Compressibility factor (Z) of a real gas I was wondering whether the compressibility factor of a real gas (given by $Z = V_{\mathrm{real}}/V_{\mathrm{ideal}}$) is supposed to be measured while keeping pressure constant?
I was attempting to answer the question below and got stumped at part (iii) because I had trouble finding out which was the dominating force being that when I tried calculating the compression factor I got a value of 1 (because $V_{\mathrm{real}}$ and $V_{\mathrm{ideal}}$ are the same, but the pressures are significantly different).

3.0 mol $\ce{CO2}$ behaving as a van der Waals gas, when it is confined under the following conditions: at 750 K in $\pu{150 cm^3}$. For $\ce{CO2}$ van der Waals coefficients, a: $\pu{3.610 atm L^{2} mol^{-2}}$, b: $\pu{4.29 \times 10^{-2} Lmol^{-1}}$. The universal gas constant R is $\pu{8.206 * 10^{-2} L atm K^{-1}mol^{-1}}$
i) Calculate the gas pressure (in atm), [5]
ii) What is the molar volume of the gas (in L/mol), [1]
iii) What is the pressure when it is ideal gas and use it to state the dominating force in the above gas?

From the given data, the molar volume $V_\mathrm m$ is $$V_\mathrm m=\frac{150\ \mathrm{cm^3}}{3\ \mathrm{mol}}=50\ \mathrm{cm^3/mol}=0.05\ \mathrm{L/mol}$$

In terms of molar volume, the van der Waals equation reads:
$$\left(p+\frac{a}{V_\mathrm m^2}\right)(V_\mathrm m-b)=RT$$

So, solving for the pressure gives:
$$p=\frac{RT}{V_\mathrm m-b}-\frac{a}{V_\mathrm m^2}$$

So, to get the pressure you don't need to know the compressibility factor.  Which of these two terms dominates numerically in determining the pressure?

$$\left(p+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT$$

$$pV-pnb+a\frac{n^2}V-a\frac{n^3b}{V^2}=nRT$$

cancelling $n$ on both sides

write $pV$ as $znRT$ where $z$ is compressiblity factor
$$(z-1)RT=pb-a\frac nV+ab\left(\frac nV\right)^2$$

it is behaving like ideal gas hence $z=1$ and right hand side is $0$

$$pb-a\frac nV+ab\left(\frac nV\right)^2=0$$

substitute the values and get the result


How does an oscillating reaction work? I watched a video showing an orange solution that goes to clear and then back to orange (and so forth.) The reaction goes through a number of cycles before it will no longer oscillate. This tells me that the overall energy in the system dissipates over time, but how does such a reaction work?

What is happening that causes this oscillation in concentration?

I worked on a project using this reaction fairly recently. There are a significant number of intermediate species involved in the reaction with a significant number of autocatalytic steps.

In order to obey the rules of free energy, the reaction may not oscillate about its equilibrium as a pendulum does- there is always a monotonic decrease in reactant concentration and an increase in product concentration, but the multitude of intermediates oscillate in concentration as the reaction switches its reliance on certain sections of the mechanism (dependent upon the instantaneous concentration of stuff in the mixture). The indicator used dictates the colour changes- I used ferroin, and so when the concentration of Ce(4+) (the catalyst used) is high the mixture is red, and when the concentration of bromate is high the mixture is blue.

It should be noted that the reaction will always come to a stop unless you use a continuously stirred flow reactor in which products are removed and reactants added as the reaction proceeds.

There are several mechanisms such as the FKN, MBM, GTF and Oregonator. The Oregonator is by far the most simple but lacks chemical accuracy, the former 3 display real chemical conversions yet still fail to perfectly model the reaction. The mechanisms can be modelled using coupled differential kinetic equations, which I did during my project in Mathematica.

The simplest general reaction scheme for a chemical oscillator is given by the Brusselator:

A → X
2X + Y → 3X
B + X → Y + D
X → E

In this scheme, A represents the reactions and E the products. X is autocatalytic (it catalyses its own production).

Another interesting example comes into play in population dynamics, the Lotka-Volterra equations (also known as predator-prey equations). They can be used to quantify the changes in populations in a predator-prey system such as foxes and rabbits. The reaction itself has also been suggested as a very primitive form of chemical order- interesting stuff.

Disclaimer: Sorry if this seems a little off-topic; the details of the specific BZ mechanisms are far less interesting than the concept of the oscillators themselves. If you want to look into the actual mechanisms though, start with the FKN model, it's fairly easy to digest and even model.

Perhaps the simplest oscillating ''Gedankenexperiment'':

1:  A -> B
2:  B -> C
3:  C -> A

Now we say that A inhibits reaction 2, B inhibits reaction 3, and C inhibits reaction 1. If you start with any pure substance A,B or C, the next step never starts before the previous is fully completed.

Of course this needs an energy source, otherwise the last step could never run:

1:  A     -> B
2:  B     -> C
3:  C + O -> A + OO


Ca energy release when gaining an electron Why does $\ce{Ca}$ have a negative electron affinity, i.e. energy is released when it gets an electron (ca. $\pu{-2 kJ/mol}$), when $\ce{Be}$ and $\ce{Mg}$ have positive values?

I know that the electron will be settled in the $\ce{3d}$ orbital which has a lower principal quantum number than $\ce{4s}$, but still, why would energy be released?

Two sign conventions: (1) the more common one states that a positive
  electron affinity value represents energy release when an electron is
  added to an atom; (2) the other states that a negative electron
  affinity represents a release of energy.

http://www.rose-hulman.edu/~brandt/Chem251/GenChem_Review_v3.pdf

  Recommended values for these electron affinities, in the units
  commonly employed in introductory texts and with the sign convention
  used here, are 2.37, 5.03, and 13.95 kJ/mol for Ca, Sr, and Ba,
  respectively. The endothermic electron affinities often quoted for Be
  and Mg are also too large and should be reported simply as ">0". An
  argument for a return to the original sign convention for the electron
  affinity is presented in this paper.

http://pubs.acs.org/doi/abs/10.1021/ed074p123

  OP: I know that the electron will be settled in the 3d orbital

Actually, as explained in the above article, the electron occupies 4p.

  OP: but still, why would energy be released?

Very little energy is released and the lifetime of Ca- is on the microsecond scale in practical experiments.  The reference Contributions to the electron affinity of calcium and scandium may be the best source for theoretical models that do and don't predict a slight energy release.  According to Atomic negative ions: structure, dynamics and collisions, core electrons need to be considered in addition to the valence electrons in calculating the stability of Ca-.  Early calculations that considered only valence electrons predicted that Ca- was unstable.  Configuration-interaction study of differential correlation energies in Ca-, Ca, and Ca+ presents detailed calculations of energy levels.

Please verify your data.

$\ce{Be}$ has negative electron affinity (precisely -2.4 eV).

Also calcium has positive electron affinity (precisely 0.0245 eV).

Check the Wikipedia page Electron Affinity Values

I assume you got confused between electron affinity and gain enthalpy. Electron gain enthalpy is negative of electron affinity.

EDIT 1: DAVE PhD, you misunderstood me. Electron affinity has absolute definite values. Only one convention is used. Electron affinity can assume both positive and negative values. But beryllium has negative electron affinity, by any convention you can't assume it to be positive. You can even check the Wikipedia page.


How does SO2 have 2 π bonds? The hybridization of $S$ in $SO2$ is sp2 which means it has only one unhybridized p orbital to form a $\pi$ bond. How does it form two?
Could it be one electron from a $p$ orbital getting promoted to a $d$ orbital leading to one $d-p$ and one $p-p$ $\pi$ bond. Is this even possible or are $\pi$ bonds only between p orbitals? Could the electrons possibly be delocalized in the $d-p$ and $p-p$ $\pi$ bonds?

Yes, you are right. One electron from 3p orbital is unpaired and excited to a 3d orbital. The hybridization of sulfur atom is sp2 hence a lone pair and two bond pairs(due to sigma bonding) reside in these hybrid orbitals.

The unpaired electrons are 3p and 3d hybridized orbitals are used in pi bonding with oxygen's unhybridized 2p orbitals. Hence two pi bonds are formed which are p-p pi and d-p pi bonds. Because of delocalisation these bonds are identical and have properties intermediate to the two different pi bonds.

To understand how the electron is able to make a jump with high energy difference, you will have to understand that d orbitals usually participate in bonding only when central atom is joined to many highly electronegative atoms. Because of polarization, there is a slight positive charge on the sulfur atom due to which orbital contract. This contraction is very very significant in d orbitals. Because of this electrons can be excited to the less energetic and tightly held d orbital. Also, the bonds formed will have a greater extent of overlap because d orbitals are not that diffused anymore. This is the reason that PF5 exists but PH5 does not.

Also, note that we cannot use dz^2 orbital for pi bonds as its lobes have the same phase and hence net overlap during pi bond will become zero.


Is there any software to simulate a laboratory? I wanted to do some experiments with some chemicals but unfortunately I do not have access to them. Is there any software I can use to get the products?

I don't want to do very complicated things, I just want to gain some experience in basic chemistry, with simple chemicals like $\ce{H2O2}$, some salts, sulfuric acid, potassium permanganate, etc.

I tried to search on the net but most of the software I found is too complicated for me.

Do you know about any such software? Or do you have any alternatives?

There is a product called Yenka Chemistry which simulates basic chemicals/apparatus, and allows basic manipulations (for instance to show the pH change as more acid is added to base, or the amount of gas evolved).

The program evolved from another piece of software called Crocodile Clips, which was aimed at GCSE science in the United Kingdom. For this reason it is incredibly basic, mainly focusing on inorganic reactions (acid/base, titrations, reactivity of group 1 and 2 metals), with no real capacity for demonstrating organic reactions.

I would recommend standardizing on Mathematica. Mathematica is the best general software to do physics and chemistry. You can use Mathematica absolutely free from the web with Wolfram Alpha. Instead of learning a specific chemistry program, if you learn Mathematica you can do any kind of science with it. Its the single most valuable tool you can learn to do math and science.

Here is an example, let's say you want to know what happens when you combine hydrochloric acid and calcium hydroxide. Just type it in to Wolfram Alpha:

It tells us the result: calcium chloride and water. Mathematica can do anything you would need to do in computational chemistry, stoichiometry, rate equations, molecular modeling, you name it.

Mathematica is free or inexpensive for students.


Is matter really divided in two categories? Page 1 of almost every Textbook start with the ordinary divison of matter in 2 categories: Substances and Mixtures.
Substances, or Pure Substances sometimes, are defined

as a form of matter that has constant chemical composition and
characteristic properties.It cannot be separated into components
by physical separation method.

This  is according to Wikipedia, at least.
Mixtures always citing wikipedia, instead are

materials made up of two or more different substances which are mixed
but are not combined chemically....
Mixtures are one product of a mechanical blending or mixing chemical
substances such as elements and compounds, without chemical bonding or
other chemical change, so that each ingredient substance retains its
own chemical properties and makeup.

Is this separation so radical?
Is there something in between? What about alloys?

Matter....

Matter is generally categorized as either mixtures or pure substances.  Pure substances are elements or compounds.  Mixtures are either heterogeneous or homogeneous.  For heterogeneous mixtures one should be easily able to discern the different components, while if the mixture is homogeneous it appears uniform.  Solutions and alloys are homogeneous mixtures.

