Arrange in order increasing reactivity towards nucleophilic substitution Arrange the following compounds in order of increasing reactivity towards SN2 reaction

According to me, the order should be P>Q>S>R as more hindrance should result in lesser reactivity but answer given in the book is P>R>Q>S. Why is the reactivity of R greater than that of Q and S even though the leaving group is on a $2^\circ$ carbon.

The reactivity of the SN2 reaction is determined by the stability of its transition state. More stable the transition state (hereby referred to as TS), lower will the activation energy be for the reaction, and therefore, the reactivity increases.
Let's first have a look at the transition state of an SN2 reaction.$^{[1]}$

What you see inside the square brackets is the transition state, Nu represents the incoming nucleophile, while X represents the outgoing leaving group (both bearing a partial negative charge).
In this transition state, the carbon atom in the middle has a p orbital that shares one pair of electrons between the old and the new bonds. This suggests that the transition state for an SN2 reaction has a more or less planar carbon atom at the centre with the nucleophile and the leaving group arranged at $180$° to each other.
The point to be noted here is that the p-type orbital of carbon in the TS stays directly perpendicular to the plane formed by its three substituents, which is the same as the plane of the ring (in this case, all the three hydrogens shown in red). This means that sterically bulky groups which are present in the plane of the ring, and directly protrude towards the site of reaction (which is still perpendicular to this plane) and therefore will have a much bigger effect in inhibiting the entry of the incoming nucleophile as compared to those outside it.
This will be a general idea for the stability of TS (and hence the reactivity towards the SN2 reaction) in other cases as well.
Now, although all the options have adjacent benzyl groups, it might be worthwhile to see how benzyl and allyl groups stabilize the TS. These compounds react rapidly by the SN2 mechanism because the π system of the adjacent double bond can stabilize the transition state by conjugation.
The p orbital at the reaction center (shown below in brown, and corresponding to the brown orbital in the previous diagram) has to make two partial bonds with only two electrons — it is electron deficient, and so any additional electron density it can gather from an adjacent π system will stabilize the transition state and increase the rate of the reaction. This is shown here for allyl bromide:$^{[2]}$

The benzyl group acts in much the same way using the π system of the benzene ring for conjugation with the p orbital in the transition state. This stabilization will be present in all the four options,and I am making no attempt to compare them here
So, the geometry of the transition state and steric consideration will put S at the end due to the highly directed bulkiness of the di-ortho substituted benzene group, while P will jump ahead due to the absence of such clashes. S will still have benzylic interactions, but the incoming nucleophile will be hindered by the methyl group which points towards the perpendicular p orbital
Out of R and Q, R might achieve favourability over Q because again, a steric clash in the plane of substituents by a benzene ring having a methyl group directed towards the site of reactivity (the p-orbital perpendicular to plane of the ring) will generate much more hindrance than a single methyl group smaller than o-methyl benzene, which is directed away from the site.
Finally, another small reason that may tip the scales in the favour of R could be an interaction between the σ $\ce{C-H}$ bonds and the p orbital of carbon, in a kind of hyper-conjugative fashion similar to what happens in a carbocation (although this interaction would most probably be weak) as follows:

This weak σ $\ce{C-H -> p+}$ might lead to a lowering of the energy of the p-orbital, and thereby slightly increase the feasibility of attack by lowering the activation energy.
The visualization for this interaction may be seen in a similar fashion as hyperconjugation for a carbocation:$^{[3]}$

Such an interaction is not quite feasible in the TS of Q, where any interaction with the ortho-substituted methyl group will require complete delocalization over the benzene ring, which is not very probable as the TS is a short-lived species, and is not a full-fledged carbocation (especially since the leaving group is chloride, which is a good leaving group and doesn't lead to excessive charge development by a very early or late departure)
Hence, you get the order P>R>Q>S.
References:
$[1]$ Organic Chemistry Second edition, Jonathan Clayden, Nick Greeves and Stuart Warren,
ISBN $9780199270293$ Nucleophilic substitution at the carbonyl group, [p.$340$]
$[2]$ Organic Chemistry Second edition, Jonathan Clayden, Nick Greeves and Stuart Warren, Nucleophilic substitution at the carbonyl group, [p.$341$]
$[3]$ Molecular orbitals and organic chemical reactions, Fleming


Free chemical reactions database I'm creating a program, which predicts reactions of compound. So, I need a database of chemical reactions. And I need both inorganic and organic reactions. But I can't use Scifinder or Reaxys because they are not free. Is there any free analogs?

In the current form the question is addressed, and assuming you took a look at a listing of software already in the field like here, I speculate the sheer number of organic reactions (just some) alone renders such a task as a one-man / one-woman project too large.  There is a lot of work associated to create such a database, or software. Organic Synthesis is one open-access reference for organic reactions, yet the normal output is a html / pdf of the reaction procedure, and not a *.sdf (or other chemically relevant *.xml-like) format.  Inorganic Synthesis is a publication with similar intent as the former, yet not openly accessible.

Nevertheless, if you are interested in this topic, I suggest to consult the recent publications Computer-aided synthesis design: 40 years on (DOI 10.1002/wcms.61) as well as the just about a year old Computer-Assisted Synthetic Planning: The End of the Beginning in Angewandte Chemie (English version DOI: 10.1002/anie.201506101).


Finding time t in a first order reaction (JEE Adv 2019) The decomposition reaction $\ce{2N2O5(g) ->[\Delta] 2N2O4(g) + O2(g)}$ is started in a closed cylinder under isothermal conditions at an initial pressure of $\pu{1 atm}$. After $\pu{Y \times 10^{3} s}$, the pressure inside the cylinder is found to be $\pu{1.45 atm}$. If the rate constant of the reaction is $\pu{5 \times 10^{-4} s-1}$, assuming ideal gas behavior, the value of Y is?

I used the equation given here for a first order reaction, $[A]=[A]_oe^{-kt}$
Using $k = \pu{5 \times 10^{-4} s-1}$, I am getting 4.606 but the answer given is 2.3.
Why is this wrong?
The question is from JEE Advanced 2019 paper 2.

The basic law of chemical kinetics states for $\ce{aR -> bP}$, that
$$A = kR = -\frac{1}{a}\frac{\mathrm dR}{\mathrm dt} = \frac{1}{b}\frac{\mathrm dp}{\mathrm dt}$$
(where $A$ is activity and $k$ is called rate constant1)
Note that, in our case, the stoichiometric coefficient of reactant (i.e., $\ce{N2O5}$) is 2, therefore the differential rate equation becomes,
$$-\frac{1}{2}\frac{\mathrm dR}{\mathrm dt} = kR$$
or,
$$-\frac{\mathrm dR}{\mathrm dt} = (2k)R$$
On integrating and re-arranging, we get,
$$R = R_0 \mathrm e^{(2k)t}$$
which means, $$t = \frac{2.303}{2k}\log_{10}{\frac{R_0}{R}}$$
This factor of $\frac{1}{2}$ was what you missed, I think.

1: As per my first equation, it's clear how rate constant is specific to a certain reaction. The following is an extract from Concentration and Rates (Differential Rate Laws) (under the heading "Reaction Orders"):

The proportionality constant (k) is called the rate constant, and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular rate constant value under a given set of conditions, such as temperature, pressure, and solvent.


