text:
Pi electron stacking, how does it work?
I've come across the term base-pair stacking (with reference to B-DNA) in my school text book, and I had posted a question in that regard on Bio.SE.
I've also seen a similar (albeit brief) version of my question on Chem.SE.
I looked up the term online, and after checking out a couple of links, I'm led to believe that this term stacking refers to the forces of attraction between aromatic rings on account of the de-localized pi-electron clouds on either side of their (the aromatic rings) planes.
Now I'm not sure if I skimmed over some important details when I read up on 'stacking', but I'm unable to find a simple and solid explanation as to how these interactions are brought about.
If 'stacking' is the interaction between pi-electron clouds, then shouldn't these interactions be repulsive (on account of 'like-charges repel')? But I'm told they're attractive interactions.
I thought about it for a bit, and I realized that if you consider the fact the the pi-electrons, despite being de-localized all over the plane of the aromatic ring, can only be found at one particular location at any given instant of time, resulting in a 'partial' negative charge being formed there. The other regions of the electron cloud probably acquire a 'partial' positive charge on account of the fact that the electron is not present there at that instant. It's these opposite, partial charges formed that results in the attraction. (Essentially the same idea that governs London Dispersion Forces)
But the thing thing is I'm not sure if my 'hypothesis' is correct in the present case. If this isn't what happens then I've no idea what makes 'stacking' attractive interactions.
So my question stands: How exactly does stacking give rise to attractive interactions between aromatic ring? Why don't they result in repulsive interactions?

Constructive overlap of orbitals increase the concentration of electron between the two nuclei, thereby lowering the potential energy.

It is essentially the same with forming a bond, which is an exothermic process.

I thought you said that the interactions are between pi-orbitals...

Also isn't stacking a non-covalent 'phenomenon'? Kinda confused here because you mentioned that it's the same as forming a bond.......would you mind making it an answer?

Pretty sure hydrogen bonds are also not covalent

@AaronAbraham This might be interesting for you: https://www.researchgate.net/publication/255765109_Rethinking_the_Term_Pi-Stacking

It's not mostly about covalent or electrostatic bonding at all: most of the interactive forces between pi-systems will be van der waals interactions which are higher between large delocalised electron clouds.

The wikipedia page would be a good place to start, and also read this: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1301220/pdf/11159389.pdf

You going along the right track. The stacking attraction occurs when two molecules with $\pi$ orbitals come face to face with one another, typically their separation is 0.34 nm. However, for the interaction to be attractive the two molecules have to be displaced slightly, this prevents direct electron repulsion between $\pi$ electrons on one molecule with those on the other. The attraction is in general between the $\pi$ electrons on one molecule and the $\sigma$ framework on the other$^1$ and vice - versa.
The model assumes that a $\pi$ orbital can be split into three charged parts, separated by a distance $\delta$. The lobes of the orbital have charge -1/2 each and at the position of the atom's nuclei is placed a charge of +1. Thus there are also three interactions $\sigma - \sigma , \pi - \sigma $ and $ \pi - \pi$ and three positions of charge, making 9 interactions in all for each atom. The interaction is taken to be electrostatic, i.e. Coulomb and each term has the form $E_{ij} = q_iq_j/r_{ij} $ for atoms i and j and is summed over all atoms.
In addition a van-der-Waals term is added to the calculation of the total interaction, this has the form $E_{vdw} = \Sigma _{ij}(A_{ij}exp(-\alpha_{ij}R_{ij})-C_{ij}/r_{ij}^r))$ where A, $\alpha$ and C have standard values$^2$. In solution the value of this energy is much reduced vs. its value in the crystal making the $\pi-\pi$ interaction the dominant one.
The charges are shown schematically below for carbon atoms. For other atoms, e.g. nitrogen the charge is given as 1.5 with 1.5 electrons associated with them. (All images are from ref 1 )

charge distribution

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

The contour plot for $\pi-\pi$ interaction and van-der-waals (vdw) energy as a function of their centre to centre offset as shown above (energies in $\pu{ kJ mol^{-1}}$). The black squares are geometries from $\pi-\pi $ stacking in crystals. The shape of the contour plot with or without the vdw interaction is the same as with just the $\pi-\pi$ interaction alone, i.e this interaction determines geometry but  the vdw generally lowers the energy and more clearly show regions of low energy as negative values.
$^1$ The model was developed by Hunter & Sanders J.Am. Chem. Soc. 1990, v112, p 5525-5534
$^2$ Van der waals parameters Caillet & Claverie, Acta Crystallogr. sect A1975, v31, p448

Oh, the figure was clear as soon as I realised that I should have been looking at negative values. I was totally confused by the positive bit and looked for positive ones. Side note: Apart from once calculating two benzene molecules approaching each other at the MP2 level of theory I don’t really know crap about pi stacking ;)

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

You are very close to our current understanding what most people call "π−π interactions/stacking". There are a several molecular parameters that determine "π−π interactions/stacking":

Polarizability of the molecules
Electronic nature, i.e. electron deficient and electron rich
Solvent

What drives "π−π interactions/stacking" is actually much more complicated. There are several papers that actually try to measure the attractive energies for "π−π interactions/stacking" and it was found that sometimes the "stacking" picture i.e. the rings are face-to-face, does not always work, e.g. 1, in defining "π−π interactions/stacking". The solvent can dictate the nature of the "stacking". Aromatic electrostatic effects also play a crucial role in determining the "attractive forces" of "stacking". If you have an electron deficient and electron rich aromatic systems that are within proximity, then electrostatics becomes significant. In DNA, these "π−π interactions/stacking" is driven by hydrophobic and electrostatic interactions, as well as solvation. The attractive forces that drive "stacking" can be due to several factors that need not to invoke "orbital overlaps".
The term "π−π interactions/stacking" is a misnomer as it generally explained as face-to-face interactions, π−π cloud interactions, and the ubiquity of dispersion forces, which is not yet proven. In reality, aromatic-aromatic interactions (my preferred terminology) is much more complex than what it seems to be. It is sufficient for now that the attractive forces in aromatic-aromatic interactions is driven by electrostatics.
See: Rethinking the term "pi-stacking" for more details

The stacking that occurs between the base pairs in the DNA double helix occurs between a "donor" base, having relatively higher electron density, and a "acceptor" base, having a relatively lower electron density. Such donor-acceptor pairs can be formed, for example between Adenine and Guanine, Adenine being the donor because of an electron donating amine group and guanine being the acceptor molecule, because of a electron withdrawing C=O group. Such interactions contribute significantly to the stability of DNA, along with the usual H bonding and London dispersive forces, as you have observed.

All combinations of bases can be adjacent to one another and stack with each other, so classifying some as "donor" and some as "acceptor" makes little sense.


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Skeletal formula for carbon with two double bonds
How do I depict an sp-hybridised carbon with two double bonds?
$$\ce{R=C=R}$$
If I just leave it out, like I would for a normal $\mathrm{sp^3}$ or $\mathrm{sp}$ carbon, it's difficult to see that it's there.

In chemical structure diagrams, carbon atoms that are bonded to at least two other atoms are usually left unlabeled. The carbon atom is indicated by the angle in the bonds.
However, according to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), a carbon atom with two double bonds should be explicitly labeled:

(…), any carbon atom with two identical collinear bonds should always be explicitly labeled, to remove the possibility of the two bonds being misinterpreted as one long bond.

The use of a dot in place of an explicit carbon atom label is acceptable in allenes and related molecules with three consecutive carbon atoms. It is not acceptable to use a dot to represent a carbon atom when either of its adjacent atoms is other than a carbon atom.

When a dot is used to represent a carbon atom, it must be depicted clearly and unambiguously. The dot should be large enough to be clearly seen by the reader, and must be visibly separated from the adjacent bonds.

And then there’s also the variant of one continuous line and two non-connecting double-bond lines which is probably badly inacceptable by IUPAC ;)

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Carbon-based life and helium
Help me please with a question about carbon-based life development. I'm almost sure I know answer, could anybody with more solid knowledge in the area confirm it. Could carbon-based life develop on a planet without helium? Did helium help the process on any stage?
As I understand, yes could, because helium is noble gas which does not react with other elements. Or maybe I am wrong?

You see much helium around here? Not really. But helium is constantly produced  via alpha decay, so it would be hard to not have any of it. Why are you even asking this?

You are right, helium plays no role in biochemistry and could just as well not be there; that would make no difference.

Helium plays no part in chemistry never mind life. A more interesting question would be whether life could work in the absence of radioactivity (which often produces helium).

As much as I would love to see life use helium somehow, it seems too hard. The neutral atom shares a spot with neon for the least reactive substance in all of chemistry, whereas the other commonly-found form of helium, the alpha particle ($\ce{He^{2+}}$) from radioactive decay is arguably the most reactive substance in all of chemistry. Life can only work with supremely delicate balancing of reactivity in the middle of the scale, not such crazy extremes.

Without helium there would not be floating balloons. Without floating balloons, people would be unhappy. Unhappy people are bad for society, which would implode in the long run. And that's only because there is no helium. So you see, it is quintessential and human life would not be possible without helium.

Helium in stars is an initial product in the fusion reactions that go on to make "more interesting" elements.  In that sense we do need helium to get life going.
Once the more complex atoms are made, however, life appears to have no further use for helium.  As Nicolau mentioned, it seems too hard to see life use helium. The neutral atom shares a spot with neon for the least reactive substance in all of chemistry, whereas the other commonly-found form of helium, the alpha particle (He2+) from radioactive decay is arguably the most reactive substance in all of chemistry. Life can only work with supremely delicate balancing of reactivity in the middle of the scale, not such crazy extremes.

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How to set up equation for buffer reaction?
For buffer equations, how can you tell what to react with water?
For example, if I have a 1 M acetic acid solution and 1 M sodium acetate solution (a conjugate acid-base pair) mixed together and am asked to find the pH, how do I know how I should write the equation?
Do I react acetic acid with water to give me the products, or sodium acetate with water to give me a different set of products?

Hint here.

