Does H2O2 break down into H2O + O or H2O + O2? In hydroponics, Hydrogen Peroxide is used to sanitise nutrient solution from pathogens. Allegedly it does not take sides, oxidising aerobic as well as anaerobic, including damaging the roots, so I was told. I believe this argument, since O as a free radical is highly reactive. I would appreciate your input on this. However, this isn't the main question.

The main question is that manufactures also claim that it will provide DO, dissolved oxygen, to the solution. I find this hard to believe. Reason being, DO refers to dissolved $\ce{O2}$ (Dioxide) not Monoxide. When $\ce{H2O2}$ breaks down it breaks down to $\ce{H2O + O}$ (free radical), right?

Can you please put some clarity into this?

$\ce{H2O2}$ -> $\ce{H2O +1/2O2}$

This might make it easier to understand.

$\ce{H2O2}$ will homolytically cleave for form two $\ce{.OH}$ radicals. Radicals are very reactive and will start a chain reaction, but ultimately you will end up with water and oxygen products from $\ce{H2O2}$.

(Why does it form two hydroxy radicals? because the O-O single bond is weak and unstable - see MO theory)

How do we know this? Look at a bottle of $\ce{H2O2}$ that has been left out of the fridge for too long. There will be a slight pressure build up of $\ce{O2}$ gas and the % concentration $\ce{H2O2}$ will be down due to formation of $\ce{H2O}$and loss of $\ce{H2O2}$.

For sake of completion,
You can alternatively balance the equation thus:
$\ce{2H2O2 -> 2H2O + O2}$

Hydrogen peroxide is unstable under standard conditions and breaks down into $\ce{H2O + O2}$. Although you are not left with a free radical, diatomic oxygen is still an extremely powerful oxidizing agent, which is the same reason rust forms on metals. So even though it is not a free radical, it still has the ability to oxidize your organic compounds.


Is rearrangement possible in cyclic bromonium ion? I wonder what the product of the above reaction is, that is, bromination of the given alkene in non polar solvent carbon tetrachloride (proceeds through cyclic bromonium ion intermediate).

Once the cyclic bromonium ion is formed, can the ring expand as shown or is the usual product formed? Five remembered rings happily expand to six membered ones if possible - so what's going on here?

It'd be great if you could quote sources with your answer - and help me understand whether or not species involving cyclic bromonium ions can rearrange in a reaction, and why?

Thanks!

This is apparently not a likely rearrangement, even in more strained circumstances. There are a number of reactions out there in the literature like the one below.

This result makes sense if we consider the relative stability of the bromonium ion vs. your proposed rearrangement. I've redrawn the mechanism to show all the formal charges. The bromonium should be much more stable (positive charge delocalized over two carbon atoms, all atoms have full octet) compared to a secondary carbocation.

Bromonium rearrangements are not entirely impossible though, check out this fun example (J. Org. Chem., 1998, 63, 2646 DOI: 10.1021/jo9722055)

Here is a reference.
http://research.cm.utexas.edu/nbauld/teach/alkenes3.html#bromination
Read the part “electrophilic addition of bromine and other halogens”. Technically, to refine my argument, the positive charge is mainly on the bromine.


How can one reclaim isotopic cadmium from a solution of acid, i.e. H2SO4 or HCl etc.? I have dissolved isotopic cadmium-114 with $\ce{H2SO4}$ and or $\ce{HCl}$.  Now that I have the cadmium-114 in solution, precipitating out $\ce{CdSO4}$ and or $\ce{CdCl2}$, how can I reduce these simply to leave the pure cadmium metal isotope?  I have dissolved $\ce{^{114}Cd}$ with acid in very complex electronic tubes and I wish to reduce the precipitated compound(s) to pure metal for reuse.

$\ce{^{114}Cd}$ or is running about $ 1,500 per gram.  So extremely useful to reclaim it. Can anyone tell me a simple way to do this?

Thank you everyone for the discussion.  Sorry, I have been away for a few days, so was unable to follow up.
Reading the ideas, it is a toss up on a procedure.  Scraping the Cd off of an Aluminum electrode might work, but Al contamination would be very bad for the laser oscillation.  It might work for awhile, but would die from contaminants in short order.
The Al oxide, though only a molecular layer, is extremely tough and can withstand a great deal of heat and abuse.  I make Aluminum edge-wound voice coils for my speaker company and we depend on the oxide layer to act as an insulator from winding to winding.  Our coils can withstand 600 degrees and will melt before they fail.
Stripping the oxide gives you only a few seconds before it reforms in the normal atmosphere.  Any other thoughts?
Perhaps reduction of the salt by heating?
Thanks
Dale

From a solution of an aqueous cadmium salt ($\ce{CdSO4}$ or $\ce{CdCl2}$), try inserting a metal plate of say Aluminum.

With time and warming of the solution, a coating of cadmium should appear on the metal sheet.

Try scrapping off the deposited cadmium metal, per the replacement reaction:

[EDIT]$$\ce{2 Al(s) + 3 CdSO4 (aq) -> 3 Cd (s) + Al2(SO4)3 (aq)}$$

Note: depending on your tools for scrapping, you may induce some metallic impurities.

[EDIT] Importantly, my standard chemistry approach (based on a single-displacement reaction) is different from an electrolysis or electrochemical cell approach, employing an Aluminum electrode, as I suspect, that the electrode may quickly become coated and effectively inoperative.  Also, an electrochemical cell with a large surface area anode (here Al) would require an even larger surface area cathode to operate efficiently. Further, the ability to facilely recover the Cd metal is key to the cadmium recycling process. My proposed process may fulfill these requirements.


Why does a NOx container get cold when NOx is released? and What is the relationship of the vapor pressure curve with the liquid-vapor dome? The temperature at which $\ce{NO_x}$ is in equilibrium with its liquid and vapor phase at 1 atm is about –84 °C. Does that mean $\ce{NO_x}$ exists as a liquid at that temperature? What about it's vapor phase then, because it's vapor and liquid phase should be at equilibrium, right?

Also, it's that the reason why compressed $\ce{NO_x}$ when released from its closed and pressurized container (e.g. 50 bar) at 20 °C becomes a gas while its temperature drops? Therefore, the compressed $\ce{NO_x}$ was in it's liquid state at 50 bar, right?

If that's the case, then what is the relationship between the vapor pressure curve and the liquid-vapor dome? How would you use the saturated liquid-vapor tables to predict the vapor pressure for your substance? And which specific volume would you chose to define your substance: its liquid specific volume or its vapor specific volume?

The vapor pressure curve describes the relationship between temperature and pressure for which the pure substance can exist as liquid and vapor, together in equilibrium.  So it is not necessarily all liquid or all vapor.  Both can be present in different proportions.  As far as specific volume is concerned, your tables should list both the vapor specific volume and the liquid  specific volume.  The specific volume of the combination of vapor and liquid is proportional to the mass fraction of each.  (So you need to know the mass fractions of vapor and liquid).

Every gas at decompressing cools down (adiabatic expansion).

See for example:
https://physics.stackexchange.com/questions/14140/why-does-the-gas-get-cold-when-i-spray-it


Are there colorful metals besides gold and copper? Are there yellow metals besides gold? Are there metals with color different than yellow and grey/silver or reddish?

Osmium has a bluish-gray tint.
Well; slightly.
Cesium is silvery-golden!,
But don't wear it.

It seems everyone forgot bismuth, the most colorful of metals!
(Only when crystallized, but that is an extremely easy process)


Which is more acidic BF3 or AlCl3? Which of the species $\ce{BF3}$ and $\ce{AlCl3}$ is more acidic, both are electron deficient.

In case of $\ce{BF3}$ one can argue that there is extensive p-π-p-π back bonding and as a result the electron deficiency is less.
That means that the positive charge on the overall molecule is less.

In case of $\ce{AlCl3}$ one can argue there is p-π-d-π back bonding and as a result the electron deficiency is more in case of aluminiuim.
Therefore the overall positive charge on the molecule is more, meaning it is more acidic.

So $\ce{AlCl3}$ is more acidic that $\ce{BF3}$ is conclusion of the above factors. Is this plausible?

I also faced the counter logic that $\ce{BF3}$ being a small compound tends to react more readily and therefore is a strong acid.

Actually, there are three reasons why BF3 is more acidic than AlCl3.

BF3 is smaller in size and can easily attract the incoming pair of electrons.
Fluorine is more electronegative than chlorine. So boron has less ionisation potential as compare to aluminium thus forms anion easily.
Boron is more electronegative than aluminium, therefore boron has more tendency to attract electrons. It means that BF3 behaves more like Lewis acid than AlCl3.

This photo has been taken from this article.

$\ce{BF3}$ is less acidic than $\ce{AlCl3}$ as there much stronger pi-backbonding between the Boron and Fluorine atoms while there is no backbonding in AlCl3 as well. The tendency of boron to accept electrons from the base, is way lesser than that of Al.
And, There is no backbonding in AlCl3, check out this link:
https://byjus.com/questions/why-alcl3-does-not-show-back-bonding/


What is the reason for the anomalous electronic configuration of palladium? The p block elements generally have the general valence shell configuration as $$n\mathrm{p}^{1-6}\,n\mathrm{s}^{1-2}.$$ Expected electronic configuration for palladium is $$\ce{^46Pd} = \mathrm{1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^8}.$$ But I find that it's valence shell instead is $$\mathrm{4d^{10}\, 5s^0}$$ I first thought, for achieving fully filled d-subshell, 2 electrons go to the d-subshell from s-subshell. This electron transfer should happen for all members of its group, yet only palladium demonstrates this. Why is this so?

The Aufbau principle states that, hypothetically, electrons orbiting one or more atoms fill the lowest available energy levels before filling higher levels (e.g., 1s before 2s). In this way, the electrons of an atom, molecule, or ion harmonize into the most stable electron configuration possible.

Electron behavior is elaborated by other principles of atomic physics, such as Hund's rule and the Pauli exclusion principle. Hund's rule asserts that even if multiple orbitals of the same energy are available, electrons fill unoccupied orbitals first, before reusing orbitals occupied by other electrons. But, according to the Pauli exclusion principle, in order for electrons to occupy the same orbital, they must have different spins.

The order in which these orbitals are filled is given by n+l rule known as Madelung Rule or diagonal rule.Orbitals with a lower (n+l) value are filled before those with higher (n+l) values where n represents principal quantum number and l represents azimuthal quantum number.

The fact that most of the ground state configurations of neutral atoms fill orbitals following this n + ℓ, n pattern was obtained experimentally, by reference to the spectroscopic characteristics of the elements.

The Madelung energy ordering rule applies only to neutral atoms in their ground state,even in some cases there are examples which show different electronic configuration than those of experimentally determined.Copper,Chromium and Palladium are common examples of this rule.