A concept map which I give to my students: https://c7chemistry.wikispaces.com/file/view/CW%20matter%20concept%20map%20new%20rev%20091211.pdf/300804074/CW%20matter%20concept%20map%20new%20rev%20091211.pdf


Why does potential chemical energy decrease with increasing oxidation of carbon? I saw an interesting spectrum in my biology class the other day. On the left end was methane - the least oxidized form of carbon - and the most energetic form of carbon (at least relative to what else was on the slide). On the right end was carbon dioxide - the most oxidized form of carbon - and the form with the lowest energy.

Now, I do agree with the slide - we can combust methane but we can't combust carbon dioxide - at least I don't think we can.

However, is there an intuitive explanation of why increasing oxidation state correlates with lower potential energy, at least for carbon?

Does increasing oxidation state correlate with lower potential energy for all elements?

Why does it seem that oxygen forms rather strong bonds with other elements in many cases?

is there an intuitive explanation of why increasing oxidation state
  correlates with lower potential energy, at least for carbon?

The heat of formation of methane, carbon dioxide and carbon tetrafluoride are -74.9, -393.1 and -927.2 kJ/mol respectively (the trend remains unchanged even if you divide by MW and compare per unit mass).  As we shift electron density from electropositive elements to electronegative elements we stabilize electrons, their energy is lowered.

  Does increasing oxidation state correlate with lower potential energy
  for all elements?

I haven't checked extensively, but the trend seems to hold up.  I'd expect it to hold up.

$\ce{BeH_2}$, $\ce{BeF_2}$  127.2, -796.7 kJ/mol

$\ce{CaO}$, $\ce{CaF_2~~}$      39.1, -789.1 kJ/mol

$\ce{NaH}$, $\ce{NaF~~}$      126.2, -288.0 kJ/mol

  Why does it seem that oxygen forms rather strong bonds with other
  elements in many cases?

I'm not sure what you mean by "strong bonds".  The C-O bond in carbon dioxide is much more reactive than the C-H bond in methane.  Perhaps you're alluding to molecular stability which takes us back to the top, e.g. compounds with polar bonds typically have stabilized electrons, electrons in lower energy orbitals, as discussed above.

This scenario has a very simple explanation. The reduction of any electron to a more electronegative atom will release energy. To understand this, imagine a highly reduced methane molecule; the C-H bonds are unstable and the electrons in the H atoms is itching to escape the methane molecule to pair with an oxygen. In this example the reduction of oxygen by the hydrogen releases energy thereby lowering the reducing agent's( methane) free energy. Oxidation usually will result in a lower free energy state, due to the energy being released in the reaction.

TL DR;  High energy reducing agent ---- donates high energy electron to oxidizing agent. Reaction releases energy which the reducing agent supplied.

oxidizing substance becomes more stable b/c it lost a high energy electron.
Reducing substance becomes more stable b/c it attained an electron for an electronegative atom and the surplus energy is released.


In halogenation of ethene using bromine for example, why does the lone pair of bromine attack the C atom? In the presence of the electron rich (pi bond) of ethene, the bromine molecule becomes a temporary induced dipole, one end being electron deficient and acting as the electrophile. (The pi bond in) ethene attacks the electrophile.

At the same time, one of the lone pair of electrons in bromine attacks one of the C atom. Why does this happen?

Normally, I'm familiar with the same process, but with the carbocation being formed without the lone pair of electrons of bromine attacking the other C atom:

Does this have something to do with resonance as shown by the diagram below?

There are too many bond lines to explain the reaction; the question suggests that we have to explain the bond lines!

Let's start from scratch, with a different viewpoint. Ethylene isn't the attacker; it doesn't burst into flame in oxygen or do anything aggressive. It may be reactive, but in a "normal" range. Bromine, on the other hand, is a potent oxidizer and does go around interacting with relatively inert materials - it's a powerful bleach.

So, when bromine comes near the ethylene pi bond, it attacks the electron pair and grabs it, but the second bromine steals it away and departs as Br-. After all, bromine is more electronegative than carbon; the first bromine is just a link in the chain of events that allows the second bromine to escape with an electron. The first bromine winds up stuck to the C-C double bond; the two carbons and the bromine share the positive charge. The bromine certainly gets the lion's share of electron density, leaving the carbons with most of the positive charge. That's why another bromine can attack the carbons from the rear to give a dibromide. (If a Br- attacked the bromine attached to the double bond, it would just regenerate the reactants.) I bet that the second bromine to bond to the ethylene is not from the same bromine molecule as the first bromine. (But since all bromine molecules look the same, I can't prove it.) Doing the bromination in a solution with an overwhelming abundance of another nucleophile would probably give a mixed product.

A C-C double bond is about 1.34 Angstroms long; a bromine atom (in Br2) is about 2.29 Angstroms in diameter. Rather than envisioning the cation as a triangle of three grapefruit, I see it as a cantaloupe sitting on two large oranges. Just consider the empty p orbital of bromine and the pi lobe of the carbons; no need or benefit from making a three-membered ring.


Equilibrium Constants For Phenol Tautomers I encountered the following equilibrium constant values for the keto-enol interconversions of phenol in Organic Chemistry (Solomons, Fryhle, Snyder, 3rd Edition):

I wonder what could explain the differences (though I'm not sure whether they're significant) between the equilibrium constants of the two keto-enol interconversions wherein in one case the double bond of carbon with the oxygen and the hydrogens are oriented in ortho position and in para position in the other.

Edit: Can this difference concern the medium the reaction is catalyzed in or directive influence of groups?
Do explanations for equilibrium constant differences exist for other tautomer interconversions or are such queries quite speculative?

I believe that's because in the first step, phenol regains its aromaticity which makes the $K_{eq}$ substantially ($1000$ times) larger than that in the second step where it loses its aromaticity.


Converting Chlorine Dioxide to Free Residual Chlorine Our water treatment facility used Chlorine Dioxide to disinfect the produced drinking water. The analysis results indicate concentrations of 0.2 mg/l of chlorine dioxide at the source. Can we analys residual chlorine in the distribution network? We dont have water quality standards for chlorine dioxide, only residual chlorine. Can I calculate the concentration of chlorine in the water from the concentration of chlorine dioxide?

Some chemistry, first add H2O2 and NaOH:

$\ce{2 ClO2 + H2O2 + 2 NaOH -> 2 NaClO2 + O2 (g) + 2 H2O }$

Source: https://chemiday.com/en/reaction/3-1-0-1114
Continue adding H2O2, as:

$\ce{2 NaClO2 + 2 H2O2 ->  2 NaClO + 2 O2 (g) + 2 H2O }$
$\ce{2 NaClO + 2 H2O2 -> 2 NaCl + 2 O2 (g) + 2 H2O }$

which implies a net reaction of:

$\ce{2 ClO2 + 5 H2O2 + 2 NaOH -> 2 NaCl + 5 O2 (g) + 6 H2O }$

So, slowly add H2O2 until the bubbling ceases. Collect the volume of oxygen and convert to moles. Divide by 2.5 to determine moles of ClO2 and your chlorine therefrom.
Here is a related reference, Titrimetric and photometric methods for determination of hypochlorite in commercial bleaches, which cites the use of hydrogen peroxide and acidic permanganate.


What causes the DNA fragments to stop moving in gel electrophoresis? I'm currently studying VCE BioChemistry, and we're studying the separation of DNA strings of different lengths via gel electrophoresis.

(This involves having 'clumps' of DNA at one end of a gel medium and applying an electric current, pushing the DNA strands across the gel. Different sized strands move at different speeds through the gel)

At some point, the strands stop moving such that there is a distribution of strands across the gel. What causes the strands to stop?
Surely they'd continue until the end of the gel.

If suppose the electric current could be stopped, thus causing the strands to stop moving, but how would one know when to stop the current?

There are electrophoretic methods where the molecules stop automatically like Isoelectric Focusing, but in the common agarose gel electrophoresis of DNA you mention this is not the case.

The DNA is negatively charged, it will move along the electric field you set up until you stop applying any current. It will happily move out of your gel if you let it run for too long.

To see the progress of your gel some dyes are usually added to the sample like bromophenol blue or xylene cyanol. They don't interact with the DNA, they just provide you a reference on how far your gel has run. You still need to have a rough idea how your samples run compared to those dyes if you want to use the full length of your gel.

DNA Molecules may stop to move in agarose gel due to high molecular weight , the molecular weight of DNA  affect its mobility in gel , the smaller the molecular weight the higher the speed of migration and long distance . large molecules tend to migrate slowly and they move short distance.


SN1 or SN2 reaction? Here, this is a 1° allylic halide. If it goes by SN1, the intermediate will be resonance stabilised.

Also, since it is 1° Halide, it can also go via SN2. Which one will be preferred?

This reaction will happen by both SN1 and SN2 mechanism, apart from this SN1' and Sn2' (allylic mechanism ) will also occur. Thus from SN1 and Sn2 one will get primary cyanide and from SN1' and SN2' one will get secondary cyanide .( read organic chemistry by jerry march for the detailed explanation).
I am attaching a screenshot of clayden which explains this beautifully.

Well, Your Question seems  a bit off as the type of solvent has not been mentioned.
Is it:-

Non-polar
Polar (protic or aprotic)

Obviously, SN1 can occur only in Polar protic solvents, and in reactions where the alkyl halide is a conflicting one (reaction can occur both by SN1 and SN2), the mentioning of the said solvent is essential.


Does solubility of acetic acid depend on the pH? Does the solubility of acetic acid increase or decrease in water if I adjust the pH?

Conversely, a water-insoluble molecule (like retinol), can its solubility be increased by adjustment of the pH?

I would say no (in both cases) because the protonating or deprotonating these molecules does not change their overall hydrophilic or hydrophobic character.

You are correct in both cases, though your reasoning isn't quite on target.

Regardless of the pH, retinol will not be protonated or deprotonated to any significant degree (which would greatly affect it's solubility if it were). It's just a nonpolar molecule that will stay that way regardless of pH.

Acetic acid also would not be protonated to any significant degree at low pH, but that's a mute point as it's fully miscible in water anyway.

The one situation in which you would see deprotonation of the compound is that of acetic acid at high pH. For the most part, both acetic acid and the acetate ion produced by the basic conditions should remain highly soluble. It is possible however, that if you added enough strong base, say NaOH, to a concentrated solution of acetic acid, you could eventually precipitate sodium acetate.

Summary, TL;DR:
You are correct that the answer to both questions is no. Acetic acid is miscible with water (so it's solubility definitely can't be increased) and it's solubility simply won't be affected by pH. If retinol could be protonated or deprotonated, that would increase it's solubility, but it can't be protonated or deprotonated to any significant degree so it is insoluble in water, regardless of pH.

In an aq medium solubility of a weak electrolyte depends on its ionisation constant which increases on dilution.But conversely if a little amount of strong acid is added to the medium the ionisation of acid is suppressed due to common iron effect and hence its solubility.Protonation is highly subjective to the ions available in solution.


Is there a cheap, safe liquid that evaporates EXOthermically around room temperature? The heating and AC system in the building where I occasionally work, works like this (basically a thermostat):

The administrators set some temperature that is maintained automatically. Say, in the winter, the outside temperature is 0 °C. The admins will set the desired inside temperature to 20 °C. There is some thermometer behind a little steel panel. It measures it to be cold in the room; therefore the heat comes on. I can trick the heat into coming on even more, by rubbing some alcohol or acetone on the steel panel. The acetone evaporates endothermically, making the steel panel REALLY cold; then the heater thinks that it's REALLY cold in the room and pumps the heat in.

