How do we get NaClO3 in the reaction between Cl2 and NaOH? From problem no. 4080 in a Russian example exam set:

$$\ce{6NaOH + 3Cl2 -> NaClO3 + 5NaCl + 3H2O}$$

How do we get that $\ce{NaClO3}$? I can't seem to guess. I found this explanation on YouTube, but there the products are different:

$$\ce{Cl2 + 2NaOH -> NaCl + H2O + NaClO}$$

In both reactions, there is twice as much sodium chloride as there is chlorine. How do the two additional oxygens connect to the chlorine?

P.S. A related question - thanks, Nilay Gosh.

I find that when I'm faced when problems having to do with why the products are formed I make lewis structures.

if you make a lewis diagram of the $$\ce{NaClO3}$$

First off, you will see that a single Cl will be bonded to three oxygen molecules whom all carry full valance shells which were striped from the hydroxide ions in the reactants. (Remember that in solution the Na and OH will be around as free spectators.)

From the structure you will see the reason why there was a formation of the three waters. Since there were three oxygen stripped from the hydroxides and then recombination with more OH to make a reaction like this:

$$\ce{OH- + H+ -> H2O}$$

Also notice that on the products side for every molecule of NaClO3 made there will be a molecule for NaCl made. From there you can speculate that when the Cl-Cl bond breaks three oxygen atoms and a sodium atom will surround one of the Cl while another is tied to a Na molecule simultaneously.


Valence electrons in helium? I am having trouble understanding how valence electrons are counted in Helium.

My references are:

Here where it appears that Helium has 0 valence electrons because the shells are finished
Here where it appears that Helium has two valence electrons (which is what I thought)

How many valence electrons does helium have?

Helium has two electrons in total, and according to the aufbau principle, it adopts the electronic configuration $\mathrm{1s^2}$. This means it has two electrons in s orbitals with a principal quantum number of 1.

The last (and only) level of helium's electronic configuration is $\mathrm{1s^2}$, and therefore $\ce{He}$ has 2 valence electrons.

The outermost shell of an atom can accomodate a max of 8 electrons. It was observed that the atoms of elements,having a completely filled outermost shell show little chemical activity. In other words their combining capacity is 0. Of these inert elements, the helium atom has 2 electrons in its outermost shell and all other elements have atoms with 8 electrons in the outermost shell. Their combining capacity of the atoms of other elements,that is their tendency to react & form molecules with same or different elements,was thus explained as an attempt to attain a fully-filled outermost shell.


Permanent cloudy colloid with water I need to make a cloudy white liquid where the particles won't eventually settle out. It will be exhibiting the Tyndall effect for a very long time.

I'd use milk, but I don't want it to become rancid. I've tried white inks and fabric softener, but both eventually tend toward the bottom half of the container. I think it's a matter of particle size.

What substance would hit the sweet spot - small enough to stay dispersed, large enough to scatter light, and not reactive?

Starch in water might work, perhaps with table salt or other bacterial/fungal growth inhibitor. Follow directions in most recipes: mix a bit of starch in cold water and then add gradually to hot water to make a more stable "soup".

BTW, blasting a colloid with ultrasound (e.g. from an ultrasonic humidifier) homogenizes it and may delay separation.

As a first approximation, there are no stable colloids. I suggest you look for kinetically stable (i.e. metastable) colloids. The academic community is divided between those who insist that no colloid system is stable and those who offer theoretical reasons why certain systems are thermodynamically stable. I used to believe no colloids are stable (thermodynamically) but changed my mind (based on the scientific literature at the time). Unfortunately, that was decades ago, and I'm not able to give you chapter and verse rationale why certain colloids may be stable. To be "stable" the particle needs to both not fall (match density to its media so that thermal motions (and mechanical vibrations, possibly) overwhelm any tendency to fall) and more difficultly have a lower free energy in the highly dispersed state, specifically it must be electrostatically repulsive towards other particles and have virtually no tendency towards Ostwald Ripening. This limits your options to organic (charged) polymers and if you can "structure" the water, that's all to the better. If I recall certain acrylates were able to be synthesized into particles fitting those requirements, but it was a long time ago. Permanent is a long time, I really doubt you need permanent. I'd bet 50, 100, or maybe 200 years would do just fine.


Bond angles in NH3 and NCl3 The bond angle in a molecule is inversely proportional to the electronegativity of the surrounding atom if the central atom is same. This also happens with $\ce{NH3}$ and $\ce{NF3}$, as bond angle in $\ce{NH3 > NF3}$.

Why then is the bond angle in $\ce{NCl3 > NH3}$?

As Tan Yong Boon stated, it is Bent’s Rule with which we can explain the lower bond angle of $\ce{NF3}$ when compared to $\ce{NH3}$. The rule as stated by Henry Bent:

  Atomic s character concentrates in orbitals directed towards electropositive substituents.

Fluorine is more electronegative that hydrogen, and the $\ce{N−F}$ bond would have greater p character than the $\ce{N−H}$ bond. And more s character leads to large bond angles. Thus, the bond angle is greater in $\ce{NH3}$ than in $\ce{NF3}$.
Now, consider $\ce{NCl3}$. Clearly, $\ce{Cl}$ atom islarger in size than the central atom, nitrogen. Hence the higher bond angle here is due to the steric crowding caused by $\ce{Cl}$ atoms.(More pronounced than the electrongativity of $\ce{Cl}$ atom). They repel eachother and hence bond angle increases. This is not the case with small $\ce{H}$ atoms around the nitrogen in $\ce{NH3}$. Therefore, the $\ce{Cl-N-Cl}$ bond angle of $\ce{NCl3}$ is greater than $\ce{H-N-H}$ bond angle of $\ce{NH3}$.

Some search on bond angle data for NCl3 and NH3.
Many places, it is mentioned that  bond angle of NCl3 > NH3 but no references are given. I looked around to find some experimental data on this.
To my search bond angle in NCl3 (107) < NH3 (107.8).
Here are the references for my analysis.
In case of NCl3:
(1) http://www.chm.bris.ac.uk/motm/ncl3/ncl3js.htm
(2) https://en.wikipedia.org/wiki/Nitrogen_trichloride#cite_note-2
In case of NH3 See book: Chemistry of Elements by N N Greenwood and Earnshaw (second edition)


Is the boundary of an atom well defined? That question might seem weird, but that's what I have been imagining lately; going through Bohr's theory on the hydrogen atom. He clearly defines the radius of an orbit with an exact mathematical expression. But what do they mean when they say any atom (say $\ce{Na}$) has a radius $r$. Are we to consider atoms as solid, rigid spheres? If that's the case, what about water?

If all the atoms are defined as solid, rigid, spherical balls, how come an aggregation of so many solid objects be a liquid? How can it flow? So, in a nutshell what is the boundary of an atom made up of, or even this: does it have a defined boundary?

This troubles me a lot. I think maybe the modern atomic model has dealt with this problem. I haven't read it in detail, and I would be highly thankful to any references.

No, the boundary of an atom is not well defined. According Quantum mechanics the electron density around an atoms decreases exponentially, thus getting asymptotically close to zero.
Considering atoms as rigid spheres is only an approximation. The criteria used to decide on the actual value of the atomic radius is arbitrary. You can for example choose the expectation value $\langle\hat r\rangle$ of the electronic wavefunction. For the Hydrogren atom, this results in the bohr radius $a_0$. But you might as well choose the maximum of the electron density ($\frac{3}{2}a_0$ for $\ce{H}$), or something based on experiments.
Now the aggregate state is something completely different and not directly  related to atomic radius and whether they are rigid sphere or not. Solid means the molecules/atoms of a substance are fixed in their positions. They can vibrate, but do not change their average position with respect to each other. In gases molecules may move freely. And a liquid is something in between, you have some order if you look at a small group of molecules, but not on a larger scale as for solids.

Seriously? Bohr's Atomic Model? It was known by 1920 that that model failed (badly) to explain observational data. That is, it hasn't been taken seriously for nearly 100 years. Perhaps you should invoke the 4 humors and reason from there. (The only atom the the theory did a pretty good job for is hydrogen. One down, 91 elements to go....)  There are lots of different models which are useful for different purposes. One thing we know about all of them, is they're all wrong. So, since the "right" model isn't known, picking a "best" model requires you to know what your purpose is, what the context is. The issue is deciding which model is most useful, not which model is correct. That turns out to depend on context. What I was taught as an undergraduate chemistry major (almost 50 years ago) was that an electron bound in an atom (or molecule, or ion, or radical) is best thought of as a probability wave. This means that the location of the electron is "smeared out" in space and that for any distance, r, from the nucleus (center), there is a non-zero probability of the electron being at that distance. This means, to answer your question, that atoms do not have "hard surfaces". The reason why some Physical Chemistry textbooks still cover the Bohr Model is because it helps the student transition between the physics of everyday objects to the non-intuitive world of quantum mechanics.  The best book for you depends on what your mathematical knowledge is, how much effort you want to spend, and how detailed you want to be. It is a mature science, and many popular books have been written about it. Concepts such as "radius" are mostly nonsensical at the quantum scale. For instance, the radius of an atom depends on how you measure it. Consider the simple case of shooting electrons at it. Then shooting protons at it. Then shooting neutrons at it. Then microwaves. Then (visible) light waves. Then x-rays. Each will give you a different radius. In some ways it is similar to the question of how long is a coastline. If you measure it with a 100 meter cable, you'll get one answer (ignoring differences in sea level, tides, wind, rain, etc.), if you measure it with a meter stick, you'll get a larger value. Measure it centimeter by centimeter, and it will be even larger, and just imagine if you could measure it atom-by-atom, it would be very difficult to determine where the liquid began and where the land ended. Coastline is a macroscopic concept, with little meaning at the micrometer scale (let alone the nanometer scale). Very much like your concept of radius. (Although to be fair, we "steal" the word and use it in various contexts which often have similarities to its macroscopic use.)


Which process would separate the mixture of toluene, cyclohexane and water? toluene: $\ce{C6H5-CH3}$
cyclohexane: $\ce{C6H12}$

I think that answer would be:

Toluene and cyclohexane do not dissolve in water (since they are nonpolar), so I would first separate the water from the mixture with a separator funnel.

A mixture of cyclohexane and toluene would be separated by fractional distillation. Cyclohexane, which has a lower boiling point than toluene, would be eliminated from the mixture.

Is there anything wrong in my answer? Would there also be anything to add?

Distillation
The most effective and easiest way to separate the mixture of toluene and cyclohexane is simple distillation. Cyclohexane has boiling point is about 81°C, and toluene’s boiling point is 111°C.
References:

http://origin-www.thermoscientific.jp/content/dam/tfs/ATG/CAD/CAD%20Documents/Product%20Support%20Bulletins/NMR%20Support%20Documents/pS45-pS80-Simple-Distillation-of-Cyclohexane-and-Toluene.pdf
https://www.sas.upenn.edu/~dbalmer/chem502/Labs/lab2_distillations/balmerexp2.pdf


Formation of Hydrogen - from a proton and an electron If I place a proton and an electron close to each other, can they form an atom of hydrogen? Why or why not? Please explain the scenarios where this is not likely/unlikely.

Personally I doubt it, since the electron would have to spin around the proton to call it an atom.

Can it be well-explained by classical mechanics?

How does quantum mechanics explain this sort of a possible bonding? I mean,an electron and a proton can't collide with each other because they would have had infinite attraction between them (that is they would act like magnets), and we know that this does not happen.

