) that I learned about states of matter. I have asked many of my teachers, some of them told me it's a gas some that it's a plasma. Can anyone answer my question?

Recently I've learned that the plasma state is obtained when all the electrons from the atom are removed. Obviously it's present in the sun due to high temperatures.

A candle flame is the product of the oxidation of wax to $\ce{CO2}$ and other small organic compounds. Since these small compounds are energetically lower, energy is released during the reaction, in the form of light and heat. So really, we start out with gas.

However, flames can also conduct electricity, and the conductivity varies with the location in the flame. The linked science fair project summary states that the lowest resistance (highest conductivity) was found at the edge of the bright yellow part of the flame the conductivity was the largest, meaning that the concentration of ions was the highest. The lowest resistance value found was $80~\Omega$, which is still very large in comparison to the resistance of plasma, which is generally considered to be $0$.

So, summarizing my answer: A flame is a gas with lightly ionized portions, but does not qualify as a real plasma.


Why is an alkoxide released instead of hydroxide in the attack of trialkyloxyborane by hydroxide? In the mechanism of a hydroboration-oxidation, the oxidation part includes a nucleophilic attack of a hydroxide on a trialkyloxyborane ( $\ce{B(OR)_3}$ where $\ce{R}$ is an alkyl group).

It creates a tetrahedral intermediate consisting of a negative center of boron with four groups attached to it (the hydroxide nucleophile and the three alkoxides). This step is followed by the loss of an alkoxide as the leaving group. However, to my understanding, the alkoxide is a stronger base than the hydroxide. Thus, why doesn't the nucleophilic attack fail?

Edit: Is it because of steric comparisons between the size of the alkoxide versus the hydroxide (since boron is a relatively small atom)?

Both hydroxide and alcoholate can be displaced from the tetracoordinated intermediate. The difference is that displacing the hydroxide is just the back reaction of the inital attack while displacing the alcoholate will move the reaction forward.

You cannot argue that $\ce{X}$ is a stronger base than $\ce{Y}$ and therefore $\ce{X}$ will always get displaced. At best, you can argue that $\ce{X}$ is ten times stronger so the probability of $\ce{X}$ begin displaced is higher than that of $\ce{Y}$. For all intents and purposes of the argument, the basicities of (primary) alcoholates and hydroxide are too similar to make a real difference.


Why do some materials turn black and others white after combustion? Most solids objects turn black after they are burnt.
But some turn white.
Why is that ?

The objects which turn black are usually carbon based/organic compounds. When burned/combusted they follow one of the following unbalanced reactions:
\begin{align}
\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &\ce{->} \ce C_\text{(solid)} + \ce{H_2O}_\text{(liquid)} + {\ce{CO_2}}_\text{(gas)}\\
\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &\ce{->}  \ce{H2O}_\text{(liquid)} + {\ce{CO2}}_\text{(gas)}\\
\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &\ce{->}  \ce {H2O}_\text{(liquid)} + {\ce{CO2}}_\text{(gas)} + \ce{CO}_\text{(gas)}
\end{align}

You'll notice that some of the reactions will produce solid carbon, which is the black thing you see.

On the other hand, inorganic compounds begin to decompose into their ions and/or release their  "trapped" water molecules. As you'll notice, most ionic compounds are white solid salts.


Pi electron stacking, how does it work? I've come across the term base-pair stacking (with reference to B-DNA) in my school text book, and I had posted a question in that regard on Bio.SE.

I've also seen a similar (albeit brief) version of my question on Chem.SE.

I looked up the term online, and after checking out a couple of links, I'm led to believe that this term stacking refers to the forces of attraction between aromatic rings on account of the de-localized pi-electron clouds on either side of their (the aromatic rings) planes.

Now I'm not sure if I skimmed over some important details when I read up on 'stacking', but I'm unable to find a simple and solid explanation as to how these interactions are brought about.

If 'stacking' is the interaction between pi-electron clouds, then shouldn't these interactions be repulsive (on account of 'like-charges repel')? But I'm told they're attractive interactions.

I thought about it for a bit, and I realized that if you consider the fact the the pi-electrons, despite being de-localized all over the plane of the aromatic ring, can only be found at one particular location at any given instant of time, resulting in a 'partial' negative charge being formed there. The other regions of the electron cloud probably acquire a 'partial' positive charge on account of the fact that the electron is not present there at that instant. It's these opposite, partial charges formed that results in the attraction. (Essentially the same idea that governs London Dispersion Forces)

But the thing thing is I'm not sure if my 'hypothesis' is correct in the present case. If this isn't what happens then I've no idea what makes 'stacking' attractive interactions.

So my question stands: How exactly does stacking give rise to attractive interactions between aromatic ring? Why don't they result in repulsive interactions?

You going along the right track. The stacking attraction occurs when two molecules with $\pi$ orbitals come face to face with one another, typically their separation is 0.34 nm. However, for the interaction to be attractive the two molecules have to be displaced slightly, this prevents direct electron repulsion between $\pi$ electrons on one molecule with those on the other. The attraction is in general between the $\pi$ electrons on one molecule and the $\sigma$ framework on the other$^1$ and vice - versa.

The model assumes that a $\pi$ orbital can be split into three charged parts, separated by a distance $\delta$. The lobes of the orbital have charge -1/2 each and at the position of the atom's nuclei is placed a charge of +1. Thus there are also three interactions $\sigma - \sigma , \pi - \sigma $ and $ \pi - \pi$ and three positions of charge, making 9 interactions in all for each atom. The interaction is taken to be electrostatic, i.e. Coulomb and each term has the form $E_{ij} = q_iq_j/r_{ij} $ for atoms i and j and is summed over all atoms.

In addition a van-der-Waals term is added to the calculation of the total interaction, this has the form $E_{vdw} = \Sigma _{ij}(A_{ij}exp(-\alpha_{ij}R_{ij})-C_{ij}/r_{ij}^r))$ where A, $\alpha$ and C have standard values$^2$. In solution the value of this energy is much reduced vs. its value in the crystal making the $\pi-\pi$ interaction the dominant one.

The charges are shown schematically below for carbon atoms. For other atoms, e.g. nitrogen the charge is given as 1.5 with 1.5 electrons associated with them. (All images are from ref 1 )

charge distribution

optimal geometry for Zn porphyrin $\pi-\pi$ interaction. The molecules are displaced (in x and y) relative to one another at a separation of 0.34 nm.

The contour plot for $\pi-\pi$ interaction and van-der-waals (vdw) energy as a function of their centre to centre offset as shown above (energies in $\pu{ kJ mol^{-1}}$). The black squares are geometries from $\pi-\pi $ stacking in crystals. The shape of the contour plot with or without the vdw interaction is the same as with just the $\pi-\pi$ interaction alone, i.e this interaction determines geometry but  the vdw generally lowers the energy and more clearly show regions of low energy as negative values.

$^1$ The model was developed by Hunter & Sanders J.Am. Chem. Soc. 1990, v112, p 5525-5534

$^2$ Van der waals parameters Caillet & Claverie, Acta Crystallogr. sect A1975, v31, p448


Electrochemical cell-SRP value On what basis can the SRP value of different elements in the periodic table be explained?
For example why does Cu have a greater tendency to accept an electron and Zn have a tendency to lose electrons in a Daniel cell?

There are many factors and the ideal way to understand this would be with the quantum mechanical description of atoms, which gets too complicated for answering such questions. But in general, following are few simple-to-understand face that helps in understanding the SRP values-

Electronegativity and Electron gain enthalpy: Higher the electronegativity, greater is the tendency to accept electrons, be reduced and hence greater (more positive) is the Reduction potential. For example, reduction potential of fluorine is greater than chlorine and that is greater than bromine. As we move down the halogen group, electronegativity decreases.
Ionization potential: This is a consideration for metals, which usually loose electrons and hence electronegativity is not the right quantity to consider. The lesser the IP, easier is the oxidation, and more negative (lower) is the SRP. In gas phase, the SRP of cesium is the most negative (in the group) and it becomes less negative as we move upwards.
Hydration: This is an important fact to consider since most electrochemical reactions occur in aqueous phase and hence the commonly cited SRP values corresponds to the aqueous phase. Greater the hydration of the ion formed after oxidation/reduction, greater is the tendency to lose/gain electrons and hence lower/higher is the SRP values. Lithium ion has very high hydration enthalpy and hence lithium has a strong tendency to lose an electron to form an ion which can be excellently solvated. This lends it one of the most negative SRP in the periodic table. The same holds true for fluorine, which is excellently solvated in the ionic form by water and hence has a high SRP.
Size, EC etc: Several factors like ionic size, it's electronic configuration, the inner orbital structure and the nuclear force of attraction determines how the incoming electron will be received and consequently the SRP values. Higher atomic size, higher inner electron repulsion, almost-stable configuration(just 1 electron more) and lower nuclear force, all favour lower SRP values.

All of these consideration, in varying proportions can be applied to the transition elements, but they act differently and to different extent in different compounds. Hence, predicting the SRP order without experimental data is very difficult and not trustworthy.


The difference between CASSCF and MCSCF methods I'm not sure the exact difference between the MCSCF and CASSCF methods. I know that the CASSCF method proceeds by optimizing both the CSFs and MOs (I mean the coefficients), but I don't know if this is also the case in MCSCF calculations.

My guess is that in MCSCF only the coefficients of CSFs will be optimized, but the HF reference MOs would be the same.

Both methods optimize orbitals and CI coefficients. MCSCF is the more general term, while CASSCF is a special type of MCSCF.

As the name Multi-Configurational Self Consistent Field already implies, it combines the Configuration Interaction method with the SCF method (Hartree-Fock). Since this combination increases complexity of the wave function, one usually applies restrictions to both sets of parameters (CI and orbital coefficients). For example core orbitals may be kept fixed (frozen) or only a small subset of configurations is considered.

Regarding the selection of configurations, there are various possible approaches. For example:

hand selecting configurations using chemical intuition or some other a priori knowledge
selecting a subset of orbitals, which should at least cover the HOMO and LUMO. In this active space one considers all possible configurations, hence Complete Active Space Self Consistent Field (CASSCF)
restricting configurations only to certain excitation levels (e.g. Singles and Doubles) within the active space: Restricted Avtive Space Self Consistent Field (RASSCF)

So CASSCF is a special variant of MCSCF, therefore there is not really a difference between the two. Although not every MCSCF is necessarily  a CASSCF, the later is the most common type.


On Spontaneity of the Redox Reactions According to this Mc Graw-Hill link during a spontaneous redox reaction Gibbs Free Enthalpy must decrease and at the same time the change in the Gibbs Free Enthalpy is the maximum electric work that the reaction can do:
$$\Delta G = W_\text{max}=-n F E_\text{cell}$$
in which $F$ stands for the Faraday constant, the amount of charge carried by one mole of electrons, and $n$ represents the number of moles of electrons to be transferred from the reducing agent to the oxidizing agent
according to the balanced equation, and $E_\text{cell}$ is the electromotive force of the cell, which equals $V_\text{cathod} - V_\text{anode}$, or alternatively, $E_\text{oxidizer} - E_\text{reducer}$ wherein $E$ here stands for the reduction potential. As $\Delta G$ must be nagative in the spontanous direction so $E_\text{cell}$ must be positive.