The same system is in place in the summer, but in reverse. However, I can't trick the panel in the same way. If I put acetone on the panel, it'll cool off, and then the A/C unit thinks the room is just fine, because the panel is cool. I'm wondering if there is some liquid that will evaporate exothermically, and heat up the panel, in order to make the system think I need more A/C?

I don't believe that exothermic evaporate exists; here is a list of heat of vaporization for various substances; all positive (endothermic). http://www.engineeringtoolbox.com/fluids-evaporation-latent-heat-d_147.html
But maybe I'll be surprised and it does exist!

I have been holding my laptop's hot backside up to the panel instead, but that's a bit too manual for my taste.

Edit: To be a bit clearer, let me specify that this is just a thought experiment. I'm not actually expecting to find something which doesn't make sense like an exothermically evaporating liquid.

No such liquid, safe or otherwise, can exist. Evaporation is a strictly endothermic process in all cases.

The change in state from liquid to gas is marked by the individual particles gaining enough translational kinetic energy to overcome the mutual attractions present in the liquid phase to "fly free" in the gas phase. It is logically inconsistent for a substance to increase its internal energy and release energy to the surroundings as heat in the same process.

In order to achieve both evaporation and a release of energy, one would have to find a liquid that reacts to (a) release heat and (b) form gaseous products.  The energy required to move from the liquid to the gas phase is substantial; more than likely the only reactions exothermic enough to provide a net release of heat are combustion reactions.

I somehow doubt dumping a flammable liquid into your thermostat, inserting a wick, and lighting it on fire is a satisfactory solution for you.

To avoid having to stand there or wire up the panel, I would aim some low power laser to the thermostat cover. This can be started and stopped remotely. A dark sticker would enhance energy absorption.


Does nitric acid release hydrogen only when reacting with Mg and Mn? $\ce{HNO3}$ always reduces its anion to give nitrogen containing compounds ( $\ce{NO2}$, $\ce{NO}$, and even $\ce{N2O}$) when reacting with metals, but with $\ce{Mg}$ and $\ce{Mn}$ it releases $\ce{H2}$. Is there a specific reason for this variation?

Actually, it usually doesn't. It produces all types of products ranging from $\ce{NO2}$ to $\ce{NH4NO3}$ with exact products depending on concentration and conditions.

The reason for it is high oxidation power of nitric acid, exceeding that of hydrogen cations, so nitric acid reduces first. As result, to produce some nitrogen, the oxidation power must be brought below that of hydrogen cations. I'm not aware if it is actually possible, but possibly high dilution may help.

As for Mg, it reacts with hot water, so it's unclear if the reaction of $\ce{Mg}$ with diluted nitric acid counts. Not aware about $\ce{Mn}$ though.

Because reaction of Metal+Acid= Salt + hydrogen
$$\ce{Mg +2HNO3 -> Mg(NO3)2 +H2}$$

$\ce{Mg}$ has a charge of +2 while nitric acid ($\ce{HNO3}$) has a charge of -1

Metals always react with acids to form salt and hydrogen gas


What will happen when a metal reacts with an acid which is neither dilute nor concentrated? We have always studied about metals reacting with dilute and concentrated acids, but I do not understand what will happen if the same metal reacts with an acid that is neither concentrated nor dilute?

This question is a bit vague, but briefly, the reason some acids behave differently at high and low concentration is because changing the activities of components in the solution changes the redox potential of the various possible reactions.

Let's take the reaction of iron with sulfuric acid as an example. This is not necessarily the only factor and I'll make a few assumptions that are likely quite inaccurate to simplify the problem, but it will illustrate the situation:

Let's assume that the half-reaction with iron is always the same—iron is being oxidized:

$\ce{Fe_{(s)}<=> Fe^2+_{(aq)} +2e-}\ E^\circ=+0.44\ \mathrm{V}$

Now, if you look at a table of standard reduction potentials, you'll see that there are a couple of potential half-reactions that could be reduced to drive the iron oxidation:

$\ce{2H+_{(aq)} +2e- <=> H2_{(g)}}\ E^\circ=0\ \mathrm{V}$

$\ce{SO4^{2-}_{(aq)} +4H+_{(aq)} + 2e- <=> SO2_{(g)} +2H2O_{(l)}}\ E^\circ=+0.17\ \mathrm{V}$

If all these reagents were at unity activity, then the second reaction would be more favoured and we would see mostly $\ce{SO2}$, but the activities of the substances in solution are also important. By the Nernst equation, $E=E^\circ -\frac{RT}{zF}\ln Q$, we see that the potential of a complete cell or a half-cell reaction is dependant on the reaction quotient $Q$. Because we're assuming that only the one reaction happens to the iron, we can just look at the reduction half-cell. If we assume that the sulfuric acid dissociates completely to $\ce{SO4^2-}$, then $[\ce{H+}]\approx 2\cdot[\ce{SO4^2-}]$. If we assume the gaseous products of the reaction don't affect the equilibrium, for our two reactions, $Q$ is $\frac{1}{[\ce{H+}]^2}=\frac{1}{(2[\ce{H2SO4}])^2}$ and $\frac{1}{[\ce{H+}]^4[\ce{SO4^{2-}}]}=\frac{1}{(2\cdot[\ce{H2SO4}])^4[\ce{H2SO4}]}=\frac{1}{16\cdot[\ce{H2SO4}]^5}$, respectively.

Now, if we substitute these into the Nernst equation for different concentrations of acid, we can see which reaction is more favourable: In dilute acid (1 mM), we get −160 mV for hydrogen formation and −238 mV for sulfur dioxide. The picture changes as the acid gets more concentrated. In 1 M acid, the hydrogen formation half-cell has a potential of 18 mV, but sulfur dioxide is now up to 206 mV.

So you can see how the concentration of acid changes which reaction is more favourable. To answer your question about what happens in the middle, one need only do the calculation for a moderate concentration and find that at some point, the two half-reactions become close in potential and we may see a significant mixture of both reactions occurring.

This is a very simplified picture and the Nernst equation is only technically valid under conditions of no current flow, meaning that other factors may influence what is observed experimentally, but I hope it's apparent that there's not really a solid line between dilute and concentrated. Many different reactions can occur simultaneously, even if a single reaction is dominant under certain conditions.

Some metals react explosively with water (Alkali metals, Group 1, sodium, potassium), and water is neutral. Some metals do not react even with very strong and consentraded acids (platinum).

Some acids are strong, some are weak. Even vitamine C is an acid, but so also are sulferic acid and nitric acid. Consentration also plays a role in this picture.

Magnesium reacts slowly with water, and is oxidized. Gold does not react. Iron reacts. and so on. The nobler the metal, the harder it is for an acid to oxidize it, and the stronger and / or more consentrated the acid will have to be to break down the metal. Alkali metals are 'weak' enough to be oxidized by clean air.

So this question is far from simple to give a short answer to.


Decomposition through thermolysis, electrolysis, photolysis What decides whether a compound will dissociate (break-up) under thermal decomposition or photolysis or electrolysis i.e if I am provided with a compound and I'm asked to dissociate it then how will I get to know that it would dissociate under thermal decomposition or photolysis or electrolysis.

Please try to provide an easy-to-understand as I'm in just 10th grade

You should analyse the chemical bindings in your compound, e.g. ion bindings will easily break-up in water, stronger bindings may require You to use heating or electric current to dissociate the molecule. Also if you know the structure of the compound, look for some similar structures on the internet, the more similar the structures, the more similar properties they have, so similar factors will dissociate them.


Do Ligands electrons fill the d-orbitals of the metal ion? We know that ligands make coordination bonds with d-orbitals of transition metals forming complexes. My question is, if that happened, all d-orbitals will be filled and no electron transition could happen between eg and t2g orbitals to be responsible for magnetism or colour formation.

I need clarification.

You are mistaken. The electron pairs donated in the coordinate bonds by the ligands do not fill the d orbitals of the metal ion.

The basis of crystal field theory is the electrostatic repulsion between the d orbitals of the central metal ion and the electron cloud of the ligands. This repulsion then leads to the breaking of the degeneracy of the orbitals, giving rise to disparity in energy between different sets of d orbitals. This arises due to the different spatial orientations of the 5 d orbitals (i.e. $xy, yz, xz, x^2-y^2, z^2$). Thus, they experience different amounts of repulsion from the ligands when the ligands are oriented in different geometric configurations (i.e. octahedral, square planar, tetrahedral etc.) around the central metal ion.

For more details on crystal field theory and an alternative explanation known as ligand field theory (based on molecular orbital theory), you can refer to the following discussion:

In an octahedral complex, what happens to the electrons donated by the ligand?

Since the d orbitals are not filled up by any ligand electrons, the electronic transitions between the orbitals of different energy levels (such as in the case of $t2g$ and $eg*$) are certainly possible.

Now, you may wonder: Where do the donated electrons go?

Well... based on valence bond theory, they are used in the formation of covalent bonds between the central metal ion and the ligands. These covalent bonding interactions are described as the overlap of the ligands' atomic orbitals with the electron pairs with the hybrid atomic orbitals of the central metal ion.

For example, in the case of a hexaaquairon (III) complex, the iron (III) ion has 5 electrons, each singly occupying one d orbital. As the d orbitals are occupied, the vacant 4s, 4p and 4d orbitals are hybridised to give 6 $sp^3d^2$ orbitals to be used to overlap with the atomic orbitals of the ligands.

Thus, the donated electrons actually are "shared" between the valence atomic orbitals of the ligand and the valence hybrid orbitals of the metal ion.

Hope I have clarified your doubts.


How to calculate the pH of the neutralisation of HCN with excess of KOH? Write the reaction between $\pu{0.05 mol}$ $\ce{HCN (aq)}$ and  $\pu{0.08 mol}$ $\ce{KOH (aq)}$ with $V = \pu{500 mL}$ and then calculate the pH.

So I wrote the reaction:
$$\ce{HCN (aq) + KOH (aq) <=> KCN (aq) + H2O}$$

Begin:
\begin{align}
  n_\mathrm{begin}(\ce{HCN}) &= \pu{0.05 mol}\\
  n_\mathrm{begin}(\ce{KOH}) &= \pu{0.08 mol}\\
  n_\mathrm{begin}(\ce{KCN}) &= -\\
  n_\mathrm{begin}(\ce{H2O}) &= -\\
\end{align}

End:
\begin{align}
  n_\mathrm{end}(\ce{HCN}) &= -\\
  n_\mathrm{end}(\ce{KOH}) &= \pu{0.08 mol} - \pu{0.05 mol} = \pu{0.03 mol}\\
  n_\mathrm{end}(\ce{KCN}) &= \pu{0.05 mol}\\
  n_\mathrm{end}(\ce{H2O}) &= \pu{0.05 mol}\\
\end{align}

The product of the reaction is a salt of a strong base ($\ce{CN-}$) completely dissociated in water:
$$\ce{KCN (aq) -> K+ (aq) + CN- (aq)}$$

And
$$\ce{CN- (aq) + H2O -> HCN (aq) + OH- (aq)}$$

Knowing this I calculated the
$$[\ce{KCN}] = \frac{\pu{0.05 mol}}{\pu{0.5 L}} = \pu{0.1 M};$$

Then I used the equation:
$$[\ce{OH-}]_1 = \sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}} \cdot [\ce{KCN}]} = \pu{1.27x10^-3 M}$$

Since I still have some $\ce{KOH}$, it dissociates completely creating $\ce{OH-}$ ions, so:
\begin{align}
  [\ce{OH-}]_2 &= [\ce{KOH}]\\
  [\ce{OH-}]_2 &= \frac{\pu{0.03 mol}}{\pu{0.5 L}} = \pu{0.06 M}
\end{align}

Finally
\begin{align}
[\ce{OH-}] &= [\ce{OH-}]_1 + [\ce{OH-}]_2 \\
&= \pu{1.27x10^-3 M} + \pu{0.06 M} = \pu{0.06127 M}\\
\ce{pOH} &= -\log[\ce{OH-}] = -\log(0.0617) = 1.213\\
\ce{pH} &= 14 - \ce{pOH} = 12.787\\
\end{align}

Is it right to calculate the $[\ce{OH-}]$ in this way?