For Palladium,according to Madelung's rule, 5s^2 has lower (n+l) value i.e. (5+2=7) and 4d^8 has higher (n+l) value i.e. (4+8=12),so the 5s orbital  fills up first before 4d orbital.

Hence, Palladium shows anomalous electronic configuration.


Why are nitro compounds with at least one alpha-H soluble in NaOH? Are all the nitro-compounds having at least one alpha-$\ce{H}$ soluble in $\ce{NaOH}$? And what does solubility has to do with alpha-$\ce{H}$?

I was reading about tautomerism today and got stuck on this thing. Is it due to its conversion to the enol form? Does the keto form also show this? What could be the possible reasons?

The compounds containing nitro groups when dissolved in basic medium, the following things can happen.The base can take the $\ce{$\alpha$-H}$ of the nitro-group, which will create a negative charge on that $\ce{$\alpha$-C}$, which can further delocalise with the Nitro-group (similar to Nitro-Acinitro tautomerism).
So, the whole compound will be in the state of a Resonance Hybrid where there will be partial double bond character between $\ce{$\alpha$-C, N }$ and two $\ce O$ atoms. and also there will be a $\delta$- charge on the $\ce{$\alpha$-C }$ atom as well as on the $O$ atoms. , and $\delta$+ charge on the $N$ atoms.Due to such salt like compound formation, it becomes soluble. This $\delta$- charge created on the Carbon atom is stable if the side chain is short, but starts to get destabilised as the side chain grows longer. So, I think the solubilities of the Nitro-compounds decrease as the side chain grows longer, and the compound with $10-12$ C atoms may not be at all soluble in base because of such large Hydrophobic part and destabilisation of the $\delta$- charge.


Does acetylene give a positive Tollens' test? If not, why is this so? One of the uses of the Tollens' reagent is to confirm the presence of a terminal triple bond in a compound. Does acetylene give a positive Tollens' test? If not, why is this so?

Tollen's reagent (ammoniacal silver nitrate) reacts with terminal acetylenes to form the silver acetylide which precipitates out of solution

$$\ce{RC#CH + AgNO3 + NH4OH -> RC#CAg v + NH4NO3 + H2O}$$

In the case of acetylene itself, since both ends of the molecule are (if you will) terminal acetylenes, a bis-silver acetylide salt is formed and precipitates out of solution (see page 3, item 6b in a PDF from www.sakshieducation.com/ or via the Internet Archive)

$$\ce{HC#CH + 2AgNO3 + 2NH4OH -> AgC#CAg v + 2NH4NO3 + 2H2O}$$

Tollen's reagent is not very stable, therefore it needs to be freshly prepared before use.  The reagent has a number of other uses in chemistry (test for aldehydes and α-hydroxy ketones), anatomical staining, and silver mirroring.  This Wikipedia article provides a nice, concise overview.

Terminal hydrogens in acetylene are acidic and are easily removed to form acetylide ions, and in this case, elementary silver will be displaced from the ammoniacal silver nitrate and will form silver mirror on the surface of the reacting vessel.


Which one is the best notation for sodium acetate I have seen two different notations for sodium acetate. The first one is:

$$\text{NaCH}_3\text{COO}$$

The second one is:

$$\text{CH}_3\text{COONa}$$

Now I'm confused, which one is the best to use?

There is nothing wrong with either formula. And you can use even more:

$\ce{NaC2H3O2}$
$\ce{C2H3NaO2}$

It really depends on which point you want to bring across.

$$\ce{NaCH3COO}$$

This formula, being analogous to formulae like $\ce{NaCl}$ stresses the inorganic salt view more. It shows that there is a cation ($\ce{Na+}$) and an anion ($\ce{CH3COO-}$) which form a salt crystal together much like the anion $\ce{Cl-}$ would do with the same cation. Inorganic nomenclature prefers cations to be written first.

$$\ce{NaC2H3O2}$$

Is basically the same except for saying ‘I don’t care how those seven atoms combine to form the anion, all I care for is what’s in there.’

$$\ce{CH3COONa}$$

This one stresses a more organic-chemical point of view where it’s relevant where the cation is actually bound to within the molecule. Oftentimes, organic chemists would even write structural formulae with a bond between oxygen and sodium as if it were hydrogen. The reasoning behind this is ‘I don’t care if the structure ends up being a salt, all I need to know is that the sodium somehow connects to the carboxyl group.’

Note that you can combine the first’s and the third’s idea to give $\ce{NaOOCCH3}$ which works, but is a lot less used than either of them.

$$\ce{C2H3NaO2}$$

This is the Beilstein-type lookup formula. You just know the elemental composition of your compound and now want to look it up — you can’t know whether it’s sodium acetate or sodium hydroxyethanalate (if that name is even correct). Or maybe you just don’t care. The reasoning behind this is $\ce{C}$ first, $\ce{H}$ second, the remaining elements following in alphabetic order.

Finally, there are also shortened formulae. The most common (the one I actually use most) would be $\ce{NaOAc}$. Here, $\ce{Ac}$ is an abbreviation for $\ce{H3C-C=O}$ where whatever follows or precedes is bound to the carbonyl carbon. Note that this formula is not considered standard, and would need to be included in a list of abbreviations or defined otherwise if you decide to use it.

Both are correct, but the second one is more commonly used. Actually, the way of writing a molecular formula signifies the structure of the molecule.

Let me explain: There are compounds that have the same molecular formula but have different structures. Such molecules are called structural isomers. An example for a carbon compound:

As you can see, even if both have the same formula they have different structures (they are structural isomers to say it precisely). Generally formulae of many compounds or molecules are written in a way so that they can signify the structure or arrangement of atoms in molecules.

Now look at that image again. If I want to write chemical formulae of both compounds that can signify their structures, I would write:
$\ce{CH3CH2CH2Cl}$ for the first one, and the second one is $\ce{CH3CHClCH3}$.

I guess now you can understand that the way of writing a formula defines the structure of the compound, so always write the formula in a way that can define or give some hint about the structure.


Lewis Structure For H2SO4 When I was drawing the lewis structure for $\ce{H2SO4}$ I got this configuration, now I know this isn't the best configuration but I don't see whats wrong with it. The number of valence electrons is right, the free electrons are on the most electronegative element and the formal charges are all 0. I'm not sure if there's a rule I'm missing.

As Jan mentions in the other answer, in general, a Lewis structure is preferred if it retains the octet rule. There can be more complicated bonding, and sulfur and phosphorous are two common examples (e.g., $\ce{SF6}$).

I find that many students can come up with "non-traditional" Lewis structures that, like yours, satisfy the number of valence electrons and minimize the formal charge. I usually consider these correct responses on an exam.

The "missing piece" is ring strain, which is not typically discussed until organic chemistry courses.

Note that your diagram as two O-S-O three-membered rings. These are extremely high in energy, because the O-O-S angle won't be anywhere near the expected 109.5°.

I tried a few quick calculations using the PM7 method and MOPAC. The best ring structure I could find looks like this (a distorted octahedral shape) and is estimated to have a $\Delta H_f^0$ = +33.05 kcal/mol. The O-O-S ring angles are ~65°.

The lowest energy geometry is the traditional $\ce{H2SO4}$ Lewis structure, estimated at $\Delta H_f^0$ = -177.88 kcal/mol. While PM7 isn't a highly accurate method, the difference in stability isn't close.

You are missing the octet rule. In your proposed structure, sulphur would have a total of twelve valence electrons which just doesn’t happen.


Why is Aluminium oxide amphoteric? I was reading my book, in which it mentioned that aluminium and gallium oxides are amphoteric and and indium and thalium oxide are basic in their properties.
But no explanation was given.
My main question is we know that mostly metal oxides are basic and non metal oxides are acidic. But we see some metal being amphoteric, what is the cause of this difference?
I tried to search online but found no reliable piece to read.

I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements
To start off, please go through my answer to this question: Why is chromate stable in basic medium and dichromate stable in acidic medium? to read about the preface to this method, including the potential concerns.
Essentially, the book talks about determining the type of reactivity of inorganic molecules by looking at their structures. While this may not reflect with 100% accuracy the actual reaction pathway, we can take it to be a pretty good approximation of what happens in reality. While the completely correct mechanism can only be found out experimentally, we can get a probable mechanism by employing an arrow-pushing approach, and that should at least mirror the actual pathway to a sufficient extent, if not entirely.
Now, if we look at the structures of Aluminium oxide and Gallium oxide ($\ce{Al2O3}$ and $\ce{Ga2O3}$), we can see that they possess a similar bridged dimer-kind of structure as shown below for aluminum oxide(the same structural skeleton is valid for gallium oxide as well, just by replacing aluminum with gallium below) :

This structure now possesses some unique properties. It contains an $\ce{Al-O-Al}$ bridge with an $\ce{Al=O}$ group on each side.Now,the $\ce{Al=O}$ group can be said to be structurally similar to the carbonyl group from organic chemistry.
Hence, like the carbonyl group, it can be said that there will be high partial positive charge on the $\ce{Al}$ atoms due to the electron-withdrawing effect of the $\ce{Al=O}$ groups. Hence, they can serve as the ideal electrophilic site for the attack of a nucleophile like $\ce{OH-}$ in basic medium, and lead to bond cleavage(and further reaction) in alkali as follows:

In a basic medium, the $\ce{AlO2H}$ moiety will also be deprotonated, and thereby you will end up with two moles of $\ce{AlO2^-}$ . If the base being used was $\ce{NaOH}$, then these products are better reported as $\ce{NaAlO2}$ for aluminum. In an analogous case, you would get $\ce{NaGaO2}$ if the we had started out with gallium oxide.
In excess of alkali, the reaction can proceed further, as further additions of hydroxide are still possible at the $\ce{Al=O}$ site as follows:

The reaction can either finish after the formation of the hydroxide, or a continued process may lead to the addition of yet another $\ce{OH^-}$ to form the complex ion $\ce{[Al(OH)4]^-}$ , which might be called the aluminate ion here.In another analogous case,we would have ended up with $\ce{[Ga(OH)4]^-}$ (called the gallate ion) if we had started out with gallium oxide.
Now, for the acidic medium, let us suppose that $\ce{Al2O3}$ reacts with $\ce{HCl}$. Here, the central oxygen of the $\ce{Al-O-Al}$ bridge represents a weak link that can be easily cleaved after protonation(because it has moieties that can function as good leaving groups on either side). Also,the counter-ion of the acid(that is, $\ce{Cl^-}$), can serve as a nucleophile to break down the dimer skeleton after protonation as follows:

The $\ce{AlO2H}$ can again be protonated to form an oxonium ion that can expel water and take up a $\ce{Cl^-}$ to form another mole of $\ce{AlOCl}$(shown in the next diagram).
Following that, the $\ce{AlOCl}$ can be engaged in a repetitive process of protonation, attack by chloride and removal of water to yield the final products as follows:

The products for these reactions can be verified for aluminium and gallium oxides from their Wikipedia pages respectively.
So we can see that one major reason for the amphoteric nature of oxides can be their bridged structures, which allows them to react with both acids and bases as illustrated above.
Now, coming to indium oxide, the formula of this compound as stated here is $\ce{In2O3}$, which bears a resemblance to the formula of $\ce{Al2O3}$ and $\ce{Ga2O3}$ which we stated above. So, I would expect this oxide to atleast have some amphoteric properties. This is indeed the case, as stated here for indium hydroxide(since hydroxides are simply hydrated forms of their oxides, we can expect a similar nature for indium oxide as well):

The result indicated conclusively that In(OH)3 is only slightly amphoteric, being much more basic than acidic.