Is there something in QM that defines a minimum distance of some sort that is relevant to the proton-electron forming hydrogen?

Both in classical and quantum mechanics you have to get rid of a (rather large) amount of energy, i.e. the difference in potential energy from zero (=infinite distance) to the bound state (= e.g. Bohr radius), also called the binding energy.

Nature has no way to do that without a third particle taking the excess energy with it. In quantum mechanics there is the additional problem with the spin, but also classically that doesn't work if you assume elastic interaction of proton and electron.

So the answer is no.

(Inelastic interaction requires a zoo of additional assumptions, which I let fall victim to Occams razor here. Inverse photodissociation is a possibility, if overall energy and momentum are conserved.)

yes

This is exactly what happens when hydrogen is ionised in the gas phase and then returns to an unionised state.

When it is less likely to happen is when the proton has very high energy (eg. having been emitted by nuclear decay) such high energy protons are  either absorbed by proton capture into another atomic nucleus or the lose their energy by inelastic collisions until they are able to capture an electron.

In the same way, most of the helium on earth was formed by alpha particles from nuclear decay capturing electrons.


Why do atoms need 8 electrons to stabilize? As the title says. I have surfed all of the net but could never find the answer to this question. Why do atoms need 8 electrons to stabilize? I mean why not 7 or 5 or 10 electrons? Why specifically 8? And what does stabilization of atoms even mean? Are the atoms going to burst without 8? Why do they need 8 electrons?

This is a question which is not often discussed in chemistry and I need an answer which really makes any sense. Not just `it is because the atoms need to fill their octet.

The valence orbitals of atoms are composed of suborbitals (s and p)  there is 1 s suborbital which is spherical and can hold 2 electrons (one with up spin and one with down spin). There are 3 p suborbitals which are dumbbell shaped (look like two balloons tied together at the ends) and align along the x,y&z axies and hold a total of 6 electrons (2 per axis, 3 with up spin and 3 with down spin).  Since an atom is energetically satisfied when all of the electrons are paired and 2+6 = 8, an atom must have 8 electrons in its valence shell to pair all of the electrons.

The octet rule does not apply to all atoms though, for transition metals and later elements there is a d suborbital which can hold an addition 10 electrons, meaning that now an atom may have 18 electrons in its valence shell to be satisfied.  This allows for materials such as metal complexes and the sulfate poly-atomic-ion to exist despite having more than eight electrons in the valence shell.

Please excuse my rather layman reply, my specialism is not in chemistry or physics, but I felt this question is asking more why than how.

Why do atoms need 8 electrons to stabilize? I mean why not 7 or 5 or 10 electrons? Why specifically 8?

I suspect this more has to do with the fact that force is equalised. Even numbers. An odd number would suggest an unpaired electron which would suggest an unstable state. Think of it like having two cars pushing against each other evenly, or two equally attracted magnets. I don't think anyone knows why it's specifically 8, as opposed to 6 or 4 or 2.

And what does stabilization of atoms even mean?

It means they are in a low energy state. A high energy state is where an atom holds a potential to react. Take for example sodium - when introduced to water it reacts violently and explodes. That is because, in simplistic terms, it transfers electrons between sodium and water. When an atom achieves 8 electrons, there's no more electrons for it to gain or lose (IE if it was less than 8 it would have to gain additional electrons from some other material, and more than 8, it could lose some - essentially a reaction would occur).

So when we say stabilisation has occurred, we are essentially saying that conventionally the atom won't react. It has become stable.

Are the atoms going to burst without 8?

The atom itself won't burst, per se, but like in the example, the sodium will.

Edit: Not sure why the downvotes. No feedback, no correction. Electron transfer produces reactions:
https://en.wikipedia.org/wiki/Electron_transfer
https://en.wikipedia.org/wiki/Redox


Why does ligand substitution occur when OH- is a better ligand than NH3? $$\ce{[Cr(H2O)3(OH)3](s) + 6NH3 -> [Cr(NH3)6]^3+ + 3H2O + 3OH-}$$

When excess $\ce{NH3}$ is added to triaquatrihydroxychromium(III), why does ligand substitution occur when $\ce{OH-}$ is a better ligand than $\ce{NH3}$?

Wouldn't the 3 $\ce{H2O}$ molecules be deprotonated to $\ce{OH-}$ and therefore form $\ce{[Cr(OH)6]^3-}$?

This also seems to occur when excess $\ce{NH3}$ is added to tetraaquadihydroxocobalt(II):

$$\ce{Co(H2O)4(OH)2(s) + 6NH3 -> [Co(NH3)6]^2+ + 4H2O + 2OH-}$$

This is an equilibrium reaction. Using LeChatelier's principle, you can figure you that if you increase the concentration of NH3 sufficiently, you shift the equilibrium so far to the right that you get the hexamine as the main product. At neutral pH in water, the concentration of hydroxide is quite low (100 nanomolar). You would expect this reaction to be pH-dependent, and of course depend on the concentration of ammonia in solution.


Synthesis of chloroquine from 4-amino-7-chloroquinoline Can we get chloroquine from 4-amino-7-chloroquinoline (7-chloroquinolin-4-amine) by adding 4-chloro-N,N-diethylpentan-1-amine $(\ce{CH3CHCl(CH2)3N(Et)2})$ ?

My intuition tells me that the nitrogen in the ring is more nucleophilic than the exocyclic amine. A quick-and-dirty electrostatic potential calculation using B88LYP/6-31G(d) on the PM3 geometry shows a large electron density (blue and violet) on the ring nitrogen and lower density (yellow) on the exocyclic amine:

Are there conditions that might work? Yes. Should you try it, if you need that molecule? Yes. Will it be difficult? Yes.

yes, it is quite possible as aromatic anines nitrogen being moore nucleophilic further the secondary chloro reacts easily, the reaction may either be SN1 or SN2. I have done couple of reaction of this type, if you need help feel free to contact me.


Why does sp hybridised orbital have more directional character than p orbital? It states so in my textbook. But since s orbital is non directional, shouldn't sp hybridized orbital be less directional in character than p orbital?

Ignore everything else in image , this image is just for shape comparison.

As you can see in image that so orbital is a lobe with unsymmetrical shape this leads to its more directionality.

P orbital possess equal Electron Density on both sides of the orbital , but in case of sp orbital electron density isn't equal and that's the reason why sp is more directional than p .

That's also the reason for their different Electronegativity.


How to determine the concentration of a base through indirect titration? $5.267~\mathrm{g}$ of $\ce{Na2CO3}$ was dissolved in $250.00\ \mathrm{mL}$ of water. $10.00\ \mathrm{mL}$ of this solution was titrated with $\ce{HCl}$. The end-point occurred at $21.30\ \mathrm{mL}$. This acid was titrated with $25.00\ \mathrm{mL}$ of $\ce{Ba(OH)2}$. The end-point occurred at $27.10\ \mathrm{mL}$. Calculate the concentration of the $\ce{Ba(OH)2}$.
  [Answer: $0.1013~\mathrm{M}$]

Using the mass they gave me ($5.267~\mathrm{g}$) I worked out the moles of $\ce{Na2CO3}$ which was $0.0497\ \mathrm{mol}$.
I then worked out the concentration of $\ce{Na2CO3}$ using the moles I just worked out and the volume they gave me ($0.25~\mathrm{L}$) and I got $0.1988~\mathrm{M}$.
The question says that only $0.01~\mathrm{L}$ was used, so I worked out the number of moles this would be which was $0.001988\ \mathrm{mol}$.
Using stoichiometry, I multiplied this value by two to get the moles of $\ce{HCl}$ that would have reacted with this, which was $0.003976\ \mathrm{mol}$.
It says that the end-point occurred at $21.30~\mathrm{mL}$, which if you subtract the volume of $\ce{HCl}$ which was already there, I worked out that $11.3~\mathrm{mL}$ or $0.0113~\mathrm{L}$ of $\ce{HCl}$ must have been used.
From here, I worked out the concentration of $\ce{HCl}$ (using $0.0113~\mathrm{L}$ and $0.003976\ \mathrm{mol}$) and got $0.35186~\mathrm{M}$.
It says that for the next reaction, $25~\mathrm{mL}$ of $\ce{Ba(OH)2}$ was used and the end-point occurred at $27.10~\mathrm{mL}$. From this information, I worked out that $0.0021~\mathrm{L}$ of $\ce{HCl}$ must have been used in this reaction.
By knowing the volume of $\ce{HCl}$ used, I worked out the amount (moles) of $\ce{HCl}$ used in the reaction, using the previously found concentration. I got $0.0007389\ \mathrm{mol}$.
Using stoichiometry, I worked out the amount of $\ce{Ba(OH)2}$ which must have reacted with this many moles of $\ce{HCl}$. I divided the $\ce{HCl}$ moles by $2$ and got $0.00036945\ \mathrm{mol}$.
By using the volume of $\ce{Ba(OH)2}$ used which they give me ($0.025~\mathrm{L}$) and the moles I just found ($0.00036945\ \mathrm{mol}$) I finally worked out the concentration of $\ce{Ba(OH)2}$, which was $0.0148~\mathrm{M}$.

My answer, $0.0148~\mathrm{M}$ is different to the answer the question gives ($0.1013~\mathrm{M}$), and I'm not sure why.

It says that the end-point occurred at $21.30\ \mathrm{mL}$, which if you subtract the volume of $\ce{HCl}$ which was already there, I worked out that $11.3\ \mathrm{mL}$ or $0.0113\ \mathrm{L}$ of $\ce{HCl}$ must have been used.

I think, this is where you went wrong. If a question says something along the lines of

  The end-point occurs at $\dots~\mathrm{ml}$

then you should generally assume that that volume of whatever you are titrating with was used. In this case, that would mean a full $21.3~\mathrm{ml}\ \ce{HCl}$. The same thing of course happened at your point 7, where you write:

  It says that for the next reaction, $25\ \mathrm{mL}$ of $\ce{Ba(OH)2}$ was used and the end-point occurred at $27.10\ \mathrm{mL}$. From this information, I worked out that $0.0021\ \mathrm{L}$ of $\ce{HCl}$ must have been used in this reaction.

Working through this bit by bit, and starting at your point 5:

$$c(\ce{HCl}) = \frac{n(\ce{HCl})}{V(\ce{HCl})} = \frac{0.003976~\mathrm{mol}}{0.0213~\mathrm{l}} = 0.1867~\mathrm{M}$$

For the titration of $\ce{Ba(OH)2}$:

$$n(\ce{HCl}) = c(\ce{HCl}) \times V(\ce{HCl}) = 0.1867~\mathrm{M} \times 27.10~\mathrm{ml} = 5.059~\mathrm{mmol} = 0.005059~\mathrm{mol}\\
n(\ce{Ba(OH)2}) = \frac{n(\ce{HCl})}{2} = \frac{5.059~\mathrm{mmol}}{2} = 2.529~\mathrm{mmol} = 0.002529~\mathrm{mol}\\
c(\ce{Ba(OH)2}) = \frac{n(\ce{Ba(OH)2})}{V(\ce{Ba(OH)2})} = \frac{2.529~\mathrm{mmol}}{25~\mathrm{ml}} = \frac{0.002529~\mathrm{mol}}{0.025~\mathrm{l}} = 0.1012~\mathrm{M}$$

The difference still there can be attributed to a rounding error somewhere along the way.

