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How can transition metals form so many bonds with ligands?
For instance, copper can form $\ce{[Cu(H2O)6]^2+}$ so it accepts 6 electron pairs from $\ce{H2O}$ molecules to form dative covalent bonds. However it has one electron in the 4s subshell and 10 in the 3d subshell. To which shell does it accept these electrons? Is it to the 4d subshell? And how are these accepted electrons arranged?

Copper is  in +2 oxidation state so it has configuration $\ce{[Ar] 3d^9}$. Since 4s, 4p and 4d orbitals are vacant, Cu will show $\ce{sp^3d^2}$ hybridisation and will accept these elctron pairs in its 6 $\ce{sp^3d^2}$ hybridised orbitals. Read more about VBT here http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/valence.php


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Why does small value of van der Waals gas constant "b" ensure easier liquefication?
Larger value of b signifies that the size of the molecules is larger and hence the forces acting should become larger due to increased surface area.This should ensure easier liquefication. But my book says that smaller value of b corresponds to easier liquefication. Where am going wrong?
The book says "Chlorine is more easily liquefied than Ethane because 'a' for  chlorine is greater than 'a' for ethane but 'b' for fluorine is less than 'b' for ethane. Van der Waals constant 'a' is due to force of attraction and 'b' due to finite size of molecules. Thus, greater the value of 'a' and smaller the value of b, larger the liquefaction"

The $b$ in the Van der Waals equation of state signifies the volume that is effectively taken out per molecule by intermolecular repulsive interactions. You can imagine that a small $b$ will help the gas to liquefy, since in most ordinary liquids molecules are packed closer together than in a gas.
As written in Atkins Physical Chemistry:

For an ideal gas b = 0, does that mean it is very easy to liquify an ideal gas? But, ideal gases cannot be liquified, right?

@AyushPateria I don't think you can resume assuming ideal gas when the gas is about to/in the process of liquify/liquefaction.

What do you meam by "taken out per mole by intermolecular repulsive interactions?"

That it is the volume that cannot be occupied by other molecules due to repulsive interaction between them. It is how the Van der Waal equation is derived. I updated my answer with an excerpt from Atkins.

I'll point out that the a and b constants are not really independent of temperature or pressure. The constants are "best fit" over some range of experimental data, say 0.100 moles of gas at some various temperatures and pressures. So if you change the range of fit, then the constants would change. Thus a and b provide temperature and pressure corrections to ideal gas law. The a term is mostly a pressure correction, and the b term mostly a volume correction.

@Jori You have quoted Atkins but how does that quote help justify your argument. From what I can see, there is no sentence in that quoted part which makes any slight indication to "smaller $b$ leads to easier liquefaction". Please clarify! Thanks!

@AyushPateria Though for ideal gas b=0, a is also equal to 0, which means that the attractive forces are absent. So how could liquefication then be possible?

The virial equation of state reads:
$$Z = 1 + \left(b - \frac{a}{RT}\right)\frac{1}{V_\mathrm{m}} + \left(\frac{b}{V_\mathrm{m}}\right)^2 \cdots$$
Going by this equation, it seems like increasing the value of $b$ will increase $Z$, which means repulsive forces will increase, making the gas liquefaction difficult. Also for $a = 0$ and $b = 0$, the gas becomes an ideal gas whose liquefaction is not possible, so I think what you are saying is incorrect.
The question you are talking about, that why chlorine is more easily liquefied than ethane? It is because $a$ of $\ce{Cl2}$ is greater than $a$ of $\ce{C2H6}$, which means $\ce{Cl2}$ is easier to liquify. $\ce{C2H6}$, being greater in size than $\ce{Cl2}$, has a higher value of $b$. It's wrong to say "the smaller the value of $b$, the larger the liquefaction".

Increase in the value of z would simply imply increase in the volume of the gas. It does not mean that liquefication would become difficult due to increas in volume.

And does Vm represent ideal volume or real volume?

Z > 1 signifies positive deviation. Which means the gas will occupy more volume than what an ideal gas would have. But why is that? It's because of repulsive forces that molecules are not able to come closer and hence have larger volume. Thus,  the gas is more difficult to compress than an ideal gas.

Vm represent the volume measured, or simply the volume which an ideal gas would have occupied.

This isn't the Van der Waals equation. There are a zillion gas equations with various fudge factors. Each different equation has experimentally derived coefficients which are based on some rationalization. See Real Gas https://en.wikipedia.org/wiki/Real_gas at Wikipedia.

b value has no contribution to the liqufication/critical temperature of a gas. Whereas ‘a’ value has implication over the critical temp.. Higher the ‘a’ value easier liquification / high critical temp. because ‘a’ value is a direct measure of intermolecular force.
Krypton and Nitrogen has nearly equal ‘b’ value. Whereas ‘a’ value of Krypton is higher than the Nitrogen. Hence the critical temp of Krypton is higher than the Nitrogen. Intermolecular force decides the critical temperature, not the volume factor. Neon has much lower ‘b’ value than Krypton and Nitrogen, but lower ‘a’ value keep its critical temp at 44 kelvin much much lower than critical temp of Nitrogen 126K and Krypton 209K.

I'm not sure I would call $a$ a "direct Measure".

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A unique demo for singlet oxygen?
I have long known that mixing bleach and peroxide produces oxygen gas:
$$\ce{NaOCl + H2O2 -> O2 + H2O + NaCl}$$
However I only recently learned that the specific form of oxygen produced by the reaction is singlet oxygen:
$$\ce{NaOCl + H2O2 -> O2(a^1\Delta_{g}) + H2O + NaCl}$$
Is there a simple chemical demonstration that can show the difference between singlet oxygen and triplet (ground state) oxygen?  Will singlet $\ce{O2}$ relight a splint "faster", or decolorize a dye faster when bubbled through solution, or otherwise be distinguishable from triplet oxygen in a home lab?

Admittedly, I'm having a professional photochemical bias when it comes to the generation of singlet oxygen!
A rather safe and simple approach - even in home experiments - is the use of air, sunlight and a sensitizer (rose bengal, methylene blue, or even chlorophyll from natural resources, such as Spirulina).
Here, the sensitizer is irradiated and excited to the first singlet state, undegoes intersystem crossing (ISC) and transfers the energy to ground state triplet oxygen, upon which singlet oxygen is formed:
\begin{align}
\ce{Sens &->[h\nu]\ ^1Sens^{\ast}}\\
\ce{^1Sens^{\ast} &->[ISC]\ ^3Sens^{\ast}}\\
\ce{^3Sens^{\ast} +\ ^3O2&->\ Sens +\ ^1O2}\\
\end{align}
A prominent reaction of $\ce{^1O2}$ is the concerted, diastereoselective addition to 1,3-dienes in Diels-Alder fashion.

A classical example for this reaction is the synthesis of the natural peroxide ascaridole (2) from $\alpha$-terpinene (1). G. O. Schenck, the former director of the Max Planck institute for Radiation Research, performed the reaction in batteries of flasks in his garden and published the results in 1944 (Naturwissenschaften 1944, 32 (14-26), 157–157). Ascaridole is a natural anthelmintic (= against parasitic worms), hence the name.
Maybe not really suitable for home experiments due to the chromatographic workup envolved but worth to be mentioned is the Schenck ene reaction of allylic alcohols:

The reaction products can be converted to 1,2,4-trioxanes by peroxyacetalization, as shown by Griesbeck and coworkers (Beilstein J. Org. Chem. 2010, 6 (61))

Note that the 1,2,4-trioxane moiety is the key element in the natural anti-malaria drug *artemisinin.

There is a great amount of relevant information in Chemistry of singlet oxygen. IV. Oxygenations with hypochlorite-hydrogen peroxide J. Am. Chem. Soc., 1968, 90 (4), 975–981.
The lifetime of the singlet oxygen in solution will be on the order of a mircosecond.
Reactions of the formed singlet oxygen with tetramethylethylene, tetraphenylcyclopentadienone and several other compounds are described.
For tetraphenylcyclopentadienone, it says "The reaction was followed by loss of the intense color of tetraphenylcyclopentadienone.", which sounds good for a demonstration.
Effects of temperature, concentration and solvent are discussed.
Solvent selection is apparently very important.

The lifespan of singlet oxygen in the presence of $\ce{H2O}$  is milliseconds. It reacts almost instantly to produce trioxidane (hydrogen trioxide if you prefer). Much of the reactivity of singlet oxygen is due to this molecule.  $\ce{O2* + H2O  \rightarrow H2O3}$. To capture active oxygen you need strictly anhydrous conditions, which you cannot obtain by mixing hydrogen peroxide with sodium hypochlorite. The best way to get singlet oxygen is to run pure oxygen through a tube coated with dye and illuminated with strong visible light. The active oxygen generated by this method can last for an hour. The best way to see it is to concentrate it which causes the molecules to  collide with each other and release their excess energy as a photon of red light as they revert to triplet oxygen. There is a good pic pf this in the Wikipedia article on singlet oxygen. It shows a flask with an eerie deep red glow.
I stand corrected! I found a YouTube video that shows beautiful crimson chemoluminescence from singlet oxygen using common aqueous reagents! https://www.youtube.com/watch?v=QHJfBMDHwD4 You need hydrogen peroxide (probably 30%) a comparable sodium hydroxide solution, and chlorine gas from a gas generator. The hydrogen peroxide solutions are chilled to near freezing and added to the reaction bottle and blended together, Add the reagents to generate chlorine (I recommend inexpensive calcium hypochlorite for swimming pools and hydrochloric acid also for swimming pools) to the generator and promptly connect this to the reaction bottle and turn out the lights. You should get a conspicuous but transient red glow that will however persist until you stop pumping chlorine into it. I am guessing that the use of vhilled reagents and breaking it into a 2 part reaction gives the singlet molecules enough time to bump into each other and release their excited energy as a photon of red light. Now if only there were a simple way of proving the existence of tripxidane! $\ce{H2O3}$

Wikipedia seems to say that trioxidane is unstable, and reverts to singlet oxygen and water in milliseconds in aqueous environments.  I guess there is an equilibrium.

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Reason for the formation of azeotropes
Why is it that some liquid mixtures (that exhibit positive/negative deviation from Raoult's law) form azeotropic mixtures at certain compositions? What is the physical reason behind this; are there any extra forces holding the two components together? If so, please elaborate on this.