There is a general way to do these sorts of problems. The idea is to consider equilibria of both acid/base and of the water.
When there is solution of a weak acid and it salt, or just the weak acid or just the salt, i.e. pure HA, NaA + HA or pure NaA, then there is no distinction between these types of solution because of the equilibria involved. ( NaA just represents any salt ). The following calculations apply for each or any of these solutions.
Start by defining the equilibrium constants for the reactions. A molecule dissociates as $\mathrm{HA \leftrightharpoons H^+ + A^-}$, but to be general we let the base be [B] instead of $\mathrm{[A^-]}$ and so write $\mathrm{HA \leftrightharpoons H^+ + B}$
$$K_a=\mathrm{[H^+]_e[B]_e/[HA]_e} \tag{1}$$
where the concentrations are those at equilibrium.
There is also the equilibrium $\mathrm{H_2O \leftrightharpoons H^+ + OH^-}$ to consider and
$$K_w=\mathrm{[H^+]_e[OH^-]_e}$$
We know the amounts of acid $c_a$ and base $c_b$ added at the start of the reaction. To obtain the pH at equilibrium the amount $\mathrm{[H^+]}$ has to be calculated. To do this work out the concentration of acid and base in terms of the initial amount and the ionised species.
The total concentration of HA is
$$c_a=\mathrm{[HA]_e+[H^+]_e - [OH^-]_e}\tag{1a}$$
and for the base
$$c_b=\mathrm{[B]_e -[H^+]_e + [OH^-]_e\tag{1b}}$$
The e subscripts are now dropped for clarity. [Note that some authors use mass and charge balance instead to work out the concentrations.]
These values can be substituted into the equilibrium constant equation (1) and using $K_w=\mathrm{[H^+][OH^-]} $. Substitution leads to the general equation
$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - K_w/\mathrm{[H^+]} }{c_a - \mathrm{[H^+]} + K_w/\mathrm{[H^+]}  }\tag{2}$$
which is a cubic equation in $\mathrm{[H^+]}$ that is best solved numerically in the most general case. This equation can be simplified under different conditions as the examples below demonstrate.
Note that when $c_b=0$ equation (2) describes the case of pure HA, when $c_a=0$ it describes the hydrolysis of pure NaA solution.
The figure shows how the pH changes for a weak acid/conjugate base with the p$K_A$ shown and $c_a$ = 0.01 M over a range of $c_b$ concentrations. The curve was calculated by numerically solving eqn 2.  The straight line (coloured grey) is the Henserson-Hasselbalch equation described below in (A). It is clear where this approximate equation fails.

Figure 1. pH vs concentration of base $c_b$. Red line full calculation, straight line (grey) the approximate Henderson-Hasselbalch eqn. The dashed lines show the p$K_A$ and the acid concentration $c_a$ used. The region where the Henderson-Hasselbalch eq. is a good approximation is clear; $c_b$ should be no less than $\approx 0.1c_a$. The plot uses the values in example (B).

Example (A) The Henderson-Hasselbalch eqn is an approximation but limits should be tested.
When the concentrations $c_a$ and $c_b$ are far larger than $(\mathrm{[H^+]} + K_w/\mathrm{[H^+]})$ or $(\mathrm{[H^+]} - K_w/\mathrm{[H^+]})$ these terms  can be ignored without error. This produces
$$\mathrm{[H^+]} = K_Ac_a/c_b,  \qquad \mathrm{pH}=\mathrm{p}K_A - \log_{10}(c_a/c_b)  $$
which is the formula often used. The log form is called the Henderson-Hasselbalch eqn.  It is often the case that $c_a/c_b=1$ then $\mathrm{pH}=\mathrm{p}K_A$.
Example B. What is the pH of a buffer solution consisting of 0.01 M of a weak acid with $K_A$ = 1.5$\cdot$ 10-4, and 0.01 M of its conjugate base?
We could use eqn. 2 but this means solving a cubic which is not necessary. Realising that in acid solution $\mathrm{[H^+] \gg [OH^-]}$ then
$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - K_w/\mathrm{[H^+]} }{c_a - \mathrm{[H^+]} + K_w/\mathrm{[H^+]}  } \to \mathrm{[H^+]}\frac{ (c_b + \mathrm{[H^+]}) }{ (c_a - \mathrm{[H^+]} )}   $$
This equation has to be solved and not approximated to the Henderson-Hasselbalch eqn. since $\mathrm{[H^+]}$ may not be very different to $c_b$.
Letting $x =\mathrm{[H^+]}$ then $x^2+(c_b+K_A)x-K_Ac_a=0$ or
$$x=\frac{-(c_b+K_A)+\sqrt{(c_b+K_a)^2+4c_aK_A}}{2} $$
which gives $\mathrm{[H^+]}=1.45\cdot 10^{-4}$ or a pH = 3.84. This answer is close to that of the HH equation which gives $\mathrm{pH = p}K_A =  3.82$, so as it turns out the HH equation was good enough.

Can we simplify cubic equation$(2)$ to the quadratic equation when we have a strong monoprotic acid solution? thank you for your powerful effort.

Then $[HA]_e$ will be zero. The answer here https://chemistry.stackexchange.com/questions/100346/calculate-ph-of-a-mixture-of-a-strong-base-and-acid-knowing-only-the-ph-wt-v/100355#100355  explains the case of a strong acid/base.

How can I reform equation $(2)$ to derive equation describes the case of pure monoprotic strong acid?

You cannot, you should use the method in the  link above.

I found the equation below in the textbook which derived from equation $2$.  $$[\ce{H+}]^3
+(C_\mathrm{b} + 1)[\ce{H+}]^2
-(K_\mathrm{w}+C_\mathrm{a}K_\mathrm{a})[\ce{H+}] -K_\mathrm{w}=0

$$   Is it correct ? or like this equation :
$$
[\ce{H+}]^3
+(C_\mathrm{b} + K_\mathrm{a})[\ce{H+}]^2
-(K_\mathrm{w}+C_\mathrm{a}K_\mathrm{a})[\ce{H+}] -K_\mathrm{a}K_\mathrm{w}=0

$$

rearranging eqn 2 I obtained the second equation you list.

How can we use mass and charge balance to derive equation (1a) and equation  (1b)?

material balance on $A^-$ is $c_A+c_b=[HA]+[A^-]$, neutrality is $(Na^+ , or, c_b)+[H_20^+]=[A^-]+[OH^-]$.

Thank you for the interest, the source of my confusion that I   missed  $\ce{Na+}$ in the charge balance equation.

Buffer equations work on two main assumptions:

The acid/base in consideration (here it is $\ce{CH_3COOH}$) is weak and has a low $K_\mathrm{a}$.
The amount of acid/base ($\ce{CH_3COOH}$ here) and conjugate base ($\ce{CH_3COONa}$ in this case) is large.

In your situation, you start with $\pu{1M}$ $\ce{CH_3COOH}$ and $\pu{1M}$ $\ce{CH_3COONa}$ in water. The following equilibrium should be established:
$$
\ce{CH3COOH + H2O <=> CH3COO- + H3O+}
$$
where
$$
K_\mathrm{a} = \frac{[\ce{CH_3COO}^-][\ce{H_3O}^+]}{[\ce{CH_3COOH}]} = \pu{1.75 \times 10^{-5}M}
$$
and $\ce{CH_3COONa}$ completely dissociates to give $\ce{CH_3COO}^-$ and $\ce{Na^+}$.
In the beginning, the ratio is 0 as $[\ce{H_3O^+}]=0$, so some $\ce{CH_3COOH}$ will dissociate to give $\ce{CH_3COO^-}$ and $\ce{H_3O^+}$. But since the concentrations of $\ce{CH_3COOH}$ and $\ce{CH_3COO^-}$ are very large, their concentrations won't change.
So essentially $\pu{1.75\times10^{-5}M}$ of $\ce{CH_3COOH}$ will react, making the ratio:
$$
\frac{(1)(1.75\times10^{-5})}{(1)} = 1.75\times10^{-5} = K_\mathrm{a}
$$
so the $\mathrm{pH}$ will simply be
$$
\mathrm{pH} = -\log_{10}[\ce{H_3O+}] = -\log_{10}(1.75\times10^{-5}) = \mathrm{p}K_\mathrm{a} = 4.7
$$
Suppose we start with different concentrations of $\ce{CH_3COOH}$ and $\ce{CH_3COONa}$, the logic remains the same. Concentration of $\ce{CH_3COOH}$ and $\ce{CH_3COO^-}$ don't change as their starting quantities are very large. Then $[\ce{H_3O^+}]$ satisfies
$$
K_\mathrm{a} = \frac{[\ce{CH_3COO^-}][\ce{H_3O^+}]}{[\ce{CH_3COOH}]}
$$
$$
-\log_{10}K_\mathrm{a} = -\log_{10}[\ce{H_3O^+}] -
\log_{10}\frac{[\ce{CH_3COO^-}]}{[\ce{CH_3COOH}]}
$$
$$
\mathrm{p}K_\mathrm{a} = \mathrm{pH} - \log_{10}\frac{[\ce{CH_3COO^-}]}{[\ce{CH_3COOH}]}
$$
$$
\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log_{10}\frac{[\ce{CH_3COO^-}]}{[\ce{CH_3COOH}]}
$$
where $[\ce{CH_3COO^-}] = [\ce{CH_3COONa}]$

The answer might be surprising because when setting up a chemical reaction we usually mix reactants that then react to yield products. When making a buffer by adding a conjugate acid/base pair to water, one ingredient is the reactant and the other is the product (and there is very little reaction, but sufficient to set the pH).

For example, if I have a 1 M acetic acid solution and 1 M sodium acetate solution (a conjugate acid-base pair) mixed together and am asked to find the pH, how do I know how I should write the equation?

With one ingredient as reactant and the other as product:
$$\ce{CH3COOH(aq) <=> CH3COO-(aq) + H+(aq) }\tag{1}$$

Do I react acetic acid with water to give me the products, or sodium acetate with water to give me a different set of products?

We already wrote an equation showing the major species in (1), and it is the reaction governed by the acid dissociation constant. If you want to show the minor species, you can also show to auto-dissociation of water:
$$\ce{H2O(l) <=> OH-(aq) + H+(aq) }\tag{2}$$
You can mix and match these two (e.g. letting the hydroxide react with the acetic acid), but the system is sufficiently described by these two chemical equations along with their equilibrium constant.
As described in the other answers, when the buffer concentration is high and the pH is near neutral, you can neglect the second reaction and still get a good estimate of the pH from the ratio of initial concentrations (conjugate base to conjugate acid).

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Bones of Strontium
Elements in the same column of the Periodic Table have similar chemical properties to one another; for example, Group 1 elements such as sodium and potassium react rather vigorously when put in contact with water.
This being the case, is it possible to substitute one important element from an object or system, for instance the calcium in our bones, with another element from its respective group, such as strontium, and still have it function in the same/similar way?
Would there be any benefits or draw-backs to replacing calcium with strontium in the human body (if it is indeed possible) and how would it effect the chemistry/appearance of our bodies?

Everything will change drastically if you do that.There are physical and chemical properties unique to each element.

Shortly, yes, but it may interfere with other body functions in unexpected ways. For example, different solubility of a metal ion can result in different morphology of bones or can form bones / dissolve them on long term. Nothing is really static in the body, so even small fiddling can have dramatic effect on equilibriums.

Yes, such substitutions are indeed possible and can possibly happen.

This being the case, is it possible to substitute one important element from an object or system, for instance the calcium in our bones, with another element from its respective column, such as strontium.

There are some strontium based supplements on the market (strontium citrate, ranelate etc.) that are being touted as potential treatment for diseases like osteoporosis, and such like, however, evidence supporting their efficacy is tenuous at best. It is chemically similar enough to calcium to compete with calcium uptake.
Strontium-90, is a radioisotope of strontium that is described as a bone seeker ( an element that tends to accumulate in the bones of humans and other animals when it is introduced into the body).
It is typically a product of nuclear fission, and is found in spent nuclear fuel and radioactive wastes. It does have some industrial and medical applications.
There is some dispute over its half-life in the human body, primarily because of its complex metabolism in the human body. The elimination rate of strontium from the human body shows variance between individuals, based on age, and sex.
As strontium has an affinity to the calcium-sensing receptor of parathyroid cells that is similar to that of calcium, the increased risk of liquidators of the Chernobyl power plant to suffer from primary hyperparathyroidism could be explained by binding of strontium-90.