This is how I work out the answer which is 0.1012 M.


Preparing an amine from an azide Is there any mechanism available for reduction of alkyl azides to amines.
One of the reactions is conversion by $\ce{Na}$ in ethanol or by $\ce{LiAlH4}$.
But how do these proceed?
$\ce{LiAlH4}$ will produce a hydride ion which attacks where?

I googled it for a while and this was the best thing I found (Source):

  This is a tough question, that is probably not known with any certainty.
  Even Jerry March (Adv Organic Chem, 4th ed, page 1219) doesn't commit himself to a mechanism.

  You'll have to speculate.   Add a hydride on the N next to the alkyl group, then
  grab a proton on the terminal N.    Add a hydride on the terminal N, then grab a proton on the middle N.  After that, it gets a little sketchy for breaking the N-N bond.

  Try a SN2 attack by hydride on the middle N, giving hydrazine and  RNH-, which is a lousy SN2 leaving group, but we're stuck with it.
  Protonation of the RNH- gives the amine and hydrazine.   Or, you could do a SN2 on the N next to the alkyl group, and kick out a NH2NH- anion, which is also a lousy leaving group for the SN2.
  You can see why most books conveniently leave out a mechanism for this reduction.


What gas is released during dissolving NaOH What gas is released during dissolving solid $\ce{NaOH}$ in a high purity water? Water more than 10 M$\Omega$/cm and standard quality $\ce{NaOH}$.

I don't think any gas is released.  You are probably experiencing dust or mist that contains NaOH itself, or tiny amounts of gas phase NaOH.

There is related discussion at a soapmaking website

http://www.tikvah.com/medical/lye.html

See also "Pneumothorax following inhalation of caustic soda fumes" Archives of Emergency Medicine, 1988, 5, 45-47

Ammonia is possible in the small amount.

$$\ce{NH4+ (aq) + H2O (l) <=> NH3 (aq) + H3O+ (aq)}$$


How do I name this alkane with a branch that has bromine? I identified the orange portion as the longest carbon chain which is the first step.

Next, I can't call it 3-ethylhexane as the bromine is a branch from a methyl branch.

This is the problem I am facing. Since the methyl branch is on the 3rd carbon, and bromine branch is in the "branched" 2nd.

So, is it called as 3-(2-bromomethylhexane)? Or have I got it wrong?

3-(bromomethyl)hexane

You can't have a 2-position on a methyl group, there is only a 1-position and we omit redundant 1-enumerations of position.

The IUPAC name for this compound would be 3[ 1' - bromomethyl] - hexane

//Edit:
Where 1' refers to the carbon in the methyl group attached to the 3rd carbon in the primary chain.
Also if you had an ethyl group attached and if the Bromine was attached to the 1st or 2nd carbon in that chain then it would be 1' -bromoethyl.... And 2' -bromoethyl.... . Try drawing this structure of this compound.

3-[1'-bromo - 2'- chloro- 3' - flouro- 4' - [1"- methyl ethyl] hexyl] hex- 3 - ene.

Apply the same concept, i.e., the carbons in the longest chain of the substituent group will be numbered 1', 2', 3' and so on and the carbons in substituent group of the substituent group would be 1", 2", 3" and so on.


What happens when FeF3 is reacted with benzene? My text book says that halobenzenes can be prepared by aromatic substitution on benzene ring by using Fe or FeX3 as Lewis acid but this method isn't useful for preparation of fluoro benzene because of it's high reactivity.
My question is what will be the product if FeF3 is reacted with benzene, I tried searching a bit on internet but couldn't find anything helpful.

Iron catalyst will not be needed as Fluorine will react directly with benzene explosively,  the probable product formed would be hexaflurocyclohexane , which can predicted by looking at similar reaction of chlorine with benzene in presence of UV rays which is mentioned in NCERT class 11 chemistry  Textbook.

(Replace $Cl$ with $F$  and since fluorine is more reactive UV light is not needed.)


Is the rate of desorption dependent on pressure? While I was learning about Langmuir's adsorption isotherm in my chemistry class, my teacher talked about the situation when the rate of adsorption and desorption will become equal. However, he mentioned that the rate of adsorption is dependent on pressure but the rate of desorption is not and when he gave an explanation I wasn't satisfied with it.
I would imagine that greater the external pressure on the adsorbent lesser will be the tendency of the adsorbate particles to leave the surface. But that isn't what my teacher said. Please explain.

Without knowing exactly what you were told its hard to be exact, but here is a description of the situation.

The Langmuir model makes assumptions (a) adsorption is complete when the surface is filled with one gas molecule  per site, (b) all adsorption sites are equivalent (i.e. the same) and (c) adsorption and desorption are separate processes, i.e. one does not depend on the other, i.e if one site is occupied it does not affect adjacent sites an any way. Note that this is a model and this determines how we look at the problem.
The model is then $$M(g) + S(surface) = GM(surface) $$ where G is gas and S surface and rate constant for adsorption is $k_a$ and for desorption $k_d$.

We let $\theta$ be the extent of adsorption which varies from 0 to 1, i.e. how full the surface is, then the rate of change of $\theta$ depends on the rate constant for adsorption , pressure and fraction of vacant sites or $N(1-\theta)$ for N vacant sites. Thus for adsorption $$ \frac{d\theta}{dt} = k_aPN(1-\theta) $$

and for desorption $$ \frac{d\theta}{dt} = -k_dN\theta $$

At equilibrium we make the rate of change equal to zero, i.e. rate of adsorption equals rate of desorption and with a bit of algebra find that $$\theta=\frac{KP}{KP+1}$$  where K is the equilibrium constant or $K=k_a/k_d$ This equation has the fancy name as the Langmuir Isotherm.

You can see that the rate going onto the surface must be equal to the pressure (no gas, no adsorption) and the area free to adsorb onto $(1-\theta)$. (We assume in this model than no molecule double adsorbs; this is dealt with in more advanced treatments). The rate leaving has to be proportional to the amount already there , i.e. proportional to N the total number of surface sites multiplied by the fraction covered.

Ask more questions in the comments if this is not clear.

Desorption is inverse of adsorption.since pressure is directly proportional to adsorption so desorption is inversely proportional to pressure(provided surface does not reach to a saturation level)


What is the proper Lewis structure for HCOOH? I tried to draw the Lewis structure of $\ce{HCOOH}$ but I didn't get the correct representation.

As you can see in the picture below, at no. 1, I put $\ce{C}$ as the central atom because it has less electronegativity than $\ce{O}$ ($\ce{H}$ cannot be a central atom). I drew the structure, but it is wrong because it violates the octet rule.

Then I came up with another structure as shown at no. 2,
and it seemed correct to me, but when I searched on Google for the correct structure, I found out that I am wrong.

Why is the structure that I drew in no. 2  wrong, even though the formal charges are zero?

How do I figure out the correct order of atoms in such molecules like the one above. What I mean is how do I know that the $\ce{C}$ will have $\ce{H}$ on the left and one $\ce{O}$ up and $\ce{O, H}$ on the  right side?

With familiarity you will recognize that $\ce{COOH}$ in a formula generally refers to a carboxylic acid group. Sometimes you will also see $\ce{CO_2H}$. Either is acceptable. Vinegar, or acetic acid, is a carboxylic acid. Its formula is $\ce{CH3COOH}$ or $\ce{H3CCOOH}$.

It is important to realize that a carboxylic acid group is NOT a peroxide. Peroxides have  involve O-O bonds. You may also generalize that the oxygens in peroxides have a negative one oxidation state. A common peroxide is hydrogen peroxide, $\ce{H2O2}$ or $\ce{HOOH}$.

Also note that carbon is generally tetravalent - i.e. it is commonly found in stable molecules as having an octet of electrons. In addition, carbon generally exhibits no overall formal charge. So this suggests that most of the time you will see carbon forming four bonds - and this is the case the majority of the time. This configuration gives carbon no net formal charge and fills its octet.

Of course, you may find carbocations or carboanions, but carbocations are highly unstable and exist only ephemerally. Carboanions also exist; see the cyanide ion for an example.

The above, in addition to keeping in mind that a carboxylic acid is a Bronsted/Lowry acid - i.e. proton donor - should enable the construction of an acceptable Lewis structure.

Oxygen can have less than an octet. Here's why:

In the triple bond resonance structure of $\ce{CO}$ the $\ce{O}$ has 5 electrons around it giving it a +1 formal charge. The same thing is true for the $\ce{O}$ to the right in structure 1.

$\ce{O}$ has the equivalent of 3 lone pairs of electrons (that is 6 electrons) to form a bond with. If it uses 1 lone pair out of the 2,2,1,1 arrangement for $\ce{O}$ to form a single bond (which it does) than the $\ce{O}$ has used 2 electrons to form a bond leaving 4 nonbonding electrons. If you separate the $\ce{O}$ and the $\ce{C}$ by breaking the bond the $\ce{O}$ is going to have its original 6. It is exceeding an octet that is invalid for period 2, not less than an octet.

If $\ce{O}$ has less than an octet than yes it is unstable but it is still possible. This is why positive ions of $\ce{O}$ exist as well as $\ce{O^-}$. There is also a small amount of neutral atomic oxygen (which is a biradical). Carbocations exist for the same reason that oxygen ions with less than 8 electrons do. That reason is because having less than an octet is not invalid.


Does the enthalpy of formation include translational and rotational energy? I can look up the enthalpy of formation of water vapor and get -241,826 KJ/mol. I want to use this to compute the total energy of water vapor at STP from the energies of O2 and H2 gas. What I don't know is whether I should use the electronic energies of H2 gas and O2 gas or the total energies of H2 gas and O2 gas, which include the contributions from the translational and rotational partition functions.
Basically, which of these two is the correct usage of Enthalpy of Formation?

ETotal(H2O) = Eelectronic(O2) + Eelectronic(H2) -241,826 KJ/mol + ETranslational(H2O) + ERotational(H2O)
ETotal(H2O) = ETotal(O2) + ETotal(H2) -241,826 KJ/mol

To add to @Kexanone's answer above, the enthalpy of formation refers to how much heat is evolved or adsorbed from the environment in a reaction. So it's the total energy, and in an ideal gas it has the form: dft + translational + rotational + vibrational + PdV, where the PdV refers to the energy needed to make space for the gas (grand canonical ensemble; in the canonical ensemble (fixed volume) this does not show up). At room temperature the vibrational and PdV terms are typically small compared to DFT energy errors (order 1 millihartree), with the PdV term only coming into play at pressures within an order of magnitude or two of those found in the detonation of thermonuclear bombs.


Which diagram shows the effect of catalysis on chemical equilibrium? My course book has the diagram in figure (1) to represent the effect of a catalyst on chemical equilibrium. I drew the second diagram in figure (2).

When reviewing the diagram in figure (1), I observe the following mistakes:

The rate of catalyzed and uncatalyzed forward reaction  at equilibrium will be the same, which seems to be wrong because the value of rate constant of catalyzed forward reaction ($k_f$ catalyzed) more than the value of rate constant of uncatalyzed forward reaction ($k_f$ uncatalyzed), so the rate of catalyzed forward reaction at equilibrium = ($k_f$ catalyzed) $[A][B] > $ ($k_f$ uncatalyzed) $[A][B]$, so  it isn't the same as the diagram.
The rate of catalyzed forward reaction before equilibrium less than the rate of uncatalyzed forward reaction before equilibrium, which seems to be wrong too.