Also, on the Wikipedia page for indium oxide it clearly states that this oxide is soluble in acids but not in alkali. This is pretty much expected because when you move down a group, the elements tend to become more metallic in nature, and hence their oxides tend to become more basic as well. Hence, an amphoteric oxide slightly tending towards the basic end seems reasonable for indium(III) oxide. The reaction between acid and $\ce{In2O3}$ should follow the general scheme that is illustrated above for $\ce{Al2O3}$ and $\ce{HCl}$, and should yield similar products. This indeed is the case as can be seen here
As for thallium, due to the inert pair effect, it generally forms an oxide in which thallium is in the +1 oxidation state, with the formula $\ce{Tl2O}$ as stated here. This doesn't fit in with the general formulation of $\ce{Al2O3}$ and $\ce{Ga2O3}$ that we discussed above. Hence, one would expect this oxide to simply become more basic than the other group members(as it is an oxide of the lower-most element of the group) and not show much signs of an amphoteric nature. This matches with what is stated here.
Before wrapping up the answer, I would like to re-iterate a point I made at the beginning of the answer:
All these pathways are simply possible mechanisms, and there is no way of knowing whether they are completely correct or not without performing experiments. However, they are highly probable propositions, and you can expect these pathways to represent the actual mechanisms to a good approximation

Metal hydroxides are basic in water. Non-metal oxides are acids in water. The limit between metals and non-metals is a sort of staircase going through the periodic table from the middle of the first line to the lower corner at the right-hand side. Aluminium touches this staircase. This is why Aluminium behaves like a metal and like a non-metal, and why aluminium oxide can react with acids and with bases.
Tin Sn has the same property.


How many atoms are represented in this equation? $\ce{Pb(NO3)2 + 2KI -> PbI2 + 2KNO3}$

The answer says 26, but shouldn't it be 13? Because actually, the number of atoms involved in the reaction are 13, and once the reaction is completed, again, number of atoms is 13.
Please explain.

The referenced atoms on both sides of the arrow are the exact same atoms? I like your thinking!

But, chemical equations do not refer to individual atoms.

The formula $\ce{Pb(NO3)2}$ can be read as "A molecule of this substance consists of six atoms of O, two atoms of N, ...". That is correct.

But the equation $\ce{Pb(NO3)2 + 2KI -> PbI2 + 2KNO3}$, strictly speaking, does not talk about single molecules. The numbers (including the 1's that are nor written) are called "stoichiometric coefficients". The equation means: "Any number of $\ce{Pb(NO3)2}$ molecules react with the double amount of $\ce{KI}$ molecules to yield ...".

Wikipedia supports my view:

  In lay terms, the stoichiometric coefficient (or stoichiometric number in the IUPAC nomenclature) of any given component is the number of molecules that participate in the reaction as written.

  ...

  In more technically precise terms, ...

... the stoichiometric coefficient is the number of molecules divided by the progress variable or extent of reaction.

  ...

  The (dimensionless) "units" may be taken to be molecules or moles. Moles are most commonly used, but it is more suggestive to picture incremental chemical reactions in terms of molecules.

they didn't specify counting either the reactants or products side alone. You were to count both the reactants & products side


5-methylhexa-1,3-diene reaction with HBr What will be the major product of the reaction?
I guess that rearrangement will take place and give 5-bromo-5-methylhex-2-ene as a product, but I'm completely not sure

In this reaction if peroxide is not present then reaction follows markovnikov addition in which carbonation is formed as intermediate. By stability due to resonance and 2° carbonation
5-methylhexa-1-ene cation with C+ at 3 is formed. Hence 3-bromo-5-methylhexa-1-ene is formed as product.


Is iron the most stable element in the periodic table? According to the binding energy per nucleon vs mass number graph, it is observed that iron-56 has the maximum value of binding energy per nucleon ($\pu{8.75 MeV}$). It means that iron-56  is the most efficiently bound nucleus meaning that it has the least average mass per nucleon. This is the approximate basic reason why iron and nickel are very common metals in planetary cores since they are produced profusely as end products in supernovae and in the final stages of silicon burning in the stars. So, in one word, iron is quite stable.

But, what about helium and other noble gases? They are considered the most stable elements in the whole periodic table. But their binding energy per nucleon value is less than iron-56. So, they are not stable as iron-56. Is that true? Is iron the most stable element in the periodic table both structurally and chemically?

Yes, $^{56}\ce{Fe}$ has the most stable nucleus, and $\ce{He}$ is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in a man. As for structural stability, there is no such thing in chemistry (there is one in architecture and another in mathematics, but those are out of scope of this question).

Well, Iron has the most stable isotope(Iron-56), it has a large half-life which means it takes a fairly large amount of time to decay into the half of what was initially there. Noble gases have a low reactivity, which means a low electronegativity and a high ionization energy. Noble Gases should be identified as inert and not stable. They are not the same.


If sulphuric acid barely evaporates, why we can smell it? If sulfuric acid barely evaporates, why we can smell it? Or does it give off sulfur dioxide gas?

There are at least two plausible explanations for why an odorless compound with very low vapor pressure could nevertheless generate an odor:

Sulfuric acid is highly reactive.  If you are pouring or dispensing sulfuric acid, you are probably generating some mist. This is a concern in industrial settings where sulfuric acid is used:

https://www.cdc.gov/niosh/docs/81-123/pdfs/0577.pdf

Thus, even if the sulfuric acid itself has neither a significant vapor pressure nor an odor, as the mist enters your nose it could react with substances in your nasal passages to create compounds that have an odor.  Some studies have reported the perception of odor from sulfuric acid mists:
https://www.epa.gov/sites/production/files/2014-09/documents/sulfuric_acid_interim_dec_2008_v1.pdf

And even without misting, particulates in the air, as they fall into an open container of sulfuric acid, could react with sulfuric acid to generate compounds that create a smell.  This effect is seen with metals -- the metallic "smell" is not from the metals themselves (which have an extremely small vapor pressure at room temperature), but from reaction products of the metals:

Why can we smell copper?

It may reach our nose due to a varietyof reasons. The sensation of smelling is it probably because sulfuric acid absorbs water from the nose and irritates it. Similar to how eyes are irritated while cutting onions due to formation of sulphuric acid. Also, it isn't necessary for it to vaporize. It can be just fine droplets that reach the nose and cause irritation.


Insoluble Group I/ammonium salts Most solubility references (e.g. this one) say that there are "few" insoluble salts of alkali metals.

Are there any exceptions?

Lithium carbonate has poor water solubility.  Cesium triphenylcyanoborate is insoluble in water.

There are some cations whose Group I/ammonium salts are not soluble in water. For example, (thanks Nicolau) the diuranate of ammonium is insoluble in water.

Additionally, (as I just found) the hydrides (H-) of alkali metals are not soluble, it seems, in any room-temperature liquid solvent. They hydrolyze violently (http://goo.gl/Dl8Bd4), and the hydride ion has never been observed in solution.


How can metals like magnesium and sodium burn underwater? Please answer with as much detail as possible. I can't find the answer anywhere else and I need it for an essay.

Any websites I have found have conflicting answers.

One said that they don't actually burn underwater, rather than separate hydrogen from oxygen and burn at the surface. Another actually said that they do burn underwater because they take the oxygen source from the water. I do not have a specific definition for "burn" (the question itself is my assignment- write an essay answering that question!), so I know about as much as you do concerning the wording of the assignment.

I do know that alkali and alkaline earth metals are generally very reactive with water, and usually result in exothermic reactions.

Your last sentence was the key.  Metals like sodium and magesium react with water according to the following equations
$$\ce{2Na + 2H2O -> 2NaOH +H2 + heat}$$
$$\ce{Mg + 2H2O -> Mg(OH)2 +H2 + heat}$$
Both reactions produce hydrogen gas and a lot of heat.  As the hydrogen gas bubbles out of the water it mixes with atmospheric oxygen. The intense heat ignites the hydrogen-oxygen mixture and the mixture continues to burn (possibly explosively depending upon the quantities involved) as long as hydrogen continues to be generated.

I specialize in environmental problems, namely waste-water. From what I can remember, sodium and magnesium react with water, very violently. Sodium is actually used in nuclear submarines to detect steam/water leaks. Cesium, as far as I know, is stored in some form of oil to keep it from reacting with moisture. I was in the 5th grade when my science teacher showed me a demonstration of sodium and how oxygen and hydrogen implode to form water - just like a rocket-ship engine works.


Can I determine the rate law order from stoichiometric coefficients? I've read in two books, and seen in three separate videos courses, that you cannot determine the rate law order from stoichiometric coefficients.

Then, I have seen homework and "practice" problems that ask you to do exactly that.

Needless to say, I am confused.

This is how I am presently making sense of the issue: Given a balanced equation, and absent any other information, you can (and must) use the stoichiometric coefficients to determine the rate law. So, for example, given the equation $2A+B \rightarrow C + 3D$, the rate law would be $rate = k[A]^2[B]$.

But, this would only hold true if this were an "elementary reaction" (i.e., no equilibrium).

So, I wonder why the books and videos emphasize that you can only determine order by experiment and not from the stoichiometric coefficients. I am guessing that they do so because, in practice, you do not know if the reaction is an elementary reaction or not and therefore experiment is the only way to make that determination. And, because you don't really know if it is an elementary reaction, you can only determine the rate law experimentally.

The bottom line:

as a practitioner, determine rate law experimentally;
as a student, given no information other than the balanced equation, determine the rate law from the stoichiometric coefficients;
as a student, given rate and concentration data, determine the order (and rate law) from the data.

That is a pragmatic, but unsatisfying, solution to the problem but I would prefer a clearer explanation of what is going on.

There are multiple issues here:

You should not assume that there is a simple rate law. Chemistry is complicated, and even simple reactions may proceed through multiple pathways in parallel.
More generally, you should not assume that you are given enough information to determine the solution. Just because the stoichiometric coefficients are there, does not mean that you can use them to determine the answer. So, if you just have a balanced reaction, you know nothing. If you want to become a scientist, you should learn to be comfortable with not having the answer.