Great answer. Partial metal substitution can happen with other metals, too, eg cadmium incorporated in bones as a result of cadmium poisoning (itai itai dissease)


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Discovery of elements
How did the first chemists know that a sample contained so and so elements?
If they tried to find through reactions,they should know what elements make up the other reactant(s), and how it reacts beforehand (which is absurd as elements weren't discovered & studied before that)

The notion of chemical element in modern sense arised at the end of 19 century.
Before this, the term "element" was obviously too fuzzy to discuss here, as well as methods and proofs involved.
After this, the optical spectroscopy was method of choice to find new elements and prove their existence,
because of distinct and reproducible spectra.
See the history of helium, for example.
Of course, besides of spectroscopy, chemists proved the finding of new elements by separation of individual substances.
For example, metals are expected to form oxides and salts with (at least roughly) expected properties and mutual transformations.
In particular, salts, even in microamounts, may be recrystallized to form visually distinct crystals of individual substance.
If such a substance possessed properties not seen before, this was indication/proof of the presence of new element.

It's actually easier than you think. Elements don't break down, so once you find reactions that cause substances to combine, the combination is not an element and any degradation reactions mean that the reactant is not an element. So before the modern era, you had a fairly good idea that something was an element when you couldn't find reactions that broke down the substance. And once we were good enough to weigh carefully the reactants and products, we can determine the quantitative relationships to further eliminate substances as being elements.


text:
Transition state - multidimensional surfaces
I've been reading an explanation about transition states in Smith's Organic Synthesis:

A transition state is an energy maximum along the reaction coordinate. However, a chemical
  reaction where the reactant contains N atoms actually involves movement over an energy surface
  with 3N-6 dimensions, and a transition state is actually an energy minimum in each of the remaining
  (3N-7) dimensions. Liken a chemical reaction to a drive across a mountain range. The “goal” is to
  go from one valley to another, just as the goal of a chemical reaction is to go from one minimum
  on the energy surface (the reactants) to another minimum (the product). The road moves about a
  two-dimensional surface whereas the energy surface for a chemical reaction may involve dozens
  or even hundreds of dimensions. There is no need to climb to the top of a mountain (a maximum
  in both dimensions) to go from one valley to another. Rather, is sufficient to go through a pass
  between mountains (a maximum in one dimension but a minimum in the other). Similarly, it is
  only necessary to pass through an energy maximum in one coordinate (the reaction coordinate)
  for a chemical reaction.
  Because a transition state is an energy maximum (albeit only in one of many dimensions) it cannot
  function to “trap” a molecule, let alone collect a population of “trapped” molecules. This means
  that a transition state cannot even be observed experimentally let alone isolated or characterized.
  Quite simply, a transition state does not exist in the sense that a molecule exists.

This multidimensional explanation completely confounds me. Could someone explain this in a more understandable language? What is this surface with 3N-6 dimensions? How did the author use it to conclude that TS can't be observed experimentally?

related: http://chemistry.stackexchange.com/questions/51903/how-should-i-think-about-potential-energy-surfaces-for-reactions-of-complex-syst You may get some results if you search for "potential energy surface", since that is exactly what this paragraph is describing. We can only depict them in 3D, but if you get the hang of the idea, then it is not to difficult to convince yourself that it can be generalised to 4D, etc.

Those formidable math formulas don't help me much. Is there an intuitive explanation as I don't understand the one in the quoted text?

The picture below shows a simple potential energy surface with a transition state at the maximum point on the reaction path, which is at the 'saddle point'. The reaction path is the solid black line. (Oscillations in this show that the reactant and product are vibrating). The contours are at different energy with large negative being lowest. This picture is based on a calculation using known experimental parameters.
Along the reaction path the energy is as small as it can be, even though it increases as we go along, because by moving to the right or left (perpendicular to the general direction of the path)  the energy is even higher. So the saddle point is just like a pass between high mountain peaks. At the highest point on the reaction path the only way to lower the energy is to go down, either to return on the original path or to continue to products. The time it takes to cross the transition state is definitely sub picosecond, < $10^{-12}$ s and possibly up to a 100 times faster.
It is not possible to visualise a multidimensional reaction path, so we have to make do with 2D plots, as below, of one reaction coordinate vs. the other with contours at different energy. In principle many similar plots could be needed for a complex reaction but very often most of the atoms in the molecule are bystanders and it possible then to only look at the separation and energy of the atoms in bonds being broken and made.

Trajectory of $\ce {Cl + H2}$ in a gas phase reaction. The H atom approaches the chlorine molecule at the right. The yellow circle shows the transition state.

Could you elaborate a bit? What is the curved black line? Are the negative numbers, energies? What does the green dashed line represent? Why is the TS situated at that single position?

And why are there 3N-6  dimensions?

The black line is the trajectory, and the oscillations are the effect of vibrations at the molecule moves. The numbers are energies. The Cl approaches the $\ce{H2}$ in the deep valley on the lower right of the picture and moves leftwards. After reaction the HCl is vibrating more than the $\ce{H2 }$ was. Note that the plot is of separation $\ce{Cl + H2}$ on the lower axis and of $\ce{H + HCl}$ on the other.

The $3N-6$ are the number of internal degrees of freedom in a non-linear molecule of N atoms. The 6 is to remove whole molecule translational motion plus whole molecule rotations along/around each of x, y & z axes.

First, the question may pose itself why $3N-6$ at all? And for that, we should take a step back and ask ourselves: Why $3N$?
These are the degrees of freedom of a molecule. If you break a molecule down into its atoms, each atom would have three degrees of freedom: $x, y$ and $z$. In a molecule, these degrees of freedom are generally retained; however, you must modify them. It is commonplace to say that three degrees of freedom of the entire molecule represent translation (into $x, y$ and $z$ directions) while another three (of a non-linear molecule) represent rotations ($R_x, R_y$ and $R_z$). That leaves us with $3N-6$ degrees of freedom which are typically called vibrational. The key is that the $6$ nonvibrational ones do not affect the interior of the molecule while the other degrees do move the atoms relatively to each other.
As an example, let’s assume $N = 3$, thus $3N-6 = 3$. This allows us to keep an analogy of a three-dimensional something; you would just need to expand the thought experiment to more dimensions. Furthermore, I’m just going to say that one of the three atoms is a central one and the movement of the other atoms can be expressed as relative location to that central one. (I’m effectively ignoring a degree of freedom here, which I would like to define as vibration of the central atom towards its partners so that nothing productive can happen. It is a strech, I know.)
You can now imagine a 3D-plot where the $x$ axis is one atom’s position in arbitrary units, $y$ is the second one’s and $z$ is energy. Different arbitrary $x$ and $y$ positions, so positions of atom $1$ and $2$ lead to a different energy. Two spots of this $x,y$ grid must be minima, though: they correspond to the reactant and the product. To react, the atoms must somehow take a path across these arbitrary units from one minimum to the other.
This is now where the mountain range analogy comes in. As somebody from close to the alps, I have no problems imagining mountain ranges, valleys, summits and passes. I don’t know if people from areas as flat as pancakes would, though.
Across that path I just mentioned, there must be a transition state. This is the highest altitude you reach on your way from basin 1 (or valley 1; the reactant) to basin 2 (or valley 2; the product). Of course you need to go up and back down but is there anything else you can say about the transition state? Let’s go back to the dimensions, $x$ and $y$. Let’s assume the ‘best’ path is following a straight line abiding to $x+2y =0$. Of course, neither dimension $x$ nor $y$ tell us much about the transition state — it is a maximum along both dimensions —, however, we can mathematically transform $x$ and $y$ until one of the transformed coordinates, $p$, lines up nicely with the reaction direction and the other one, $q$, is orthogonal to it. If we express the path as functions of our new coordinates (= dimensions) $p,q$ then:

the transition state is still a maximum in $p$ direction; but
it is a minimum in $q$ direction!

Why is it a minimum in $q$? Well, the transition state is a pass. The lowest possible spot to cross a ridge. So if you dissect the mountain range along said ridge, you can find a minimum on that ridge — this corresponds to your pass or your transition state.
What we just found out for $p$ and $q$ holds true for all $3N-6$ dimensions: when defined correctly, there is always one in which the transition state is a maximum; call it the reaction coordinate. And for all the other $3N-7$ dimensions it must be a minimum (if they are defined correctly). If it doesn’t work out of the box, ‘transform dimensions’.

Well, of course it's a stretch. It sounds like the symmetric stretch to me...


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Identical particles: Why is triplet oxygen more stable than singlet oxygen?
Quoting from Eisberg Resnick Quantum Physics:

If we consider space variables of two electrons (identical particles) to have almost the same values, then their wavefunctions are 'almost' identical if they are in the same quantum state, ie, $\psi_{a}(1)~ \simeq~\psi_{a}(2)$ and $\psi_{b}(1)~\simeq~\psi_{b}(2)$ [the label 1 and 2 denote the spatial co-ordinates of the electron '1' and '2' i.e. ($x_1,y_1,z_1$) and ($x_2,y_2,z_2$), and the labels a and b for the wavefunction denote the three quantum numbers $n,l,m$ of two different quantum states].
In this case, the antisymmetric space eigenfunction describing the system of two electrons is
$$\frac{1}{\sqrt2}[\psi_{a}(1)\psi_{b}(2) - \psi_{a}(2) \psi_{b}(1)]\simeq\frac{1}{\sqrt2}[\psi_{b}(1)\psi_{a}(2) - \psi_{b}(1) \psi_{a}(2)]\simeq 0$$
and the symmetric space eigenfunction is
$$\frac{1}{\sqrt2}[\psi_{a}(1)\psi_{b}(2) + \psi_{a}(2) \psi_{b}(1)]\simeq\frac{1}{\sqrt2}[\psi_{b}(1)\psi_{a}(2) + \psi_{b}(1) \psi_{a}(2)]\simeq\sqrt2\psi_{b}(1) \psi_{a}(2). $$

Clearly symmetric space eigenfunction has more probability density, implying when two electrons have 'almost' same spatial co-ordinates, symmetric space eigenfunction and antisymmetric spin state(the singlet state) is more favourable as opposed to antisymmetric space eigenfunction and symmetric spin state(the triplet state), since the total eigenfunction has to be antisymmetric. This implies it is energetically more favourable for electrons to have symmetric space eigenfunctions which is why in solids, when atoms are brought together from infinite distance apart, energy splitting takes place such that in the energy band that forms, the lowest energy is that of symmetric space eigenfunction.