Before getting to the azeotropes, it's worth reminding ourselves of the relevant assumption that we make about ideal mixtures: that the intermolecular interactions have constant energy regardless of the mixture.
Given the important role that intermolecular interactions have in determining the boiling point of a pure liquid, it's surprising that so many mixtures behave approximately ideally, in line with this assumption. There are two main reasons for this.
First, liquids are only miscible if they have somewhat similar types of intermolecular interactions with similar magnitudes of energy. For example, water and hexane have very different intermolecular interactions, so surrounding water with hexane or hexane with water is quite unfavorable, and the two do not mix. Thus, any actual mixture has interactions between the two types of molecules that are not hugely different from the interactions within the pure liquids.
Second, substances that are liquid are room temperature generally do not have especially strong intermolecular forces, so the small differences in the intermolecular interactions of two miscible liquids is usually dwarfed by the impact of the difference in molecular weight that is a major determinant of boiling point. Water is quite unusual in having such a low molecular weight and still being a liquid at room temperature, a symptom of the exceptionally strong role of intermolecular interactions in determining its boiling point.
With that in mind, we can talk about azeotropes. To get a minimum boiling azeotrope, the deviation from Raoult's Law requires that the vapor pressure of the lower boiling component be reduced from what is predicted for ideal behavior and that the vapor pressure of the higher boiling component be increased. In order for that to happen, we need two liquids with substantially different intermolecular interactions whose mixing changes those interactions in opposite directions.
In the example of the water:ethanol azeotrope, the water is able to increase the amount of hydrogen bonding between ethanol molecules by increasing the density of hydrogen bond donors and acceptors per unit volume. At a low water concentration, the water also does not interfere greatly with the hydrophobic interactions between the ethyl ends of the molecules. As a result, the average intermolecular interaction for the ethanol molecules is stronger than in pure ethanol, and we have the necessary decrease in the vapor pressure of ethanol. For the water, when it is at low concentration in mixture, each water molecule is surrounded by ethanol molecules, so it has fewer hydrogen bonding interactions than in pure water. As a result, its average intermolecular interactions are weaker than in pure water, and we have the necessary increase in vapor pressure. Many low molecular weight alcohols form azeotropes with water for the same reasons.
When the water concentration gets below what is required for the azeotrope, enough of the alcohol molecules interact only with other alcohol molecules that the deviation from ideality decreases.
For the maximum-boiling azeotropes, the requirement is that the vapor pressure of both components is decreased. This is quite common in acid:water mixtures where the degree of ionization of the acid in the mixture is higher than when it is in pure form, which means that there are more ionic interactions, which are quite strong. If the water concentration gets below what is required for the azeotrope, the amount of ionization decreases and the deviation from ideality decreases with it. When the water concentration is higher than in the azeotrope, there may be a great deal of ionization of the acid, but there is enough water that is not interacting with the acid (and conjugate base) molecules that the deviation from ideality is less.

A solution that has a maximum or minimum vapour pressure (vs mole fraction) is called an azeotrope. The liquid is in equilibrium with the vapour and mole fractions  in the liquid are the same as in the vapour at a given temperature. As the composition of liquid and vapour are the same the $dp/dx_A=0$ where $p$ is pressure and $x_A$ mole fraction of species A. (The mole fraction of species B is $x_B=1-x_A$). The cause of this effect is intermolecular interaction between species A and B. This may be hydrogen bonding but need not be since benzene/cyclohexane and chloroform/hexane and many others form an azeotrope. Other intermolecular interactions are often generally called van-der-waals interactions and  are dipole-dipole and pi-pi interactions among others.
In a perfect solution the total pressure $p$ of a binary mixture varies linearly with mole fraction of either component, (Raoult's law)i.e.  $\displaystyle p=x_Ap^0_A+x_Bp^0_B = x_Ap^0_A+(1-x_A)p^0_B$ where $x$ are the mole fractions and $p^0_A,p^0_B$ the partial pressures of the pure liquids. Experimentally this 'law' is found to occur only when the species A and B are very similar, such as ethylene bromine and propylene bromide, or benzene and dichlorobenzene. Clearly such mixtures will not form azeotropes. What is needed is to modify the partial pressures to account for interaction between molecules.
The result of allowing pairs of molecules to interact with energy $\Delta E$ is
$$p=p^0_Ax_Ae^{(1-x_A)^2\Delta E/k_BT}+p^0_B(1-x_A)e^{x_A^2\Delta E/k_BT}$$
which now depends on the A-B interaction energy and predicts variations from Raoults law for energies comparable or larger than $k_BT$ where $k_B$ is the Boltzmann constant.  The detailed calculation is shown below and the figure shows how the vapour pressure deviates from the idea. The numbers show different values of $\Delta E/k_BT $. A value of zero corresponds to Raoult's law. The total pressure expressed in this way has a maximum or minimum depending on the sign and magnitude of the interaction energy.

A figure showing ratio of vapour pressure vs mole fraction for species A. The numbers correspond to $\Delta E/k_BT$. The curve with  $\Delta E/k_BT=1$ is close to that for ethanol in water. When the total pressure for species A and B is plotted a maximum or minimum is produced just as in an azeotrope.

Derivation of vapour pressure curve equation.
The aim is to calculate the chemical potential using statistical arguments and the interaction energy between molecules. The chemical potential is given by $\mu=\mu^0+RT\ln(a)$ where $a$ is the activity, and the partial pressure $p=p^0a=p^0\gamma x$ where $\gamma$ is the activity coefficient and $x$ the mole fraction. The chemical potential is the partial derivative of the Gibbs energy with number of moles $n$, for species A for example  $\displaystyle\mu_A= \left( \frac{\partial G}{\partial n_A}\right)_{T,p,n_B} $. Notice that $T,p$ and $n_B$ are the temperature, pressure, and the number of moles of species B, all of which should be held constant. The Gibbs energy is $\Delta G = \Delta H-T\Delta S = \Delta E+p\Delta V-T\Delta S$ and as we shall make the volume constant $\Delta G = \Delta H-T\Delta S = \Delta E-T\Delta S$ thus the energy and entropy changes have to be determined.
A regular solution is considered to be a random mixture of type A and B molecules in which specific interactions, such as hydrogen bonding, are absent but in which there is nevertheless interaction between molecules of either type and between different types, i.e interactions of the form AA, BB and AB and these need not be equal to one another. The entropy of mixing is assumed to be that of an ideal solution.   Some simplifying, but not entirely realistic, assumptions  are made so that a tractable solution can be obtained and are (i) that the molecules are situated on a lattice (ii) the molecules are of approximately the same size, which will allow random mixing and no change in volume, and (iii) intermolecular forces are only between nearest neighbours. This is called the Bragg-Williams model.
As we assume $z$ nearest neighbour interactions only, the intermolecular potential energy of solution A with molecules at their equilibrium positions is $U_{AA}=N_Azw_{AA}/2$ for $N_A$ molecules with average potential energy $w_{AA}$. The 1/2 arises because there are $N_Az/2$ pairs of molecules. There is a similar energy for solution B. In the mixture there are three types of pairs AA, BB and AB and the total energy of the mixture is then
$$U_{AB}=N_{AA}w_{AA}+N_{BB}w_{BB}+N_{AB}w_{AB}$$
and the change in energy on mixing is
$$\Delta E_{mix}= U_{AB}-U_{AA}-U_{BB} $$
To continue the calculation $N_{AA}$ and $N_{BB}$ are needed in terms of $N_{AB}$.
Each A molecule has $z$ neighbours and so $zN_A$ in total. This number of neighbours is also $N_{AB}+2N_{AA}$ producing $N_{AA}=(zN_A-N_{AB})/2$ and similarly for B, $N_{BB}=(zN_B-N_{AB})/2$. The factor of 2 arises because AA pairs are counted twice and AB pairs only once. Substituting gives
$$\Delta E_{mix} = N_{AB} \left( w_{AB}-\frac{ w_{AA} }{2}-\frac{ w_{BB} }{2}  \right)$$
A statistical argument is now used to find the number $N_{AB}$. As $N = N_A+N_B $ the first random choice to pick any molecule can be made in $N$ ways, the next can be made in $N-1$ ways hence to pick two $N(N-1)/2 \approx N^2/2$ as $N$ is very large. (The 1/2 arises are there are two ways of choosing, i.e. it does not matter which is chosen first.) This gives the number of all pairs. If, however, we must choose molecule A first and then B the number is $N_AN_B$ then the fraction of these AB pairs is $2N_AN_B/N^2$. As each of the $N$ molecules has $z$ neighbours, the total number of neighbouring pairs is $Nz/2$ thus
$$N_{AB}= \frac{zN}{2}\frac{2N_AN_B}{N^2}=\frac{zN_AN_B}{N}$$
The energy is now
$$\Delta E_{mix} = z\frac{N_AN_B}{N_A+N_B} \left( w_{AB}-\frac{ w_{AA} }{2}-\frac{ w_{BB} }{2}  \right)$$
Letting  $n_A,\,n_B$ be the number of moles of A and B respectively, and $N_0$ Avogadro's number and making $\displaystyle w= w_{AB}-\frac{ w_{AA} }{2}-\frac{ w_{BB} }{2} $ gives
$$\Delta E_{mix} = N_0 z w\frac{n_An_B}{n_A+n_B}$$
As we assume that the mixing is random the entropy change on mixing is that for an ideal mixture, namely
$$\Delta S_{mix} = -R\left(n_A\ln(x_A)+n_B\ln(x_B)\right)$$
where $x$ are the mole fractions. The derivation of the entropy change can be found here Deriving the entropy of mixing of a non-ideal solution.
Combining equations gives
$$\Delta G_{mix} = N_0 z w\frac{n_An_B}{n_A+n_B}  -RT\left(n_A\ln(x_A)+n_B\ln(x_B)\right)$$
The free energy is $G=G_A+G_B+\Delta G_{mix} $ and differentiating to give the chemical potential gives
$$\mu_A=\mu_A^0 + \left( \frac{\partial \Delta G_{mix}}{\partial n_A}\right)_{T,p,n_B} $$
where $G_B$ differentiates to zero because $n_B$ is a constant when differentiating with $n_A$. By definition $\partial G/\partial n_A=\mu^0$.  Performing the differentiation $\Delta G_{mix}$ with $n_A$ gives
$$\mu_A=\mu_A^0 +RT\left(\ln(x_A)+\frac{N_0zw}{RT}x_B^2\right) $$
simplifying by letting $\beta = N_0zw/RT$ gives $\displaystyle \mu_A=\mu_A^0 +RT\left(\ln(x_A)+\beta x_B^2\right) $. There is a similar equation for $\mu_B$.
Since  $\mu_A=\mu^0_A+RT\ln(a_A)$ then $\ln(a_A)=\ln(x_A)+\beta x_B^2$ and as $a_A=\gamma_A x_A$ then $\gamma_A=e^{\beta x}$ and so
$$p_A=p_A^0\gamma_A x_A = p_A^0x_Ae^{\beta x_B^2}$$ and also for $p_B$
$$p_B=p_B^0x_Be^{\beta x_A^2}$$
as $\beta = N_0zw/RT=zw/k_BT$ then as $zw$ is an energy this partial pressure  becomes equal to $p^0_Ax_Ae^{(1-x_A)^2\Delta E/k_BT}$ which is the equation for the partial pressure we sought.
The azeotrope can be determined by  calculating the derivative of the total pressure and setting it to zero, $\left( dp/dx_A \right)_T =0$ giving $\displaystyle x_A^{azeo} =\frac{1}{2} \left( 1+ \frac{k_BT}{\Delta E}\ln\left( \frac{p_A^0}{p_B^0} \right) \right)$

What is the origin of the exponential term containing $\Delta E$ and $kT$ in the first equation defining $p$ vs. $p°$ ?

From the partition function for randomly mixing interacting molecules and their interaction energy. Then the Helmholtz free energy, then chemical potential, and finally the vapour pressure. It's quite involved.

