Why isn't cyclopropane considered unsaturated? Looking up "cycloalkanes" on wikipedia (link), I found this:

  In organic chemistry, the cycloalkanes (also called naphthenes, but distinct from naphthalene) are monocyclic saturated hydrocarbons.

Of course, that implies that cyclopropane is saturated. Later I looked up "saturated and unsaturated compounds" on wikipedia (link). Here's the definition they mentioned:

  In chemistry, a saturated compound is a chemical compound (or ion) that resists the addition reactions, such as hydrogenation, oxidative addition, and binding of a Lewis base.

Now the definition mentions that a saturated compound "resists addition reactions". I am aware that cyclopropane does undergo addition reactions, as mentioned here (link):

So finally, to me it seems these two wikipedia pages are contradicting each other. My question is - why is cyclopropane considered saturated if the above definitions are correct?

In short, there are two definitions with separate purpose.

IUPAC defines alkanes as

  "Acyclic branched or unbranched hydrocarbons having the general formula $\ce{C_nH_{2n+2}}$, and therefore consisting entirely of hydrogen atoms and saturated carbon atoms."

and notes notes about cycloalkanes, contrasting e.g., to unsaturated cycloalkenes:

  "Saturated monocyclic hydrocarbons (with or without side chains), e.g. cyclobutane. Unsaturated monocyclic hydrocarbons having one endocyclic double or one triple bond are called cycloalkenes and cycloalkynes, respectively. Those having more than one such multiple bond are cycloalkadienes, cycloalkatrienes, etc. The inclusive terms for any cyclic hydrocarbons having any number of such multiple bonds are cyclic olefins or cyclic acetylenes."

Thus, as in the earlier comment by @MaxW, cyclopropane equally is a saturated alkane because each carbon atom is connected with the next carbon atom only by a $\ce{C-C}$ single bond.  Or, as MSU condenses,

  "Alkanes and cycloalkanes are termed saturated, because they incorporate the maximum number of hydrogens possible without breaking any carbon-carbon bonds."

This set of definitions is not the one serving to determine organic structures from spectra, e.g. in Field's training book, where

  "[...] the molecular formular is reduced to $\ce{C_nH_m}$ and the degree of unsaturation is given by $$\mbox{Degree of Unsaturation} = n - \frac{m}{2} + 1$$ The degree of unsaturation indicates the number of $\pi$ bonds or rings that the compound contains."

(4th edition, p. 4; emphasis added by mine.)

I don't think it has something to do with cyclopropane in particular. Indeed, cycloalkanes are naturally unsaturated, but they are associated with alkanes, which are saturated. The fact that the article specifically mentioned "monocyclic" offered enough details about the degree of unsaturation. However, just for the fact that they are the result of cyclization of saturated compounds, many sources assign cycloalkanes a somewhat saturated character.

TL;DR Cycloalkanes are the saturated hydrocarbons from the family of monocyclic compounds (which may contain other unsaturation-causing factors)


Melting points of NaCl and RbCl Why is the melting point of $\ce{NaCl}$ more than that of $\ce{RbCl}$ ? Using Fajan's Rule of polarizing power, $\ce{RbCl}$ has more ionic character and thus should have a higher melting point.

No doubt about it: it is due to the ionic radii and thus the distance between ions. The Rb-Cl bond is not as strong because of the distance between them. Na has an E.N. value of 0.9 Paulings while Rb is at 0.8; thus no significant difference in the magnetic forces of attraction.

LiCl has an even lower melting point due to excessive polarization (this is when Farjan's rules come into play), and the result, along with electronegativity again, is still an ionic compound, but with more covalent character.

See as you know they are ionic compounds and have some force of attraction. To break this energy is needed and the attraction between Na and Cl is more compared to Rb and Cl. Think about it like breaking a box when it is arranged in a lattice structure.

http://www.google.co.in/imgres?imgurl=https://upload.wikimedia.org/wikipedia/commons/e/e9/Sodium-chloride-3D-ionic.png&imgrefurl=https://en.wikipedia.org/wiki/Lattice_energy&h=948&w=1000&tbnid=Gm4SV7MURH7-UM:&docid=bdbhKWpu1E4z3M&hl=en-in&ei=DtkwVvSZCIjH0gTV0a4g&tbm=isch&client=safari&ved=0CCAQMygDMANqFQoTCPTY_9-t5cgCFYijlAod1agLBA


Why is the speed of ideal gases mass dependent while the kinetic energy is not? The rms speed of an ideal gas is $v_{\text{rms}}$ = $\sqrt{\frac{3RT}{M} }$ and the kinetic energy is $E_\text{k} = \frac32RT$. From this, it is concluded that the speed is mass dependent, while the kinetic energy isn't.

This doesn't make sense to me. I know the speed and kinetic energy for things other than ideal gases are surely mass dependent. The heavier an object is, the slower it is and will have a bigger kinetic energy. Why isn't this true for ideal gases?

As the mass increases, the velocity decreases, yes that is true. But how does the kinetic energy remain the same?

The increment in the mass is balanced by the decrement in the velocity. In other words, the velocity changes as the mass changes in such a manner that the kinetic energy of the molecules of the gas will remain to be same no matter what. It's something like the multiplication of two variables always giving the same constant value. Keeping in mind that this is true in an ideal environment.

The answer can be
$$Speed \propto1/√m$$
And$$ K.E. = 1/2mv^2$$.
Finally m will cut, and therefore K.E. does not depend on mass.


How many hydrogen bonds can methanol make? I believe that it can form 2 hydrogen bonds, one with oxygen and the other with hydrogen but a classmate said that oxygen can form 2 hydrogen bonds? Is this true? I can't find anything on google to support or refute that claim.

Although this was an MD study (and might be said to be somewhat dated), quoting from Matsumoto and Gubbins (Ref. 1):

  Most of the molecules are in the two-bonding state, but a
  small fraction of molecules are in the one- or three-bonding
  state, as already pointed out in previous studies. 7
  ,8

According to this study most of the methanol molecules form 2 H-bonds but ~4-7% form 3.

References

Matsumoto, M., Gubbins, KE. The Journal of Chemical Physics 93(3):1981-1994 (doi 10.1063/1.459076).

If we consider n methanol molecules such that intermolecular H bonding exists between all( value of distance has value that allows intermolecular H bonding).
One molecule has 1 O atom that bonds with n-1 H atoms. This case is for n atoms. So, number of bonds should be n(n-1). The H atom on 1molecule is counted as this bond is formed by other molecules having O.


Keeping one component concentration constant and varying another in a solution This is a simple question but it has been so long since I have done any chemistry. Basically I would like to keep the concentration of A constant and the concentration of B variable and vice versa.

Assuming we want to make a 50 ml solution with constant concentration of A and variable concentration of B.From my knowledge I would use C1V1=C2V2 in order to dilute A to its fixed concentration lets say from 4% to 3%. But how do i go about keeping it constant and B concentration variable for a final 50 ml solution.

EDIT:
Ok. So lets say A had an initial concentration of 4%. I want a constant A concentration of 3% in a volume of 50 ml. So I calculate my starting volume according to C1V1=C2V2 and get V1 to be (3*50)/4 and a starting volume of 37.5 ml. My remaining volume is 50-37.5 and get 12.5 ml. I want to make 5 solutions with 50 ml end volume. I put 12,5 ml of A in each flask giving me a constant A concentration of 3%. How do i finish up my 50 ml with different B concentrations?

2nd EDIT:
Apologies, this is my first time using this site and was told to edit my question and add my comment. I didn't put values as I believed it was clear enough but I guess not.

@ y chung that is what I have carried out in order to create the solutions, so I'm guessing I did it correctly then.

@ bon thanks for the comment. I will tell you exactly what Im trying to accomplish. I have two solutions one is A with a concentration of 9% and one is B with a concentration of 7%. I want to create 5 solutions each 50 ml with a constant 5% concentration of A and varying concentration of B from 1-5 %.

What I have done is as follows:
I reduced my A concentration to the required 5% by C1V1=C2V2 for a 50 ml solution getting 28 ml. I put 28 ml in each volumetric flask of 50 ml capacity.

I then calculated how much volume of B with the 7% concentration I would need to get it to say the 15 concentration based on my remaining volume of 22 (50-28). For the 1 % concentration this yielded a volume of B at 7% of 3.14 ml. I added that to my flask contain the 28 ml of A and made up the solution to 50ml.

The question is, is this correct? from what I gathered from y chung ,it is.
Thanks

Add varying amount of water and a standard B solution on top of the constant amount of A you have added. Then you will finish with several solutions with different B concentration.


Why does lead(II) acetate does not ionize appreciably? As this page in this book says, lead(II) acetate is soluble, but does not ionize appreciably. It seems contradictory to me, since, to call a substance soluble means it has high $K_\mathrm{sp}$ (when compared with other substances with the same elemental ratios in their formula units). But how can the book say it doesn't ionize appreciably? Why call it soluble then?

EDIT: I forgot that it is wrong to say that "to call a substance soluble means it has high $K_\mathrm{sp}$", so I revised it accordingly.

Here's the picture of the actual paragraph for those who can't access the link:

I recognize that this question actually reflects my inability to distinguish between dissociation and dissolution, now that I have cleared it from me, I rephrased the question to reflect my actual question. I did not erase the previous version to be able to show the reference of MEL Science's and alphonse's answers.

ACTUAL QUESTION: Why does lead(II) acetate not dissociate appreciably in water, considering that the ionic forms of acetate and lead(II) ion can be hydrated well due to the presence of partial charges in the water molecule?

A source in my native language says that lgK of PbAc2 is 4.08. It seems that lead acts as a central atom with two acetates as two-dentant ligands and two molecules of water acting as ligands too.
That means that lead acetate has its lattice ruined during dissolving, but it still forms a coordination compound.
 I am not really good in coordination chemistry of lead, but I just found out that lead wants to have octahedral geometry around it, so I would assume the structure of that would be

All bonds from lead ion are not really covalent rather than coordination bonds. Mind the charges!
P. S. aq=water
P. P. S. People in the comments are saying lead is too large for acetate to be bidentate,thus I would assume acetate just forms coordination compound with lead just like a simple monodentate ligand.

How can a book say?  A book can say that giraffes are purple, six-legged and can fly. The better question is: what is the book saying? I don't know. A cursory glance at the Wikipedia articles (you apparently aren't aware of the need to clearly communicate the identity of the chemical you wish to discuss. I have no idea whether "lead acetate" refers to the Pb(II) or Pb(IV) compound, or something else) shows that both acetates are soluble in polar organic solvents. This indicates that both are capable of dissolution without dissociation. Confusing dissociation (which generally occurs in solvents with high dielectric strength, such as water) with dissolution (which can occur in almost any liquid) is a newbie mistake. Whether the book you refer to (your link doesn't work) is actually claiming this or something else, I don't know. You don't provide enough information and I am not familiar enough with the solution chemistry of lead to make any judgement. Lead(II) acetate is used to precipitate anions (in aqueous solution) so I'd guess that the book is simply wrong - IF - it meant to imply that there was negligible dissociation of the dissolved salt.