I based my calcualtions on this equation:
Knowing that the amounts of substance in moles are defined by $Z = Z_1 + Z_2$ in $K$ liters of solution, we can consider this concentration to be
$$[X] = \frac{Z}{K} = \frac{Z_1 + Z_2}{K} = \frac{Z_1}{K} + \frac{Z_2}{K} = [X_1] + [X_2]$$

While you've generally done a good job, you've made an important error. The hydrolysis of the cyanide ion should be written as an equilibrium reaction:

$$\ce{CN-(aq) + H2O(l) <=>  HCN(aq) + OH-(aq)}$$

That's because $\ce{HCN}$ is a very weak acid and its conjugated base a weak base (homework: look up the $K_b$ of $\ce{CN-}$).

Now, as you duly noted, there's excess $\ce{KOH}$ in the solution, which dissociates completely. The $\ce{OH-}$ resulting from that dissociation now push the equilibrium:

$$K_b=\frac{[\ce{HCN}]\times[\ce{OH-}]}{[\ce{CN-}]}$$

to the left. As a result, the contribution of the $[\ce{OH-}]$ coming from the hydrolysis of the cyanide is negligible and can be ignored. Only the $[\ce{OH-}]$ from the excess $\ce{KOH}$ is to be counted. So $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$.

We can confirm that numerically by assuming $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$ and with:

$$\frac{[\ce{HCN}]}{[\ce{CN-}]}=\frac{K_b}{[\ce{OH-}]}\approx 0.0027$$

So neglibibly little cyanide is present as the acid $\ce{HCN}$.

As regards the matter of adding molarities, in some instances that is allowed.

Say we add $n_1$ moles of $X$ to $V\ \mathrm{L}$, that would give a molarity $M_1=\frac{n_1}{V}$. At a later stage we add $n_2$ moles of $X$, that molarity would be $M_2=\frac{n_2}{V}$.

The total molarity would be:

$$M=M_1+M_2=\frac{n_1}{V}+\frac{n_2}{V}=\frac{n_1+n_2}{V}$$

But if we were to mix volumes of solutions it would be:

$$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$$

Now  they are no longer simply additive.

First of all your calculation is based on the assumption that HCN is a strong acid, which it's not. Check for ionisation constant of HCN.
And yes you can add moles directly because moles are discrete particles, even if they are in solution. But you can't add concentration or molarity because it's volume dependent and when you consider volumes, you'll eventually end up adding moles again.


Why does neighbouring group participation happen? Can someone please explain why neighbouring group participation happens?

Does it stabilise the carbocation? If so then is NGP dominant over an intermolecular SN2 reaction (especially in aprotic medium)?

There may be some confusion over what neighbouring group participation (NGP, also sometimes referred to as anchimeric assistance) is. Carey puts it quite succinctly:

  When a molecule that can react by nucleophilic substitution also contains a substituent group that can act as a nucleophile, it is often observed that the rate and stereochemistry of the nucleophilic substitution are strongly affected. The involvement of nearby nucleophilic substituents in a substitution process is called neighboring- group participation.

  Source: Advanced Organic Chemistry (Part A). Francis Carey.

Rather than a discrete carbocation forming, which is then stabilised, NGP usually (though not exclusively) occurs via two consecutive SN2 processes, with overall retention of stereochemistry.

The reason why the NGP happens, rather than an external nucleophile just doing the substitution is generally due to the fact that intramolecular reactions happen at a faster rate than intermolecular ones, since this removes the necessity of encounter between nucleophile and electrophile (the two are essentially tethered together). The initial internal attack forms a ring (usually 3 or 5 membered for reasons that should be familiar to you) which may then be opened by the external nucleophile.

A classical example is of a sulfide displacing a halide, with subsequent opening of the reactive 3-membered intermediate:

  Source: Organic Chemistry. Clayden/Warren.

In sulphur act as neighboring group it is less electronegative,it is large size,it is more polarizable.hence N.G mechanism participate leaving group.formed any memberd ring(3 or 6)due to angle strain..


Reaction mechanism of combustion of hydrogen After a discussion about the usefulness of hydrogen fuel for cars with a friend of mine, I wondered what the reaction mechanism for combustion of $\ce{H2}$ was. I only study (well almost) organic chemistry, so I have no idea if the mechanisms from organic chemistry can be applied at all to non-organic molecules.

So we have the following reaction:

$$2\ce{H2 + O2 -> 2 H2O}$$

In most textbooks this would be described as a redox-reaction with the two half reactions:

$$2\ce{H+ + 2 e- -> H2}$$
$$\ce{O2 + 4 H+ + 4 e- -> 2 H2O}$$

But I wonder what the real mechanism is. Based on my organic thinking I thought the following (it's probably completely wrong, but I think it is important to state because someone can correct me much better if I express my thinking than if he has no idea where the misconception started):

If the temperature is high enough a long pair of oxygen could break the $\ce{H-H}$ (the lone pair will be donated to the $\sigma^{*}$ bond) bond and form $\ce{2 H-}$ (each H gets one electron because of their equal electronegativity). $\ce{H-}$ is very unstable and one will immediately react with the empty oxygen orbital, the other will break the $\ce{O=O}$ $\pi$ bond. The electrons in the $\pi$ bond will go to the positively charged oxygen atom (the one that donated its electrons in the first step). Now one oxygen atom is neutral and the other has a positive charge (because the electrons of the $\pi$ bond went for the other oxygen), where the $\ce{H-}$ will attack forming hydrogen peroxide ($\ce{HO-OH}$). The previous process will now repeat, but instead, the $\sigma$ bond between the oxygen atoms will be broken, resulting in two water molecules, $\ce{2 H2O}$.

This post deals with the mechanism that is observed in the gas phase. It is of course not as simple as the equation might suggest and you did suspect that already.
$$\ce{2H2 + O2 -> 2H2O}$$

This will be divided into many different elementary sub reactions. Any mixture of oxygen and hydrogen is metastable (stable as long as you do not change the conditions). If you provide sufficient energy to overcome the activation barrier, the reaction will take its cause:
$$\ce{H2 <=>[\Delta T] 2 H.}$$

The resulting hydrogen atoms are very unstable and will react with anything in reach, but most importantly with $\ce{O2}$ (which is a triplet biradical), forming hydroxyl radicals. This reaction is endothermic (it requires energy).
$$\ce{H. + O2 -> HO. + O}$$

The resulting oxygen radicals can again react with $\ce{H2}$ to form more hydrogen radicals. This reaction is also endothermic.
$$\ce{O + H2 -> HO. + H.}$$

The hydroxyl radicals can also react with $\ce{H2}$ to form more hydrogen radicals, which is a slightly exothermic (releases energy) reaction.

The net result of those equations leads to a slightly exothermic sum, with high potential:
$$\ce{3H2 + O2 -> 2H2O + 2H.}$$

In principle, the hydrogen radicals react as a catalyst. However, this reaction is a highly branched chain reaction, including a lot of radical reactions. Due to this, more and more hydrogen radicals will be produced.

This scheme will eventually end when the concentrations of $\ce{O2,H2}$ will become lower forcing the excess radicals to react with each other.
$$\ce{HO. + H. -> H2O}$$

Another possibility is forming as a byproduct hydrogen peroxide, which is a very exothermic reaction:
$$\ce{H. + O2 -> HO2.}\\
\ce{2HO2. -> H2O2 +O2}$$

(Also happening but not as important: $\ce{HO. + HO. <=> H2O2}$)

As long as there are hydroxyl and hydrogen radicals present, the peroxide will take part in the chain reaction. However, this reaction also provides the necessary energy to cleave more $\ce{H2}$. peroxide is also easily cleaved again, or reacts with each other:
$$\ce{2H2O2 -> 2H2O + O2}$$

This all results in the main product water.

Please keep in mind, that there are many factors, that influence these reactions. It is strongly dependent on pressure, temperature, and surroundings. Surfaces and/ or catalysts involved in this reaction may change it completely.

Further reading:

How explosive is hydrogen gas?
Eugene P. Dougherty1 and Herschel Rabitz1, J. Chem. Phys. 72, 6571 (1980) most comprehensive (theoretical) study on the $\ce{H2-O2}$ system.
Wiberg, Nils: Lehrbuch der Anorganischen Chemie Kap. 3.2 (german) The mechanism is in part taken from this source.
http://arrow.utias.utoronto.ca/~ogulder/ClassNotes3.pdf Thanks to Silvio Levy

The combustion of hydrogen can occur both in the gas phase and on the surface of a metal, it can also occur on the surface of platinum. This is used in the hydrogen recombiners which are present in the containment of many nuclear reactors.

Another example of a flameless combustion on a metal surface is the oxidation of solvents such as acetone on the surface of copper metal. Here is one of the many films on youtube showing the reaction. The oxidation of acetone with air using a copper metal surface is much easier to do than the combustion of hydrogen with a platinum surface.

The hydrogen will adsorb onto the metal surface first by physical adsorption and then by chemisorption. This will result in the platinum surface having hydroegn atoms on the surface.

Oxygen molecules from the air will also adsorb onto the platinum, then on the surface a reaction between hydrogen atoms and the oxygen will occur which forms water.

TPD, XPS and LEED were used by Yuichi Ohno and Tatsuo Matsushima to show that oxygen on a platinum surface can be present both as molecules and as atoms, Surface Science, 1991, volume 241, page 47.

If you want to know more then a good refernece would be Study on chemisorption of H-2, O-2, CO and C2H4 on Pt-Ag/SiO2 catalysts by microcalorimetry and FTIR , X. Wang et. al., JOURNAL OF THERMAL ANALYSIS AND CALORIMETRY, 2005, volume 82, page 103.


How do I calculate the degree of dissociation in equilibrium? In my textbook, for calculating the percentage dissociation of $\ce{HF}$ for the given equation:
$$\ce{HF + H2O <=> H3O+ + F-}$$

The solution is:

Initial Concentrations
$$[\ce{HF}] = 0.08~\mathrm{M}, \: \ce{[H3O+]} = 0, \:\ce{[F- ]}= 0$$
Equilibrium concentrations
$$[\ce{HF}] = 0.08~\mathrm{M} - x, \: \ce{[H3O+]} = x, \:\ce{[F- ]}= x$$

I am not able to undestand why $x$ is subtracted from 0.08 and not $cx$ [ where $x$ is the degree of dissociation]

I tried solving the same problem taking $cx$ but not able to get the solution, can anybody explain the difference to me? I have tried asking a similar question earlier too but it is really hard for me to get my head around this concept.