My question then, is, why is ortho oxygen (triplet oxygen) energetically more favourable (source: Wikipedia)? The reason I am asking is that ortho and para states of oxygen form from the unpaired electrons and not from the nucleons. So then why is it not the case that singlet state is more favourable in case of oxygen molecule when their electrons have 'almost' the same spatial co-ordinates?

PS: I did read about the molecular orbitals and I know the Hund's rules but the contradiction still bugs me.

Possible duplicate of What is the physical basis for Hund's first rule?

And to directly answer your issue: electrons repel each other more when they have almost the same spatial coordinate. This repulsion is a destabilising effect - so we want to avoid it as much as possible. So, the singlet is higher energy.

So simple coulumb repulsion is the answer? What about all i quoted about the symmetric and antisymmetric space wavefunctions playing a role?

Yes, simply larger Coulomb repulsion in singlet state, because electrons have higher probability of being close to each other (as can be seen from the eqns you wrote). Your sentence "...implying when two electrons have 'almost' same spatial co-ordinates, symmetric space eigenfunction... is more favourable as opposed to antisymmetric space eigenfunction..." is wrong. It is precisely the other way round, because same spatial coordinates means greater repulsion and that is not something we like having. Greater value of PDF does not imply greater stability, it means it is more likely to happen.

Much clearer. one question though. I got that greater probability density does not mean greater stability. But when it does mean it is more likely to happen, why isnt it happening? I mean, in other words, that configuration which has less energy is 'more likely to happen'

I don't see that as a duplicate to the linked question, but I will close this questions to resolve issues with cross-posted questions. You can remove the question on physics or here. If you chose to delete it on physics, flag this post for reopening.

This just adds a picture to aid what has already been said in the comments by @orthocresol. The picture show the spatial extent of the wavefunction in a singlet & triplet state in a 2D particle in a box. You can see that the electrons in the triplet 'avoid' one another whereas those in the single pair up.

I still dont get it. The coulumb repulsion in the singlet (symmetric spatial wavefunction) makes the case for electrons being in the triplet state (antisymmetric spatial wavefunction). But probability density of two electrons being close together is more in case of symmetric spatial wavefunction(singlet) meaning they are more likely to be found in this configuration in contrast with antisymmetric spatial wavefunction(triplet). So why is the latter argument not obeyed? Where am i wrong?

To save having to duplicate, have a look at http://chemistry.stackexchange.com/questions/58625/what-is-the-physical-basis-for-hunds-first-rule   for more complete explanations of this effect. But note the spin and spatial parts of the wavefunction are important.


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How does aqua regia dissolve gold?
I recently learned that aqua regia can dissolve less reactive or noble metals like gold. Aqua regia is a mixture of $\ce{HCl}$ and $\ce{HNO3}$, but separately, neither of these acids dissolve gold.
So why don't concentrated $\ce{HCl}$ or $\ce{HNO3}$ dissolve gold, though combined in aqua regia they can?

related: http://chemistry.stackexchange.com/a/24750/16683

Aqua regia is a concentrated mixture of nitric and hydrochloric acids. The two acids react to generate soluble nitrosyl chloride, water and nascent chlorine.The mixture reacts with the noble metals like gold.

You might be interested in the excellent Aqua Regia dissolves Gold - Periodic Table of Videos.

The wikipedia article of aqua regia has beautifully mentioned the mechanism of gold dissolving in aqua regia:

Aqua regia dissolves gold, though neither constituent acid will do so alone, because, in combination, each acid performs a different task. Nitric acid is a powerful oxidizer, which will actually dissolve a virtually undetectable amount of gold, forming gold ions ($\ce{Au^3+}$). The hydrochloric acid provides a ready supply of chloride ions ($\ce{Cl−}$), which react with the gold ions to produce tetrachloroaurate(III) anions, also in solution. The reaction with hydrochloric acid is an equilibrium reaction which favors formation of chloroaurate anions ($\ce{AuCl4−}$). This results in a removal of gold ions from solution and allows further oxidation of gold to take place. The gold dissolves to become chloroauric acid. In addition, gold may be dissolved by the free chlorine present in aqua regia. Appropriate equations are:
$$\ce{Au + 3HNO3 + 4HCl <=>> AuCl4- +  3NO2 + H3O+ + 2H2O}$$
$$\ce{Au + HNO3 + 4HCl <=>> AuCl4- + NO + H3O+ + H2O}$$
If the aqua regia solution only contains gold, solid tetrachloroauric
acid may be prepared by boiling off excess aqua regia, and removing
residual nitric acid by repeatedly heating with hydrochloric acid.

You failed to mention an important point. Aqua regia itself is unstable and decomposes. So you must make the mixture just prior to using it.

is potential oxidation for Pt is more negative than NO3
but NO3 still oxidize Pt?


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what surface would liquid wax adhere to best?
I want to make a very thin sheet of wax by pour hot wax over a surface and letting it cool. What surface would wax spread out on best? I.e. which one would give the best wettability?
Many thanks everyone
THanks guys - will let you know how this goes!

Do you want to remove the wax layer from the surface after setting? If yes, then a good wettability will make that difficult.

Hi Vapid. Good point and yes I would need to. The important thing though is that its a very thin sheet I create - hence why I'm thinking the surface needs to adhere well to the wax to decrease the surface tension and allow it to spread out better. Perhaps once I've got this I could then heat up the surface itself very gently just enough that the wax sheet slid off?

I think that you need a surface with very low wettability, because as you said in the comments, you want to remove the solidified wax layer afterwards. It is possible to make a very thin layer of wax by spreading it between two hard plates pressed against each other. I would try something like this:

You spread the molten wax between the two sheets made from an appropriate material placed between two plates (glass or metal). To precisely regulate the thickness of the wax layer you use some separators placed near the edges. After setting you remove the plates, then carefully remove the sheets. The main problem is to choose a suitable material for the sheets. You can experiment with various materials. I think that a sheet made from, or covered with a silicone rubber would be perfect. You can try a parchment paper, aluminum foil, glassine and so on. If you need a really thin and fragile wax layer, you can try to use a tin or aluminum foil, and then remove it chemically using diluted hydrochloric acid.
EDIT: Now it hit me. There is another, simpler method you may try. If you want a wax plate with a uniform thickness and smooth surfaces use the procedure above. However, if you don't have those requirements, then simply pour the molten wax on the surface of hot water in a large pot. The wax should spread evenly on the whole surface. Leave it to cool down and collect the wax plate. It may be a bit thicker near the edges and its surface will be matte, but maybe it'll suit your needs.

I want to add to this that the thin sheet should be flexible enough to allow for bending, so that it can more easily be separated from the wax plate by "rolling back" of the support method.

Hi Vapid, thats brilliant. Sorry I've been away for a while and just picked this up, but will definitely try it asap! Cheers

@OliverWalters That's OK, let us know how did it go. I am sure that it will work, but I can't say if the resulting product will fulfill your requirements.


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Bond energy and s-character
While studying the basics of chemical bonding, I've often come across the following statements:
1. The bond energy increases with the increase in the s-character in a hybrid orbital, i.e sp3 sp2-sp2, and so on.
I can't seem to correlate the two. If a hybrid has a higher bond energy, shouldn't it also form stronger bonds?

Let us start with the following most important part of the IUPAC definition for the bond energy.

bond energy (mean bond energy)
The average value of the gas-phase bond dissociation energies (usually
  at a temperature of 298 K) for all bonds of the same type within the
  same chemical species.

I think it should be immediately clear now that the higher a bond energy is, the stronger is the bond, so that there is no any contradiction in a bond energy increase with the increase of the $\mathrm{s}$-character in hybrid $\mathrm{sp}^n$-orbitals.
Indeed, $\mathrm{2s}$-orbitals are lower in energy that the $\mathrm{2p}$ ones, and thus, the more $\mathrm{s}$-character a hybrid orbital has, the lower is its energy: $E_{\mathrm{sp}} < E_{\mathrm{sp^2}} < E_{\mathrm{sp^3}}$. The lower an atomic or hybrid orbital energy is, the lower is the energy of a molecular orbital in formation of which it participates, $E_{\sigma_\mathrm{sp}} < E_{\sigma_\mathrm{sp^2}} < E_{\sigma_\mathrm{sp^3}}$, and consequently, the higher is the corresponding bond dissociation energy, as well as, the average of them, i.e. the bond energy. So, the bond energy indeed increases as the s-character grows, say $E_{\ce{C(\mathrm{sp})-C(\mathrm{sp})}} > E_{\ce{C(\mathrm{sp^2})-C(\mathrm{sp^2})}} > E_{\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}}$.
Below is the MO diagram for $\sigma$ $\ce{C-C}$ bond between two $\mathrm{sp}$-, two $\mathrm{sp^2}$-, and two $\mathrm{sp^3}$-hydridized $\ce{C}$ atoms that illustrates the logic above.

@DHMO: if the sp hybrid orbital is stronger than the sp3 orbital, why is the sp-sp bond weaker than the sp3-sp3 bond?

@Kōhai, first note that orbitals aren't stronger/weaker relative to each other, but rather, they are lower/higher in energy. Now, the lower a bonding orbital is, the stronger is the corresponding bond. Finally, who  told you that $\ce{C(\mathrm{sp})-C(\mathrm{sp})}$ bond is weaker than the $\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}$ one?

A quick Google search results in 347–356 kJ/mol for bond-dissociation energy (at 298 K) for $\sigma$ $\ce{C(\mathrm{sp})-C(\mathrm{sp})}$ bond in alkanes and 837-839 kJ/mol for the triple bond in alkynes, out of which the $\sigma$ $\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}$ bond contributes 369 kJ/mol.

Okay, just to ensure that I've understood this correctly, lower the hybrid orbital energy, higher is the dissociation energy of the bond it forms, right? I think I messed up between the energy and the bond energy.

@Kōhai, yes, the lower a bonding orbital is, the more energy is required to break the bond. In our case the bonding orbital is formed from a hybrid orbital of C atom and some orbital of another atom X. And for a given orbital of atom X, the lower the hybrid orbital of C atom is, the lower a bonding orbital will be. I think I can try to draw this...


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Why isn't the side chain of arginine totally protonated at low pH?
I was wondering why arginine cannot be protonated at $\mathrm{pH\ 0}$ on all the $\ce{N}$ atoms in the side chain, as they all have a free electron pair. As this image shows, it can only be protonated once.

Why don't the other two nitrogen atoms in the side chain get protonated? Does it have something to do with resonance with the $\ce{C=N}$ double bond?

See the picture I just added showing the structure at the different pH levels @DHMO

If they all accept hydrogen ions, the molecule would be too positive...

Related: Protonation of Guanidine and (the answers to) Is the conjugate base of trinitromethane aromatic (What is Y-aromaticity)?

basically this

So far I as can see the N atom on the double bond is favoured or is this overgeneralized? @DHMO

It may be hard to see from a 2D-drawing, but there are simply no more available lone pairs on arginine than the ones protonated in form $\ce{A}$.
The guanidine group ($\ce{H2N-C(=NH)-NHR}$) is isolobal to a urea group, and indeed guanidine behaves much like urea except for the common differences between $\ce{C=O}$ and $\ce{C=NH}$. The lone pairs of $\ce{-NH2}$ and of $\ce{-NHR}$ are actively taking part in resonance with the $\ce{=NH}$ group’s double bond much like an amide structure. One can equivalently say that those lone pairs are delocalised into the $\ce{C=N}$ $\pi^*$ MO. Hence, only the $\mathrm{sp^2}$ hybridised lone pair on the $\ce{=NH}$ nitrogen is basic in any way and can be protonated.