However, according to the first and second law we have

\begin{align}
&\delta Q-\delta W=\mathrm dU=T\mathrm dS-p\mathrm dV+\sum_i\mu_i\mathrm dN_i\\
&\Rightarrow\;\delta W-p\mathrm dV+\sum_i\mu_i\mathrm dN_i=\delta Q-T\mathrm dS\le0\qquad\because\text{2nd law}\\
&\Rightarrow\;\delta W-p\mathrm dV+\mathrm dG\bigr|_{p,T}\le0\\
&\Rightarrow\;\delta W-p\mathrm dV\le-\mathrm dG\bigr|_{p,T}\\
\end{align}

If $\delta W=p\mathrm dV$ then $\mathrm dG\le 0$, that is, the condition $\mathrm dG\le 0$ coincide with the 2nd law only id the only work done by the system is the pressure work.

Also if $\delta W=p\mathrm dV+\text{other works}$, then $\text{other works}\le-\mathrm dG\bigr|_{p,T}$. This means the change in the "minus Gibbs function" is the maximum work attainable from the system beside the pressure work. This extra work can be positive or negative, in the form of electric work or friction work or etc.

Now here I have some problems with the content of the link provided at the beginning of my question, as follows:

First of all, the maximum work that a system can do on its surrounding equals minus the change of the Gibbs Free Enthalpy, so the equality $\Delta G=W_\text{max}$ doesn't hold!?
If $\Delta G=W_\text{max}$ is valid, that is, I mean if the work attained is the maximum possible, then the process must be assumed as reversible , but if the process is reversible thenwe will need $\Delta G=0$ and not $\Delta G<0$, so that the process should be at equilibrium and spontaneous in no direction!?
That the Gibbs Free Energy must decrease in a closed system in the constant temperature and pressure is true only when the only type of work which is done is in the form of pressure work in expansion of the system's volume, not if other works like friction or electric works are also available, so that again using at the same time $\Delta G=W_\text{max}$ and $\Delta G\le0$ is misleading!?
The electric work is an internal work for a system containing the redox reaction and shouldn't enter the above formulation at all, as by work in the 1st law formulated above we mean only those work-based energy transport that occur at the boundaries of the system. So the electric force can never enter this formulation unless we change the control system under study, isn't it right?
And the main question. if I was wrong in analyzing the content of the provided link (which is very probable) where is my mistakes and if I was correct then how can I obtain the same result (which is certainly correct), that "in a spontaneous redox reaction we should have $E_\text{oxidizer}-E_\text{reducer}\ge0$"?

Question 1:

You are mixing the sign conventions for work from physics and chemistry. Sadly, they are not the same. In chemistry, the work done by the system on the surroundings is negative to signify energy leaving the system. Any transfer of energy (heat, work, etc.) from the system to the surroundings is denoted with a negative sign to ensure the energy change of the system is negative (energy is lost). Chemistry is concerned ultimately only with the system (usually the chemical reaction under investigation), and not the surroundings (apparatus, equipment, experimenter, laboratory, and the rest of the universe).

Thus, for pressure-volume work:

$$W=-p\Delta V$$
$$\delta W=-pdV$$

We need this negative sign. The system does pressure-volume work on the surroundings when the system expands (think a piston in an automotive engine) so $dV>0$. Without the negative sign, we would assume that the system gains energy and the surroundings lose energy when the system expands.

Thus, in the chemistry formulation of the first law (assuming only pressure-volume work) is:

$$dU=\delta Q + \delta W =TdS - pdV$$

In physics, some consideration is given to the surroundings (and in some cases the surroundings may be more interesting than the system, such as in the proper definition of entropy). Especially in the early days of physics, work leaving the system was useful in that it could be utilized. Thus, in physics $W=p\Delta V$ means work is being done by the system on the surroundings. Thus, the first law as formulated in physics is:

$$dU=\delta Q - \delta W = TdS -pdV$$

Note however, regardless of the sign convention in the definition of $W$ as a function of $p$ and $\Delta V$, we still end up with $$dU=TdS-pdV$$

The sign of $W_{\text{max}}$ in chemistry needs to be negative to follow the same conditions of chemistry. Thus, the equation in your chemistry textbook is correct. An equivalent definition in physics would be:

$$W_{\text{max}}=nFE_{\text{cell}}$$
$$\Delta G_\text{cell}=-W_\text{max}=-nFE_\text{cell}$$

In chemistry the negative sign goes in the definition of work. In physics the negative sign goes in front of the work term. Don't mix and match conventions, and you will get consistent answers.

  Question 2:

$\Delta G = 0$ is the definition of equilibrium not the conditions for reversibility. A reaction can be reversible (and most electrochemical reactions are) and still be far from equilibrium at the initial conditions.

  Question 3:

Temperature is not required to be constant here. Also, note that the term $W_\text{max}$ refers to the maximum possible electrochemical work. Usually less electrochemical work is done because heat is also generated (which changes the temperature). Also, most electrochemical systems are not closed. Even a battery is not a closed system (by the strict physics definition). It's hard to maintain constant pressure in a closed system.

  Question 4:

Technically the formulation of the first law that you used is explicitly for pressure-volume work done by closed gaseous systems. As most redox reactions occur in solution in open systems, that formulation is invalid for a different reason.

  Question 5:

Given that your initial equation is correct, then for a spontaneous reaction:

$$\Delta G < 0$$
$$\therefore -nFE_\text{cell} < 0$$
$$nFE_\text{cell} > 0$$
$$E_\text{cell} > 0$$
$$E_\text{red} - E_\text{ox} > 0$$


Ductless fume hood for acetone? Administration on my campus has become worried about the lack of a fume hood when using acetone in the 3D Print Club. I'm helping in the process of picking one out.

They do acetone vapor smoothing on their prints, but I know they have a box to do that in. Other than that, it's acetone in a bottle for dissolving support material and plastic residue.

Does there need to be a fume hood with acetone?
If I can get a ductless fume hood, what am I looking for to be sure that it can effectively take care of the acetone problem?

I need to prove to administration that the club is fixing the problem, and I frankly have no idea what to look for in fume hoods.

Does there need to be a fume hood with acetone?

Unless you're using very large quantities of it or in a confined space, usually not. I've spent a fair bit of time washing glassware out with acetone at the sink (using a bin to capture the runoff for transfer to a waste container) and I never had issues. Though I don't recommend doing this yourself, I've also had several errant splashes directly on my skin when working with stuff from the hardware store--again, no problems.

Acetone is staggeringly nontoxic: the NIH suggests an 8500 ppm EEL, meaning that in emergency situations, it's considered acceptable to breathe air that is nearly one full percent acetone for up to one hour. Hell, the air in my house contains less water than that sometimes.

  If I can get a ductless fume hood, what am I looking for to be sure that it can effectively take care of the acetone problem?

Look at this link. If the acetone is heated in the box, you should be able to meet all the criteria for being able to use a ductless hood:

A limited number of different chemicals should be used.

Acetone only. Check.
No extreme heating should be carried out in the hood, i.e. acid digestion applications.

You only need to boil acetone. That's 56C, not even close to the boiling point of water. Check.
Modest chemical volumes should be used, around 500 mls or less per chemical.

Even if you need more than this (you shouldn't, 500mL of acetone is quite a bit), it'll mostly be sealed in a box, so you shouldn't have nearly that much escaping into the hood. Check.
Moderate chemical exposure times should be maintained.

I don't know exactly how long it takes to smooth a print, but I don't think it can really be more than a few hours. You're probably okay on this one.
Will the chemicals involved in your application be effectively filtered with the available carbon filters?

Yes. Acetone is so common I can all but guarantee that every company will have a filter that can capture it (probably whatever is used for other general organic compounds as well). In Labconoco's case, that's the organic vapor filter.
How often will you have to replace your filters?

I don't know, that'll be your club's responsibility.

Finally, make sure someone doesn't get the idea that the hood will stop all fumes and try to do a perchloric acid experiment using that hood--that will go badly.

Now, all this being said...

What you really need isn't to make this safe, what you need is to convince administration that you're doing things safely. This may have nothing at all to do with whether things are actually safe or not.

At my old school (where I was briefly president of the local ACS chapter), I could have put an experiment into a fume hood with no filter, not plugged the thing in, and gotten an okay from the administrators because they believed that fume hood = safe and no fume hood = unsafe no matter what else was going on.

I say this not to encourage you to try to trick your way out of this, but to point out that, unless your administration is scientific (i.e. actually understands science), then it's fairly likely you're going to have to jump through a few silly hoops in order to satisfy them.

Best of luck!


Finding the concentration of a solution Q.) Determine the concerntration of a solution made by dissolving 10.0g of sodium chloride in 750.0ml of solution.

I did the following; 1mol of sodium chloride = 58.44

58.44*10 - 584.4g
Concentration = number of moles/volume
584.4/0.750 = 779.2 - and this is the wrong answer.

Any pointers would be appreciated.

The molar mass $M$ of $\ce{NaCl}$ is $58.44\ \mathrm{\frac{g}{mol}}$ indeed:

$\displaystyle\text{molar mass}(\ce{NaCl}) = \frac{\text{mass}}{\text{amount of substance}} = 58.44\ \mathrm{\frac{g}{mol}}$

$\displaystyle\Rightarrow \text{amount of substance} = \frac{\text{mass}}{\text{molar mass}}=\frac{10.0\ \mathrm g}{58.44\ \mathrm{\frac{g}{mol}}} \approx 0.171\ \mathrm{mol}$

$\displaystyle\text{concentration} = \frac{\text{amount of substance}}{\text{volume}}=\frac{0.171\ \mathrm{mol}}{0.75\ \mathrm L}=0.228\ \mathrm{\frac{mol}{L}}$

Assumptions: Adding $\ce{NaCl}$ to these $750\ \mathrm{ml}$ does NOT change the volume noteworthy.


Energy required to break the "delocalised electron cloud" of benzene I learned that when benzene undergoes hydrogenation, $\pu{208KJ}$ of energy is given off and thus $\pu{152KJ}$ less energy than what would be given off if Kekule's structure was correct. My confusion is that my book says that this $\pu{152KJ}$ of energy is required to break the delocalized electron cloud - this does not make sense to me.

How is the energy discrepancy of value the energy required to break the delocalized electron cloud in the benzene?

Let's sum it up :)

When cyclohexene is hydrogenated, $\mathrm{-120\, kJ\cdot mol^{-1}}$ is released.
When 1,3-cyclohexadiene is hydrogenated, $\mathrm{-232\, kJ\cdot mol^{-1}}$ is released. That's pretty close to 240, so it looks that the energies for the hydrogenation of the double bonds sum up. So far, we're cool.

Based on Kekule's model of alternating single and double bonds in benzene, we would now assume that the energy for hydrogenation is $\mathrm{ 3 \times -120 = -360 \, kJ\cdot mol^{-1}}$.

But as your textbook says it's only $\mathrm{-208\, kJ\cdot mol^{-1}}$.

We must conclude that Kekule was almost right, but the bonding situation in bezene is special. The energy of the molecule is much lower than that of a hypothetical cyclohexatriene. And when it's lower in energy, less is released upon hydrogenation. The difference of $\mathrm{152\, kJ\cdot mol^{-1}}$ is attributed to resonance stabilization.

