What makes Kleenex Cool Touch tissues cool to the touch, and is it safe? I love Kleenex Cool Touch tissues.  I'm just wondering what makes them actually feel cool.  For some reason it makes me think of refrigerant but I'm sure that's not in them.

Is it the combination of common ingredients that somehow gives it a moist feel, making it feel cooler than normal tissues?

Are the tissues safe to put to your face, if you are potentially inhaling the ingredient(s)?

Ingredients:  Stearyl heptanoate, stearyl alcohol, polyethylene, mineral oil, coconut oil, aloe

My first guess as to the nature of this behaviour is similar to water. Hypothermia is a realistic risk if you are drenched in water. This is because water can absorb large amounts of energy (it has a large specific heat capacity) to only rise in temperature slightly. Water itself evaporates into steam spontaneously, if some water is spilled then it will evaporate readily, regardless of the temperature. Likewise, the solids in the tissue are probably coming closer to melting temperature, which I believe is higher than your body temperature, but quickly absorbing a lot of energy from your body in the process.

Something similar to properties of water are probably characteristic to the tissue. Leave the tissue in the open in a cool room, along with a glass of water. After awhile, touching the glass of water and the tissue should invoke the same sense of coolness. This is because your body is at a higher temperature than the items in the room.

There is likely another property of matter besides specific heat capacity that is a measurement of how quickly heat is transferred, this is likely to play a role as well. I do not know what this property is called and it was never covered in my classes so far, but take metal for example. Metal has a very small specific heat capacity, it can not hold much energy. However, touching a metal object invokes a sense of coolness. I believe this is because heat is transferred very quickly through metal, although it warms very quickly due to its limited heat capacity. If you put a metal cup and a porcelain cup on the stove, the metal will burn you much quicker if you grab it than grabbing the porcelain/ceramic cup.

If you were very interested and inquistive you could test the specific heat of the tissue through experimentation. Also, try holding the napkin for a few seconds to try and find out how quickly the tissue warms to the touch. Chances are, it warms relatively fast and can absorb a lot of energy, giving you the impression that it is 'cooler' than most objects, to which marketers will be overwrought with the joy of hearing.

Here is the answer, but I don't understand it.

http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=2&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=86&f=G&l=50&co1=AND&d=PTXT&s1=%22kimberly+clark%22&OS=%22kimberly+clark%22&RS=%22kimberly+clark%22


Synthesis of oxalyl chloride that doesn't involve phosphorus I have been synthesizing TCPO (bis(2,4,6-trichlorophenyl) oxalate) for chemiluminescent reactions, using 2,4,6-trichlorophenol and oxalyl chloride.

Since I am tired of paying for oxalyl chloride (and hazard class 3 shipping), I wish to find a synthetic route to produce my own stock.

The most common synthesis of oxalyl chloride uses oxalic acid and phosphorus pentachloride. This would reduce the cost, but is also problematic because phosphorus chlorides are considered to be drug precursor (21 CFR 1310.02), still too expensive, and hazardous to ship.

Can anyone recommend a synthetic route to oxalyl chloride that does not use phosphorus? Or an alternate (safer / easier / less expensive) route to TCPO?

Most amide coupling agents are suitable for esters. Feast your eyes on: Amide bond formation - beyond the myth of coupling reagents.

This lists old favourites such as CDI, carbodiimides, etc. A lot of it is overkill. The water soluble carbodiimides might be preferred for ease of urea removal later.

The likes of CDI or $\ce{Ac2O/DMAP}$ can react with your phenol when over-stoichiometric amounts are used.

Alternatives could be: boronic acid derivatives e.g.  phenylboronic acid catalysis; scandium or yttrium triflate catalysts (all requiring water removal, but a Dean-Stark could be used if a water immiscible solvent is used); dimethyl oxalate (already suggested)  but in conjunction with Takasago's ZnTAC24 catalyst. MeOH removal by distillation is required, which is quite trivial for such a volatile component.

i think you should use oxalic acid and convert it to oxalyl chloride by thionyl chloride.take a few mL of thionyl chloride and put oxalic acid into it and reflux for about three to four Hr, evaporate the remaining thionyl chloride. i used this method for the preparation of benzoyl chloride


Is every substance in equilibrium with its vapours? In one of my previous questions, I got an answer from user Jan:

The smell of a compound as we perceive it usually depends on which receptors in the nose will bind the molecule (or not).
These receptor, being proteins, are usually rather specific for certain shapes, not for bonding patterns. So chlorobenzene might fit into similar receptors as toluene which would lead to at least partly a similar smell, but that is it.

This means that a substance must be in the vapour phase in order to smell it, which brings me to my question:

Do all substance (at least those which have a characteristic smell) maintain an equilibrium with their vapours (if the substance is in closed container)?

But if it's true for those solids, then it must be true for all other solids, as well as nonvolatile liquids that do not have a characteristic smell to humans. Is this not the case? Also, if this substance is placed in open air, will it evaporate totally?

If they were maintaining an equilibrium with their vapour you wouldn't be smelling it.

In theory any equilibrium in the case of vapor pressure occurs in a closed container, in reality a vapor pressure equilibrium is rarely if ever reached.


Condensation of Ethylene Glycol I'm looking to reflux a 250 mL of ethylene glycol solution. Would an Allihn condenser be suitable; if so what length?

Ethylene glycol was a boiling point of 193 °C using any water cooled condenser will crack. Use a air cooled condenser. It is just a glass rod. The length of tube depends on how much you want to condense and how quickly you want it to come over. Remember visible condensation should always take place in the condenser. If it goes further down the tube then some hot vapours will be lost to atmosphere if your system is open. If under vacuume the ethylene glycol will boil at a lower temperature but you should use a cold trap at minus 40 °C or below to catch vapours which come over. Remember liquid ethylene glycol is toxic and the vapours are more dangerous check the MSDS. I suggest that you should work with a closed system in a fume cupboard for this job or you might never see your grandchildren. Good luck.


Why is sulfur the central atom in the Lewis structure for SO2? So apparently the Lewis structure of $\ce{SO_2}$ is

It was my understanding that the central atom is the one that is more electronegative. And an atom is more electronegative the closer it is to Fluorine (top-right). Oxygen is definitely closer to fluorine than sulfur is. Then, why is sulfur the central atom?

Also, one thing that has been bugging me for a while: if there are two or more atoms for the central element, how do you make the Lewis structure?

I think you messed up something. Because it is exactly opposite.

"Central part of atom is normally least(not the most) electronegative atom i.e. S in this case.

For polyatomic molecule least electronegative polyvalent atom acts as central atom and by using this we wlll get most plausible  structure  (structure having least magnitude of formal charge) for polyatomic molecule.


What would the name of this organic compound be? I am a little lost in naming this compound.

I think it would be something like 2 bromo 4 methylene pentene??

Is that close?

It is a five carbon chain, so pentane, with a Br on carbon two, a methylene substituent on carbon 4, and a alkene on carbon 1.


The impossibility of 100% ionic bond Recently, I read the definition of  oxidation state on Wikipedia. It read that a 100% ionic bond is impossible. So what does a 75% ionic and 25% covalent bond mean at all?

A "100% ionic" bond would be a bond whose bonding electron(s) were never in the vicinity of the cation, but rather always in unperturbed valence orbitals of the anion.  That's not far from the truth in a paradigmatic ionic compound like $\ce{NaCl}$, but no matter how electronegative the anion is, the bonding electrons will still experience some attraction toward the positive charge on the cation, and so a detailed model of their wave functions will show a small but nonzero amplitude near the cation.  That's just basic electrostatics; you have a positive charge and a negative charge not that far away from each other, they will attract.

Despite that, we can still talk about 100% ionic bonds as a theoretical limit and an acceptable approximation for modeling some compounds.  It's kind of like how we can use absolute zero as the theoretical zero point of a temperature scale even though nothing in the universe can ever be that cold.

There cant be a 100% ionic bond because even if an element has more electronegativity than another the element with lower electronegativity would still attract the bonding pair.


Does a change in oxidation number in a redox reaction mean that an actual electron transfer occured? Oxidation numbers are fictitious charges that pretend the entire molecule is an ion i.e. it artificially localizes electrons onto atoms within a single molecule. In a redox reaction, we find that an oxidation state of one molecule changes compared to a different molecule. Does this mean an actual electron was transferred given that oxidation numbers are fictitious? I'm trying to rationalize this, my guess is that the answer is yes because oxidation numbers are a formal way to localize electrons within a molecule, but differences in oxidation number across molecules only happen with physical electron jumps.

Oxidation numbers are fictitious charges that pretend the entire molecule is an ion i.e. it artificially localizes electrons onto atoms within a single molecule.

Sort of. In calculating oxidation numbers we pretend that covalent bonds are ionic. Perhaps it is valid to say we are pretending the entire molecule is an ionic compound.

In a redox reaction, we find that an oxidation state of one molecule changes compared to a different molecule.

More accurate, we find the that oxidation states of atoms change going from reactants to products. Charge conservation, coupled with the way we assign oxidation states, guarantees that the sum of oxidation states is constant. So if one atom shows an increase in oxidation state, at least one other has to show a decrease in oxidation state.

Does this mean an actual electron was transferred given that oxidation numbers are fictitious?

It is not clear what "an actual electron" means. Electrons are indistinguishable, you can't paint one red and follow its path during a reaction. If the electron transfer (according to oxidation numbers) is intramolecular, it becomes a bit of a philosophical question whether there was an electron transfer. For example, is there an electron transfer during keto-enol tautomerism?

I'm trying to rationalize this, my guess is that the answer is yes because oxidation numbers are a formal way to localize electrons within a molecule, but differences in oxidation number across molecules only happen with physical electron jumps.

It is not clear what "physical electron jumps" are.
For some reactions, it is clear that you have an electron transfer because the only thing that distinguishes reactants from products (pairwise) is the number of electrons:
$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$
If this reaction is run as two half-cells, this is even more clear-cut.
For other reactions, the changes are more subtle. If a iodine radical combines with a methyl radical (as part of a chain termination reaction), it is formally a redox process:
$$\ce{I. + .CH3 -> I-CH3}$$
However, the electronegativities of iodine and carbon are quite similar, so it is not intuitive that an electron transfer happened. As Maurice points out in the comments for a similar case, you would have trouble writing this in terms of half-reactions.
In the reaction of methyl lithium with a carbonyl, the oxidation numbers change. You might not see it as a electron transfer and would rather think of it as methyl transfer. However, because carbon is more electronegative than lithium, the oxidation numbers of the carbonyl carbon and the methyl anion change when a bond is formed between them.

Does a change in oxidation number in a redox reaction mean that an actual electron transfer occured?