What are physical properties and chemical properties? As the question title says, I would like to know what the physical and chemical properties are. If possible, please include examples. Since it seems to be related, I would also like to know what physical and chemical changes are. I believe that things such as crushing a can is a physical change, but I am not sure what would constitute a chemical change (perhaps doing something that rearranges the molecules?)

Edit: I originally wanted to ask if cooking an egg constitutes as a chemical or physical change, but I thought that that would be closed as too vague. If possible, please explain which one it is too. Thanks!

This is a tricky question.

I have to concur with Paul - it's difficult to give a truly hard and fast definition of the distinction between a chemical change and a physical change as there is a fairly smooth continuum between 'physical' and 'chemical' interactions. However, a good rule of thumb is that chemical changes involve changes in electronic structure, usually in the form of chemical bonding or redox. The problem is that traditionally 'physical' properties may have big implications for chemistry - I'll give some examples.

The Van der Waals and dipole-dipole interactions are mediated respectively by 'instantaneous' and permanent polarisations of electron density with respect to positive charges. Dipole-dipole interactions such as those found in water are quite significant and give rise, for instance, to water's frankly freaky liquid range as well as water's ability to mediate a wide range of chemical processes. These interactions don't involve significant changes in electronic structure however they have chemical implications in, for example, making a choice between liquid water or liquid nitrogen as a solvent.

The atomic mass (and hence the isotope) of an element is typically considered a physical property, however it has implications for reaction kinetics. This can be illustrated (expensively) by giving living organisms nothing but heavy water to drink. They will eventually experience toxicity due to the slowing of key biochemical reactions.

The magnetic properties of an atomic nucleus are typically considered to be physical, however they can be exploited in the observation that e.g. different isotopes of uranium have slightly differing hyperfine splittings and hence different ionisation energies. This allows isotopes to be enriched by selective ionisation.

Regarding the egg:

The dramatic changes that occur when you cook an egg are the result of denaturation of the proteins in the egg. Proteins are long chains of amino acids that take on their characteristic active conformations through a process called 'folding', so called because that's what it looks like (if you've heard of Folding@Home, that simulates the folding of proteins). If you raise the temperature much higher than physiological temperature, the protein degrades through unfolding into a different conformation. This involves the Van der Waals and dipole-dipole interactions holding together the secondary/tertiary/quaternary structure of the protein being broken en mass as well as the formation and/or scission of covalent bonds.

Physical changes are reversible, whereas chemical changes are irreversible. For example, boiling water is a physical change because you can reverse it by cooling it down, but setting fire to wood or boiling an egg cannot be undone.


Static electricity with calcium carbonate particles We are working with small calcium carbonate particles (microfossils) but one of the problems is they stick to the plastic or metal tools we are using, presumably due to static charge. For example, small particles on plastic tray will stick to the surface even if it is upside down.

I'm guessing the static potential is related to how much calcium carbonate wants to gain or give away electronics?
What is this called in this context?
How do I choose a material that has a low potential for static electricity (i.e. sticking) with calcium carbonate particles?

Some links of a commercial equipment to avoid problems of static electricity in powders during weighing:

https://www.mt.com/us/en/home/products/Laboratory_Weighing_Solutions/Accessories/anti_static_devices.html

http://lab.mt.com/antistatic/

https://www.mt.com/dam/mt_ext_files/Editorial/Generic/6/Anti-static_0x000246700004f0770006d93b_files/ion_e.pdf


Why does hydrogen bonding require a lone pair? Why does hydrogen bonding require a lone pair of electrons on the partially negative atom? They would still be partially negative even without the lone pair right? Or does a bond exist, but is much more negligible compared to when a lone pair is there?

You don't need a lone pair. Water can hydrogen bond to methane, attacking the back side of one of the carbon-hydrogen bonds in the methane. In a molecular orbital the sigma bonding orbitals in the methane are sufficiently delocalized to serve in place of lone pairs, overlapping the antibonding orbitals in the water [1]:
Abstract (from [1])

Quantum chemical calculations were performed at different levels of theory (SCF, DFT, MP2, and CCSD(T)) to determine the geometry and electronic structure of the HOH···CH4 complex formed by water and methane molecules, in which water is a proton donor and methane carbon (sp 3) is an acceptor. The charge distribution on the atoms of the complex was analyzed by the CHelpG method and Hirshfeld population analysis; both methods revealed the transfer of electron charge from methane to water. According to the natural bond orbital (NBO) analysis data, the charge transfer upon complexation is caused by the interaction between the σ orbital of the axial С–H bond of methane directed along the line of the O–H···C hydrogen bridge and the antibonding σ* orbital of the О–H bond of the water molecule. Topological analysis of electron density in the HOH···CH4 complex by the AIM method showed that the parameters of the critical point of the bond between hydrogen and acceptor (carbon atom) for the O–H···C interaction are typical for Н-bonded systems (the magnitude of electron density at the critical point of the bond, the sign and value of the Laplacian). It was concluded that the intermolecular interaction in the complex can be defined as an Н bond of O–H···σ(С–H) type, whose energy was found to be 0.9 kcal/mol in MP2/aug-cc-pVQZ calculations including the basis set superposition error (BSSE).

Reference
1. A. N. Isaev,"O–H···C hydrogen bond in the methane–water complex", Russian Journal of Physical Chemistry A 90(2016), 1978–1985.

Hydrogen bonds occurs when a H atom is bound to a highly electronegative atom like oxygen. In such a bond O-H, the 1s electron of hydrogen is mainly localized between the H and O nuclei. So, at the outer part of the bond, the H nucleus is nearly not surrounded any more by an electron layer. The H nucleus is nearly "naked". Nothing prevents it from touching and attracting electrons from the vicinity, and specially these electrons which are not engaged in a bond, like those in outer layer of the Oxygen atoms. This electrostatic attraction is called a Hydrogen bond.


Gas chromatography analysis - number of peaks? This is an example in a Princetown Review prep book. I am not understanding the statement "his separation was only partial (because two different peaks are recorded)." This is implying that Compounds A and B did not separate, but um... there were two peaks? Doesn't that imply that it was separated?

Unless there is a solvent peak? But I did a lot of Googling, and didn't find any indication that there should be a solvent peak under normal circumstances.

Why did the example say that the 2 compounds weren't separated despite the fact that there are 2 peaks?

The GC is the instrument used (in this case, although for very small amounts, there is no reason GC couldn't be used to separate materials - assuming collection after detection (via, say, refractive index - for obvious reasons flame ionization, etc. wouldn't be usable)) to determine if his primary separation technique was effective. It wasn't.


Why is the stoichiometric coefficient in the denominator in the rate expression? For the reaction:  $$\ce{2A + B → C}$$

The rate expression is given as:

$$\ce{-1/2 * Δ [A]/Δt}$$  = $$\ce{-1 * Δ [B]/Δt}$$ = $$\ce{Δ [C]/Δt}$$

I can't get my head around why it is $$\ce{-1/2 * Δ [A]/Δt}$$  and not $$\ce{-2 * Δ [A]/Δt}$$

If A is consumed at twice the rate that B is consumed, why is the rate for A half (and not double) the rate for B?

The only answer I could find on the internet is "To show a standard rate of reaction in which the rates with respect to all substances are equal, the rate for each substance should be divided by its stoichiometric coefficient." Source: http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Kinetics/Rate_Laws/The_Rate_Law

But this doesn't make much sense to me. Why does the rate with respect to all substances have to be equal when one substance (A) is used up twice as fast as another substance (B)?

From the stoichiometric coefficients $\nu_{i}$ in your reaction equation

$$\ce{2A + B → C}$$

you can see (as you already correctly stated) that for each molecule of $\ce{B}$ consumed, $\nu_{B} = -1$, two molecules of $\ce{A}$ must be consumed as well, $\nu_{A} = -2$. Similarly for each molecule of $\ce{B}$ consumed one molecule of $\ce{C}$ must be formed, $\nu_{C} = +1$ (and analoguously for $\ce{A}$ and $\ce{C}$).
This will also be reflected in the changes of the amounts of those species, $\Delta n_{i}$, during the reaction, e.g. if the amount of $\ce{B}$ changes by $2 \, \mathrm{mol}$ you can expect a change in the amount of $\ce{A}$ by $4 \, \mathrm{mol}$. So, the ratios of the $\{\Delta n_{i}\}$ and the $\{\nu_{i}\}$ will be equal, i.e.

\begin{align}
  \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{B}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{B}}} \ ; && \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{C}}} \ ; && \frac{\Delta n_{\ce{B}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{B}}}{\nu_{\ce{C}}}
\end{align}

Rearranging these equations a bit leads you to:

\begin{align}
  \frac{\Delta n_{\ce{A}}}{\nu_{\ce{A}}} = \frac{\Delta n_{\ce{B}}}{\nu_{\ce{B}}} = \frac{\Delta n_{\ce{C}}}{\nu_{\ce{C}}} \ .
\end{align}

Now, divide these equations by the reaction volume $V$ and use $\frac{\Delta n_{i}}{V} = c_{i} \equiv [i]$ to introduce concentration changes

\begin{align}
  \frac{\Delta [\ce{A}]}{\nu_{\ce{A}}} = \frac{\Delta [\ce{B}]}{\nu_{\ce{B}}} = \frac{\Delta [\ce{C}]}{\nu_{\ce{C}}} \ .
\end{align}

Finally, if you divide now by the reaction time $\Delta t$ and use the values for the stoichiometric coefficients, you get the equations from your question:

\begin{align}
  \frac{1}{\nu_{\ce{A}}} \frac{\Delta [\ce{A}]}{\Delta t} &= \frac{1}{\nu_{\ce{B}}} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{\nu_{\ce{C}}} \frac{\Delta [\ce{C}]}{\Delta t} \\
  \frac{1}{-2} \frac{\Delta [\ce{A}]}{\Delta t} &= \frac{1}{-1} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{1} \frac{\Delta [\ce{C}]}{\Delta t} \\
-\frac{1}{2} \frac{\Delta [\ce{A}]}{\Delta t} &= -1 \frac{\Delta [\ce{B}]}{\Delta t} = 1 \frac{\Delta [\ce{C}]}{\Delta t} \ .
\end{align}

The rate is just a characteristic of a reaction, not how fast one element's concentration is increasing or not. You can simply multiply it by stoichiometric coefficient to find the rate of one element's consumption/creation.

So the answer: "To show a standard rate of reaction in which the rates with respect to all substances are equal, the rate for each substance should be divided by its stoichiometric coefficient." is the right one which was shown in a detailed way by philip:

  From the stoechiometric coefficients νi

in your reaction equation

2A+B→C

you can see (as you already correctly stated) that for each molecule of B
consumed, νB=−1, two molecules of A must be consumed as well, νA=−2. Similarly for each molecule of B consumed one molecule of C must be formed, νC=+1 (and analoguously for A and C). This will also be reflected in the changes of the amounts of those species, Δni, during the reaction, e.g. if the amount of B changes by 2mol you can expect a change in the amount of A by 4mol. So, the ratios of the {Δni} and the {νi}

will be equal, i.e.