Edit 1 If that's still a bit unclear with the signs of the energies, we can do some reverse experiments in mind.

Let's take cyclohexane and rip adjacent hydrogen atoms off to generate unsaturated molecules.

We need $\mathrm{120\, kJ\cdot mol^{-1}}$ to make cyclohexane.
We need $\mathrm{232\, kJ\cdot mol^{-1}}$ to make 1,3-cyclohexadiene.
If benzene would be cyclohexatriene, we would need $\mathrm{360 \, kJ\cdot mol^{-1}}$.

But we need less energy to make benzene, because it is more stable than a hypothetical cyclohexatriene. Actually, it is even more stable than 1,3-cyclohexadiene!

Edit 2

  Benzene's actual structure is stable by 152 KJ w.r.t. Kekule's structure. That 152KJ is used for breaking down the delocalised electron cloud of benzene.

Guess what? The teacher is just repeating with other words what we have already worked on our own.

Imagine that you want to convert benzene to the hypothetical Kekule structure.

It's a process that would "consume" energy, $\mathrm{152\, kJ\cdot mol^{-1}}$ to be precise.

Why is that? It's because benzene has the delocalized $\pi$-system that aromatic molecules use to have. If you want to overcome that and push the stable molecule uphill and turn in into the less stable (= not having the delocalisation) cyclohexatriene, you have to do some work.


Is there any relationship between ultrasonication and temperature In chemistry we use ultrasonication for cleaning vessels ,synthesis of chemical compounds (sonochemistry) etc.I know that during ultrasonication high temperature and pressure is formed "locally".But I want to know whether there is any equations which relates between ultrasonication and temperature .

The temperature is highly nonlinear with sound pressure and dependent on external pressure, tensile strength of liquid etc. Cavitation, http://en.wikipedia.org/wiki/Cavitation, can produce local shock waves and temperatures of thousands of K, causing sonoluminescence. See also http://www.scs.illinois.edu/suslick/sonochembrittanica.html.

Because of that nonlinearity and all the dependencies, I do not know if any equation for calculating the peak temperatures exists, and there is still research going on to see how much of sonoluminescence is due to black-body radiation, the recombination of radicals etc.

BTW, you can view a mantis shrimp with a sonoluminescent punch at https://www.youtube.com/watch?v=sNzI8K-lCqo and https://www.youtube.com/watch?v=ti2Uoc1RXuQ, and another species at https://www.youtube.com/watch?v=QXK2G2AzMTU.


Is there a type of detergent/surfactant that evaporates at STP? Are there any type of detergent/surfactant chemicals that would be good for removing dirt and grease from fabric that also would evaporate from the clothes within, say, 24 hours leaving no residue?

Googling "volatile surfactant" turned up fluorocarbons and something called Surfynol 61, which is really 3,5-Dimethylhex-1-in-3-ol as possible candidates. I don't know if these would be safe for laundry.

Volatile surfactants

There are many molecules that are both surfactants (in water) and volatile, but of the ones I can think of, none can be used safely in home laundry applications.

Hydrocarbon derivatives.  The examples you found by googling are in this class.  3,5-dimethyl-1-hexyn-3-ol is the principal component of Surfynol 61 (probably named Surfynol 61 because it is a surface-active (surf) alkyne alcohol (ynol) with six carbons (6) and the alkyne moiety at carbon one (1).  Other examples would be medium-chain fatty alcohols like hexanol, octanol, cyclohexanol, etc.  These molecules reduce surface tension of water and have OK-ish cleaning power (not as good as larger but non-volatile alternatives like SDS), but their volatility is also their downfall: their use in home environments creates a strong fire and explosion hazard when used in automatic heated dryers.
Fluorocarbon derivatives.  I found the same article you probably did about volatile fluorocarbon surfactants.  The most powerful at reducing the surface tension of water seems to be nonafluoro-tert-butyl alcohol, which reduces water surface tension by 78%!  However this experiment required saturating the water with the vapor of this compound.  And 1 gram costs >$80.  And toxicity is a concern, despite this compound apparently being researched as a component of or precursor to artificial blood mixtures in the 1970s.
Other halocarbon derivatives.  These include solvents like perchloroethylene, chlorobutanol, etc.  These are all highly toxic, and with some of them, there is a flammability/explosion risk as with hydrocarbon derivatives.

Non-surfactant use

You might consider just using organic solvents that are not (very) surface active as a way to degrease fabric.  However, these may leach dyes out of the fabric or otherwise damage them.  Isopropyl alcohol (IPA) is accessible to home users, and is relatively safe.  Be sure to do the soaking in a well-ventilated area, and do not use heated dryers to remove the IPA!  Instead, rinse out the IPA thoroughly with water.  Depending on local regulations, you will have to dispose of all of the dirty IPA and IPA-contaminated rinsing water as solvent waste.

The best option

Pay to have your clothes dry cleaned!


Calculating the pH of the endpoint in a titration of weak acid and strong base Show that the pH at the endpoint in the titration of $\mathrm{0.100\ mol\ dm^{-3}}$ ethanoic acid with $\mathrm{0.100\ mol\ dm^{-3}}$ sodium hydroxide is 8.72. The dissociation constant Ka of ethanoic acid is $\mathrm{1.80\ x\ 10^{-5}\ mol\ dm^{-3}}$.

I managed to obtain the value of Kb using the formula (Ka x Kb = Kw = $\mathrm{10^{-14}}$).

I worked out the chemical equation at the endpoint which is:

$$\ce{CH3COO- + H2O <--> CH3COOH + OH- }$$

However, I am unsure as to what the concentration of $\ce{CH3COO-}$ should be.

I understand that if I find [$\ce{OH-}$] then I would be able to find pOH and by extension pH.

Would it be correct to assume that the concentration of $\ce{CH3COO-}$ to be the concentration of ethanoic acid that was present at the start of the titration ($\mathrm{0.100~mol~dm^{-3}}$), given that at the equivalence point, all $\ce{CH3COOH}$ had reacted to form a solution containing only the conjugate base $\ce{CH3COO-}$?

You can easily rearrange the $\text{K}_\text{a}$ equation to solve for the information that you need to answer that question.

$$\text{K}_\text{a} =\frac{\ce{[H^+][A^-]}}{\ce{[HA]}} $$
or
$$\frac{\ce{[HA]}}{\ce{[A^-]}} = \frac{\ce{[H^+]}}{\text{K}_\text{a}} = 1.06\times10^{-4}$$

Since $1.06\times10^{-4}$ is insignificant compared to 0.100 you can assume that essentially all the HA is converted to $\ce{A^-}$.

NOTE: This does not mean that $\ce{[HA] = 0}$


Nuclear reactions involve naked nuclei only. True or false? In nuclear reactions are all the species involved, naked nuclei that is, are they all atoms that have been completely stripped of their electrons? If not, then shouldn't an alpha decay result in the formation of an anion and shouldn't a beta decay result in a cation?

Nuclear reactions involve energies far greater than anything in chemistry. Whether or not the nuclei are stripped of their electrons, does not matter at all (except for electron capture, of course; to capture an electron, you must have one around in the first place).

So yes, a beta decay of a neutral atom does indeed result in a cation. As for alpha decay, here we have a massive particle leaving, so the remaining nucleus would be shoved in the opposite direction with considerable force, no doubt enough to break it free of any chemical bonds and strip it of quite a few outer electrons. If not for this factor, the process would leave behind an anion.


Is it possible for the forward and reverse reactions to have different reaction orders? For example, in the case of non-dissociative adsorption, the rate of the forward process is given by

$$\frac{\mathrm{d}\theta}{\mathrm{d}t} = kCN(1−\theta)$$

where $C$ is the concentration, $N$ is the number of surface sites, and $\theta$ is the fraction of filled sites.

The reverse reaction is given by

$$\frac{\mathrm{d}\theta}{\mathrm{d}t} = −kN\theta.$$

Hence, it seems like the first reaction is second-order, while the second is first order. This seems strange to me.

Why is this strange?

Consider:

$$\ce{2A <=>[$k_1$][$k_{-1}$] A_2}$$

Forward rate is second order: $k_1[\ce{A}]^{2}$.

Reverse rate is first order $k_{-1}[\ce{A2}]$.


Methylene "anti" Jahn-Teller effect Yesterday a Reddit user posted a page from Morrison's Organic Chemistry in which it is said that singlet methylene is less stable than triplet methylene. Another user asked basically the same I'm asking here, as he got no answer and probably won't get any.
I'm used to see radicals (and therefore diradicals) as extremely reactive and lone pairs as stable, when not reacting in some Lewis base fashion.
Also, Jahn-Teller effect explanations usually say that unpaired electrons configurations with each electron on a degenerate orbital are less stable than the paired configuration with some deformed structure, and therefore the structure deforms to "de-degenerate" the orbitals and achieve a lower energy configuration, hence the effect.
What we see here is exactly the opposite, isn't it?
What is it then that makes the triplet state more stable?
Wikipedia confirms the observation and adds a value to the energy difference: 38 kJ.
Wikipedia also says it is normal for carbenes to be more stable in the triplet state.

The Jahn-Teller effect is not relevant here. The triplet state is observed in many molecules to be of lower energy than the singlet. The Pauli exclusion principle, makes the states symmetric or anti-symmetric to the exchanging electrons.
A triplet state has unpaired spins while in a singlet they are paired. The triplet is lower in energy than the corresponding singlet. One reason for this is that the electrons try to ‘avoid’ one another in the triplet; (see Salem 'Electrons in Chemical Reactions', publ Wiley 1982 for a detailed discussion). This means that the spatial part of the triplet wavefunction is asymmetric to exchange of the electrons;
$$\varphi_T = \psi_1(1)\psi_2(2) − \psi_1(2)\psi_2(1) $$
and the singlet spatial part is symmetric which means that the electrons tend to crowd together raising the energy
$$\varphi_S = \psi_1(1)\psi_2(2) + \psi_1(2)\psi_2(1) $$
The numbers in brackets refer to the electrons’ coordinate. As the total wavefunction is anti-symmetric to exchange of electrons the corresponding spin parts have the opposite symmetry to the spatial parts.
If the spins are on two separate atoms then the singlet and triplet energies can be very close to one another and the order can be changes by external magnetic fields. There is more detail on this effect in this answer How does spin flipping of triplet carbenes occur?


Which alkene has a higher Cahn-Ingold-Prelog priority in (2Z,5E)-hepta-2,5-dien-4-ol? How do I determine whether the central carbon of this compound is (R) or (S)?

On both the sides of carbon, the groups are same. They differ only in the relative position of hydrogen atoms. On what basis do I assign priority in these groups?

Does the R/S nomenclature for the middle carbon really depend on E/Z nomenclature of the alkenes, in such cases of stereoisomerism? Or is there an entirely different argument in such a case?

In order to arrive at an unambiguous description of stereoisomers, the priority for the groups attached to the chirality center is established by the application of ‘Sequence Rules’. In the hierarchical Sequence Rules given in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), we find

P-92.1.3.3 Sequence Rule 3
When considering double bonds and planar tetraligand atoms ‘seqcis’ = ‘Z’ precedes ‘seqtrans’ = ‘E’ and this precedes nonstereogenic double bonds.