Let us first define the terms needed here.

Degree of dissociation (DOD)
Degree of dissociation is the fraction of a mole of the reactant that underwent dissociation. It is represented by $\alpha$.
$$\alpha = \frac{\text{amount of substance of the reactant dissociated}}{\text{amount of substance of the reactant present initially}}$$
Number of moles dissociated
It is defined as the product of the initial concentration of the reactant and the degree of dissociation

Now suppose you have a reaction like this
$$\ce{A->B + C}$$

The initial state of A is always the concentration of A (should be given in the question) while initial moles of B and C are zero (if anything else is not specified).
The final state of A is always defined as $\text{(amount of substance initially present) - (amount of substance dissociated)}$ while for B and C it is just $\text{amount of substance of A dissociated)}$.

Writing our equation again,
\begin{array}{lcccc}
 & \ce{A &-> &B &+ &C}\\
\text{Initial amount of substance}& a && 0 && 0\\
\text{Final amount of substance}&   a - a\cdot(\mathrm{DOD}) && a\cdot(\mathrm{DOD}) && a\cdot(\mathrm{DOD})\\
\end{array}

x is subtracted from reactants , because the reaction is forward, so it will change in to ions, or we can say the concentration of reactants will decrease therefore we write minus x, and the acid you mention here is weak acid so it will not dissociate completely, so we donot know directly how much it will dissociate thus we use a variable x ,to find the value of x  we make equation ,eg c-x=x+x then kc =x2/c-x ,then we solve for x , and get the amount which dissociated,


Why is nitrous acid prepared in situ? Why is nitrous acid prepared in situ?

I am not sure about the above question. First of all, I don't exactly understand what is meant by "in situ". Secondly, how would preparing nitrous acid in situ, in the diazotisation of amines with nitrous acid, make it less toxic or unstable?

Well, if you don't make it in situ, you would have to add the nitrous acid you have already kept in your laboratory shelf, and surprise! It's not there! Because, it has already decomposed.

Nitrous acid ($\ce{HNO2}$) has an intermediate oxidation state of +3, and hence, it tends to disproportionate into more stable oxidation states:

$$\ce{2HNO2 -> NO2 + NO + H2O}$$

$\ce{NO2}$ itself dissolves in water to give $\ce{HNO3}$ and $\ce{HNO2}$. This gives an overall reaction in aqueous solution as:

$$\ce{3HNO2 -> HNO3 + 2NO + H2O}$$

This should give you a good idea on why we make nitrous acid within the reaction mixture. You can't store it beforehand. It must be made when it's time to use it.

In the language of chemistry,in situ means The reaction mixture.Here,in case of HNO2,it is too unstable.Here,if we see the reaction,
HNO2→HNO3+NO+H2O
The oxidation no of central atom N:
in HNO2-+3
in HNO3-+5 (oxidation,as oxidation no increased)
in NO-+2(reduction,as oxidation no decreased)
If we observe,we can see that,here at the same time, the oxidation and reduction of N occurs.When in any compound,the oxidation and reduction of any element occurs at the same time,we call it Disproportionate reaction instead of oxidation-reduction reaction.
HNO2 has an intermediate oxidation state of +3.Thats why,it tends to be disproportionate into more stable oxidation state.
Instead of this,to prepare  Nitrous acid,we take the mixture of NaNO2 and HCl.
NaNO2+HCl→NaCl+HNO2
Through this process,HNO2 is prepared in the labratory in need


Distance between successive tetrahedral voids in FCC What is the distance between successive tetrahedral voids in a FCC
crystal structure?

The distance between a face centered atom and a tetrahedral void is $\sqrt{\frac{3a}{4}}$ if the cube's edge length is $a$. Now If I know the angle made, between the line joining a tetrahedral void and the face centered atom, and the successive tetrahedral void and face centered atom, I can find the distance.  But how do I calculate that distance since I don't know the angle.

As shown in figure, if we divide a FCC unit cell into 8 small cubes, then each small cube has 1 Tetrahedral void at its own body centre.
Thus, there are total 8 Tetrahedral voids in one unit cell.
It can also be seen from the figure that the nearest distance between two Tetrahedral voids is a/2.

Answer Is Given In This Image With Complete Explanation... Hope It Works...


Significance & applications of polarizability for nucleophiles One of the reasons given for the greater nucleophilicity of iodine ion compared to fluorine ion is the greater polarizability of the large electron cloud of iodine. The other reason given is that it is less crowded by polar protic solvent molecules.

However, when the solvent is changed to a polar aprotic solvent the nucleophilicity order is reversed.

If polarizability had any significant role to play, the order would've not been completely reversed. It seems to me that the only reason iodine ion is more nucleophilic than the fluoride ion is because it is less crowded by the solvent.

Now, this leads me to believe that polarizability is not a very important factor and should only be used when comparing nucleophiles of the same basicity, but of different size (and hence polarizability) in a polar aprotic spent.

But why then, for uncharged nucleophiles does polarizability dominate over basicity in determining the order of nucleophilicity (regardless of the solvent used)?

The examples the site you mentioned gave of uncharged nucleophiles were: the hydrides of Group V and VI. They say that the bigger atoms can have a more effective overlap of atomic orbitals and thus, the bigger the uncharged nucleophile, the stronger it is.

I would like to dispute this reasoning and answer your question regarding this trend.

How we should see a nucleophile should always primarily be in terms of electron donation because a nucleophile donates electron density to electrophiles to form bonds. This is a very important consideration.

For bigger atoms, the valence electron pair, which is being donated to the electrophile, is further away from the nucleus and is more shielded from the nuclear charge. Thus, it is easier for donation to take place. This should be the reason why uncharged nucleophiles with bigger atoms are more nucleophilic; that is, they are more capable of donating the electron pair of theirs. My explanation of the trend has nothing to do with polarisability.

The reasoning stated is incorrect because although bond formation is easier for bigger atoms, it does not mean that the overlap of atomic orbitals is more effective. In fact, it is likely less effective because the atomic orbitals used are more diffused (Recall that as the principle quantum number n increases, the orbitals become more diffuse in terms of electron density). Thus, their reasoning is nonsensical.

Hope I have clarified everything you were not sure of.

It seems to me that both trends end up pointing the same best uncharged nucleophiles, that means, polarizability increases down the periodic table (bigger electron clouds have a better overlap in the Sn2 transition state) while basicity decreases (the electron density is more disperse around bigger atoms). So polarizability does not dominate over basicity anyway, because it would be unfair to relate a property which is considerably increasing with another which is decreasing down the PT, it is just that polarizability has a more pronounced effect then basicity in this situation. If you compare analogous alcohols and thiols you can see this trend.


Selectivity in organic reactions I am finding great difficulty in understanding the differences between stereospecific, stereoselective, regiospecific and regioselective nature of reactions. Can you explain it in simple terms without mixing them with each other?

Yours is a common source of confusion. Reactions are can be selective but they are not specific. What's specific is a mechanism for the reaction. If the reaction only proceeds via one mechanism, then it is extremely selective.

Stereospecific. The mechanism provides a specific stereochemical outcome or proceeds from a specific stereochemical configuration. Examples:

a. Pd-catalyzed hydrogenation and hydroboration proceed via syn addition. These mechanisms are stereospecific because they indicate always adding to the same side of an olefin.

b. $\mathrm{E}_{2}$ elimination processes require that the hydrogen and leaving group are anti-periplanar. This mechanism is stereospecific.

c. Acid-catalyzed hydration proceeds via a planar carbocation intermediate where previous stereochemistry might be lost. This mechanism is not stereospecific.
Stereoselective. The reaction provides more of a product with a specific stereochemistry than any of the other possibilities.

a. Allylic alcohol directed epoxidation reactions frequently yield different amounts of the two product epoxides that result from reacting on the different faces of the olefin. This reaction is selective.

b. Asymmetric reactions that rely on pathways through diastereomeric transitions states are stereoselective.

c. Acid-catalyzed hydration could be stereoselective because the molecule can have other stereocenters that provide diastereoselectivity, even though the base mechanism disregards stereochemistry.
Regiospecific. The mechanism indicates a certain connectivity that should be obtained in the product.

a. Exo- versus endo- intramolecular reactions have regiospecific mechanisms.

b. A bit of a stretch, but carbonyl addition happens at the carbon. The mechanism is regiospecific.
Regioselective. The reaction favors one connectivity pattern over the other.

a. Hydrobromination favors the Markovnikov product, while hydrobromination in the presence of peroxides favor the anti-Markovnikov product.

b. In intramolecular ring closing reactions, 5-exo-tet products are preferred over 6-endo-tet products.

c. Carbonyl addition happens at the carbon. The only mechanism in operation is the one mentioned above, so the reactions are regioselective.

Importantly, reactions should not be stereospecific or regiospecific, even if they give only one product. This is a concrete case of a reaction proceeding exclusively (or almost exclusively) via a single mechanism that is stereospecific or regiospecific, respectively. The reaction itself is, respectively, stereoselective or regioselective.

In short, "xyz selective" is the more general case with one preferred outcome of the reaction, over the other one.  On contrast, "xyz specific" describes a subset of the former, where only one -- but not the other -- outcome occurs, and consequently is the only observed.

Regarding the regioselectivity, you may encounter for example the reactions of nucleophiles on $\alpha,\beta$ insaturated ketones.  A hard nucleophilic reagent (in terms of HSAB) will show preference to attact on the carbonyl carbon, a soft one in $\beta$-position.  If there is a preference for a reaction to occur on one site over other sites, it is regioselective.

The beauty (or, the hassle) with organometallic reagents is that depending on the hardness / softness of the metal deployed, you may tune the regioselectivity of your reagents.  (Other conditions like additional spectator ligands, Lewis basicity of the solvent, temperature etc., have some influence, too. Take these two examples with a grain of salt, please.)

To account for NotBaran's comment, IUPAC' Gold Book specifies regioselectivity as this:

  regioselectivity (regioselective)

  A regioselective reaction is one in which one direction of bond making or breaking occurs preferentially over all other possible directions. Reactions are termed completely (100%) regioselective if the discrimination is complete, or partially (x%), if the product of reaction at one site predominates over the product of reaction at other sites. The discrimination may also semi-quantitatively be referred to as high or low regioselectivity. (Originally the term was restricted to addition reactions of unsymmetrical reagents to unsymmetrical alkenes.) In the past, the term 'regiospecificity' was proposed for 100% regioselectivity. This terminology is not recommended owing to inconsistency with the terms stereoselectivity and stereospecificity.

  See also: chemoselectivity

  Source: PAC, 1994, 66, 1077 (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1160

Regarding stereoselectivity, consider as an example a reduction of a ketone.  It may be that a reagent is performing this reaction like lithium borohydride, and the product is a mixture of both stereoisomers possible; in this case (absence of Cram conditions) even to a racemate.  With help of a suitable chiral borane (derived from $\alpha$-pinene, as one example), however the reaction may yield preferentially one stereoisomer, over the other. (Here, the substrate already is prochiral.) Hence, it is a stereoselective reaction.

Note: I do not remember if in the instance of the last reaction of acetophenone the ($R$)- or the ($S$)-configurated enantiomer is the correct one.  The point to retain is, yielding one (or the other) product makes a difference, energetically / kinetically speaking.