If I answer this like a certain politician: it depends upon what the meaning of the phrase "actual electron transfer" is. So it is mostly a question about how we talk about redox reactions rather than a question about the underlying chemistry.

Yes, the electron transfer does take place...oxidation is the loss of electrons and reduction is the gain of electrons. In a redox reaction, the oxidizing agent gains electrons and the reducing agent loses electrons, i.e, actual electron transfer takes place.


Is the 'n' in Delta G = -nFE, ever negative? Is it possible for $n$ in $\Delta G=-nFE$ to ever be negative? Trying to figure out where I went wrong on some thing and want to rule that out.

No. This result occurs from the Nernst equation.

The "n" is the number of electrons transferred. If no electrochemical reaction occurred, then n = 0.

Otherwise "n" is positive.

Having a negative number of electrons transferred would be impossible.

$n$ cannot be negative.

$$\Delta G = -nFE$$

Transfer of $\ce{e-}$ cannot be negative.


Is it possible to melt a diamond? I am school student. In my school text book there is a fact about diamond. That is

  Each carbon atom is covalently bonded to four other carbon atoms. A lot of energy is needed to separate the atoms in diamond. This is because covalent bonds are strong, and diamond contains many covalent bonds. This makes diamond's melting point and boiling point very high.

If a diamond is a single molecule, how does a single molecule melt? Is that fact correct? Please tell me the answer!

The melting point of diamond is certainly an exotic topic. Diamond is slightly less stabile than its allotrope, graphite, and will in most circumstances turn to graphite way before its melting point. If you have enough pressure it will stay diamondlike and it will at some point melt.

A diamond is not a single molecule unless you stretch the definition of molecule. It is usually defined as the smallest fundamental unit, and for diamond I'd call the tetragonal carbon unit cell the smallest fundamental unit, or maybe even a carbon atom. Diamond, having one of the highest values for heat transfer among substances would certainly melt almost like a single molecule though. But in theory, one and one carbon decides to leave behind it's friends bound by the diamond structure boundaries behind to seek their fortune in the hippie camp that is called liquid.

It's true that the high melting temperature of diamond is caused by its structure. Usually a substance melts when you overcome the intermolecular attraction forces, whereas for diamond the forces involved are interatomic, which are several orders of magnitude stronger. However, one could argue that fusion is a physical process, and in this case we're breaking chemical bonds so it's more of a chemical process which would make it, technically, not fusion.


Why change in enthalpy is negative? I noticed that the enthalpy of the products of an exothermic reaction (bonds formed are stronger than bonds broken) is lower than the enthalpy of the reactants. Enthalpy equals internal energy plus $PV$. In a constant pressure environment, change in enthalpy equals to change in internal energy as $PV$ in constant. Internal energy consists of $E_k$ (kinetic energy) of particles and $E_p$ (potential energy) of bonds between particles. When stronger bonds are formed, $E_p$ is bigger. And the energy of the heat released to environment equals to the energy gained in form of potential energy, so why is the internal energy lower?
It would be helpful if someone shows me my mistake :)

Bonds actually have negative potential energy. Think about it this way: in order to break a bond you have to expend energy so when a bond is created, energy must therefore naturally be released. Hence internal energy is lower.

The change in enthalpy is negative in an exothermic reaction because energy is "lost" through the reaction (because there is more energy on the products side than on the reactants side).

Another way to think about this is by calculating the enthalpy before and after a reaction, for example - and this is a synthesis [and exothermic] reaction:

$$\ce{2H2 + O2  <=> 2H2O}$$
The ΔH [change in enthalpy] is -484 kJ (you can calculate this using standard enthalpy of formation numbers).

You can also think back to the definition of enthalpy - disorder: When two molecules that were 'freely' moving around, come together, there is less disorder [so the disorder in the system has decreased, and consequently is negative].


How to plate zinc evenly on a threaded bar I need to set a number of threaded bars into cement, to hold a new gatepost. When I cut the bar into shorter pieces, the metal near the cut became red hot, presumably hot enough to evaporate some or all of the zinc with which the entire rod was plated.

Now I want to plate zinc back onto the ends, and have had mixed success so far. I dissolved some zinc oxide with hydrochloric acid and hooked it up to a little PV panel I have, and after a while it seems there is a light-coloured deposit on the cathode. The pencil-lead (6B) anode is no longer as smooth as it first was, but it doesn't seem to be losing (much) of its bulk. I assume it to be zinc, because... well what else can it be? It plated thick enough that I had to (ab)use a nut to re-cut the thread, leaving a light-coloured shiny surface that gives me more confidence that it is, in fact, zinc.

Problem: the edges of the threads seem to be taking more zinc than the valleys. In some cases the thread's valley doesn't show any evidence of plating. The fractal-like deposit on the edges is quite pretty actually, pity I can't take a (good) closeup photo.

Is there anything I can do (perhaps using some other common(ish) kitchen chemicals?) that will result in a more uniform layer of zinc? Is it worth trying lower / higher current or pulsed current? Am I overthinking things - is it even necessary for the whole stud to have a layer of zinc?

I think the easiest way is to sputter coat the bars, if you can access one. One that can rotate your bars such that the rotation is vertical to the direction of sputtering would be the best. If not, switch sides manually.

Depends on the composition of your bars, you might want to coat something else before coating zinc. A wild guess: nickel.  I suggest so because I don't recall zinc being sticky. So the coating might come off easily.

And I'd try sputter slow but for a long time. I think low sputtering rate tends to produce more densely packed and uniform coating. I am not sure. But dense and uniform coating provides good foundation of the zinc coat.

For electric coating, I think low voltage and uniform coating is still nice. I don't think you want formation of gas at the bar electrode, as that disturbs the structure of the coating. Not sure what's influencing the nucleation sites on the bars. I would guess that the current is not uniform on the bars, say there more on the hill tops. Still consider coating something else first. I would try whatever is available. I would try to get the solutions as pure as possible though. Organic compounds can act as surfactants which tend to aggregate the particles.

Whichever your method is, search for some journal articles characterizing similar processes. That might help.

I see this is an old question ; however, in the real world steel rebar is protected with polymer coating or stainless steel rebar is used. Generally protection is unnecessary except for things like seacoast bridges.


What are Lattice Points? What are lattice points exactly? For what purpose are they used in solid state chemistry?

There are different types of lattices. A lattice point is a point in any of these lattices. Lattices are used to describe  highly ordered systems such as crystals and some supersolids. I am unsure whether lattices can only describe periodic systems (see quasicrystals). However, I will be assuming that we are talking about periodic lattices. I will also be assuming that we are dealing with crystals.
Crystal lattice
The crystal lattice is used to describe the lattice of a real crystal. For example, in NaCl, a lattice point in a crystal lattice represents the position of a sodium ion or a chloride ion.
Bravais lattice
Bravais lattices are more mathematical and abstract than crystal lattices. They are pretty much the same as crystal lattices. Unlike the crystal lattice, however, lattice points in the Bravais lattice no longer represent a position of a particular atom. Instead a lattice point represents a position in which an atom can be placed. In other words, a lattice point in a Bravais lattice is a point, which is equal and indistinguishable from any other another point. What matters in a Bravais lattice are not the points themselves, but how they are arranged (i.e. symmetry).
Reciprocal lattice
The reciprocal lattice is the Fourier transform of either the crystal lattice or a Bravais lattice. More often than not, it is used to refer to the Fourier transform of the Bravais lattice. More of that is explained in Physics.SE. However, I personally think that this video explains it better. A lattice point here can represent an atom or a point, depending on the context.
Finally, as mentioned by other posts, these points can be vacant.

The points in a crystal lattice, representing the positions of structural units(atoms,molecules or ions) in the representative crystal are called lattice points .
Each lattice point in the crystal has the same environment as the structural units have in the real crystal.
How many lattice point in the unit cell, depend on the type of crystal .


Esterification of a carboxylic acid and the production of water In an esterification reaction between a carboxylic acid and an alcohol, a carboxylate ester and water is produced.

I found the following diagram online, and my textbook has a similar one as well

Question: In forming the water in an esterification of a carboxylic acid, why does the hydrogen ion come from the alcohol and the hydroxide come from the carboxylic acid and not the other way around?

I think it might be because carbon's bond to the OH- is weaker than the R bond to the H+ ion and requires less energy to remove those two ions from their respective compounds.

The mechanism of the reaction doesn't involve the direct exchange of ions in the way that, e.g., a metathesis/double replacement reaction might. Take a careful look at the reaction mechanism from the Wikipedia page. Notice that one of the intermediates (the product after step three) is a tetrahedral complex with two hydroxyl ($\ce{-OH}$) groups. One of those hydroxyls is indeed the original hydroxyl of the carboxylic acid moiety, but the other one is actually formed from the protonated oxygen of the carbonyl. Either one of them could, with equal likelihood, be protonated in a subsequent step to provide the water molecule that serves as a leaving group. In the image that you linked, it's convenient to circle those specific atoms, but that shouldn't lead you to believe that the mechanism of the reaction is in fact a simple exchange of ions.

The question that you didn't ask directly, but perhaps implied, is: why is protonation of the alcohol by the carboxylic acid, followed by nucleophilic attack by the carboxylate on the carbon of the protonated alcohol, not a viable mechanism? First, a protonated alcohol is more acidic than a carboxylic acid by six or seven orders of magnitude, which means that the initial proton transfer would be slow and unfavorable. Second, the resulting ester is apt to be hydrolyzed eventually (albeit slowly without catalysis, but then the whole reaction proceeds very slowly), yielding the original carboxylic acid and alcohol.

Esterification involves ANDN mechanism which is basically attack of one nucleophile and the departure of another from the same molecule.here OR- group attack the carboxylic acid at the carbonyl carbon causing OH- group to leave. OH- will combine with the H+ ions present in the system to form water.


Why is Kc not affected by change in pressure? Why is $K_c$ not affected by change in pressure? I know the mathematical explanation, but I don't really understand the reason when only looking at $K_c$. The explanation I know is with the reference to $K_p$, but is it possible to explain without $K_p$?

Equilibrium is reached when the forward and reverse reaction rates are equal. These rates are determined by the frequency of successful collisions, which is turn is determined by concentration (seen in the rate laws), temperature, activation energy, and how often molecules collide with the correct orientation (together these last three determine the rate constant, $k$, as shown in the Arrhenius equation).

The equilibrium constant, $K_c$ is the ratio of the rate constants, so only variables that affect the rate constants can affect $K_c$. Pressure doesn't show in any of these relationships.

$$K_c=\frac{k_\text{forward}}{k_\text{reverse}}=\frac{[\text{products}]^y}{[\text{reactants}]^x}$$

$$\text{rate}_\text{forward}=k_\text{forward}[\text{reactants}]^x$$

$$\text{rate}_\text{reverse}=k_\text{reverse}[\text{products}]^y$$

$$k=A\mathrm e^{-\frac{E_\mathrm a}{RT}}$$

It may also be useful to think about different ways pressure can be changed.