ΔnAΔnB=νAνB ;ΔnAΔnC=νAνC ;ΔnBΔnC=νBνC

Rearranging these equations a bit leads you to:

ΔnAνA=ΔnBνB=ΔnCνC .

Now, divide these equations by the reaction volume V
and use ΔniV=ci≡[i]

to introduce concentration changes

Δ[A]νA=Δ[B]νB=Δ[C]νC .

Finally, if you divide now by the reaction time Δt

and use the values for the stoechiometric coefficients, you get the equations from your question:

1νAΔ[A]Δt1−2Δ[A]Δt−12Δ[A]Δt=1νBΔ[B]Δt=1νCΔ[C]Δt=1−1Δ[B]Δt=11Δ[C]Δt=−1Δ[B]Δt=1Δ[C]Δt .


Entropy Change of Resistor A very large swimming pool filled with water of temperature equal to $20\ \mathrm{^\circ C}$ is heated
  by a resistor with a heating power of $500\ \mathrm{W}$ for $20$ minutes. Assuming the water in the
  pool is not in any contact with anything besides the resistor:

  Is the change of entropy of the resistor positive, negative, or zero?

My Attempt

We know that entropy is given by the formula:

$$\mathrm dS = \frac{q_\text{rev}}{T}$$

My first thought was that since the resistor is giving off heat energy to the water, $q$ must be negative and therefore the entropy change must be negative. However, does the fact that there is a continuous current of electrons passing through the resistor complicate this simple calculation? I am thinking that the current of electrons increase the entropy of the resistor and offsets the decrease in entropy due to heat loss. Am I correct?

  The answer in the book is that the entropy change is $0$.

Here entropy change will be zero because resistor convert the electrical energy into heating so it take as work not heat so entropy change will be zero


Oxidation and reduction in organic chemistry Why is it that in order to make an carboxylic acid from an alcohol you oxidise it twice...

ie alcohol + 2[O] ---> carboxylic acid

...but to get back to the alcohol you reduce it four times?

carboxylic acid + 4[H] ---> alcohol

My understanding was that oxidation is the opposite of reduction, and in organic chemistry [O] means addition of an oxygen atom or removal of a hydrogen atom, with [H] meaning the opposite.

Addition of oxygen and hydrogen can be useful to predict oxidation and reduction in most of the cases but this criteria fails to classify a reaction as oxidation or reduction when there are no oxygen or hydrogen atoms involved. This method is very primitive.

A better perspective :

Oxidation : Increase in oxidation state (by loss of electrons) of central atom.

Reduction : Decrease in oxidation state (by gain of electrons) of central atom.

If you already know how to find the oxidation state of the central atom (in your case carbon), then I hope you will be able to get your answer now.

The answer to the question why oxidation of alcohol to carboxylic acid takes 2 oxidation reactions, lies in the reaction mechanism. These mechanisms cannot be challenged, they are what they are. We can only observe and provide a acceptable explanation to why they happen like the way they do.

in reaction dont see how much O or H you are using.Just see how your derired substance is oxidized.During first oxidation to carbonyl alcohol loses hydrogen in case of methanol 1 hydroge.And nest on to convert to acid adds one oxygen.
Similar is the case whil reducing acid first adds remove one Oygen atom and then in carvonyl to alcohol adds one H atom.


Nitrogen molecule vs carbon-carbon triple bond stability Why does the triple bond in the nitrogen molecule convey stability, whereas a triple bond between carbon atoms conveys instability?

You might already know that elements like to follow the octet rule, which means that atoms tend to combine in a way that each atom can have eight valence electrons. Nitrogen can exist diatomically because through the triple bonds, each nitrogen will have one lone pair and three shared pairs. As seen in the picture, each nitrogen satisfies the octet rule.

Carbon, on the other hand, cannot satisfy the octet rule diatomically. Since carbon has one less valence electron than nitrogen, a triple bond would not work because the carbon does not have enough electrons to bond and satisfy the octet rule. It would leave each carbon atom with three bonded pairs of electrons and only one lone electron, giving each atom a configuration of only seven electrons per atom, not eight. Sorry I don't have a picture of a triple carbon bond. Hope this answers your question though!


Is methoxide a better nucleophile than hydroxide? What is more nucleophilic, $\ce{OH-}$ or $\ce{CH3O-}$?
When I look it from the viewpoint of steric hindrance, the answer seems to be $\ce{OH-}$, but when I see it from the viewpoint of extra electron density provided by hyperconjugation of $\ce{CH3}$, the answer seems to be $\ce{CH3O-}$.

Alkoxides tend to be stronger bases than the hydroxide ion. Adjacent atoms in an alkane chain will donate electron density to oxygen, allowing it to be more polarizable. An increase in polarizability will allow oxygen to more easily donate electrons.

The oxygen in a hydroxide ion does not experience the inductive effect. Therefore its valence electrons do not "reach" out as far.

Methoxide is a relatively unhindered base, as is hydroxide. But as explained above, methoxide experiences induction whereas hydroxide does not. So one would expect methoxide to be a better nucleophile than hydroxide because it is a stronger base than hydroxide and still remains unhindered.

My answer would be same as the above mentioned.  I would just add that of the reduction in nucleophilic character of methoxide due to steric factors is 10 then an increase due to inductive factor is 40 which makes a net +30 better than hydroxide.
Note- above values have no physical significance they are just made up to explain the comparative difference.


What is the most common oxidation state of gold? I had a test this morning which had the following question:

  What is the most common oxidation state for gold?
  (a) +1  (b) -1  (c) +2  (d) +3

Since the electronic configuration of $\ce{Au}$ is $\ce{[Xe] 4f^{14} 5d^10 6s^1}$. After removing the $\ce{6s^1}$ electron and gaining $+1$ charge, $\ce{Au}$ should have noble gas configuration and hence this should be the most stable configuration. So the answer has to be +1.
Could someone verify my reasoning?

It's not obvious, but common oxidation state for gold is +3. It caused by destabilization of the $5d^{10}$ orbital. Detailed explanation you can find in The Chemistry of Gold, in Chapter 1.1.3.

Oxidation is nothing but an atom loosing electrons in chemical compound. Oxidation state works as an indicator and sometime it could be negative or positive or Zero.
Most common Oxidation states of Gold Include
+1 (gold(I) and +3 (gold(III)

If you want to know the Less common oxidation states of gold
then it includes
−1, +2, and +5.

You can know more about
gold by exploring its Properties!


Entropy change of surroundings and system The entropy change of the surroundings can be calculated by the equation $dS_{sur}=\frac{dq}{T_{sur}}$ regardless of the path (irreversible or reversible).

The argument is that because the surroundings may be approximated as either constant volume or constant pressure, the heat absorbed/released by it is equal to the internal energy or enthalpy change, respectively. As both are state functions, it does not matter if the path taken was reversible or not, the changes would be the same for both, so $dU_{sur,rev}=dU_{sur,irrev}$ and $dH_{sur,rev}=dH_{sur,irrev}$. However, $dH=dq$ for constant pressure, and as per definition $dS=\frac{dq_{rev}}{T}$, the initial equation follows.

This is somewhat the way my book derives the formula. It seems to be inconsistent to me because it equals $\frac{dq_{rev}}{T}=\frac{dq_{irrev}}{T}$, which some pages after the book says that is not the case and uses the fact that $dq_{rev}\geq dq_{irrev}$ to derive the Clausius inequality.
I do understand the Clausius inequality and its derivation, however I do not understand why is it that it is not used to calculate the entropy change of the surrounds, only the system. The arguments for its derivation seem to apply to the first case as well.

Similar question that has no answer: Entropy change of surroundings

In evaluating the entropy change of the surroundings, you need to first look exclusively at its initial state and its final state, before and following the irreversible process.  The only difference between the initial state and the final state is a difference in enthalpy of the surroundings, equal to amount of irreversible heat that flowed to the surroundings.  To get the entropy change for the surroundings, you next need to separate it entirely from the original system, and subject it to a new reversible process while in contact with its own new surroundings (i.e., a second set of surroundings).  No matter what new (alternative) reversible process you devise, you will always end up with the same enthalpy change and same beginning and ending pressures, and its entropy change will always be the same, and equal to the irreversible heat it received in the original process divided by its absolute temperature.

On the other hand, for the original system, if you follow the same procedure (separating it from the original surroundings and putting it into contact with a third surroundings), the reversible process you devise will give you a different value for the integral of $dq/T_{boundary}$ than the integral of $dq/T_{boundary}$ for the original irreversible process.

The most important concept to understand is that entropy is a state function.
Best explained by this image:

The change in entropy is just based on the final state and the initial state. Therefore we model all processes as a reversible process because it has a simple equation. The reversible process is the one in blue above. That allows us to use the equation for the change in entropy for reversible process:
$$\mathrm dS =\frac{\mathrm dq_\mathrm{rev}}{T}.\tag{1}$$
Hint: use the first law to get the equation for $q.$
So if I want to calculate change in entropy for the system I integrate from state 1 to state 2. It will give you the same answer if you want to calculate $ds$ for any process.
Nuance of Surroundings:
If I want to calculate the change in entropy of the surroundings it changes, slightly, because the temperature of the surroundings is constant. That is because the surrounding is a reservoir. So we end up with
$$\int^{2}_{1}\mathrm dS_\mathrm{surr} = \int^{2}_{1}\frac{\mathrm dq_\mathrm{rev}}{T_\mathrm{surr}},\tag{2}$$
which will simplify to
$$S_2 - S_1 = \frac{q_\mathrm{rev}}{T_\mathrm{surr}}.\tag{3}$$


what is the difference between a redox reaction and an ionic bond? I am new into chemistry...
I'd like to know what the difference is between an ionic bond: which involves a bond with an electron transfert, and a redox reaction; which seems to be exactly the same to me.

Thanks

An ionic bond is formed by electrostatic attraction of electrons between one atom and the other whereas in redox reactions the electrons are transferred from one atom to the other. Hope this helps.


Sulfuric acid fumes from cleaning shower drain This morning a workman came by to my newly rented property to fix a few things, among which a slightly blocked bathtub drain.

He let some mild water run continuously and he poured down a chemical, and it wasn't before I saw a huge vapor coming out and an intense boiling-like sound, I realized something was wrong. Basically he let down an entire bottle of >95% sulfuric acid down the drain just like that, without even wearing safety goggles or gloves. A pungent odor emanated through the whole flat and I had to open every window (he hadn't advised to open them beforehand).

Now, I don't have much experience with chemicals of this sort, so here are my questions:

Did I inhale toxic fumes?
What should I do to make the bathroom/shower usable again? I noticed that the mirror and several water tiles that are not in the close proximity of the drain but are on the way to the nearest window opening, show condensation streaks, such as those left by drops of water that run down. They were definitely not there before. Is that condensed sulfuric acid vapor? Do I need to clean with a sodium bicarbonate solution?
Needless to say, I also had towels, toothbrushes, and other items in the room, do I need to take any precaution before using them (cleaning, rinsing, etc?).

In short, how do I clean up this mess?

If the workman did use concentrated sulphuric acid (which is not the best choice for drain cleaning) then the best solution if to flush the drain with large quantities of water and wash the surfaces with water.