Again, to account for @NotBaran's comment, IUPAC Gold Book describes the current definition here:

  stereospecificity (stereospecific)

  A reaction is termed stereospecific if starting materials differing only in their configuration are converted into stereoisomeric products. According to this definition, a stereospecific process is necessarily stereoselective but not all stereoselective processes are stereospecific. Stereospecificity may be total (100%) or partial. The term is also applied to situations where reaction can be performed with only one stereoisomer. For example, the exclusive formation of trans-1,2-dibromocyclohexane upon bromination of cyclohexene is a stereospecific process, although the analogous reaction with (E)-cyclohexene has not been performed.
  The term has also been applied to describe a reaction of very high stereoselectivity, but this usage is unnecessary and is discouraged.

  Source:
  PAC, 1994, 66, 1077 (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1167
  PAC, 1996, 68, 2193 (Basic terminology of stereochemistry (IUPAC Recommendations 1996)) on page 2219

Both the references of IUPAC's Gold Book, as their references in Pure and Applied Chemistry may be accessed freely:

the recommendations by 1994 DOI 10.1351/pac199466051077
the recommendations by 1996 DOI 10.1351/pac199668122193


Why is hydrogen sulfate put together as it is? I recently looked up hydrogen sulfate ($\ce{HSO4-}$) to see how it's put together, and found this image of methyl hydrogen sulfate:

I'd assume that hydrogen sulfate looks the same, but without the methyl group attached. However I'm confused, how can it be constructed like that? This shows the sulfur having a total of 12 valence electrons, but shouldn't the maximum be 8?

I might be asking something that'll be obvious to me later on in chemistry, but this (and other groups that act as atoms) appear early on. I probably don't need to know it yet but I'd prefer to know how it works if I can as opposed to just accepting it.

Hypervalency is an obsolete concept - it was used to explain bonding situations of molecules, that seem to exceed the octet rule. It was accomplished by the help of $\ce{spd}$ hybrid orbitals. It is now known, that for almost all molecules that were described this way the contribution of $\ce{d}$ orbitals is negligible (<1%).

Focussing on the question at hand, the bonding situation of the core $\ce{SO4^{2-}}$ anion is isoelectronic to the phosphate anion $\ce{PO4^{3-}}$.

For the methylsulfate this situation may be extended. A qualitative analysis of the natural charges (BP86/cc-pVDZ) reveals no difference. All of the oxygen atoms carry a negative charge, while sulfur is highly positive charged. Further more, NBO analysis confirms that the terminal oxygens have three lone pairs, while the other oxygens have two lone pairs. A more accurate Lewis structure would represent that fact and only use single bonds.

I think it's time for you to know about Octet Expansion rule. The molecules that expand their octet are known as Hypervalent Molecules.

  Normally, atoms in group 2 have 8 electrons and can thus form 4 bonds.  This is mainly because they have s and p orbitals. However, in elements of group 3, they have extra orbitals called d orbitals that can be used to form more bonds.  when forming extra bonds, the atom promotes electrons into the d orbitals and allows the atom to from more than 4 bonds.

Same thing happens in the case of sulfur. Sulfur belongs to 3rd period and therefore it has a vacant 3d orbital and thus can accommodate more than 8 electrons in its valence cell.


Disproportionation of Au(II) $\ce{Cu(II)}$ and $\ce{Ag(II)}$ complexes are known to exist while $\ce{Au(II)}$ is unstable and disproportionates to $\ce{Au(I)}$ and $\ce{Au(III)}$.

My guess as to possible reasons to this effect is that $\ce{Au(II)}$ like $\ce{Cu(II)}$ and $\ce{Ag(II)}$ cannot undergo Jahn-Teller Distortion, the cause of their stability, because of the extremely diffuse nature of $\mathrm{d}$ orbitals of $\ce{Au(II)}$ owing to relativistic effects. Could this be the right reasoning?

Also, what is the cause of the stability of $\ce{Au(I)}$ and $\ce{Au(III)}$ over $\ce{Au(II)}$?

I think this could be a counter answer. In contrast, I found a literature evidence that disproportionation of $\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ has had happens Ref.1). The abstract states that:

  Interaction of $\ce{Ag+}$ salts in anhydrous liquid hydrogen fluoride, $\ce{aHF}$, with $\ce{AgF4-}$ salts gives amorphous red-brown diamagnetic $\ce{Ag^IAg^{III}F4}$, which transforms exothermally to brown, paramagnetic, microcrystalline $\ce{Ag^{II}F2}$ below $\pu{0 ^\circ C}$. $\ce{Ag^IAu^{III}F4}$ prepared from $\ce{Ag+}$ and $\ce{AuF4-}$ in $\ce{aHF}$ has a tetragonal unit cell and a $\ce{KBrF4}$ type lattice, with $a = \pu{5.788(1) Å}$, $c = \pu{10.806(2) Å}$, and $Z = 4$. Blue-green $\ce{Ag^{II}FAsF6}$ disproportionates in $\ce{aHF}$ (in the absence of $\ce{F-}$ acceptors) to colorless $\ce{Ag^IAsF6}$ and a black pseudotrifluoride, $\ce{(Ag^{II}F+)2Ag^{III}F4^-AsF6-}$. The latter and other $\ce{(AgF)2AgF4MF6}$ salts are also generated by oxidation of $\ce{AgF2}$ or $\ce{AgF+}$ salts in $\ce{aHF}$ with $\ce{F2}$ or in solutions of $\ce{O2+MF6-}$ salts ($\ce{M = As, Sb, Pt, Au, Ru}$). Single crystals of $\ce{(AgF)2AgF4AsF6}$ were grown from an $\ce{AgFAsF6/AsF5}$ solution in $\ce{aHF}$ standing over $\ce{AgF2}$ or $\ce{AgFBF4}$, with $\ce{F2}$ as the oxidant. They are monoclinic, $P2/c$, at $\pu{20 ^\circ C}$, with $a = \pu{5.6045(6) Å}$, $b = \pu{5.2567(6) Å}$, $c = \pu{7.8061(8) Å}$, $β = 96.594(9)°$, and $Z = 1$. The structure consists of $\ce{(AgF)_n^{n+}}$ chains ($\ce{F−Ag−F} = 180^\circ $, $\ce{Ag−F−Ag} = 153.9(11)^\circ $, $\ce{Ag−F} = \pu{2.003(4) Å}$), parallel to $c$, that enclose stacks of alternating $\ce{AgF4-}$ and $\ce{AsF6-}$, each anion making bridging contact with four $\ce{Ag(II)}$ cations of the four surrounding chains “caging” them. There is no registry between the ordered array in one “cage” and that in any neighboring “cage”. The $\ce{F-}$ ligand anion bridges between the anions and, with the $\ce{Ag(II)}$ of the chains, generates a trifluoride-like structure. $\ce{(AgF)_2AgF4AsF6}$ [like other $\ce{(AgF)_n^{n+}}$ salts] is a temperature-independent paramagnet except for a Curie “tail” below $\pu{50 K}$.

However, I cannot find a counter example for $\ce{Cu(II)}$ complexes. Yet, $\ce{Cu(I)}$ is known to disproportionate to $\ce{Cu(II)}$ and $\ce{Cu(0)}$, which has limited $\ce{Cu(I)}$ chemistry. For example, if you react $\ce{Cu(I)}$ oxide ($\ce{Cu2O}$) with dilute sulfuric acid (hot), you'd get a brown precipitate of $\ce{Cu(0)}$ and a blue solution of $\ce{CuSO4}$ (a $\ce{Cu(II)}$ salt), because the disproportionation reaction dominates. You would not get the expected copper(I) sulfate solution from this reaction.

$$\ce{Cu2O + H2SO4 -> Cu + CuSO4 + H2O}$$

However, if you use certain ligands to form $\ce{Cu(I)}$ complexes, they would stabilizes the $\ce{Cu(I)}$ oxidation state. For example, both $\ce{[Cu(NH3)2]+}$ and $\ce{[CuCl2]-}$ are $\ce{Cu(I)}$ complexes, which don't disproportionate.

References:

Ciping Shen, Boris Žemva, George M. Lucier, Oliver Graudejus, John A. Allman, Neil Bartlett, "Disproportionation of$\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ in Fluoride Systems and Syntheses and Structures of $\ce{(AgF+)_2AgF4^-MF6-}$ Salts ($\ce{M = As, Sb, Pt, Au, Ru}$)," Inorg. Chem. 1999, 38(20), 4570-4577 (https://doi.org/10.1021/ic9905603).
Also read for the Disproportionation of$\ce{Au(II)}$: Khaldoon A. Barakat, Thomas R. Cundari, Hassan Rabaâ, Mohammad A. Omary, "Disproportionation of Gold(II) Complexes. A Density Functional Study of Ligand and Solvent Effects," J. Phys. Chem. B 2006, 110(30), 14645-14651 (https://doi.org/10.1021/jp062501y).

It is seen that Au(III) and Au(I) complexes are more stable than Au(II).I have an explanation but that is my observation-

Let's consider their electronic configuration.

Au → 5d10 6s1

Au+ → 5d10

Au2+ → 5d9

Au3+ → 5d8

So, if we split d-orbital into t2g and eg, we get

For Au →

For Au+ →

For Au2+ →

For Au3+ →

So, As you can see,

Au (I) - Fully Filled Configuration
Au (III)- Half-Filled Configuration

Therefore, both of them are more stable than Au(II) because of the fact that fully filled and half-filled orbitals render more stability than any other configuration.


Explanation for the trends in nucleophilicity—Orbital interactions or electrostatic attraction The trend of halide nucleophilicity in polar protic solvents is $$\ce{I- > Br- > Cl- > F-}$$

The reasons given by Solomons and Fryhle[1], and by Wade[2] are basically as follows.

Smaller anions are more solvated than bigger ones because of their 'charge to size ratio'. Thus, smaller ions are strongly held by the hydrogen bonds of the solvent.
Larger atoms/ions are more polarizable. Thus, larger nucleophilic atoms/ions can donate a greater degree of electron density to the substrate than a smaller nucleophile whose electrons are more tightly held. This helps to lower the energy of the transition state.

Clayden et al.[3] give the reason that in SN2 reactions, the HOMO–LUMO interaction is more important than the electrostatic attraction. The lone pairs of a bigger anion are higher in energy compared to a smaller anion. So, the lone pairs of a bigger anion interact more effectively with the LUMO σ*. Thus, soft nucleophiles react well with saturated carbon.

What I don't understand is the trend of halide nucleophilicity in polar aprotic solvents.

$$\ce{F- > Cl- > Br- > I-}$$

In aprotic solvents, only the solvation factor is absent. Bigger anions are still more polarizable and can interact in a better way with the LUMO. Then, what is the reason for the trend reversal in aprotic solvents? Solomons and Fryhle[1] say

  In these[aprotic] solvents anions are unencumbered by a layer of solvent molecules and they are therefore poorly stabilized by solvation. These “naked” anions are highly reactive both as bases and nucleophiles. In DMSO, for example, the relative order of reactivity of halide ions is opposite to that in protic solvents, and it follows the same trend as their relative basicity.

Shouldn't the orbital interactions matter more than the electrostatic attractions even in polar aprotic solvents? Why does nucleophilicity follow the basicity trend in polar aprotic solvents?