Finding value of equilibrium constant I tried to find the value of equilibrium constant several times.
I failed..

the value of equilibrium constant at 500K
has enthalpy 250kJ/mol and entropy 48J/mol*K

the answer is $ 2.45*10^{-24} $

which formula do I have to use?


Why is the difference between the melting point and boiling point large for some compounds, and small for others? I was looking at a list of melting and boiling points of various compounds and I realised some had very large differences while some have very less differences between their melting and boiling points.

Example:

for cyanogen, boiling point is -21.2°C while melting point: -27.9°C
for propane, boiling point: -42 °C while the melting point:  -144°C

Why does this happen?

well thinking that the same factors influence the melting and boiling point of materials is possibly wrong.
high melting point is due to strong packing(highly symmetric molecules) while  high boiling point is due to large surface area
so if a compound has good packing and less surface area will end up to have very less difference in melting and boiling point and sublime


How much UV lamp power is needed to initiate radical polymerization? I'm looking to purchase a UV lamp for initating some radical polymerizations. Is a hand lamp like the type for examining TLC plates (5-10W, like this) enough power to do this, or do I have to buy something a lot more expensive?

In my case, I'm looking to polymerize a fairly small sample (1mm thick, about 100 mm$^2$ of area, encased in 1mm-thick lab slides) and the chemistry will probably be MMA (initiated with AIBN). I'd like to know for my particular case, but it would be great to know if there's some rules of thumb involved (does it depend on the chemistry? does it depend on the amount? Will different power levels give different final results?)

Sounds interesting! These hand lamps typically have two tubes with emission maxima around $\lambda$ = 254nm and 360 nm, respectively. The latter should be just right to excite AIBN and isn't cut off by Pyrex or comparable glas.

More power is obtained from a high pressure mercury lamp, such as the classical HPK 125W, but these lamps are normally used in cylindrical irradiation vessels with immersion cooling.

If you want a complete system with top- or side irradiation, replacable tubes, metal housing and ventilation have a look at the portfolio of companies like LuzChem.


How does a Frost diagram reproduce the solutions to the wave equation? I came across a Frost diagram for cyclic compounds in my book, and all my book had to offer was that it geometrically reproduces the solutions of the wave equation, and can therefore determine the relative energies of each pi molecular orbital.

How, exactly, does a Frost diagram work? My book was of no use.

Algebraically, the levels of cyclic polyenes may be derived using simple
Hückel theory (see also: Pi molecular orbitals of polyenes). The general result for the energy of the $j$-th level for a cyclic system containing $N$ atoms is:

$$e_{j} = \alpha + 2 \beta \cos\left(\frac{2j\pi}{N}\right)$$

where $\alpha$ is the energy of each carbon $\mathrm p_{\pi}$ orbital before interaction (Coulomb integral), $\beta$ is the interaction energy between two adjacent $\mathrm p_{\pi}$ orbitals (the resonance integral) and $j= 0, \pm 1, \pm 2,\ldots, \pm \frac{N - 1}{2}, +\frac{N}{2}$ for even $N$, and $j= 0, \pm 1, \pm 2, \ldots, \pm \frac{N - 1}{2}$ for odd $N$. The very simple form of this equation leads to a useful mnemonic for remembering the energy levels of these molecules. Draw a circle of radius $2\beta$ and inscribe an $N$-vertex polygon such that one vertex lies at the bottom position. The points at which the two figures touch define the Hückel energy levels. And that is what is called a Frost diagram.


Relation between enthalpy and Cv MY ATTEMPT

We already know that $\Delta U=nC_V\,\Delta T$
which is for constant volume

Now we know that $\Delta H = \Delta U + p\,\Delta V$

so $\Delta H = nC_V\,\Delta T + 0$ (as at constant volume $\Delta V = 0$)

$$\Delta H=nC_V\,\Delta T$$

But I have never such an equation anywhere.

Please verify.

For an ideal gas, your equation for $\Delta U$ applies not only to constant volume processes, but to all processes.  It's just that, in experiments to measure Cv, it is very convenient to use a constant volume path because, under those circumstances, no work is done, in which case $\Delta U$ is just equal to the amount of heat added.  So this gives you a direct measurement of Cv.

As far as $\Delta H$ is concerned, the correct equation is $\Delta H=\Delta U+\Delta (PV)$.  But, from the ideal gas law, $\Delta (PV)=nR\Delta T$.  So what does that give you for $\Delta H$?


Are there any substances that react very slowly with water to create heat? I know there are things like the alkali metals that react very violently with water. I was wondering if there is anything that produces a slower more controlled reaction?

Also, is there anything that reacts with water to form heat that is safe for humans to touch?

I'm trying to think of ways to melt snow besides salt.

Hand- and foot-warmers are made containing iron filings, with a bit of paper pulp and salt, and sometimes a sealed water packet. These rely on the (surprisingly rapid) rusting of iron in contact with air, catalyzed by the water and salt (the pulp is just to hold the water and keep it in contact with iron). This reaction is a fairly safe way to melt a bit of snow, with some caveats:

It's more expensive than using freezing point depression of salts such as $\ce{CaCl2}$, $\ce{Mg(C2H3O2)2}$, or $\ce{NaCl}$ (table salt). Even adulterated wine containing ethylene glycol has been used for runway deicing.
The rust will eventually leave brown streaks on any surface below the deicer, such as the walls of a building if used to melt snow on the roof.


What happens when we connect a external emf (having emf greater than that of emf of a galvanic cell) to a galvanic cell? I read a statement given in the book and I did not understand it

"If any external battery is connected whose emf is more than the emf of galvanic cell ,then it will act as Electrolytic cell"

Also can you please explain what happens if we connect a external battery having emf less than that of galvanic cell?

Think of a generic redox pair:
$$\ce{Ox + ne- -> Red}$$
and use Nernst equation (and a few simplifications) to calculate its potential:
$$ E = E^\circ + \frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]}{[Red]}\right)$$

So, now you know what potential you would measure on such a half-cell. However, what would happen if you applied some arbitrary potential $E_{w}$ ?

Well, if $E_{w} > E$, then that also means that
$$\frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]_w}{[Red]_w}\right) > \frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]}{[Red]}\right)$$

Finally, you get that this situation implies an increase of $\ce{Ox}$ at the expense of $\ce{Red}$, which is what you would expect from an electrolytic cell - the supply of electrical energy causes electrochemical reactions.

If you apply a potential that is below $E$, the only difference would be that $\ce{Ox}$ would decrease and $\ce{Red}$ would increase, but the overall treatment remains the same.


Overall effective activation energy for parallel reactions Today, I was solving a question from Chemical Kinetics.

Consider the following parallel, first order reactions:

$$\begin{align}
\ce{A -> P} \tag{1} \\
\ce{A -> Q} \tag{2}
\end{align}$$

with rate constants $k_1$ and $k_2$, and activation energies $E_1$ and $E_2$ respectively.

I was asked to calculate the overall activation energy $(E)$ for the decay of A.

In my book a formula for $E$ is given to be :

$$E = \frac{E_1k_1+E_2k_2}{k_1+k_2}$$

This formula works well and provides correct answer, but I was trying to derive this expression using the Arrhenius equation. But I didn't succeed. I could deduce that,

Since:

$$ k_1=A_1 \mathrm e^{-E_1/RT}~; ~ k_2=A_2 \mathrm e^{-E_2/RT}$$

and

$$k_{\text{eff.}}= k_1+k_2 = A_{\text{eff.}}\mathrm e^{-E/RT}; $$

Using these both:

$$k_1+k_2= A_1 \mathrm e^{-E_1/RT} + A_2 \mathrm e^{-E_2/RT}$$

Now, I have no clue what to do.