@Maurice further to my comment. As the chemical potential for a perfect gas is $\mu_i=\mu_i^\text{o} +RT\ln(p_i)$ rearranging for $p_i$ shows that the partial pressure rises exponentially with $\Delta E/RT$ and as as the pressure is the sum of partial pressures, intuitively we get the form of equation you ask about.

 Is there any extra force holding the two components together?

   Correct. Namely, hydrogen bonding. As a quick refresher, hydrogen bonding is an  electrostatic attraction between an electropositive hydrogen atom on one compound being (weakly) attracted to a highly electronegative atom on another, nearby molecule, typically oxygen, although it can happen between a hydrogen and another electronegative atom such as oxygen or fluorine.
   Azeotropes form when there is a 'deviation from Raoult's law':
   A solution that shows positive deviation from Raoult's law forms a minimum boiling azeotrope. A very well-known example would be the azetrope that forms from a ethanol and water mixture when it is approximately 95% ethanol by volume. Once this composition has been achieved, the liquid and vapor have the same composition, and no further separation occurs. Methods for breaking this azetrope are typically either via chemical means, such as adding a small amount of benzene to the azetrope, or by a drying agent such as $\ce{H2SO4}$, magnesium sulfate, or calcium chloride. Interesting thing to note: You do not have to dry all the way to 100% ethanol, but simply BREAK the azetrope by getting it beyond that 95.6% concentration. After you get it beyond ~96%, you can distill the rest of the way to 99% via the normal distillation method.
   A solution that shows a negative deviation from Raoult's law forms a MAXIMUM boiling azeotrope, or as it is often stated: A negative deviation from Raoult's law is where the adhesive forces between different components are stronger than the average cohesive forces between like components. In consequence each component is retained in the liquid phase by attractive forces that are stronger than in the pure liquid so that its partial vapor pressure is lower. Nitric acid and water is a classic example of this class of azeotrope which forms when the solution has an approximate composition of 68% nitric acid and 32% water by mass.

-1: I am sorry, but I feel that this a terrible answer, mainly because it doesn't fully ddress the OP's question. You have written about many definitions, for which the OP did not ask. It will be best if u can elaborate on these hydrogen bonds; you have not explained how these hydrogen bonds are formed at a given composition so that the mixture becomes an azeotrope in the first place. This is what the O.P was looking for, I think.

Incomplete, "breaking the azeotrope" is nonsense, hydrogen bonding can be involved, but is by no means necessary, definitely not with minimum boiling azeotropes.

There is always a force between molecules, because unlike vapor liquid is a condensed phase. The question is mainly about whether the intermolecular interaction is stronger or weaker than the average of AA and BB. Ttherefore, this is not determined by hydrogen bonding, although hydrogen bonding can be one type of intermolecular interaction.

I like the wikipedia answer https://en.wikipedia.org/wiki/Azeotrope

in other words: Raoult's law predicts the vapor pressures of ideal mixtures as a function of composition ratio. More simply: per Raoult's law molecules of the constituents stick to each other to the same degree as they do to themselves. For example, if the constituents are X and Y, then X sticks to Y with roughly equal energy as X does with X and Y does with Y. A so-called positive deviation from Raoult's law results when the constituents have a disaffinity for each other – that is X sticks to X and Y to Y better than X sticks to Y. Because this results in the mixture having less total affinity of the molecules than the pure constituents, they more readily escape from the stuck-together phase, which is to say the liquid phase, and into the vapor phase. When X sticks to Y more aggressively than X does to X and Y does to Y, the result is a negative deviation from Raoult's law. In this case because the molecules in the mixture are sticking together more than in the pure constituents, they are more reluctant to escape the stuck-together liquid phase.
When the deviation is great enough to cause a maximum or minimum in
the vapor pressure versus composition function, it is a mathematical
consequence that at that point, the vapor will have the same
composition as the liquid, resulting in an azeotrope.

Why do things stick to each other?
I like to think of  atoms as having dispersion and electrostatic attractions/repulsion. This is a simplification, but that is not a bad thing for a stack exchange answer. The core of an atom interacts with the core of other atoms, a simple model of this is the Lennard-Jones model. Far away they attract, really close they push away from each other. Around the core are electrons. They interact with other atoms and their electrons too.
Hydrogen is interesting because when it is part of a molecule, usually the rest of the molecule hogs the electrons, so the hydrogen becomes positively charged. This happens in other molecules too, so they have areas that are more positive and also other areas on them that are more negative. Thus these now positive hydrogens become attracted to the overly negative areas of other molecules and they "stick" to them but only for a short while because they are after all covalently bonded to their own molecule, and this is much stronger than any short "flings" hydrogen may have with other molecules.


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Comparing SN2 reaction rates
I've read in a book that the main factor for determining SN2 reaction rate is steric hindrance. The lesser it is, the faster the reaction.
So consider this question:

$\ce{KI}$ in acetone undergoes SN2 reaction with each of $\ce{P,Q,R}$ and $\ce{S}$. The rates of reaction vary as:

I do not understand how $\ce{S}$ is the fastest. In my opinion it should be the slowest as it is very sterically hindered. What is going on here?

Source: Joint Entrance Exam (JEE) 2013 India

There are a number of factors that can influence the rate of an $\mathrm{S_{N}2}$ reaction.  Solvent, leaving group stability, attacking group nucleophilicity, steric factors and electronic factors.
In the series of compounds you've presented, all of these parameters are held constant except for the steric and electronic factors.  Considering only steric factors for a moment we would order the compounds as P>S~R>Q.  But now let's consider electronic factors.
Considering the allyl system in compound R, the transition state looks something like this

image source
The p orbitals in the adjacent double bond overlap with the p orbital at the $\mathrm{S_{N}2}$ reaction center (remember that $\mathrm{S_{N}2}$ transition states are roughly $\ce{sp^2}$ hybridized) and stabilizes the transition state through resonance delocalization.  Note that the transition state is electron rich, there is a full unit of negative charge in the transition state.
When a carbonyl is placed next to the $\mathrm{S_{N}2}$ reaction center, we replace the $\ce{C=C}$ double bond in the above drawing with a $\ce{C=O}$ double bond.  Again there is overlap and resonance stabilization of the transition state.  But in the carbonyl case the stabilization is even greater than with the $\ce{C=C}$ double bond because the carbonyl carbon is positively polarized and inductively stabilizes the electron rich (remember that negative charge) transition state even more.

image source
While we would expect the allyl case (R) to be faster than the ethyl case, in your series of compounds apparently it is not faster than the methyl case.
(Note: often the allyl system will react a bit faster than the methyl analogue, but the two rates are usually close)
However, the additional inductive stabilization from the carbonyl allows it to react faster than the methyl compound so our final reactivity order becomes S>P>R>Q.

You beat me by a few seconds :-) . I wanted to add that the iodine electrons are diffuse and hence the partial charge of the attacked carbon does not matter that much. It's only about the transition state stabilization. I wouldn't say that the allylic stabilization is in general less important than the reduction of steric hindrance when going from methyl to primary substituted carbon. Or am I wrong in thinking that? +1

Hey , thanks for the answer. Okay I'm now confused. Rajesh's answer says S is the best because of inductive effect which actually gives that carbon a slight positive charge. You're saying that because of orbital overlap it gets a slight negative charge. What is correct?

@Jori  No you're right Jori, often the ally compound will react faster than the methyl analogue.

Or am I misunderstanding and you're actually saying that transition state is stabilized because its extra negative charge is 'taken away' from it thus stabilizing it?

@Helpneeded  I'm saying two things. 1) orbital overlap stabilizes the transition state through resonance and 2) because the polarization in a carbonyl group makes the carbonyl carbon electron deficient the carbonyl carbon further stabilizes the transition state through an inductive effect.

Ron , so you actually mean that the carbon attached to chlorine doesn't like the negative charge and so it is stabilized by resonance and inductive effect right? I'm sorry for misunderstanding.

Yes, that's what I'm saying.

Thank you. But why are those orbitals SP2 hybridized?

Just to elaborate @Jori's last comment:  Delocaization of the negative charge is preferable to having it localized on 1 atom.  So as you said, "the carbon attached to chlorine doesn't like the negative charge" - Correct, it "likes" it better delocalized.

A key part of the SN2 reaction is that the TS is roughly sp2 hybridized, hence the availability of a p orbital in the TS.  It comes from the inversion of configuration in the reaction mechanism.  We start with an sp3 bond to the leaving group, the TS has the p orbital and bonds to both the leaving and attacking group, finally the "umbrella" inverts and we have an sp3 bond to the attacking group.

Aah I see. By the way I just checked Jerry march's Advanced organic chemistry and it gives rates as following - Methyl 30 Isobutyl 0.03
Ethyl 1 Neopentyl 10^-5
Propyl 0.4 Allyl 40
Butyl 0.4 Benzyl 120
Isopropyl 0.025
So it means Allyl is faster than methyl. What do you think? Do you know of any other resource we can check?

Generally allyl is a bit faster than methyl, but they are close, so it depends on the specific series your considering as to which one is actually faster.

@Helpneeded Ron's answer is based on the T.S while mine is about the nucleophilic attack . Both would provide the same answer .

@ron Hi, one thing in your answer confuses me. How can a transition state be stabilized by resonance effect? As such, the negative charge on the transition state is a $\delta^-$ and not a unit negative charge, so is it possible for even $\delta^-$ to delocalize? Thanks for your answer!

@GaurangTandon why partial negative charge cannot be stabilized by resonance?

@ron Excellent answer. I just want to ensure that; is it correct to always consider the attacked centre ( means the roughly sp2 hybridised carbon) of Sn2 transition state as dominantly negative charged ? But I have seen in many places that ; that carbon is electrophilic. Aren't these two contradictory?

@Shiva  No, they are not contradictory; carbanions exist.  In this case the SN2 center has some negative charge on it, the full unit of negative charge is spread out over several atoms.

The reaction rate of SN2 depends primarily on the leaving group involved. Since in your question, $\ce{-Cl}$ is common in all 4 options we move on to the next important criteria.
The partial positive charge on carbon (or the electrophilicity of the carbon) is considered more important than the steric effect that you mentioned. Look at Wikipedia's first line.
The nucleophile attacks the more electrophilic carbon. This can be easily compared by the Inductive effect of the substituents.
This gives us the correct answer as option B.

Nice , I'm glad that you told me about that factor because it is not in any book.

If you'll look at the transition state in SN2 mechanism, you'll find that there's a partial negative charge on carbon.
In option S, The leaving group is attached to a 1 degree carbon and there will be a -M by the carbonyl group making the negative charge most stable.
In P, we have a 1 degree carbon with minimum hindrance.
In R, we have a 1 degree carbon and on which the negative charge can be stabilized by resonance.
In Q, we have a 2 degree carbon means steric hindrance, further the +I effect of CH3 will make the negative charge on carbon even worse.
So correct answer is B.


text:
Does sodium form complexes like transition metal ions?
I realise that there is a similar question here Difference between sodium ion and a transition metal ion dissolving in water? and it seems to answer my question, however I was reading about how aluminion ions form complexes in water, and aluminium is not a transition metal. So how can aluminium form complexes like transition metal ions, but sodium does not?