Sulphuric acid is not toxic but is very corrosive when concentrated. But it is highly water soluble and not harmful when it is dilute enough. So dilution with water will solve whatever problems it seems to cause. If it got splashed around, wash everything thoroughly and there will be no further problem. If you use enough water then it won't even be necessary to neutralise the result as any remaining acid will be swept through the drain and highly diluted by the flow of water.

Sulfuric acid seems like an odd choice to clean a drain. Typically drain cleaners are strong alkali mixtures.

Concentrated sulfuric acid ionizes on contact with water with a very exothermic reaction. It could easily boil or splatter up from the drain.

Rinsing with water would be sufficient to clean up any mess. To be on the safe side, I'd wash linens in bathroom before using again.


Why does a reaction stop when turning into a solid? I recently was looking at a question about the volume of carbon dioxide produced when reacting calcium carbonate with dilute sulfuric acid versus using nitric acid. In the answer it says that calcium sulfate is produced, which is a solid, which stops the reaction, but I don't get why. Calcium nitrate produces an aqueous solution which causes the reaction to carry on reacting.
The reaction equations are:
$$\ce{CaCO3(s) + H2SO4(aq) -> CaSO4(s) + H2O(l) + CO2(g)}$$
$$\ce{CaCO3(s) + 2HNO3(aq) -> Ca(NO3)2(aq) + H2O(l) + CO2(g)}$$

The reason is due to the surface area of calcium carbonate available to react.

When reacting a block of solid calcium carbonate with dilute sulphuric acid, solid calcium sulphate is produced which is deposited on the surface of the calcium carbonate block (since that is where the reaction occurs). The calcium sulphate is insoluble in dilute sulphuric acid and so you end up with calcium carbonate coated in a layer of calcium sulphate, which stops the acid from getting at the carbonate and so stops the reaction.

In contrast, calcium nitrate, produced from the reaction with nitric acid, is highly soluble in dilute nitric acid and so no solid layer forms around the carbonate block and the reaction can continue until one of the reagents is used up.

Reactions such as these take place in an ionic form where Ca2+ and CO3 2- are formed in the aqueous medium and react with H2 2+ and SO4 2-. When CaSO4 is formed, it precipitates as it is insoluble. Thus it covers the surface of the CaCO3 and the reaction stops.
In the second reaction CaNO3 is not insoluble so the reaction continues and CaNO3 solution keeps being formed.

Hope this answers your question!


Balancing a reaction with unknown products I'm given this equation in word form:

$$\ce{sodium (s) + water (aq) -> unknown~products (s/g/aq)}$$

and I have to convert this to a chemical equation. I understand how to covert this normally, that equation can hence be represented by:

$$\ce{Na + H2O -> unknown~products\:(s/g/aq)}$$

This is indeed a single displacement reaction so it should become (according to what I know so far) $\ce{Na2O + H2}$ but after doing some research, I got that it should be:

$$\ce{Na + H2O -> NaOH + \frac{1}{2}H2}$$ or $$\ce{2Na + 2H2O -> 2NaOH + H2}$$

I'm lost now, where did $\ce{NaOH}$ come from? Especially the H since there is no reason for hydrogen to be present.

Please explain why and how to balance this reaction or convert this and what factors to look at when dealing with these types of equations.

Water is normally a mix of hydroxide and hydrogen, at very dilute levels.  For example, neutral water (pH of 7) is 10-7 of OH- and H3O+ ions.  Add sodium and it combines with the OH-.  The loss of electrons by sodium are available to the hydrogen ions, which combine to make hydrogen gas.  Sodium in metal form is so reactive, it will sustain this reaction even at the low hydroxide levels in pure water, and even in slightly acidic seawater.


Resonance structures of 4‐aminobut‐3‐en‐2‐ylium Problem

How many resonance structures can be drawn for the following molecule?

(a) 1
(b) 4
(c) 3
(d) 2

Answer

 (d) 2

Question
I think that the molecule can be drawn in three ways, which contradicts the book's answer. Can anyone explain?

Just shift the pi-bond from the double bond to the right side. The carbocation then shifts to the left. That's the second resonance structure. The first one is already shown, making a total of two different structures (D).

You might be wondering about the allylic lone pair on the Nitrogen atom. Well, that could have resulted in another resonance pair, but if you shift the lone pair to the adjacent C-C bond on the right, there will be a total of 3 charges on the resonance structure. This is a highly unstable form, so it's not a significant resonance structure. Thus, it can be neglected.


Types of bonds in a molecule For example in dinitrogen pentoxide, $\ce{N2O5}$, covalent as well as coordinate bonds (type of covalent bonds) are present, but it appears that it contains only covalent bond.

What is a proper method to find out which type of bonds are present in a molecule?

Electrovalent bonds are easiest to identify. If a compound is made up of a metal and non-metal/non-metallic radical (like carbonate), then, 99.99% times, it contains electovalent bond. If a compound is made up of 2 or more non-metals/non-metallic radicals, then it contains covalent bond. Coordinate covalent bonds appear mostly with compounds containing Hydrogen element. To identify the coordinate covalent bonds, you can draw the branched structural formula of the compound and see if the shared pair of electrons are coming from the same molecule.


Calcium reactivity vs electronic configuration Calcium is an alkaline earth metal, so it is reactive. But, it has two valence electrons. Don't those two electrons fill up the first energy level? Isn't an atom with a full energy level considered stable? Why is it reactive then?

The electronic configuration of Ca is $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2}$

It has a total 4 Energy levels. The outermost energy level has 2 electrons called valance electrons. As, Calcium is metal and it can easy donate two electrons and becomes cation (Ca+2). Now, this cation is stable.

Calcium is reactive because it can easy donate electrons by providing energy to it called ionisation energy.


Can methanol be oxidised by PCC? Pyridinium chlorochromate (PCC) in $\ce{CH2Cl2}$ can be used to oxidise primary alcohols only to aldehydes. I can't however, find any reference of it oxidising methanol (which is strictly speaking not a primary alcohol). My question is, does PCC oxidise methanol to formaldehyde and if not, what can I use to achieve this reaction?

Look at the examples given here:

(Ignore the stuff written with pen; I have a habit of taking notes when reading)

PCC is taking two hydrogen atoms – one from the carbon atom attached to the -OH group and the other one from the -OH group itself, and then making a C=O double bond. Methanol has both of these. So, PCC should be able to oxidise it to formaldehyde.


What is Molar Conductivity of an electrolytic solution Please explain the physical meaning it carries ? And how is it that it can increase with dilution ?

And why is it called Molar ?

And what is equivalent conductivity ?

I will answer your question about the definition of molar conductivity and why it increases with decreasing concentration.

As Jerry pointed out, you can probably learn what you need about equivalent conductivity using a Google search.

The molar conductivity of an electrolyte solution is the conductivity of the solution divided by the molar concentration of the dissolved electrolyte.

Conductivity ($\kappa$, in $\text{S/cm}$) is proportional to the square root of the concentration of ions ($\sqrt{[\text{ions}]}$, withy $\ce{[ions]}$ in $\text{mol/L}$). In the following equation, A is an empirical proportionality constant with units needed to convert from $\text{M}^{1/2}$ to $\text{S/cm}$, namely $\frac{\text{S/cm}}{\text{M}^{1/2}}$

$$\kappa = A\sqrt{\ce{[ions]}}$$

The ion concentration is related to the electrolyte concentration ($C$) by the van't Hoff factor ($i$).

$$[\ce{ions}]=iC$$

$$\therefore \ \kappa = A\sqrt{iC}$$

The molar conductivity $\kappa_M$ is the conductivity divided by the molar concentration of electrolyte ($C$).

$$\kappa_M=\frac{\kappa}{C}=\frac{A\sqrt{iC}}{C}=A\frac{\sqrt{i}}{\sqrt{C}}$$

As $C$ decreases, so does $\sqrt{C}$, and thus $\frac{1}{\sqrt{C}}$ increases, while $A$ remains constant. The van't Hoff factor changes a little with concentration, but not dramatically. ($A$ probably also varies a little with concentration. It certainly varies with temperature.)

For example, if you have a series of solutions of sodium chloride, concentration decreases faster than conductivity, so conductivity divided by concentration increases. Note that the conductivity data are fabricated with an arbitrary value of $A$. I could not find the specific value of $A$ for sodium chloride. The van't Hoff Factors come from the "Colligative Properties" Chapter of Chemistry: Atoms First by Burdge and Overby, McGraw Hill, 2012.

Molar conductivity of an electrolyte is defined as the conductivity of a volume of solution containing one mole of electrolyte.


Do the dimensions of the cathode and anode matter in electrochemistry? I'm curious whether the dimensions of the cathode and anode matter in an electrochemical process. Let's take copper electrorefining as an example. For instance, should the cathode and anode be of the same length and be in alignment with each other or does the process not care?

From my perspective, the cathode doesn't 'see' the anode, it only sees the $\text{Cu}^{2+}$ ions and therefore the anodes dimensions and placement relative to the cathode don't have an effect. Note, let's ignore the increase in resistance as the cathode and anode are placed further away and other similar effects. The essence of what I'm trying to understand is how their placement/alignment effects the occurrence of the reaction.

The dimensions of your electrodes should not effect the open circuit voltage, since this is determined solely by the identity and concentrations of the redox species.

Once the cell is running, the dimensions may affect how much the voltage drops off as you draw current, known as "overpotential". If your cell is limited by its ability to deliver ions to the surface of the electrode, or to transfer charge into the electrode, then the higher surface area of a large electrode is likely to result in a more efficient battery.

If you examine the Nernst Equation, there is no indication that the cathode or anode have any effect. What does effect the current is the resistivity of the solution. The further apart the electrode are, the more resistance the solution offers to the current. If the electrodes get too close, a short can occur.

Also, a high current between the electrodes can deplete the carriers due to a lack of diffusion or electrostatic repulsion.

At lower currents, you will see the same amount of current for any electrodes with the same surface area at the same distance.

If the electrodes are not aligned, then the electrodes look like tapered cones due to the increasing resistance from the closest point to the ends.


Arrangement of elements in the reactivity series I have seen a variety of reactivity series of elements online and also have consulted some books regarding the arrangement of the elements. I understand that the arrangement of elements is based on degree of reactivity of elements with cold water, steam, oxygen, dilute acids etc.
Lithium is less reactive with respect to the criteria mentioned above yet it is placed on the top of more reactive elements in the reactivity series that I have observed.
What is the reason for the position of Lithium above let us say potassium and sodium which are more reactive.
Link to one of the references:
https://kaiserscience.wordpress.com/chemistry/metal-reactivity-series/

The reason that lithium is on the top of the activity series is due to its behavior in aqueous reactions with water and acid. Due to the lithium atom's very small size (only two energy levels), the lithium ion interacts well with water forming ion-dipole interactions.  The lithium atom's small size compared to sodium and potassium is more entropically favorable in the formation of ion-dipole interactions with water and consequently so are reactions involving lithium ion formation favored to to other alkali metals.