I need feed back about the diagram or both are wrong and about my reasoning.

If (as suggested in the comments) the y-axis is supposed to represent concentration, such that the upper curve is "concentration of $\mathrm{A+B}$" and the lower curve is "concentration of $\mathrm{C+D}$", then Figure 1 is correct, albeit horribly mislabeled.

Your Figure 2 is correct if we want the y-axis to represent a rate-of-conversion instead, i.e., if the upper curve is "number of parcels of $\mathrm{A+B}$ being converted to $\mathrm{C+D}$ per second", and the lower curve is vice-versa. As you point out, a catalyst increases the conversion rate, and does so in both directions. So, exactly as you have drawn in Figure 2, those catalyzed rate curves should both be higher than their uncatalyzed siblings.

("Rate of reaction" is an awfully ambiguous term here – since at equilibrium, the rate of reaction must be zero!)

Figure 1 shows the rate as $k_f [A][B]$ whereas Figure 2 shows only $k_f$. This is why Figure 1 has the equilibrium at the same rate for the 4 reactions.


Organic Nomenclature: But-1-en-3-yne or But-3-yne-1-en Just wondering whether "but-1-en-3-yne" or "but-3-yne-1-en" follows the correct IUPAC naming convention for organic molecules. I'm going through a text book that uses the former in an example, but in the answer to one of the questions it uses the latter, so is it interchangeable or is one just wrong?

The IUCN Blue Book says that:

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride [...] is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

Therefore the double bond gets priority whatever be its locants (numbering) i.e., '-ene' comes before '-yne'. And in this case it's "but-1-en-3-yne".


Aren't hydrogens on less substituted groups more electrophilic? I am having some trouble understanding the behaviors of some Alkene reactions that I have been studying. I took two semesters of O-chem a couple of years ago and am trying to refresh before an advanced class, but I have found two reactions in my studies that aren't making sense.

At the top of page 13 of the e-book found here: http://crab.rutgers.edu/~alroche/Ch08.pdf ... I see that the water molecule is attacking the carbon with two methyl groups on it... but wouldn't the carbon with only one methyl group be more electrophyllic? Because of induction, I would think that the carbon with two methyl groups would be more stable than the other, and so the carbon with less substituents would be more electron-poor.... So why does the water attack the more substituted carbon?

Secondly, in the image below (sorry it's turned), where the second proton transfer is taking place, you see the water molecule pulling the hydrogen off of the more substituted carbon. But wouldn't the hydrogens on the methyl group adjacent to the positive charge be way more electrophyllic, thus creating a double bond on the end of the chain?

I appreciate any help on this. Thanks for your time and effort!

Water can attack both carbocation and hydrogen atom. Sometimes it attacks C, sometimes it attacks H.

Whenever it attacks C alcohol forms so the reaction goes back to reactants. When it attacks H the product forms. We are interested in product, so we show only this path.


Why are rates of reaction faster for ionic compounds than for covalent compounds? Why is it that ionic reactions (i.e., reactions involving ionic compounds) are faster than covalent reactions (reactions involving covalent compounds), even though the ionic bond is stronger than the covalent bond and hence more time will be required to break the bond?

Not all ionic reactions are not faster, i.e. have a larger rate constant than those between neutral species, however, at similar activation energies there are effects that only ionic reactions suffer. There are two considerations, the dielectric constant of the solvent used, water, hexane, ethanol, acetone etc., and the relative charges on the ions, similar or opposite as well as their value, $\pm1,\pm 2 $ etc.

For slow reactions, that is not diffusion controlled, (see below) the 'primary kinetic salt effect' is important. This shows that the rate constant can increase or decrease depending on the ionic strength of the solution and the charge on the ions. It is based on the Debye-Huckel theory of ions in solution which relates activity coefficients in dilute solution to the ionic strength.

The rate constant for a bimolecular reaction  has the form
$$ \log_{10}\left(\frac{k_2}{k_2^0}\right) = 2\alpha z_az_b\sqrt I$$
where I is the ionic strength, the z the charges on the two ions, $(-1,+1),(-1,-1) $ etc. and $\alpha \approx 1.6 \cdot 10^{-2} \pu{ m^{3/2} mol^{-1/2}}$ is a constant originating in Debye-Huckel theory. At low ionic strength ($ \lt 0.5 \pu{mol dm^{-3}}$ ) the rate constants increase or decrease linearly with $\sqrt I$.

The effect of increasing the dielectric constant (relative permittivity) is to lower the rate constant, thus rate constants are lower in water ($\epsilon = 78$) than in cyclohexane ($ \epsilon = 2.2$) for the same reaction. This is due to the fact that, at a given distance, in a high dielectric solvent the electric field of the ion is greatly reduced compared to its value in a low dielectric solvent. Thus in a high dielectric the ions are far more isolated from one another and so rate constants are only affected when ions come in relatively close contact. Conversely in low dielectric solvents the ions 'feel' the presence of one another at a large separation, this can considerably increase or decrease the rate constant depending on relative charges.

The rate constant is given by
$$ \ln(k_2) =\ln(k_2^{\infty}) - \frac{z_az_bq^2 }{4\pi \epsilon _0 \epsilon d_{ab}k_BT}$$
where $k_2^{\infty}$ is the limiting rate constant for a solvent of infinite dielectric constant. The permittivity of free space is $\epsilon _0$, the relative permittivity $\epsilon$, q is the charge on the electron, $d_{ab}$ the separation of the ions at reaction and $k_B$ the Boltzmann constant. A plot of the rate constant vs. $1/\epsilon$ yields a reasonably straight line  for all but the lowest $\epsilon$ values.

When a reaction's activation energy  is low, less than $\approx 10 \pu{kJmol^{-1}}$ the reactants react so rapidly on contact that the rate constant in determined by how quickly they can approach one another in solution; the rate constant then becomes diffusion controlled and thus depends on the solution viscosity.

Using the Stokes-Einstein equation and assuming equal sized molecules the rate constant has the simple form
$$ k_2= \frac{8RT}{3\eta}$$
where $\eta$ is the solvent viscosity. Water has a value $\approx 1\cdot 10^{-3} $ Pa s or $1$ centipoise (cP) at room temperature. In fluid solution such as hexane the rate diffusion controlled constant is typically $10^{10}$ to $10^{11} \pu{dm^3mol^{-1}s^{-1}}$.

The effects of ions changes the diffusion controlled rate constant for the reasons described above and is now
$$ k_2= \frac{8RT}{3\eta}\left(\frac{x}{e^x -1}\right)$$
where
$$ x=\frac{z_az_bq^2 }{4\pi \epsilon _0 \epsilon d_{ab}kT}$$
and is the term used before.

When the ionic charges are the same and the dielectric constant small the rate constant is greatly reduced, and conversely it is enhanced when the charges are opposite in sign to one another. In high dielectric solvents these changes are greatly reduced. These effects are similar to those described above.

covalent compounds are formed by sharing of electrons between two atoms . they form very strong bond and they do not contain ions . their reactions are slow because reactions involve breaking of existing bonds and formation  of new bonds. and additional energy require to break bonds therefore reactions are slow                          ionic compounds formed by transfer of electron from one atom to another so they contain ions. they easily ionized they do not to expend energy to break bonds because they are already dissociated so they show faster reaction                                   hope i helped...??


Geometry of AsF5 molecule Which of the following is the best description of the arragement of fluorine atoms around the arsenic atom in a molecule of $\ce{AsF5}$?
  (a) trigonal bipyramid
  (b) octahedron
  (c) tetrahedron
  (d) square pyramid
  (e) planar pentagon

The correct answer is (a).

I know that the answer can't be (b), (c), or (e) since octahedron requires 6 ligands and tetrahedron requires 4 ligand, and planar pentagon is only observed in $\ce{[IF5]-}$ and $\ce{[XeF5]-}$. I recall that first row transition metals prefer trigonal bipyramidal configurations, thus trigonal bipyramidal is the more likely geometry, but I'm not sure why this is true?

Any help would be greatly appreciated.

There are no lone pairs in $\ce{As}$ after it forms a bond with each of the five $\ce{F}$.

Hence, the number of lone pairs of electron around the central atom ($\ce{As}$) is zero, and the number of sigma bonds the central atom forms is five (a single bond has one $\sigma$-bond only).

Hence the hybridization number of $\ce{As}$ in $\ce{AsF5}$ is $0+5=5$ and it follows $\ce{sp^3d}$ hybridization geometry with no change in actual shape because of absence of lone pairs.

So, it has a trigonal bipyramidal structure: two $\ce{F}$ ions in the axial positions and three others in equitorial positions.


How does the polarity of the eluent and sample affect the Rf value in thin layer chromatography? A three-component mixture (X, Y, Z) was separated using normal phase TLC. Three eluents of varying polarity were used. Select the TLC plate that was developed using the least polar eluent (click on the corresponding TLC number 1, 2, or 3 on the bottom of the plate). On TLC plate 3, select the least polar compound (click on X, Y, or Z spot on plate 3).

Does normal phase TLC mean silica gel TLC? If this is the case then plate 3 has the least polar eluent because the compounds are at the top. For the same reason compound Z would be the least polar.

The term eluent means solvent doesn't it.

Never in my years in the lab have I seen a more polar eluent that causes a lower $R_\mathrm{f}$ value than a less polar eluent. It also does not make sense theoretically.

TLC 2 is obviously the one with the least polar eluent and compound $z$ is the least polar compound.

I fully disagree with half of MaxW’s proposed answer.

You must learn the terminology to follow along in chemistry. If in doubt Google the term...

Eluent - As Zhe noted "Eluent is a fancy name for solvent in chromatography." It can be a gas or a liquid. For TLC or HPLC a liquid is used. For GC a gas is used.
Normal phase TLC - For normal phase TLC the stationary phase is polar and the mobile phase is typically non-polar or only moderately polar. Polar molecules are attracted to the stationary phase while non-polar compounds are carried along by the mobile phase.
Reverse phase TLC - For reverse phase TLC the stationary phase is non-polar and the mobile phase is typically moderately polar to strongly polar. Non-polar molecules are attracted to the stationary phase while polar compounds are carried along by the mobile phase.

The normal designations is because the first HPLC columns used alumina and silica packing, which are both polar, in columns. With setup The least polar compounds were eluted first and the most polar compounds were eluted last.

Switching to non-polar packing was then noted as reverse phase  since it reversed the order of elution. The most polar compounds were eluted first and the least polar compounds were eluted last.

ANSWER

This is normal phase TLC so the stationary phase is polar.

Assuming the compounds are "relatively" polar

Plate 2 has the LEAST polar eluent because the compounds have lower $\text{R}_{\text{f}}$ values than they do on the other two plates.

Plate 3 has the MOST polar eluent because the compounds have higher $\text{R}_{\text{f}}$ values than they do on the other two plates.

On plate 3 the least polar compound would be Z.

Assuming the compounds are "relatively" non-polar

Plate 2 has the MOST polar eluent because the compounds have lower $\text{R}_{\text{f}}$ values than they do on the other two plates.

Plate 3 has the LEAST polar eluent because the compounds have higher $\text{R}_{\text{f}}$ values than they do on the other two plates.

On plate 3 the least polar compound would be Z.