So, I would say that your bottom line is incorrect.

The experimental rate law should be determined experimentally.

Given a putative mechanism with elementary steps and knowledge of rate constants for those steps, you can propose a rate law that for that mechanism. Do not assume it has any bearing on reality, since your putative mechanism may be wrong.

Frequently, the approach is to take the experimental rate law and see if the derived rate law for a proposed mechanism is consistent with the experimental result.

Nobody can determine the order of a reaction with just the stoichiometric coefficients.

For example look at the reaction $I_2 + Acetone -> ICH_2COCH_3 + HI$. It has a rate corresponding to the law : -$d([I_2])/dt = k[H^+][Acetone]$. The concentration $[I_2]$ does not appear in the rate law. It may seem incredible, unbelievable. Nobody can guess it. It is an experimental result.

On the other hand, if a reaction A + B --> .....   has a rate law like : - d[A]/dt = k[A][B], in this case one can say that the reaction is an elementary reaction.


Difference between Resonance Effect and Mesomeric Effect While studying electron displacement effects in organic chemistry, I read that the Resonance Effect and Mesomeric Effect are the same. Every source I checked used the heading or opening "The Resonance (or Mesomeric) Effect is..."

However, I was able to find the difference between Resonance Structures and Mesomeric Structures is and found out that the two differ in how they depict conjugated pi bonding. The Mesomeric Structures are all the compounds with all possible arrangements or the pi bonds drawn separately. The Resonance Structure is one diagram that uses dashes and lines to denote the partial double bond characters in the compound.

So my understanding from this difference between the structures is that the effects must only differ in how they are depicted. But I was unable to find a source that could help confirm this. Can someone please confirm or explain it for me?

The IUPAC gold book (IUPAC are the organisation which 'define' standards for things such as chemical nomenclature) describes them as being essentially synonymous. The following definitions are given (link):

Resonance  In the context of chemistry, the term refers to the
representation of the electronic structure of a molecular entity in
terms of contributing structures. Resonance among contributing
structures means that the wavefunction is represented by 'mixing' the
wavefunctions of the contributing structures. The concept is the basis
of the quantum mechanical valence bond methods. The resulting
stabilization is linked to the quantum mechanical concept of
'resonance energy'. The term resonance is also used to refer to the
delocalization phenomenon itself.
Mesomerism Essentially synonymous with resonance. The term is particularly associated with the picture of π-electrons as less localized in an actual molecule than in a Lewis formula. The term is intended to imply that the correct representation of a structure is intermediate between two or more Lewis formulae.

There does seem to have been some historic difference that has largely been reconciled into a single definition however one use of Mesomerism that still gets commonly used is the definition of groups as having a positive (+M) or negative (-M) effect to describe the effect of a group on a conjugated system.

The Mesomeric Structures are all the compounds with all possible arrangements or the pi bonds drawn separately. The Resonance Structure is one diagram that uses dashes and lines to denote the partial double bond characters in the compound.

This statement is not entirely correct.  The 'Mesomeric' structures you talk about are usually refered to as resonance structures/forms. The depiction with the dashed lines and partial bonds is a resonance hybrid.

Yes, they're almost synonymous. The only minute difference between them is this:

Mesomerism = donation/withdrawal of electrons

Resonance = delocalization of electrons


When to regard water as a strong base Question:

Classify each of the following ions according to whether they react with water to give a neutral, acidic, or basic
solution:

As a general rule:

Strong acid + Strong base $\rightarrow$ Neutral solution
Strong acid + weak base $\rightarrow$ acidic solution
Weak acid + Strong base $\rightarrow$ Basic solution

According to my solutions manual.

a) $\ce{F-}$ is the conjugate base of a weak acid, therefore  the solution is basic.
b) $\ce{Br-}$ is the anion of strong acid, therefore the solution is neutral.
c) $\ce{NH4+}$ is the conjugate acid of a weak base, therefore the solution is acidic
d) $\ce{[K(H2O)6]+}$ is the conjugate base of a weak acid, therefore the soltuion is basic.
e) $\ce{SO3^2-}$ is the conjugate base of a weak acid, therefore the solution is basic
f) $\ce{[Cr(H2O)6]^3+}$ is an acidic solution, therefore the solution is acidic.

However, all these answers imply that water is acting as a strong base. How is this possible when water pH is $\sim 7$?

Question:

Classify each of the following ions according to whether they react with water to give a neutral, acidic, or basic
solution:
As a general rule:

Strong acid + Strong base $\rightarrow$ Neutral solution

Strong acid + weak base $\rightarrow$ acidic solution

Weak acid + Strong base $\rightarrow$ Basic solution

That's not true. The only time in which these rules are true is when you have an equimolar concentration of the acid and base in question, and the acids and bases must both be monoprotic and monobasic, or diprotic and dibasic, etc. Otherwise there are too many variables to consider for these guidelines to always ring true. Consider mixing 1.0 molar $\ce{H2SO4}$ with 0.10 molar $\ce{NaOH}$ ... I guarantee you that the pH of that solution is not "neutral."

d) $\ce{K(H2O)6+}$ is the conjugae base of a weak acid, therefore the soltuion is basic.

I didn't know that potassium hydrated to a significant extent in water; even so, what the solution manual has written down is not the conjugate base of hydrated potassium cation - the weak acid.

However, all these answers imply that water is acting as a strong
base. How is this possible when water pH is $\sim 7$?

How do these answers imply that water is acting as a strong base?
Also regarding how water can be a strong base in general - it just depends on your solvent system.

One more thing that you have to keep in mind is that water acts as a strong base ONLY with weak acids.

PECULIAR nature of water is that with weak bases, water acts as an acid.
So, it can be said that it is amphoteric in this regard.


How to calculate the amount of substances product produces from the masses and amounts of starting materials? Calculate the amount of carbon dioxide formed when we burn 1 mole of carbon in the presence of 16 g of dioxygen.
  $$\ce{C + O2 -> CO2}$$

1 mole of carbon means 12 g. Therefore amount of carbon dioxide is 12 + 16 = 28. Am I correct?

From the equation,
$$\ce{C + O2 -> CO2},$$
we can see that one mole of $\ce{C}$ reacts with one mole of $\ce{O2}$.

Therefore, let us calculate the amount of each reactant present.

The amount of $\ce{C}$ is already given: $n(\ce{C}) = \pu{1 mol}$.

The molar mass of $\ce{O2}$ is $M(\ce{O2}) = \pu{32 g//mol}$, so $\pu{16 g}$ of $\ce{O2}$ are $\pu{0.5 mol}$.

Since one mole of $\ce{C}$ only reacts with one mole of $\ce{O2}$, there remain $\pu{0.5 mol}$ of unreacted $\ce{C}$, while $\pu{0.5 mol}$ of $\ce{C}$ react with $\pu{0.5 mol}$ of $\ce{O2}$.

In formal terms, oxygen is called the limiting reactant as it limits the amount of product formed.

From the equation, one mole of $\ce{CO2}$ is produced per each mole of $\ce{C}$. Therefore, $\pu{0.5 mol}$ of $\ce{CO2}$ is produced.

The one mole of carbon would react with one mole of dioxygen to give one mole of carbon dioxide.

$\pu{12 g}$ of carbon would therefore react with $\pu{32 g}$ of $\ce{O2}$ to give $\ce{44 g}$ of $\ce{CO2}$.

As someone else has pointed out, it would be tricky to get absolutely all the carbon to burn in oxygen unless you set it up in a particular way to do so. Probably in real life some would not burn efficiently and you might end up with carbon monoxide, which is poisonous if you breath it in. Some carbon might also stay as carbon (soot).


Why does electronegativity create a stronger acid, but so does greater electron distribution? I just learned that:

Acids get stronger with:

Increasing electronegativity of atom bonded to $\ce{H}$.
Size of atom bearing negative charge.
Hybridization of negative charge (more s character, $\ce{e-}$ closer to nucleus, more acidic).
Resonance stabilization (like size, greater distribution of electrons).

So basically, the  more concentrated the negative charge (think $\ce{F-}$, $-1$ and very small), the tighter it's going to "hold on" to the proton. With resonance and large negative groups, the charge is less concentrated, so it's not going to "hold on" to the proton as well.

However, electronegativity is the tendency for an atom to "hog" electrons. Therefore, the more electronegative, the more electrons there are in a similarly sized space.

More concentrated electrons = more stable base = weaker acid.

But electronegativity = more concentrated electrons (even though size trumps it).
And electronegativity = stronger acid.

Feel free to edit if you see what I'm trying to say but can say it better!

The main point about strong acids is ‘how stable is the conjugate base?’

The better stabilised the negative charge on a conjugate base of a neutral acid is, the happier the species will be to loose a proton and thus the more acidic the proton is. All of the points you noted just emphasise that.

Therefore, the only wrong assumption you are making is:

  more stable base = weaker acid.

The correct way is:

  more stable base = stronger acid.

Increasing EN means less acidic.
Refer to the $\mathrm{p}K_\mathrm{a}$ values below.

$$
\begin{array}{l|rrrr}
\hline
\ce{HX} & \ce{HF} & \ce{HCl} & \ce{HBr} & \ce{HI} \\
\hline
\mathrm{p}K_\mathrm{a} & 3.2 & -7 & -9 & -10 \\
\hline
\end{array}
$$


What is meant by the "probability density" of finding an electron? Could someone please explain to me in layman's terms what probability density of finding an electron means, just as probability means chances of finding an electron. With due respect, please don't answer the formula or the distribution curves as I understand and can plot those, I just need the definition.

In quantum mechanics we often need to find the probability at some position $x$. However, $x$ is a continuous variable with an infinite number of values so it does not make sense to ask what is the probability of being at some exact position , say $x=1/4$, as this will be vanishingly small. Instead we imagine that the probability is that of being at position x to $x + \mathrm{d}x$ where $\mathrm{d}x$ has some very small value. If $f(x)\,\mathrm{d}x$ is that probability then $f(x)$ is the probability density which must be a real non-negative number. It turns out that quantum mechanics postulates that the probability density is given in terms of the wavefunction is $|\psi|^2$.

So imagine you have a friend who's an alcoholic, and your phone is broken. You know he's not home, so he's probably at his favorite bar. If you had an aerial map of the area and had to guess where he was, you'd draw a dot on his favorite bar stool [most likely place he is], a circle around the bar building [still could be here], and a dotted line encompassing all of the shrubs, taco truck, and parking lot around the bar [he MIGHT be getting a taco, beat up in the parking lot, or passed out in the bushes].
So in this little parable, your friend is an electron and you have drawn a probability density.
Extra credit: the reason we have to use a probability density is that, for complex mathematical reasons [Heisenberg uncertainty principle] you can't know your friend/electron's velocity AND position, so you're making a mathematical guess about where your friend/electron is, and then describing that mathematical guess in terms that are confusing to us plebes. But really it's just a heat map. Also, depending on what element/number of electrons/orbital you're dealing with, you can get different shapes. Don't be confused - it's just like knowing more friends each with their own different bar, and they don't like each other so they don't overlap.