The volume of the reaction can be changed. This will affect concentrations of reactants and products, and the reaction will likely have to shift left or right to reestablish equilibrium. If volume were decreased, the rates of the forward and reverse reaction will each increase due to higher concentration. If the exponents in the rate laws are different, the amount of increase for each will be different. Although the rates change, the rate constants do not, so $K_c$ doesn't change either.
The temperature can be changed. This will affect the rate constants of the forward and reverse reactions according to the Arrhenius equation. An increase in temperature will increase $k_\text{forward}$ and $k_\text{reverse}$, but by different amounts because their activation energies are rarely equal. This changes $K_c$, which is the ratio of the two rate constants.
More gas can be added without changing the volume of the container.

a. If the gas is one of the reactants or products, this would affect one of the concentrations, and the reaction will have to shift to reestablish equilibrium. Again, this changes one of the rates, but does not affect the rate constants, so $K_c$ is unaffected.

b. If the gas is inert and doesn't react with anything involved in the equilibrium, the concentration of each reactant is unchanged. Neither the rates nor the rate constants are affected, so the system should still be at equilibrium and $K_c$ is unaffected.

Consider the equilibrium: A+2B (reversible reaction arrow) 2C.
If [C] and [B] are 2 moldm^-3 while [A] is 1 moldm-3 originally: Kc = [C]^2/([A][B]^2) = 2^2/(12^2) = 0.8 mol^-1dm^3

If pressure increases and nothing happens (which cannot happen in actual): [A], [B] and [C] increase by say, 3/2 times because an increase in pressure can only be due to an increase in the number of particles per unit volume unless there is a change in temperature, which is considered another factor as it results in a change in the value of Kc. The new Kc value will then be: Kc = 3^2/(1.5*3^2) = 0.0.667 (3.s.f.), which is lower than the actual Kc value. Hence, this is not possible as a pressure change in the system cannot result in a change in the Kc value, according to Le Chatelier's principle.
If pressure increases and and Le Chatelier’s principle is applied (which happens in actual):
[A], [B] and [C] increase by say, 3/2 times because an increase in pressure can only be due to an increase in the number of particles per unit volume unless there is a change in temperature, which is considered another factor as it results in a change in the value of Kc. However, unlike scenario 1, Le Chatelier’s principle is applied. When Le Chatelier’s principle is not applied, Kc value would decrease as the numerator ([C]^2) would increase by a smaller proportion than the denominator ([A]+[B]^2). However, when it is applied, some A and B is converted into C, thereby increasing the proportion by which ([C]^2) increases and decreasing the proportion by which ([A]+[B]^2) increases. Hence, according to Le Chatelier’s principle, the Kc value will increase back to it’s normal value at 0.8 mol^-1dm^3 from 0.783 mol^-1dm^3 when pressure of the system increases.


Bond order for carbonate ion for resonance I saw in a textbook that for carbonate ion, there are 3 resonance forms and the bond order is 1 and 1/3. So in general, how should we calculate the bond order for resonance structures?

Is there any definite way to do this?

There definitely is an easy way to do this.

One of the easiest examples that comes to mind is benzene.  If we look at the structure:

we see that each of the carbon bonds are either single bonds to the adjacent carbon or double bonds to the carbon in the other direction.  If we draw the other resonance structure these bonds shift by 1 atom to look like the following:

Thus, the overall structure can be obtained by looking at those two structure and averaging the bond order between them.  For benzene one resonance form for each bond will be single and the other double $(2 + 1)/2$. The "$2$" on the bottom of the equation is the number of resonance structure being considered.  Thus, for benzene the bond order is $1.5$.

Applying this same method to the carbonate ion, we have 3 resonance structures with bond orders of 2, 1, 1 when considering the bond between carbon and a single oxygen.  Since there are 3 structures the answer is $(2+1+1)/3 = 1.33333$ or $4/3$.

There is a formula for this too... it is = 1×[(no. Of pi bonds b/w resonating atoms of resonating structures)/(no of sigma bonds  b/w resonating atoms of resonating structures)]....for example: in benzene, hydrogen atoms do not participate in resonance so bond order = 1+(3/6) =1.5; in carbonate ion, it is 1+ 1/3=1.33 hope this helps....


Nucleophillic addition vs SN2 The following question came in Paper 2 of JEE Advanced 2014-

Here we have $\ce{CH3-}$ as the nucleophile. There are 2 possible attacks:

SN2 type attack on the chlorine group.
Nucleophilic addition on the carbonyl.

The answer given is option D, which will happen by the 2nd pathway(nucleophilic addition).

Why is nucleophilic addition given a preference over SN2? I feel that nucleophilic addition should rather decrease the reaction rate because the carbon goes from sp2 to sp3, decreasing the bond angle and hence increasing the repulsion

To begin this short discussion one has to differ between Sn2 and addition to the carbonyl group, as the former is driven by orbital interactions while the latter is driven by coulomb interactions.
This can be explained with a closer look at the carbonyl function.
due to the highly polarized π-bond between carbon and oxygen, carbon is rendered quite positive. It results that any nucleophile bearing a negative charge will be attracted to attack it. In addition the reaction of a carbanion and a ketone or aldehyde is irreversible!
there is fundamentally no way in which the carbanion could ever be considered as a leaving group.

On the other hand Sn2 reactions appear on saturated carbon (in the best case methyl or primary) with at least one good leaving group (e.g. a halide or tosylate). Often the sigma bond connecting carbon and the leaving group is not even polarized, at least not significantly. Just consider the electronegativity of iodine, the BEST leaving group for SN2.
That is why orbital interactions are more dominant in thess kind of reactions. It can be easily noticed that a potential nucleophile has to donate its electrons into a sigma antibonding orbital in order to  react with the carbon electrophile. These kind of orbital can be considered as very high in energy as it is the counterpart of the sigma bonding orbital which is quite stable.
Therefore the best nucleophiles for nucleophilic attack on saturated carbon is usually a second or third row nucleophile like sulphur or phosphorus (Wittig reaction).

Therefore for a carbanion the irreversible addition to the carbonyl group happens much faster, resulting in a alkoxid. The new formed alkoxid is in close proximity to the chlorine substituted carbon and therefore likely to form the five membered ring .

Any references that support everything written above can be found in Clayden, Organic chemistry second edition.
Specifically the chapter nucleophilic substitution at saturated carbon will help.

Best regards,
John


How is work converted to heat? My book says work can be converted into heat but I want to know how work affects kinetic energy of molecules which, when transferred, is called heat?

Work can be understood as a variation of energy during which the transporters (may they be molecules) are spatially aligned or bonded, meanwhile heat is a variation of energy in which the transporters aren't aligned (therefore heat dissipates energy).
With this in mind, any machine that uses an ordered energy transfer (like your hand waving) can induce heat in a fluidic means, as particles in fluids are initially moving randomly and are not likely to move far on the orientation of your hand.
In this example, the induced work is little and restricted to the analysis on the surface of the moving hand, but if the analysis is made from further on a wider system, we can assume you're doing work to generate heat alone.

Work is the change in kinetic energy, and in some cases, say you do some work to move a bunch of molecules; if it results in them colliding, thereby releasing heat in the process, due to their kinetic energy during collisions converting into heat.


Covalent vs Ionic bond in Silicon dioxide I was wondering why silicon dioxide tends to be a covalent bond instead of an ionic bond. Based on lewis structures, the octet rule seems to work for silicon dioxide so shouldn't it also be an ionic bond?

Because silicon and oxygen are nonmetals on the period, hence they form covalent bonds. For ionic bonding, it must be between a metal and a nonmetal.


Conversion from t-butanol to alkene How can I accomplish the above conversion?

My thoughts:

One way of introducing the double bond, would be using the appropriate Grignard Reagent after oxidation of t-butanol (somehow)
I think that initial conversion to carbonyl compound is necessary - only then can we follow up with nucleophilic addition.

P.S. The question was in the organometallics section of my assignment, possibly hinting that reagents apart from the above species may be used. However, won't it be great if we could design a mechanism using only the starting species and commonly available reagents?

t-butanol + allyltrimethylsilane + Lewis acid

This is called the Hosoni-Sakurai Reaction (see J. Org. Chem. 2006 71 8516)

Can only find an image for the more common addition to carbonyl compounds but there are examples of additions to alcohols (see the JoC reference)

What you could do is convert the t-butanol into a tertiary bromide by heating it with concentrated sulfuric acid and then treating it again with hydrobromic acid to get the tertiary bromide. After that, convert into a Grignard by oxidative addition of a Magnesium atom. Then you can just react it with ethylene oxide, and after working it up one more time, we can again heat it under concentrated sulfuric acid to eliminate the hydroxyl and form the double bond.


Why is it important to minimize impurities when doping a semiconductor? I was reading "Electronic Devices and Circuits" by Boylestad, and in there the author talks about how the semiconductor industry seeks to add as little impurities as possible when manufacturing semiconductors.

What I don't understand is why it is so important to keep the material's purity? What characteristics are improved by doing that? If the objective of doping is improve the conductivity of the material (in a controlled manner), wouldn't reducing the ammount of doping material decrease that efficiency?

Semiconductors are based on impurities.  They just call it doping.  The addition of group 3 (Boron group) or group 5 (Nitrogen group) impurities create a hole or an extra electrno, respectively.  These form charge carriers which are necessary for a semiconductor to be used in a circuit.  However, the dopants need to be added in highly controlled and precise amounts to get the necessary mobility.  The silicon also needs to be highly pure for regular electronic response, for regular electronic properties over arbitrary distances.  Another important consideration is that the dopant allows some degree of tuning of the electrical conductivity of the semiconductor.

Also, highly pure crystalline silicon  fractures nicely along the crystal planes, meaning it can be cut with high precision.  Since most mass-produced electronic processors are made on 12- or 16-inch silicon wafers with dozens of devices manufactured at once, and initially all physically connected, the ability to precisely excise each tiny circuit is critical for high through put and yield in manufacturing.

If you add too much dopant to silicon the silicon will become conductive. In order other words it becomes a resistor, not a semiconductor.


Conversion of propane to methane I had a question in my textbook to convert propane to methane. The basic idea was chlorination in presence of sunlight. However, two products are obtained in this case - namely 2-chloropropane and 1-chloropropane. So it didn't work out at all.

I came up with another idea i.e. dehydrogenation, but it has not been taught to us at all. In addition the dehydrogenation reactions I searched on google were too complex. So, I hope there's another method for this.

Can anybody help me to sort this out?