References:

Solomons, T. W. Graham; Fryhle, C. B. Organic Chemistry, 10th ed.; Wiley: Hoboken, NJ, 2011; pp. 258–260.
Wade, L. G. Organic Chemistry, 8th ed.; Pearson Education: Glenview, IL, 2013; pp. 237–240.
Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012; pp. 355–357.

This is a rather intellectually-stimulating question and one that is also very difficult to answer. You have constructed a very good case for why the nucleophilicity order would not be expected to reverse in polar aprotic solvents, i.e. the order of intrinsic nucleophilicities of the halide ions should be $\ce {I^- > Br^- > Cl^- > F^-}$, based on the concepts of frontier molecular orbital interactions, as well as electronegativity and polarisability of the nucleophilic atom.

Affirming your viewpoint

Admittedly, the saturated carbon atom is a soft centre and based on electronegativity considerations, the energy of the $\ce {HOMO}$ of the halide ions should increase from $\ce {F^-}$ to $\ce {I^-}$, i.e. the hardness of the the halide ions as nucleophiles decreases from $\ce {F^-}$ to $\ce {I^-}$. Based on the hard-soft acid-base (HSAB) principle, we would expect the the strength of the interaction between the halide ion and the carbon centre to increase from $\ce {F^-}$ to $\ce {I^-}$.

We can provide more quantitative justification for this using the Klopman-Salem equation$\ce {^1}$:

$$\Delta E=\frac{Q_{\text{Nu}}Q_{\text{El}}}{\varepsilon R}+\frac{2(c_{\text{Nu}}c_{\text{El}}\beta)^2}{E_{\text{HOMO}(\text{Nu})}\pm E_{\text{LUMO}(\text{El})}}$$

The first term is indicative of the strength of the Coulombic interaction while the second term is indicative of the strength of the orbital interactions. In the reaction with the soft saturated carbon, it is actually expected for the reactivity to increase from $\ce {F^-}$ to $\ce {I^-}$ (ref 1, p. 117).

At the moment, all the evidence presented points us towards the conclusion that you have posited. However, we have overlooked the importance of a thermodynamic factor - the strength of the $\ce {C-X}$ bond.

A not-particularly-relevant add-on

Actually, if you think about it the considerations made in the Klopman-Salem equation are essentially kinetic considerations. We cannot conclude anything about reaction energetics from such considerations. The strength of the $\ce {HOMO-LUMO}$ interaction or that of the Coulombic attraction tells us nothing about how the electrons would organise themselves subsequently, what would be the strength of the resultant bonds etc.

A comment made on this by Olmstead & Brauman (1977)

On this topic of explaining the relative order of intrinsic nucleophilicity of the halide ions, Olmstead & Brauman (1977) had this to say$\ce {^2}$:

  The intrinsic nucleophilicities follow the reverse order of the polarizabilities (e.g., $\ce {CH3O^-}$ > $\ce {CH3S^-}$ and $\ce {F^-}$ > $\ce {Cl^-}$ > $\ce {Br^-}$). This could be due to a stronger interaction between the more concentrated molecular orbitals of the anion with the carbon center. It could also be simply a reflection of the greater thermodynamic methyl cation affinities of the smaller anions.

It is important to note that Olmstead & Brauman are merely postulating the reasons. However, I do feel that the point on thermodynamics leads us on the right track to answering this question. On the other point of "more concentrated MOs", I am unclear of how it can be interpreted.

Let us focus now on this point.

How does thermodynamics even influence kinetics?

Well... It certainly does here. What are we concerned with when dealing with nucleophilicity? As it is a kinetic phenomenon, we are essentially concerned with the rate of the substitution reaction. This makes the consideration of the activation energy ($E_\mathrm{a}$) a particularly important one because the higher the $E_\mathrm{a}$, the lower the rate of the substitution reaction.

Here, the $E_\mathrm{a}$ can be taken to be the difference in energy between the reactants and the transition state. Thus, we can decrease the $E_\mathrm{a}$ by either raising the energy level of reactants or decreasing the energy level of the transition state. The reaction energy profile for an $\ce {S_N2}$ reaction is shown below, taken from ref. 3 (p. 394).

There are two ways to see how the strength of the $\ce {C-X}$ bond can affect the $E_\mathrm{a}$. I am not sure if they may be considered equivalent perspectives. Nonetheless, I will present both.

The first perspective is well-explained by Francis & Sundberg (2007) with regard to the $\ce {S_N2}$ reaction$\ce {^3}$:

  Because the $\ce {S_N2}$ process is concerted, the strength of the partially formed new bond is reflected in the TS. A stronger bond between the nucleophilic atom and carbon atom results in a more stable TS and a reduced activation energy.

Another perspective (which may be somewhat similar to the first, or even possibly the same) is that the transition state structure bears resemblance to the product structure, based on the Hammond's postulate. However, this would be more applicable for an endothermic $\ce {S_N2}$ reaction where the transition state structure does indeed bear more resemblance to the product. When the product is more stable, i.e. has bonds that form more exothermically, the energy of the transition state would thus be lowered, and the $E_\mathrm{a}$ would also decrease, although not necessarily proportionally.

Conclusion

The argument you have presented failed to consider the important thermodynamics factor. Thus, it resulted in the predicted reactivity of the halide ions in the reversed order. The takeaway here is that the application of the $\ce {HSAB}$ concept alone does not always produce the correct prediction.

Additional information

More examples are cited for which the HSAB concept fails. These include:

The favourable combination of $\ce {H^+}$, a hard acid, and $\ce {H^-}$, a soft base (discussed here as well) (ref. 4).
The fact that the rate of reaction between $\ce {Ag^+}$, soft acid, and an alkene, soft base, is slower than that between $\ce {Ag^+}$ and $\ce {OH^-}$, a hard base (ref. 1, p. 110)

References

Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Student Edition). John Wiley & Sons, Ltd. United Kingdom, 2009.
Olmstead, W. N.; Brauman, J. I. Gas-phase nucleophilic displacement reactions. J. Am. Chem. Soc. 1977, 99(13), 4219–4228.
Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.
Pearson, R. G.; Songstad, J. Application of the Principle of Hard and Soft Acids and Bases to Organic Chemistry. J. Am. Chem. Soc. 1967, 89(8), 1827-1836.

In polar aprotic solvent, there is no possibility of formation of a hydrogen bond, to reduce the nucleophilicity of a nucleophile.

The order of basicity of $\ce{F-,Cl-,Br-,I-}$ are,
$\ce{F- > Cl- > Br- > I-}$ (since the order of acidity of the hydrogen halides are $\ce{HI > HBr > HCl > HF}$).

So, among halide ions, $\ce{F-}$ has the highest ability to donate its electron pair, and thus, act as a nucleophile.

Also, $\ce{F-}$ being the smallest faces less steric hindrance as a nucleophile.

So, the order of nucleophilicity in polar aprotic solvent is, $$\ce{F- > Cl- > Br- > I-}$$


Is there a simple and inexpensive way to generate nitrogen gas? It's fairly easy to generate a significant yield of hydrogen gas simply by reacting zinc metal with $\ce{HCl}$ or aluminum with $\ce{NaOH}$ solution. But is there a simple and inexpensive chemical reaction that generates a significant yield of pure nitrogen gas?

Per Wikipedia, we have this laboratory method:

$$\ce{NH4Cl(aq) + NaNO2(aq) → N2(g) + NaCl(aq) + 2 H2O (l)}$$

Also per Wikipedia, this method generates small quantities of impurities of $\ce{NO}$ and $\ce{HNO3}$, which can be eliminated by passing the gas through potassium dichromate $\ce{(K2Cr2O7)}$ dissolved in sulfuric acid $\ce{(H2SO4)}$

Both ammonium chloride and sodium nitrite are reasonably available chemicals, and the other products (table salt and water) are undeniably safe. Sulfuric acid might a bit tougher to get depending on where you live, and potassium dichromate will be the most difficult given that the use of hexavalent chromium is discouraged (and you should research procedures to reduce the $\ce{Cr2O7^{-2}}$ to $\ce{Cr2O3}$). But otherwise, this looks like the best approach.

Thermal decomposition of ammonium nitrite:
$$\ce{NH4NO2 -> N2 + H2O}$$


Eutectic Phase Diagram and Lever Rule A solution having composition $p$ (left side of graph) is cooled to just above the eutectic temperature (point $s$ is at $0.18$ $x_{\ce{Si}}$ and the corresponding tie line intersects the liquidus curve at $0.31$ $x_{\ce{Si}} $); calculate the composition of the solid that separates and that of the liquid that remains.

The liquid phase is obviously 31 % silicon and 69 % gold as given in the question. The solutions manual also gives the weight percentage of the solid phase. Using the lever rule,

$$ m_\text{solid}(0.18)=m_\text{liquid}(0.31-0.18)$$
$$ \frac{m_\text{solid}}{m_\text{liquid}}=\frac{0.13}{0.18} $$
$$ \frac{m_\text{solid}}{m_\text{solid}+m_\text{liquid}} = \frac{0.13}{0.13+0.18}=0.42$$

However, the solution manual states that the wt% of solid phase is 58 %, and that of the liquid phase is 42 %. Is the solutions manual wrong, or am I losing my mind?

You are correct. Another way to calculate it is: $$X_{\mathrm{L}} = \frac{s-b}{e-b} = \frac{0.18-0}{0.31-0}=0.58$$
You can also inspect the diagram visually and see that the $bs$ line is longer than the $se$ line. The lever rule dictates that the line length is proportional to the percentage of the phase opposite from the line. Thus, it is not possible that there is more solid than liquid at this point.

There is definitely a mistake in your textbook's solutions.

There is no mistake in the solutions of your book. I hope to help any other students that may come across this post. The formula for the weight percentage of the solid phase is
$$\text{wt% of solid} = \frac{\text{wt% of solid}}{\text{wt% of solid + wt% of liquid}}.$$

Applying this formula you get: $$\frac{0.18}{0.13 + 0.18}= 0.5806$$

Thus the $\text{wt% of solid}$ is $58.06\%$.

It is a common mistake to get the values backwards since it appears to be backwards from what it should be. I urge you to look up the formula for yourself to verify as well as check out a derivation of this formula.


Why is MnO2 not a peroxide? As I know, $\ce{MnO2}$ is called manganese(IV) oxide.

Why can't we name it manganese peroxide since manganese(II) exists too?

In other words, how do we know that the oxidation number of manganese in $\ce{MnO2}$ is $4$ and not $2$?

Peroxides contain two oxygens connected by a single bond. X-ray or neutron diffraction will show that the oxygens in $\ce{MnO2}$ are too far apart to be bonded, and therefore it is not a peroxide.

Peroxide is well known as a ligand to transition metals (though whether it is better described as peroxide or superoxide is sometimes not always clear, e.g. in some cobalt species). An example which you might come across in the lab is chromium(VI) oxide peroxide, $\ce{CrO5}$.

Peroxides are compounds in which the oxidation state of Oxygen atoms is -1.
Since you know that MnO2 is Mn(IV) oxide, which means Oxygen atoms are in -2 oxidation state. Hence it is not called a peroxide.


Is the kinetic energy change of a system (q) equal to the enthalpy change of a system (∆H)? When silver nitrate and hydrochloric acid are put together in a container, they form silver chloride. If you're given information about the reactants (including temperature change, heat capacity, and mass of reactant aqueous molecules), then you can find the kinetic energy change of the system.