So the problem would seem to be ambiguous. We don't know that any of the solvents is "absolutely" non-polar like hexane, or that one might be "absolutely" polar like water.


Work done by the System In thermodynamics work is defined as $~-p_{ex}\Delta V~$ for an ideal gas. If the total work is a negative value, that is in case of expansion of gas, we say that the work is done by the system and is negative.

I don't understand that why we use the pressure exerted on the gas in expression of work that is the external pressure and say that the work is done by the system ? If work is done by the system we should use internal pressure and if we use that in case of expansion the work done by the system will be positive which is contradicting. Please provide an explanation

Well, think of a gas that is expanding against a piston shown below:

We want to find the total work which is given by: $$w = F\Delta d$$ So we need to know what is the force that is applied by the gas onto the piston. To do this, we must realise that, the force that is stopping the gas from expanding freely into the atmosphere is the external atmospheric pressure. It is this pressure which is pushing downwards onto the gas, preventing it from expanding. Therefore, the force that is required for the gas to expand must be equal to the total pressure pushing down it and has nothing to do with its internal pressure.

Consider this example. I want to push a box across my room which is covered in carpet. To move the box, obviously I need to apply a force to it. The force that I need to apply to must be equal or greater than the resisting force which is the friction between the box and carpet. This force that is required has nothing to do with the speed of the box.

So now that we know the force, we get that $$w = -p_{ex}A\Delta d$$ but the area times the displacement is actually equal to the volume, therefore we get: $$w =-p_{ex}\Delta V$$

So, as you can see, we have derived the equation for work done by a gas. The main thing that you need to recognise is that the internal pressure is irrelevant when considering the force required for the gas to expand. All you need to consider is the force that is preventing it from expanding which is the external pressure.

This is an argument I have with the sciences. An expanding system does work onto the surrounding atmosphere. Consider the expanding system as a subsystem of our atmosphere, hence as it expands it does work onto the surrounding atmosphere i.e. it is as if the atmosphere has expanded, hence the work onto the atmosphere is [(Psys)(dVsys)]. Since the system's final pressure is 1 atm and the volume change is such that (dVsys)=(dVatm), we can write that the work done onto the atmosphere is [(Patm)(dV)atm].

The confusion originates from not writing the enthalpy equation correctly because simply writing  H=E+PV does not tell us what is happening where. So clarity is given by writing

dH=(dE)sys + (PdV)atm   and since the work is done exterior to the system then we can rewrite it as
    dH=(dE)sys-(PdV)sys
 But the above can be confusing. Imagine we are heating water bring it to a boil, and dQ is the energy in (due to heating. Now we would write

dQ= (dE)sys + (PdV)atm  = (dE)sys + (W)atm

The reason is that the heating must do two things
1) change the system's internal energy
2) do work onto the surrounding atmosphere

The above equation is really latent heat of vaporization with the systems energy change being it changes in the bonding energy i.e. (dE)sys = (dU)sys

Now I argue (this is published in peer review [1,2,3] but remains controversial) that the work done onto the surrounding atmosphere: (W)atm = (PdV)atm is lost work because it is irreversible i.e. in order for the atmosphere to work on a system, then the atmosphere must be at a higher pressure than the system. This is controversial because it can become a challenge to the second law, namely lost work into the atmosphere means that the second law is no longer necessary to explain why we cannot have perpetual motion, or why energy is lost by expanding systems like a steam engine or why the Carnot cycle is ideal rather than realistic.

[1] Mayhew, K.W. "Second Law and Lost Work" Physics Essays 28,1 (2015) pg 152-56
[2] Mayhew, K.W. "Resolving Problematic Thermodynamics" Hadronic Journal vol 41 no 3 (2018) pg259


Why do sulfur and oxygen form sulfur dioxide? Both sulfur and oxygen have 6 electrons in their outermost shell. Since both only need two electrons to stabilize, why doesn't just one atom of oxygen react with one sulfur atom, forming sulfur monoxide SO instead of sulfur dioxide $\ce{SO2}$?

One atom of oxygen is a fictional species outside of a plasma.  Oxygen is naturally diatomic, and the process by which it oxidizes sulfur is radical-based, leading to both oxygen atoms reacting.  The fact that SO2 forms instead of SO is a result of the [redundant answer] thermodynamics favor SO2 over SO.

Given that simple fact here is a quantum mechanical rationalization: Sulfur can expand its octet, which means in SO2 it has 12 electrons 'around' it.  This allows resonances forms where O- has a significant contribution to resonance, while S still has at least 8 electrons around it.  In SO no such resonance forms exist, meaning the highly electronegative oxygen must share its electrons with with sulfuer to avoid exposing a six-eletron sulfur species.  Hence SO2 should be a better structure than SO overall.

Its just like answering formation of $\ce{CCl4}$ against $\ce{CCl2 }$
for $\ce{SO}$ and $\ce{SO2}$

It is highly unstable
It becomes more stable after forming a double bond with O ( maybe resonance )

Structures:

 ,

It has an electronic configuration of $\ce{[Ne]} \: 3s^2\: 3p^4$

It has 4 bonds due to uncoupling of p electrons (a little energy is spent), but due to the formation of four bonds, a large amount of energy is released.


How to calculate the concentration of the elements after dissolution of iron(III) chloride? I was presented with the following problem:

  In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

  What is the molarity of the solution?
  What is the concentration of iron(III) cation in M?
  And what is the concentration of chloride anion in M?

I know how to do the first part of the problem, the answer is $\pu{0.406 M}\ \ce{FeCl3}$. But when I go to balance the equation, the computer system tells me the equation is wrong.

My equation:
$$
2\ce{FeCl3 -> 2 Fe + 3Cl2}
$$

The computer system:

$$
\ce{FeCl3 -> Fe^3+ + 3Cl-}
$$

I don't understand the computer systems' answer. Simply because when by itself, chlorine is a diatomic molecule therefore, should be $\ce{Cl2}$ not just $\ce{Cl}$. Am I wrong to think this? Am I correct or the computer system? If the computer system is correct, why?

When answering such problems, I recommend writing everything know down, then use the appropriate formula to solve for the unknown. For the first part you already did that correctly.

  In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

\begin{align}
  m(\ce{FeCl3 (s)}) &= \pu{24.7 g}\\
  M(\ce{FeCl3}) &= \pu{162.3 g//mol}\\
  V(\ce{FeCl3 (aq)}) &= \pu{375 mL}\\
\end{align}

  What is the molarity of the solution?

Molarity is another word for amount concentration, so you are looking for $c(\ce{FeCl3 (aq)})$.
\begin{align}
  c &= \frac{n}{V}\\
  n &= \frac{m}{M}\\
  c(\ce{FeCl3 (aq)}) &=
    \frac{m(\ce{FeCl3 (s)})}{M(\ce{FeCl3})\cdot V(\ce{FeCl3 (aq)})}\\
  c(\ce{FeCl3 (aq)}) &=
    \frac{\pu{24.7 g}}{\pu{162.3 g//mol}\cdot\pu{375E-3L}}\\
  c(\ce{FeCl3 (aq)}) &=
    \pu{0.406 mol//L}
\end{align}

  What is the concentration of iron(III) cation in M?
  And what is the concentration of chloride anion in M?

You are looking for the ion (amount) concentrations in your solution, i.e. $c(\ce{Fe^3+ (aq)})$ and $c(\ce{Cl- (aq)})$.
For these two questions you need to know how the substance dissolves. The wording here is already very explicit, as it says iron(III) cation, i.e. $\ce{Fe^3+ (aq)}$, and chloride anion, i.e. $\ce{Cl-}$. Otherwise, with some more experience you will know that this compound is a salt and it dissociates into ions when dissolved in polar solvents like water. Therefore the reaction equation is
$$\ce{FeCl3 (s) ->[H2O] Fe^3+ (aq) + 3 Cl- (aq)}.$$
From this you can derive the actual relationships between the ion concentrations.
You can clearly see that there is one iron(III) ion and three chloride ions per formula of $\ce{FeCl3}$; therefore
\begin{align}
  c(\ce{Fe^3+ (aq)})&= c(\ce{FeCl3 (aq)})
  &&=\pu{0.406 mol//L},\\
  c(\ce{Cl- (aq)}) &= 3\times c(\ce{FeCl3 (aq)})
  &&=\pu{1.22 mol//L}.
\end{align}

The other equation would probably be appropriate when you (thermally) decompose the substance (in absence of everything) as it is the reverse equation which can be used to prepare the molecule:
$$\ce{2 Fe (s) + 3 Cl2 (g) -> 2 FeCl3 (s)}.$$

you finish question 1 and you need to finish question 2 and 3
but why are you making a equation for that. it seem like that was what stopped you working in the first place.

but to make it balanced... you said dissolved in volumetric flask. if it dissolved its ionic.
and that means there should be a "cation +" and a "anion -"
hence the $$\ce{fe^3+ + 3cl- }$$
and the $\ce{fe^3+}$ is third oxidation state floating around in solution it could had been a $\ce{fe^2+}$ if you started with $\ce{fe(oh)2}$

so the pc is correct....

1 mole of $\ce{fecl3}$ will yield's 1,5 mole of cl2 therefore it has more $cl$ then $fe$ ions.

concentration of $fe$ ion in $M = 0.406$

concentration of $cl$ ion in $M = 1.218$

$final\ equation\ =$ $$molarity\ of\ \ce{fecl3} * number\ of\ fe\ ion's\ in\ the\ molecule$$
$$\ce{fe^3+}=\  0.406M * 1 = 0.406M$$
$$molarity\ of\ \ce{fecl3} * number\ of\ cl\ ion's\ in\ the\ molecule$$
$$\ce{cl^-}=\    0.406M * 3 = 1.218 M$$


The Lewis structure of HNO3 So I was checking the structure of nitric acid in Wikipedia, however I couldn't quite fathom why it looked like that because it seemed to contradict the following statement:

  A Lewis structure with small or no formal charges is preferred over a Lewis structure with large formal charges.

If we draw it like the one on the right, we get rid of formal charges and the structure is said to be more "stable". Why does this concept not work in this case?

Nitrogen is in the second row with no $d$ orbital in the valence shell.  It obeys the octet rule and cannot have more than 8 electron.

There are exceptions to the octet rule.  Having less than 8 electron is less preferable but still possible, and is commonly seen in free radicals and cations. On the other hand, having more than 8 electron is extremely unfavorable for second period atoms. Such electronic structures can be found in extremely unstable species or excited states, such as the CH5 radical.

As a comparison, first shell atoms obey the duet rule while atoms in the 3rd shell and beyond may obey 18-electron rule, 12-electron rule or 8-electron rule.   However, 18-electron rule and 12-electron rules are much less strict than 8-electron rule and violations are common place.

According to valence bond theory, the electronic structure of a molecule is a combination of all possible resonance structure that you can write down, including structures with all possible formal charges and strange electron counts.  However, their contributions are not even.  some of them are more favorable than others.

For HNO3, in order to satisfy the octet rule, the nitrogen atom would form 1 double bond and 2 single bonds.  Based on octet rule alone, there are 3 possible resonance structures that are favorable.

However, the first two resonance structures are significantly more favorable than the third, because they have smaller amount of formal charges.  As a result, we usually only write the two dominant structures. The bond between OH and N is close to a normal single bond. The other two N-O bonds have bond order near 1.5.