This probably defies the purpose, but still. Two-step process with steam reforming and Fischer–Tropsch synthesis

$\ce{C3H8 +O2 +H2O -> CO + H2 + CO2}$

This step is usually performed with methane over nickel catalyst, but should work with propane as well. Remaining water is condensed and carbon dioxide is separated using triethanolamine as absorbent. Then a reverse reaction is performed

$\ce{CO + 3 H2 -> CH4 + H2O}$

depending on catalyst, and exact conditions it is possible to get methanol, linear alkane/alkene mix or methane. AFAIK, $\ce{Ni}$ catalysts favor methane formation, while cobalt and iron favor linear alkanes and terminal linear alkenes.

If you are trying to do a dehydrogenation, dehydrohalogenation of both your products would get you to propene, even though it takes two steps.

Good luck getting to methane. Of course, if you were in industry you would just heat the crap out of it and watch the money roll in... Well, maybe not with methane as the product...


Why is change in gibbs energy not zero through the gibbs free energy equation? Change in enthalpy of the system is equal to its temperature multiplied by its change in entropy (as change in enthalpy is equal to heat absorbed /given out at constant pressure and change in entropy is heat absorbed/given out divided by its temperature). So, through the Gibbs free energy equation, the change in Gibbs free energy should be zero?

This question might seem messy, as I don't know how to write equations on Stack Exchange.

Please help. The problem is in the elementary concepts, I think.

The change in Gibbs free energy for a closed system is given by:
$$\Delta G = \Delta H - T \Delta S$$
What you are describing is:
$$\Delta H = T \Delta S$$
which is only true for systems at equilibrium, or when $\Delta G = 0$
Why this is true is a little bit difficult to explain. It is essentially a form of the Clausius Inequality at constant temperature and pressure. The Clausius Inequality is really just a statement of the second law of thermodynamics, and it says that at best, for any possible process you can get a zero change in the entropy of the universe, and usually it will be a positive change. From the perspective of the system you can write that as:
$$\Delta S \ge \oint \frac{dQ}{T}$$
which is basically just saying the entropy change of the system is equal to or greater than the sum of the negative entropy changes of the rest of the universe, for each step in a closed cycle.
If you integrate that at constant temperature you get
$$\Delta S \ge \frac{\Delta Q}{T}$$
At constant pressure (and in the absence of non-expansion work) $\Delta Q = \Delta H$, so we can write:
$$\Delta S \ge \frac{\Delta H}{T}$$
...and a little rearranging gives:
$$\Delta H - T \Delta S \leq 0$$
So this says that any process which can occur on its own (without adding work or heat to it) will obey this inequality. That's a pretty useful thing to know, so we let
$$\Delta G = \Delta H - T \Delta S$$
to give us an easier way to talk about it. Processes will happen (on their own) if $\Delta G$ is negative, they won't happen if it's positive, and when they are at equilibrium, $\Delta G$ will be zero.

if the second equation of thermodynamics holds good then dH-dST is greater than or equal to zero. A system cant have negative entropy because of the internal energy of a system, intermolecular forces of attraction aswell as the due to the fact that stable systems have minimum energy(fermi energy) and zero forces. futher explanation can be given when we analyse the atomic structure of atoms interms of quantum attraction forces and the forces of repulsion between atoms.
so the energy released when a molecule moves from a high chemical potential region to a lower one is the same for the simultaneous phase transitions but the energy of the two systems is never reduced to zero.


What are radicals in chemistry? I read somewhere that radicals are atoms, molecules or ions that have unpaired electrons, but how do ions have unpaired electrons when an ion has all its electrons paired either by loosing or gaining electrons to attain the nearest noble gas configuration?

but how do ions have unpaired electrons when an ion has all its electrons paired either by loosing or gaining electrons to attain the nearest noble gas configuration?

(Quote from your question, emphasis mine)

Your fallacy is the word I emphasised. Surely it is often taught like that—especially in introductory chemistry—that when ions are generated the noble gas configuration (vulgo: octet or dublet in the case of He configuration) is reached. That is not the whole truth, though.

Especially when you form a polycharged anion such as oxide $\ce{O^2-}$, you can (and should!) consider the electrons being added (or removed) stepwise. Thus, going from an oxygen atom you would initially generate an oxide radical anion $\ce{O^.-}$ before arriving at the classical oxide anion $\ce{O^2-}$. If you look at it energetically, the first electron addition is sometimes (but not always!) exothermic but the second one is always endothermic, so it makes energetic sense to separate like this.

These monoatomic radical ions are not commonly encountered as they typically have very short half-lifes. They often rapidly react to either give the full noble gas ion or the original, neutral atom. However, the vast majority of known ions are molecular or polyatomic ions such as phosphate. Within these, one very common species of radical anion forms rather stable compounds: the superoxide anion $\ce{O2^.-}$.

Structure-wise, it is similar to the peroxide anion $\ce{O2^2-}$ which can be written as $\ce{^-O-O-}$ except with one electron removed. However, it is usually generated  starting from dioxygen ($\ce{O2}$) with one electron added, which turns dioxygen’s double bond ($\ce{O=O}$ in Lewis formality) into a single bond. when the heavy alkaline metals (potassium, rubidium or caesium) are burnt in air, the corresponding superoxide is the product:

$$\ce{K + O2 -> KO2}\tag{1}$$

If lithium or sodium are reacted identically, the product is the oxide or the peroxide, respectivesly:

$$\begin{align}
\ce{4Li + O2 &-> 2Li2O}\tag{2}\\
\ce{2Na + O2 &-> Na2O2}\tag{3}\end{align}$$

From the Gold book http://goldbook.iupac.org/html/R/R05066.html

"A molecular entity such as .CH3 , .SnH3 , Cl. possessing an unpaired electron. (In these formulae the dot, symbolizing the unpaired electron, should be placed so as to indicate the atom of highest spin density, if this is possible.)"

Go to the Gold book for more details.


Do I need to include lattice energy when calculating the energy required to vaporize NaCl? I understand that the lattice energy is the energy released when a crystal forms and that's the reason it takes the negative sign. But do we need to add this energy again when we are trying to vaporize an ionic compound ?

For example, let's assume we want to completely vaporize a mole of NaCl starting at room temperature, how much energy do we need ?

I don't know neither the specific heat nor the latent heat of vaporization of NaCl, but I understand we would need:

1- Energy to raise the temperature of mole of NaCl from 298 °K to its boiling point.

2- Energy equal to the latent heat of vaporization of a mole of NaCl.

I don't know whether NaCl would go through the liquid state or not, but if it does then we would need to add the latent heat of fusion too.

Now, according to wikipedia, NaCl has a lattice energy of −756 kJ/mol. So my question here is: do we need to add this lattice energy to the calculated steps above to account for all the energy required to vaporize a mole of NaCl ?

Lattice energy

  Now, according to wikipedia, NaCl has a lattice energy of −756 kJ/mol.

First, we have to understand the term lattice energy.
Here is the textbook explanation (Fleming: Physical Chemistry): The lattice energy is the energy required to separate the ions in an ionic lattice so that they are at infinite distance (but still ions). This would be difficult to do experimentally, but the value may be determined by using a Born Haber cycle. As the diagram below shows, we know the enthalpy of reaction for:

$$\ce{Na(s) + 1/2 Cl2(g) -> NaCl(s)}$$

We can get the same energy in a thought experiment (path 2 in the diagram), first turning all species into gas, then into atoms, then transferring the electron and - finally - forming the lattice from the individual anions and cations. All these processes are experimentally accessible except for the last, so you can determine the lattice energy this way.

Source: diagram adapted from the cited textbook.

Enthalpy of vaporization

  But do we need to add this energy again when we are trying to vaporize an ionic compound?

You problem is not well-defined. What are the initial and final temperatures, what is the pressure? What does it mean to vaporize solid NaCl or any ionic compound, i.e. what is the product? If we boil sodium chloride at 1 atm pressure, it produces mostly NaCl monomers in the gas phase, see https://chemistry.stackexchange.com/a/14560. If you are interested in this process, you could write it as:

$$\ce{NaCl(s) -> NaCl(g)}$$

So to get the required energy, you could first separate all ions (lattice energy) and then calculate the energy of forming individual pairs of sodium and chloride ions (assuming the interaction is purely ionic).

$$\ce{NaCl(s) -> Na+(g) + Cl-(g)}\tag{1}$$

$$\ce{Na+(g) + Cl-(g) -> NaCl(g)}\tag{2}$$

It should be obvious that process (1) cost much more energy than what you get out of process (2) because you are going from 6 nearest neighbors to just one. If you look up the atomic distance of $NaCl(g)$, you could estimate the enthalpy of (2) pretty accurately and get a numeric answer.

  [Comment from https://chemistry.stackexchange.com/a/14177]: Roughly, somewhere around half of the lattice enthalpy of a salt comes solely from the binding of the smallest electrically neutral agglomerate; that is, it takes about as much energy to break a macroscopic solid NaCl crystal into a gas of ion pairs as it does to break all the ion pairs and create a true plasma. Thus, it is rather unlikely that, at reasonable temperatures, an ionic gas will break down any further than the smallest possible electrically neutral aggregates. – Nicolau Saker Neto Jul 8 '14 at 16:39

According to the comment above, the heat of vaporization should be about half of the lattice energy (with opposite sign).

  I don't know whether NaCl would go through the liquid state or not, but if it does then we would need to add the latent heat of fusion too.

Because enthalpy is a state function, it does not matter if you choose the actual (realistic) path or take a different path. You are right that if you step through what actually happens, you need to know heats of fusion and vaporization as well as heat capacities of all phases for the relevant temperature ranges.

The energy to vapourize one mole of NaCl is the negative of its lattice energy, assuming you are starting at 25 degrees C.
Can be calculated via born habor cycle-
The Born-Haber Cycle can be reduced to a single equation:

Heat of formation= Heat of atomization+ Dissociation energy+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy

*Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive.

Rearrangement to solve for lattice energy gives the equation:

Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities


Why is it obligatory to cool down the container of a sample to measure its mass in a lab? We were doing an experiment about hydrated crystals and more precisely how to determine $n$ in $\ce{CuSO4.nH2O}$. After we heated the crucible we were to cool it down using a Desiccator. Then this question came to my mind: Why must we cool it down? By doing researches in the Net I were only able to come up with these suggestions, but none seem very applicable for me:
Using Einstein's formula:
His famous equation:$$E = mc^2$$
This means that matter is able to be converted to energy. Thus, more energy should mean more mass because $C$ is constant. However, I argue that since $C$ is a very large number, normal and observable-in-normal-labs changes in energy aren't able to modify the net change of the mass dramatically (to a degree that the balance is able to detect)
A gut feeling, no backup research:
Anyone sane, and with regards to the value the balance has in the lab, would give slightest chances that the hot container would harm the surface of the balance and therefore, damage it. But chances are that I'm wrong too.
So, why must we cool the crucible in order to measure the mass of the sample in it using a balance?