The reasoning behind this is that reactions favor a greater state of entropy (disorder) in the universe. With larger cations, more water molecules have to become ordered around the cation to form the appropriate ion-dipole interaction.  However, due to the lithium cation's small size, less water molecules are required to form the ion-dipole interaction, thus less order, being more favorable in terms of entropy.


What is the difference between ∆G and ∆G°? In Brady's Molecular Nature of Matter, I read that $\Delta_\mathrm{r} G^{\circ}$ is $\Delta_\mathrm{r} G$ at $25~^\circ\mathrm{C}$. But later, it gives a value for $\Delta_\mathrm{r} G^\circ$ at another temperature.

What is the meaning of $\Delta_\mathrm{r} G^\circ$?  Specifically:

Does it only refer to $25~^\circ\mathrm{C}$?
Does the $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T \Delta_\mathrm{r} S^\circ$ equation only refer to $T=298~\mathrm{K}$?
If at $25~^\circ\mathrm{C}$, a system that has reached equilibrium ($\Delta_\mathrm{r} G=0$), and thus

$Q=K$
$\Delta_\mathrm{r} G^\circ=0$
$\Delta_\mathrm{r} G^\circ=0=-RT\ln K$
$RT\ln K=0$
$\ln K=0$
$K=1$

But doesn't $K$ vary from reaction to reaction? It shouldn't always equal 1, should it?

Short answer

  Does it need to be at $25~^\circ\mathrm{C}$?

No. $\Delta_\mathrm{r} G^\circ$ can be defined at any temperature you wish to define it at, since the standard state does not prescribe a particular temperature. If you change the temperature, $\Delta_\mathrm{r} G^\circ$ will change.

  Does $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T\Delta_\mathrm{r} S^\circ$ always use $T = 298~\mathrm{K}$?

No. You use whatever temperature you are running your reaction at.

  (...maths...)

Yes, at equilibrium, $\Delta_\mathrm{r} G = 0$ and $Q = K$.

However, everything after the first bullet point is wrong. You cannot conclude that $\Delta_\mathrm{r} G^\circ = 0$, nor can you conclude that $K = 1$. The equation $\Delta_\mathrm{r} G^\circ = -RT \ln K$ does not analogously translate into $\Delta_\mathrm{r} G = -RT \ln Q$! The accurate relation is:

$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$

Setting $Q = K$ and $\Delta_\mathrm{r} G = 0$ in this equation does not tell you anything about the value of $K$. In fact, if you try doing it, all you will find out is that $\Delta_\mathrm{r} G^\circ = -RT \ln K$ - no surprises there!

Long answer

Any book that writes that $\Delta_\mathrm{r} G^\circ$ is the "special case" of $\Delta_\mathrm{r} G$ at $T = 298~\mathrm{K}$ is wrong.

The Gibbs free energy of a system is defined as follows:

$$G = H - TS$$

Under constant temperature and pressure (from now on, I will just assume constant $T$ and $p$ without stating it), all systems will seek to minimise their Gibbs free energy. Equilibrium is reached when $G$ is minimised. When $G$ is at a minimum, any infinitesimal change in $G$, i.e. $\mathrm{d}G$, will be $0$. Therefore, this is equivalent to saying that the condition for chemical equilibrium is $\mathrm{d}G = 0$.

Clearly, we need a way to relate this quantity $\mathrm{d}G$ to the actual reactants and products that are in the system. This can be done by using the Maxwell relation (see any physical chemistry text for details):

$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T + \sum_i \mu_i\,\mathrm{d}n_i$$

Under constant $T$ and $p$, $\mathrm{d}p = \mathrm{d}T = 0$ and therefore

$$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i$$

where $\mu_i$ is the chemical potential of species $i$, defined as a partial derivative:

$$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{n_{j\neq i}}$$

So, we now have a refined condition for equilibrium:

$$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i = 0$$

We can go further by noting that the values of $\mathrm{d}n_i$ for different components $i$, $j$, etc. are not unrelated. For example, if we have a reaction $i + j \longrightarrow k$, then for each mole of $i$ that is consumed, we must also use up one mole of $j$; this means that $\mathrm{d}n_i = \mathrm{d}n_j$.

This can be formalised using the idea of a stoichiometric coefficient $\nu_i$, which is defined to be positive for products and negative for reactants. For example, in the reaction

$$\ce{3H2 + N2 -> 2NH3}$$

we have $\nu_{\ce{H2}} = -3$, $\nu_{\ce{N2}} = -1$, and $\nu_{\ce{NH3}} = 2$.

By stoichiometry, if $1.5~\mathrm{mol}$ of $\ce{H2}$ is consumed, then $1~\mathrm{mol}$ of $\ce{NH3}$ has to be produced. We could write $\Delta n_{\ce{H2}} = -1.5~\mathrm{mol}$ and $\Delta n_{\ce{NH3}} = 1~\mathrm{mol}$. These quantities are proportional to their stoichiometric coefficients:

$$\frac{\Delta n_{\ce{H2}}}{\nu_{\ce{H2}}} = \frac{-1.5~\mathrm{mol}}{-3} = 0.5~\mathrm{mol} = \frac{1~\mathrm{mol}}{2} = \frac{\Delta n_{\ce{NH3}}}{\nu_{\ce{NH3}}}$$

The quantity $0.5~\mathrm{mol}$ is a constant for all chemical species $\ce{J}$ that participate in the reaction, and it is called the "extent of reaction" and denoted $\Delta \xi$ (that is the Greek letter xi). If the reaction is going forward, then $\Delta \xi$ is positive, and if the reaction is going backwards, then $\Delta \xi$ is negative. If we generalise the above result, we can write

$$\Delta \xi = \frac{\Delta n_i}{\nu_i}$$

and if we make $\Delta n_i$ smaller and smaller until it becomes an infinitesimal, then:

$$\begin{align}
\mathrm{d}\xi &= \frac{\mathrm{d}n_i}{\nu_i} \\
\mathrm{d}n_i &= \nu_i\,\mathrm{d}\xi
\end{align}$$

If we go back to our condition for equilibrium, we can substitute in the above to get:

$$\mathrm{d}G = \sum_i \mu_i\nu_i\,\mathrm{d}\xi = 0$$

Now, $\mathrm{d}\xi$ is no longer dependent on $i$, since we have established already that $\Delta \xi$ (and by extension $\mathrm{d}\xi$) is a constant for all chemical species. So, we can "divide through" by it to get:

$$\Delta_\mathrm{r} G \equiv \frac{\mathrm{d}G}{\mathrm{d}\xi} = \sum_i \mu_i\nu_i = 0$$

where $\Delta_\mathrm{r} G$ is defined to be $\mathrm{d}G/\mathrm{d}\xi$.

Note that $\Delta_\mathrm{r} G$ is an intensive property and has units of $\mathrm{kJ~mol^{-1}}$, since $\mathrm{d}\xi$ has units of $\mathrm{mol}$. This ensures that the units we use are consistent: since we know that $\Delta_\mathrm{r}G = \Delta_\mathrm{r} G^\circ + RT\ln Q$, $\Delta_\mathrm{r}G$ must have the same units as $RT$.

How do we interpret the physical significance of $\Delta_\mathrm{r} G$, or in other words, what does it even mean? There are two ways, each based on a different mathematical expression.

We have $\Delta_\mathrm{r}G = \sum \nu_i \mu_i$. This means that $\Delta_\mathrm{r}G$ is simply the difference between the chemical potentials of the products and the reactants, weighted by their stoichiometric coefficients. For the reaction $\ce{3H2 + N2 -> 2NH3}$, we have:

$$\Delta_\mathrm{r} G = \sum_i \mu_i\nu_i = 2\mu_{\ce{NH3}} - 3\mu_{\ce{H2}} -\mu_{\ce{N2}}$$

We have $\Delta_\mathrm{r}G = \mathrm{d}G/\mathrm{d}\xi$. This means that it is the slope of a curve of $G$ against $\xi$:

Note that up to this point, we have not stipulated any particular temperature, pressure, amounts of species present, or any conditions whatsoever. We have only said that the temperature and pressure must be constant.

It is important to realise that $\Delta_\mathrm{r}G$ is a well-defined quantity at all $T$, all $p$, and all possible values of $n_i, n_j, \cdots$! The shape of the curve will change when you vary the conditions. However, no matter what the curve looks like, it is always possible to find its gradient ($= \Delta_\mathrm{r}G$) at a particular point.

What exactly, then, is $\Delta G^\circ$? It is just a special case of $\Delta G$, where all the reactants and products are prepared in a standard state. According to IUPAC, the standard state is defined as:

For a gas: pure ideal gas when the pressure $p$ is equal to the standard pressure $p^\circ$.
For a liquid or solid: pure liquid or solid at $p = p^\circ$
For a solution: ideal solution when the concentration $c$ is equal to the standard concentration $c^\circ$.

$p^\circ$ is most commonly taken to be $\pu{1 bar}$, although older texts may use the value $\pu{1 atm} = \pu{1.01325 bar}$. Since 1982, IUPAC has recommended the value $\pu{1 bar}$ for the standard pressure (Pure Appl. Chem. 1982, 54 (6), 1239–1250; DOI: 10.1351/pac198254061239). However, depending on the context, a different value of $p^\circ$ may prove to be more convenient. Likewise, $c^\circ$ is most commonly – but not necessarily – taken to be $\pu{1 mol dm-3}$.

Note that in the above definitions, no temperature is specified. Therefore, by defining the standard Gibbs free energy, we are fixing a particular value of $p$, as well as particular values of $n_i, n_j, \cdots$. However, the value of $T$ is not fixed. Therefore, when stating a value of $\Delta_\mathrm rG^\circ$, it is also necessary to state the temperature which that value applies to.

When a reaction vessel is prepared with all its substances in the standard state, all the components of the system will have an activity of exactly $1$ by definition. Therefore, the reaction quotient $Q$ (which is a ratio of activities) will also be exactly equal to $1$. So, we could also say that $\Delta_\mathrm{r}G^\circ$ is the value of $\Delta_\mathrm{r}G$ when $Q = 1$.

Returning to the graph of $G$ against $\xi$ above, we note that at the left-most point, $Q = 0$ since there are only reactants; at the right-most point, $Q \to \infty$ as there are only products. As we move from left to right, $Q$ increases continuously, so there must be a point where $Q = 1$. (In general, the point where $Q = 1$ will not be the same as the equilibrium point.) Since $\Delta_\mathrm{r}G$ is the gradient of the graph, $\Delta_\mathrm{r}G^\circ$ is simply the gradient of the graph at that particular point where $Q = 1$:

The gradient of the graph, i.e. $\Delta_\mathrm{r}G$, will vary as you traverse the graph from left to right. At equilibrium, the gradient is zero, i.e. $\Delta_\mathrm{r}G = 0$. However, $\Delta_\mathrm{r}G^\circ$ refers to the gradient at that one specific point where $Q = 1$. In the example illustrated above, that specific gradient is negative, i.e. $\Delta_\mathrm{r}G^\circ < 0$.