Of course, you can further protonate arginine theoretically. However, it would take superacidic media.

Technically, it's not a $\pi$ system because this isn't a 2 atom MO. The HOMO and LUMO should both have $E$ symmetric from the $C_{3}$ point group.

@Zhe What?$%comments~$

Sorry, I think I got too technical. The $pi$ designation doesn't apply here unless you ignore the two nitrogen atoms that with the lone pairs. A $pi$ system is for two atoms. But that's clearly part of the conjugated system. For accuracy, you'd need to look at the frontier molecular orbitals. And for 3 nitrogen atoms around a central carbon, the p orbitals would form a point group that doesn't have anything with $pi$ symmetry. The group is actually $D_{3}$ (I was wrong before because we're looking at p orbitals).
Please ping if I need to clarify.

@Zhe Oh, that’s what you mean; I understand. But note that at least in organic chemistry, pi system is used to denote a series of atoms with collinear p-orbitals that form an MO antisymmetric with respect to the $\sigma_\mathrm{h}$ plane (that traverses the set of atoms).

Yeah, we even call it an extended $\pi$ system, but there's no way that thing has $\pi$ symmetry in aggregated... :/

The amino group of arginine at the terminal of side chain are more expose than the embeded amino group so the chance for that terminal amino group is more to be protonated and second is
the charge of these terminal amino group is than stabilized by resonance which make it more stabilized when it get protonation


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Why a higher s character increases a carbon atom's electronegativity?
To make us understand certain phenomenon like the higher acidity of alkynes than alkanes and alkenes, our teacher told us to learn this as a general rule:

Higher s character leads to a more electronegative carbon atom.

Note: By s character I mean the percentage of s in the hybridisation of the carbon atom (e.g., $25\%$ s character in $\ce{sp^3}$ hybrid carbon)
Though this well explains the above mentioned phenomenon (and certain others like the decreased $\ce{C-Cl}$ bond length in chlorobenzene), but I was unable to figure out the reason for this increased electronegativity.
This did not provide me a satisfactory answer. Electronegativity is after all the ability of an atom to attract the shared pair of electrons towards itself. So, how would one explain this using the concept of orbitals or molecular orbital theory?
I would also like to know if the same argument is valid for other p-block elements?

higher s character -> lower potential energy -> stabler electrons

@DHMO then this should be : stabler electrons->more electron attracting power. Right?But why?

because electrons want to lower their potential energy.

@DHMO then for a C-H bond, electrons should prefer to move towards the H atom(100% s character). Is it something that I'm missing?

But you are comparing with another C-H bond where the C is sp2 hybridized. Comparatively more electrons go to C. Also, although you are right this time, you cannot compare different orbitals of different atoms. For example, the 2s orbital of N has a lower energy than the 2s orbital of C; the 1s orbital of H has a higher energy than the 2s orbital of F.

@DHMO that's fine. But if we compare between C and H, is there some other factor to be accounted for because clearly H atom due to its very lower potential energy electrons should attract the shared electrons more. But we know that carbon attracts more than hydrogen.

@DHMO sorry for firing too many questions. But then what about fluorine? an electron would suffer much stronger repulsion there.(very small size and very high electron density)

Electronegativity is the power of an atom to attract bonding pairs of electrons to itself. It clearly depends on the nuclear charge: the larger it is, the more strongly the nucleus attracts electrons and the closer to the nucleus they stay, so the larger is the electronegativity. Besides, electronegativity also depends on the number of other electrons present in the atomic shells ahead of valence electrons participating in chemical bonding. These electrons shield the valence electrons from the positively charged nucleus decreasing its effective charge and consequently lowering the electronegativity of the atom.
Now, back to the question. $\mathrm{2s}$ orbitals are lower in energy than $\mathrm{2p}$ ones, so that electrons occupying the former are (at average) closer to the nucleus than that occupying the later. Thus, the more $\mathrm{s}$ character a hybrid orbital has, the lower its energy is and the closer (at average) electrons occupying it are to the nucleus. The lower the energy of a hybrid orbital of an $\ce{X}$ atom is, the lower is the energy of the molecular orbital in formation of which it participates, and consequently, the closer (at average) the bonding pair of electrons stay to that $\ce{X}$ atom.
So, with respect to a bonding pair of electrons that occupy a molecular orbital to which a hybrid orbital of an $\ce{X}$ atom contributes, the $\ce{X}$ atom indeed appears to be more electronegative as the $\mathrm{s}$ character increases. Specifically for $\ce{C}$ atoms, the larger the $\mathrm{s}$ character of a hybrid orbital is, the lower in energy is a bonding $\sigma_{\ce{C-C}}$ orbital formed from the hybrid ones: $E_{\sigma_\mathrm{sp}} < E_{\sigma_\mathrm{sp^2}} < E_{\sigma_\mathrm{sp^3}}$. So that electrons occupying the $\sigma_{\ce{C-C}}$ orbital stay closer to $\ce{C}$ atoms (at average) as the $\mathrm{s}$ character of hybrid orbitals increases.

With respect to the question OP asked in comments,

But if we compare between $\ce{C}$ and $\ce{H}$, is there some other factor to be accounted for because clearly H atom due to its very lower potential energy electrons should attract the shared electrons more.

the factor OP is looking for is the nuclear charge, which is 6(!) times larger for $\ce{C}$ atom comparing to the $\ce{H}$ one.

" Besides, electronegativity also depends on the number of other electrons present in the atomic shells ahead of valence electrons participating in chemical bonding. " Do you mean the intermediate shells?

@Kaushik, I'm not familiar with that term, but it very likely may be the right one if the meaning of it is "shells preceeding a valence one (counting from a nucleus)".

Yes I meant the shells between the nucleus and valence shell.

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


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Why does an increase in alkyl group substitution cause downfield shift H'NMR?
Since alkyl groups are electron donating, wouldn't an increase in alkyl group subsitution mean the protons would feel a lower magnetic field? Because they feel a lower magnetic field, wouldn't this cause an upfield shift? Is it just that alkyl groups are less electron donating than a hydrogen bound to the carbon?

The short answer to this question is that C is more electronegative than H (H 2.2, C 2.5), and therefore one would expect a trend in increasing chemical shift with increasing carbon substitution.
The longer answer....
As you point out, alkyl groups are electron donating, and you are on the right track in surmising that alkyl groups are less electron donating than protons. Remember that all shielding is relative. Looking at shielding effects at the α-substituent;

a bare proton nucleus (H+) would come at δ35,
H-H comes at δ7.4, and
H-CH3 comes at δ0.23.

A clear upfield shift due to the electron-donating alkyl group. So, we need to look deeper at the effect of β-substituents:

H-CH3 (δ0.23),
H-CH2CH3 (δ0.86),
H-CH(CH3)2 (δ1.33), and
H-C(CH3)3 (δ1.56).

This longer answer involves looking at the magnetic anisotropy around the observed nucleus, and questioning some age-old depictions of the C-C bond. Historically, the C-C magnetic anisotropy was depicted as below, with shielding at the side, and deshielding along the ends. Have a look in your NMR textbook, and you'll probably find this. Taken in isolation, this model conforms with the observed trend of increasing shift with increasing substitution with C.

However, this model contradicts somewhat the notion that alkyl groups are electron donating, and indeed what is observed with α-substitution. And with good reason. Recent studies by Baranac-Stojanovic (DOI: 10.1002/chem.201204267) show that this anistropy model should be reversed, and that there are shielding zones at the end of the C-C bond. Why then do we observe the increase in chemical shift with increased C substitution? Again, it is because the shielding anisotropy of the contributing C-C bond component is considerably less than contributions from the geminal C-H anisotropy contributions.


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Chemicals that react with blood
Is there a chemical or chemicals that can be infused in fabric that will react when exposed to blood that would produce a bright or glowing color other than red?

Something like luminol which reacts with the iron in the blood ?  What is the motivation for this question ?

I don't think this question is too broad or unclear.  It already has a reasonable answer.

The "infuse" part is rather unclear, though. OP should give more background information about the intended use of the "infused" fabric.

Luminol (IUPAC name: 5-Amino-2,3-dihydrophthalazine-1,4-dione) displays chemiluminescence when it is oxidised, and glows a blue colour. It is used in forensic science to detect iron (in haemoglobin) and also to detect copper in biological material.
The process leading to luminol emitting light is explained in this image.
As for whether luminol can be "infused" in fabric, this is difficult to answer, as you have not explained what you mean by "infused". As it is soluble in water, it will easily be washed out, and is thus not suitable for long-term use. Moreover, it will react with a variety of substances (apart from blood), for example bleach.


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Why there is hydrogen bonding involved in salt bridge formation between glutamic acid and lysine?
I saw this figure on Wikipedia:

After seeing this image I got really confused about the difference between these two, or maybe the similarities. Of course, I understand that a hydrogen bond can be seen as +/- interaction (because of the dipole). However, I don't understand why this Lys + Glu interaction would also make a hydrogen bond. If it would, I would expect the hydrogen bond between the double bounded oxygen and the $\ce{N}$ atom from the $\ce{NH3+}.$
I hope someone can explain me what this figure means in respect to both of the interactions shown.

Electrostatic interaction: the interaction between the negatively charged oxygen atom and the positively charged nitrogen atom.

Hydrogen bonding: the interaction between the oxygen and the hydrogen.

it does: "salt bridge is a combination of two noncovalent interactions: hydrogen bonding and electrostatic interactions " @DHMO

An electrostatic interaction is one between a centre of positive charge and a centre of negative charge. In your example, instead of a primary ammonium cation ($\ce{H3\overset{+}{N}R}$) you could also consider the fully methylated quarternary ammonium cation $\ce{Me3\overset{+}{N}R}$ which would still be able to build up an electrostatic interaction with the acid anion. The centre of positive charge is somewhere between the three ammonium hydrogens and the centre of negative charge somewhere between the two carboxylic oxygens.
A hydrogen bond, on the other hand, is a directional interaction of the type $\ce{X-H\bond{....}Y}$, where $\ce{X}$ and $\ce{Y}$ are electronegative elements and $\ce{Y}$ must feature a lone pair. Thus, hydrogen bonds always contain three atoms and only those three. One one of the three ammonium hydrogens can form a hydrogen bond with only one of the two oxygen atoms.

For a hydrogen bond to be present, the lonepairs of the H-bond acceptor must be pointed and interact with the antibonding orbital of the donor hydrogen. Electrostatic interaction has no such requirement.