In the given example, however, the stereochemical configuration at the central carbon atom is not defined. Note that solid wedges should be used to indicate bonds that project above the plane of the paper and hashed wedges to indicate bonds that project below the plane of the paper.


Is it possible to create a salt with multiple anions with only one cation(e.g. CaBrCl)? I know that usually more electronegative anions would displace less electronegative anions but is there a way to get different anions to "share" a cation?

A cursory search showed some interesting cases.

The oldest reported examples seemingly are mercury halides $\ce{HgXY}$.

$\ce{HgClI}$ was prepared by iodination of solid $\ce{HgCl2}$
$\ce{HgBrCl}$ was obtained from the chlorination of an ethanolic suspension of $\ce{HgBr2}$
$\ce{HgBrI}$ was prepared by mixing ethanolic solutions of $\ce{HgBr2}$ and $\ce{HgI2}$

The crystal structure of $\ce{Cs6Cl3[Fe(CN)6].H2O}$ has been reported.

Passing $\ce{H2S}$ through solutions of oxidosulfido molybdates and tungstates let to various mixed salts with halide, hydrogen sulfide and thiosulfate as additional anions beside the trisulfidometalates $\ce{[MoOS3]^{2−}}$ and $\ce{[WOS3]^{2−}}$.

But the concept isn't limited to inorganic chemistry.

The crystal structures of mixed salts of amino acids with different anions were reported.

From the field of organic conductor research, salts of BEDT-TTF (bis(ethylenedithio)tetrathiafulvalene) with mixed anions of $\ce{[CuBr2]-}$ and $\ce{[CuCl2]-}$ are known.


Does dissociation happen in gases? I understand that dissociation is when something dissolve. I heard this mostly only happen in liquids. But now I heard it can also happen in gases. Is this true? If so, when and why does this happen, and does it only happen to certain gases? I tried to do research but didn't find too much information. The only real information I found was on Wikipedia and it was confusing.

The information is confusing precisely because your question itself is confusing. There are different kinds of dissociation, and different degrees thereof. Moreover, there are different areas of study, each with its own concepts of what is significant and what is not. For example, if one molecule in a trillion splits into ions, this is nothing for chemistry, but probably quite a something for atmospheric physics.

As for the question you supposedly intended to ask: no, electrolytic dissociation of dissolved substances never happens in gases the way it happens in water.

Also, welcome to Chem.SE.


Whats the symmetric, anti-symmetric and non-symmetric part of the direct product of irreducible representations? I am trying to understand how to see if a vibrational mode is Jahn-Teller active or not. According to the group theoretical description of the Jahn-Teller effect one needs to check if the symmetric part of the direct product the irreducible representation (=irrep) of the electronic state with itself contains the irrep of the distortional mode in its symmetric part. And this is where I have trouble to understand. I can understand that one can decompose tensors (tensor products) into a symmetric and an anti-symmetric part, but I fail to understand how to apply that to the the direct product of two irreps, as they are in most cases one-dimensional. Can someone explain to me what exactly means "symmetric, anti-symmetric and non-symmetric part" in this context and how to see that in specific cases?

The symmetric and antisymmetric part of the direct product here refers to the fact that certain irreducible representations in the direct product are symmetric while others are antisymmetric (w.r.t some operations like $\sigma_v$, $C'_2$). So, for instance, when we write:

$$\Pi \times \Pi = \Sigma^+ + [\Sigma^-] + \Delta$$

the $[ \ ]$ simply means that the irreducible representation $\Sigma^-$ is the antisymmetric part of the direct product. It doesn't mean that you are somehow decomposing $\Sigma^-$ into a symmetric and antisymmetric part, and then selecting the antisymmetric one. This won't be possible since $\Sigma^-$ is a one-dimensional representation.


Does a Grignard reagent react with enones to give the 1,2- or 1,4- product? Cyclohexenone, an α,β-unsaturated ketone, is electrophilic at both the carbonyl carbon, as well as the β-carbon. So, it can either undergo 1,2-addition to give the allylic alcohol 1, or 1,4-addition to give the ketone 2.

Where does the Grignard reagent add to?

The correct answer is the 1,2-addition product (i.e. the allylic alcohol 1). In general, Grignard reagents and organolithium reagents add directly to the carbonyl carbon, while organocuprates (organocopper reagents) add to the beta-position of an unsaturated ketone.

This exact transformation was reported by Akai and coworkers recently (Org. Lett. 2010, 12, 4900) and they obtained compound 1 in 95% yield (the supplementary information (PDF) is free).


How do electrolytes allow electrolysis to occur? From what I know, when two electrodes are places in a electrolyte (say $\ce{H2SO4}$, which decomposes into $\ce{H}^{+}$ and $\ce{SO4}^{2-}$ in aqueous solution), the negative ion is attracted to the positive electrode and is oxidized while the cation is attracted to the negative electrode and is reduced.

This transfer of electrons from the negative electrode to the cation and from the anion to the positive electrode "completes" the circuit and electricity is conducted through the wire.

What I don't understand is how the electrolyte allows for the electrolysis of water, since the electrons are used in the redox reaction of the ions in the electrolyte in which water isn't even involved.

In the electrolysis of $\ce{H2O}$, a small amount of sulphuric acid is added. Why? Because pure $\ce{H2O}$ doesn't ionize much on its own ($K_w=10^{-14}$ approximately). So traces of an electrolyte ($\ce{H2SO4}$ in this case) is added so that it can help break one $\ce{O-H}$ bond and $\ce{H}^{+}$ released forms a coordinate bond with one $\ce{H2O}$ to form $\ce{H3O}^{+}$ ion. So now the ions present in the solution are: $\ce{H3O}^{+}$, $\ce{OH}^{-}$ and $\ce{SO4}^{2-}$. Of these, the oxidation potential of $\ce{OH}^{-}$ is higher than that of $\ce{SO4}^{2-}$ so at the anode $\ce{OH}^{-}$ gets oxidized to $\ce{O2}$. In this case only one cation goes to the cathode so it gets reduced to $\ce{H2}$. In case you use an electrolyte that gives a cation $\ce{M}^{n+}$ other than$\ce{H}^{+}$ then $\ce{H2}$ would be formed only if the reduction potential of $\ce{M}^{n+}$ is lower than that of $\ce{H3O}^{+}$. So, if the electrolyte used is $\ce{ZnSO4}$ then $\ce{H2}$ will be formed but if $\ce{CuSO4}$ is used then $\ce{H2}$ is not formed, but $\ce{Cu}$ is deposited at the cathode.

So the ionization of water doesn't have anything to do with the electrons used in the redox process. The ionization depends simply on the electrolyte used.


Separating sodium sulfate from chromium trioxide solution I was looking at the industrial process for the production of chromium trioxide by treating sodium chromate or the corresponding sodium dichromate with sulfuric acid. This results in the formation of chromium trioxide as well as sodium sulfate. I think both of them would be mixed as a solution; so what way is employed to separate them? I read about the process in Wikipedia.
I thought of using methods which I know like distillation, but I don't know how can I check if they work practically and are employed industrially. I searched the web as well and couldn't find anything satisfactory.
I am in 12th grade and I haven't really studied this topic, so I didn't know how to approach this question.

The overall reaction is:
$$\ce{H2SO4 + Na2Cr2O7 → 2 CrO3 + Na2SO4 + H2O}$$

How to separate chromium trioxide and sodium sulfate from solution?

Since chromium trioxide is soluble, it exist as chromic acid in the solution. So, in the reaction pot, there are compounds like  $\ce{H2CrO4, H2Cr2O7}$ along with our target compound $\ce{CrO3}$, all in equilibrium. But with the addition of a large excess of concentrated sulfuric acid, water is withdrawn from the hydrated forms and the chromium trioxide separates from the solution in the shape of red needles. Successively, the reaction pot is made to stand overnight for the chromium trioxide to precipitate slowly in form of needles and then filtered off from the solution using Buchner filter. The synthesis procedure of chromium trioxide can be found in prepchem.com:

100 grams of sodium dichromate are dissolved in 250 ml of water and
the obtained solution is filtered. 200 ml of concentrated sulfuric
acid are dropwise added with constant stirring until a slight
permanent precipitate of chromium trioxide is formed. The mixture is
cooled for half an hour or longer, then slowly, while stirring, 200 ml
of concentrated sulfuric acid are added dropwise. The reaction mixture
is left to stand over night in order that the crystal meal may become
somewhat coarser. The larger crystals (and purer) of chromium trioxide
could be obtained by heating the solution with crystal meal with
stirring to $\pu{100 ^{\circ}C}$ and allowing to cool slowly. If this process is
repeated once or twice, a more satisfactory product will be obtained.
The obtained crystals of chromium trioxide are filtered with a Buchner
funnel with a sintered glass disc instead of the usual paper filter
(chromium trioxide is a strong oxidizing agent and will oxidize
organic material).  After draining the crystals completely and
pressing the surface and treating with 15 ml of concentrated (65-70%)
nitric acid. The nitric acid is removed by suction and the operation
is repeated twice with 10 ml of nitric acid each time. Finally, the
red crystals of chromium trioxide are dried by suction as much as
possible, transferred in a dry evaporating dish and by gentle heating
the remaining nitric acid is removed. Dried product is stored in a
stoppered bottle. Yield is 60%.

Note: chromium trioxide is toxic, corrosive, a powerful oxidizer and a potent carcinogenic. Sulfuric acid is corrosive and can burn your skin if you are careless. Handle those chemicals with care. Perform this synthesis only after you become a qualified chemist. This answer is only for educational purpose only.


How to find number of moles of HNO3 behaving like acid in a redox reaction? In the reaction $(1)$, how to find the number of moles of $\ce{HNO3}$, which is behaving like acid?

The reaction is given as follows:
$$\ce{Sn + HNO3 -> Sn(NO3)2 + NH4NO3 + H2O} \tag1$$
To find number of moles of $\ce{HNO3}$ behaving like acid, I first balanced the reaction using oxidation number method which gave me:
$$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$
But I could not proceed any further. I am unable to find any hint. How should we proceed to reach the answer? Answer mentioned in book is 9.  This is a question from chapter "Redox Reactions".