Why are solutions electrically neutral? I keep reading that solutions are electrically neutral, but I've never read an explanation. I've read after the fact justifications, such as one problem which asked me to calculate the force of repulsion between two beakers with a tiny, tiny bit of excess charge (it ended up being over 10,000 pounds).

So why are solutions electrically neutral? Not interested in after the fact explanations like the above example.

It is more acurate to say that they aren't very far from being electrically neutral. If you take a charged sphere and drop it into a beaker of some electrolyte sitting on an insulating counter, the result is a charged solution, albeit slightly.

Also, in spite of your amusing ceiling analogy (it made me chuckle), the argument about electrostatic attraction and repulsion does justify why solutions (and solids) can't be too far from being electrically neutral.  The flaw in the analogy is that humans don't need ceilings to exist; but you can' t have two charged beakers without a force between them. If each has a charge of 1 C -- which is a mere $10^{-5}$ times the charge of a mole of electrons -- and they're sitting 1m apart the force between them is about a million tons -- clearly this situation won't last very long!

Because the solute is neutral when its solid so it must me neutral when it's aqueous.


The solubility of haloalkanes in water Haloalkanes aren't very soluble in water because they can't form hydrogen bonds, and the energy required to break hydrogen bonds in water etc is higher than the energy released when new bonds between the haloalkane and water are formed.

But as fluorine forms hydrogen bonds why couldn't a fluoroalkane form hydrogen bonds with water thus become more soluble in water?

I think the answer lies with the fact that alkanes are basically non-polar, even if they have halogen substituents. Water, on the other hand, is polar, so the solvent and solute are incompatible.

I felt that fluorine would be able to form hydrogen bond only when it is initially bonded to hydrogen atom so that it gets significant electron cloud, in our case alkyl group is attached to fluorine so it becomes difficult for fluorine to hogg electrons now when fluoroalkane is dissolved in $\ce{H2O}$ attractive interactions between fluoroalkane and $\ce{H2O}$ is not strong enough to break hydrogen bond which binds $\ce{H2O}$ molecules. Thus fluoroalkene is not much soluble in water as it won't be able to form bonds with $\ce{H2O}$ molecules.


How can I determine the structure of complexes with ethylenediamine as a ligand? I am a bit confused about how to solve this problem.  I can interpret coordination complexes like $\ce{[CoCl6]^4-}$ but I just came across a coordination complex that I am unsure of how to solve.

It is $\ce{[Mn(en)3]^3+}$ where en stands for ethylenediamine.  The problem first asks that I draw the structure (not sure how to do this as ethylenediamine is a large molecule).  I would also like to know whether this complex would be tetrahedral, octahedral, trigonal planar, etc.  I'm just kind of confused as to how to solve this problem when its a large molecule as the ligand rather than a molecule like $\ce{Cl6}$.

In principle, there is no difference between having monodentate ligands such as chlorido ligand or more complex ligands such as ethylenediamine.

The first step is always to draw out the structure of the ligand by itself. Then, identify how many coordinating entities it contains. Ethylenediamine is $\ce{H2N-CH2-CH2-NH2}$ and both nitrogens can form coordinate bonds.

In the next step, identify how many coordinating entities you have in total and thus what the maximum coordination number is. As mentioned, each $\ce{en}$ has two potentially coordinating nitrogens for a total of six dative bonds. Thus, a hexacoordinated octahedron is the maximum possible.

Continue by looking at the metal. What do you know about its properties in complexes? Does it prefer any type of complex? In this case, manganese(III) is absolutely fine with having an octahedral geometry, so that’s taken care of.

Once you’ve done all this, you will notice that the metal enjoys octahedral coordination and there are enough Lewis bases for octahedral coordination to occur — the complex will likely be octahedral. If the complex had been $\ce{[Ni(en)2]^2+}$, we would have arrived at ‘square planar’ by the same methodology.

Next, draw the complex. Start with the metal and the octahedral geometry and put a nitrogen at the end of each line. Then, try to connect the nitrogens with $\ce{CH2-CH2}$ fragments. There should be two possible, enantiomeric outcomes.

Basically $\ce{en}$ is a strong field ligand and is a neutral one so the $\ce{Mn}$ in $+3$ state would have $3\mathrm{d^4}$ configuration and $\ce{en}$ would pair them up in the first two orbitals so there would be an inner orbital complex formation with octahedral geometry having $\mathrm{d^2sp^3}$ hybridization.


In the addition of HBr to isoprene why does the less stable carbocation form? According to me, the answer is B as the tertiary allylic carbocation should be more stable. But the answer given is D and in the given solution, the tertiary allylic carbocation rearranges to primary allylic carbocation to yield Option D.

My question is why does rearrangement from tertiary allylic carbocation to primary allylic carbocation take place?

Source: JEE Advanced 2015 Paper 1 Q31

The carbocation will be formed as per stability rules but as it is conjugated system the cabocation resonates and as the less stable carbocation is more reactive  it is attacked by Br-


Looking for a non-toxic substance that evaporates very quickly Firstly I am not a chemist so please be kind. I am a creative technologist working on a large art installation. The challenge is to make people perceive rain without it raining. I am looking for something we can very occasionally drop from above that will feel like a rain drop but almost instantly evaporate leaving the person completely dry when they leave (they will only experience a few drops whilst in the room — the rest of the effect uses lights and sound).

My initial thought was liquid nitrogen but I assume that whilst a tiny occasional drop on the back of the hand would be fine, one in the eye for example would probably be dangerous. Plus the room would need to be heavily ventilated to prevent the risk of asphyxiation.

I'm assuming the answer is nope but if anyone has a genius idea for some way i could achieve this effect I would love to hear it. Over to you.

I'm assuming the answer is nope but if anyone has a genius idea for some way i could achieve this effect I would love to hear it.

Nope ;-)

Your concerns are justified and I wouldn't expose the audience to anything but water either.

you can

use water from all above

and let the audience get wet
and hand out rain coats

use water and only let it rain to the left and right of a marked path or
apply a more sophisticated technique, as used by the random international art studio in their rain room. Here, it rains thoughout a whole room, but the visitors are tracked by cameras and whereever they move, the water is selectively switched off.

You can look into Galden Fluid, such as HT70 or HT55. It's priced around $800 a gallon so if money is a concern, look elsewhere.


Laboratory setup for the distillation of sulfuric acid Purification of Laboratory Chemicals, 5th Edition, by Armarego, reads (p. 53):

Reagents such as water, ammonia, hydrochloric acid, nitric acid, perchloric acid, and sulfuric acid can be purified via distillation (preferably under reduced pressure and particularly with perchloric acid) using an all-glass still.

as well as (p. 479):

Sulfuric acid, and also 30% fuming $\ce{H2SO4}$, can be distilled in an all-Pyrex system, optionally from potassium persulfate. Also purified by fractional crystn of the monohydrate from the liquid. Dehydrates and attacks skin—wash immediately with $\ce{H2O}$.

I've never seen sulfuric acid fume (only spread with steam when heated), and Wikipedia states a boiling point of $\pu{337 ^\circ C}$, along the fact that it slowly decomposes to $\ce{SO3 + H2O}$ above $\pu{330 ^\circ C}$, so I'm wondering, mostly from a theoretical standpoint:

How far lower can the boiling temperature effectively get under reasonably strong vacuum?

Does the vacuum not also bring down the temperature level at which sulfuric acid degrades?

What would be an ideal laboratory (not industrial!) setup and process for the safe distillation of sulfuric acid in fair enough amounts for laboratory use (say producing in a batch the equivalent of $\pu{5L}$ at 30% concentration)?

Nota Bene: I know it's a better idea to buy sulfuric acid at the desired level of purity, but that is not the question.

Just on the purification of sulphuric acid, well the one he gets has a dark green tinge, and what is in the sulphuric acid is anticorrosive inhibitors, etc, and they are all organic in nature, and the simple heating up a beaker of sulphuric acid, and before it gets to hot, slowly add 100 ml of 3% grocery store bought food grade hydrogen peroxide, and stir whilst slowly heating, well before long the acid will be crystal clear, any fumes are the skeleton structures of the organic additives, and concentrating it back to 98% sulphuric acid is either density determined, temperature, to insure water has been removed hence both will get a much purer, crystal clear sulphuric acid. Anything organic will be consumed by the peroxide and what doesn't will turn into a gaseous product that evaporates of. The product described has served his associates and himself, for many years.


Diastereomers - cyclic compounds Why in this problem there isn't a diastereomer with the $\ce{-CH3}$ in the back (dash) and the $\ce{-CH2CH3}$ in front (wedge)?

To answer this question I would first assign the R-S configuration to the original molecule and the molecule proposed by you and show you that both are identical. At the end I will also show you a simpler way to tell that both are identical.
Original molecule:

Proposed molecule:

As you can see, both the molecules have the same R-S configuration and thus are identical.
Now to see this easily and intuitively just assume you are looking at the original molecule from inside the screen. What do you see? You will observe that the original molecule will look just as same as the proposed molecule which means that both are identical. This method works because our frame of reference doesn't change the optical properties of a molecule until and unless we change the configuration of the molecule by doing some reflections.

Actually for this question if we draw the isomer with methyl group in the back and ethyl group in the front as in:

we would end up getting the enantiomer of the given compound in the question (you could visualise it by flipping the above compound (plane of compound) by 180 degree and then rotate the compound through an angle of 180 degree in the same plane)
And if we draw the isomer like this:

we would end up getting the same initial compound!!( you visualise it in the same way as mentioned before)
So in either case the compound with methyl in front and ethyl at back would not yield a diastereoisomer.
Hope it helps!!!


Why doesn't tetrachloridotitanate(II) obey the 18-electron rule? I have worked out that the complex comes to 10 electrons, however I do not understand why it does not obey the 18-electron rule? What are the factors that account for this?

There is more to complex chemistry than the 18-electron rule. It states one condition when complexes may be particularly favourable, namely when all metal d-orbitals are fully populated and the ligands favourably interact with all metal s- and p-orbitals leading to all paired spins and hopefully a large energy difference between populated and unpopulated orbitals. The list of complexes that do not obey the 18-electron rule may well be larger than that of those that obey it. In no particular preference that includes $\ce{[MnO4]-}, \ce{[CrO4]^2-}, \ce{[Fe(CN)6]^3-}, \ce{[Mn(H2O)6]^2+}$ and many more. (A few of those are not typically described as complexes but they are nonetheless.)