It is also common to write a mixture of resonance structures as the hybrid form

Note that this representation is not a single Lewis structure, but a convenient way to represent many resonance structures in the same figure. It gives no information about the exact bond order or formal charges on individual atoms.  The dotted lines indicates that in some structures where that bond is a single bond and in others a double bond, and the order is somewhere between 1 and 2.

Three oxygens are bounded to one nitrogen atom.Oxygen  being more electronegative attracts the pair of electron more strogly.One of the oxygen binds with an atom of hydrogen and share an electron with nitrogen to complete its octet and the remaining oxygen atom share an other electron with nitrogen atom. In this way electron pairs are partially donated and coordinate covalent or dative bond is formed


Test for diethylether or tertiary alcohol? What is a simple chemical test to tell the difference between a tertiary alcohol (e.g. 2-methylpropan-2-ol) and an ether (specifically diethylether/ethoxyethane)?

Neither of them can be oxidised by reagents such as potassium dichromate, Tollens reagent or fehlings reagent.

I came up with a test in which a less than stoichiometric amount of ethanoic acid is added and left to react, then some sodium carbonate is added and if there is no effervescence that means an ester has formed hence it must have been the alcohol, otherwise it was probably the ether. However I am unsure about the extent to which this reaction completes and it's simplicity.

As mentioned in the comments, using Lucas' Reagent is the best option in this case because you're dealing with a tertiary alcohol vs an ether, but let's consider your proposal as well.

In principle, it should work because an ether will not react with ethanoic acid whereas an alcohol would. However, the formation of an ester simply by the addition of ethanoic acid to alcohol is not likely to complete (you generally need a dehydrating acid like $\ce{H2SO4}$ to drive the reaction). In other words, when you put in the bicarbonate you will most likely see effervescence.

If you're thinking of looking for effervescence, using thionyl chloride $\ce{SOCl2}$ might be an option since it will react with the alcohol to give a sulfur dioxide gas and HCl:

$$\ce{R-OH + SOCl2 -> R-Cl + SO2(g) + HCl}$$

You have to make sure the solvent is dry though.

In the specific case of t-Butanol vs diethyl ether, though, you could simply heat the solution slightly to about 40 degrees Celsius. If it boils, then you've got diethyl either. If it doesn't, it's t-Butanol.

Diethyl ether is minimally soluble in water (6 grams per 100), but t-butyl alcohol is completely miscible with water, according to Wikipedia.

Selecting between these two compounds can be made by mixing 5 mL of H2O with 5 mL of the unknown. A single phase means t-butanol; two phases means diethyl ether.


Nucleophile in a reaction How can we determine which atom is a nucleophile in a compound in a reaction?

I know we see resonance, but I am little confused between electronegativity, formal charge, solvent, and size.

Here are some rules i remember from organic chemistry in Collage.
When trying to distinguish between nucleophile and base,

Look for resonance
If atom is involved in resonsance, this makes ato/group more nucleophilic,
if electrons are more localized its more basic.
Eg. A carboxylate is a nucleophile, the -ve charge is delocalized (i.e - charge is  spread out across carbonyl and oxyanion. Compared to nitrogen on an amine,  electron are restricted.

Sterics,
more sterics less nucleophilic.
Smaller radius atoms, valence electrons are more localized in space, thus more basic. Big radius atoms, poor localization relative to small, thus more nucleophilic.
More electronegastive, less basic
Electronegativity is a measure of the nuclear positive charge effect on near -ve charges.
The more electronegative the less basic. Because the atom doesnt want to give up its electrons. Why? Because it has high electronegativity
Conduction effects
If there are electronegative atoms within 3 bonds (taking into account electronegativities, and proximity) more nucleophilic less basic.

Nitrogens are usually bases, oxygens nucleophiles. I swear it was always like that. Halogens are weak nucleophiles.

Will edit for legibility when i get to computer


Why the bond angle of PH3 is lesser that that of PF3? We can explain why the bond angle of $\ce{NF3}$ (102°29') is lesser than $\ce{NH3}$ (107°48') by the VSEPR theory, since lone pair lone pair repulsion is greater than lone pair bond pair repulsion. Then for $\ce{PH3}$ and $\ce{PF3}$, also, it is expected that the bond angle of $\ce{PF3}$ will be smaller. But I found that the thing is just reverse. Why is this so? Is it an exception?

VSEPR works by accident in the cases of $\ce{NH3}$ and $\ce{NF3}$. In reality, there are much more things to consider as shown in this answer. All four compounds should have bond angles of $90^\circ$ if there were no other effects present.

For both nitrogen compounds, the effects are the short $\ce{N-X}$ bonds which lead to steric clash of the three substituents. Thus, the bond angle is broadened from the original, hypothetical $90^\circ$ to whereever it may seem comfortable to the molecule. Since fluorines are larger atoms than hydrogens are, the $\ce{N-F}$ bond is longer than the $\ce{N-H}$ bond ($137~\mathrm{pm}$ versus $102~\mathrm{pm}$) and therefore the atoms can agree on a smaller bond angle (less crowding).

For phosphorous, we can initially assume that the bond angle of $\ce{PH3}$ is close enough to $90^\circ$ for no hybridisation to be necessary (semi-proven by this answer of Martin) and likewise for $\ce{PF3}$. The bond lengths are $142~\mathrm{pm}$ and $156~\mathrm{pm}$, respectively. The difference in bond lengths is only half of that of the nitrogen compounds ($14$ versus $35~\mathrm{pm}$). Therefore, we can assume the bond length to be mainly dictated by the central phosphorous atom, while the bond angles must be dictated more by the outer atoms. Since again fluorine is larger than hydrogen, this time the better way to stabilise the molecule is not to lengthen the bonds but to expand the angle.

In PH3 and PF3 bond angle of PF3 is greater as in PF3 back bonding takes place.
Back bonding is possible in PF3 as P has vacant d orbital(as its atomic no. is 15 therefor its electronic configuration is 1s(2e) 2s(2e) 2p(6e) 3s(2e) 3p(3e)
this shows it has vacant 3d orbital which can perform back bonding with fluorine as fluorine has pair of electrons.
as backbonding takes place the lonepair of P will be used in doublebond formation with F. and obviously if their is no lonepair then their is no one to repell. there for more bond angle
also in PH3 no backbonding can take place as hydrogen has no pair ofelectron(e) for backbonding to take place.
May this solution help you.:)


Confusion regarding $\delta$ charge compared with electronegativity I heard on a lecture of an online site that, let us say, we have two orbitals (1 and 2). Now, it will be like a spherical cloud. Now, what they say is that

Now, we can notice that for orbital number 1. There is a dark region and a light region (right side). Now, what they say is that the darker region means the electron finding probability is more there. Similarly, for the number 2, electron finging probability is more on left side.
Therefore, we say that 1 develops $\delta +$ charge and 2 develops $\delta -$ charge. Therefore, the two will attract each other.
Now, my teacher on the other hand told it a bit differently:

The more electronegativity means more $\delta -$ charge. But, I have a confusion in it. Between $\ce{H}$ and $\ce{F}$. I have drawn a line. Fluorine has 7 electrons. My teacher says that $\ce{F}$ and $\ce{H}$ share that one electron where $\ce{H}$ has one and fluorine 7th electron. Now, since $\ce{F}$ is more electronegative. The electron (by hydrogen) is more pushed towards fluorine. My confusion is that if $\ce{F}$ is more electronegative than $\ce{H}$. Then, the electron should be present more towards the $\ce{H}$ side.

Although, OP indicated the confusion regarding $\delta^\pm $ charge compared with electronegativity, the text body indicated whole lot of confutations over interatomic/intermolecular forces. Thus, I start with the electronegativity.
According to IUPAC Gold Book, electronegativity is the concept introduced by Linus Pauling as the power of an atom to attract electrons to itself. There are several definitions of this quantity. However, Wikipedia gives a simple to understand version better suited for OP:

Electronegativity is the measurement of the tendency of an atom to attract a shared pair of electrons (or electron density).

The relative electronegativity of $\ce{H}$ atom has given a value of $\pu{2.1 eV^{-\frac{1}{2}}}$ and the electronegativity for each of other atoms in the periodic table have been calculated relative to this value for $\ce{H}$ atom (IUPAC Gold Book). These calculations are beyond OP’s field of study so should be disregarded. The important point for OP to consider is the higher the associated electronegativity, the more an atom or a substituent group attracts electrons. For instance, the electronegativity of $\ce{F}$ is $\pu{4.0 eV^{-\frac{1}{2}}}$ and hence, shared electron fair in $\ce{H-F}$ bond stays more closed to $\ce{F}$ than to $\ce{H}$. As a result, $\ce{F}$ carries permeant $\delta^- $ charge while $\ce{H}$ carries   permeant $\delta^+ $ charge in the $\ce{HF}$ molecule, creating a permanent dipole. Hence, $\ce{HF}$ is a polar molecule. Therefore, among $\ce{HF}$ molecules, there are always dipole-dipole interactions (the image 2 of OP's text). Since the positive end of these dipole is $\ce{H}$ atom, the relevant dipole-dipole interaction is called $\ce{H}$-bonding, the strongest intermolecular force among dipole-dipole interactions. These dipole-dipole interactions occur only in molecules.
Now, I think I have addressed the OP's main confusion:

My confusion is that if $\ce{F}$ is more electronegative than $\ce{H}$, then, the electron should be present more towards the $\ce{H}$ side.

According to the above explanation, PO's understanding of electronegativity is not correct. In reality, if an atom is relatively more electronegative than that of the other atom it attaches to, that means it attracts the shared pair of electrons, increasing its electron density compared to its protons (gaining $\delta^- $ charge). Meanwhile, the other atom feels decreased electron density compared to its protons (gaining $\delta^+ $ charge). This is completely opposite of what OP's indicated understanding.
On the other hand, the London dispersion force is also one of the intermolecular forces, but is the weakest among them and occurs between atoms as well. For example, why do you think gaseous $\ce{He}$ become a liquid when the temperature is lowered sufficiently? You can find the answer in chem.purdue.edu:

The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms/molecules occupy positions that make the atoms/molecules form temporary dipoles. Thus, this force is sometimes called an induced dipole-induced dipole attraction. London forces are the attractive forces that cause nonpolar substances to condense to liquids and to freeze into solids when the temperature is lowered sufficiently.

An atom or a molecule can develop a temporary (instantaneous) dipole when its electrons are distributed unsymmetrically about the nucleus because of the constant motion of the electrons (e.g., the image 1 of OP's text). The explanation can be depicted in an image:

For instance, assume two helium atoms approaching each other. When they get closed to each other, the electrostatic attraction force between the nucleus of first $\ce{He}$ atom with two positive charges and the two electrons in the second atom causes each atom to polarize  (creating an induced dipole in each atom). Note that at the same time, there is  an electrostatic repulsion acing as well between electrons of each other as depicted in the image above.
Dispersion forces are present between all atoms or molecules. In molecules, these forces are present regardless of whether they are polar or nonpolar. London dispersion forces depends on few factors:

Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a larger atom or molecule, the valence electrons are farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles (e.g., $\ce{Cl2}$ is a gas at room temperature and $\pu{1 atm}$ pressure, while $\ce{Br2}$ is a liquid and $\ce{I2}$ is a solid under the same conditions).
The strength of London dispersion forces varies according to molecular shapes. For example, $n$-$\ce{C5H12}$ and $neo$-$\ce{C5H12}$ are both pentanes, but former is more cylindrical shape while the latter is more spherical shape. Although, both have the same molar mass, $n$-pentane is a liquid (average boiling point of pentanes: $\pu{36.1 ^\circ C}$) at room temperature and $\pu{1 atm}$ pressure, while $neo$-pentane is a gas (boiling point: $\pu{9.5 ^\circ C}$) under the identical conditions.