The activation energy $E_\mathrm a$ should, in general, be viewed as an empirical quantity that characterises the sensitivity of the rate to temperature. From this perspective, it may be obtained via the equation

$$E_\mathrm a = RT^2 \left(\frac{\mathrm d \ln k}{\mathrm dT}\right) \tag{1}$$

This relation can be "derived" from the Arrhenius equation. (Quotation marks because the Arrhenius equation itself is only an empirical relation.)

$$\begin{align}
k &= A\exp\left(-\frac{E_\mathrm a}{RT}\right) \\
\ln k &= \ln A - \frac{E_\mathrm a}{RT} \\
\frac{\mathrm d \ln k}{\mathrm dT} &= \frac{E_\mathrm a}{RT^2}
\end{align}$$

If a reaction rate has Arrhenius-type behaviour, i.e. $k = A\exp(-E_\mathrm a/RT)$, then equation $(1)$ simply returns the usual interpretation of the activation energy. However, equation $(1)$ is useful in that it allows us to define an "activation energy" in the case where the rate constant does not have the mathematical form $A\exp(-E/RT)$. The parallel reactions that you have described fall exactly into this category. The sum of two exponential functions generally does not return an exponential function, so the overall rate constant $k = k_1 + k_2$ should not be expected to conform to an Arrhenius description.

If we substitute

$$k = k_1 + k_2 = A_1\exp\left(-\frac{E_1}{RT}\right) + A_2\exp\left(-\frac{E_2}{RT}\right)$$

into equation $(1)$, then we get:

$$\begin{align}
E_\mathrm a &= RT^2 \cdot \frac{\mathrm d}{\mathrm dT}\left\{ \ln\left[A_1\exp\left(-\frac{E_1}{RT}\right) + A_2\exp\left(-\frac{E_2}{RT}\right)\right] \right\} \\[8pt]
&= RT^2 \cdot \frac{A_1(E_1/RT^2)\exp(-E_1/RT) + A_2(E_2/RT^2)\exp(-E_2/RT)}{A_1\exp(-E_1/RT) + A_2\exp(-E_2/RT)} \\[8pt]
&= \frac{A_1E_1\exp(-E_1/RT) + A_2E_2\exp(-E_2/RT)}{A_1\exp(-E_1/RT) + A_2\exp(-E_2/RT)} \\[8pt]
&= \frac{E_1k_1 + E_2k_2}{k_1 + k_2}
\end{align}$$

as desired.


How to Dissolve AgNO3 in ZnAc Solution? In the preparation of $\ce{Ag}$ doped $\ce{ZnO}$ thin films by sol-gel method, I dissolved zinc acetate in 2-propanol while adding DEA and the solution is ready.

After that I added different concentrations of $\ce{AgNO3}$ to the solution, stirring on a stirrer, starting with $0.005$ concentration, but I found that it didn't completely dissolved even after $20$ minutes, and it gets harder for higher concentrations. Can I get any help for this?

We all learn right away (Chem I) that most silver salts are insoluble - except nitrate - in water, but how soluble are they in isopropanol?  That lonely alcohol functional group may not give the same "umph" that water would for the dissolution of the Mighty Silver Cation.

Normally, you'd start with a ZnOAc dihydrate salt, so there's gonna be some water in your mix to dissolve the nitrate, but if you're going for an {aq.} soluble salt in ethanol, did you consider just adding DI to the mix?

What about changing your starting components to include the silver and zinc salts together in dissolution before adding the alcohol?  Add them both gravimetrically ratioed as salt to your alcohol solvent and use the DEA to assist in solvation.  There's docs on this, let me know if you can't find them.

Update:(moved from a second answer, ed.)

So we actually just tried using water to help dissolve the ZnOAc and AgNO3 in the ethanol solution and we found that adding DI water dropwise did help to dissolve the ZnOAc, but after aging overnight a cloudy precipitate had developed, after googling around a bit, I saw that water in ethanol could react with ZnOAc to form ZnOH. In order to test this, I added ~3M Nitric Acid dropwise, and it cleared up the solution. Unfortunately, the solution then failed to stick to our slides when spin-coating.

The good news is, the solution with the cloudy ZnOH actually produced similar XRD results to the solution without water added. We have been using monoethanolamine as a stabilizer to help dissolve the ZnOAc. MEA acts as a bidentate ligand to hold the Zn+ ions in solution. I should have a parametric analysis on a ZnO sol-gel process published by the end of semester - will post it when available.

Additionally, note from Ahmed's comment below.  Silver salts tend to be extremely photosensitive, so the smallest amount of exposure of their solutions may cause discoloration.  We have had improved success by performing as many steps as possible in darkness, and/or by covering glassware with aluminum foil.


What is the liquid in Mathmos lava lamps? How can I identify the liquid in Mathmos lava lamps?
I have two of them with broken caps, I need to get rid of them. Can I flush the stuff down the toilet?

It smells funny.

The company is not very forthcoming with the ingredients in their FAQ; however, they do state in regards for disposal:

  Q. How should I dispose of my lava lamp bottle when it reaches the end of its life?

  A. Please make a hole in the metal cap with a sharp point. Then pour
  away the liquid. The bottle may then be placed in the glass
  recycling.

However, personally, I go by a general rule that if I do not know what substance is, I do not pour it down the sink.  The authors of the HowStuffWorks webpage How to Dispose of Lava Lamps state the actual ingredients are a trade secret, but the first patent for lava lamps states the ingredients as being:

  "solidified globule of mineral oil, paraffin and a dye as well as paraffin wax or petroleum jell, preferably Ondina 17 with a light paraffin, carbontetrachoride, a dye and the paraffin wax or petroleum jelly."

as for disposal, the author of that website made contact with a company, who advised that

  They told me that I could wrap the lava lamp in newspaper and throw it in the trash. I was told that it was non-toxic. They warned me not to pour the liquid down the sink because it contained wax.

The wax could make a mess of the pipes from your sink if it were to accumulate and solidify, and there is more than enough rubbish and pollutants in the water ways as it is.

Best to consult your local area's rules and regulations as to disposal.


Will this beta-branched secondary alkyl halide undergo SN1 or SN2? What is the correct mechanism here? Would SN1 occur here? I chose SN2 because we have a secondary alkyl halide with a weak base $\ce{CN-}$.

However, the solution says SN1 is major. Can you explain why?

My friend got full credit for this:

Secondary alkyl halides are on the borderline of $\ce{S_{N}2}$ and $\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more stable tertiary carbocation, which leads to the observed product.

Full mechanism:

I am of the opinion that it's a difficult question given the conflicting factors, and I don't like it as an exam question because the relative influence of those factors isn't usually taught or easily predicted.
I imagine that the solvent plays a large role here. Cyanide is a typical nucleophile for $\ce{S_{N}2}$, but in water its nucleophilicity will be reduced by hydrogen bonding. In order for the $\ce{S_{N}2}$ mechanism to take place for this hindered electrophile, the nucleophile will need to be as strong as possible.