It is always easy to balance redox equations by considering two half reactions. The equation to be balanced is:
$$\ce{Sn + HNO3 -> Sn(NO3)2 + NH4NO3 + H2O}$$
Thus, the two relevant half reactions to this redox equation are:
$$\ce{Sn (s) -> Sn^2+ (aq)} \\
\ce{NO3- (aq) -> NH4+ (aq)}$$
First balance elements other than $\ce{O}$ and $\ce{H}$. In this case they are $\ce{Sn}$ in first equation and $\ce{N}$ in second equation. Then, balance $\ce{O}$ and $\ce{H}$ in both equations with water and $\ce{H+}$, respectively. This is because the reaction is in aqueous acidic medium:
$$\ce{ Sn (s)  -> Sn^2+ (aq)}\\
\ce{NO3- (aq) + 10 H+ (aq) -> NH4+ (aq) + 3 H2O (l)}$$
Finally, balance the positive charges by electrons so that each equation has no net charges:
$$\ce{ Sn (s)  -> Sn^2+ (aq) + 2e-} \tag1$$
$$\ce{NO3- (aq) + 10 H+ (aq) + 8 e- -> NH4+ (aq) + 3 H2O (l)} \tag2$$
These are your mass and charge balanced oxidation ($(1)$) and reduction ($(2)$) half-reactions. Now, you can add these two equations in order to cancel electrons. To do so add $4 \times (1) + (2)$:
$$\ce{ 4 Sn (s) + NO3- (aq) + 10 H+ (aq) -> 4 Sn^2+ (aq) + NH4+ (aq) + 3 H2O (l)} \tag3$$
This is your balanced ionic equation. As evidence, you need nine $\ce{NO3-}$ ions to balance as counter ions in right hand side, so you need to add $\ce{9NO3-}$ to both side to keep the mass and charge balance.:
$$\ce{ 4 Sn + NO3- + 10 H+ + 9 NO3- -> 4 Sn^2+ + NH4+ + 9 NO3- + 3 H2O} \tag4$$
When simplify, the equation $(4)$ would look like:
$$\ce{ 4 Sn +10 H+ + 10 NO3- -> 4 Sn(NO3)2 + NH4NO3 + 3 H2O} \tag5$$
Since $\ce{ +10 H+ + 10 NO3- = 10 HNO3}$, your complete balance equation is:
$$\ce{ 4 Sn +10 HNO3 -> 4 Sn(NO3)2 + NH4NO3 + 3 H2O}\tag5$$
Therefore, your answer is correct. Answer in your textbook must be a misprint or accidently for got that one $\ce{HNO3}$ is needed to be reduced. Other $\ce{9NO3-}$ is to balance the products (as counter ions). All $\ce{10H+}$ ions needed to complete the redox reaction.


API formulations - are excipients required I have the following API's
Itraconazole, Ivermectin, Fenofibrate, Pyrvinium Pamoate,  Sulfasalazine, Artesunate,  All-Trans Retinoic Acid, Doxycycline,  Atovaquone,  Extremestane, Syrosingopine and Mebendazole
Why is it that API's require excipients?
Are they necessary for the active pharmaceutical ingredients bioavailability?
Or are the excipients used more as stabilises, preservatives, flavourings, binders and fillers so that as a finished pharmaceutical drug, it can hold its integrity/ shelf-life / taste etc.?

The ICH guidelines give some details about the presence of excipients, and the forms they may take (along with acceptable types/classes of excipients);
“The excipients chosen, their concentration, and the characteristics that can influence the drug product performance (e.g., stability, bioavailability) or manufacturability should be discussed relative to the respective function of each excipient. This should include all substances used in the manufacture of the drug product, whether they appear in the finished product or not (e.g., processing aids). Compatibility of excipients with other excipients, where relevant (for example, combination of preservatives in a dual preservative system), should be established. The ability of excipients (e.g., antioxidants, penetration enhancers, disintegrants, release controlling agents) to provide their intended functionality, and to perform throughout the intended drug product shelf life, should also be demonstrated. The information on excipient performance can be used, as appropriate, to justify the choice and quality attributes of the excipient, and to support the justification of the drug product specification (3.2.P.5.6).”
The major use of excipients is often just as bulk material in tablets. Many modern drugs are active at doses of 10-100 mg — these quantities are often so small that fillers are needed to make a tablet of acceptable size to be acceptable to the patient (compare to some older drugs where doses were often in the gram range).


Ring contraction in pinacol rearrangement of cyclohexane-1,2-diol I came across this question in three different textbooks.

The question was: What is the major product obtained on dehydration (treatment with concentrated $\ce{H2SO4}$) of cyclohexane-1,2-diol.

I figured it would undergo a pinacole-pinacolone rearrangement. I expected the product to be cyclohexanone but to my surprise the answer was 1-cyclopentylmethanal.

What can be the driving factor for the ring to preferentially contract? Shouldn't the intermediate carbocation be stable in a six membered ring rather than a five membered ring?

Background

You don't specify whether we are starting with cis- or trans-cyclohexane-1,2-diol, but it doesn't matter since both isomers will interconvert under the reaction conditions.  Therefore we will need to consider the reaction products from both isomers.
These reactions run in strong acid are never clean.  I'd bet that some cyclohexanone was also formed, but that it is the minor product.  Here is a link (see section 2. Pinacol Rearrangement, example #3) showing that 1,2-dimethylcyclohexan-1,2-diol produces >90% of the corresponding cyclopentyl analogue and around 6% of the cylcohexanone analogue.
Predominant formation of the cyclopentyl carbaldehyde indicates that our reaction must be kinetically driven. 5-membered rings have ~6 kcal/m more strain than 6-membered rings.  Therefore, if the reaction were thermodynamically controlled the 6-membered ring would predominate.
In the case of cyclohexane-1,2-diol, if a protonated hydroxyl group departed, a seconday carbocation would form.  Secondary hydrocarbon carbocations are not particularly stable, so it is likely that rather than generate a secondary carbocation there is some migrating group migration at the same time the protonated hydroxyl is departing. Such migrations have a strong preference for occurring by way of an anti-periplanar geometry.

Solution

In the figure below I've drawn the 3 isomers of cyclohexane-1,2-diol. The isomer on the left is the axial, axial trans isomer, the isomer in the middle is the equatorial, equatorial trans isomer, and then on the right is the axial, equatorial cis isomer.  I've also labeled the hydroxyls (A and B) for clarity.  In an equilibrium mixture of these isomers we would expect the equatorial, equatorial isomer to predominate because both hydroxyl groups are in the sterically less encumbered equatorial position and both hydroxyls can hydrogen bond with each other in this conformation.

Examining the isomer on the left, neither the $\ce{C-H}$ bond nor the $\ce{C-C}$ bond is aligned anti-periplanar to either hydroxyl A or B, consequently this isomer should be relatively unreactive.

Examining the middle isomer only $\ce{C-C}$ bonds are aligned anti-periplanar to the hydroxyl groups.  Rearrangement of this isomer can only produce the cyclopentyl derivative.

Examining the isomer on the right we see that a $\ce{C-H}$ bond is aligned anti-periplanar to hydroxyl A, while a $\ce{C-C}$ bond is aligned anti-periplanar to hydroxyl B. Hence, this isomer can produce both the cyclopently carbaldehyde and cyclohexanone.

We see that there are 3 pathways leading to the 5-membered ring product and just one leading to the 6-membered ring product. This suggests that the 5-membered ring product should predominate.  If we further consider that, as explained above, the middle isomer should predominate at equilibrium, then our product mixture should be even further enriched in the 5-membered ring product just as you reported.


Maximum temperature attainable by thermite reaction In the thermite reaction
  $$\ce{2Al(s) + Fe2O3(s) -> 2Fe(l) + Al2O3(s)}$$
  what is the maximum temperature attainable by the products? Assume reactants at $25\ \mathrm{^\circ C}$ and $1\ \mathrm{atm}$ and that all heat capacities remain constant.

I started by using Hess's law to find the enthalpy change of the reaction. I couldn't find a heat of formation for liquid iron so I assume I have to find that out. I just wrote out the equation
$$\ce{Fe(s) + heat + enthalpy of fusion -> Fe(s)}$$
and figured the heat of formation for liquid iron at $25\ \mathrm{^\circ C}$ is around $-302.71\ \mathrm{kJ/mol}$.

To find the max temperature I used
$$\Delta H_\mathrm r=\Delta T\cdot C_\mathrm p(\text{products})$$
and came out with an equation like
$$-14445\ \mathrm{kJ/mol}=0.147\ \mathrm{kJ/(mol\ K)}\cdot T_\text{max}$$
This gives me a negative temperature, so I assume it is wrong. Even if I were to take the absolute value, the answer seems awfully high. I was expecting something around a few thousand degrees not near ten thousand.

According to Wikipedia the maximum temperature is approximately 2500 °C, limited by the boiling point of aluminum, which is 2519 °C.  The article provides some discussion of how the temperature is limited, and how to calculate it using Hesse's law.


Role of heat in combustion What role does heat play in the combustion process.Specifically, does the amount of heat  determine whether complete or incomplete combustion occurs, and is the amount of heat impacted by the amount of oxygen

Yes, temperature and heat do affect the spontaneity and the equilibrium of reactions. The term heat can be referred to enthalpy, which can be quite confusing in context, as I believe you are asking about raising the temperature.

Let's first take a look at the equation for the Gibb's Free Energy and raising temperature:

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$

The Gibb's Free energy does not only relate to the spontaneity of a reaction, the magnitude of Gibb's Free energy describes how much "free energy" is available, or how much work is possible.

Since, entropy, enthalpy are state functions, we can calculate them by using Hess's Law and using tabulated values for standard thermodynamic data:

$$\Delta H^\circ_{rxn} = \Sigma \Delta H^\circ_{products} - \Sigma \Delta H^\circ_{reactants} $$

$$\Delta S^\circ_{rxn} = \Sigma \Delta S^\circ_{products} - \Sigma \Delta S^\circ_{reactants} $$

For simplicity's sake, I will use the combustion of methane that mannaia has so kindly provided and run the reaction at STP using heats of formation:

$\Delta H^\circ_{\text{comb of} \ \ce{CH_4}} = -890 \text{kJ} \cdot \text{mol}^{-1}$

$\Delta H^\circ_{\text{f of} \ \ce{H_2O} \ \text{(l)}} = -285.83 \text{kJ} \cdot \text{mol}^{-1}$

$\Delta H^\circ_{\text{f of} \ \ce{CO_2} \ \text{(g)}} = -285.83 \text{kJ} \cdot \text{mol}^{-1}$

$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$
$$ \Delta G^\circ  = -604.53 \text{kJ} \cdot \text{mol}^{-1} - (298.15 \text{K}) *97.39 \text{J} \cdot \text{mol}^{-1} = -633.7 \text{kJ} \cdot \text{mol}^{-1} $$

If we use the enthalpy of combustion for methane:

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$
$$ \Delta G^\circ  = -1569.34 \text{kJ} \cdot \text{mol}^{-1} - (298.15 \text{K}) *97.39 \text{J} \cdot \text{mol}^{-1} = -1598.38 \text{kJ} \cdot \text{mol}^{-1} $$

What would occur if we raise the T term? Obviously, the Gibbs Free Energy would become larger in magnitude and more negative, thus being more spontaneous. Inversely, if we lower the temperature to such a degree that $T\Delta S^\circ = \Delta H^\circ$, the reaction would not occur at all, and if $T\Delta S^\circ > \Delta H^\circ$, we would need to input energy in order for the reaction to proceed forward.

In general:

If $\Delta H<0$ and (-) and $\Delta S>0$ (+), the reaction is spontaneous at any temperature.

If $\Delta H>0$ and (+) and $\Delta S<0$ (-), the reaction is NOT spontaneous at any temperature.

If $\Delta H<0$ and (-) and $\Delta S<0$ (-), the reaction is spontaneous at low enough temperatures.

If $\Delta H>0$ and (+) and $\Delta S>0$ (+), the reaction is spontaneous at high enough temperatures.

It is possible to think of heat as a product or reactant. Combustion reactions are highly exothermic. Le Chatlier's principle includes raising the temperature as a stress. Alternatively, we can think about the equilibrium of this reaction to see how much product will be formed by looking at the magnitude of the equilibrium constant, $K$.