Further constraints that you need to be aware of include, but are not limited to:

steric stress, i.e. too many large ligands around too small a metal centre,
charge density, i.e. surrounding a weakly charged cation with too many anions,
the actual electronic configuration, which can render octahedral, Jahn–Teller-distorted octahedral or square planar complexes more favourable.

In the case of a chloridotitanate(II), we first need to realise that we are surrounding a moderately small cation (titanium(II)) with rather large anions (chloride) which may generate steric stress. We also need to observe that titanium(II) is only weakly charged and attempting to generate an octahedral configuration would give us an anion with $-4$ charge — not exactly favourable (exceptions may apply). Also, chlorido ligands are known to be weak-field ligands, so one would not expect an overly large crystal field split.

The two greatest factors are probably sterics and electronics. Consider tetrachloridocobaltate(III) $\ce{[CoCl4]-}$: while potentially six chlorides may fit around the rather small cation, even though it is $+3$ the electronic situation favours only tetracoordination. This results in a 14-electron complex rather than an 18-electron one. In analogy, I am tempted to say that the even lesser charged titanium(II) ion cannot accommodate for a total of six chlorides, allowing only four of them in its vicinity. Taking electronics into account, square planar coordination is out of the question due to lack of stabilising interactions and the complex would be tetrahedral.

In fact, the tetrahedral geometry allows for a mildly stabilised triplet state: both $\mathrm{e_g}$ orbitals are semipopulated which is generally a tad better than two of three $\mathrm{t_{2g}}$ orbitals as it would be in the octahedral case.

[TiCl4]2- is a weird one because rather than just using one effect for why its an exception it has two. It is both a pi donor ligand, and a high spin complex. I would go into more detail but its only one mark on the tutorial, so whats the point :/


Phase Diagrams and Equilibrium In this link

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/phase.php

It says (in the latter part ) that for all combinations of Pressure and Temperature along line BC

the rate of boiling of liquid to form a gas= the rate of condensation of gas to form a liquid.

And C is the Critical Point.  Similar thing has been said  for other lines also

Now 100℃ (or boiling point of the liquid) will lie somewhere on BC

So according to the link rate of boiling of water to form gas is the same as the rate of condensation of gas to form water.

But I have read that at boiling point all the liquid changes to gas or there is no equilibrium at that point.

So why then it says- the rate of boiling of liquid to form a gas= the rate of condensation of gas to form a liquid.

I am not good at chemistry so this might not be a good question. But I was trying to understand phase Diagrams for the first time !

In the phase diagram of a pure substance there are regions of a single phase e.g. solid, liquid or vapour. The boundary lines you refer to are where the substance exits as $2$ two phases, say solid and vapour in equilibrium with one another.

The Gibbs phase rule states that the number of degrees of freedom, F,  which is the number of (intensive) variables that can be varied is given by
$$F = C-P+2$$
where C is the number of components, $1$ for a pure substance, and P the number of phases, which is $1$ anywhere in the diagram except on a boundary line. Therefore the number of degrees of freedom is $2$, (except on a boundary line) which means that pressure and temperature can both be varied independently.

The boundary lines indicates the condition of P and T where there are two phases in equilibrium, say solid and vapour. As there are $2$ phases there is now only $1$ degree of freedom so that temperature and pressure are no longer able to be varied independently. Thus on the liquid /vapour curve, if the pressure is increased some vapour is condensed (heat is given out) and the temperature increases to a new value but one that that is still on the boundary line, i.e. vapour is still in equilibrium with the liquid. Along any boundary curve the two phases are in equilibrium and so, e.g., the rate of melting is equal to rate of freezing. (Any point along the liquid/vapour boundary line can be considered to be boiling, but we define the normal boiling point to be the boiling temperature at $1$ atm pressure.)

At the triple point, where the three boundary curves meet, there are no degrees of freedom and this point is fixed. At and above the critical point the density of the liquid and vapour are the same and the liquid's meniscus  disappears. The super-critical fluid looks very much like the swirling foam that one sees for a short while on pouring a bottle of Guinness into a glass.

The boundary lines can be calculated for a substance using the Clausius and Clausius-Clapeyron equations.

An excerpt from the link you provided:

  The figure below shows an example of a phase diagram, which summarizes the effect of temperature and pressure on a substance in a closed container...

When heating a liquid in a rigid, close container, no boiling* will occur. The pressure (and therefore density) of the vapour will rise as the temperature is rised. If you carry this on for long enough, the temperature on the liquid surface will reach what is called a critical temperature, $T_{c}$. At $T_{c}$ the pressure of the system is called critical pressure, $p_{c}$. At this point the container will be filled with a supercritical fluid.

*  Boiling, in this case, refers to free vapourization throughout the liquid.


What is the difference between ΔG and ΔG "with an o on top"? What is the difference between $\Delta G$ and $\Delta G^\circ$? I couldn't figure it out.

Ex:

$\Delta G^\circ = -RT\ln K$
$\Delta G = \Delta G^\circ + RT \ln Q$

What is the difference, is it that $\Delta G^\circ$ is the value predicted by the standard free energies and that $\Delta G$ is just the actual value?

A simple explanation would be that the "o on top" denotes standard state and the one without the "o on top" denotes conditions that are not standard state. However, that may not allow you to "get it" so let's look at the equations.

$\Delta G^{\circ} = -RT\ln K$
$\Delta G         = \Delta G^{\circ}  + RT \ln Q$

The first one allows us to find the Gibbs free energy at the standard state, as was mentioned. However, it would be useful to review what exactly the standard state is:

For the standard state of formation:

  The partial pressure of any gas involved in the reaction is 0.1 MPa.
  The concentrations of all aqueous solutions are 1 M.
  Measurements are generally taken at a temperature of 25 °C (298 K).
  The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference between the free energy of a substance and the free energies of its constituent elements at standard-state conditions.
  (source)

Or, very simply:

Partial Pressure of gases = 0.1 MPa
Concentration of Aqueous solutions are = 1 M (Molar)
Temperature usually 298 K

So if it's so nice and well, why would we need the one without "o on top"? Well, if we meet a problem or situation without the concentrations or pressures as stated above, we need to use our handy equation to relate the standard state which may have been given to the actual situation, with perhaps 0.1 M or some other concentration. This is powerful because it allows you to view different states,perhaps with more or less reactants or products, and realize how one can affect equilibrium.

An add : the Gibbs energy is at its standard state only when the extent of reaction ( greek letter Xi ) is equal to zero. That counts also for the standard state of enthalpy.


How do you calculate the concentrations of multiple reactants in a certain point in time? For $\ce{A + B -> C + D}$, given that $v = k [\ce{A}] [\ce{B}]$. I know the derivation for $\ce{A->B + C}$ , being $\ln[\ce{A}]=\ln[\ce{A}]_0−νkt$, but I'd like to know how you can, given a point in time, calculate the concentrations of A and B in the first equation, given the initial concentrations.

$$\frac{dA}{dt}=-kAB$$
$$\frac{dB}{dt}=-kAB$$subject to initial conditions $A=A_0$ and $B=B_0$ at time t = 0.  Let x = number of moles of A that react = number of moles of B that react.  Then,$$\frac{d(A_0-x)}{dt}=-k(A_0-x)(B_0-x)$$and$$\frac{d(B_0-x)}{dt}=-k(A_0-x)(B_0-x)$$Therefore, from both equations  $$\frac{dx}{dt}=k(A_0-x)(B_0-x)$$ subject to the initial condition x = 0 at t = 0.  This differential equation can readily be integrated using separation of variables and partial fractions.

Given dX/dt = k[A][B] and A0 and B0 where you failed to define what X is, along with no value for k? It can not be done. see https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions (and probably hundreds of other sites) for the integration over time. In short, a parameter x is inserted so that [A]= [A]0-x & [B]= [B]0-x, etc. and we integrate dx/dt which requires two cases: [A]0 = [B]0 and [A]0 ≠ [B]0 but I won't repeat the rest here. If they're equal [A] = [A]0 ÷ ([A]0kt + 1) and if they're not equal then generally we consider the ratio of [A]:[B] but determining either [A] or [B] is simple ; i.e. don't replace [A]0 - x, etc. with [A], etc. at equation 1.22 in the above link and then solve for x. Once you have x, you know [A]0 - x, etc.


Why does bond angle decrease in the order H2O, H2S, H2Se? I know that bond angle decreases in the order $\ce{H2O}$, $\ce{H2S}$ and $\ce{H2Se}$. I wish to know the reason for this. I think this is because of the lone pair repulsion but how?

Here are the $\ce{H-X-H}$ bond angles and the $\ce{H-X}$ bond lengths:
\begin{array}{lcc}
\text{molecule} & \text{bond angle}/^\circ & \text{bond length}/\pu{pm}\\
\hline
\ce{H2O} & 104.5 &  96 \\
\ce{H2S} &  92.3 & 134 \\
\ce{H2Se}&  91.0 & 146 \\
\hline
\end{array}

The traditional textbook explanation would argue that the orbitals in the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions, thereby forcing the $\ce{H-X-H}$ angle to contract slightly. So instead of the $\ce{H-O-H}$ angle being the perfect tetrahedral angle ($109.5^\circ$) it is slightly reduced to $104.5^\circ$.  On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ have no orbital hybridization.  That is, The $\ce{S-H}$ and $\ce{Se-H}$ bonds use pure $\ce{p}$-orbitals from sulfur and selenium respectively.  Two $\ce{p}$-orbitals are used, one for each of the two $\ce{X-H}$ bonds; this leaves another $\ce{p}$-orbital and an $\ce{s}$-orbital to hold the two lone pairs of electrons.  If the $\ce{S-H}$ and $\ce{Se-H}$ bonds used pure $\ce{p}$-orbitals we would expect an $\ce{H-X-H}$ interorbital angle of $90^\circ$.  We see from the above table that we are very close to the measured values.  We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two $\ce{X-H}$ bonds the angle opens up a bit wider.  This explanation would be consistent with the $\ce{H-S-H}$ angle being slightly larger than the corresponding $\ce{H-Se-H}$ angle. Since the $\ce{H-Se}$ bond is longer then the $\ce{H-S}$ bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum.  The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the orbitals in the two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.

We know that as the electronegativity of central atom increases, the bond angles also increase. The relevant electronegative order is $$\ce{O > S > Se}\,,$$ hence the bond angle order of $$\ce{H2O>H2S>H2Se}\,.$$


Shape and polarity of the iodate ion I am trying to figure out the shape of the iodate ion or $\ce{IO3^-}$. No matter how many times I draw it, I keep getting iodine in the centre, and 3 oxygen atoms surrounding it, 2 with single bonds, and 1 with a double bond. This would make it a trigonal planar shape. It would mean that it is not a "bent" shape.