Salt bridge in galvanic cell When we make a galvanic cell using zinc rod (in zinc sulphate solution) and copper rod (in copper sulphate solution), there is a flow of sulphate ions from the copper compartment to the zinc compartment through the salt bridge. Once the sulphate ions reach the zinc compartment, what happens to them?

According to my opinion i think you are wrong because the sulphate ion does not enter the other compartment. This is because the function of the salt bridge is to maintain electrical neutrality by either allowing the sulphate ion to reside in it or by allowing the cations inside the salt bridge to enter the compartment so as to balance the ions.


Term symbols: counting electrons. How? I am having some difficulty in writing the correct term symbols for diatomic molecules in their ground states.

For instance, the $\ce{S2}$ molecule has a MO configuration in its valence shell $\ce{3s\sigma^2_u 3s\sigma^2\mbox{*}_g 3p_z\pi^2_u 3p_y\pi^2_u 3p_x\pi^2_u 3p_y\pi^1\mbox{*}_g 3p_x\pi^1\mbox{*}_g}$

Now, for its term symbol I would say (erroneously) that I have two unpaired pi electrons, hence $\ce{\Lambda}=2$, so I would obtain a $\ce{\Delta}$ symbol.

As for the spin multiplicity, it would be 3. And the spin-orbit coupling term would be therefore S+2=3. Concerning simmetry, I would say that it's g, since the last occupied shell is g (still, I don't understand why Ar2 should be g if the last shell is u. Maybe because having wto electrons in a u shell I should consider u*u=g? If that is the case, it's correct to say that every molecule with an even number of electrons in the last energy level, and hence every homonuclear diatomic, is g?)

Actually, it looks like the correct term symbol for the $\ce{S2}$ molecule is $\ce{^3\Sigma _g}$, but I don't understand why. Can you help me please?

This example is probably described quite thoroughly on the internet for the case of $\ce{O2}$ (which is isoelectronic), so it may be worth searching a bit for alternate explanations.

You cannot have a $^3\Delta$ term symbol, because of the Pauli exclusion principle. The π* orbitals come as a set of two: essentially, one of those two is associated with +1 angular momentum, and the other is associated with −1 angular momentum.

Here, to be specific, "angular momentum" refers to the projection of the angular momentum onto the internuclear axis. The "angular momentum" quantum number of each MO is denoted $\lambda$, so in other words, the π* orbitals come as a set of $\lambda = +1$ and $\lambda = -1$.

The quantum number that measures the "angular momentum" of the term is denoted $\Lambda$, and because $\Lambda$ is a projection quantum number, the value of $\Lambda$ is simply the sum of the individual $\lambda$ values. The Greek letter that represents it in the term symbol depends on the absolute value of $\Lambda$: if $\Lambda = 0$ then it is a $\Sigma$ term, if $|\Lambda| = 1$ then it is a $\Pi$ term, and if $|\Lambda| = 2$ it is a $\Delta$ term.

Now, in the ground state configuration of $\ce{S2}$, we can ignore all the orbitals below the π* orbitals because all those $\lambda$'s sum to zero. The only thing of interest is the π* electrons.

If you want to have a triplet state, i.e. the electrons have the same spin, then you cannot put them into the same π* orbital: one will have to go into the $\lambda = +1$ orbital, and the other will have to go into the $\lambda = -1$ orbital. Hence you get $\Lambda = 0$, which leads to a $^3\Sigma$ term symbol, not a $^3\Delta$.

Finding the parity of the term symbol is nothing more than multiplying all the individual parities of the electrons (i.e. for doubly occupied $\mathrm{g}$ MOs you need to multiply by $\mathrm{g}$ twice). This is useful, because $\mathrm{g \otimes g = u \otimes u = g}$, which allows you to immediately discard all doubly occupied MOs. You only need to look at the two π* electrons, which are both $\mathrm{u}$, leading to an overall $\mathrm{g}$ parity and a $^3\Sigma_\mathrm{g}$ term symbol.

For $\Sigma$ term symbols there is an additional detail in that you need to specify the symmetry with respect to reflection in a plane containing the molecule. In this case it turns out that the molecular wavefunction is antisymmetric with respect to this reflection, so the proper term symbol is $^3\Sigma_\mathrm{g}^-$.


Unusual Friedel–Crafts alkylation with pivaloyl chloride In class we were told of an anomalous Friedel–Crafts reaction with a tertiary acid chloride — $\ce{(CH3)3CCOCl}$, or pivaloyl chloride. When this reacts with benzene in the presence of $\ce{AlCl3}$, instead of effecting the normal acylation, it instead breaks down into the tert-butyl carbocation, which then alkylates benzene to form tert-butylbenzene.

I could find no satisfactory explanation for this, however, except for the fact that a tertiary carbocation is a very stable carbocation — and nor did I find any mention of this reaction on searching. Could anyone please detail this further?

You were on right path; tert-butyl carbocation is quite stable$\ldots$ so stable that the acylium cation, which normally reacts as an electrophile itself, instead decomposes via decarbonylation (loss of stable carbon monoxide molecule).

The t-butyl carbocation isn't as stable as the acylium cation (which is stabilised by resonance), but the difference is small enough that with the help of the non-reversibility of CO loss (which is removed from the mixture as a gas) and reduction of the steric strain (which is lower in t-butyl than in acylium), it is enough to make it the main reaction route instead of a side route.


why don't carbons use sp3 instead of sp2 orbitals to make double bonds? I don't have a strong chemistry background, but I was thinking why do carbons have to make different kinds of orbital hybridization for double and triple bonds?
Why doesn't a carbon make two sigma bonds into a double bond?

Only one electron can be in an orbital at a time. There are three p orbitals, one in each of the x, y, and z axes. A pi bond is the overlap of one p oribital from each atom. Like the p orbital, a pi bond also can be rotated from the x axis to the y or z to produce a different unique pi bond. Because a sigma bond has rotational symmetry, rotating it does not give a unique bond. Therefore, you cannot have two of them between the same pair of atoms.

Edit: It appears this question has been asked before and there are much better answers than mine (Why can there not be more than one sigma bond in a set of bonds?)


Is it possible to make deuterated organic compounds? Organic compounds always have a $\ce{C-H}$ bond. But can we make compounds having $\ce{C-D}$ bonds where  D is the deuterium part like $\ce{CD4}$, $\ce{C2D5CHO}$, $\ce{C6D6}$  and other complex organic compounds? Overall, can we make organic compounds containing deuterium? Also, can we consider those compounds organic? Are the compounds stable? Give some example of preparatory reaction of those compounds.

Deuterium occurs at around the 0.02% level in nature (natural abundance level).  So all compounds that contain hydrogen will already contain a low level concentration of the deuterated isomer(s).  So in fact, deuterated compounds do exist in nature.

Despite this low level of natural deuterium incorporation, the deuterium can still be detected by various spectroscopic techniques (mass spectrometry and nmr for example).

Further, organic compounds can be synthesized with deuterium in specific locations. These deuterated compounds are still considered organic compounds. Since deuterium is a stable nucleus (e.g. does not undergo radioactive decay), the compounds containing deuterium are also stable.  Because the $\ce{C-D}$ bond is a bit shorter and therefore a bit stronger than the corresponding $\ce{C-H}$ bond, there is an isotope effect in reactions where these bonds are broken. Because the $\ce{C-H}$ bond is weaker, it will break up to 7 times faster than the corresponding $\ce{C-D}$ bond. This can have a significant effect on biological processes.