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Dipropylamine vs n-dipropylamine
Are dipropylamine and n-dipropylamine the same thing? What are the differences or similarities in their chemical structures?
I think dipropylamine is $\ce{CH3CH2CH2NHCH2CH2CH3}$, but I am unsure about n-dipropylamine.

That would be N,N-dipropylamine. n is for linear hydrocarbons.

Well, the n- could potentially refer to the propyl group. It's not perfectly clear from the question, but I suspect it is not to indicate the propyl groups are on the N. There's no other place to put them...

Chemspider as well as Pubchem list the names “dipropylamine” and “n-dipropylamine” as synonyms. Hence, it is likely that both names have actually been used in the literature for the same compound. However, that doesn’t mean that such names are generally acceptable or in accordance with any past or present nomenclature rules.
The name “dipropylamine” corresponds to the traditional method that was in accordance with the IUPAC recommendations from 1979 as well as 1993.
The name “n-dipropylamine” is not in accordance with IUPAC nomenclature. It was probably used to emphasize the difference to “diisopropylamine”.
According to Subsection P-62.2.2.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), functional parent names like “dipropylamine” are now deprecated. Nevertheless, in general nomenclature, similar names for amines may be formed by prefixing the names of the substituent groups $\ce{R-{}}$, $\ce{R'-{}}$, and $\ce{R''-{}}$ to the parent hydride name ‘amine’, but the prefixes are set off by parentheses to distinguish the names from the above-mentioned deprecated names. Therefore, a correct name for the compound that is given in the question is “(dipropyl)amine”.
The preferred IUPAC name (PIN) is generated substitutively using the name of the parent hydride and the suffix ‘amine’ with further N-substitution. Therefore, the PIN for the compound that is given in the question is “N-propylpropan-1-amine”.


text:
R & S Stereochemistry of an unsaturated ring system

Using the R- & S- nomenclature assign a configurational symbol to decalin 4-methyl (above the plane), 4-hydroxy (below the plane), 7-isopropyl (above the plane), 1,9 diene. (Numbering the carbons in decalin ring-system: ring-junction carbons are 1 (top), 6 (bottom))

Carbon 7—holding the isopropyl group: I think this is S (isopropyl $ > \ce{C-C=C} > \ce{CH2} > \ce{CH}$).
Carbon 4—holding $\ce{CH3}$ & $\ce{OH}$: I think this is S ($\ce{OH}> \ce{CH3} > \ce{CH2=CH2}$).
Is my reasoning correct?

I have added a guessed structure diagram based on your description. However, your numbers of locants are not in accordance with this structure. In particular, the bridgehead atoms of decalin are actually not numbered 1 and 6.

Guess things have changed in 30-years. Numbering decalin, as I did, the 1-double bond is on the ring junction, otherwise the diagram is correct. Don't have a drawing package; and, given the silly prices of Chemdraw (a monthly subscription) am not likely to acquire one. All I require is the comfort of knowing my assignments are correct--if they are. Cheers.

@tony I changed the picture. Please check and confirm or rollback my edit.

Loong:yes, that's it: guess I'd better do something about presentation! Thanks. Cheers.

There are a number of free chemical drawing programs; just check the corresponding Wikipedia sites (if need really be I can find a link). But even if they are too much of a hassle, drawing on paper and scanning/taking a photo works, too.

@Loong Why didn’t you edit isopropyl in my answer, too? =C

I don’t really know how you arrived at your reasoning, but it seems to fail as soon as double bonds are involved. If you encounter a double bond, it must be expanded with ghost atoms to two single bonds in the following manner:
$$\ce{(R2)C=C(R'2) -> (R2)C(C)-C(C)(R'2)}$$
The carbon atoms in bracket are ghost atoms not connected to anything else. These ghost atoms are treated just like normal atoms when determining priorities. You can read more on Wikipedia, which also includes a better graphic example. (But since your case is rather simple, we can do without.)
Let’s first determine the tertiary alcohol’s carbon’s absolute configuration:

$\ce{O}$ from the hydroxy group.
$\ce{C}$ (along the ring to the left)

$\ce{C}$ (the quarternary carbon)

$\ce{C}$
$\ce{C}$ ← This is the first point of difference.
$\ce{(C)}$ — this is the double bond’s ghost atom.

$\ce{H}$
$\ce{H}$

$\ce{C}$ (along the ring upwards)

$\ce{C}$

$\ce{C}$
$\ce{H}$ ← This is the first point of difference.
$\ce{H}$

$\ce{H}$
$\ce{H}$

$\ce{C}$ from the methyl group.

$\ce{H}$
$\ce{H}$
$\ce{H}$

Thus, the priorities are $\ce{OH} > \ce{CH2C_{(q)}} > \ce{CH2CH2R} > \ce{CH3}$, because the quarternary carbon that one of the two $\ce{CH2}$ groups is bonded to beats the secondary carbon at the top of the right-hand ring. That makes that carbon’s absolute configuration (S).
Similarly, for the carbon attached to iso-propyl:

$\ce{C}$ (the quarternary carbon)

$\ce{C}$
$\ce{C}$
$\ce{C}$

$\ce{C}$ (the iso-propyl group)

$\ce{C}$
$\ce{C}$
$\ce{H}$

$\ce{C}$ (the secondary carbon)

$\ce{C}$
$\ce{H}$
$\ce{H}$

$\ce{H}$

As you can see, we only need go down to the second level to find a difference: $\ce{C(C3)} > \ce{C(C2H)} > \ce{C(CH2)} > \ce{H}$ Thus, that carbon’s absolute configuration is (R).

I think the carbon with the isopropyl group on it should be (R). The carbon at the ring fusion has highest priority because it has two real carbons and a phantom carbon from the double bond. This is higher priority than the two carbons and a hydrogen of the isopropyl group.


text:
Why do NMR samples require a solvent?
I am running a proton NMR on a liquid organic product. I understand that we wouldn't want to use anything that contains hydrogen atoms as a solvent, but I'm not sure why we need a solvent at all.

Well, solid-state NMR does exist, and in theory any liquid can be a ‘solvent’.

There is a minimal volume you need to fill into the NMR tube to be able to get a good shim, this is around 500 microliters with regular 5mm tubes. Not using a solvent would require a lot more of your actual sample just to fill the tube. In many cases you might not even have enough of your product to fill an NMR tube.
The solvent does have some effect on the spectra, it could be harder to compare spectra of pure product to the more typical spectra measured in a common solvent. You're measuring in a different environment, product in product is different than product in solvent.
Modern NMR spectrometers use the deuterium signal from the solvent for the lock, which corrects a drifting magnetic field. You could measure without the lock if your experiment is short or your signals are very broad anyway. But this is not something that every NMR facility has just set up for all users, you might have to manually set this up at the spectrometer.
I suspect that the automatic shimming methods typically used now would not be able to deal with a sample without solvent. But I never tried that.

The shifting effect of the solvent is usually not too bad, but can be confusing sometimes. It mostly depends on polarity. Water is the extreme, it comes at 0.4 in benzene, and 4.8 in water. Methylene group in ethanol drifts from 3.3 to 3.7, acetone from 1.5 to 2.2

The issue with amount of substance is probably the primary reason. You usually need to fill the tube with 5 cm of solvent to get something that is relatively uniform. Otherwise, anisotropic effects make it hard to get a spectra that has clean sharp lines.

Just to add some clarity to some already acceptable answers by others....
For running solution (liquid) state NMR, a solvent is not necessary. There are a number of applications where dilution with a solvent is unwarranted and counterproductive. So, why use a solvent for running liquid state NMR?

The main reason a solvent is used is to improve the resolution of signals.

And by main reason, I mean completely overriding factor if resolution is of any importance at all. All other factors are secondary. If resolution is of little importance (broad signals are expected or of no concern), then diluting with a solvent is unnecessary, and can simply be wasting valuable spectrometer time.
Diluting in a low viscosity solvent significantly increases the tumbling rate of molecules in solution, and decreases signal broadening through T1 mechanisms. This is why the most common NMR solvents are small molecules with low viscosity; chloroform, acetone, methanol, etc, and not something like glycerol. Most neat organic liquids that are made in the lab have very high viscosities, compared to common solvents, and the slow molecular tumbling will lead to efficient T1 relaxation, and hence very broad linewidths. In order to see the effects of solvent viscosity on linewidths of observed signals, compare the solvent linewidths of residual CDCl3, acetone-d6, dmso-d6.
-edit-
As an example - look at the following spectra of propylphenylketone in various solvents, and neat. Image 1 shows a standard spectrum of 1% PPK in acetone-d6. An expansion of the highlighted region is shown for various conditions in image 2.

For all samples below, standard 1H parameters were used; 1 scan, 64k data points over 20ppm, processed with 64k points without any apodization. All samples were shimmed using automatic gradient shimming routines.

As can be seen very clearly, the optimum resolution is obtained in the sample with the lowest viscosity - acetone. As the viscosity of the sample increases, the resolution decreases. In fact, the level of resolution obtained in acetone-d6 and that for acetone-d0 is almost identical. The spectrum below shows that excellent results for resolution can be obtained without lock and without deuterated solvent for a 1% sample.

The linewidth for the signals in non-deuterated acetone is about half that for neat PPK. This really drives home the importance of suitable solvents for running high resolution NMR. There is a balance that needs to be struck with NMR. It is an inherently insensitive technique, and it is important to get as much material in the coil volume as possible, but still critical that it is low in viscosity.
-end edit-
Never mind arguments about shimming and sample depth and sample amount. There are a squillion ways around these problems. Different size probes, different size tubes (running 3mm tubes inside a 5mm probe is a very common method of reducing the amount of required sample volume for instance), susceptibility plugs, shigemi tubes, sample depth adjustments are all standard applications for routine NMR.
The advantage of providing a spectrometer lock is secondary; an internal lock is not necessary for the vast majority of NMR experiments. In fact, most modern 2D experiments use a lock-hold switch which actually turns off the lock to allow gradient pulses to be used. For the vast majority of 1H experiments, simply turning off the field sweep will allow a stable enough field for a few minutes of data collection. Over the course of an hour or so, though, this would lead to significant broadening of signals for 1H - but perfectly fine for 13C experiments.
Even the argument of signal dampening by intense solvent peaks is not valid. Modern spectrometers can deal with suppression of multiple intense solvent peaks without any concern. Even the application of digital quadrature detection filtering allows for a very specific region of the spectrum to be observed without collecting signals for intense solvent peaks. Shaped pulses for selective excitation and even selective methods such as Hadamard are all well enough developed methods for getting around the problem of dynamic range conflict due to large solvent peaks. It is simply the case that, for the majority of the samples, we simply make use of the fact that we have dissolved our sample in a suitable solvent that excludes troublesome solvent peaks. In some cases, this is not so. In most protein work, deuterated solvent (D2O) for lock is added at a maximum of 10% to ensure exchange with labile amides is limited. These types of systems are typically analysed in 90%H2O/10%D2O. Metabolomic screening methods of blood plasma, urine, juice etc etc are typically run in non-deuterated conditions, or with minimal addition.
Let's debunk some of the statements on this page:

"I suspect that the automatic shimming methods typically used now would not be able to deal with a sample without solvent"

Not the case. There are multiple methods of automatic shimming. The most common, and most effective, uses a gradient pulse method to establish field homogeneity plot. This can be used on either 2H or 1H. All that is required is the shift of a well refined peak (singlet works best, but are not necessary). I use this method on samples of neat ionic liquids and it gives excellent shims above and beyond what I can ever achieve by hand.