Again, I reiterate that the temperature has nothing to do with this. If you were to change the temperature, you would get an entirely different graph of $G$ versus $\xi$. You can still find the point on that graph where $Q = 1$, and the gradient of that graph at the point where $Q = 1$ is simply $\Delta_\mathrm{r} G^\circ$ at that temperature.

We have established the qualitative relationship between $\Delta_\mathrm{r} G$ and $\Delta_\mathrm{r} G^\circ$, but it is often useful to have an exact mathematical relation.

$\Delta_\mathrm{r} G^\circ$ is exactly the same as $\Delta_\mathrm{r} G$ except for the imposition of the standard state. It follows that if we take the equation

$$\Delta_\mathrm{r} G = \sum_i \mu_i \nu_i$$

and impose the standard state, we get

$$\Delta_\mathrm{r} G^\circ = \sum_i \mu_i^\circ \nu_i$$

Thermodynamics tells us that

$$\mu_i = \mu_i^\circ + RT\ln{a_i}$$

where $a_i$ is the thermodynamic activity of species $i$. Substituting this into the expressions for $\Delta G$ and $\Delta G^\circ$ above, we obtain the result:

$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$

where the reaction quotient $Q$ is defined as

$$Q = \prod_i a_i^{\nu_i}$$

When equilibrium is reached, we necessarily have $\Delta_\mathrm{r} G = 0$ (see the discussion above). The equilibrium constant $K$ is defined to be the value of $Q$ at equilibrium. Therefore, at equilibrium, $Q = K$. Plugging this into the equation above gives us the famous equation:

$$\Delta_\mathrm{r} G^\circ = -RT\ln K$$

Again, no temperature is specified! In general, $K$ depends on the temperature as well; the relationship is given by the van 't Hoff equation.

Just adding some notes here.

$\Delta G$ is a measure of the energy of your system . It is expressed as $\Delta G = \Delta H - T \Delta S$, so it is temperature-dependent.

The point of using $\Delta G$ is that if you have a system at a well-defined temperature and pressure (eg. an open cylinder at the lab), then the equilibrium will be at the MINIMUM of $\Delta G$. This does NOT imply $\Delta G = 0$, so it does not imply $K = 1$. It only implies minimum, which can be essentially any number.

Oh and another misconception here: there is no such thing as  (well, there is, but we can not define it or measure it). So we always write $\Delta G$, which means free energy change to a relative zero value. This is not connected to differentiation. You often see things like $\mathrm{d}(\Delta G)$, which is differentiating it. So that's why I said minimum does not imply $\Delta G = 0$.


Does fluorine ever form a double or triple bond? Does fluorine ever form a double or triple bond?
I wonder if seeming lack of such higher order bonds is related to the electronegativity of fluorine.

The compound $\ce{NF}$, which is isoelectronic to $\ce{O2}$ is known, has been isolated in matrices and characterised and subjected to calculations. Like for dioxygen, three different states of this molecule are known: one triplet and two singlet states. Without performing any sophisticated analysis of its orbitals, we can expect a bond order of 2, and thus a double bond.

Harbison performed calculations on this compound. He came to the conclusion that the most stable triplet state is best described using only a single $\ce{N-F}$ bond. The two singlet states which require full electron pairing, however, display a much shorter $\ce{N-F}$ bond distance and are thus better described by an $\ce{N=F}$ double bond. Adding formal charges would lead to:
$$\ce{\overset{-}{N}=\overset{+}{F}}$$

Reference:

G. S. Harbison, J. Am. Chem. Soc., 2002, 124, 366–367. DOI: 10.1021/ja0159261.

For most purposes, i.e. writing Lewis Structures and such things, fluorine always forms single bonds. In higher level chemistry, it may be possible (like in the comments). Fluorine, like you said, is very electronegative, and therefore it doesn't like to "share" its electrons, leading to almost always making single bonds.


Electron defficient molecules Electron defficiency is a common feature in some class of compounds of Group 13 elements (B,Al,Ga). In some covalent compounds of Be too the same nature prevails. To compansate this electron defficiency they undergo dimerization forming bridges('3c 2e' or coordination) or in some cases they get stability through backbonding (halides of group 13 elements). Hydrides of them are known to stay in dimeric form through bridging.
Halides of B are found to remain in monomeric form due to backbonding where as 'Be, Al'-halides highly favour bridging structures through coordination of halide. Why is such difference? In Be,Al, vacant p orbital is present too to favour backbonding.
Then why Be,Al favour dimerization (or polymerization) over backbonding in their halides?

I believe this may be a duplicate of this question:
Why is BCl3 a monomer whereas AlCl3 exists as a dimer?
From what I remember, the stronger the backbond, the smaller the tendency to undergo dimerisation, however, way back in the dark ages when I was a student and asked my second-year chem prof, his answer was that the simple reason why none of the tri-halides of boron undergo dimerisation is steric hinderance - i.e. the boron atom is just too small to accommodate the fourth larger Cl/Br/I atom. He then pointed me to several sets of further reading.
However, you will probably find a better answer at one of the following links:

Mulliken J. Chem. Phys. 3, 635 (1935); https://doi.org/10.1063/1.1749567
Lappert, et.al. Chem. Soc., Faraday Trans. 2, 1976,72, 539-55 https://doi.org/10.1039/F29767200539


Why do alkali metals have an exothermic electron affinity? Alkali metals should have positive electron gain enthalpy as they are electropositive elements and also there atomic size is big in their periods so they should be reluctant to take electrons but they show negative electron gain enthalpies. Why?

I think you mean to ask about the electron affinity of the alkali metals. That is, the enthalpy gain or lost when adding an electron to an atom.

While the electron affinity for alkali metals are much smaller than the halogens, they are slightly positive. That is, quoting from Wikipedia:

  Electron capture for almost all non-noble gas atoms involves the release of energy and thus are exothermic.

The catch, is that a positive electron affinity (Eea) refers to an exo-thermic process:

  Confusion arises in mistaking Eea for a change in energy, ΔE, in which case the positive values listed in tables would be for an endo- not exo-thermic process.

You refer to the alkali metals as "electropositive" -- in this case, you're referring to their electronegativity relative to, say, halogens. Yes, a sodium atom is likely to become a sodium cation and give up an electron. But the electron configuration is $3s^1$ and consequently, can also gain an electron and become the more stable $3s^2$ state.

It's more likely that a sodium atom will lose an electron, and become a cation, but the electron affinity reflects the modest gain in stability as $3s^2$ after attaching an extra electron.

All alkali metals have negative electron gain enthalpy.
Ionisation of sodium(or any atom) is endothermic process (495.8 KJ/mol), meaning, in isolated gaseous state, neutral atom is more stable,
Electron gain process of Na+ is obviously an exothermic process (reverse of ionisation) due to attraction of Positively charged, Na+ for a free electron.
If an atom has spontaneous tendency, i.e., a positive tendency , to gain electron, then the electron-gain process is exothermic & hence, its electron gain enthalpy is said to be negative and if the atom is reluctant to gain an electron, i.e., it has a negative tendency to gain an electron and is forced to accept it, the electron-gain process is exothermic & hence its electron gain enthalpy value is positive.
Electron gain enthalphy of Na+ is (-495.8) KJ/mol. [process is Na+(g) + e= Na(g)]
Note that here, obviously process is related to gaseous phase. So, any misconception about , like
non-spontaneous electrolysis of NaCl (is due to compensation of ionisation energy of sodium by greater stabilisation of ion by hydration) or,
Stability of sodium compoounds in which sodium ion is present (Ionisation energy is compensated by lattice energy of the solid) is common.


How does an anode metal come into solution if it cannot react with an electrolyte? From what I understand, an anode material is chosen so that it does not react with the anode electrolyte solution.

I was just wondering, then, that when you have a galvanic cell which has a Mg metal anode submerged in sodium chloride solution, how the magnesium metal gives two of its electrons and come into solution (as magnesium ion) even though it does not react with (and cannot react with) sodium chloride?

What I mean is, in a typical single displacement reaction, there is a metal that reacts with the solution and comes into solution. For example, with $\ce{Zn}$ and copper sulfate, the zinc replaces the copper and comes into solution as zinc ion. However, magnesium doesn't react with sodium chloride (because the activity series predicts it can't). How then can we say that we have magnesium ion in solution?

Is it that the metal is more stable in its ion state and thus readily gives up its electrons?

Traditionally, the anion is the one responsible for giving electrons: the chloride ion in the sodium chloride solution should come into contact with magnesium anode and give its electrons, which the magnesium metal should then carry. And thus the reaction should be

$$
\ce{ 2Cl- \!(aq) -> Cl2\!(g) + 2e-}
$$

and not

$$
\ce{Mg\!(s) -> Mg^2+\!(aq) + 2 e-}
$$

However, my books don't say this: they say that magnesium gives up two of its electrons and enters the solution.

What am I understanding wrong?

I was just wondering then that when you have a galvanic cell which has a $\ce{Mg}$ metal anode submerged in sodium chloride solution, how the magnesium metal gives two of its electrons and come into solution (as magnesium ion) even though it does not react with (and cannot react with) sodium chloride?

Your misunderstanding stems from assuming that the electrode reaction has to be the same sort of single-displacement redox reaction that would occur with no electrodes present at all.

At a dissolving magnesium anode$^\dagger$, the only reaction that needs to occur is the one you wrote involving magnesium:

$$
\ce{Mg\!(s) -> Mg^2+\!(aq) + 2e-}
$$

The potential applied to the electrode pair is what enables the above reaction to occur without a paired reduction reaction at the anode surface.  Instead of being taken up in another chemical reaction, the electrons are drawn into the anode and on into the connected electrical circuit. Those electrons are then "pushed" into the cathodic reaction, whatever it might be.

  From what I understand, anode material is chosen so that it does not react with the anode electrolyte solution.

This is true, but does not have to strictly hold. MaxW is right: $\ce{Mg}$ does react directly with water. But, the reaction is slow enough that it's perfectly safe, and that one can conduct useful electrochemical operations even while it's going on (e.g., see here and here).

  Is it that the metal is more stable in its ion state and thus readily gives up its electrons?

Despite what one reads from a table of standard reduction potentials or a metal reactivity series, in practical electrochemistry, the above is true for most metals, except for the noblest of the activity series. In my experience, it's usually a lot easier to dissolve metals than to electrodeposit them.  You may have to use additional chemistry to make it happen (e.g., including $\ce{HF}$$^\ddagger$ to strip the passivating oxide from $\ce{Ti}$, $\ce{Nb}$, $\ce{Ta}$, etc.), but once the metals are oxidized/dissolved they tend to stay that way.

 $^\dagger$ Operating at $100\%$ faradaic efficiency. Other reactions that might occur in this system, depending on the applied potential, include the chloride oxidation reaction you noted and water electrolysis to oxygen gas.
 $^\ddagger$ $\ce{HF}$ is very dangerous, do not use at home!