$\ce{D2O}$ and $\ce{D2}$ gas are relatively common and inexpensive materials that can be used to introduce deuterium into other molecules. Here are 2 examples involving the use of these materials for deuterium incorporation.  In the first example $\ce{D2O}$ is used as the source of deuterium, in the second example deuterium gas is used to introduce the deuterium.  If you google "organic preparation of deuterated compounds" you'll find many more examples.

image source

image source

Specifically deuterated compounds are widely used in many areas of research and industry.


How does the anti-bonding eg orbital (eg*) affect the bonding of ligands? In the molecular orbital for octahedral metal-ligand complexes, there is an eg orbital and an eg* orbital. When the d orbitals on the metal split, it will fill up some of the eg* orbitals I believe. However, my question is that won't these orbitals cancel out with the eg orbitals and make fewer bonds? For example, $\ce{Ni^{2+}}$is a $\ce{d^8}$ metal and will split in the presence of ligands and will partially fill the eg* orbitals. The $\ce{Ni^{2+}}$ metal can then form complexes like $\ce{[Ni(H_2O)_6]^{2+}}$ with 6 ligands attached. My question is how would the filling of the eg* orbitals affect the bonding and chemical properties of the compound? Here is the diagram I am referencing to:

As you would expect from increasing occupancy of antibonding orbitals, the result is weaker metal-ligand bonding. This is one reason why the octahedral geometry is not as strongly favored for metals with greater numbers of d electrons. For a d8 metal such as the Ni(II) in the example, square planar and octahedral complexes are both common, and planar structures tend to be favored if the ligand is strong enough to result in a low-spin planar complex (as is often the case with Ni(II)). Only a weak field ligand like H2O will tend to form octahedral complexes with Ni(II).
A semi-quantitative approach to this issue is found in "structure energy" approximations and determination of "structure preference energy". Although that approach is not rigorously quantitative, it will show you trends.
It is important to note, though, that even with full occupancy of the eg antibonding orbitals (for example in a d10 metal like Zn(II)), octahedral complexes are possible. One can think of the sigma bonding in that case as being essentially entirely via the s and p metal orbitals, similar to a p-block central atom such as S in SF6.


Synthetic steps for the interconversion between aldehydes and ketones I am learning about carbonyl chemistry at the moment. So far, we have been able to make aldehydes and ketones from alcohols (via oxidizing agents). I can't help but wonder if there is possibly any way to make aldehydes from ketones or vice-versa through a series of synthetic steps. I would appreciate it if the synthetic steps are in the level of an undergraduate organic chemistry course.

Question: Synthetic steps for the interconversion between aldehydes and ketones?

You can easily convert aldehydes to ketones and vice-versa by using simple organic synthetic steps which are very much well known.

Converting aldehydes to ketones

For converting aldehydes to ketones there are numerous pathways. Here I am mentioning the simplest two ways, one of which is by using Grignard reagent and other by using thioacetal intermediates.

You can react aldehydes with Grignard reagents ($\ce{R^2 -MgBr}$) and perform acidic workup to generate secondary alcohols. Then you can oxidise the alcohol to get a ketone by commonly used oxidising agents like PCC (pyridinium chlorochromate).

The other way is to react the aldehyde with a propane-1,3-dithiol to generate a cyclic thioacetal . Now, interestingly, the proton attached to the carbon now can be removed by strong base like n-Butyllithium, and generate  Carbanion which can perform a $S_N2 $ reaction with an alkyl halide to generate cyclic thioketal. After that the thioketal can be hydrolysed using $\ce{HgCl2/CdCO3/H2O}$  to generate your desired ketone.

Converting ketones to aldehydes

The easiest step for this conversion is to perform a Bayer-Villiger Oxidation reaction on the ketone to get an ester. Then that ester can be reduced by Diisobutylaluminium Hydride (DIBAL-H) and an aqueous workup will generate desired aldehyde (along with an alcohol). (If you wish you can further oxidise that alcohol also in case you want aldehyde of that part).


How to perform IR spectroscopy on silica particles at various relative humidities when they are passed through a tube using pressurized air? Surface bonds of silica particles are susceptible to humidity and frequently change between silanol and siloxane with changing RH. I want to measure how these surface bonds change with the humidity using IR spectroscopy when the particles are flown using pressurized air through a tube.

Despite your question is a bit vague about the sample, I assume «when they are in motion» refers to particles passing a tube, are close to each other -- in contrast to particles floating in air.

If so, than you should consider to couple infrared spectroscopy (about the energy band) with a probe based on attenuated total reflection.  Roughly speaking, classical optics predicts you a threshold incident angle where light of a certain frequency no longer is able to leave an optical dense material (high refractive index, for IR for example Silicon, Germanium, or a small diamond) to enter a less dense material (for IR, for example air, most of organic materials).  Recognizing light equally has character of a wave however «allows» that light at this interface still interacts sufficiently enough that one may extract information to calculate an IR spectrum.

What you see in the lab as an ATR-FT-IR spectrometer typically is a gate-like construction

(source)

but there are ATR-IR probes which you may immerse into fluids to monitor the advancement of reactions (e.g., Mettler's reactir)

(source)

where the probe window is at the side or on the tip of the probe itself:

(source)

Potentially -- compared to classical transmission IR -- you truncate any sample preparation (no KBr pill, nor nujol) and may record data of not-too wet samples, too.  Don't forget an ATR correction, as the small penetration depth $d_p$ of light across the phase boundery between window material and sample depends both on $n_c$ (window / crystal material), $n_s$ (sample), the incident angle $\theta$ and wavelength $\lambda$ applied.

$$ d_p = \frac{\lambda}{2\pi n_c \sqrt{\sin^2 \theta - \left(\frac{n_s}{n_c}\right)^2 }}$$


How to determine the configuration of an propargylic alcohol? I am trying to determine whether the stereocenter of 2-methyl-1-penten-4-yn-3-ol is R or S.

With the ghost atom approach I am determining S:

My supervisor said the stereocenter is actually R configuration. Could someone please explain this and tell me where I went wrong?

I think your working is correct, and the molecule should be assigned as S, with the alkyne taking priority over the disubstituted alkene such that the priority goes:

alcohol
alkyne
alkene
hydrogen

This leads to the S configuration, as you propose.

There is a chance that your supervisor is confused because in IUPAC naming nomenclature, the alkene often (and indeed, in this case) takes priority over the alkyne, which would of course lead to assignment of the molecule as R.


If a reactant appears on both sides of an elementary equation, can it be cancelled? Me and a friend were debating the following. For the elementary reaction

$$\ce{A + B -> A + C}$$

my friend says that it is unimolecular because $\ce{A}$ does not "participate" in the reaction. So, he says that the rate law is

$$\text{rate} = k[\ce{B}].$$

But, I think that that since it is an elementary reaction step, you cannot cancel $\ce{A}$ from both sides. I think that it is a bimolecular reaction with

$$\text{rate} = k[\ce{A}][\ce{B}].$$

Who is correct?

If the elementary step is indeed written

$$\ce{A + B -> A + C}$$

and assuming that you haven't just written in $\ce{A}$ for fun, i.e. $\ce{A}$ is actually a participant in the step, then yes, this would be considered bimolecular. This sort of step, featuring the same chemical species on both sides of the reaction, does pop up in (for example) the Lindemann mechanism.