Although less common than transition metal complexes, sodium does form complexes with some ligands, particularly oxygen based ligands.
Aqua complexes are formed in aqueous solution, the most common being $\ce{[Na(H2O)6]+}$.
Sodium forms many complexes with crown ethers, cryptands and other related ligands. For example, 15-crown-5:

A sodium ion has no outer electrons, and only has a charge of one. The complexes it forms are mostly defined by the structure of the ligand(s) and electrostatics.
Simply speaking, a pure aluminium 3+ ion would be too highly charged to be allowed in condensed matter. The hughe electrostatic field would just rip off electrons from any unsuspecting neighbour. Complexes of Al3+ ions are partially covalent, involving ligand electrons "donated" into the outer (formally unoccupied) orbitals, making directional bonds to the ligand or to the eg. oxygen atoms in alumina.
Complexes of transition metals work the same, only they have more outer orbitals, which also have some electrons, so there are even more possibilities to form complex complexes. :-)


text:
How surface passivation works?
Surface passivation is done to deal with "dangling bonds" (for example, $\mathrm{SiO_2}$ layer on top of a $\mathrm{Si}$ layer). But isn't the new surface will have dangling bonds on its own at the new formed surface?

Surface passivation creates different chemistry on the surface than the chemistry of the bulk solid. Imagine bulk Si with a clean surface (presumably in a vacuum and freshly made). It can be thought of as having "dangling bonds", though this is not necessarily the best way to think about it. Think more generally that the structure on the surface is different to the bulk structure. There may be different elements present and the bonds may be "satisfied". Or they may be strained compared to the bulk material.
Expose the Si to, for example, O2 and the surface layer (your dangling bonds, if you like) will react to give some SiO2 just on the top layer. This new layer may be much less reactive than bulk silicon in normal air, hence can be said to passivate the surface.
If the chemical and physical properties of the new layer are right, the layer will protect the bulk from further reaction: this is the definition of passivation. A similar situation occurs naturally for aluminium, for example, where the Al2O3 layer that forms is both strong and relatively impermeable to air preventing further reaction (aluminium with the layer removed will react so rapidly with oxygen it gets warm and disintegrates). Iron, on the other hand, forms rust which doesn't protect the surface as it is flaky and weaker than the bulk metal. To passivate iron you need to add other elements which encourage the formation of stronger passivating layers (this is how many steels are made not to rust).
Chemical manipulation of the surface can create specific types of passivation that serve a particular purpose rather than relying on the natural reactions of air and the surface. This is common in the semiconductor industry and can also be used to generate the coloured layers on aluminium and titanium products where the goal is aesthetic as much as protective.

So there would still be "dangling bonds" for the new layer, only less reactive than the previous surface?

@Sparkler IF you insist on thinking in terms of dangling bonds then, yes. But this is a simplistic way of thinking as some surfaces won't have any meaningful equivalent to "dangling bonds" especially when passivated.

So the type of required passivation depends on the target environment, right?

@Sparkler yes. different passivation will work in different environments.

What other thinking than "unsatisfied bonds" can I adopt?

@Sparkler Think more generally that the structure on the surface is different to the bulk structure. There may be different elements present and the bonds may be "satisfied". Or they may be strained compared to the bulk material.

So in your example, the SiO2 surface structure will be in fact different from the bulk, but more stable than Si?


text:
Heat Capacity temperature dependence and Gibbs energy
I got this problem in school as a tool to get ready for International Chemistry Olympiad, but I have some problems with this challenge.
So, the problem is to calculate as precise as possible the Gibbs free energy for $\ce{CaCO3}$ decomposition at 1200 K.
I know that $\mathrm{d}G=\mathrm{d}H-T\cdot\mathrm{d}S$.
But $\mathrm{d}S$ and $\mathrm{d}H$ are temperature dependant, so
$$\mathrm{d}H=\mathrm{d}H_0+\int_{T_0}^TC_p\mathrm{d}T=\mathrm{d}H_0+\mathrm{d}C_p\cdot\left(T-T_0\right)$$
and
$$\mathrm{d}S=\mathrm{d}S_0+\ln\left(T/T_0\right)\cdot\mathrm{d}C_p$$
But molar heat capacity at constant pressure is also temperature dependant, and the equation is
$C_p=a+b\cdot T+c/T^2$, where a, b, c coefficients for each compound at specific phase.
But the problem is with both solids, because this equation can not be used for them, because in 1200 K they are not gases and Einstein’s Quantum Mechanics and Debye models do not work, because Einstein’s approximates are maxed at $C_p\approx3R$, which is about 24.9, but $C_p$ at 298 K, 1 atm for both solids in tables are > 24.9.
Where is the problem? I would appreciate any possible lead to mistake/ correct answer.

At constant pressure, $dG=-SdT$.  Also, the definition of G is $G=H-TS$, so $S=\frac{H-G}{T}$.  Substituting,
$$dG=\frac{G-H}{T}dT$$
This equation can be manipulated into the form:
$$\frac{d(G/T)}{dT}=-\frac{H}{T^2}$$
So,
$$\frac{d(\Delta G/T)}{dT}=-\frac{\Delta H}{T^2}$$
I leave the rest up to you.
A math error was corrected on 10/16/15

The answer given by Chester Miller is completely correct but I think it just causes old problems on new places and does not solve the problem.
In the last (van't Hoff) equation an enthalpy is assumed that is independent from temperature. In general it is:
$$
\frac{\partial \Delta G/T} {\partial T} = \frac{\Delta H (T)}{T}
$$
This leads to:
$$
\left. \frac{\partial \ln K}{\partial T} \right|_p = \frac{ \Delta H^0(T)}{RT^2}
$$
Also in this case you end up searching for the temperature dependence of the enthalpy hence the heat capacity.
I would just recommend to get experimental data for the polynomials, perhaps CRC handbook. I do not know any theoretical method better than Debye modell that you could still apply just with calculator and pen.

The method I gave certainly does solve the problem.  All you need to do is use Hess's law to determine the heat of reaction as a function of temperature, using measured values of the temperature-dependent heat capacities of the products and reactants (analytically fit to appropriate functional forms).  Such analytical fits are readily available in the literature.  You then integrate the right hand side of the equation with respect to T to get the standard free energy change at the desired temperature.


text:
Experimental difficulty with the iodine clock reaction
I did the iodine clock reaction using sodium metabisulfite and potassium iodate. As I understand it, increasing concentration is supposed to speed up your reaction rate. I did 10 trials. The first 5 I varied the concentration of metabisulfite and kept the iodate constant. Trials 6-10 I varied the potassium iodate and kept the meta constant. I don't understand why as I increased concentration, the reaction rate became slower? For example in trial 6 I had 2 ml of sodium metabisulfite and 4 ml of potassium iodate, that only took 7 seconds. In trial 10 I had 2 ml of meta and 10 ml of potassium iodate and that took 17 seconds? Does anyone know what I did wrong? Increasing concentrations should have increased collisions and made the rate faster, but it did the opposite! Please help me figure out why things went wrong!

The theory is correct. Increasing concentrations increase the number of effective collisions per unit time and the rate increases. So accordingly even in the experiment it should happen like that.
So in visualizing your situation, it must have been an experimental error and  not the theory. (Say there must have been less amount of $\ce{H2O2}$ in the mixture).
In this reaction what happens first is the conversion of $\ce{I-}$ to $\ce{I2}$ (have to add $\ce{H2O2}$). This $\ce{I2}$ will then react with metabisulphite or (even sodium thiosulphate) to produce $\ce{I-}$. Since you have added a constant value of meta in trials 6-10 there is a definite amount of $\ce{I2}$ removed in each trial by meta then the remaining $\ce{I2}$ will react with starch to give the color change.


text:
Is graphite more stable than diamond?
The change in enthalpy of carbon in diamond form to graphite is negative. This suggests that graphite should be more stable than diamond. Is it true?
What I thought was that graphite has van der Waals forces while diamond has covalent forces so diamond should be more stable. But the melting point of graphite is more than diamond which implies that it is thermodynamically more stable. I am thoroughly confused. So which is more stable?

At normal room temperature and pressure graphite is (slightly) more stable than diamond. But the melting point is not a good indicator of this.
Melting points are determined by the bonding structure of the solid and any potential liquid (indeed there may not be a liquid phase at some pressures). They don't necessarily tell you whether a different bonding structure is more thermodynamically stable.
The thermodynamic stability of different carbon structures depends on subtle tradeoffs among several factors in the bonding of the various solids (allotropes) than can be formed with different bonding arrangements. There is no easy way to tell which is the most thermodynamically stable other than by measurement or some very high-powered calculations about the quantum mechanics of the bonding.
The different allotropes have radically different structures (tetrahedral in diamond, a ball of hexagons and pentagons in buckminsterfullerene and flat plates of hexagons in graphite). Changing one into another requires a lot of bond breaking which isn't going to happen much at room temperature. Interconversion is possible at high pressures and temperatures when carbon is dissolved in liquid rock or metal and the carbon allotrope that forms is more likely to be whatever is stable at that whatever the pressure is (diamond is more stable at higher pressure which is why the earth or industry needs very high pressures to create diamond).
So, for most practical purposes, it doesn't matter than graphite is the more stable allotrope: your diamonds are not going to spontaneously turn to ash during your, or the universe's, lifetime.

This questions calls for an answer from thermodynamics.

The figure provided above, lifted from here, is what we call a phase diagram. On the abscissa is temperature in units of Kelvin and on the ordinate is the pressure given in units of Gigapascals. (For your own reference, 1 GPa is nearly 10,000 times the pressure we live under, the earth's core is estimated to be over 300 GPa!) This diagram represents the input of experimental work and extrapolation with mathematical tools provided by thermodynamics. What this figure shows very clearly is that for a given temperature and pressure which phase is thermodynamically preferred.
A line between the diamond and graphite phases represents a phase boundary, and tell us precisely what temperature and pressure are required for these two phases to be in equilibrium. The phase boundary between liquid carbon and graphite represents the melting temperature for graphite; the same can be said for the diamond/carbon line, this again is the melting temperature of diamond.
Our melting temperature therefore depends on phase and pressure. So if you only have two melting points, does that really tell you which is thermodynamically preferred at room pressure (0.0001 GPa)? No, it does not. It is more subtle than that.
So how would we determine the relative stability of graphite and diamond at room temperature and pressure? For that we compute the standard Gibbs energy of reaction (at STP), this requires we use some values from a table: $\Delta G_f^0 (diamond) = 2.90 (kJ/mol)$ and $\Delta G_f^0 (graphite) = 0 (kJ/mol)$.
The reaction we want to calculate is: $C(diamond) \rightarrow C(graphite)$
Now we know that $\Delta G_{rxn} = \Sigma_{products} \Delta G_{f} - \Sigma_{reactants} \Delta G_{f} = 0 - 2.90 = -2.90 (kJ/mol)$. So $\Delta G_{rxn} < 0$ tells us at 298 K and 1 atm of pressure, diamond would spontaneously form graphite if it could; however, for this transition carbon atoms would have to change their location within a lattice and the rate at which this happens in a solid tends to be very slow at low temperatures. Hence, we call diamond a metastable phase under STP.
Now if you provide a sufficient energy to get over this activation barrier, then you will be able to see what thermodynamics prefers. Personally, I find this video very compelling.