Hmmm. Electrochemistry can definitely be confusing. Context is quite important. Anyone who makes a broad claim that anodes don't react is just plain wrong. It depends on the set-up, including the applied current (and/or voltage). Anyone who makes the statement that in measuring electrolytic reactions both electrodes should be inert is going to be right a lot (that is, more often than someone will be able to find exceptions).
Magnesium is a VERY reactive metal! The only reason it doesn't explode into flame upon contact with water is that the coating of MgO slows down the reaction. Anyone who thinks putting a piece of Mg metal into a salt solution is a good idea, needs to be very well aware of the hazards.
When thinking about electrochemical reactions, you have to be aware of all of the possible oxidations and reductions that can go on. Running current through pure water can generate H2, O2, as well as several other compounds (in trace amounts, usually). As you probably know, the concentration of ions in pure water is very low [H] = [OH] = 1E-7. Which means that the amount of current (work) you can do is quite limited. So, NaCl (or some other electrolyte) is added. This complicates things since potential reactions (is that punny or what?) include Na(+)→Na° (not going to happen), Cl(-)→Cl2 ( one way to make chlorine gas), and things like Cl(-) + H2O→ OCl(-)+... (and a bunch more). Only if you know what the reaction potentials are for all of the possibilities can you make good predictions. Some electrode materials can act as catalysts (Pt electrode, for instance). These lower the EP necessary while remaining "inert".
It's been so long since I did any electrochemistry (decades) that I don't recall which electrode is the "anode" and which is the "cathode"...doesn't it depend on the type of cell involved? I've forgotten. I do remember that Leo says Ger (Leo is another name for lion, and Grrr is the English for the sound a threatened (or threatening) animal makes.) LEO = Loss of Electrons is Oxidation. GER = Gain of Electrons is Reduction. You have written one oxidation and one reduction. In a driven electrolytic cell, they will NOT occur at the same electrode!!!!!!!!


Why does iron have an abnormally high ionization energy? Along a period the ionization energy should increase because the atomic number is increasing, but there is negligible increase in shielding. However, $\mathrm{IE}_\ce{Mn} < \mathrm{IE}_\ce{Fe} > \mathrm{IE}_\ce{Co} > \mathrm{IE}_\ce{Cu}$, so why does iron have such a high ionization energy?

I've considered that iron has a $\mathrm{4s^2 3d^6}$ configuration that goes to a $\mathrm{4s^1 3d^6}$ configuration upon ionization, while $\ce{Co}$ has a $\mathrm{4s^2 3d^7}$ that goes to $\mathrm{4s^0 3d^8}$ configuration, $\ce{Ni}$ has a $\mathrm{4s^2 3d^8}$ that goes to a $\mathrm{4s^0 3d^9}$ configuration and $\ce{Cu}$ has a $\mathrm{4s^2 3d^9}$ that goes to a $\mathrm{4s^0 3d^{10}}$ configuration (Lang, J. Chem. Ed., 2003).  In the Long paper, the authors explain the increase of IE at iron is due to "electronic structure" and I would like to know exactly what about the electronic structure it is that causes iron to have a high IE than expected based on a general periodic trend.

I was thinking that adding an electron to the $\mathrm{3d}$ orbitals might be energetically favorable enough to lower the ionization energy of $\ce{Ni}$, $\ce{Co}$ and $\ce{Cu}$, so rather than iron having a high ionization energy the surround metals just have a low ionization energy because the ion is stabilized by having another $\mathrm{d}$-electron added? Does that sound reasonable?

Another thing I was thinking was that iron is the first transition metal to have a paired $\mathrm{d}$-electron, could that effect the shielding and increase the ionization energy?

"I mean "abnormal" in that it defines a general trend" makes no sense to me at all. Did you mean "defies"?  Take a look at the chart here: https://en.wikipedia.org/wiki/Ionization_energy.
By my count, for the s-block elements the "trend" always holds.  For the p-block elements, it holds in 5/6ths of the cases. And for the d-block, I can't tell from the chart, but claiming it is increasing for them seems agross exaggeration to me; I'd say they might have a very slight tendency to increase, but they're so close to being flat-lined, that it seems to be not a useful claim. If you count the exceptions to the rule, and you expect your "rules" to hold at least 90% to 95% of the time (this isn't the Social Sciences, after all), then it is a just plain lousy rule. Nothing unusual about that: I think the Periodic Table is best when used to "predict" Atomic Number (they increase moving right and down, without exception), and so untrustworthy with the other periodic trends as to not be something in my "mental toolbox". Use it at your own peril. (The at. no. "trend" was a joke, of course)


How can I calculate the charge distribution of a water molecule? I would like to obtain the 3D charge density of a water molecule for plotting purposes (i.e. not necessarily super-accurate). I know this is a tough problem, but I'd like to give it a try. Since I'm currently using python, I've identified PyQuante and PySCF, and gather that I should learn how to run density functional theory (DFT) calculations. Before I do that, I'd like to find out if there is something else out there I should use instead. Multiple suggestions are welcome.

I've installed PyQuante, which according to it's README.txt may be more accessible to understanding.

For example, the color plots in this answer give a hint of what I'm talking about, though I'll ultimately plot the field strength contours and some electric field lines.

I'd like to obtain a large table of values (or expressions/expansions?), so I can create and work with surfaces and field line plots myself.

Additionally is it possible that there exist supplementary data in published articles. Where I might find that? Although this is about existing data, I actually do plan on putting in the time to have at least a minimal understanding of the calculation itself. If DFT is it, then I will be spending time studying that.

Two suggestions: MolCalc: Calculate Properties > Polarity and Solvation gives you a 3D surface plot with the electrostatic potential superimposed.

Another suggestion is Avogadro/GAMESS/MacMolPlt. See for example here

If you only need the electron density distribution of a water molecule, it doesn't make much sense for you to leanr DFT method. Although I don't think there will be such data out there, it's pretty easy to run such a calculation for people who do DFT calculation regularly. It will take several minutes. My suggestion is that you find someone to help you with the calculation. The calculation could give you the data describing electron density in the 3D space. Then you can play with the data by yourself.


What is the mechanism of chlorosulfonation of benzene? The above reaction looks very similar to a conventional sulfonation reaction but if the reaction pathway is analogous to sulfonation I do not see how the $\ce{SO_2Cl^+}$ would form. Surely $\ce{Cl^-}$ would leave rather than $\ce{H_2O}$ in the formation of the electrophile? Please help me find a mechanism.

Also, I would like to approach questions, such as this one, like I would a puzzle rather than calling upon my memory. Is there a series of steps that I should follow? (i.e: 1. Identify electro/nucleophile, 2. Identify acidic protons ...).

Rob, IMO this is a difficult problem, so don't feel bad about not "seeing" the solution right off.

Chlorosulfonic acid (or chlorosulfuric acid) is both difficult and hazardous to work with so it's reactions aren't discussed all that frequently.  It can also give rise to different products at different temperatures.  At higher temperatures it generates $\ce{SO3}$ which will behave as an electrophile and react with aromatic compounds to produce sulfonation products.

image source

At lower temperatures (your case), chlorosulfonic acid reacts by way of the following equilibrium (who knew this?)

$$\ce{3 ClSO2(OH) <=> SO2Cl^{+} + 2 SO3Cl^{-} + H3O^{+}}$$

The equilibrium generates $\ce{SO2Cl^{+}}$ which is the electrophile of interest in your reaction.  $\ce{SO2Cl^{+}}$ will react with an aromatic nucleus to produce the corresponding sulfonyl chloride by way of a typical electrophilic aromatic substitution reaction.

If you'd like to read more about the complexities of reactions involving chlorosulfonic acid see here starting around page 11.

The difficult part here was identifying the source of the electrophile. Looking at the final product you get an idea of what the electrophile is, but how is it produced?  Sometimes you have to do a little googling to sort things out.

Again, IMO this was a tough one.

The answers provided so far are actually incorrect.

First, the benzene reacts with chlorosulfonic acid to produce benzenesulfonic acid and HCl.  The chloride acts as a leaving group, not the OH.  Second, the benzenesulfonic acid reacts with a second equivalent of chlorosulfonic acid to produce benzenesulfonyl chloride and sulfuric acid.  Thus, in net you need two equivalents of chlorosulfonic acid to convert benzene to benzenesulfonyl chloride.

Edit update:
I should add that the arrow-pushing mechanism of sulfonation with SO3, provided above by Ron, is actually incorrect too.  Modern research has shown that the traditional organic chemisty mechanism of electrophilic aromatic substitution, in this case, is actually not true.  The intermediate (sometimes call the sigma complex or the Wheland intermediate) in which SO3 is bound to the arene leaving a positive charge on the arene, actually does not exist.  Instead, modern computational research, as well as kinetics studies that are much older, indicate a termolecular pathway.  Typically, three molecules are involved.  In concentrated SO3 or oleum, two molecules of SO3 form a transition state with the arene... it is a concerted mechanism.  In sulfuric acid, the termolecular complex involves the arene, sulfuric acid, and SO3.  In fact, sulfonations with higher SO3 concentrations (e.g. oleum) actually don't produce the sulfonic acid product; they first produce sulfonic anhydrides which are later hydrolyzed during work up.

For whoever downvoted me, go look up the various papers by Cerfontain for the 2nd order in SO3 data, and look up various recent computation papers (Shi, Computational Theoretical Chemistry, 2017, 1112, 111; or, Schaefer, 2016, Acc. Chem Res., 49, 1191).  You will see that the simplified world of organic chemistry's golden age is unfortunately not perfect or correct.

The mechanism provided for chlorosulfonation in which ClSO2+ is generated... this is very hand wavy.  You should at least provide a reference justifying this sort of thing.  It's hard to believe that equilibrium (if even real) out-competes the equilibrium of ClSO3H with SO3 and HCl.

Response to Ron:
For the chlorosulfonic acid mechanism, the ref your provided is a review book which links to a journal from the Russian Journal of Organic Chem from 1975.  I unfortunately can only access the abstract because that year does not appear to be electronically uploaded and I don't have access to the physical copy of the journal.  Anyways, the abstract indicates kinetics that are third order in chlorosulfonic acid.  This is great and pretty neat, but does not necessarily support the crazy ion-pair mechanism that is proposed.  A lot of these crazy old mechanisms from the 'golden era' of organic chemistry (40s to 80s approx) are often wrong.  Good data kinetic data was often obtained, but the mechanisms are often later found (often with computational support) to be imprecise.

One of the best examples is SO3, which you have listed above.  The mechanism of sulfonation with sulfuric acid varies depending on conditions due to the wide variety of species present (see the ref you listed; the chlorosulfonic acid book).  But, no one does sulfonations this way, because you need forcing conditions to drive the reaction to completion; people do them in highly concentrated sulfuric acid or with oleum.  Under these conditions SO3 is often the sulfonating agent, or pyrosulfuric acid is, which is what you get when you react sulfuric acid with SO3 (e.g. oleum).  Under these realistic conditions, the reaction is typically termolecular.  Please read the refs I listed.  The intermediate you listed above is actually too far uphill energetically, and modern academics have abandoned it.  This is why I am extremely doubtful of your listed chlorosulfonation mechanism, and why the SO3 mechanism you listed is not currently supported.


How do I compare the C=C bond lengths in this cumulene? How do I arrange these chemical bonds in an increasing/decreasing order based upon their lengths?