With regard to kinetics, however, it is worth noting that steps of this kind can sometimes indicate a reaction that is catalytic with respect to $\ce{A}$. If that is the case, then the concentration of $\ce{A}$ is essentially constant and you would obtain pseudo-first-order kinetics.


Can there be oxygen dissolved in water in a hydrogen atmosphere? I'm trying to determine (for a science fiction story) whether a certain type of planet could have a significant amount of oxygen dissolved in water (e. g. produced continuously by some aquatic organism) if its atmosphere was made mostly of hydrogen. Is there a way, or will any oxygen just "pop" out of the water into the atmosphere and react with the hydrogen? I'm assuming the hydrogen pressure would be in the order of 10-100 atm.

This question is very good as it brings together many disciplines$-$astronomy, chemistry, atmospheric physics. Regretfully, as a consequence it is hard to compentently answer by a member of one field. I look forward to a healthy discussion in the comment section.

Hydrogen

A scientifically sound story ought to be such in most aspects. Specifically,

how does that planet possess hydrogen in high concentrations?

Hydrogen is a molecule with small molecular mass. As such it occupies the high end of Maxwell-Boltzmann speed distribution:

$$f(v)\begin{align}&=4\pi\sqrt{{\left(\frac M{2\pi RT}\right)}^3}v^2\exp\left(-\frac{M{v^2}}{2RT}\right)
\end{align}.$$

$\ce{H2}$ would thus escape the atmosphere over time since they often reach escape velocity

$$v_e\approx\sqrt{\frac{2GM_{planet}}{r}}.$$

Possible explanations why this would not have occured include:

1. Relatively young planet

A newly-formed celestial body is very likely to be hot. As such, it will not contain liquid water, definitely not to the extent of forming oceans.

One study by Tian et al. 'A Hydrogen-Rich Early Atmosphere' ($2005$) suggest the possibility of oceans and a concurrent hydrogen mixing ratio of $30$%. However, this claim has been contested.

  Tian et al.’s argument that hydrogen escape was inefficient on early Earth assumes that the upper atmosphere would be cold as on Venus, because the atmosphere is composed mostly of CO2. Their claim is controversial because their model depends on gases other than hydrogen to provide the radiative cooling, but their model does not actually include gases other than hydrogen. In particular, the key assumption that the upper atmosphere was cold is not obvious and needs to be addressed quantitatively. [----]

  Visconti’s calculations contrast markedly with the cold thermospheres in Tian et al.’s pure hydrogen escape models.

  source: Zahne et al. 'Earth’s Earliest Atmospheres'. (2010) link

2. Massive size and a magnetoshpere

Massive size would mean it has not really lost any atmosphere. Unless it is close to a star, in which case hydrodynamic escape would become important. These planets are mostly gas giants. We do not really know what is going on at their center, but it is almost certainly not a liquid ocean.

Without a magnetosphere the planet's atmosphere would be washed away by solar wind, similarly to that of Mars.

3. Hydrogen-producing microorganisms, e.g., a hyperactive, modified Chlamydomonas reinhardtii

This would be your best  bet. I am not qualified to discuss the likelihood of simultaneous thriving of both hydrogen-producing and oxygen-producing organisms. At face value$-$and this is speculation$-$it is probably not viable. Especially with so high pressures of hydrogen (high solubility).

What if?

Whether the reaction between $\ce{O2}$ and $\ce{H2}$ actually takes place (with considerable speed) depends on numerous variables. The ratio of these gases, temperature, pressure.

  Can there be oxygen dissolved in water in a hydrogen atmosphere?

Short answer is yes.

  [Is it] a significant amount?

Probably not. There ought to be some equilibrium amount but negligible. Even leaving aside the reaction between hydrogen and oxygen, the oxygen would be used to oxidise minerals. After these are, in a sense, saturated and when aquatic organisms adapt to more oxygen-rich water will the concentration of $\ce{O2}$ grow. It is now that the high hydrogen content becomes an issue.

Extra discussion requires actual calculation with supplied parameters, and the definition of what constitutes a significant amount. I have provided some sources for the kinetics of $\ce{O2/H2}$ mixture if this a route you wish to undertake.

Marcus Ó Conaire et al.. 'A Comprehensive Modeling Study of Hydrogen Oxidation' link
Alekseev et al.. 'The effect of temperature on the adiabatic burning velocities of diluted hydrogen flames: A kinetic study using an updated mechanism' Combustion and Flame, 162, 5, 1884$-$1892. (2015). link
Martin Hersch. 'Hydrogen-Oxygen Chemical Reaction Kinetics in Rocket Engine Combustion'. NASA Technical Note. (1967). link

Rogue planets

In a paper by David J. Stevenson 'Life-sustaining planets in interstellar space' (Nature vol. 400, page 32, 1999, link) he proposes that planets which form sufficiently far from their star or are gravitationally accreted may be sent on a course outside their star system. These planets could retain their hydrogen because the effective temperature would be fairly low: estimated to be about $30 \ \mathrm{K}$. According to Stevenson,

  [P]ressure-induced far-infrared opacity of $\ce{H2}$ may prevent these bodies from eliminating internal radioactive heat except by developing an extensive adiabatic (with no loss or gain of heat) convective atmosphere.

Thus surface temperatures may be enough for liquid water to exist. Assuming the planet collided with enough asteroids and comets at a stage where water was retained we arrive at the conditions:

hydrogen-rich atmosphere, Stevenson estimates that the pressure could be $100$ to $10000\ \mathrm{atm}$, so your intuitive estimate was pretty adequeate;

Sidenote: Pressure at the bottom of oceans on Earth is (at the deepest) about $1000\ \mathrm{atm}$. So we would have the same amount of pressure on the surface of the ocean.

liquid oceans.

A possibility for life would be anaerobic chemoautotrophs: they do not need oxygen, and would be able to be their own source of organic molecules. They do require a geothermal energy source. Volcanic activity at the bottom of the ocean floor suffices. Through random genetic mutation these organisms might transition from being obligatory to being facultative anaerobs. Remember: if $\ce{O2}$ is your product, $\ce{CO2}$ is likely the raw material the organism is utilising.

I have no idea how things could proceed from here. We could have a cosy ecosystem in such an ocean (especially near vents) but how and why there would be evolutionary pressure toward oxygen-rich life at other depths... I simply do not know. Only way is to decrease hydrogen content but, well, we need that to keep warm :-). Perhaps a gradual enrichment or replacement of hydrogen by other gases without lowering temperature further is possible? (see suggestion below)

As Stevenson notes,

  If life can develop and be sustained without sunlight (but with other energysources, plausibly volcanism or lightning in this instance), these bodies may provide a long-lived, stable environment for life (albeit one where the temperatures slowly decline on a billion-year timescale). The complexity and biomass may be low because the energy source will be small, but it is conceivable that these are the most common sites of life in the Universe.

Suggestion

If you wish for more information, consider follow-up questions on our

biology,
physics,
world building,
astronomy site.

I suggest against asking in all sites at once. Consider taking a concrete point (for example the transition from anaerobic to aerobic on biology.SE). Read their guidelines to be sure it is on-topic, and check for possible duplicates. Link to this question when asking again.

Good luck!