Also, London dispersion forces are depend of the molecule's polarizability (The ease with which the electron distribution around an atom or molecule can be distorted is called the polarizability). London dispersion forces tend to be stronger between molecules that are easily polarized, but weaker between molecules that are not easily polarized.

Now , since F is more electronegative . The electron(by hydrogen ) is more pushed towards Fluorine. My confusion is that if F is more electronegative than H. Then , the electron should be present more towards the H side.

Electronegativity is a measure of an atom or an ion to attract electrons in a covalent bond and by definition both electrons of the covalent bond will be closer to F.

Now , we can notice that for orbital number 1. There is a dark region and a light region(right side). Now , what they say is that the darker region means the electron fining probability is more there. Similarly , for the number 2. Electron fining probability is more on left side.

This is not a London Dispersion Force.We use London Dispersion Forces to understand intermolecular forces of molecules which are non-polar.


How is the Nitrate Ion (NO3) formed? I understand how the nitrite $\ce{NO2}$ ion can be formed and have a negative charge, but the nitrate $\ce{NO3}$ ion is confusing me. I made some simple drawings to try to explain what I don't understand:

Does the "electron from outside" that the oxygen receives, as indicated in the drawing, comes from the nitrogen atom? If so, then the N atom would be left with only 4 valence electrons instead of 5, thus being able to form another covalent bond with the extra oxygen, forming $\ce{NO3}$. Is this the correct line of thinking?

An organiker would look at it this way.

An inorganiker might look at it as literal N(+5) with two double-bonded oxygens.  It does redox as N(+5). Coordination structures are varied,

your NO2- structure is correct ,  the 3rd Oxygen which is trying to be added in NO2- has 6 electrons in valence shell , that means it needs 2 more. N has 1 lp means 2 unused electrons. N will share its lone pair with Oxygen (oxygen will not use its 6 e, only 2 e of N will be shared by both O and N) , this type of bonding is known as coordinate bond. As Nitrogen is sharing its electron it will get +ve charge , and O is taking Nitrogen's electrons it will get -ve charge. Both -ve and +ve neutralize each other and the resulting compound is NO3-.  If any confusion left, you can check some videos of coordinate bonding , then you will surely understand.


Which base pair is more stable, U:G or I:C? As a result of damage (usually deamination), DNA will sometimes contain uracil and inosine bases. I am trying to determine which unnatural base pairing combo would be more stable: uracil paired with guanine or inosine paired with cytosine. See the images below.

One can discuss the strength of H-bonds, but that would be half of the picture. It is more important how well (or poorly) this pair fits in the whole helical structure. Stacking interactions (arene-arene interactions) are playing important role in stabilization of DNA. If the pair disturbs backbone shape the binding will be poor. Google scholar search for relevant papers might be the most productive way to go.


How to estimate the equilibrium constant of the reaction between NaF and HCl Since $\ce{HCl}$ is more acidic than $\ce{HF}$, the left hand side of a reaction has a stronger acid than the right hand side.

Since $\ce{NaF}$ is more basic than $\ce{NaCl}$, the left hand side has a stronger base relative to the right hand side.

$$\begin{align}\text{Stronger Acid/Base} & \ce{->} \text{Weaker Acid/Base}\\ \text{(Less Stable)} & \ce{->} \text{(More Stable)}\end{align}$$

So shouldn't the equilibrium constant be larger than $1$?

My textbook says the equilibrium constant is less than $1$.

We can entirely ignore the sodium cation; it is only a spectator ion. This means, the reaction we are observing is:

$$\ce{HCl + F- <=> Cl- + HF}\tag{1}$$

The equilibrium constant for this reaction is:

$$K = \frac{[\ce{Cl-}][\ce{HF}]}{[\ce{HCl}][\ce{F-}]}\tag{2}$$

We can expand this equation:

$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]}\tag{2'}$$

And then we realise that that is nothing else than:

$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]} = \frac{[\ce{Cl-}][\ce{H+}]}{[\ce{HCl}]} \cdot \frac{[\ce{HF}]}{[\ce{H+}][\ce{F-}]} = \frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{HF})} \tag{2''}$$

So the equilibrium constant is the fraction of the acidity constants of $\ce{HCl}$ and $\ce{HF}$. We know that $\ce{HCl}$ is a much stronger acid than $\ce{HF}$. This is reflected by the inequation:

$$K_\mathrm{a}(\ce{HCl}) > K_\mathrm{a}(\ce{HF}) \tag{3}$$

Since the numerator is larger than the denominator, the value of the fraction must be larger than $1$. Therefore, the product side is preferred.

If your book arrives at any other conclusion, it is disregarding experimental results and should be replaced.

I guess it's late, but better late than never.

Technically, HF should be the stronger acid, as its parts are already strongly partially charged. So strong in fact, that even after dissociating into water the iones are present rather as ion pairs (H3O+.F-) than as seperate ions. This however inhibits fluoridic acids property as acid.

The proper reasoning for these two acids therefore would be simply per electronegativity difference between their compounds, as both acids have only 2 atoms, of them being hydrogen and the other a member of the halogen class.


Why is CaCl2 called calcium chloride? Doing a first year chem class.
Just read through the molecular naming of compounds and now I'm confused as to why $\ce{CaCl2}$ is called calcium chloride and not calcium dichloride?

The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) reads as follows:

(…)
The stoichiometric name of the compound is then formed by combining the name of the electropositive constituent, cited first, with that of the electronegative constituent, both suitably qualified by any necessary multiplicative prefixes (‘mono’, ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., …). The multiplicative prefixes precede the names they multiply, and are joined directly to them without spaces or hyphens. (…)

So far, the systematic name of $\ce{CaCl2}$ should be calcium dichloride.
However,

Multiplicative prefixes need not be used in binary names if there is no ambiguity about the stoichiometry of the compound (…).

Therefore, the customary name calcium chloride is correct.

Calcium chloride is an ionic compound so cation is named first followed by anion. Do not get confused with carbon dioxide and sulfur dioxide. They are not ionic compounds.


How to determine number of structural isomers? I have come across many questions where I'm asked to give the number of possible structural isomers. For example number, structural isomers of hexane is 5, while the number structural isomers of decane is 75.

How can I determine the possible number of structural isomers of a given organic compound?

It isn't easy but it is an interesting research topic

Determining the number of possible structures for a given range of chemical formulae isn't simple even for saturated hydrocarbons. The number of possible structural isomers rises rapidly with the number of carbons and soon exceeds your ability to enumerate or identify the options by hand. Wikipedia, for example, lists the numbers of isomers and stereoisomers for molecules with up to 120 carbons. But the counts are getting silly even at 10 carbons where there are 75 isomers and 136 stereoisomers.

It has been an interesting research topic in computational chemistry and mathematics. This old paper (pdf), for example, list some formulae for simple hydrocarbons among other simple series. Part of the interest arises because of the relationship to the mathematics of graph theory (it seems that chemistry has inspired some new ideas in this field of mathematics partially because enumerating possible isomers of hydrocarbons is strongly related to drawing certain simple trees which is intuitively obvious if you use the standard chemical convention of drawing just the carbon backbone and ignoring hydrogens).

You can look up the answers on the fascinating mathematics site OEIS (the online encyclopaedia of integer sequences). The sequence for simple hydrocarbons is here.

But the mathematical approach oversimplifies things from the point of view of real-world chemistry. Mathematical trees are idealised abstract objects that ignore real-world chemical constraints like the fact that atoms take up space in three dimensions. This means that some structures that can be drawn cannot exist in the real world because the atoms are too crowded and cannot physically exist without enough strain to cause them to fall apart.

Luckily, computation chemists have also studied this. There is, unfortunately, no obvious shortcut other than trying to create models of the possible structures and testing them to see if they are too strained to exist. The first two isomers that are too crowded are for 16 and 17 carbons and have these structures:

If you have any intuition of the space filling view of these, you should be able to see why they are problematic. A research group at Cambridge University has produced an applet to enumerate the physically possible isomers for a given number of carbons which is available here if your Java settings allow it. The results are discussed in a paper available in the Journal of Chemical Information and Modelling.

We can calculate the number of structural isomers by double bond equivalent i.e by using the formula:

$$1+ \sum_i \frac{n_i (v_i - 2)}{2} $$

where $n$ repesents the number of atoms $i$ of a given valency, and $v_i$ represents these atoms’ valency. Example:

$$\ce{C4H6}: 1+ \sum_i \frac{n_i (v_i - 2)}{2}  = 1+ \frac{4(4-2)}{2}+\frac{6(1-2)}{2}  \\
= \frac{8-6}{2} +1 = 2 $$
Thus the compound may contain two double bonds or a triple bond and there are two possible isomers.


Why is the temperature graph sigmoidal in simple distillation of components with very different bps? I'm writing an organic chemistry lab report on distillation and I thought I understood this, but the more I look at my lab textbook, I realize I don't.  My textbook 1 explains that the temperature of a mixture will rise continuously while it boils and distills if it is composed of two components with similar boiling points, but take on a sigmoidal shape when they are

When a liquid mixture is distilled, often the temperature does not
remain constant but increases throughout the distillation. The reason
for this is that the composition of the vapor that is distilling
varies continuously during the distillation (see Figure 14.3B).

I don't understand this sentence: "The reason for this is that the composition of the vapor that is distilling varies continuously during the distillation." I understand why it is true, but not why it is an explanation for the shape of these graphs. Let's say I'm distilling volatile component B from non-volatile component A. If the boiling point of the mixture is the point at which the vapor pressure of the mixture equals the atmospheric pressure, what does the composition of the vapor have to do with anything?  The relevant pressure is the atmospheric pressure, right?  I would think that as component B is boiled off, the vapor pressure of the mixture would decrease, because the mixture has less B, by Raoult's law, Ptotal = PA + PB = PA°NA + PB°NB, and as its vapor pressure goes down, its boiling point goes up.  I don't see why this isn't happening continuously.
Is the relevant pressure not the atmospheric pressure, but some additional pressure exerted by the vapor in the flask?  (It's been a long time since I took general chemistry.)

Pavia, D. L., Lampman, G. M., Kriz, G. S., & Engell, R. G. (2011). A small-scale approach to: Organic laboratory techniques. Belmont, CA: Thomson Brooks/Cole

You must understand... in an experiment, the sigmoidal shape will never have infinite slope... This would be a discontinuity... the sigmoidal shape (rightmost) and the "curve" (middle image) are essentially representing identical phenomena... there is a transition from one pure component to another. The only difference is how abrupt the change is... and this is determined by the endpoints/boiling points for each component. The more abrupt the change from one to another pure component, the steeper the slope of the change. I hope this helps. I can speak to this further, if necessary.


How to balance the equation for the reaction of aluminum bicarbonate and hydrochloric acid? Here's what I have to balance:

  Solid aluminum bicarbonate and hydrochloric  acid combine to form an aluminum chloride solution and carbonic acid.