I went to double check on the internet, and I am now confused. On Wikipedia it says it is in fact polar and a pyramidal shape.

Why does iodine have a pair of lone electrons in its shell? Why would it not combine with one of the oxygen atoms to make a double bond?

Looking at the formal charges on the atoms, each oxygen has a charge of -2. With the added electron, the iodine has a formal charge of +5. This leaves 2 electrons in the valance shell for the lone pair. So there are 4 items (3 oxygens and a lone pair) to place around the iodine atom, giving it a tetrahedral shape. Since each oxygen has the same charge of -2, there must be 2 bonds to each oxygen. The extra electron is placed on the molecule, not the iodine atom. Since you have a tetrahedral shape with a different corner (lone pair), it has a dipole (like ammonia).

In iodate ions there are three double bonds between iodine and oxygen and there is presence of one lone pair.iodine is central atom.the shape of molecule can be seen as pyramidal with corners of base triange be oxygen atoms and iodine at top.And there is no meaning for polarity for ions as they already have a charge and hence polar.


What happens chemically to 5'-radiolabeled aqueous thyroxine (thyroid prohormone) when its I-131 undergoes beta+gamma decay? Related to this question about $\ce{^{14}C}$ beta decay in $\ce{CO2}$, what happens when the $\ce{^{131}I}$ in radiolabeled thyroxine (thyroid prohormone, or "T$_4$") undergoes beta-plus-gamma decay to form $\ce{^{131}Xe}$?

In order to narrow the scope of the question to something tractable, assume that despite its low aqueous solubility the thyroxine is in dilute solution in deionized water, and that it bears a single $\ce{^{131}I}$ at the $5'$ position (click image to enlarge):

As well, as noted above, assume that the $\ce{^{131}I}$ undergoes its most common mechanism of tandem electron & gamma decay:

$$
\ce{^{131}_{53}I -> ^{131}_{54}Xe + \beta^-~+ \gamma}
$$

I assume the resulting chemical species remaining will start out as a (very transient) xenon-bearing cation:

What happens next?

When the beta decay occurs it is likely that the C-I (or now C-Xe) bond will break. . The electron emitted in the beta decay will fly away from the molecule, the positively charged xenon atom will now pick up an electron from the aqeuous solution.

The Szilárd-Chalmers effect needs to be considered, it is typcial that when a radioactive decay occurs that the bonds to the atom at which they decay is will be broken. The recoil energy from the beta emission should be able to break the C-I bond.

I suspect that the bond will break to form a carbon centred aryl radical. This is a very high energy organic radical which will be likely to react with dissolved oxygen to form a Ar-O-O. radical. This in turn will abstract a hydrogen from another organic molecule to form Ar-O-O-H.

When this decomposes the O-O bond will break thus forming ArO. and HO. The ArO. will abstract a hydrogen from another organic solute to form a phenol. I know that 8-hydroxy-20-deoxyguanosine (8-OHdG) in urine is associated with exposure to radiation in humans. See S. Kato et. al. JOURNAL OF PEDIATRICS, 2015, volume 167, pages 1369 onwards. Also see the work of Martin Tondel et. al., ARCHIVES OF ENVIRONMENTAL CONTAMINATION AND TOXICOLOGY, 2005, volume 48, pages 515-519.

This work on this DNA marker for radiation damage does to my mind support the hypothesis that the autoradiolysis of the thyroxine will create hydroxy compounds. When S. Cefova et. al. irradated emulsions of nitrobenzene and water they did observe the formation of nitrophenols which lends further weight to the idea that the product will be a phenolic compound.

My guess would be the following:
The electron generated by the decay would most likely go either towards the O or the H. In either case, the oxygen (due to its electronegativity) will atract the electron forming an ion, and the hidrogen will be displaced.
The oxygen will form a double bond and stabilize. The hydrogen will attack the xenon and take it's place, as the molecule will be more stable that way. The xeonon will just float off.

That's what seems logical to me that will hapen in an aqueous solution.
What will happen in the human body might be different, as different molecules will most likely be in the vicinity of the thyroxine molecule.

Below is a crude drawing in paint in order to help you visualize.


Cyanogenic glycosides and HCN I'm researching cyanide for a biology project and noted that one of the largest sources of carbohydrates in the tropics is the cassava root -- which I previously researched as a food that contains a cyanogenic glycoside that is metabolized into hydrogen cyanide.

Most of the sources I reviewed claimed that, while boiling the cassava, the cyanogenic glycosides were broken down into components such hydrogen cyanide, which made the plant safer to consume. I have a limited understanding of chemistry and can't resolve why this is considered a safe way to cook the plant.

My research indicates that inhalation of hydrogen cyanide is significantly more toxic than the oral route, so why would this method be considered beneficial? Not sure why this is considered safer, unless the information I've found is not correct. Any help in putting this together would be appreciated. Thanks!

OK, I was too lazy to look up LD50 and LC50 of $\ce{HCN}$ for you ;)
But if you have a chance, have a look at

G. Padmaja & Keith H. Steinkraus,
Cyanide detoxification in cassava for food and feed uses,
Critical Reviews in Food Science and Nutrition, 1995, 35(4), 299-339 [DOI]

Seemingly, the processing of cassava (manioc) involves several steps, such as drying in the sun, grating, watering and cooking, each of which either facilitates the enzymatic or thermal cleavage of the linamarin (i.e. the toxic compound) to glucose or supports the release of $\ce{HCN}$.
In total, the processing minimizes the risk of exposition through inhalation. None of the steps is perfomed in confined space, watering helps to wash out the $\ce{HCN}$ which in water dissociates a bit ($\ce{HCN}$ is a weak acid).
Thus, digestion of the untreated roots is likely to be fatal, while the slow processing is apparently not.

Ok, I digged a bit and maybe you have you have access to this one:

Bala Nambisan
Strategies for elimination of cyanogens from cassava for reducing toxicity and improving food safety Food and Chemical Toxicology, 2011, 49, 690–693 [DOI]

And we can use google too @Greg !!

"The cyanide ion (CN-) is the toxic moiety in hydrogen cyanide. The toxicity of simple cyanide salts, such as potassium and sodium cyanide is, therefore, similar to that of hydrogen cyanide." and from link "Toxicities: Cyanide is the least toxic of the "lethal" chemical agents. The LCt50s of AC and CK by inhalation have been estimated to be 2500-5000 mg"min/m3 for AC and about 11,000 mg"min/m3 for CK. LD50s for hydrogen cyanide have been estimated to be 1.1 mg/kg for intravenous administration and 100 mg/kg after skin exposure. The oral LD50s for sodium and potassium cyanide are about 100 and 200 mg/kg respectively." AC - the term refers to the anion CN-, or to its acidic form, hydrocyanic acid (HCN).
Have a look at link absorption, metabolism and toxicity of cyanides


What is the product of a copper and vinegar reaction? I put a copper wire through a hole in the cap of a water bottle, filled the water bottle ~ half way full with vinegar, and left it over night.  The part of the wire inside the bottle that wasn't in vinegar turn black.  Really black.  Why is this?!

UPDATE: I wiped the black stuff off on paper, and it now looks green on the white paper.  Seems like it oxidized over night.  Why so fast?

If it looks green/blue on paper, you likely have copper (II) acetate. You likely have the anhydrous version $\ce{Cu(OAc)2}$ which is darker than the hydrous $\ce{Cu(OAc)4(H2O)2}$

Wikipedia Images include:

The rate of formation is a property inherent to the reactants and the products, but it also depends on temperature.

Just read Wikipedia article on Verdigris again, and it is clear per the preparation path described by Nuygen above, that the product is classic (except for the use of a copper wire in place of a copper plate, with a varying height separation from the 5% vinegar), and accurately detailed by Wikipedia at https://en.wikipedia.org/wiki/Verdigris . To quote, in parts:

"Verdigris is the common name for a green pigment obtained through the application of acetic acid to copper plates[2] or the natural patina formed when copper, brass or bronze is weathered and exposed to air or seawater over time. It is usually a basic copper carbonate (Cu2CO3(OH)2), but near the sea....If acetic acid is present at the time of weathering, it may consist of copper(II) acetate. "

And:

"Verdigris is a variable chemical mixture of compounds, complexes and water.... All the components are in an ever-changing and complex electrochemical reaction equilibrium that is dependent on the ambient environment."

All of which confirms my answer above and challenges the cited 'correct' answer that the product is simply copper(ll) acetate and not too likely, in my opinion, as was claimed on the three-sentence 'correct' answer, to be just dry (anhydrous) cupric acetate.

Actually, I agree with Nuygen on several of claims likely based directly on his observations (namely, a complex mix with an apparent presence of CuO, a shift from light (basic copper acetate per Wiki) to more colored salt, possibly cupric acetate (as also mentioned by Wikipedia).

Here is a video with vertically hung copper sheets at https://www.youtube.com/watch?v=jq5MB1H_hVc . Note the product color and compare to Nuygen's description.

As a side note, vinegar is not pure dilute acetic acid. Depending on the source materials and manner of processing, impurities may be present.


Why does water volume decrease when salt is added? Why does water volume decrease when salt is added? Our teacher asks us in the class but I don't find any strong reason.

What you are experiencing is the phenomenon of partial molal properties - components of a mixture contribute unequally to the properties of the mixture.

Sometimes the components of a mixture do not contribute linearly to the properties of a mixture. A well-known example is the water-ethanol system, which has negative excess volume. When 5 mL of water and 95 mL of ethanol are mixed, the final volume is less than 100 mL. See the following graph constructed by Wikipedia user Wilfried Cordes:

Similar behavior is occurring in your sodium chloride and water system. Now, as for why these phenomena occur, a lot of it has to do with intermolecular forces. For example, you could consider that the hydrogen-bonding and dipole-dipole interactions between water and ethanol are perhaps stronger than between water and water or ethanol and ethanol. Additionally, water molecules are smaller than ethanol molecules. There is also a thermodynamic consideration. This state of affairs must represent minimum free energy (maximum entropy).

How can you relate the interactions of sodium and chloride ions with water to the observation of negative excess volume?

(Volume decreases so) shorter bonds between
ion and water molecules
Water molecules more tightly arranged/pack
better/occupy less space
Water molecules more ordered/ clustered
(around the ions).


Why is the boiling point of a solution higher than that of a pure solvent? Why is the boiling point of a solution higher than that of a pure solvent?
I think when a non-volatile solid solute is added to a solvent, some of the liquid solvent molecules are replaced by solute molecules at the surface. The lesser the number of solvent molecules at the surface that fewer molecules evaporate, and thus the vapor pressure of the solution will be lower than the pure solvent.

Is this correct?