Is the reaction of conversion of graphite to diamond a phase change? I am asking this because when water turns from ice to liquid the transition is fast whereas for diamond it takes years. Is the conversion of two allotropes a phase change?


text:
Can decarboxylation be done under carbon dioxide?
From Why is the decarboxylation of 2-hydroxybenzoic acid (and related compounds) done 'under nitrogen'? I confirmed that nitrogen is used to exclude oxygen - in order to prevent oxidation of the material.
Can $\ce{CO_2}$ be substituted for nitrogen, or will the presence of carbon dioxide slow down the decarboxylation reaction (which is causing carbon dioxide to be released)?

Carbon dioxide is relatively reactive in comparison to nitrogen and argon. You cannot assume your $\ce{CO2}$ blanket won't interfere with your reaction.

@Jori "You cannot assume your CO2 blanket won't interfere with your reaction."  ..hmm.  Not good.  But are you saying that you know it will interfere with decarboxylation reaction, or more that it is (simply) rash to presume it will be OK?  BTW - how did you make that 2 as a subscript in a comment?  I use <sub>..</sub> for the question.

Because nitrogen is the standard inert gas in lab practice. If you are "rich" You can use argon, but carbon dioxide is seldom used, because it won't work on alkaline solutions.

or will the presence of carbon dioxide slow down the reaction

First, what you presumably mean is influencing the equilibrium unfavourably, don't you? You should learn to discern kinetics and equilibrium.
Answer: not really. Whether you have a carbon dioxide atmosphere or not will shift the decomposition temperature not more than some degrees.

Excellent.  I wanted to try CO2 since it is a lot easier (& cheaper) to thaw some dry ice and use the resulting gas than it is to get a compressed bottle of nitrogen (or argon) and a regulator.  And no, I'm not rich..  For nitrogen it would be 140 dollars for the regulator, and around  85 for 1 cubic meter of scientific grade nitrogen.  Now I just have to decide between heating the material to be decarboxylated using a liquid bath, or heating the CO2 before it goes into the reaction chamber..


text:
Having trouble understanding the iron-carbon phase diagram

At room temperature, alpha and alpha+$\ce{Fe3C}$ are present (depending on carbon content), which is ferrite and pearlite. In the lab we polished and etched samples of carbon steel to observe the microstructure, and saw both ferrite grains AND pearlite grains. Not unlike the following:

However, from my understanding of the phase diagram, only one phase should be present. For any steel with less than 0.02% carbon we should see only ferrite, and for steel with >0.02% carbon we should see only pearlite. Though this was not the case, and rather there was a mix of pearlite and ferrite, with the higher carbon content steel having a higher proportion of pearlite.
It would be much appreciated if someone could explain to me what is happening here. Links to further reading would be helpful but not required.

Perhaps the sample was formed at a higher temperature and then cooled quickly preventing the natural equilibrium from being reached.

read about quenching in application to steel.

For any steel with less than 0.02% carbon we should see only ferrite, and for steel with >0.02% carbon we should see only pearlite.

You have got a small misunderstanding here. The maximum solubility of carbon at $723\ \mathrm{^\circ C}$ is 0.02%. At room temperature it is very low ($0.005%$ at $0\ \mathrm{^\circ C}$). Also for carbon level between 0.02 and 0.8 you wont see only pearlite, you will see a combination of ferrite and pearlite for sure. Only at 0.8% you can observe pure pearlite(100%). Also note that the phase diagram is in such a way that if you cool from the ferrite+ austenite region at $800\ \mathrm{^\circ C}$ with carbon content < 0.02% we can still get a ferrite + pearlite + even austenite/martensite.
Phase diagrams give an ideal picture. In real situations, it is very tough to get very slow cooling rates to get the equilibrium structure.

The phase diagram shows two phases, ferrite/iron, and iron carbide . It does not define the morphology; the form of the carbide. It could be in particles although we usually find the carbide in layers/ lamellae = pearlite ( with typical air cooling). Cooling rates , reheating and other thermal ( and mechanical) actions will have profound influence of the microstructures. TTT diagrams  or Isothermal Transformation Diagrams explain the affect of cooling rates but they may be more than you want.


text:
Chemical formula of a compound crystallized in a cubic lattice
Question

In a cubic lattice of $\ce{XYZ},$ $\ce{X}$ atoms are present at all corners except one corner which is occupied by $\ce{Y}$ atoms. $\ce{Z}$ atoms are present at face centres. What is the formula of compound?

Answer

 $\ce{X2YZ24}$

My solution
Let the number of atoms in a unit cell be $N.$ Then
$$
\begin{align}
N(\ce{X}) &= 7 × \frac{1}{8} = \frac{7}{8}\\
N(\ce{Y}) &= 1 × \frac{1}{8} = \frac{1}{8}\\
N(\ce{Z}) &= 6 × \frac{1}{2} = 3
\end{align}
$$
Therefore, the ratio would be
$$N(\ce{X}):N(\ce{Y}):N(\ce{Z}) = \frac{7}{8}:\frac{1}{8}:3,$$
which is the same thing as $7:1:24.$ So, the formula is $\ce{X7YZ24}.$
What am I doing wrong?

Your solutions looks fine. May be the answer given is wrong,

You need to take into account which atoms are shared with other unit cells. A face centre is shared with one, an edge centre with 4 other cells etc.

The answer given is wrong and the question is poorly worded; a periodic cubic lattice requires that all the corners in a unit cell be of the same atom. If one of the corners of a cell is made of a different atom, then you're not actually looking at the unit cell.

I would like to say your answer is right, but the question is wrong.
Remember that a lattice is "infinitely" repeated units in 3 dimensional space so you should be able to expand the unit cell in x, y and z axes, indefinitely.
Now imagine your cubic lattice with an $\ce{Y}$ atom in just 1 (out of 8) corners.
Expanding the unit cell in the x-axis will necessarily duplicate the $\ce{Y}$ atom in the x-axis because all expanded unit cells must be identical in composition. Similarly, expanding the unit cell in the y-axis will duplicate the $\ce{Y}$ atom in the y-axis, and expanding the unit cell in the z-axis will duplicate the $\ce{Y}$ atom in the z-axis.
In the end you will find the having one corner as $\ce{Y}$ atom will necessarily have ALL corners as $\ce{Y}$ atoms. In other words, $\ce{X}$ IS $\ce{Y}$, and the formula is necessarily $\ce{XZ_3}$ or $\ce{YZ_3}$
Unfortunately, I don'k think the designer of the question saw his fatal contradiction because this very question is seen in the (mock) Joint Entrance Examination (JEE) in India.

And $\ce{X_2YZ_24}$ is not in the options even in the wrong question.

There is a way to interpret the unit cell that matches the question.  Assume $X$ and $Y$ occur randomly at all corners with on average one corner of a cube having $Y$.  Contrived, yes, but workable for cases such as $\ce{Nb(C,N)}$ which appears as a precipitate in the hot rolling of some steels.

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


text:
Are the energies of orbitals that are not filled in an atom, equal?
In a hydrogen atom only 1s orbital is filled. Does this mean that the energies of all other orbitals are equal? If so,why?

Do you think that your hypothesis (all unoccupied orbitals have equal energy) is also valid for a hydrogen molecule? If so, how would you then explain the Lyman and Paschen series in the UV spectrum of this molecule?

A better question is: Which of these is real - the orbital or the electron? What has energy, the orbital or the electron in the orbital? If it is the electron that has energy, then the energy of an empty orbital is an undefinable quantity.

In response of the comment above, the energy is of the system, not of electrón

Sure, if they are unoccupied they contribute nothing to the total energy of the system. Remember that orbitals are mathematical constructs and only have meaning with regard to their energy when being physically occupied by an electron that adheres this mathematical description. The energy required to fill such an orbital however differs (vastly) from orbital to orbital and nucleus to nucleus (see for example http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf).


text:
What is the electronic configuration of Fe(II) ion?
The electronic configuration of Fe is $\ce{[Ar] 3d^6 4s^2}$.
So after removing two electrons the configuration becomes:
$\ce{[Ar] 3d^6}$
But why can't the electrons rearrange themselves to give a more stable
$\ce{[Ar] 3d^5 4s^1}$ configuration?

When researching the explanation for this phenomenon, I found this:

It is natural to question why one or two electrons are usually pushed into a higher energy orbital. The answer is because $\mathrm{3d}$ orbitals are more compact than $\mathrm{4s}$, and as a result any electrons entering $\mathrm{3d}$ orbitals will experience greater mutual repulsion. The slightly unsettling feature is that although the relevant $\mathrm{s}$ orbital can relieve such additional electron-electron repulsion, different atoms do not always make full use of this form of sheltering because the situation is more complicated than just described. One thing to consider is that nuclear charge increases as we move through the atoms, and there is a complicated set of interactions between the electrons and the nucleus as well as between the electrons themselves. This is what ultimately produces an electronic configuration and, contrary to what some educators may wish for, there is no simple qualitative rule of thumb that can cope with this complicated situation.

On a semi-related note, according to one source I found (the source sites many others as well), it is in fact the $\mathrm{3d}$ electrons that fill first, followed by the $\mathrm{4s}$ electrons. This more satisfactorily explains why the $\mathrm{s}$ electrons are first lost in the ionization of the $\mathrm{d}$-block transition metals.

But wont this configuration explain why Fe2+ is more stable than Fe+ even  when Fe+ has half filled orbital but Fe2+ does not?

Your mixed up with the use of half orbital. When iron become to $\ce{Fe^2+}$, it needs to donate two valance electrons away which is at the highest energy (orbital $\ce{4s}$). So, that is why we write $\ce{Fe^2+}$ as $\ce{[Ar] 3d^6}$.

But given that half-filled orbitals have a special stability, the OP is asking, why isn't a 4s1 3d5 configuration preferred over a 4s0 3d6 configuration.


text:
Why use a salt bridge at all in an electrochemical cell?
I was reading about electrochemical cells and I really don't understand why a salt bridge is needed. Can't you just combine the solutions? I've found a lot of webpages explaining electrochemical cells but strangely enough none of them seem to find this important enough to explain. All thoughts appreciated :)

Can you give an example of cell's material ? (electrolyte and electrode)

I'm going by this: http://www.chem1.com/acad/webtext/elchem/ec2.html so a Zn and Cu electrode, and Zn[sup]2+[/sup] and Cu2+ ions (and NO3-)

What do you think would happen if you put a zinc plate into a copper ion solution?

I think I understand, the copper ions would combine with electrons from the Zn because copper apparently attracts electrons much harder, and then Zn would dissolve to give Zn2+. So you can't bring the solutions together because then your electrode would dissolve.