Observations when mixing potassium Triiodide and Sodium thiosulfate I have a reaction equation, it's all balanced but I don't understand what the physical change is. It is as follows:

$$\ce{2Na2S2O3 + KI3 \to 2Na2S4O6 + 2NaI + KI}$$

It was used to prove if $\ce{S2O3^2-}$ was in the solution or not. I'm not sure but I think it changes color to violet and back to colorless due to $\ce{I-}$ but if someone could definitively say what you are supposed to observe it would be helpful.

The keyword you should search is iodometric titration. Triiodide solutions are dark brown, just like the tincture of iodine. If you were to add thiosulfate ion (say from a buret) to  the triiodide solution would become colorless. The change is not that sharp. Therefore this is not the way end point is detected. A small drop of starch is added. Starch makes a deep bluish purple complex with free iodine. The moment all the iodine (in the form of triiodie $\ce{I2.I-}$ ) is consumed by thiosulfate, there is a very sharp transition from bluish purple to a colorless solution.


Why do tealights burn violently when nearly burnt out I'm not entirely sure if this is the right place for this question, but I decided that it's better than physics.se (and I had no other ideas). I hope this is a good match, otherwise please move the question where it would belong. :)

I often use a mildly scented tealight in my apartment to cover the lingering smell of food. I usually light them using a matchstick, and sometimes I would toss the matchsticks into the tealight after lighting it up.

Usually, the matchstick would burn out after a short while, and for most of the life of the tealight nothing interesting happens. However, when the tealight is very close to burning out with the remains of the matchstick inside, it starts burning very violently, with the flame forming fantastic shapes (it looks like the whole remaining volume of the wax is burning).

Once the flame was so violent it broke the heat resistant glass container I kept it in (I was careless and I let the side of the cup touch a thinner part of the glass, rather than just the bottom).

Why does this happen?

There are two factors that probably matter here. One is that, for most of the life of the tea light, only a small amount of wax is liquid and actually burning. Most of the flame is liquid wax burning in the wick with a small pool of liquid wax below the wick. But later, when most of the solid wax has gone all the wax will be liquid. The second is that the matchstick can act as a second wick.

When the wick is still vertical (and held there by the presence of solid wax) the area of melting is kept small as the flame is vertical and the area of melting is constrained. Later, the wick may fall over as there isn't enough solid wax left. Also the heat can spread wider as the volume of wax decreases as it is also conducted more effectively through the aluminium container (wax isn't a good conductor of heat). The net result is more liquid wax near the end of the candles life.

Then two other factors may intervene. If the wick falls over, there is now a larger area of wick where burning can happen and the whole wick (not just the tip) can burn. Sometimes the matchstick will now also act as a wick, increasing the area of flame. Both these effects make more flame and more heat in a sort of runaway reaction which might also cause enough heat to reach the flash point of the wax leading to an even bigger flame. Even if the flash point isn't reached, the flame from a bigger wick area and a match acting as a wick can be a lot bigger than the flame form a normal vertical wick.


Can "Common Ion Effect" suppress the dissociation of water molecules in acidulated water? $$\begin{align}
\ce{H2O &<=> H+ + OH-} \\
K &= \frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]} \\
\ce{H2SO4 &<=> 2H+ + SO4^2-}
\end{align}$$

In acidulated water water dissociates into its respective ions(though the dissociation is very less) and simultaneously acid dissociates into its respective ions.

Now as the acid (I have taken $\ce{H2SO4}$ here) also dissociates to give $\ce{H+}$ ions. So the $\ce{H+}$ ion concentration increases which in turn affects the equilibrium of $\ce{H2O}$ and favours the backward reaction forming $\ce{H2O}$ back. So can we say common ion effect has suppressed the dissociation of $\ce{H2O}$?

And if this is the case then why do we use acidulated water while performing the electrolysis of water? (I mean the common ion effect can suppress dissociation of water and we would get less $\ce{H+}$ and $\ce{OH-}$ ions in the medium which would lead to a lower yield if $\ce{H2}$ and $\ce{O2}$ gas.)

You have a couple of concepts mixed together in the wrong way.

(1) The ionization of water is typically written as

$\text{K}_\text{w} = \ce{[H+] \cdot [OH^-]}$

The assumption is that the ionization is so low that the concentration of water remains a constant. For a solution which is half water and half acetic acid that wouldn't be true of course.

(2) The dissociation and recombination of water molecules is a very fast reaction and occurs continuously in water.

$\ce{2H2O <=> H3O+ + OH-}$

(3) In reality the autoprotolysis of water is dependent on the activities of the $\ce{H+}$ and $\ce{OH-}$. So the relation is more properly written as

$\text{K}_\text{w} = a_\ce{H+} \cdot a_\ce{OH^-}$

In dilute solutions the activity is essentially the same as the concentration, but in more concentrated solutions the activity drops as the concentration increases because of the formation of clusters of ion pairs in solution.

Also the activity drops as the overall ionic strength of the solution increases for the same reason. So if you add $\ce{NaCl}$ to the solution, then the ionic strength of the solution increases and the activity of the hydronium  ions and hydroxide ions decreases. So in concentrated $\ce{NaCl}$ the $\ce{Na+}$ and $\ce{Cl-}$ ions aren't really just "spectators."

(4) If you add $\ce{H2SO4}$ to distilled water then the acid will dissociate according to the following reactions:

$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}\quad \quad \text{pK}_{\alpha1} = -3$

$\ce{HSO4- + H2O <=> H3O+ + SO4^{2-}}\quad \quad \text{pK}_{\alpha2} = 1.99$

The final pH will depend on the final concentration of the sulfate species which is commonly expressed as the molarity of the sulfuric acid. So if we diluting concentrated sulfuric acid we typically use

$\text{m}_1 \cdot \text{V}_1 = \text{m}_2 \cdot \text{V}_2$

where $\text{m}$ is the molarity of the solution and $\text{V}$ is the volume of the solution. This ignores the fact that in dilute solution the sulfuric acid isn't just $\ce{[H2SO4]}$ but that the overall sulfuric acid concentration is the sum of $\ce{[H2SO4]}$, $\ce{[HSO4^-]}$, and $\ce{[SO4^{2-}]}$.

(5) Acidifying an electrolysis solution isn't really about ion concentration since adding NaCl isn't as effective as adding HCl. The gist is that protons can move through solution very quickly since a water molecule can accept a proton from one direction to form hydronium and kick off a different proton in the other direction to form water again. So a net charge can move through solution quickly as a result of the chain reaction of such transfers. Thus any individual proton doesn't have to travel far.

CONCLUSION

The overall notion would be better expressed as that adding sulfuric acid to water increases the $[\ce{H+}]$ and reduces the $[\ce{OH-}]$ thus shifting the equilibrium for the autoprotolysis of water.


Friedel-Crafts Alkylation Using Alkenes: Mechanism What is the mechanism by which benzene, cyclohexene and $\ce{AlCl3}$ form cyclohexylbenzene?

I believe that cyclohexene will form a bond with Al with its double bond. However, how do we get rid of the $\ce{AlCl3}$ now attached to the previous cyclohexene after the benzene attaches to the module?

The Friedel-Crafts alkylation is another example of an electrophilic aromatic substitution. I've drawn the mechanism below showing the formation of cyclohexylbenzene.  The reaction is typically run in the presence of an acid and the aluminum chloride serves as a catalyst to enhance the electrophilic properties of the proton, at the end of the reaction the aluminum chloride has been regenerated.