Accordingly the van't Hoff isochore equation is written as:

$$ln \frac{K_2}{K_1} = \frac{\Delta H}{R}(\frac{1}{T_1} - \frac{1}{T_2}) $$

In summary, an increase in temperature favors the formation of products if the reaction is endothermic. Inversely, a decrease in temperature favors the formation of products if the reaction is exothermic.

As, Uncle Al has stated:

  Ambient oxygen makes no difference

The atmosphere is composed of about 70% nitrogen and 20% oxygen. If this reaction was to be observed in nature, there would be almost a limitless amount of oxygen for the reaction to rpoceed. Temperature does play a role in the stoichiometry of the reaction and thus whether or not the combustion is complete or incomplete. However, it is possible to limit one of the reactants as mannaia has done, specifically oxygen in a laboratory setting.

As related to the kinetic molecular theory of matter, increased temperature will increase the frequency of collisions of particles with sufficient energy to cleave bonds and reform them, driving the reaction further to the right. This is commonly done with the commercial production of ammonia using Van Haber process. Although this is thermodynamically unfavorable, when run at high temperatures, the rate of formation of ammonia is much faster.

In regards to say, an open environment, the temperature would increase the kinetic energy of the oxygen molecules, and so the particles would be spaced further apart. Describing this quantitatively... well, an experiment should be carried out. I would assume if one was so inclined to find if that would increase the equilibrium, rate, or completeness of combustion I'm sure they would find differences.


How to calculate the heat released when sodium hydroxide is dissolved in hydrochloric acid solution? How to calculate the heat released when sodium hydroxide is dissolved in hydrochloric acid solution? Here is the data I gathered from a lab experiment:

$50\ \mathrm{mL}$ of hydrochloric acid solution
$2.00\ \mathrm g$ of sodium hydroxide (in solid form)
Initial temperature is $24\ \mathrm{^\circ C}$
Final temperature is $44\ \mathrm{^\circ C}$

The chemical equation of the reaction is:

$$\ce{NaOH(s) +H+(aq) + Cl- (aq) -> Na+(aq) +Cl- (aq) + H2O(l)}$$

This is the ONLY information I can use and I cannot search up anything online.

I calculated:

$$
\begin{align}
q&=mc\Delta T\\
q&=\left(50\ \mathrm g\right)\left(4.18\ \mathrm{J/(g\ ^\circ C)}\right)\left(20\ \mathrm{^\circ C}\right)\\
q&=4180\ \mathrm J
\end{align}
$$

So calculating the heat released by the reaction, I assumed that the mass is $50\ \mathrm g$ since hydrochloric acid is a solution of hydrogen chloride in WATER and the reaction produces water as well. Therefore, since the density of water is $1\ \mathrm{g/mL}$, $50\ \mathrm{mL}$ must be approximately $50\ \mathrm g$. Also, the specific heat capacity must be similar to water as well which would be about $4.18\ \mathrm{J/(g\ ^\circ C)}$.

So my question is: Is my assumption correct? If not, why?

If that's all the information you have (no concentration or mass or density for the HCl solution, no mass for the final solution), and you weren't instructed to obtain any of that information, then what you did (assuming 50g and heat capacity of water) is ok for an introductory lab class.

There will be many sources of error:

The density of HCl solution is greater than water, so the initial solution will be more than 50g.

The heat capacity of HCl solution is less than that of water.

Without knowing the concentration of the initial solution, it is unknown if the final solution is NaCl solution, a solution of HCl and NaCl, or a solution of NaCl and NaOH.

The heat capacity of the final solution will be less than that of water.

The final solution will be somewhat more than 52g.

Just dissolving NaOH in water is exothermic. It can't be assumed that all the NaOH reacts without knowing the concentration of HCl.


How is charge density a useful measure of polarising power and polarisability? How do you use charge density ($Z/r$) as a measurement of polarising power and polarisability, and hence hard/soft acid character?

So far I've been looking at different polarising power and polarisability and how that affects a substance in whether it is a hard or soft acid/base, but I don't understand how charge density comes into play, and how we can use it to understand the characteristics.

What does higher charge density of a cation means? It means that the positive charge on a cation is concentrated in a small volume. Small, concentrated charge can cause greater distortion in the anion and therefore they have greater polarising power, or greater power to cause attract the negative electron cloud of the anion and hence greater distortion. The greater the positive charge on a cation, the larger the charge density will be (it is directly proportional to $Z$). Greater positive charge will cause greater distortion in the electron cloud of anion.

How do small and concentrated charge cause greater distortion? What will affect you more? A force applied on a very very small area of your body, or the same amount of force applied on whole of your body? Of course, the former one will affect us more, i.e., a small and concentrated force.

What about the anion? When the electron cloud of the anion is small, it is closer to the nucleus and experiences a greater attraction from the nucleus, therefore it is not easy to distort it. But, if the electron cloud is large, the inward force experienced by it from the nucleus is less, hence it can be polarised much easily. The inference is that the greater the radius, the larger the polarisability of the anion will be. What does greater radius mean? The greater the radius, the smaller the charge density will be (it's inversely proportional to $r$, or proportional to $1/r$).


Geometry of the pentachloridoindate(III) anion According to Brown et al.,1 the crystal structure of $\ce{[NEt4]2[InCl5]}$ reveals that the pentachloridoindate(III) anion, $\ce{[InCl5]^2-}$, adopts a square pyramidal $C_\mathrm{4v}$ geometry (VSEPR theory would predict trigonal bipyramidal $D_\mathrm{3h}$):

According to Greenwood and Earnshaw,2

  It will be noted that $\ce{[InCl5]^2-}$ is not isostructural with the isoelectronic species $\ce{[SnCl5]-}$ and $\ce{SbCl5}$, which have the more common $D_\mathrm{3h}$ symmetry. […] $\ce{[NEt4]2[TlCl5]}$ is isomorphous with $\ce{[NEt4]2[InCl5]}$ and presumably has a similar structure for the anion.

However, the structure of $\ce{[PPh4]2[InCl5].MeCN}$ was reported to have the usual trigonal bipyramidal structure for the anion,3 which throws a spanner in the works. I can't understand much of the paper as it's in German. Since it's a report of a crystal structure, though, I don't really expect it to say much about the chemistry of the compound.

Can the reported differences in molecular geometry here be rationalised?

References

Brown, D. S.; Einstein, F. W. B.; Tuck, D. G. Tetragonal-pyramidal indium(III) species. Crystal structure of tetraethylammonium pentachloroindate(III). Inorg. Chem. 1969, 8 (1), 14–18. DOI: 10.1021/ic50071a004.
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth–Heinemann: Oxford, U.K., 1997, p 238.
Bubeinheim, W.; Frenzen, G.; Müller, U. Die Chloroindate $\ce{[PPh4]2[In2Cl6]}$ und $\ce{[PPh4]2[InCl5].CH3CN}$. Acta Crystallogr., Sect. C: Cryst. Struct. Commun. 1995, 51, 1120–1124. DOI: 10.1107/S0108270194011789.

TL;DR Initially published crystal structure of $\ce{[NEt4]2[InCl5]}$ [1] according to the further investigations [3], is not valid. The $\ce{InCl5^2-}$ ion does not have $C_\mathrm{4v}$ symmetry, and VSEPR theory pretty much explains formation of numerous slightly distorted trigonal bipyramidal $\ce{InCl5^2-}$-containing complexes according to the most recent single-crystal structure experiments.

From the other side, the bond lengths and geometry index refer to the dominating character of square pyramidal environment, which is in this case can be dictated by the crystal packing, towards which $\ce{[InCl5]^2-}$ is more sensible than $\ce{[SnCl5]-}$ due to the noticeable size difference ($\ce{[InCl5]^2-}$ is bulkier). Either way, it looks like another crystallographic experiment (preferably at lower temperatures to decrease the size of thermal ellipsoids) is needed to determine the angles more precisely.

As a general advice, always pay attention to the $R_1$ and $R_2$ values of the crystal structure. A rule of thumb: a good structure refinement of small molecules should lead to $R_1 < 0.05$ and $R_2 < 0.12$.

Crystal structure of $\ce{[NEt4]2[InCl5]}$, determined in 1969 [1], has been subsequently criticized in several publications, mainly involving additional symmetry vibration analysis of Raman spectra for $\ce{MX5^2-}$ anions [2]. It's been established that $\ce{InCl5^2-}$ ion does not have $C_\mathrm{4v}$ symmetry (but retains $C_\mathrm{2v}$), and that the crystal may not be centrosymmetric.

The results are summarized in [3]:

  Deconvolution of the single-crystal data was again performed with the aid
  of the low temperature data, revealing $11$ $\ce{InCl5^2-}$ bands between $300$ and $\pu{100 cm-1}$. Two overlapping A stretches are clearly evident at $292$ and $\pu{286 cm-1}$, but no B stretch is apparent. The two E symmetry stretches expected for the anion of a $C_2$ site, but not for a $C_4$ site, are evident at $281$ and $\pu{271 cm-1}$. [...] With the aid of low-temperature data, the four E modes predicted for $\ce{InCl5^2-}$ at $C_2$ sites are found at $144$, $136$, $122$, and $\pu{103 cm-1}$ in the single-crystal data. With this reinterpretation of the spectrum it is no longer necessary to assume arbitrarily the presence of a lattice mode in this region.

  In summary, the vibrational data for the tetraethylammonium salts of $\ce{InCl5^2-}$ and $\ce{TlCl5^2-}$ indicate that these complexes reside on sites lacking full $C_4$ symmetry. The most straightforward conclusion from the vibrational data is that the $\ce{InCl5^2-}$ ions have local $C_2$ symmetry, which would be inconsistent with the previous structure determination [1]. [...]

  The structure of $\ce{[(C2H5)4N]2[InCl5]}$, as previously reported, consisted of centrosymmetrically related $\ce{InCl5^2-}$ ions situated about the
  Wyckoff c positions of $C_4$ symmetry in space group $C_\mathrm{4h}-P4/n$.
  Coaxial with the $\ce{InCl5^2-}$ ions are disordered $\ce{(C2H5)4N+}$ ions
  also situated about the Wyckoff c positions. The N atoms of the two remaining $\ce{(C2H5)4N+}$ ions in the unit cell occupy the centrosymmetrically related Wyckoff b sites of $S_4$ symmetry, the cation as a whole again being disordered. There are two disturbing features of this solution and its subsequent refinement: the disorder of the $\ce{(C2H5)4N+}$ ions and the significant residual electron density in the region of the basal Cl atoms of the tetragonal-pyramidal anion.

It is demonstrated further that for the ordered model the most satisfying crystallographic interpretation coexists with $C_2$ rather than $C_4$ symmetry. A correlation of allowed point symmetries implies switching from $P4/n$ to $P\bar{4}$ space group, thus reducing $R$-factor ($R_1$) from $9.5\%$ to $6.7\%$.

  During the refinement in $P\bar{4}$, small but significant changes
  occur in the basal plane of the $\ce{InCl5^2-}$ ion. The basal Cl atoms
  have moved by $\pu{-0.5 A}$ from their fourfold symmetric positions in $P4/n$, leading to variations in the bond distances and angles about the In atom (Figure 3) that are qualitatively sufficient to explain the vibrational results.