Correct. The zinc electrode would partially dissolve, and most of the copper ions in solution would deposit on it as a copper metal plating. Since the battery needs copper ions to work, by causing most of them to deposit directly on the zinc electrode, you're basically discharging the cell without collecting any electrons through the external circuit.

This before-and-after video https://www.youtube.com/watch?v=LahawEMMvvY shows what happens in the voltaic cell before the bridge salt is put in place.

The use of a salt bridge is to maintain the electrical neutrality throughout the cell.
A salt bridge will allow the movement of ions.
Electrons flow from anode to cathode.Due to this a negative charge will be developed in the cathode and a positive charge in the anode.This charge developed will oppose the electron flow (current from anode to cathode) and at one level the electron flow will be ceased.So the cell's action will be stopped.
So in order to prevent this, salt bridges are used.
There is a similar question in this site.Please refer that.
Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

There are a lot of cases where you can work without a salt bridge. If the two reactions(reduction and oxidation) don't interfere with each other, you can just let the solution be in a single pot.
It is only there to separate the two solutions and still maintain ionic contact.


text:
Frequency, the only real characteristics of electromagnetic radiation ??
My book says: frequency is considered as the only REAL characteristics of electromagnetic radiation, because it is unchangeble if the radiation comes up with some other environment, while the wavelength and the wavelength number do change. How can wavelength change and the frequency not ? The formula tells us very clear that while frequency increases, wavelength decreases and vice versa ?!

Recall that frequency $f$, wavelength $\lambda$ and the speed $v$ of a wave are all related as follows
$$
    v = f \lambda \, .
$$
So presumably, your book is talking about a situation when an electromagnetic wave crosses boundaries between different media. In such cases frequency indeed remains constant, while the speed of the wave, and consequently, its wavelength change.

Frequency of an electromagnetic wave is the characteristic of solely the source of the wave. A charge's frequency of oscillation would determine the frequency of the wave it produces. Refraction doesn't cause change in frequency, only wavelength, which results in a change in velocity as well.


text:
Integration of the first order rate equation
Given the simple first order reaction $\ce{A -> P}$ derive the integrated rate law.
$$\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -k[\ce{A}]$$
Collect terms:
$$\frac{\mathrm d[\ce{A}]}{[\ce{A}]} = -k\,\mathrm dt$$
Now for the bit I need help with, the integration:
Apparently the integrated form of $\frac{\mathrm d[\ce{A}]}{[\ce{A}]}= \ln[\ce{A}]$ but I'm struggling to see exactly how.
The next step (usually omitted)I think should be to split the fraction before integration
$$\frac{\mathrm d[\ce{A}]}{1}\frac{1}{[\ce{A}]}=\mathrm d[\ce{A}]\frac{1}{[\ce{A}]}=-kt$$
$$\int{\mathrm d[\ce{A}]\frac{1}{[\ce{A}]}}=\int{-kt}$$
$$\int{\mathrm d[\ce{A}]}\int{\frac{1}{[\ce{A}]}}=\int{-kt}$$
Now, to evaluate this I use the general integration law for a constant $\int{[\ce{A}]\,\mathrm dt} = [\ce{A}]t + C$ and reciprocal  $\int{\frac{1}{[\ce{A}]}=\ln[\ce{A}]+C}$. Therefore $$\int{\mathrm d[\ce{A}]}\int{\frac{1}{[\ce{A}]}}=[\ce{A}]t + C_1+ \ln[\ce{A}]+C_2=\int{\frac{\mathrm d[\ce{A}]}{[\ce{A}]}}$$
I know this is wrong because the textbooks say something different but I don't know how to correct it. Can anyone help?

Klaus and Binary Geek have already made the main points but I think it would be helpful to see a full explanation.
The differential equation we are trying to solve is: $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k[\ce{A}]$$
Separating the variables: $$\int{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int{-k\,\mathrm{d}t}$$
Substituting in the limits; initial concentration $[\ce{A}]_0$ at time $t=0$ and final concentration $[\ce{A}]$ at time $t$: $$\int_{[\ce{A}]_0}^{[\ce{A}]}{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int_0^t{-k\,\mathrm{d}t}$$
Integrating both sides: $$\bigg[{\ln[\ce{A}]}\bigg]_{[\ce{A}]_0}^{[\ce{A}]} = -k\bigg[{t}\bigg]_0^t$$
Evaluating: $$\ln[\ce{A}] - \ln[\ce{A}]_0 = -k(t-0)$$
Condensing the logarithms: $$\ln{\left(\frac{[\ce{A}]}{[\ce{A}]_0}\right)} = -kt$$
Solving for $[\ce{A}]$: $$[\ce{A}] = [\ce{A}]_0 \mathrm{e}^{-kt}$$

You can't write $\int \frac{\mathrm{d}[A]}{[A]} $ as $\int \mathrm{d}[A] \int\frac{1}{[A]}$.
$\int \frac{1}{[A]}$ doesn't make any sense. Integration is essentially summation of infinite infinitely small terms and so each term must be infinitely small. $\frac{1}{[A]}$ isn't infinitely small. $\frac{\mathrm{d}[A]}{[A]}$ is.
$\int \frac{1}{[A]} \neq \ln([A]) + C$ instead
$\int \frac{\mathrm{d}[a]}{[A]} = \ln([A]) + C$
Also as Klaus Warzecha has stated, you forgot to put the limits.

You omitted the limits for the integration and therefore miss the practical (and most important) aspect of the whole exercise!
Remember why all this integration is done:
It is about measuring  $[\ce{A}_t]$ over time, starting with $[\ce{A_0}]$ at $t = 0$, in order to confirm a hypothesis on the rate law.
If you work the limits for the integration in, you'll reach a useful functional relation between concentration of the starting material and time - useful for a chemist and a biologist ;)


text:
How do atoms combine?
Do atoms combine with the nearest atoms possible because of lower delta E (less energy needed to form a stable nucleus)?

If atoms combined with the nearest atom, then the air would be composed of $\ce {N2, O2, NO, NO2, ...}$. There are reasons that determine which compounds are preferred over other compounds.

Nucleus?? Are you talking about nuclear reactions?

"Combining with the nearest atom"? How do you know they will? Also, is this about a chemical or nuclear reaction?

Let's say we have an atomic soup that contains atoms (what follows also applies to molecules) $\ce{A, B}$ and $\ce{C}$.  From these reactants it would be possible to form products $\ce{A-B, A-C}$ and $\ce{B-C}$.  Just which of these molecules, if any, will form depends on several factors including the $\Delta E$ that you mention.
Background
We usually talk about a reaction coordinate, like the one shown below.

image source
We have our reactants on one side and products on the other, and a barrier connecting the two.  The height of the barrier is referred to as the activation energy ($E_{a}$).  The Arrhenius equation is often used to describe the rate (k) for this process, where A is the pre-exponential factor (more on this in a minute), R is a constant and T is the temperature (°K)
$k = Ae^{-E_a/(RT)}$
Back to the question
Just because atoms collide doesn't necessarily mean they will react to produce a product.  Sometimes they must collide from a certain direction.  Think of molecules with complex shapes, for them to react we need the other reactant to collide with a certain part of the larger molecule, maybe too from a certain angle or direction.  These geometrical factors are contained in the pre-exponential factor.  Also the collision must have enough energy to get the reactants over the energy barrier.
So just because molecules are close doesn't guarantee that they will react.  More important is how often molecules collide (concentration), the geometry of the collision (pre-exponential factor) and the energy (E) they carry.


text:
What is the significance of enthalpy definition not under constant pressure?
Under constant pressure (and only PV work), change in enthalpy can be shown to equal change in heat. If enthalpy is not meaningful outside of constant pressure, then why not use heat instead? So I think it must be meaningful outside of constant pressure scenarios, but I couldn't figure it out myself. Can someone help?

Do you know abt internal energy?

There are many processes that take place at constant pressure, such as the change of phase from a liquid to a vapor for a pure substance at constant temperature.  Also, the heats of chemical reactions and heats of formation of compounds are typically cataloged at constant temperature and pressure.  So the enthalpy comes into play in both these instances.
The first law of thermodynamics for a closed system is expressed conveniently in terms of the change in internal energy of the system.  However, for an open system (in which mass is flowing in and out) operating at steady state, the first law is more conveniently expressed in terms of the enthalpy, not the internal energy.  And, of course, all industrial applications of thermodynamics involving continuous processes are operating at steady state.
Chet


text:
Determining concentration of phosphate buffer from pH?
I bought a phosphate buffer from Sigma/Fluka (73212-1L) containing potassium dihydrogen phosphate ($\ce{KH_2PO_4}$) and disodium hydrogen phosphate ($\ce{Na_2HPO_4}$).
It is at pH 8.0 but they do not provide the phosphate concentration. Can I determine it from pH via titration?
This table shows the concentrations for a 0.2M solution:
http://www.sigmaaldrich.com/life-science/core-bioreagents/biological-buffers/learning-center/buffer-reference-center.html#phosphate

I don't get the product when I enter the catalog number on Sigma-Aldrich. How long have you had the bottle?

The closest product # I could find was 73208, http://www.sigmaaldrich.com/catalog/product/fluka/73208?lang=en&region=US, for sodium potassium phosphate buffer at pH 7.00, not 8.  The concentration of potassium phosphate monobasic in that product is listed as 7.08 g/L.  If your catalog # is correct, you should be able to contact Sigma Aldrich to get certificates of analyses for your product, which should include the confirmation of concentration.

I've had the bottle since the beginning of last year. Now it looks like it's discontinued. Back then, I called them and they said there was 0.004 mole/0.63 mole, but they sounded unsure and weren't able to give me any documents.  I have already decided to make my own, but my previous experiments have been failing and I wanted to find out if incorrect phosphate concentration was the cause.

The pH of the buffer can easily change since it is at the end of its range of it's buffering. Being more than 6 months old, you should check the pH of the buffer before using it. If it's critical that the pH is held near 8.0, I would see if a TRIS (Trizma) buffer would work. If it doesn't, there are several other chemicals (in the table above) that have a pKa near 8.

I just checked the pH and it's still at 8.0. Anyways, my experiments were done a while ago. Confirming the phosphate concentration experimentally could help with troubleshooting old experiments. Just wondering if there are too many unknowns.


text:
Definitive Guide To Removing Stains From Baby Clothes
I'd like someone to explain, scientifically, what makes some baby clothes' stains so hard to remove, and what the right way to treat the stains is, chemistry-wise. A lot of internet sites have solutions (pardon the pun) that are 'my experience' and anecdotal, which is better than nothing, but not always transferable (and sometimes may damage clothes).
So yeah... how do you get out the hard to remove orange stains (e.g. from iron, or carrot/sweet potato soup)? How about other things? Why? I'd love someone to make this the definitive guide to removing stains from baby clothes, if you can :D. (Feel free to expand beyond my stated question above so there's just a one-stop reference question here.)