A container that is not at ambient temperature will generate air currents around it.  If you place such a non-ambient container in a balance, air currents will develop around the container as it heats or cools to ambient.  These air currents will cause the balance to read incorrectly.

While these are all very good answers, I have a much simpler one. These far overthink the question. The crucible has to cool in the desiccator to avoid adding moisture to it. The experiment you're doing deals with hydrates, therefore getting as much moisture out as possible is important as it will reduce your percent error.


In which cases does the 3d subshell take precedence over the 4s subshell? I've learnt in Chemistry that for example in Chromium, the 4s subshell only has one electron in favor of making the 3d subshell half-full. Also, for Copper the same happens.

What other circumstances does this happen under? Do ions obey the same pattern, e.g. Mn+?

If we take the definition of "3d subshell takes precedence over the 4s subshell" as being that the configuration has zero 4s electrons, yet at least one 3d electron, then all such cases are ions.

+1

V: 3d^4
Cr: 3d^5
Co: 3d^8
Ni: 3d^9
Cu: 3d^10

2+

All elements from Sc to Zn

3+

All elements from Ti to Ga

Etc.

I agree with @Nicolau Saker Neto that it's complicated. You might want to look at these two papers, if you're really interested in more of the details:

R. N. Keller, J. Chem. Ed. 39(6), 1962. PDF available here
Robin M. Hochstrasser, J. Chem. Ed. 42(3), 1965. PDF available here

The links might be paywalled. Sorry. :-/


What is the solublity of water in liquid carbon dioxide at low temperature? The solubility of water in liquid carbon dioxide has been investigated by Hosmer W. Stone in 1943 from
-29 to about 22 °C.

Is there any research done on the solubility of water in liquid $\ce{CO2}$ at still lower temperatures ?

The lower the temperature and higher the pressure, the more soluble CO2 is in water.  Pepsi and Coke have staudiesd this for years.


Why is carbon monoxide the product in reactions that produce high flame temperatures? I was looking at substances that produce the highest flame temperatures upon combustion. I note that the equations for combustion are producing carbon monoxide rather than carbon dioxide.

$$\ce{(CN)2 + O2 -> 2CO + N2}$$

Why isn't carbon dioxide produced? Why doesn't the production of carbon dioxide give a hotter flame? It seems that a higher flame temperature would be generated by complete combustion. My thinking is that the heat of formation of carbon dioxide is around −400 kJ/mol while the heat of formation of carbon monoxide is around −100 kJ/mol so the additional heat released by producing carbon dioxide should account for carbon dioxide's greater heat capacity.

This is just a guess... But I wonder if this is due to entropic considerations. In the equation you have written, there are two molecules of gas getting converted to three molecules of gas. $\Delta S$ is positive, so it decreases $\Delta G$ as the reaction temperature increases. However, in the reaction $\ce{(CN)2 + 2 O2 = 2 CO2 + N2}$ you have an equal number of gas molecules, so the entropic component of $\Delta G$ will be small.

I'm not really sure why that would matter. At very high temps, the first reaction will presumably have a very large, negative free energy. Of course, the $\Delta G$ for the $\ce{CO2}$ reaction is most likely negative, so you might have to control the feed stoichiometry carefully to ensure that you are getting only the first reaction.

EDIT:

I got the $\Delta G$ values from the NIST-JANAF website, and it doesn't look like the free energy for these reactions is much different at 4000 K.

$\Delta_f G_{(CN)_2}^{0} = 136.036\ \mathrm{kJ\ mol^{-1}}$

$\Delta_f G_{CO}^{0} = -446.485\ \mathrm{kJ\ mol^{-1}}$

$\Delta_f G_{CO_2}^{0} = -393.183\ \mathrm{kJ\ mol^{-1}}$

$\Delta G_{rxn1}^{0} = -1029.005\ \mathrm{kJ\ mol^{-1}}$

$\Delta G_{rxn2}^{0} = -922.397\ \mathrm{kJ\ mol^{-1}}$

EDIT 2:

So, I just read one of the original papers on this type of flame, and it sounds like the thermal stability of the products is an issue as well. I take it that at these very high temperatures, some of the energy can be transferred to bond-breaking processes, which would lower the total flame temperature. $\ce{CO}$ and $\ce{N2}$ have very strong covalent bonds, so perhaps they do not readily dissociate at these temps.

mainly because in high flame areas, oxygen is used up by the fire more rapidly,thus there is less oxygen available,and thus CO carbon monoxide forms.


Why is the value of "a" is positive instead of negative in Van der Waals equation? Higher value of a means higher atttraction force between molecules. Attraction force causes the pressure to be less than expected. Now the actual pressure becomes $P + \frac{n^2a}{V^2}$ .
Which means if a increases the neat pressure increases too. But isn't it the other way around?
(I assume here P is the pressure exerted on the wall of the gas container considering there is no intermolecular force)

You understand the chemistry: intermolecular attraction forces cause the pressure to be less than expected. The problem here is with the math. Though unintuitive at first sight, a positive $a$ leads to a negative correction to pressure. Let's look at a numerical example.

  Assume that the pressure inside a certain container is $50$ atm with the ideal gas approximation and $45$ atm with the van der Waals equation of state. Then $P_\text{ideal} = P_\text{vdW} + 5$.

In other words, the corrected pressure is $P_\text{vdW}$, not $P_\text{vdW} + a\frac{n^2}{V^2}$.

The formula is good because it is correct in most of the regions and the parameters make sense. It is attractive because it both describes behavior of gas and approximate behavior of a liquid.

This formula has a problem in a region that corresponds to phase transition (boiling of a liquid). To account for that region correctly you need a function that is constant within some range and changes within other ranges.

Complex analysis (a branch of mathematical analysis) has a theorem that proves that you cannot have an analytical function that is constant on one interval and changing over the other intervals.

This sort of functions exist, but they are not analytic and behave poorly (hard to work with).

As a free bonus it turns out that the "strange" region describes behavior of overheated liquid and overcooled gas. Those two have different behavior at the same temperature.


Solubility of n-butanol and n-butanamine I ran across something saying that n-butanamine is miscible in water while butanol is only partially soluble. Is this correct? Can anyone offer an explanation? I would have guessed that the decreased polarity of the amine relative to the alcohol would have resulted in lower solubility for n-butanamine.

Out of butanamine and butanol, butanol is more soluble in water due to higher electronegativity of oxygen . And as butanol  have more polar oxygen so it will be more attracted towards water and form H bond.so it is soluble but amine are miscible due to enthalpies that will release out because of electron donating nature of nitrogen.


How to calculate the volume of a 0.1 M NaOH(aq) solution necessary to make a 0.5 M EDTA solution with pH = 8? I was told that EDTA dissolves at a pH of $8$. Therefore I need to use a base to increase the pH to $8$ is what I understand. How exactly would I go about this?

In my head I if I wanted $1~\mathrm{L}$ of $0.5~\mathrm{M}$ EDTA I would take $186.12~\mathrm{g}$ of the EDTA, since its molar mass is $M = 372.24~\mathrm{g\,mol^{-1}}$.

Then I would add $800~\mathrm{mL}$ of distilled $\ce{H2O}$ to dissolve it, but since it wont be at $\ce{pH}=8$, it wont dissolve.

So I assumed I would just add it drop wise until it dissolved while spinning? I lack access to a burrette and only have a crappy pH meter to check pH.

One I observe the EDTA dissolved I simply add up to $1000~\mathrm{mL}$?

I just wish I had a way to figure out how much $\ce{NaOH}$ I needed to add to dissolve the EDTA so I wouldn't have to do it drop wise.

I saw this on the researchgate.net website:

I followed this procedure for preparing  0.5M EDTA ph8(100ml) . Initially I  warmed around 70ml of water and added an ml of 6MNaoh solution to shoot up the ph to 7.6  with the container on  a hot plate .Then I slowly added the measured quantity of EDTA and Simultaneously kept adding small volumes of 6M Naoh solution while measuring  the ph.Finally around ph 8 the EDTA was completely dissolved and the process was way too quick.."

Bhavani Nagarajan · University of North Texas HSC at Fort Worth


Le Chatelier's principle: what are the effects of decreasing volume? Which of the following is increased by decreasing the volume of the
  reaction system in the following reaction: $\ce{2H_2S(g) +3O_2(g)<=>
 2SO_2(g) +2H_2O(g)} + {\text{heat}}$

  I. Rate of Reaction

  II. Equilibrium concentration of reactants

  III. Value of $\ce{K_{eq}}$

My attempt: By the gas laws, decreasing the volume of the container will increase the pressure, so equilibrium should shift to the right, as there are less gases on the right. Thus III is true. Additionally, heat is also generated by the shifting of the equilibrium, and the increased pressure on the system due to the decreasing of volume would also speed up the reaction, so I is also true. II is not true because the Equilibrium concentration of the products increases, not the reactants. So my final answer is I and III. Yet the answer key says I only. Why is that?

I think a good place to start is with the idea that that $K_{eq}$ will not change in value unless the temperature changes (see the second answer here for a good explanation). With this in mind we can say that statement III is false because the temperature is not being changed.

Statement II is shown to be false via Le Chatelier's principle, as the equilibrium concentration of reactants should decrease rather than increase due to the decrease in volume because the products have fewer gaseous moles and so would exert less pressure on the container.

Statement I can be shown to be true via Le Chatelier's principle by the same logic as Statement II: for the reaction to shift towards the product, the rate of formation of product must increase.

Therefore, only Statement I is true.

You might wonder how $K_{eq}$ can stay constant, but the rate of forward reaction can increase. Its important to remember that $K_{eq}$ is the ratio of the forward and reverse rate constants, not the forward and reverse rates. The rate constants have no dependence on volume (or pressure) so they won't change.

concept:

You should base it on the amount of moles on both sides

so if you wanted to figure it out u know that the side with the more moles is going to be affected i.e. if you decrease the pressure the side with the more moles of gas is going to decrease and shift the other way

to clarify:

1) You said that I is true but it isn't because we have to understand that only temperature changes the rate of the reaction so for rate of reaction we don't care about the decrease in volume.

2)For equilibrium concentration of reactants, we know that there is 5 moles of gases on the left side and there is four moles on the right side with heat, because you are not chaining the heat, when you decrease the volume that means that you are forcing the moles from the left to the moles on the right side.

3)For $K_{eq}$ we will go with your explanation of why it is true. But then we can rule out number 1. But it seems to me that your textbook either has a typo that tried to say "which of these are not true" or that you have not read the problem correctly.