Geometry index $\tau_5$ for $\ce{[InCl5]}$ fragment can be used to determine formal coordination environment as follows:

$$\tau_5 = \frac{\beta - \alpha}{60^\circ},$$

where $\alpha$, $\beta$ - two greatest valence angles of the coordination center ($\angle \ce{Cl - In - Cl}$, $\alpha < \beta$).

\begin{align}
\begin{cases}
\tau_5 &= 0 \qquad &\text{square pyramidal geometry} \\
\tau_5 &= 1 \qquad &\text{trigonal bipyramidal geometry}
\end{cases}
\end{align}

For the original $\ce{[NEt4]2[InCl5]}$ structure refined in [3] I've taken angular values from their supplementary materials. It turned out that changing space group to $P\bar{4}$ still results in dominating character of square pyramidal geometry of $\ce{[InCl5]^2-}$, though one must remember that the rotation axis of highest order is $C_2$, not $C_4$:

$$\alpha = (150.6 \pm 5.0)^\circ, \beta = (154.1 \pm 6.0)^\circ, \bar{\tau_5} = \frac{154.1^\circ - 150.6^\circ}{60^\circ} = 0.06; \tau_5 \in [0.04; 0.24]$$

Paper [4] mentions that for isoelectronic $\ce{[SnCl5]-}$-contining structures with various cations only trigonal-bipyramidal geometries were reported, and suggests higher influence of crystal packing on $\ce{[InCl5]^2-}$ geometry. This can also be explained by higher flexibility of $\ce{MX5}$ fragment in $\ce{[PPh4]2[InCl5]}$ in comparison with $\ce{[PPh4][SnCl5]}$ [5] due to greater average bond length: $d(\ce{In - Cl}) = \pu{2.5 A}$, $d(\ce{Sn - Cl}) = \pu{2.3 A}$.

As for auxiliary experimental data, all crystal structures including $\ce{[InCl5]^2-}$ fragment determined in the past two decade completely support VSEPR theory that predicts trigonal bipyramidal $D_\mathrm{3h}$ symmetry (with one intermediate exception of $\tau_5 \approx 50\%$).

bis(Tetraphenylphosphonium) pentachloro-indate(III) [4], trigonal bipyramidal geometry:
$$\alpha = 124.24^\circ, \beta = 174.74^\circ, \tau_5 = \frac{174.74^\circ - 124.24^\circ}{60^\circ} = 0.84$$

$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$.

bis(Diphenyldichlorophosphonium) pentachloro-indate(III) [6], trigonal bipyramidal geometry:
$$\alpha = 120.93^\circ, \beta = 179.29^\circ, \tau_5 = \frac{179.29^\circ - 120.93^\circ}{60^\circ} = 0.97$$

$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$.

tetrakis($\mu_3$-Selenido)-tetrakis($\mu_2$-1,5-bis(diphenylphosphino)pentane)-deca-gold(I) pentachloro-indate(III) [7], intermediate geometry:
$$\alpha = 138.45^\circ, \beta = 166.50^\circ, \tau_5 = \frac{166.50^\circ - 138.45^\circ}{60^\circ} = 0.47$$

$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#FFA100}{\Large\bullet}~\ce{Se}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$;
$\color{#FFD123}{\Large\bullet}~\ce{Au}$.

tetrakis($\mu_2$-chloro)-chloro-tetrakis(triphenylphosphine-P)-di-copper(I)-indium(III) tetrahydrofuran solvate [8], trigonal bipyramidal geometry (though authors refer to it as to "quasi square-pyramidal" coordination based on a single slightly shorter $\ce{In - Cl}$ bond distance $(d(\ce{In-Cl_\mathrm{ap}}) = \pu{2.36 A})$, shortest $d(\ce{In-Cl_\mathrm{eq}}) = \pu{2.42 A})$:
$$\alpha = 123.90^\circ, \beta = 172.89^\circ, \tau_5 = \frac{172.89^\circ - 123.90^\circ}{60^\circ} = 0.82$$

$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#C88033}{\Large\bullet}~\ce{Cu}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$.

Bibliography

Brown, D. S.; Einstein, F. W. B.; Tuck, D. G. Inorganic Chemistry 1969, 8 (1), 14–18. DOI 10.1021/ic50071a004.
Adams, D. M.; Smardzewski, R. R. Journal of the Chemical Society A: Inorganic, Physical, Theoretical 1971, 714. DOI 10.1039/j19710000714.
Joy, G.; Gaughan, A. P.; Wharf, I.; Shriver, D. F.; Dougherty, J. P. Inorganic Chemistry 1975, 14 (8), 1795–1801. DOI 10.1021/ic50150a011.
Bubenheim, W.; Frenzen, G.; Müller, U. Acta Crystallographica Section C 1995, 51 (6), 1120–1124. DOI 10.1107/S0108270194011789.
Müller, U.; Siekmann, J. F. Acta Cryst C 1996, 52 (2), 330–333. DOI 10.1107/S0108270195011073.
Taraba, J.; Zak, Z. Inorganic Chemistry 2003, 42 (11), 3591–3594. DOI 10.1021/ic034091n.
Olkowska-Oetzel, J.; Sevillano, P.; Eichhöfer, A.; Fenske, D. European Journal of Inorganic Chemistry, 2004 (5), 1100–1106. DOI 10.1002/ejic.200300774.
Zhang, X.-Z.; Song, Y.-W.; HuiWu, F.; Zhang, Q.-F. Zeitschrift für Naturforschung B 2007, 62 (6). DOI 10.1515/znb-2007-0605.


Why does gallium nitride allow an AESA radar to be more powerful than a radar with elements made of gallium arsenide? AESA (Actively Electronically Scanned Array) radars use thousands of individual transmit/receive elements to create a radar.  All in service AESAs use gallium arsenide as the material in the t/r elements.  It is said that gallium nitride allows a tenfold increase in the power emitted by these elements?  What allows GaN to emit so much more radio frequency energy than GaAs?

The short answer is that GaN transistors have a much higher breakdown voltage and saturation velocity than GaAs, allowing them to operate at higher voltages at high frequency, handling more power. The large band gap (~2x GaAs) means that GaN still behaves like a semiconductor, even at relatively high temperatures, allowing for higher operating powers.

GaN has really been a breakthrough in semiconductor technology. It's what allows us to have 405 nm violet laser diodes and could potentially replace magnetrons in microwave ovens. NXP has a little document here comparing GaN with a few older technologies that's a quick read.


What is the role of NaHSO3 in syn diol formation from alkenes with OsO4? All I have found on this is that NaHSO3 reduces partially the osmate ester, so breaks off Os and leaves a syn diol. Is it possible to show this step with arrows showing the process of electron movement? I suppose this step is not the main part of the reaction, and mayn't be very important, but I would like to know if it's possible - and if it is, how it looks.

After you add the $\ce{OsO_4}$ to make the reaction, you'll need to reduce the remaining $\ce{OsO_4}$ maintaining the $\ce{pH}$ of the solution constant.

In addition:

  The hydrolysis of the osmate ester proceeds by simple substitution
  reactions. The $\ce{NaHSO3}$, a mild reducing agent, facilitates ligand
  subsitution by reducing the $\ce{Os}$ further.

The Art of Writing Reasonable Organic Reaction Mechanisms, Robert B. Grossman, p 293

And $\ce{NaHSO_3}$ can do both and so you will use it. You also can add $\ce{Na_2SO_3}$.

I found also something interesting on JACS, it's mentioned at least four times in the document that $\ce{NaHCO3}$ is used as buffer. Also there is a lot of information in the references of this article.

  While the reaction is normally run under basic conditions
  ($\ce{K_2CO_3}$, $\ce{pH} = 12.2$, aqueous layer), it is possible to
  buffer the system with $3$ equiv of $\ce{NaHCO3}$ ($\ce{pH} = 10.3$,
  aqueous layer).$^{38, 39}$ Buffering of the reaction mixture does not
  affect the ee, but it can have a beneficial effect on the yield when
  base-sensitive substrates are used or base-sensitive products are
  formed. Thus, the AD of allyl or cinnamyl halides (cf. eqs $1$ and
  $3$) should be performed under buffered conditions to minimize epoxide
  formation (see eq $1$, the yield with allyl bromide as substrate is
  between $61$ and $74%$ under buffered conditions; in the absence of
  $\ce{NaHCO3}$ only $40-50$ % of diol is obtained38). Unfortunately,
  the AD reaction does not turn over if $\ce{K_2CO_3}$ is replaced
  entirely by $\ce{NaHCO3}$.

Hartmuth C. Kolb, Michael S. VanNieuwenhze, K. Barry Sharpless
Chem. Rev., 1994, 94 (8), p 2494


Sodium Polysulfide production from NaOH and Sulfur Does the sulfur also oxidize the $\ce{OH-}$ radicals after the $\ce{Na2S4}$ chains form? Something like this:
$$\ce{9S + 4NaOH -> 2Na2S4 + SO3 + H2O}$$

If you calculate the oxidation numbers for everything before and after, you will see that the oxidation number of the oxygen atoms will always be $-2$. However, sulfur $\ce{S},\ (0)$ is being both reduced to $\ce{S4^2-},\ (\frac{1}{2})$ and oxidized to $\ce{SO3},\ (+6)$. Sulfur is both the oxidant and reductant in what might be called a disproportionation. Sodium hydroxide is thus acting as both a source of oxygen atoms in the formation of $\ce{SO3}$ and probably as the electrolyte if this is occurring in aqueous solution.

  Oxidation numbers:

Element    Reactant     Product
S           0            -1/2 in polysulfide
S           0            +6 in sulfur trioxide
Na         +1            +1
O          -2            -2 in sulfur trioxide and water
H          +1            +1

The half reactions (unbalanced) are:

Oxidation: $\ce{S -> SO3}$
Reduction: $\ce{S -> S4^2-}$

You might get some sodium hydrogen sulfate forming through the followin;
$$\ce{NaOH + SO3 <=> NaHSO4}$$


Maximum density of heavy water at 11.6 degree Celsius? I came across a question asking me the temperature at which $\ce{D2O}$ will have maximum density. I didn't really know, so I checked the answer, and I found that the answer is $\pu{11.6^oC}$. But why ? Why is the density of $\ce{D2O}$ maximum at $\pu{11.6^oC}$? I can't find the reason anywhere...

The maximum density of $\ce{H2O}$ occurs when cooled to about $\pu{4^oC}$. At this point, the thermal motion of the molecules has slowed down enough such that water molecules can start to orient themselves in a manner more resembling that found in the expanded crystalline ice structure, and density thus starts to decrease with decreasing temperature. This pre-freezing expansion happens at a somewhat higher temperature for $\ce{D2O}$ due to the greater hydrogen bonding strength of $\ce{D2O}$ at any given temperature.

Also note that $\ce{D2O}$ freezes at about $\pu{4^oC}$, so clearly the maximum density of $\ce{D2O}$ must occur at a greater temperature than that of $\ce{H2O}$. In the case of $\ce{H2O}$, the temperature of maximum density occurs at about $\pu{4^oC}$ above it's freezing point, where for $\ce{D2O}$ the temperature of maximum density occurs at about $\pu{7^oC}$ above it's freezing point.


How would you go about adding a Cl group para to a brominated cinnamic acid? We recently synthesized brominated cinnamic acid in lab, and now our goal is to add a Cl para to the R group attached to the benzene ring.