Wash the garment in warm water with an enzyme detergent and oxygen bleach combo, such as Bio Kleen Oxygen Bleach Plus. To get rid of baby food stains, dunk the clothing in cold water, then soak the stain for at least 15 minutes in equal parts water and rubbing alcohol, which is safe for most fabrics when diluted.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Note that the OP explictly asked for the chemistry behind things, so your answer does not exactly hit the question.


text:
Why is temperature regulation an important factor for producing explosives?
I was wondering why temperature regulation is an important factor for producing explosives. I know many are exothermic reactions but what would happen if you did not use an ice bath and allow the temperature to go too high (Considering the explosive substance has not already been formed in the reaction)? Would it explode? For example when mixing acetone with sulfuric acid and hydrogen peroxide to make TATP.

"Mixing acetone with sulfuric acid and hydrogen" will only drag the water from the upper organic layer, generate a bit of heat and maybe some brownish black gunk between the layers.

@KlausWarzecha   Acetone and sulphuric acid mix: there won't be an organic layer. if he means hydrogen peroxide (this is the acid catalysed reaction that produces  tetra acetone tetra peroxide) the same is true. The worst case is the reaction goes bang. As does the careless chemist doing it.

@matt_black Are you sure that counts for conc. sulfuric acid too? TATP is actually triacetone triperoxide, and the most interesting part about it is the two stable conformations at room temperature.

@KlausWarzecha Careless typing on my part (or careless memory). As far as I remember the sulphuric acid is only used in small quantities to catalyse the reaction. The liquid will be hydrogen peroxide solution and acetone (which mix). I'd look it up but I'm worried the NSA or GCHQ might flag my internet usage as a terrorism suspect (TATP was used in the london tube bombings, if memory serves!)

@matt_black It was indeed, and it was the cause of an explosion at UC Davis in 2013, if I remember correctly. Two pretty good reasons to talk people out of playing with this material!

My advice on TATP is simple:
If you're an inexperienced amateur researcher, do the synthesis on paper - and only there!
Actually, the synthesis on TATP can be performed at temperatures above 0 °C. When Richard Wolffenstein accidently found in 1895 during his works on coniine that TATP had been formed in a mixture of 10% hydrogen peroxide and acetone, he did not use any acidic catalyst and had kept the mixture at room temperature for weeks:

Ueber die Einwirkung von Wasserstoffsuperoxyd auf Aceton und Mesityloxyd, Ber. dtsch. chem. Ges., 1895, 28, 2265-2269 (DOI)

When he had collected the precipitate and recrystallized it from diethyl ether, it exploded, which Wolffenstein attributed to to the formation of peroxides in ether.
One of the first reports on the synthesis of TATP under acidic conditions was given by Milas and Golubovic:

Studies in Organic Peroxides. XXVI. Organic Peroxides Derived from Acetone and Hydrogen Peroxide,
  J. Am. Chem. Soc., 1959, 81, 6461-6462 (DOI)

Btw, the experimental part of this article starts with

These peroxides are dangerously explosive and should be handled with extreme care.

Under controlled lab conditions, the decomposition of TATP in solution seemingly has been examined:

J.C. Oxley, J.L. Smith and H. Chen,
  Decomposition of a Multi-Peroxidic Compound: Triacetone Triperoxide (TATP), Propellants, Explosives, Pyrotechnics, 2002, 27, 209-216

and the activation energy for this first-order reaction was calculated to be around $\mathrm{150\,kJ\cdot mol^{-1}}$.
What makes TATP different from most of the other known explosives is

the molecule (obviously) does not contain nitro groups. Typical detectors for explosives thus fail here. Together with the simplicity of the synthesis, this is a real problem in law enforcement!
the explosion does not generate a lot of heat, but is an entropic explosion.

The latter was througly examined by Ehud Keinan and coworkers:

Decomposition of Triacetone Triperoxide Is an Entropic Explosion, J. Am. Chem. Soc., 2005, 127, 1146-1159. (DOI)

If you don't want to go though all their calculations of a possible decomposition mechanism, the following hopefully is impressive enough:

The approximate pressure increase, assuming an
  infinitely rapid conversion that occurs in the volume of 1 mol
  of solid 1 (=TATP), is of the order of 100 bar. The associated energy
  content is approximately 2500 kcal/mol. This is the origin of
  the explosive power of solid 1.

I'm aware that I probably haven't fully anwered your question, but as a take-home message I can only suggest again to not attempt any experiment with this peroxide!

I like the amount of detail you provide, but you didn't address the key question which was why is temperature control required.

Temperature regulation is vital in producing explosives as the products are, well, explosive. Worse, many of the products are temperature sensitive explosives and the reactions themselves are exothermic and can generate a lot of heat if uncontrolled. If the heat gets high, the reaction rate will increase and the product will not only be produced faster but will detonate. In the early days of explosives many experimenters paid for this knowledge with their life or their limbs.
As a result most modern experiments are done on a very small scale in very carefully controlled environments.


text:
How can I calculate the boiling temperature of a pure substance as a function of pressure?
I want to make an algorithm that will give me the boiling temperature of a substance as a function of pressure, the critical temperature of a substance, and the critical pressure of a substance. I believe this can be done using a combination of the Peng-Robinson Equation of state and the Clayperon or Clausius-Clayperon equation of state. The issue is that I believe this requires additional input criteria, specifically the latent heat but this function is supposed to be more general. Is there any way to work this method without the latent heat, or to calculate the latent heat from the other known values?

If you know the latent heat of vaporization at one temperature, you should be able to determine the latent heat of vaporization at any other temperature from Hess' law and knowledge of the heat capacities of the liquid and vapor as a function of temperature.  You may also have to include a pressure contribution as the critical point is approached.


text:
What is the justification behind the principle of corresponding states?
Two fluids with same reduced volume and temperature have the same reduced pressure.
Two fluids with same reduced pressure and temperature have the same compression factor.
What justifies these two statements and why exactly are both saying the same thing?


text:
How many moles NH4Cl must be added to NH3 to create buffer with pH=9?

How many moles of $\ce{NH4Cl}$ must be added to $\pu{2.0 L}$ of $\pu{0.10 M}$ $\ce{NH3}$ to form a buffer with $\mathrm{pH}=9$? Assume the addition does not change the volume of the solution significantly.

The solution given is as follows:

The equilibrium between $\ce{NH3}$ and $\ce{NH4+}$ is given by-
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
We know the $K_\mathrm{b}$ value and $\ce{[OH-]}$ from the $\mathrm{pH}$ and $\ce{[NH3]}$ is given, thus we can solve for $\ce{[NH4+]}$. The solution given says that the final answer is given by this concentration multiplied by the total volume $\pu{2 L}$.

I don't follow the logic in the solution. The $\ce{[NH4+]}$ we found was the total concentration of $\ce{NH4+}$ in the solution after adding the salt. But before adding $\ce{NH4Cl}$, there was already some $\ce{NH4+}$ in the solution (we don't know how much).
Thus to find the number of moles $\ce{NH4Cl}$ we need to add, shouldn't we subtract the number of moles $\ce{NH4+}$ already in the solution from the answer we got in the textbook solution?

the amount of $\ce{NH4+}$ beforehand is negligible because ammonia is a fairly weak base

$\ce{NH3}$ is such a weak base ($K_{\mathrm{b}}=1.8\times 10^{-5}$) that the initial concentration of $\ce{NH4+}$ can be considered negligible. You can consider the initial concentration of $\ce{NH4+}$, but the answer you will get is the same to a considerable number of significant figures.
By considering the equilibrium of $\ce{NH3}$ dissociating to form $\ce{NH4+}$ and $\ce{OH-}$ it is easy to find the initial concentration of $\ce{NH4+}$.
$$K_{\mathrm{b}} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$$
Since \begin{align}[\ce{NH4+}] &= [\ce{OH-}] \tag{1}\\
K_{\mathrm{b}}\cdot [\ce{NH3}] &= [\ce{NH4+}]^{2} \tag{2}
\end{align}
Plugging in values gives:
$$[\ce{NH4+}] = 1.34\times 10^{-3} (3\text{s.f.})$$
Which equates to $\pu{2.68\times 10^{-3} mol}$ in the $\pu{2 L}$ solution.
If I'm right this should be orders of magnitude smaller than the textbook answer, and therefore negligible.

This makes sense. Can you prove to me mathematically that the initial concentration of $NH_4^+$ is negligible? How would we calculate it? Is it not simply $0.1 M$ by stoichiometry?

I added a calculation of this to my answer, but it's not just 0.1M; the concentration is governed by an equilibrium.

Oops I was assuming complete dissociation, which is not the case since (as you said) $NH_3$ is a weak base. Thanks again.

@Ivan: The amount you calculated is not neglegible. It is actually $\pu{143.5 mg}$ of $\ce{NH4Cl}$.


text:
How can I deduce the structure of my compound from this NMR spectra?

I have no idea how to determine the substituents on the aromatic ring from this NMR spectrum. There is (or might be) a triplet on my NMR around $\pu{7.59 - 7.65 ppm}$ which has 4 hydrogens in the same environment. If so, there must be two ortho substituents on two aromatic rings but I think the 4 Hs won’t be in the same environment.
I assume the substance to be the one I added here:

Welcome to Chemistry.se! Are you sure that the compound you have drawn corresponds to the spectrum you have obtained? There are two singlets. At 3.86 it corresponds to the methoxy group. I see nothing in the molecule that might correspond to the 1.56 signal. (Maybe it is just water?) All the signals in your aromatic range have little shoulders (there are no values picked in this spectrum), but you should be able to calculate the coupling constants, with that information you might be able to correspond them to the protons in you compound. In principle, you should not see any triplet.

The hydrogens in the structure you drew definitely should not be in the same environment — so while overlapping or close signals would be okay, the way your signals look isn’t. To get two signals to be identical, there needs to be some symmetrical way of corresponding the hydrogens to each other.
I can confirm a trans-configured double bond (I assume you can, too?) and two disubstituted aromatic rings. The question remains which type of substitution they have (ortho, meta or para). If you expect two rings with the substituents you drew, remember they must be pretty different in their chemical shifts!
For ortho, as you drew, one would indeed expect a set of doublet, triplet, triplet, doublet for each ring. meta would expect one set of doublet, triplet, doublet and a singlet somewhere. para, finally, would expect two doublets.
Remember that triplets must have a $1\,:\,2\,:\,1$ ratio, or they aren’t triplets. With this information in hand, go back into the aromatic region and think what you could have gotten. I have dropped enough clues, I hope.


text:
What is involved in this firey demonstration?
In this rather shaky Facebook video (sorry I can't provide a better link), a science teacher takes a burning beaker of something and spills it out onto the floor.  It then proceeds to shoot fire out in all directions, and we can see embers burning themselves out, leaving black ash of some sort.

I've seen teachers do similar stunts with alcohol fires, but those burn blue.  This one is much more explosive and stays yellow.
Is there any way to tell what the teacher is using here?

This is liquid methane, which boils at a temperature of −161 °C. (Methane can be condensed in a cold trap that is cooled with liquid nitrogen.)

The power of knowing what to look for: The fourth link in a search for liquid methane is a news article about that video.  Thanks for identifying it!


text:
What is non-expansion work?
I learnt that Gibbs free energy is the maximum amount of non-expansion work.
But the phrase non expansion work confuses me. Work is defined as pressure times change in volume. If there is no expansion, how can work be done?