What's the relation between the octet rule and quantum exclusion and how is it related to oxidation state? What in the quantum nature of electrons gives rise to the octet rule? How is it related to the fact that an element cannot have an oxidation state higher/lower than the complementary to 8?

What in the quantum nature of electrons gives rise to the octet rule?

Ultimately, the octet rule is justified by the fact that in many cases an $n s^2 n p^6$ electron configuration is the most energetically favorable one. Recall that electrons are arranged in different electron shells around the nucleus. Each shell corresponds to a principal quantum number $n$ and can hold only a certain number of electrons, namely up to $2n^2$. Each shell is composed of subshells each of which corresponds to an azimuthal quantum number $l$.

Now recall also that the binding energy of the electron decreases with both $n$ and $l$, and thus, once $n s$ and $n p$ orbitals of an $n$-th shell are filled, in many cases it is indeed energetically unfavorable to fill the subsequent orbitals: $n d$ if (there are any) or $(n+1)s$. The energy gain from binding an electron to an atom might not be enough to compensate a cost of separating it from a different atom.

  How is it related to the fact that an element cannot have an oxidation
  state higher/lower than the complementary to 8?

Suppose an atom has $x$ electrons in the outermost shell, where with the exception of transition metals $x \leq 8$. In the limiting cases the atom will either gain $8-x$ electrons to completely fill $ns$ and $np$ subshells or lose all $x$ electrons. In the first case we will end up with the atom in $-(8-x)$ oxidation state while in the second case with the atom in $+x$ oxidation state. But these are only the limiting cases, while in principle atom can gain few electrons and loose few others at the same time. Besides, atoms of the same type can evenly share electrons with each without satisfying the octet rule without gaining or loosing electrons. Thus, usually elements have few possible oxidation states, though, in the above mentioned limiting $[-(8-x), +x]$ range.

Excerpt from Wikipedia's list of oxidation states of the elements.

I'm able to give an answer only for He, Ne and Ar and the answer has to do with the magnetic dipole moment of the involved electrons, known also as the electrons intrinsic spin.

The electrons intrinsic spin is somehow abstract while the magnetic dipole moment is well imaginable as a bar magnet. And exactly this feature gives rise to the octet rule as well as to Pauli's exclusion principle.

The argumentation goes the way that exactly 2 respectively 8 (and only 2 and 8) bar magnets could be in equilibration respectively in a stable state. For two bar magnets this is not a big deal. The bar  magnets will be faced north by south poles and following this picture one get for He one the shell with two electrons with opposite spin.

How the magnetic dipole moments of the 8 electrons of Ne would be distributed? The answer lays in an octahedral structure (The second picture shows the same of 8 symmetrical distributed axis or areas or volumes.):

Imagine for every of the four red corners a bar magnet with north pole to the centre and for every of the for yellow corners a bar magnet with south pole to the centre. That is the reason why Ne as well as Ar are stable as they are.

So far the magnetic dipole moment has not been considered (or only indirectly through the intrinsic spin), however it has to be.


Why is the carbonyl IR frequency for a carboxylic acid lower than that of a ketone whilst an ester is higher than a ketone The approximate stretching frequencies are as follows:

  Acid: $1710$, Ketone: $1715$, Ester: $1730$ (all $\mathrm{cm^{-1}}$)

This would seem to suggest that the acid decreases the carbonyl bond strength and the ester weakens it. This must be as a result of the fact that in the ester inductive withdrawal is more effective than resonance and the converse is true in the acid. However, I am finding it hard to reasonably justify why. Presumably this is also linked to the relative electrophilicity of the ester and the acid.

First things first, it's important to realise that functional groups may have IR frequencies over quite a broad range. I've provided the table below, from which you can see that actually there is a great deal of overlap depending upon the actual 'environment' of the functional group in question. Ketones and acids, for example, cover a very similar portion of the spectrum.

  Stretching frequencies of common carbonyl groups (Taken from Introduction to Spectroscopy, Pavia and Lampman.

That said, the difference between the frequency range of an acid vs an ester is an interesting one, as on the surface of it they are electronically quite similar (in both cases the carbonyl is next to an electronegative oxygen).

  @RobChem: This would seem to suggest that the acid decreases the carbonyl bond strength and the ester weakens it. This must be as a result of the fact that in the ester inductive withdrawal is more effective than resonance and the converse is true in the acid.

In your statement, you may have gotten a little confused, remember that the stretching frequency increases with bond strength. Therefore using your data we can say that the ester is stronger than the ketone whilst the acid is weaker.

We can explain the ester being stronger by considering the effect of the electronegative oxygen withdrawing electron density:

This argument however, fails to explain why the carboxylic acid has a stretching frequency below that of the ketone, since, as was already said, they both have the oxygen adjacent to the carbonyl.

One rationalisation for this is that the carboxylic acid group doesn't exist in isolation but rather interacts with other carboxylic acids in a hydrogen bonding interaction, which weakens the C=O bond:

This hydrogen bonding concept is universal, for instance in methyl salicylate, the C=O bond has dropped to well below that expected for an ester:

the predominant effect of the oxygen of an ester is inductive electron withdrawal, so the resonance contributor with the single bond contributes less to the hybrid. The carbonyl group of an ester, therefore, has a C-O double-bond character than does the carbonyl group of a ketone, so the former is stronger and harder to stretch.
bond in a carboxylic acid has a partial double-bond character that is due to resonance electron donation.


Using Beer-Lambert law to calculate light intensities Supposing the mean absorption coefficient of seawater in the visible region is $\pu{0.8 m-1}$, calculate the depth at which a diver will experience half the surface light intensity.
I know that you have to use $A = -\log \left(\dfrac{I}{I_o}\right) = \epsilon cl$ but I'm not sure how to proceed

In my opinion, Beer–Lambert law is no use. I would rather write
$$\frac{\mathrm dI}{I\,\mathrm dl} = \pu{0.8 m^-1},\tag{1}$$
which leads to the integrated formula
$$\ln\left(\frac{I_0}{I}\right) = 0.8 l.\tag{2}$$
And, if $\displaystyle\frac{I}{I_0} = 0.5,$ the corresponding depth is
$$l = \frac{\ln 2}{0.8} = \pu{0.846 m}.\tag{3}$$


Why is sodium carbonate less soluble in water than sodium bicarbonate? Why is sodium carbonate less soluble in water than sodium bicarbonate? If you think about their structures, the only thing that is different is that sodium carbonate has two sodium atoms, while sodium bicarbonate has one sodium atom and one hydrogen atom instead of the second sodium atom. Sodium and hydrogen have different electronegativity values, giving sodium bicarbonate polar character.

Does the difference in solubility have to do with the dipole moment of the molecules? If so, how can it be explained in more detail?

Oops! The question contains an incorrect assumption. But chemists are amazing - they can still answer such a question:

Sodium carbonate is way more soluble than sodium bicarbonate at all temperatures above about freezing. $Na2CO3$ has several crystalline hydrates that are temperature sensitive and make the curve less smooth than most other salts. (I suppose the caffeine is in there to wake us up.)
So, I woke up to the fact that when we calculate the weight of "sodium carbonate" which dissolves at 0 C, it's the decahydrate, whereas sodium bicarbonate is anhydrous, so at 0 C, the solubility of sodium bicarbonate is 0.833 moles per liter, while the same weight of sodium carbonate decahydrate dissolved in a liter is only 0.245 moles.
Now I'm more puzzled by why sodium carbonate has such a complex hydrate system, and sodium bicarbonate doesn't.

According to the Born–Landé equation, the lattice energy is proportional to Z+Z-. Therefore in $\ce{Na2CO3}$ the Z+Z- value is 2 where as in $\ce{NaHCO3}$ the value of Z+Z- value is one.  So the lattice energy is higher in $\ce{Na2CO3}$ than $\ce{NaHCO3}$. In addition to that the size of bicarbonate ion is higher than carbonate. Therefore packing efficiency will be higher in sodium carbonate than sodium bicarbonate. Thus more energy will require to disrupt $\ce{Na2CO3}$ than $\ce{NaHCO3}$. Therefore $\ce{Na2CO3}$ is less soluble in water than $\ce{NaHCO3}$.


why are halogens considered deactivating groups while methoxyl groups are activating? Something's wrong with how I think of this:

methoxyl has the partial negative charge on oxygen which means it's activating. However, halogens are more electronegative yet they're considered deactivating. Why?

On my lecture notes, I have one section that says
"if Q is an electron withdrawing group then attack on the ring is slowed because this leads to additional positive charge on the ring".

While another section says "all meta directing groups have either a partial or full positive charge on the atom directly attached to the aromatic ring."

Because fluorine has such a high electronegativity, its inductive withdrawing effect is much larger than any resonance electron-donating effect.

The other halogens have poor orbital overlap with the pi system of an aromatic ring, which precludes any appreciable resonance electron-donation (this is why you almost never see a C-X double bond). Therefore, their inductive capabilities lead to a net deactivation of the aromatic ring.

This is why an alkoxy group is more activating  than a halogen. Oxygen does have a higher electronegativity than does chlorine, bromine or iodine. However, oxygen's orbital overlap with carbon allows it to more readily donate into the ring's pi system. The larger halogens aren't able to do this (to an appreciable extent).

All the good points been drawn, the electron withdrawing ability of oxygen is reduced by the I+ effect of the alky group of the alkoxide, thus all the more favouring electron donation to the aromatic ring.


How does concentration affect The Position Of Equilibrium? I was taught that
when the concentration of the reactants is greater than the concentration of products then the position of equilibrium is shifted towards the left.

However when I used La'Chateliers Principle then we know that if you increase the concentration of reactants then the system wants to revert that change a hence produce a higher concentration of products (and subsequently a lower concentration of reactants) but wouldn't this then mean that the equilibrium would shift to the right?

You statement of Le Chatelier's Principle is a perversion that isn't true. The equilibrium does not depend on the sum of the concentrations of the reactants relative to the sum of the concentration of the products.

Your analysis is correct.

Words confuse sometimes. "Shifting" is usually used in the context of Le Chatelier's Principle, and is a verb. K is the equilibrium ratio of concentrations of species to their stoichiometic powers, and Q is the ratio of the concentrations of species raised to their stoichiometric powers at any point in time. When Q>K, there's more products than there should be. You can think of the reaction as "tilted" toward the right. The reaction will then "shift"/move to the left, trying to reduce its Q by forming reactants from products until Q=K.


Why does stannous chloride occur despite the octet rule? Shouldn't reaching an octet be any atom's "goal"?

However, I've recently learned about cases that are either expanding octets, or have lesser than "enough" electrons for an octet abiding. e.g.:

S in sulfur hexafluoride (Expanding octet into $\text{3d}$)

Source

B in boron trifluoride (It's "hextet", instead of octet)

Source

However, both $\ce{SnCl2}$ and $\ce{SnCl4}$ are existent. The latter is explainable with octet, but not the former. Surprisingly, $\ce{SnCl2}$ is more stable!