What I want to know is: does the $q$ value you find for the system equal to the $\Delta H$ value for the system? Worded differently, does the kinetic energy change of the system equal to the enthalpy change of the system? Is this one of those obvious questions that most people understand instantly but people like me who lack common sense don't?

Under the restrictions of constant $p$, and no non-$pV$ work, we have:

$$q_p=\Delta H$$

q is the heat flow into the system (postive for flow in, negative for flow out).  However, q is not necessarily the change in the system's kinetic energy.  For instance, suppose you have an endothermic chemical reaction at constant temperature.  Heat will flow into the system.  However, some or all of that heat flow could go into increasing the potential energy of the chemical bonds.  And if the endothermic chemical reaction is accompanied by an increase in volume, some of that heat flow will go into the work needed to expand the system against the external pressure.

Indeed, if the temperature is constant, and the heat capacity of the reactants and products are the same, then there will be no change in the kinetic energy of the system.  I.e., in that case, none of the heat flow will go into changing the kinetic energy of the system.

It is right if you consider that kinetic energy may have different forms: it can be either vibration, or rotation or translation energy.


Balancing the oxidative decomposition of FeS2 Doing some practice for the GRE next year.

Please could you look at my working for this and tell me why I'm wrong?

$$\ce{? FeS2 + ? O2 + ? H2O = ? Fe(OH)3 + ? H2SO4}$$

My working so far is:

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

Tot up the $\ce{H}$ on the right: $1\times 3 + 2\times 2 = 7$. $7$ on the right so $\ce{3.5 H2O}$ on the left. ($7/2 = 3.5$)

$\ce{O}$ on the right = $1\times 3 + 2\times 4 = 11$.

This mean $11$ on the left, too. I already have $\ce{3.5 H2O}$, so we must have $11-3.5 = 7.5$ $\ce{O}$'s from the $\ce{O2}$. So $\ce{3.75 O2}$ because $7.5/2 = 3.75$.

Following the rules I've written down I arrive at:

$$\ce{1.75 FeS2 + 3.75 O2 + 3.5 H2O -> 1.75 Fe(OH)3 + 3.5 H2SO4}$$

I've multiplied these by $4$ to give integers, because that's what the question asks of me, so $7, 15, 14, 7$ and $14$.

Why am I wrong?

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

This is how I found it. Do you know linear algebra ? Because that is very useful. Anyway without to know this is linear algebra you'll be able to understand.

So we are looking about the stoichiometric coefficients $\{\ce{a, b, c, d, e}\}$ of this reaction :

$$\ce{aFeS_2 + bO_2 + cH_2O + d Fe(OH)_3 + eH_2SO_4 \rightarrow 0}$$

Now what you need to do is first to find the number of different elements which are involved in your reaction, here they're four : $\{\ce{\color{\green}{H}, \color{\red}{O}, \color{\purple}{S}, \color{\orange}{Fe}}\}$. I'm not that good in the use of $\LaTeX$ but now the idea is to make column vectors under each of the reagents of your reactions and in these vectors you put in the same order for each the number of each elements there are in and if there is not just put a zero. It will gives you this :

$$a\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{0} \\ \color{\purple}2 \\ \color{\orange}{1} \end{pmatrix}+b\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{2} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+c\cdot\begin{pmatrix} \color{\green}2 \\ \color{\red}{1} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+d\cdot\begin{pmatrix} \color{\green}3 \\ \color{\red}{3} \\ \color{\purple}{0} \\ \color{\orange}1 \end{pmatrix}+e\cdot\begin{pmatrix} \color{\green}2 \\  \color{\red}4 \\ \color{\purple}1 \\ \color{\orange}0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

It can be easily transform into a system like this :

$$\begin{cases}
2c+3d+2e=0 \\
2b+c+3d+4e=0 \\
2a+e=0 \\
a+d=0
\end{cases}$$

So here we are in a good way because we have more unknowns than equation so if the system converge it has an infinity of solutions. And we can multiply by the number we want each stoichiometric coefficients and still have the same reaction.

So here we will determine four unknowns in terms of an other, I chose $e$. And I got :
$$$$\begin{cases}
a=-e/2 \\
b=-15e/8 \\
c=-7e/4 \\
d=e/2 \\
e=e
\end{cases}$$$$

I multiply then by $8$ to have only integers which give me with $e=-1$ :

$$\ce{4FeS_2 + 15O_2 + 14H_2O - 4Fe(OH)_3 - 8H_2SO_4 \rightarrow 0}$$

Then $$\ce{4FeS_2 + 15O_2 + 14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$

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


What is the difference between "reaction pathway" and "reaction mechanism"? What is the difference between "reaction pathway" and "reaction mechanism"?

Are these expressions synonyms?

Some textbooks authors use both without making clear to the learner if they are synonyms or not.

Usually with the reaction mechanism the sequence of events (electron transfers) is indicated, while the term reaction pathway usually refers to the reaction coordinate diagram associated with the reaction, i.e. the change in energy with an extra notion on the transition states.

No, Khair Muhammabd. Go through the reaction mechanism. Actually, the reaction mechanism is about sequence and detailed understanding of elementary reactions(that are here called elementary steps in the term of Composite Reaction) for conversion of reactants to the desired products.
It is accompanied with the formation of Reaction Intermediate at each elementary step(except last one). Where the overall reaction can be obtained by adding all elementary reactions together. So as the overall rate of reaction can be obtained by adding the rates of all elementary steps.
But what happens, every composite reaction has one elementary step (Mostly 1st step) that is carried out very fast as compared to other steps. Say, T1=1sec for 1st step. and T2=0.0000000001 sec reaction.
So the total time of the composite reaction will almost be equal to first step (T(total)=T1) while T2 is neglected. Basically the first step is called Rate dertermining step(RDS).
So overall reaction rate will be taken equal to the rate of RDS.
R(r*n)=R(RDS).
While I know nothing about Reaction Pathways, still going through it.


Why is the second acid dissociation constant for sulphuric acid much lower than the first in water? In aqueous solution, $\ce{H2SO4}$ ionises in two steps :

$$\ce{H2SO4(aq) -> H+(aq) + HSO4-(aq)}; \quad K_\mathrm{a_1}=1\times10^3$$
$$\ce{HSO4-(aq) -> H+(aq) + SO4^2-(aq)}; \quad K_\mathrm{a_2}=1.2\times10^{-2}$$

The value of $K_{\mathrm{a}_2}$ caught my attention. I thought that $\ce{H2SO4}$ was a strong dibasic acid, but a very low value of $K_{\mathrm{a}_2}$ implies that the second proton is not that easily donated. Why is the value of $K_{\mathrm{a}_2}$ so less?

For the second deprotonation you are trying to remove a proton from something that is already negatively charged, a process far more difficult than removing a proton from something that is neutral.
Alternatively you can think that the $-2$ charged $\ce{SO4^2-}$ anion attracts really strongly the $\ce{H+}$ resulting in the equilibrium being towards $\ce{HSO4^-}$, making it a weak acid compared to $\ce{H2SO4}$.
It is similar thinking to why the second ionization energies are always much higher than the first.

There can always be more than one ways to explain an observed phenomenon. With this in mind, one might be tempted to believe that there is suppression of dissociation of the $\ce{HSO4^-}$ ion due to the common ion effect.  The reasoning goes along the lines of:

Note that even before $\ce{HSO4^-}$ dissociates, there is an equal concentration of $\ce{H+}$ ions already present in the solution. We know, by Le Chatelier's principle, that the increased concentration of the products favors the reactants. Hence for the second dissociation, since $\ce{H+}$ is one of the products, and is in large concentration, the second dissociation is actually favored in the reverse direction. Thus, deprotonation of $\ce{HSO4-}$ is not favorable.
Hence, the $K_\mathrm {a_2}$ of $\ce{H2SO4}$ is significantly lower than its $K_\mathrm {a_1}$.

However, this is a wrong reasoning. Note that $K_\mathrm{a}$ values are always a constant. The $K_\mathrm{a}$ value for the $\ce{HSO4-}$ ion would be the same whether we took $\ce{H2SO4}$ or $\ce{NaHSO4}$. However, in the case of $\ce{NaHSO4}$, there are no pre-existing $\ce{H+}$ ions in solution when $\ce{HSO4-}$ deprotonates. Hence, it would be wrong to invoke the common ion effect here.
Therefore, the common ion effect should not be invoked when explaining the low $K_\mathrm a$ of $\ce{HSO4-}$ ion as compared to $\ce{H2SO4}$.

The value of 2nd dissociation constant is less than 1st because in 2nd equation there is anion $\ce{HSO4-}$ which will not allow to release protons,but we know that $\ce{H2SO4}$ is strong dibasic acid.


My sample of gallium does not melt I held gallium under a heat lamp. It didn't melt. I didn't even find a thin film of liquid gallium. Can anyone explain why? Does gallium rust?

Having recently had an opportunity to play around with some gallium of my own, I have a few thoughts.

It's possible you have a different metal than what you thought you had, but frankly, I don't think that's very likely.

It takes a surprising amount of absorbed heat to melt gallium. Putting a sizable chunk of gallium in $35~^\circ\mathrm C$ water doesn't cause it to melt immediately. Even boiling water doesn't melt it instantly. Because gallium is a metal, it conducts heat fairly well, so when heated gently, it won't melt until the temperature of the whole piece gets close to the melting point.

As mentioned by user5134, a shiny metal like gallium isn't going to pick up heat very quickly from a radiative source like a heat lamp. If you wait long enough, it'll definitely melt. If you want to melt the gallium relatively quickly, then you'll want to put it into physical contact with something warm. The standard method for melting solid gallium in a container is to just immerse the closed container in hot water (not boiling water, especially if the container is glass).

Elemental gallium does not occur in free form in nature, but as the gallium(III) compounds. So it mustn't be gallium but something else.


H2SO4 with ethanoic acid reaction? Earlier today I was preparing an ester (propyl ethanoate) and I noticed that instead of putting the propanol, ethanoic acid and $\ce{H2SO4}$ in the test tube and warming it (taking time and external energy) I could instead mix sulfuric acid with the ethanoic acid, which made the test tube go very hot, and then by adding the propanol and shaking it split into two layers. On pouring this into water I ended up with an immiscible layer which appeared to be an ester (it was sweet smelling and considering the reaction it would be likely to be an ester).

I cannot think of which reaction was causing the heating up initially. The test tube was clean so the only reactants were the sulfuric acid and ethanoic acid, and it appeared that a gas was being given off (as shaking it with a rubber cap on caused it to pop out of the tube, but no effervescence could easily be seen). Can anyone help? Furthermore, would the eventual product be propyl ethanoate or something else (it was suggested that it was dehydrogenating the ethanoic acid so the actual product would be something like propyl ethenoate, but from a quick look on the internet this would seem to require an iron catalyst (or another catalyst) and a high temperature, neither of which was present)?

I suspect that the ethanoic acid contains a residual amount of water and that the concentrated sulphuric acid is reacting with the water. Since this is a very exothermic reaction the test tube gets heated up which evaporates some of the ethanoic acid, causing the bung to pop off. If it's just slight evaporation this would also explain why you couldn't see any effervescence.

I think what happened is additon of conc h2so4 led to esterification of ethanoic acid and release of water. This water must have reacted with another h2so4 to produce heat without effervescence.