The change in vapour pressure with added solute is called Raoults law. The vapour pressure of a species i above a solution is proportional to mole fraction of this species (solute) $x_i$ and its pure vapour pressure $p_i^*$ or $p_i=x_i.p_i^*$.

As the solution now contains a fraction of a low vapour pressure solute then this component will not contribute much to the overall vapour pressure which will be less (in this case) than pure solvent.

It is not correct to say that solvent molecules are replaced at the surface as both solute and solvent are distributed uniformly in the solution. Thermal diffusion moves the solute and solvent around in a random manner. It is that there are now fewer solvent molecules as some have been replaced by low vapour pressure solute (in this example) and because they are of low volatility they need more energy to evaporate than the solvent does. As the temperature is constant, by the Boltzmann (energy) distribution fewer solute molecules populate higher energy levels compared to solvent and so have a lower vapour pressure.

YES the explanation is correct.
In fact it is a qualitative property of solutions.
Elevation of boiling point /lowering of vapor pressure/lowering of freezing point all depend on same reason.


Why does SiH4 have unusually large bond angles? The bond angles of $n > 2$ hydrides are usually around $90^\circ$, such as $\ce{PH3, H2S},$ $\ce{H2Se},$ $\ce{H2Te}.$ I understand that this is from increased $\mathrm{p}$ character in the bonding with these compounds.

However, for $\ce{SiH4}$ and $\ce{GeH4}$ the bond angles are actually quite close to $109.5^\circ$. Why do $\ce{Si}$ and $\ce{Ge}$ not typically use unhybridized $\mathrm{p}$ orbitals, and instead use closer to $\mathrm{sp^3}$ hybrid orbitals for bonding?

In hydrides that contain two or three hydrogens connected to a central atom such as $\ce{PH3}$ or $\ce{H2Se}$, the number of bonding partners does not exceed the number of available p orbitals on the central atom. In an extreme application of Bent’s rule, the most ideal bonding situation would be to have pure p-type orbitals form the required bonds and at least one lone pair reside entirely in an s-type orbital. This allows stabilisation of the s lone pair (as an s orbital’s energy is lower than a p orbital’s one) and better bonding (as p orbitals are directional and thus offer greater overlap with hydrogen; furthermore, they are higher in energy which better matches hydrogen’s orbital energy and again allows for greater bonding energy).

This ideal situation already falls apart in the second period where the bond angles of ammonia and water are unusually large; this is because the central atoms are too small to accomodate two or three bonding partners at $90^\circ$ angles so the bond angles are increased in a delicate balance until the lowest overall energy is obtained. The results are hybrid orbitals with a nonzero s component.

In $\ce{AH4}$ molecules, the situation is different and worse yet again. There are not enough p orbitals to accomodate four bonds so the s orbital has to help. Symmetrically, the optimal arrangement is a tetrahedron and this just happens to almost perfectly match an $\mathrm{sp^3}$ arrangement. As silane (and methane, germane, etc.) has four identical substituents, there is no energy to be gained by reducing symmetry so the ultimate geometry is that of an ideal tetrahedron whose bond angles are $109.5^\circ$.

The geometry depends on the nature of the doublets around the central atom. If the doublets are all bonding, like in $\ce{CH4},$ $\ce{SiH4}$ and $\ce{GeH4},$ the geometry of the molecule is tetrahedral, and the angles between the bonds are 109.5°. If one or two of these doublets are non-bonding, they are more cumbersome than covalent bonds. They repel the bonding electrons, and the angles between these bonding electrons are smaller. That is the case for $\ce{NH3}$ and $\ce{PH3}$, which have one pair of non-bonding electrons. It is even worse for $\ce{H2O},$ $\ce{H2S},$ $\ce{H2Se}$ and $\ce{H2Te},$ which have two pairs of non-bonding electrons.


How does hyperconjugation explain the stability of alkenes? The more the number of alpha hydrogens, the higher is the stability of alkenes, because the more the number of hyperconjugative structures. Why is this so? Has it got something to do with the electrons being easily delocalised?

As always in chemistry one phenomenon (experimental observation) can be explained with different models, that live in different "physical model worlds", and more often than not these models are not consistent among themselves and might be even bare of a physical reality, especially if they are analyzed in rigorous way, nevertheless might be very useful tools, that take the justification of existence out of the fact that they work in many cases.

That said,  especially the concept of hyperconjugation itself can be employed at different levels. A very primitive model, but one that works astoundingly well, is to assign a positive inductive (electron releasing effect) to alkyl chains that is increasing with the degree of substitution of the ipso atom. That leads to the idea that this (however small) charge that is pushed to the sp$^2$ carbon atom stabilizes its slight positive polarization (since the majority of electrons in the double bond is above and below the plane of the atoms engaged attached to the doubly bound atoms. The physical reality of this model if often drawn into doubt, which is good to know, but still its simple and powerful and thus there is no reason to abandon it.

At a higher level of model, using a MO picture the idea is that the (empty, anti-bonding) $\sigma^*$ orbitals that come inevitably with the C-C and C-H $\sigma$ bonds can act as kind of "acceptors" for the $\pi$-electrons of the double bonds if in the right orientation. It is important to note that in this idea to actual overlap that mediates the interaction is thought to be established with the part of $\sigma^*$ orbital that is opposing the bound atom. That donor-acceptor interaction ultimately leads to a stabilization of the system. Again the physical reality is also here at least disputed, but however less than in the former case. But still both have in common that the Substituent acts as the donor and that this causes stabilization.

So yes in this higher level of description of the hyperconjugation the reason for the stability is the increasing number of possibilities to form stabilizing overlap.


How to bleach polyester I have a challenge. I recently bought a cotton/polyester coat with the intention of bleaching it to near-white. To my surprise, normal bleach with sodium hypochlorite had very little effect. It changed the color from light brown to mustard-yellow-ish, even after two soakings in a 30% mixture using water and 5% sodium hypochlorite for about 10 hours each.

My second attempt was better. I put a test piece of fabric in a solution and boiled it in the microwave. It made it even lighter, but still far away from my desired near-white (white-yellow or offwhite is what I hope to achieve).

I also tried to soak it in white spirit, with a small added extra bleaching effect.

Now, using common household chemicals, how can I achieve the desired bleaching? Can I increase the effect of the chlorine? Can I use peroxide, ammonia, or any other chemical to strip the fabric of its color? I'm willing to risk to have the fabric destroyed in the process.

The yellowish mustard color I would have to say was caused by your efforts to de-color this polyester, polyester is a decently strong polymer and thus is resistant to many forms of chemical tamper, and in all effects the bleach is damaging the fabric and leaving behind that yellow color, I would suggest using Sulfur Dioxide if you can get your hands on it as that can bleach polyester, you should be left with a white translucent fabric if the bleach hasn't already damaged the fabric too terribly.

Sulfur Dioxide part is iffy, you may choose to disregard that.

NEVER ever mix chlorine bleach with ammonia. You will get chloramine gas that is deadly if breathed in concentrated amounts, as it eats holes in your lungs that would leave you with permanent lung damage at best or give a very painful death at worst. Don't do it, whatsoever. This coat is not worth your life or wellbeing. https://www.fs.usda.gov/Internet/FSE_DOCUMENTS/stelprd3793292.pdf
That being said, probably you won't be able to get a total white color from bleaching. Polyester is a synthetic fiber, and is resistant to dying and bleaching. One of my chemistry professors tried tie-dying her polyester lab coat, and it came out in pastel colors instead of the brights she wanted. And just a couple days ago, I tried using chlorine bleach with water to create an acid-wash affect on a navy blue dress that was a cotton/polyester blend. I soaked it in the solution for 6 hours, and it came out a sage green. Probably your best bet is just buying a white coat and saving the hassle.


Redox couple notation How is the redox couple as it is written for the oxidation of $\ce{MnO4-}$, as $\ce{MnO4-,H+/Mn2+,H2O}$ correct?

I agree that $\ce{MnO4}$ is being oxidized as it loses electrons (going from -1 charge to +2) but I don't see why $\ce{H+}$ is also being oxidized. Could someone explain?

The H atom is neither oxidized nor reduced in the redox semi-equation. It remains at +1. Only Manganese reacts, as it is reduced from +7 to +2. The redox couple should be written :  MnO4-/Mn2+, without mentioning Hydrogen

And MnO4 does not lose electrons, as you say. On the contrary, MnO4 fixes or consumes 5 electrons in its reduction to Mn2+.


Reaction rate coefficient constant multiplier for equation with multiple atoms consumed? For the elementary reaction $\ce{2O -> O2}$, when calculating the rate of concentration change of atomic oxygen as $\frac{d[O]}{dt}$, would I use the standard rate coefficient $k_f$ or would I have to multiply by 2 since 2 oxygen molecules are being consumed?

$\ce{O + O -> O2}$ with $\frac{d[O]}{dt} = -2 \cdot k_f \cdot [O] \cdot [O]$

or

$\ce{O + O -> O2}$ with $\frac{d[O]}{dt} = -k_f \cdot [O] \cdot [O]$

Or would the rate constant define the dependency on atomic oxygen?

In general if the reaction is $\ce{aA + bB \to cC +dD}$ the the rate is written as $$r=\displaystyle -\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$$.

If the reaction is indeed second order in oxygen atoms as stated by the OP, then the first equation given by the OP is correct, but unusual (you could just leave the factor two in the denominator on the left hand side instead of multiplying by two on the right hand side).

When we measure  the rate of the reactant concentration change (rate of disappearance of the reactant) or the rate of formation of the product experimentally, we neglect the coefficient and equation, but we take the balance equation into consideration when calculating the rate of disappearance or formation of each species in the reaction,
so: rate of disappearance of oxygen atoms = double the  rate of formation of  oxygen molecules.

$\ce{\frac{d[O]}{dt} = - k \cdot [O]^2}=2 \cdot{\ce{\frac{d[O2]}{dt}}}$


Is cycloheptatriene aromatic or not? Is cyclohepta-1,3,5-triene aromatic?

I think it should not be, because even though the π-system possesses six π-electrons, the conjugation is not complete around the ring.

You are correct. The conjugation does not extend all the way around the ring, so the molecule is not aromatic. In addition, the single methylene group ($\ce{CH2}$) sticks out of the plane of the rest of the molecule.

(On the other hand, removing a hydride ion $\ce{H-}$ from the $\ce{CH2}$ group leaves the tropylium ion $\ce{C7H7+}$, which is planar and aromatic, with resonance structures putting the positive charge on any of the seven carbons.)

To define an aromatic compound, there are four basic rules:

P(lanarity)
D(elocalization)
H(ückel's rule): $4n + 2$ π-electrons
C(onjugation)

In the above compound, if any one of the carbon is $\mathrm{sp^3}$ hybridised, then it is not planar. In the above compound, there is an $\mathrm{sp^3}$ carbon, so it is not aromatic.