How?
Why is the phenomenon happening?

Dave pointed out that the octet nonabidingness is happening in the vapor phase. So I redirect the question to ask about the vapor phase; since that's what I'm looking for.

According to Theoretical and experimental studies of the electronic structure
of SnCl2 by photoionization with synchrotron radiation AIP Conf. Proc. 258, 60 (1992)

$\ce{SnCl2}$ has a molecular orbital structure as below (HOMO at the top)

Sigma-antibonding: 36% Sn s, 22% Sn p, 41% Cl s

Nonbonding 99% Cl p

Nonbonding 99% Cl p

Nonbonding 7% Sn p , 92% Cl p

Nonbonding 3% Sn s, 6% Sn p , 90% Cl p

Sigma bonding 14% Sn p, 83% Cl p

Sigma bonding 48% Sn s, 6% Cl s 45% Cl p

Nonbonding 3% Sn p, 96% Cl s

Nonbonding 5% Sn s, 2% Sn p, 92% Cl s

So overall it seems:

There is a total bond order of only 1.

Much of the Sn p and some Sn s electron density has been donated to Cl p orbitals.

There is little mixing of Sn s and p orbitals expect in the antibonding orbital.

additional references:

A Symmetry Rule for Predicting Molecular Structures (This reference explains how, based upon second order Jahn Teller effect, the symmetry of molecular orbitals predicts geometry, for example bent vs linear, and says that the symmetry predicts bent for SnCl2)

and Metal oxidation state effect in photoionization of gas‐phase metal halides (which further discusses the molecular orbitals of SnCl2 discussed above)

in a simple way since tin belong to group 14 the stability of lower oxidation state i.e +2,is comparitively more stable than +4 oxidation state due to inert pair effect (reluctance of s orbital electron to take part in bond formation due to poor shielding ability of d orbital which causes the outer s orbital electron to be attracted towards nucleus.)thus tin loses two p electrons only.the outer s orbital electron does not take part in bond formation might.i hope i am right.


Do enolates get protonated at the carbon or oxygen atom? An enolate has two resonance structures. When it gets protonated, where is the $\ce{H+}$ more likely to go? On the oxygen to produce an enol, which tautomerizes to the keto, or on directly onto the carbon (no enol intermediate). I know that both lead to the same product, but when drawing the mechanism this would be good to know.

Like all ions in solution, the enolate does not exist on its own in free form. It will be solvated with maximum efficiency. Depending on the solvent itself, the counter ion might be further or closer away, but charge balance needs to be maintained in the bigger picture.

Contrary to popular belief, the enolate ion is nothing that truly exists in non-gaseous phase. Therefore the question what gets protonated first turns into the question: Does the oxgen or the carbon have the stronger interaction with the solvent? This is quite hard to judge and very dependent on the whole system. there is certainly no general rule for it and as previously stated by bon, it does not really matter, because keto-enol tautomerism is fast and the ratio is entirely determined by the relative stabilities to the involved species.

There is another issue when talking about protonation. The proton itself is not free, either. In an aqueous solution the smallest model would be the hydronium ion. But this is far from being complete either. A whole branch of computational and theoretical chemists work on describing water clusters with models. It is extensive. One of the reasons for it being hard to comprehend is the high flexibility and mobility of protons even when bound to other species. The high conductivity of an acidic or basic solution is mainly governed by the Grotthuss mechanism. This kind of mechanism can in first approximation also be applied to keto-enol tautomerism.

But before we dig deeper into any mechanism, I would like to look at the properties bon outlined for the enolate. The following results were obtained at the DF-BP86/def2-SVP level of theory for the deprotonated butan-2-one resonance hybrid.
Let's look at the obvious first: resonance. With the help of the natural bond orbital theory (NBO) I performed the analysis. The first and most obvious thing is that it finds a so called hyper-bond between the oxygen and the two carbons with an occupancy of 3.95 electrons. As we will see in the refined analysis, this is in favour of the carbon-carbon bond (59.5/40.5 considering only these three atoms).
A more refined analysis, I chose a cut-off of 5%, include secondary resonance structures, seven in total, gives us the following weights for the major contributors:

From these structures we can support the highlighted arguments in the statement by bon:

  Attack on an enolate by an electrophile (in this case just a proton) is governed by the balance between electrostatic interactions, which are strongest at the oxygen atom due to its greater negative charge, and orbital interactions, which tend to be stronger at the carbon atom because the HOMO of the enolate has a greatest contribution at this atom [carbon].

The predominant contributor is is the enolate, hence in the overall description also its negative charge should be predominant. The π-system is certainly part of the HOMO in any case. Since it is mainly localised over the carbons, the highest contribution should be there.

We can further support that by looking at the natural charges of this model. More prominent is it looking at the charges obtained in terms of the quantum theory of atoms (QTAIM) in parenthesis.

The second part can be cleared up by looking at the HOMO; on this level of theory, the major contribution is C3-p of 56% and C2-p of 32%. The contribution of O1-p is below the threshold of printing, hence it is <10%. The major contribution of O1-p is found in HOMO -2 with 44% and a contribution of C2-p of 35%. It certainly looks a bit different in the picture. The contour value is set to 0.04.

All these calculations are only evaluations of the ion in the gas phase, so they can serve as an example to understand the true nature a bit more. It is in no way intended to be complete, it shall only provide a more concise educated guess of the general nature.

Now that we have a lot of information on the enolate itself, it's time to look towards a mechanism of a solvent-mediated keto-enol tautomerism. I have chosen the most simplest model possible, with one water (hydronium) explicitly included. The result is a six-membered cyclic transition state, where a proton is transferred from the oxygen of the enol to the oxygen of water, and one proton is transferred from water to the carbon of the double bond.
In theory there might well be many more water molecules involved. In principle any protic solvent is capable of doing this. The principle reaction path is the same for a base mediated mechanism, for example with ammonia.

If you look closely enough, I have borrowed heavily from the Grotthuss mechanism. A somewhat simultaneous transfer of protons. It needs to be acknowledged, that this is only an approximation of a stationary system. The animation only shows the transitional vibration mode (imaginary mode). It is not the calculated structural change or reaction path.
Unfortunately the calculations even on this low level of theory take quite some time. Currently I have no barrier height to report as an estimate, how fast this reaction would be approximately. I'll update this as soon as I can.

There is of course the possibility of a non-mediated transition state. This one, however, proceeds through a four-membered cyclic transition state. This one is highly strained, if you like the concept of ring strain, and seems unlikely compared to the above in any other medium than the gas phase.
If you are repulsed by the concept of ring strain, then you might want to look at it from a different angle. The deformation energy is high, since the carbon-carbon-oxygen is forced into a near 90° angle, hence increasing the p-content of the bonding orbitals. Higher p-orbital content means higher energy of the molecule. Therefore the activation energy will also be higher.

For a deeper insight into this, I recommend “Ab Initio Molecular Dynamics Study of the Keto–Enol Tautomerism of Acetone in Solution” (Clotilde S. Cucinotta, Alice Ruini, Alessandra Catellani, András Stirling, Chem. Phys. Chem. 2006, 7 (6), 1229-1234), which is available at researchgate.net.

First, let's look at your comment about tautomerisation.

Tautomerisation requires a proton transfer! Therefore (in basic solution) it proceeds as follows (sorry the double arrows are meant to represent equilibrium not resonance):
$$\ce{Ketone + Base- <=> Enolate + BaseH <=> Enol + Base-}$$

The Enolate is the intermediate in the tautomerisation equilibrium.

I assume you're getting the enolate from some other reaction / reagents in a synthetic pathway to the ketone?

As for which the Enolate is most likely to become, that will depend on the thermodynamics and kinetics of the two equilibria.

For example, if the solution contains 90% ketone and 10% enol, then it would be 9 times more likely for the molecule to "end up" as the ketone form.

But protonation/deprotonation at carbon is much slower than at oxygen, so you might expect most molecules to interconvert between enolate and enol several times before settling on the ketone form.

The bottom line is, all three species are in a pretty rapid equilibrium, so it doesn't matter.

If you're drawing the mechanism I would show $\ce{enolate -> ketone}$. I would ignore the enol form completely, unless you want to make some kind of point, in which case I would show it in equilibrium with the enolate, in a branch on the main mechanism pathway.


Is carbon dioxide soluble in petroleum ether, and why? I googled it, and I'm finding some forum posts that say no, but no authoritative reference.
CO2 is somewhat soluble in water (1.7 g/L at 20°C) and I understand that it's mostly NOT because it forms carbonic acid: it's just dissolved CO2 molecules. Why wouldn't the same thing happen in petroleum ether?
Has anyone tried putting petroleum ether in a SodaStream device?

Carbon dioxide is, in fact, more soluble in nonpolar solvents than in water.

From the Wikipedia page on carbon dioxide: Solubility in water
1.45 g/L at 25 °C (77 °F), 100 kPa
(Alternative source: CRC handbook. Solubility is about twice as much at 0°C.)

From IUPAC, Solubility Data Series: Carbon Dioxide in Non-Aqueous Solvents at Pressures Less Than 200 kPa, chart on page 3: mole fraction of CO2 in hexane at 101.3 kPa, 300 K = 0.012, which corresponds to 4 g/L.
(Solubility is only slightly (10-20%) higher at 0°C.)

So the forum posts I read were wrong.

As to why, I suppose it's sufficient to argue that CO2 is a nonpolar molecule, and like dissolves like.
(It's true because it's Latin: Similia similibus solvuntur! )

While $\ce{CO2}$  as a symmetric molecule has  zero dipole moment, its 2 $\ce{C=O}$ bands do have dipole moment and are polar.

Polar water molecules interact with the polar bonds what decreases   $\ce{CO2(aq)}$ chemical potential and raises solubility of $\ce{CO2}$ some 2 orders above solubility of $\ce{O2}$ or $\ce{N2}$.

That does not happen in non polar solvents likes alkanes of petroether.

BTW, why would anybody put petroether in a soda stream ??


Why is anhydrous copper(II) sulfate white while the pentahydrate is blue, even though both have one unpaired electron? We all know that $\ce{CuSO4.5H2O}$ is blue and it turns to $\ce{CuSO4}$ on heating which is white. I also learnt about coordination complexes and d–d transitions and how transition metal compounds are colored. Recently I came to know that $\ce{CuSO4.5H2O}$ is also a coordination compound and in its structure four water molecules are coordinated with $\ce{Cu^2+}$ and one is hydrogen-bonded to $\ce{SO4^2-}.$ So, I thought lets try to apply the theory here and see what causes the blue color of copper sulfate.