"Theoretically, the way that NMR is commonly taught, you don't need a solvent, but it's very simplified because a lot of the technical details are quite complicated. In practice, the way that NMR is carried out, the instrument needs the solvent as a "lock" for the magnetic field"

That's a bit like saying 'because science'. The spectrometer does not need a lock to establish a stable field. Turning off the field sweep will keep the spectrometer field stable for long enough to run all but the highest resolution experiments.

"The issue with amount of substance is probably the primary reason"

A 1.7mm microprobe uses 30uL of sample. A 1mm microprobe uses 5uL of sample. Stopflow microcoil systems use as little as several nL of sample. Sample preparation for these still require sufficient dilution in a suitable solvent for optimised resolution. For 5mm probes (the most common), you can use 3mm tubes, or susceptibility-matched inserts to significantly reduce the amount of material required. However, if the sample is too concentrated, viscosity effects (and differences in susceptibility if using plugs) will lead to poor resolution.

"The solvent does have some effect on the spectra, it could be harder to compare spectra of pure product to the more typical spectra measured in a common solvent. You're measuring in a different environment, product in product is different than product in solvent."

which is why when you report your NMR data you should state the concentration of your samples. But who does that these days? It is generally assumed that samples are measured at high dilution when being characterised. At reasonable dilution, [solute]<<<[solvent], changes in chemical shift are minimal.

Oh well, to be honest, I hardly know much about it :) When I saw the question, I was hoping you would come along. Thanks for clarifying.

First of all, are all your samples a liquid? Viscosity effects line width. Solids are run without a lock. Liquids can be run without a lock if you program the magnet drift into the pulse sequence. So basically, If you want high resolution, you have to compensate for magnet drift, hence, lock on to a strong, narrow, known peak.

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 am not an NMR expert, could you explain what you mean with lock?


text:
Find second acid dissociation constant at a given pH
In short, the problem states that:

We have a molecule that has two ionizable groups
$\mathrm{p}K_\mathrm{a,1}$ is between $5.0$ and $9.0$; $\mathrm{p}K_\mathrm{a,2}$ is outside of this range.
$100\ \mathrm{mL}$ of a $0.1\ \mathrm{M}$ solution of molecule, all in the unprotonated form
added $12\ \mathrm{mL}$ of $1\ \mathrm{M}\ \ce{HCl}$
final $\mathrm{pH}$ is $7.4$

The goal is to find $\mathrm{p}K_\mathrm{a,1}$. This is how I've approached it:

First, I converted everything to moles:
$0.012\ \mathrm{mol}$ of $\ce{HCl}$ added $\ce{->} 0.012\ \mathrm{mol}\ [\ce{H+}]$
$0.01\ \mathrm{mol}$ of molecule in solution $\ce{->} 0.02\ \mathrm{mol}$ of $[\ce{A-}]$ since the problem states that everything is in unprotonated form I think this is were I am wrong in my thinking
Next, I reason that some $\ce{A-}$ is going to be neutralized by $\ce{H+ ->} 0.02-0.012$. Leaves me with $0.008\ \mathrm{mol}$ of $\ce{A-}$
Then, I have $0.012~\mathrm{mol}$ of $\ce{HA}$ that was formed.
Im plugging this in to the Henderson-Hasselbalch equation $\ce{->} \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log ([\ce{A-}]/[\ce{HA}]) \ce{->}$
leads to the incorrect answer.

I've tried looking at similar problems but I am just having a hard time with this. I feel that the problem lies with this the neutralization. Also, I know that upper boundary of $\mathrm{p}K_\mathrm{a,1}$ is $9$, which is above the final $\mathrm{pH}$, and I know that at $\mathrm{pH}=\mathrm{p}K_\mathrm{a} 50~\%$ would be protonated/unprotonated, however, I don't know how to account for this information in my calculation.

For starters, I’m going to define your molecule as being $\ce{HA}$ and its two ionised forms to be $\ce{A-}$ and $\ce{H2A+}$. This means, we are observing the following equilibrium:
$$\ce{H2A+ <=>[H+] HA <=>[H+] A-}$$
As you correctly calculated, you have $10~\mathrm{mmol}$ of $\ce{H_nA^m+}$ and $12~\mathrm{mmol}\ \ce{HCl}$ that you add to your solution.
$12~\mathrm{mmol} > 10~\mathrm{mmol}$. Thus, there are two possibilities:

we started with $10~\mathrm{mmol}\ \ce{HA}$. We ended up with $10~\mathrm{mmol}\ \ce{H2A+}$ and $2~\mathrm{mmol}\ \ce{HCl}$
we started with $10~\mathrm{mmol}\ \ce{A-}$. We ended up with $8~\mathrm{mmol}\ \ce{HA}$ and $2~\mathrm{mmol}\ \ce{H2A+}$

$\mathrm{pH}~7.4$ is only consistent with possibility 2; in case of 1 it would mean $2~\mathrm{mmol}\ \ce{HCl}$ controlling the $\mathrm{pH}$ which would give us $\mathrm{pH} < 3$. Thus, we are well within the operating scheme of $\mathrm{p}K_\mathrm{a,1}$ and we can easily use the Henderson–Hasselbalch equation:
$$\begin{align}\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{HA}]}{[\ce{H2A+}]}\\
7.4 &= \mathrm{p}K_\mathrm{a} + \log \frac{\frac{8~\mathrm{mmol}}{V}}{\frac{2~\mathrm{mmol}}{V}}\\
7.4 &= \mathrm{p}K_\mathrm{a} + \log 4\\
7.4 &= \mathrm{p}K_\mathrm{a} + 0.60\\
6.8 &= \mathrm{p}K_\mathrm{a}\end{align}$$

You did indeed identify the spot where you went wrong. Just because a compound is dibasic does not mean you should double its concentration. As I showed above, $10~\mathrm{mmol}$ of original molecule, whether it was entered as $\ce{H2A+}, \ce{HA}$ or $\ce{A-}$ will remain $10~\mathrm{mmol}$; only the number of attached protons will differ.

Thanks. This reinforces the first answer. I've managed to work my way through similar problems but still in need of more practice.

@JVazquez You know when the teacher/professor tells you to write out the equation, and no one does it because it's an extra step? Try writing out the equations, at least until you have enough practice that you can see the stoichiometric relationships in your head. You probably could have avoided your mistake if you had written out the equation.

I think you did go wrong in that step. Why are there 0.02 moles of A–?
You should consider writing out your mystery substance as H2A. Then the conjugate base is HA–. You don't really care about the second ionization, but I think the diprotic nature of this acid is causing you confusion. You only create one mole of the monoprotic form from one mole of the diprotic form. There's no extra 2 in there.


text:
If NaCl dissociates when in water, why don't we see Chlorine gas escaping?
I've always been a little confused about the whole process of ions dissociating in solutions. When salt dissolves and splits into Na+ and Cl-, shouldn't the Cl- ion still be gaseous? And shouldn't we see it's yellow-green color? Why don't we?

You know that $\text{chlorine} \ne \text{chloride}$, right?

Yes solid sodium chloride dissolves according to the following reaction:
$\ce{NaCl(s) -> Na^+(aq) + Cl^-(aq)}$
Chlorine gas, $\ce{Cl2}$, can be produced from solution via this half-reaction:
$\ce{2Cl^- -> Cl2 ^ + 2e^-}$
Since this half-cell reaction is only part of a redox reaction, some other chemical species would need to be reduced. There isn't any other chemical species in a sodium chloride solution which will do that.

When $\ce{NaCl}$ dissolves, it is broken into $\ce{Na+}$ and $\ce{Cl-}$ because water is polar and salt is an ionic compound. These are what are known as ions and are in solution of the water. Chlorine gas, $\ce{Cl2}$ is diatomic and is not an ion (not charged) so is different to the dissolved ions.


text:
Which it is the oxidation state of indium and phosphorus?
I would like to know what the oxidation state of indium and phosphorus in a structure type rock salt. It could be +1 and -1? Yes or not? and why?
Thanks

You don't need oxidation states of +1 and -1 to make a rock salt structure.  Magnesium oxide does it with +2 and -2.  In any event indium phosphide actually does not have the rock salt structure (https://en.wikipedia.org/wiki/Indium_phosphide).

Under pression there is a phase transition from zinc blende to rock salt for this system.

I've estimated net atomic charges in rocksalt type InP by simple HF calculation in CRYSTAL06 in random basis.
Lattice parameter of 5.54 A is taken from:
K Kabita et al. First principle study on pressure-induced electronic structure and elastic properties of indium phosphide (InP). Indian J. Phys. (2015), DOI 10.1007/s12648-015-0701-0.
Results: http://pastebin.com/zdb8js8Z
Total atomic charges are 47.2861139, 16.7138861, so partial charges are:
In: +1.71
 P: -1.71

Given that oxidation state is an hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, the answer is probably better estimated as +2/-2, not 1.


text:
Why do we need to assume that gas is made up of large number of small particles in kinetic molecular theory of gases?
Among the five postulates or assumptions of kinetic molecular theory of gases (KMT), only the assumption that a gas is made up of large number of small particles doesn't seem to make any sense to me. The only reason I could of think of was due to lack of negligible pressure. I reasoned that if there aren't enough molecules in the container there will be no evidence of pressure, since most of collision would be within molecules and not between the wall of the container and the molecules. But there is something missing in this explanation or it is completely wrong, as it fails to adequately explain the above stated question.
Why do we need this assumption?

What is the sound of one hand clapping?  What is the temperature of a single atom of monoatomic gas?

Surely there have only to be enough particles for the average values of quantities to be meaningful so that the random nature of single molecule events is not observed, and in fact fluctuations due to this becomes quite negligible. Thus 'large' is determined by this consideration. Clearly Avogadro's number of particles is going to be more than enough

I'd assume that the notion of a large number of particles comes from statistics. By assuming that the use of statistics is a valid way to analyze the phenomena then as the number of particles that you're averaging over gets larger, then the statistical values calculated will be better estimators.

To echo and expand on Porphyrin's comment, the idea of having a large number of particles shows up in KMT because this is a statistical theory. That is, everything KMT predicts is some average property, or, to be rigorous, the expectation value of a distribution.
Thus, if one only had a very small number of particles, you would never arrive at something like the Maxwell-Boltzmann distribution because you would have to treat each particle individually rather than averaging over a phase-space.
So, to answer your actual question, we need this assumption because it justifies us in doing things like the derivation in this link where they take the derivative of the unknown function (eq. 9.8). That assumes this function is continuous over all space because no constraints are put on the resultant function. The only way we can do that and expect to have our theory match reality is if reality has an approximately continuous distribution of particle speeds, i.e. a lot of particles present.
As it happens, $10^{23}$ particles really isn't that many in practice, so when you get down to very low pressures and temperatures (that is the number density $\frac{N}{V}$ is small) the theory still works because something like $10^6$ probably still makes a pretty continuous function.