There are a number of conditions that are selective for p chlorination of substituted benzenes. KCl and Oxone in MeCN at rt as described in Synthetic Communications vol 31 (2001) 2021 looks easy to use and should not disturb the other substituents details here


Why is ethanol distilled off before adding water in the first step of a malonic synthesis? I am doing a malonic synthesis shown below, and I wonder why I should distill ethanol off before adding water in the first step of the reaction (First I add absolute ethanol and solid sodium).

I assume this is a real procedure for actual synthesis.

Distilling ethanol off does have an effect on the workup of this reaction. Usually one needs to separate the reaction product and educts or side products afterwards by separation between an aqueous and an organic phase. Having ethanol in the mixture prevents the two phases from separating. In order to isolate your product you will need to distill ethanol off before adding water, because water would also be distilled off, if you added it first.


Calculating pH for titration of weak base with strong acid Calculate the pH at the equivalence point for the titration of $\pu{0.130 M}$ methylamine ($\ce{CH3NH2}$) with $\pu{0.130 M}$ $\ce{HCl}$. The $K_\mathrm{b}$ of methylamine is ${5.0 \cdot 10^{–4}}$.

So I started with the equation:

$$\ce{HCl + CH3NH2 <=> CH3NH3+ + Cl-}$$

and then I knew that

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{\ce{[base]}}{\ce{[acid]}} \right)$$

So, I put ${\log \left(\frac{0.130}{0.130}\right) = \log 1 = 0}$ and then added that to the $\mathrm{p}K_\mathrm{a}$, which I got from the equation

$$\mathrm{p}K_\mathrm{a} = \frac{K_\mathrm{w}}{K_\mathrm{b}} \quad \rightarrow
 \quad \mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})$$

However, after I plugged those in to get a $\mathrm{pH}$, it turned out to be wrong and then comments said that when titrated a weak base with a strong acid, the volume is doubled at equivalence point and the concentrations are halved.

Why is this? I now know that my original equation was wrong and it should be

$$\ce{CH3NH3+ <=> H+ + CH3NH2}$$

and from there I should make an ICE table with the concentration of $\pu{0.0650 M}$.

First of all, when you titrate  a weak base (methylamine) with a strong acid, the equation of titration is:$$\ce{H+(aq) + CH3NH2 -> CH3NH3+ }$$The reaction of titration, as you can see is total and quantitative. The constant of the aforementioned equilibrium is:  $$ K=\frac{K_\mathrm b}{K_\mathrm w}= 5\times 10^{10} \gg 10^3$$
According to the stoichiometry of the neutralization equation, at the equivalence point: $n_{\mathrm{acid}}=n_{\mathrm{base}}$ $$ C_{\mathrm{acid}}.V_{\mathrm{acid}}=C_{\mathrm{base}}.V_{\mathrm{base}}$$

As the initial concentration of the base equals the initial concentration of the acid, the volume of the acid at the equivalence point equals the volume of the base:  $$ V_{\mathrm{acid}}^{equiv.}=V_{\mathrm{base}}$$
You can see that the total volume is doubled.

Now, let's calculate the $\mathrm{pH}$ at the equivalence point: We have a weak acid at the initial concentration $$C_{\ce{CH3NH3+}}=\frac{C_{\mathrm{base}}.V_{\mathrm{base}}}{V_{\mathrm{base}}+V_{\mathrm{acid}}}=\frac{C_{\mathrm{base}}.V_{\mathrm{base}}}{2V_{\mathrm{base}}}=\frac{C_{\mathrm{base}}}{2}= \frac{0.13}{2} = \pu{0.065 M}$$
It's in equilibrium with its conjugated base:
$$\ce{CH3NH3+<=>H+(aq) + CH3NH2}$$ The acid is partially dissociated:$$ [\ce{CH3NH3+}]= 0.065-x$$ $$ [\ce{CH3NH2}]=x$$ $$ [\ce{H+(aq)}]=x$$
Let's write the constant of this equilibrium:
$$K_\mathrm a= 0.2\times 10^{-10}= \frac{x^2}{0.065-x}$$
We solve this equation of second order to find $x$, the concentration of ion hydronium.
We find: $\mathrm{pH}= 5.94$


Comparison of carbocation stability in bicyclo compounds Question

Which is the most stable carbocation ?

Answer

 3.

My attempt
I answered as 1 as the first one had +M of -O- and hyperconjugation effect as the positive charge is at para position. Why is the answer 3? It just has +I and +H of the second chain and +M is stronger than +I

I think the third option as greatest number of resonance structures can be alluded to it.
The following may suffice:

In the first one, the carbocation is isolated except for the presence of a single double bond in conjugation beside it. The oxygen and the double bond beside it play no role in stablizing it.
In the second one the lone pair on oxygen, the double bond and the carbocation are in conjugation. Hence, it is somewhat stable.
In three, similar conjugation like two is present, but is it extended over another double bond i.e. the lone pair of oxygen, two double bonds and the carbocation are all in conjugation and an extended resonance comes into play. Thus, it is more stable than two, since more number of resonance structures may be drawn for it than 2.
The fourth one is the least stable among them, as no conjugation or resonace structures may be constructed. Moreover the I- effect of the neighbouring oxygen tends to destablise the carbocation.

Hence, the order of stablity:
3>2>1>4
An arrow-pushing representation of resonance in the third option can be made as follows:

Hope it helps.


Can any atom donate more than 1 lone pair as a coordinate bond to the same metal? This would mean the formation of a σ and one or two π bonds, all coordinate. I haven't heard of this happening, like a fluorine anion or water with two lone pairs on oxygen will donate only one lone pair, but I wanted to know why it's not possible.

There is a class of compounds, called carbodiphosphoranes, which formally has a divalent carbon (0) central atom. Employed as a ligand it acts as a sigma and pi donor. This capability is also one of the major features to distinguish these compounds and derivatives from carbenes, which are sigma donors and pi acceptors. The following is reproduced from Thi Ai Nhung Nguyen and Gernot Frenking, Chem. Eur. J., 2012, 18, 12733-1748:

Similarly carbodicarbenes (substitute the phosphine with a carbene), also known as bent allenes, are also double donors.

This post shall only serve as an example, I am quite certain, there are more compound with this kind of bonding motive out there. Nicolau mentioned $\ce{OsO3N^{-}}$ and other metal nitrido complexes, which he will hopefully extend in an answer.


Reasons which explain why the efficiency of a reaction increases as the average rate decreases? Reasons which explain why the efficiency of a reaction increases as the average rate decreases? in other words, why does a slow chemical reaction is more effective than quick ones?
I done an experiment and the reletionship I found was that. I added a catalyst to vary the average rate (speed) of my reaction and the slower it took to react, the most product was got. Cant find a possible reason for that. any help??

Well it depends on the reaction. A notable example is the Haber process. The Haber process is the production of ammonia gas from nitrogen and hydrogen gas: $$\ce{2N2 + 3H2 -> 2NH3 + heat}$$ This reaction is extremely important in the industry sector as ammonia gas is used for a wide range of products such as fertiliser, bleach, cleaning agents and in WW2, it was used to make explosives.

Lets go off on a tangent for a moment:

  In WW1 the Germans had very little resources due to British ships blocking Germany's allies from providing them materials. This meant that they had to produce great quantities of ammonia themselves as they desperately needed it to make explosives. Thankfully German chemistry, Fritz Haber, created the Haber process which allow the Germans to mass produce ammonia. It is estimated that WW1 was prolonged by 2 years due to the creation of the Haber process. Haber later received the Nobel Prize in 1918 for his production process of ammonia and 'contribution to agriculture'.

So the question you might ask is why was creating ammonia so hard? After all it is a pretty simple reaction. Well this has to do with your question.

You see, producing ammonia gas is actually quite easy, however the main issue is that it is very difficult to get a high yield at a fast rate. The reason is due to the extremely high activation energy of this reaction. This is because for the reaction to occur, the extremely strong triple bonds of nitrogen must first be broken, which requires quite a lot of energy. This high activation energy means that the reaction is going to be really slow. This shown by the Arrhenius equation:
$$k = A\mathrm e^{-\frac{E_\mathrm A}{RT}}$$ However we are able to increase the rate of the reaction. If we were to increase the temperature, according to Arrhenius  equation, the rate of the reaction will increase. However now there is another problem, this time about the yield of the reaction.

Note above in the chemical equation, the Haber process is an exothermic reaction, meaning that heat is released. Now if we were to increase the temperature, according to Le Chatelier's principle, the equilibrium will shift to the left. This means that less ammonia gas will be produced. This is shown by the graph below:

So the above conundrum describes your question. By increasing the temperature of the sytem, you can increase the rate of the reaction, however by doing so you are going to decrease the yield.

Note: this is only true if the reaction is exothermic. If it is endothermic, then the yield will increase with temperature.


Double Newman Projections How do I draw a double Newman projection for cyclohexane? Which bond(s) do I sight down?

From what I understand, if we have 1-methylcyclohexane, we sight C1-C2 and C5-C4, as shown in my attempt below:

Would this be correct? I wish to achieve level 4!

Note that -Cl and -CH(CH3)2 are on equatorial positions(The slant lines suggest that)

And the -H is on axial position(The perfectly vertical line suggest that)

Note that it is not just a coincidence, rather it is intentional. I think it would suffice further.


Why is change in gibbs energy not zero through the gibbs free energy equation? Change in enthalpy of the system is equal to its temperature multiplied by its change in entropy (as change in enthalpy is equal to heat absorbed /given out at constant pressure and change in entropy is heat absorbed/given out divided by its temperature). So, through the Gibbs free energy equation, the change in Gibbs free energy should be zero?

This question might seem messy, as I don't know how to write equations on Stack Exchange.

Please help. The problem is in the elementary concepts, I think.

The change in Gibbs free energy for a closed system is given by:
$$\Delta G = \Delta H - T \Delta S$$
What you are describing is:
$$\Delta H = T \Delta S$$
which is only true for systems at equilibrium, or when $\Delta G = 0$
Why this is true is a little bit difficult to explain. It is essentially a form of the Clausius Inequality at constant temperature and pressure. The Clausius Inequality is really just a statement of the second law of thermodynamics, and it says that at best, for any possible process you can get a zero change in the entropy of the universe, and usually it will be a positive change. From the perspective of the system you can write that as:
$$\Delta S \ge \oint \frac{dQ}{T}$$
which is basically just saying the entropy change of the system is equal to or greater than the sum of the negative entropy changes of the rest of the universe, for each step in a closed cycle.
If you integrate that at constant temperature you get
$$\Delta S \ge \frac{\Delta Q}{T}$$
At constant pressure (and in the absence of non-expansion work) $\Delta Q = \Delta H$, so we can write:
$$\Delta S \ge \frac{\Delta H}{T}$$
...and a little rearranging gives:
$$\Delta H - T \Delta S \leq 0$$
So this says that any process which can occur on its own (without adding work or heat to it) will obey this inequality. That's a pretty useful thing to know, so we let
$$\Delta G = \Delta H - T \Delta S$$
to give us an easier way to talk about it. Processes will happen (on their own) if $\Delta G$ is negative, they won't happen if it's positive, and when they are at equilibrium, $\Delta G$ will be zero.