$\ce{CuSO4}$ has $\ce{Cu^2+}$ ions and I wrote down the electronic configuration as $\mathrm{(3d)^9(4s)^0}.$ Since no d–d transition is possible, it should be white.

$\ce{CuSO4.5H2O}$ is actually $\ce{[Cu(H2O)4]SO4.H2O}$ and the complex ion is $\ce{[Cu(H2O)4]^2+}$. $\ce{H2O}$ is a weak field ligand, therefore the electronic configuration is $\mathrm{(3d)^9(4s)^2(4p)^6}.$ But still no d–d transition is possible, so it should also be white!? What went wrong in my estimation?

Although the question is a bit old, I think it still hasn't been answered yet. And by looking at the given answers it seems like the discussion went into a different direction at some point.

So let's compare the two compounds, here I plotted the $\ce{Cu(II)}$ centers of $\ce{CuSO4.5H2O}$ and $\ce{CuSO4}$ from their crystal structure data.

As you can see, it changes from a $\ce{[Cu(H2O)4[SO4]2]^2-}$ to a let's call it $\ce{[Cu[SO4]6]^10-}$. So all effects that are typical for $\ce{Cu(II)}$ should remain, it is still an octahedron, there is still Jahn-Teller present, we only got the change from four $\ce{H2O}$ ligands to four $\ce{[SO4]^2-}$ ligands.

Those two are similar in their ligand field splitting, where the latter one is slightly weaker, so we expect a red-shift of the absorption energy. And that is also what Nagase et al. investigated on (Ref.1):

  As shown partly in Fig. 1, the $\mathrm{d–d}$ band maxima of the hydrated and anhydrous copper sulfates were observed at $\pu{13200 cm-1}$ ($\ce{CuSO4.5H2O}$), [...] and $\pu{12500 cm-1}$ ($\ce{CuSO4}$), respectively, in their reflectance spectra. This is in agreement with the expectation from the spectrochemical series, and indicates that, in a series of complexes of comparable tetragonalities, the substitution of a coordinated water molecule with a sulfate oxygen weakens the apparent ligand field strenght about $\pu{150-200 cm-1}$. In addition to this shift of the absorption maxima, the following spectral changes were observed on dehydration: (i) the width of the $\mathrm{d–d}$ band becomes gradually narrower with the advance of dehydration; (ii) a broad and very weak absorption appears near $\ce{20000 cm^{-1}}$ in the spectrum of $\ce{CuSO4.H2O}$, and, with somewhat increased intensity, in that of $\ce{CuSO4}$; and (iii) a strong absorption (probably charge-transfer band) appears in the ultraviolet, and shifts remarkably to the lower frequency side with the advance of dehydration, indicating that the $\ce{[SO4]^2-}$ ions are getting more and more strongly bonded, as more and more sulfate oxygens are getting coordinated in the $x-y$ plane of the complex and shared among many $\ce{Cu^2+}$ ions.
  [...]
  Both the slight red-shift and the narrowing of the $\mathrm{d–d}$ band contribute to the disappearance of the blue color [...]

Or, as it is summarized in the book by K. Sone and Y. Fukuda (Ref.2):

  Therefore, the disappearance of the blue color of $\ce{CuSO4.5H2O}$ caused by dehydration is only an apparent phenomenon, which is brought about by a relatively small shift of the $\mathrm{d–d}$ band to the red end of the visible spectrum (and partially into the near infrared) and perhaps also by the spontaneous pulverization of the dehydrated product; if we had eyes which could register radiation from near IR to near UV, we should certainly feel that $\ce{CuSO4}$ is much more darkly colored than $\ce{CuSO4.5H2O}$!

  Since the ligand field strengths of $\ce{H2O}$ and $\ce{[SO4]^2-}$ are seemingly only slightly different from each other, these changes will only bring about a small shift in the $\mathrm{d–d}$ band towards the red. On the other hand, the $\ce{[SO4]^2-}$ ions, which are weakly held in the axial positions of $\ce{CuSO4.5H2O}$, are gradually brought into the equatorial coordination sphere, so that their CT interaction with the $\ce{Cu^2+}$ ion will be strongly enhanced. The observed spectral changes thus come about. Somewhat similar spectral changes are also observed when the double sulfates $\ce{M2Cu[SO4]2.nH2O}$ (where $\ce{M = Na^+, K^+, Rb^+, Cs^+}$ and $\ce{[NH4]^+}$) are heated.

References:

K. Nagase, H. Yokobayashi, K. Sone, “Spectrophotometric and thermal analytical studies on the dehydration of copper(II) sulfate and its double salts,” Thermochimica Acta 1978, 23(2), 283-291 (https://doi.org/10.1016/0040-6031(78)85070-9).
Kozo Sone, Yutaka Fukuda, In Inorganic Thermochromism (Inorganic Chemistry Concepts, Volume 10); Christian K. Jørgensen, Michael F. Lappert, Stephen J. Lippard, John L. Margrave, Kurt Niedenzu, Heinrich Noth, Robert W. Parry, Hideo Yamatera, Eds.; Springer-Verlag: Berlin, Germany, 1987 (ISBN 978-3-642-51019-9).

The blue color which arises when we dissolve $\ce{CuSO4}$ is indeed due to d–d transition. The reason behind this is the splitting effect.

The $\ce{Cu^2+}$ ion is hydrated by four $\ce{H2O}$ molecules, i.e it forms a $\ce{[Cu(H2O)4]^2+}$ complex. Due to the water molecules, the d-orbitals which were degenerate (of same energy) prior to the hydration, are split up into two discrete groups having different energy levels (they are called $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ groups). The energy difference (i.e the energy of photon which will be released due to the transition) between the groups is corresponding to the blue color seen.

Refer to crystal field theory (CFT) if you are willing to know more.


What are energies in computational methods relative to? As per my understanding, in molecular mechanics (MM) the computed strain energy is based off the strain inherent in the molecule under study relative to an idealized form. For example, cyclobutane to its acyclic form.

However, what about computed values from other methods like PM6 energies, HF energies or MP2 enthalpies. What are these computed values relative to?

Ab initio calculations give the absolute energy of the system of fixed nuclei and moving electrons (all considerations on the type of calculations are implicit such as the Born-Oppenheimer approx.). There isn't any distinction of concepts of absolute energies between softwares when ab initio methods are employed (no ambiguity is allowed as pointed out by Pentavalentcarbon). Maybe the confusion can be illustrated with an example: you can't obtain heats of formation, as an absolute energy, from an ab initio calculation but it serves as a good approximation depending on the method.


How does sulfuric acid dehydrate sugars? I know sugars have $\ce{-OH}$ groups, and that sulfuric acid is a strong acid. What I'm failing to see is how exactly sulfuric acid dehydrates sugars. Does it protonate the hydroxyl groups, making them good leaving groups? What about the case of sucrose?

For example, how are the oxygens "taken care of" by the sulfuric acid? Are they doubly protonated? Is there some established mechanism for the dehydration of sugars by sulfuric acid, because I can't find any - I just find general equations and I've looked on Google and through textbooks.

After some further research I learned that the gas, sulfur dioxide, is evolved. This suggests to me that this reaction is also redox in nature. So is this reaction both an acid-base reaction and a redox reaction? Sulfur in sulfuric acid has an oxidation state of +6; sulfur in sulfur dioxide has an oxidation state of +4. Plus the elemental carbon formed has an oxidation state of 0 while in sucrose, carbon has various oxidation states.

$\ce{H2SO4}$ protonates the alcohol to make $\ce{R-OH2+}$ and $\ce{HSO4-}$, the $\ce{HSO4-}$ then takes a hydrogen from the carbon atom adjacent to the alcohol, yielding an alkene and water.  Alcohol dehydration reactions using acid are just elimination reactions.

Additionally, at temperatures above 340 degrees Celsius, sulfur trioxide exists in equilibrium with sulfuric acid in significant concentration.  The presence of sulfur trioxide explains the presence of sulfur dioxide  (via dissociation), or some redox chemistry occurring with the sugar.

The sulfuric acid brakes up the bonds of the sugar and absorbs the water from the crystallization


Reduction half-reaction for the reduction of elemental oxygen Write the reduction half-reaction for the reduction of elemental oxygen in the presence of water, producing hydroxide ions

This does not make sense to me - a half reaction (from what I understand) is when you have an "element + some number of electrons -> element charged" or "charged element -> element + some number of electrons"

Reaction:

$\ce{HOH + O -> 2OH}$

Oxidation States: They all seem to be neutral to me - but I may be missing something.

So, how is a half reaction being introduced into this and what would it look like? Also, how can i know which element/compound is reduced/oxidized and is the reducing agent/oxidizing agent?

I am really sorry if this seems simple but I am a chem noob and I was out for a month and my teacher is behind schedule so he is giving us worksheets without teaching and i do not understand a lot. Thank you for your time

O2 + 2HO+ 4e- —> 4OH-
That is the correct answer


Is there any way to determine if the compound is oxide or peroxide or superoxide? I know that in an oxide oxygen's oxidation number is $-2$, in a peroxide oxygen's oxidation number is $-1$ and in a superoxide oxygen's oxidation number is $-\frac12$. So, how will I be able to detect that any compound is oxide or peroxide or superoxide?

For example, to get the oxidation number of $\ce{Mn}$ in $\ce{KMnO4}$ how will I determine whether $\ce{KMnO4}$ is an oxide, superoxide or peroxide? Is there any way?

generally the most universal way to prove this is X-ray crystallography. It allows to obtain crystal structure of target compound directly. It has, however, an inherent flaw: it can't reliably detect position of hydrogen atoms, so in some cases some uncertainty may remain.

In case obtaining crystal is impossible, the second option is NMR, that allows to get various info about atom's neighborhood, but interpreting NMR data is really hard. In case of superoxide, EPR may prove direct evidence.

Generally there is no way to detect such things from brutto-formula. For example, compound $\ce{CH2O2}$ may be either dioxirane (methylene peroxide) or formic acid. In practice, however, peroxides are really unstable, so in most cases if compound is not overloaded with oxygen (like, $\ce{H2SO5}$) and is reasonably stable towards thermal decomposition, it is most likely oxide and not peroxide. Peroxides are usually formed either from another peroxide or electrochemically, or (in rare cases) via cold oxidation with molecular oxygen. And there are exactly three common superoxides.

If oxygen was with the metals of main group elements, the compound is peroxide, but if it was with nonmetals or other metals except for the metals of main group metals the compound is oxide. For example $\ce{KMnO4}$ is oxide it is composed of $\ce {K+}$ and $\ce{MnO4-}$ because there is manganese which is a metal that isnt among the main group elements group in the polyatomic ion. and the oxidation number of the $\ce{Mn}$ is $7^+$.