As a totally unrelated aside, you might also wonder if the MB distribution is wrong because it allows particles to have speeds ranging from $0\rightarrow \infty > c$... I computed the special relativistic correction one time and it's so laughably small as to actually annoy you that you even took the time to do it.
I know this answer focused mostly on the MB distribution, but that is very central to KMT so I thought it would make for a representative though not exhaustive answer.

EDIT:
To be critical of my answer, however, I'm almost certain that much of KMT is derivable while only thinking about a single particle sampling its whole phase-space (perhaps all at once) and once could simply argue that the above function I mention taking the derivative of is a hypothetical distribution in which each of the infinite states need not be occupied at any given time. So basically taking a time average of the phase-space over infinite time gives this continuous distribution. That idea also lines up well with ideas to do with the partition function where every state could be equally likely but we get a non-even distribution because some states are more likely to be sampled.
Then perhaps it is just more physically instructive to think about a large number of particles.

Nice way of looking at it :) +1

I don't remember the "large" in the assumptions. I'm with you on questioning why.
Possibly the writer of this assumption was taking the perspective that because atoms are so small, it takes a lot of them to effect a measurment. That is, all the meaniful quantities of gases we typically deal with are composed of large enough numbers for humans to manipulate in the lab or commercially.
But, as far as I know, the kinetic theory does not break down at low particle concentrations. All the gaw laws should still hold. Two particles in a container will still collide with the walls exerting pressure (assuming the other assumptions are in play). Good luck measuring it.  I've never seen asterisks next to the gas laws admonishing you to not use these at low particle concentrations.
edit: I see someone has downvoted this answer, without a courtesy comment. For lack of a better focus, let me address current comments to the OP's post.
Temperature is a measure of the average kinetic energy of the prticles in a substance. This is the definition inherent in assumptions of KMT. It's the one I taught for 24 years and has appeared in every textbook I've used. This, the most useful, and accepted, definition with regard to KMT and the gas laws, does not rely on being able to use a thermometer to measure it. The average kinetic energy of the particles is equal to $3RT/2$, where $T$ is the Kelvin temperature and $R$ the gas constant. There is no imposition on the number of particles. And it is a average being used; a fraction of the number of particles, with the same average KE, still has the same temperature.
As to pressure and the implication of "fluctuations" at low particle numbers. Let's put ten particles in a container. With this small a number, there will be moments when none of those particles act to generate force on a particular chosen area. But at other moments all ten might be acting. Over very small intervals of time, the pressure at this locus will vary wildly. But over meaningful spans of time the pressure values will average out. I can't imagine why that time-averaged pressure would be any different than predicted by KMT.
I would agree that relying on "time averaging pressure" and considering the temperature of one particle are not practically useful concepts in the application of gas laws. I just checked my copy of Zumdahl (AP Chemistry), no use of "large". I just checked my Gialcolli (AP Physics), oops! He uses "large". I would postulate Zumdahl is a theoretical kind of guy and Giancolli is practical.

If it's worth anything, I agree with you. The temperature as defined in kinetic theory is simply related to $\langle v^2 \rangle$ which is a well-defined property even for a single molecule. There is no mention of such an assumption in Atkins' Physical Chemistry 9th ed. (Chapter 20).

It's worth my sanity. Thanks for the formatting edit too, I'm not yet conversant with equation formatting here.

It would seem that you have invoked the ergodic hypothesis which allows the time average to be the same as the ensemble average, which seems reasonable.

@porphyrin Thanks, I'll look that up. I've always been a bit stymied explaining to students how gas pressure arises when there is no time element in the definition of pressure yet the gas pressure depends on particle velocities.

Ah, @porphyrin I think you hit the nail on the head here. :)

Short answer: It is not absolutely necessary, the theory and the results derived from it still hold at low particle concentrations but they cease to be meaningful. It is not so much a theoretical requirement, but a statistical one (for the results to make sense)
The kinetic theory of gases attempts to describe macroscopic properties like pressure and temperature based on a microscopic particulate picture of matter.
Pressure, for instance, is described as the force per unit area exerted by the gas molecules on the walls of the container. So, if you have a large number of particles in a container, whizzing about with some velocity and bumping into the container walls all the time. The collisions exert a force, which we measure as pressure.
Now thing of a macroscopic container with just one gas molecule in it. It will still whizz about inside the container with some velocity, and will also collide with the container walls every now and then exerting a force on them.  If we invoke the ergodic hypothesis, which means that the time spent by a molecule in region of space is related to the volume of said region, it would reasonable to say that the particle spends more time in the bulk volume of the container than near its walls.
Thus, we will not have the constant bombardment of particles on the container walls like the many molecule case, and thus we cannot obtain a meaningful measurement of pressure.
Similarly, temperature is related to the rms velocity in the kinetic theory. That quantity can be calculated for a single lone molecule, however, if you stick a thermometer in the container you wouldn't be able to get a meaningful measurement.

The kinetic theory seems to implicitly assume 'large' numbers of molecules in the sense that any one molecule is in thermal equilibrium with a heat reservoir at some temperature T, which is formed of all the other molecules in the gas. This is called a 'canonical ensemble'.
In the calculation of the Maxwell-Boltzmann distribution it is assumed that the state of any molecule can be defined by position r and momentum p. The state of a molecule is defined to be in the range  $r \rightarrow r+dr$ in the volume element $d^3r=dxdydz$  and have momentum in the range $p \rightarrow p+dp$ in momentum space volume $d^3p=dp_xdp_ydp_z$. The molecule is in weak interaction with other molecules which then act as a heat reservoir, and as we assume a dilute gas we can think of this molecule as a distinct small system in contact with a heat reservoir, and thus having a canonical distribution.
As a molecule's position can be anywhere in the volume of gas and its speed from $-\infty$ to $+\infty$ this leads to the probability of being at $r\rightarrow dr$ and $p\rightarrow dp$ $$P(r,p)d^3rd^3p \approx  exp(-\beta p^2/(2m)~)d^3rd^3p$$ with $\beta =(k_BT)^{-1}$. The exponential is the canonical distribution.
A consequence of being in contact with a heat reservoir is that there is a range of molecular speeds (and hence momenta). If there are $dN$ molecules with speed $v\rightarrow v+dv$ then $dN$ remains constant at equilibrium, even though the speeds of molecules are continually changing due to collisions. This also implies that a large number of molecules are colliding.

You're trying to describe a uniform gas. You want a large number of particles so that the gas appears to be uniform. If it's not uniform, it's hard to describe properties of the gas. As an extreme case, what is the pressure exerted by a single gas particle and how much volume (of the entire container) does it take up?


text:
Apple Cider vs Apple Juice Chemical/Physical Properties in Relation to Milk
Why does non-alcoholic apple cider mix smoothly with milk, but mixing apple juice with milk causes it to curdle?

Apple juice contains organic acids which causes the proteins present in milk to coagulate (curdle). The non-alcoholic cider most likely (depends on the brand?) contains less acids, so there is no coagulation. Good thing you mentioned that there is no alcohol present in the cider - ethanol is another thing that can cause milk to curdle.


text:
How to solve a math expression with proper significant digits?
I've talked with my friend from chemistry about this question, and looked all over the internet for like hours now, and couldn't find anything that would help in my situation.
The problem is this:
$$\frac{(19-9.25)}{(7.4 \times 2.26)}$$
So right away I know that the uncertainty in the addition is with the $19$, at $\pm 1$. The numerator would have $2$ significant digits from the $7.4$, but as the final answer is $.582944989$, do I use one significant digit from the numerator and make it $6 \times 10^{-1}$? Or does the overall division cancel out the subtraction somehow, so
$$\frac{9.75}{(7.4 \times 2.26)}?$$
I looked all around, and all of the examples I've found never had a whole number without a decimal, or never had one in an equation like this. Appreciate any  help.

For the record, no scientist actually uses significant figures. They use error bars because you can quantify how big the uncertainty is beyond it is in such and such decimal place.
There should be one significant figure. Because the $9.75$ in the numerator has one significant figure (before rounding). The extra precision in 7.4 doesn't mean much because of the large uncertainty in the numerator.
Try this:
$19$ is really $19\pm 1$. $7.4$ is really $7.4\pm 0.1$.
For the uncertainty in $19$, you could have:
$$\frac{20-9.25}{7.4 \times 2.6}\approx 0.643$$ on the upper end and
$$\frac{18-9.25}{7.4 \times 2.6}\approx 0.523$$ on the lower end. From this,
you can see the answer should be $0.6\pm 0.1$.
For the uncertainty in $7.4$ you have:
$$\frac{19-9.25}{7.3 \times 2.6}\approx 0.591$$ on the upper end and
$$\frac{19-9.25}{7.5 \times 2.6}\approx 0.575$$ on the lower end.
Based on the error here, the answer would be $0.58\pm 0.01$.
Notice that the precision is much worse for assuming that the error is in the number $19$. This is why significant figures work.

As a rule you can only be as certain as your least amount of significant figures. If you are doing only a rough calculation or homework question, my teacher usually suggests that you use exact values where possible and round to 3 sf for the final answer. If you are being more accurate you need to follow the rule scritly however

This is correct for multiplication and division. The question given is more complex than this.


text:
Electrical double layer
I am seriously confuse with the sign of electric field in the electrical double layer. Say zinc is dipped into the zinc sulphate solution. Then why the electrical double layer has positive sign in the electrode side of EDL and negative sign in the electrolyte side of EDL.
I am confused that since zinc has more tendency to dissolve, then zinc should lose two electrons and retain those 2 electrons in the electrode side of the metal-electrolyte interface. This will give eletrode negative sign. Also a layer positive charge of $\ce{Zn^2+}$ develops parallel to electrode.
Where I am wrong here?

The phrase "retain those 2 electrons" is wrong.
Metal surface has positive potential in solution,
but this doesn't mean that lost electrons are concentrated somewhere.
Electrons don't form layers. You can think of electrons as much more "volatile" things than ions, they may discharge everywhere depending on potential, for example, to reduce Zn ions in solution.
So, surface is charged positively and hence first ionic layer is negative.

Great. I was confusing with the layers of electrons. You cleared major doubts. But what in case of Cu2+? How the electrode surface turns negative even if Cu2+ has to be deposited at the electrode surface as Cu. From where it takes those 2 electrons?

As electric charge associated with double layer is typically very small (picofarades),
there is not much sense to bother about it in the first place.
It may get spread over electrode/solution, or drawed/discharged into external circuitry like the hand of experimenter :), and so on.
Btw, metal surface has non-zero potential in the air too.
If you absolutely need the source of electrons for mental experiment in absolutely isolated environment, then it may be oxidation of water:
H20  -->  O2 + H(+) + e(-)