It's when I stair-climb the Willis Tower in Chicago each November for the Shirley Ryan AbilityLab. I certainly don't expand during the climb, but I do work against Earth's gravity.

You do have to push aside the air you encounter as you rise that takes extra energy. Try climbing up in a tank of water with negative buoyancy. This explanation of Gibbs free energy is very non general.

Pressure-Volume work is a type of mechanical work .
There are also non-mechanical forms of work where in pressure and volume terms are not involved .
The most famous example is electrical work . The work done to move electrical charges ( like motion of electrons in a conductor) comes under this category .

Another example is the stirring of a liquid in a tank

authors: محمد الحميري


text:
If Gibbs energy of activation is positive, how is the change to the transition state possible?
The Gibbs energy increases from the reactants to the transition state, but according to the Second Law of Thermodynamics, entropy of the universe can only increase. Why is the transition state possible then?

Oh no, not that dilemma again! Btw, welcome to chemistry.SE!

Thanks. If you know the answer, though, could please explain?

Wait...You mean there can't be an entropy decrease?

That's what I thought - doesn't the second law say that?

The thermodynamics laws that your studied are valid only for some conditions. It is needed that the system be macroscopyic. Also this law only gives probabilities. It is posible to extend the theory to other areas. Doing so, in molecular cases, it is not too much unprobable this change for one molecule. All the molecules changing (that is what your can calculate with your thermodynamics) is highly unprobable.

Is it because, though the Gibbs energy of the individual molecules as they react increases to the transition state, for the system as a whole, till equilibrium the Gibbs energy is decreasing, and we can only look at the Gibbs energy change of a macroscopic system since Thermodynamics only deals with macroscopic systems?

Also Gibbs energy or entropy are only defined for macroscopic systems - a few molecules can't be given values of entropy or Gibbs energy. The Gibbs energy of activation for that set of molecules must be, I think, extrapolating by finding it for an entire macroscopic system converting to the transition state, and then diving it by the number of molecules in that system.

Actually you can get something like Gibbs energy per molecule (its change from reactive state to TS.) Even it is posible to calculate it for the macroscopic system asuming particle independence (works good for gases). The change the state of ALL molecules is the same as the change of the macroscopic system. It is hihgly unprobable, because it is the product of all individual probabilities. So, I won't happend. But, it is much more probable the change in some small set of molecules. That is why TS theory works. Only a few proporción of molecules at a time are in this more unprobable state.

Always take in mind that how big is delta G gives a measure of how much probable is one state than the other one. Suppose that your have a system with only two accesible states. And that particles are independent each others. If one state is 10 times more probable than the other one, there will be 10 times more molecules in the that state. It can sound strange becausd in macroscopic thermodynamic courses the deltas of the examples are big enought to allow considereing only one state populated.

Also, the final state (products) does not affect this ratio of particles in reactive and ts.

...according to the Second Law of Thermodynamics, entropy of the universe can only increase.

Well, there was a nice similar question in bio.SE about this. A system is not universe. The overall entropy of the universe increases, but that isn't true for only a particular thermodynamic system.
Also, no one said the transition state is stable!

The transition state of a chemical reaction is a particular configuration along the reaction coordinate. It is defined as the state corresponding to the highest potential energy along this reaction coordinate. Wikipedia

(High potential energy = lesser stability)
This chemwiki.ucdavis page is also good to help reshape your understanding of second law of thermodynamics.


text:
Reduction of amide with LiAlH4
In reduction of carboxylic acid with $\ce{LiAlH_4}$ the first step is deprotonation.
But in the mechanism of reduction of amide with $\ce{LiAlH_4}$ (here), the first step is addition of hydride on carbonyl group. Why doesn't the hydride ion abstract a proton from $\ce{-NH2}$? Since acid base reactions are faster than nucleophilic addition.

Probably a side reaction which doesn't amount to much .

As compared to the carboxylic acid, deprotonation of the amide is negligible.
I don't recall any data for THF or 1,4-dioxane as the solvent, but F.G. Bordwell has determined a lot of $pK_a$ values in DMSO.
In DMSO, the $pK_a$ of acetic acid is 12.6 [1], while the $pK_a$ of acetamide is 25.5 [2].

So, what is the maximum pKa upto which LiAlH4 will deprotonate and not add it self? I mean is there any criteria by which we can decide? In both cases deprotonation will cause H2 to release which obviously is much less acidic than either of them. So the equilibrium should well be in forward.

In the mechanism on the page you referred to, a positive species ($\ce{HRC=NH2+}$) is formed, which is very unlikely in strongly basic conditions. I do believe the amide would deprotonate at some point, most likely in the first step, similarly to the carboxylic acid. Here is a link to a page which shows a more likely mechanism.


text:
What's the source of "2.303" in Van't Hoff equation?
I've seen a "2.303" term multiplied to $R$ in Van't Hoff equation several times.
Here is one example:
http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation#Error_propagation

$$\Delta H^\circ=2.303R\frac{\log K_1-\log K_2}{\frac1{T_2}-\frac1{T_1}}$$

What is this number and where does it come from?

Related: Help me understand the correlation between the equilibrium constant and delta G

It is the multiplier used when converting the equation to use a base 10 log instead of a natural log.
It comes out to be the natural log of 10.

To elaborate on Burak's answer, which states that 2.303 is the multiplier, the correct converison is:
$$\ln(x) = 2.303\cdot\log_{10}(x)$$
Or as a divisor to convert the natural logarithm to a base 10 log:
$$\log_{10}(x) = \frac{\ln(x)}{2.303}$$

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. I have reformatted and reworded your answer in order to reflect the current situation. Please have a look and edit as needed.


text:
Where do the bubbles in a glass of water come from?
This weekend I accidentally left a glass of water on my kitchen table for a few days while I was out of town. Unsurprisingly, bubbles formed on the walls.
As I understand from the last time I attended a chemistry class quite a while ago, this happens due to water self-ionizing (at about a ratio of $10^{-14}$ if I recall correctly).
$$\ce{H2O + H2O \rightleftharpoons H3O+ + OH-}$$
However, for some reason, the bubbles only formed up to a certain depth:

What is the reason for this? Does self-ionization happen uniformly in the glass? Do bubbles start forming from the top down and would they have reached the bottom if left untouched for longer?

Nucleation sites in the glass?

Gases are less soluble in warm water than cold water.
The bubbles are likely from dissolved gas coming out of solution as the water warmed.
The bubbles are unrelated to self-ionization of water.  The self-ionization does vary somewhat with pressure, so at an extreme depth in the ocean you might need to consider what Kw is, but not in a glass of water.
The solubility of gases increases with pressure, so on this basis it is more likely that you would have bubble appear near the top.  However, the pressure is only 1-2% greater at the bottom of the glass compare to the top.  There may be fewer nucleation sites (such as tiny scratches) near the bottom of that particular glass.  Also, if the bubbles move up along the wall sometimes, it would be logical that there are less at the bottom.

So there would be no / fewer bubbles if I left a hot glass of water to sit and cool, correct?

I wouldn't expect any

I've always assumed that the bubbles formed because they're coming out of solution after the water pressure drops. Inside your plumbing, the pressure is high and the water can hold a lot of gas, and then after you pour it into the glass, the pressure is lower and you have a supersaturated solution, and the gas gradually comes out of solution at the nucleation sites on the glass walls.
Suggested experiment: take such a glass of water after it's been standing a few days, and pour the water into a new glass. See if new bubbles form.
Another thing to try: seal the water off from the atmosphere, while leaving it at approximately atmospheric pressure. Maybe you could invert a glass full of water in a tub, taking care not to trap a bubble? I'm going to do this right now!
Update:
The inverted glass gathers about as many bubbles as the upright glass, so the gas seems to be coming from "inside" the water, and not diffusing there. I didn't notice any bias toward the top of the glass at my first check after a few hours, but now, after two days, the inverted glass has some more, and larger, bubbles near the top, where they get trapped. I'm speculating that that's just gravity doing it's thing: occasionally a bubble will detach from the walls and simply float up, although I don't have the patience to see if they ever reattach to the walls rather than simply always continuing to the top.

yes, that's a good point, the pressure in the plumbing would be an important factor.  Could be as high as 80psi for city water. 30 psi  is common for well water

You are mixing things up. The self ionization of water is not related with bubbles at all. The ionization is an equilibrium process which goes on forever under constant conditions meaning thereby the concentration of ions does not increase under constant conditions. Also ions are produced in aqueous phase not gaseous phase
$$\ce{2H2O~(l) -> H3O^{+}~(aq) + OH-~(aq)}$$
The drinking water is not pure, it is a solution of many minerals and gases. Changing temperature and pressure of the surroundings can affect the solubility of the gases. Gases are less soluble at high temperature. So as water absorbs heat gases come out in form of bubbles.
About the particular depth, solubility varies inversely with pressure. As pressure increases with depth so bubbles appear only upto a particular depth.

@Saad Try asking people in Alaska....

@Saad That's a french balcony.

Oops sorry about the balcony


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Why do ionic solids change to a structure with a higher coordination number at high pressures?
This seems counter intuitive to me since ionic radius increases with increasing coordination number. Surely the solid doesn't expand under high pressure? Can you explain?

If there is a phase transition to another allotrope under increasing pressure, it would only make sense that the phase would become more dense (the PV term of the Gibbs free energy).  The denser crystal structure is likely to have a higher coordination number than the less dense phase - to reduce the atomic volume the atom has to be closer to the ones around it, and likely is now closer to more atoms.

But changing from a sodium chloride structure to the cadmium chloride structure means going from a close packed array of anions to a non close packed array. How does that make sense?

But cadmium chloride has the halogens in a cubic close packed structure. The structures really aren't that different once you look at them (perhaps slightly cross-eyed).

Apologies, I meant caesium chloride (body centered cubic). 8 coordinate

Sodium chloride can be thought of as fcc, so coordination of 6, going to a bcc coordination of 8 under pressure?  Seems fine by me...

But ionic radius increases with coordination number!?

And high pressures are trying to squeeze it all together more tightly. The PV energy gained by a denser solid from the higher coordination plays off (and pays off) against the ionic unhappiness.

Lets put some numbers on the phase transition of NaCl from the normal NaCl structure (based on fcc, 6-fold coordinated) at atmospheric pressure, to the CsCl (based on bcc, 8-fold coordinated) structure at high pressure.
From Compounds and Alloys Under High Pressure: A Handbook, by E. Yu Tonkov (google found it) one gets that the transition occurs at 27-29GPa, with a volume change of 5%.
The density of NaCl, taken from Wikipedia, is 2.165 gm/cc, or about 27cm$^3$/mol, or 27E-6 m$^3$/mol.
Multiply on through and you get, very roughly, the PV energy gained by converting from the less dense NaCl structure to the denser CsCl structure is 40kJ/mol - this is not an inconsequential energy gain for the Gibbs free energy.  Thus, the PV energy gained by a denser solid with the higher coordination plays off (and pays off) against the ionic unhappiness.
In the literature there are also indications that the CsCl, and/or higher pressure phases, may be metallic in nature, so the assumption of purely ionic bonding may be in jeopardy.