Or in an equation form:

$$\ce{\text{__ } Al(HCO3)3 + \text{__ } HCl -> \text{__ } AlCl3 + \text{__ } H2CO3}$$

Side Note:

$\ce{Al}$ has a charge of $3+$ and $\ce{(HCO3)}$ has a charge of $1-$. To balance the ionic compound we have to change it to: $\ce{Al(HCO3)3}$.

The underscores before each compound in the chemical equation are just there for number values we can input in order to balance the equation.

My question is: what would be the reactants and products in this chemical equation? Would $\ce{HCO3}$ as a whole compound be a reactant itself?

My question is... what would be the reactants and products in this chemical equation?

You give the equation, you just need to balance it.

  Would HCO3 as a whole compound be a reactant itself?

No, $\ce{HCO3^-}$ is an anion.


Why is this reaction spontaneous in standard conditions? To study if this reaction is spontaneous I proceed in this way:
$$\ce{2H_{3}O^{+} +Fe ->H_{2} +Fe^{2+} + 2H_{2}O}$$
First I write the two semireaction:
$$reduction: \ \ \ce{2H_{3}O^{+} +2e^{-} -> H_{2}} $$
$$oxidation: \ \ \ce{Fe -> 2e^{-} +Fe^{+}} $$
Then I write the scheme of the cell:
$$\ce{Fe|Fe^{2+}||H_{3}O^{+}|H_{2}}$$
I know that:
$$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}$$
$$E^{0}_{cell}=E^{0}_{H_{3}O^{+}|H_{2}}-E^{0}_{Fe|Fe^{2+}}=-E^{0}_{Fe|Fe^{2+}}=-0.44$$
But also:
$$\Delta G^{0}=-nFE^{0}_{cell}$$
So $$\Delta G^{0}>0$$
And the reaction is not spontaneous, while my textbook says it is, so I must be wrong. Someone could find my error?

It has been a while since I took electrochemistry so I'm not sure about the formal procedure that you're using.

You wrote the correct overall reaction.

$$\ce{2H3O+ + Fe -> H2 + Fe^{2+} +2H2O}\tag{1}$$

and you almost wrote the correct half cells reactions (should be $\ce{Fe^{2+}}$ in equation 3):

$$reduction:  \ \ \ce{2H3O+ + 2e− ->H2}\tag{2}$$

$$oxidation: \ \ \ce{Fe -> 2e^{-} +Fe^{2+}}\tag{3} $$
Then you wrote the scheme of the cell: $$\ce{Fe|Fe^{2+}||H_{3}O^{+}|H_{2}}\tag{4}$$ And an equation for the overall cell potential.
$$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}\tag{5}$$

Your problem is between equations 4 and 5. To subtract $E^{0}_{left}$ the potential must be written as the reduction potential for the half cell. So either you're supposed to be:

Adding the values for the right and left reactions in equation 5

-- or--

Writing both half cells as reduction potentials in equation 4.

The gist here is that the tabular values are always written as reduction potentials. So for iron:

$$\ce{2e^{-} +Fe^{2+} -> Fe }\quad\quad E^{0}_{red} = -0.44\tag{6} $$

I was taught to reverse the sign for an oxidation reaction.

$$\ce{Fe -> 2e^{-} +Fe^{2+}}\quad\quad E^{0}_{ox} = 0.44\tag{7} $$

Then add the reduction and oxidation potentials to get the overall cell potential.

$$E^{0}_{cell}=E^{0}_{red} + E^{0}_{ox}\tag{8}$$

Your mistake is that $E^\circ_\ce{Fe/Fe^{2+}} = \pu{-0.44 V}.$ So that $-E^\circ_\ce{Fe/Fe^{2+}} = \pu{+0.44 V}.$


How do we calculate N factor and why is it useful in calulations? I don't understand the concept of N factor at all. Why is it important?

I know that it is something dependent upon the oxidation number, for example N factor of Na would be 1 since it has oxidation number of +1.

For acid and bases we look at the replaceable $\ce{H+}$ or $\ce{OH-}$ present in them.

Apart from this, I didn't really understand how I can use N factor for calculation or how to find the N factor for whole chemical reactions. Please correct me where I'm wrong.

There are some different applications.
For one, it tells you how many times an acid can lose a proton. For instance, you can have a diprotic ("with $\ce{2 H+}$") molecule, with $n=2$. For instance, $\ce{H2SO4}$, deprotonates to $\ce{HSO4-}$, which can then go to $\ce{SO4^{-2}}$.
Why is that important? Take an acidic reaction with $\ce{H2SO4}$. If the extent of reaction is strongly forward, you would treat $\ce{H2SO4}$ as if it had twice its concentration to take into account that it has two protons it can lose. If you only accounted for the first proton you would be underestimating this effect.
It's also known as equivalent concentration, which might make it sound less foreign to you.
In redox reactions, it shows how many electrons an agent can accept/donate, and in precipitation reactions it measures the # of ions that'll precipitate out of a solution. For all these situations the n-factor or equivalent concentration corrects for multiplicity, since now we know not all molecules are made equal. Some acids can deprotonate twice, some reagents will lose 4 electrons in redox while others may lose just 2. To consider these differences we use this concept.
Sources: Wikipedia

n factor is atomic weight or molecular weight by equivalent weight. n factor for acids is its basicity, ie, number of replaceable hydrogen ion per molecule. For exampleg $\ce{HCl}$ has n factor 1 n $\ce{H2SO4}$ has 2 $\ce{H3PO3}$ has n factor 2 because when $\ce{H3PO3}$ dissociates to form H+ and HPO3^-2. Same ways for bases its n factor is its acidity or no of replaceable OH ions per molecule. Eg: for $\ce{NaOH}$ it is 1 n for $\ce{Ca(OH)2}$ its 2.


What hinders SN2 more: geminal alkyl group or alkyl group on the aromatic ring? Which among these two (1-chloro-1-phenyl-ethane and 2-(chloromethyl)-1,3-dimethylbenzene) is more reactive towards SN2?

I strongly believe that it should be Y because of lesser hindrance on the electrophilic carbon.
However, the answer key of my exercise book states that X > Y. What's the reason behind that?

The correct answer is X because the 2 methyl groups on the Y are bulky enough to cause steric hindrance. Surprisingly, from what I have seen, if you only had one methyl group on Y, the SN2 reaction would proceed.


Hydrolysis of a salt B2X An aqueous solution of soluble salt $\ce{B2X}$ has concentration of 0.1 M.
Given data :

  $\ce{K_b(BOH) = 2 × 10^{–5}}$

  $\ce{K_{a1}(H2X) = 2 × 10^{–7}}$

  $\ce{K_{a2}(H2X) = 5 × 10^{–10}}$

What is the pH of the solution?

My attempt,

$$\ce{B+ + X2- -> BOH + HX-  \,\,...      (1)}$$

The hydrolysis constant of $\ce{HX-}$ comes out to be about $10^{-7}$, so I decided to neglect it (is this correct?)

$$Kh=[HX-][BOH]/[X2-][B+]$$

So  the  value  of  $$\ce{K_h=\frac{Kw}{K_{a2}*K_b}}$$

Upon substituting the values I get the Kh=1..(something i have never seen before)

So this means that for the reaction

$$\ce {B+ + X2- -> BOH + HX-}$$

We will end up with 0.05M of $\ce{B+}$ , 0.05M of $\ce{of X2-}$, 0.05M of $\ce{BOH}$ , and 0.05M $\ce{HX-}$ if we start with 0.1 M of $\ce{B+}$ and 0.1 M of $\ce{X2-}$

Now, I do know that

$$\ce{pH=7+\frac{pK_a + pK_b}{2}}$$

I also know that this is true when the hydrolysis constant is smaller than 1. Therefore I cannot use if for this problem.

(Nevertheless, I disregarded this and just used it. I got the answer to be 9.305)

So I am stuck over here.

I got another idea.

When I now look at the sum, the resulting solution looks like a mixture of two buffer solutions,

Buffer 1: $\ce{BOH/B+}$

Buffer 2: $\ce{HX-/H2X}$

I don't know whether this observation is correct. Even if it is correct, I am not sure whether it be used to solve the problem.

I am posting a solution image regarding same question. 2nd hydrolysis of X2- is neglected as you did. important thing is degree of hydrolysis is not small and should not neglected.


How does NaCl (or any inorganic salt) increase surface tension? Does the compound "prefer" to stay in the solution?
If yes, wouldn't the surface tension of the solution be equal to that of pure water, since only water is in the surface?

See: Does NaCl reduce the surface tension of water?

Or is the compound distributed evenly and because the ionic bonds are stronger than the attractions between water molecules, the surface tension increases?

Read here a reddit answer to why do ionic compounds such as NaCl increase the surface tension of water?

Explain like I'm 5 please.

I hope this picture helps.

The interaction between a molecule of water and an ion is stronger than the hydrogen bonding that occurs between two water molecules.

1+1 = 2, exactly. Other than that, use of the term "exactly" and similar terms like "absolutely" should be avoided in technical discussions. Also, while I understand what you meant, its quite unlikely that most 5 year olds are capable of understanding such abstract concepts, but I digress...

The ions increase the average particle-to-particle (p2p) attractive forces in the aqueous solution, which means that more energy is required to increase the surface area (bring a particle from the interior (bulk) to the interface. The traditional example is comparing the bond energy of a ion in a cubic closest packed (6 nearest neighbors in a "square" 3D grid) (so it has a total bond energy of $6x$, where $x$ is the energy of a single bond) to the same type of ion at the surface of a (large) lattice - at the surface, one bond will be missing ($5x$). (This is a simplification, since it ignores all electrostatic forces (both repulsive and attractive) of ions not 'nearest neighbor'.)

Of course, in the real world surface energy/tension considerations are usually concerned with at least two phases (eg liquid & gas in aerosols) and often three (eg. water wetting the side of a glass container in air). You can imagine that the measured surface tension could be different if measured in helium compared to the surface tension of the same substance measured in carbon dioxide. The link you provided to the reddit had some correct information, but then it went off the rails when it discussed detergents. It's wrong, there. The stackexchange Q&A link you provided seems to me to be a correct point of view, but I don't find it useful, do you? The same argument applies at the surface as in bulk (for small solute molecules and ions). The ion must have a solvation sheath (a coat of solvent molecules) to 'insulate' the charge, otherwise it wouldn't go into solution. (This applies to polar solvents (and water is both polar and a strong hydrogen-bonding molecule)). It should be clear that if the ion were at the surface, that it wouldn't have as "thick" a coat so it will tend to avoid it.

Keep careful track of each argument's scale. Talking about a lattice of ions is considering a much smaller scale than a coated ion in solution. By that I mean the word "surface" means different things at those two scales. In the former (crystalline solid) it is the outermost ions of the lattice, while in the latter it is a zone of multiple solvent molecule diameters depth, with the molecules and ions bouncing around, and forming only intermittent (secondary as in intermolecular, non-covalent, non-ionic) bonds.

Detergent molecules are generally much larger, and they usually have a non-polar tail which doesn't interact with water (hydrophobic) and a polar head (which has much more interaction with water (meaning it increases the secondary bonding energy of the system)). These molecules orient themselves at the surface and so for (most) detergent solutions, the surface is highly enriched with them. They use this fact to coat mono molecular films (one molecule thick) onto glass. So, it is not true that ions aren't to be found at the surface of the solution, but it is true that their concentration there will be less (and depend on temperature). The key fact is that they increase average intermolecular bond energy of the solution.