I understand that the bond (a) will be longer than (b) because (a) is a bond between sp² hybrid and sp hybrid carbon atoms, as opposed to (b) which is a bond between two sp hybrid carbon atoms. But how do we compare bond (c)?

Label the double bonds (a), (b), and (c), in correspondence with the image provided in the question. Label the carbons in the cyclopropane ring $\ce{A}$, $\ce{B}$, and $\ce{C}$, with $\ce{A}$ being the carbon with the double bond (c).

At $\ce{A}$, the usual $\ce{sp^2}$ hybridization would lead to bond angles of $120^\circ$, whereas a lack of hybridization (i.e. simple $\ce{p}$ orbitals) would lead to bond angles of $90^\circ$. The latter is not usually favored, but the high ring strain in cyclopropane leads to carbon-carbon bonds $\ce{A-B}$ and $\ce{A-C}$ having greater $\ce{p}$ character than is usual. The double bond (c) at $\ce{A}$ thus receives more $\ce{s}$ character, and is stronger than a usual $\ce{sp^2{-}sp}$ bond would be (i.e. (a)).

We conclude, therefore, that the bond strengths are (b) > (c) > (a), and hence that the bond lengths are (a) > (c) > (b).

You can draw Hyper conjugating Structures to compare bond length. More the α Hydrogen of alkene, more are the Hyper conjugating structures, more is the delocalization of π-bond and more is the bond length (since double bond character decreases, bond length increases).

Thus bond length C would be greater than A.

As there are 4 α hydrogens around C, whereas '0' α hydrogens around A.

Therefore the order of bond length would be C>A>B


How to measure saturated salinity and mineral levels for flaked sea salt? I am a sea salt enthusiast looking to make the perfect flakey sea salt. I start with sea water and boil it down to a thick brine. Then with less heat I reduce the brine more slowly and inverted salt flake pyramids are made. I am looking for tools/instruments that will help me measure/understand the content of the brine (mineral content and salinity). Sometimes I get perfect flakey salt pyramids like the photo attached and sometimes I get I get more chunky results. The temperature of the brine is always the same but I get different results at times. This leads me to think that its the content of minerals in the saturated brine solution.

Are there chemistry tools/instruments I can use to understand the actual chemical/mineral makeup of the brine that is turning to salt.

Why not use salt (NaCl) bought from a store, or even better at a defined purity grade from a chemical vendor?

When you use sea salt, you get lots of other ions but more importantly you get colloidal particles, organic substances, and other compounds that will affect the crystallization. Chunky results sounds to me like a colloid particle that started crystallization in all directions at once.

If you want perfect crystals, do it perfectly, i.e. with a pure solution, in a clean and new glassware or polished stainless container. If you want perfect crystals you have to reduce the new-age-ness of it ;)

As for understanding, nobody fully understands crystallization, but a lot of things are known. There are books on this topic, check amazon and search for "crystallization". As for instruments, x-ray fluoroscopy and diffraction wouldn't be amiss but you will get very far with an optical microscope, I know I can watch crystals through a microscope for hours and still find something interesting. Barring XRF, XRD and microscope, and within a household budget there are salinity measurement probes and conductance measurement probes which will be able to tell you something of the brine such as it is.

There's no good way to quantitatively or semi-quantitatively analyze for Na, Ca or Mg at home. You could investigate Ion Selective Electrodes and see if they'll give you the desired accuracy (ThermoFisher has Cl, Na, & Ca I.S.E.s which generally can be simply plugged in to a pH meter and then calibrated). Two other things worth mentioning: you need to filter ocean water - bacteria, algae, and all sorts of other organisms will be present in varying amounts, not to mention silt, and other colloidal inorganics. There also could be differences in dissolved gasses present. So, you need to standardize your handling of the water - from where you get it, to where and how you store it, to how you filter is, and how you process it. Don't assume that mixing it - or shaking it won't affect it. Try to figure out what you're doing to get the crystals you want. It's unlikely, imho, that the Na, Ca, Mg content varies much (although you should treat my opinion as a layman's - I've no direct experience with sea water variability.) I AM assuming you know enough to understand what near-by rivers and recent rains and obviously wind and tidal currents might do to change the composition of the surface water (not to mention air temperature and sunlight). Anyway, one of the cheapest easiest ways to determine dissolved solids is using a specific gravity bulb. (as long as temperature is held constant) You simply float the bulb into the solution and read from the scale what the density is. Finally, what container are you heating this in? Be aware that aluminum should not be used! And I'd prefer glass to stainless steel, but as long as there's no visual signs of corrosion, ss is probably ok. Also, while I'm sure you want a "natural" product, seed crystals of store-bought sodium chloride might make the crystallization more reproducible. Chemists are taught to scratch the glass side of a beaker to start crystallization. Scratches are also a site which may "catalyze" crystal formation, and since any semi-commercial (?) process can't use a new polished pan for each batch, maybe scouring the pan consistently before each batch is the way to go. HTH, these are mostly random thoughts of an industrial chemist kibitzing in an area he has little experience in.


Understanding reason behind existence of chemical equilibrium Many chemistry reactions are reversible in nature.
In such reactions initially reactants react to form products but after some time products also react to give reactants back again. And an equilibrium is established where forward and backward reactions occur with equal rates.

If products are more stable, why will they react again to give reactants?

It is best to think of reversible reactions statistically

A good understanding about equilibrium can be had by thinking about the detailed molecular level processes involved at the molecule level.

Take, for example, the simple equilibrium between liquid water and water vapour in a closed vessel. At room temperature, the equilibrium clearly favours liquid water. But there is always some water vapour present (that's what we call humidity in the wider atmosphere). Why is there any vapour if the equilibrium favours liquid? The reason is that the water molecules don't all have the same amount of energy. some are travelling fast and some some are slow. A small percentage of the fast ones can escape the liquid and become vapour (ie gas-phase water molecules not liquid phase water molecules). In the gas above the liquid, the same is true. Not all the gas molecules have the same speed. Some are travelling slowly and, if they hit the liquid surface, stay in the liquid. Both processes are dynamic and are happening fairly fast. So molecules are constantly exchanging between the vapour and the liquid.

Equilibrium is not reached when everything stops: at room temperature there is a lot of kinetic energy shared out among the molecules of the system. Equilibrium happens when the rates of evaporation and liquefaction are the same. So the number of molecules per second going into the vapour from the liquid equals the number going the other way.

The same is true for other, more complex, reversible reactions. Equilibrium is not stasis but an average state resulting from dynamic reactions in both directions.

This point of view makes it clear why, even when one product is far more stable, the reverse reaction can still happen.

It is no use comparing the stability of the reactants and products.

Maybe you would better understand with an experimental example. Suppose you prepare $\pu{10 mL}$ of an alcoholic solution of cobalt chloride hydrate $\ce{CoCl2.6H2O}$. This solution is blue due to the formation of $\ce{CoCl4^2-}$ according to the following equation:
$$ \ce{2 CoCl2.6H_2O -> [Co(H2O)6]^2+ + [CoCl4]^2-}$$
The ion $\ce{[Co(H2O)6]^2+}$ is pale pinkish, and the ion $\ce{CoCl4^2-}$ is dark blue. So the color of the whole solution is blue. Now you add two or three drops of water. You observe that the solution turns pinkish, because of the following equation, which I will show that it is an equilibrium:
$$\ce{[CoCl4]^2- + 6H2O <=> [Co(H2O)6]^2+ + 4 Cl^-}$$

This equation is written in the exothermic sense.

Now if you heat this pink solution to $\pu{60 ^\circ C}$ or $\pu{70 ^\circ C}$, the solution turns blue. $\ce{[Co(H2O)6]^2+}$ is transformed into $\ce{CoCl4^2-}$. The reaction goes to the left hand side. This is a consequence of the Le Chatelier principle: Heating a system favors the endothermic side, which is also the side where the entropy is maximum.

Cooling the hot blue solution will change its color to pink again, because $\ce{[Co(H2O)6]^2+}$ is favored since cooling favors the exothermic transformation. This can be repeated by successive heating and cooling.


Is the nomenclature "primary", "secondary", "tertiary", and "quaternary" carbon atoms limited to sp3 carbon atoms? In organic chemistry we can classify carbon atoms as primary, secondary, tertiary, or quaternary based on the number of additional carbon atoms bonded to the carbon atom of interest. The definitions in the textbook I teach from (Klein, 1st Ed. Wiley, 2011), as well as in several others I have, and from from several locations around the web, are:

Primary $(1^\circ)$ carbon atom - bonded to one other carbon atom, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{H2CH3}$
Secondary $(2^\circ)$ carbon atom - bonded to two other carbon atoms, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{H(CH3)2}$
Tertiary $(3^\circ)$ carbon atom - bonded to three other carbon atoms, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{(CH3)3}$
Quaternary $(4^\circ)$ carbon atom - bonded to four other carbon atoms, e.g. ${\bf \color{red}{\ce{C}}}\ce{(CH3)4}$

Is this labeling system limited to $sp^3$-hybridized carbon atoms? I do not see anything in the definitions that exclude $sp^2$- or $sp$-hybridized carbon atoms. Nor have I seen or read anything that explicitly excludes these atoms. However, the images in textbooks and around the web all show and discuss only $sp^3$-hybridized carbon atoms in this context. Are the following valid?

The central atom in propene $\ce{CH3 -}{\bf\color{red}{\ce{C}}}\ce{H=CH2}$ is a $2^\circ$ carbon atom because it is bonded to two other carbon atoms.

The carbon atoms in acetylene $\ce{HC#CH}$ are $1^\circ$ carbon atoms because they each are bonded to one other carbon atom.

The ipso carbon in toluene $\ce{C6H5CH3}$ (the carbon atom where the $\ce{CH3}$ is attached) is a $3^\circ$ carbon atom because it is bonded to three other carbon atoms.

To best of my knowledge, the classification primary to quaternary is used for discussing reactivity of hydrocarbons, i.e. alkanes. There the number of substituents matters, as it governs carbocation stability, steric hindrance, etc...

In all other cases, the electronic structure of the molecule rules the reactivity. As you mentioned alkenes, alkynes or aromates, there it makes no sense to discuss such subtle effect as number of neighbouring carbons, when you have $\pi$-system at hand.

The IUPAC Gold Book entry for alkyl groups supports this distinction:

  The groups $\ce{RCH2}$, $\ce{R2CH}\ (\ce{R} ≠ \ce{H})$, and $\ce{R3C}\ (\ce{R} ≠ \ce{H})$ are primary, secondary and tertiary alkyl groups, respectively.

the correct definition is
"a tertiary carbon atom has 3 bonds to ANY number of carbon atoms"

the carbon atoms in acetylene are tertiary, not primary (this came as a shock to me too the first time i heard it).

the reason for this is, you can further improve the classification by adding the hybridisation of the C atom, so in acetylene, the carbon atoms are both SP tertiary, while in isobutane, the central carbon atom is sp3 Tertiary.

if we would follow the flawed definition "a tertiary carbon is bound to 3 other carbon atoms" , then adding the hibridisation would be useless, as it would not alter the classification, which is why this definition is flawed.