In the insert box I've shown that when the reaction is carried out with n-propylchloride isopropyl benzene is produced, not n-propylbenzene.  A carbocation intermediate is generated in thr Friedel-Crafts alkylation and will rearrange to a more stable carbocation if possible.  In the cyclohexene case, rearrangement of the intermediate carbocation just regenerates the same cyclohexyl carbocation, so the rearrangement is "invisible" in this case (unless we label the starting cyclohexene).


Specific latent heat of fusion of a noble gas Could someone explain to me where the additional energy goes when applied to an object in the state of melting (such as water) so that the temperature of the object does not increase, and could someone please explain to me why Helium has a specific latent heat.

When you heat a substance (not at its melting point or boiling point), the energy supplied goes to increase the kinetic energy of the atoms/molecules in the substance.

When the same heat is applied to the substance at its melting point (or boiling point), the heat goes to increase the potential energy of the atoms/molecules in the substance. You could refer to the potential energy as the energy used to break the inter-particle bond. In ice, there are relatively strong forces of intermolecular attraction, which makes the ice hard.

Now, temperature is directly related to the kinetic energy of the atoms/molecules of a substance, and since the kinetic energy of the atoms/molecules does not increase during a change of state, the temperature remains constant.

At least, this is what happens at a macroscopic level and in the long term. But if you heat ice at one particular point (at particle level and for a short period of time), that point will melt faster and the temperature will tend to rise there, but this heat energy will then be distributed to the other parts of the ice still cold for melting.

It is only when all the ice has molten than the heat supplied will contribute once again to the increase in kinetic energy, that is, temperature.

And for the last part, every substance that undergoes a change of state has latent heat.


Why are covalent bonds directional? It is said that covalent bonds are directional, while ionic bonds are not. Why? Is it because of the orientation/directional properties of the overlapping orbitals?

The meaning of covalent bonds being directional is that atoms bonded covalently prefer specific orientations in space relative to one another. As a result, molecules in which atoms are bonded covalently have definite shapes.

The reason for this directionality is that covalent bonds are formed by sharing electrons between atoms, or, in other words, as you said, by overlapping the atomic orbitals of participant atoms. And usually only few patterns of overlap are possible, consequently, only few spatial arrangements of atoms are possible.

Ionic bonds are different: there is no electron sharing (or atomic orbital overlaps) and the number of anions surrounding a cation is limited by the charges of the ions, their sizes, and the efficiency of the lattice packing.


Effect on acidity of benzene due to electron withdrawing groups With respect to the compounds I–V, choose the correct statement(s).

Compound II becomes more acidic when it has a $\ce{-NO2}$ substituent.
The acidity of compounds follows the order I > IV > V > II > III.
The conjugate base of compound IV is aromatic.
The acidity of compound I is due to delocalization in the conjugate base.

I am not able to decide about the first statement properly. I compared the acidity by considering the conjugate base, which is the same (benzyne) even if $\ce{NO2}$ group is added.
However, the solution says that it will increase acidity because of being an electron withdrawing group.

Both the predictions are good. The $\ce{NO2}$ will actually be an effective electron withdrawing group, and the lost of that $\ce{H}$ will stabilize the compound (if there is and $\ce{NO2}$ on it). It is also important to rescue, that both effects will be stronger just on para or ortho position. You can make the different structures to check it.


How many unique molecules exist? And how many exist only by synthesis? I'm interested in

How many  unique molecules (of all types: organic and inorganic, etc), exist  in the world accessible by man (let's exclude black holes and the centers of stars etc)?
What subset of those distinct types are only in existence because of synthesis by man?

To make it simpler, ignore variants such as molecules with different atom isotopes and different bond configurations (e.g. cis, trans). E.g. standard water and "heavy water" are both $\ce{H2O}$ and count as one.

I'm looking for a theoretical estimate, so orders of magnitude only are required. I'm not looking for a count of records in some database catalogue of chemicals.

Please explain methods for determining your estimates. This isn't really helpful.

Polymers make this question more or less unanswerable.

Consider human chromosome 1, which contains about 249,000,000 base pairs.  There is nothing that says we couldn't order those pairs in any way we like, so for DNA alone, in a quantity that exists in every person on Earth, there exists the possibility for 4249000000 different molecules.  Include things like alternative nucleotides (there are almost surely at least thousands of these that are possible) and longer chains and the number gets even more enormous, and that's only including DNA (and close relatives.)

There's no real way to include all of them, since there have to be an enormous number of ways to put these molecules together that we haven't found yet.

How many exist?  An awful lot.
How many only exist in our minds?  Almost all of them (the universe only contains about 1080 particles, after all.


Intuition for the overlap integral I want to ask a question about the overlap integral.

I was studying about the MO diagram of $\ce{H_2^+}$ and we managed to derive an expression which had the following term:

$$\langle \mathrm{1s}_\ce{A}(r_\ce{A}) | \mathrm{1s}_\ce{B}(r_\ce{B}) \rangle$$

and our lecturer later explained that this is known as the overlap integral.

  The overlap integral is the magnitude to which the atoms overlap when combining AOs to produce MOs.

The trouble is, I later wanted to find a proof for this idea in order to confirm my understanding of this idea, but could not find anything that could explain this phenomenon from principle.

I did come across this question, but this only explained its main definition and not actually any derivation from basic principle.

How can the overlap integral be explained from basic principle?

Let $\Phi, \Psi$ be normalized wave functions.

Then $0 \le|\langle\Phi\vert\Psi\rangle| \le 1$, with $\langle\Phi\vert\Phi\rangle = 1$ and $\langle\Phi\vert\Psi\rangle = 0$ for orthogonal $\Phi$ and $\Psi$.

Since we can argue that any wave function completely contains itself and for two orthogonal wave functions none of them contains a part of the other, we can interpret the scalar product $\langle\Phi\vert\Psi\rangle$ as a measure for overlap.


What is the oxidation state of aluminium in NaAlO₂? I guess that sodium loses an electron so has an oxidation state of +1 and that oxygen has an oxidation state of −2. But what oxidation state does the aluminium atom have in sodium aluminate ($\ce{NaAlO2}$)?

For every compound the sum of the oxidation states of the constituent atoms must be equal to the charge of the compound. So, since your compound is neutral the individual oxidation states have to add up to $0$. And as you have already determined the oxidation states of 2 atoms in the system the last one must be $+3$, since $1 + 3 - 2 \cdot 2 = 0$.


Rate law of reactions, what about the proportions? Consider the reaction

$$\ce{A + B + C -> D + E}$$
and $\mathrm{rate} = k[\ce{A}][\ce{B}]^2.$ My book said if we double the amount of $\ce{A},$ then the reaction rate will double.

I understand this, but what about the law of definite proportion?

[OP, in a comment] oh ok, so if we double the amount of only one substance, that substance will get left?

Yes. The rate law tells you how fast product is being made. The change in the different species is still given by the chemical reaction equation (which tells you in which proportion the amount of species will change, not in which proportion they are present at the beginning or at the end of the reaction).

You can see that when you look at the definition of rate:

$$\mathrm{rate} = \frac{d[X_i]}{\nu_i dt}$$

where $\nu_x$ is the (signed, i.e. negative for reactants) stoichiometric factor of the $i$-th species. Dividing by the stoichiometric factor ensures that the law of definite proportions is followed.

  [OP in another comment] we cant simply put more of a substance to get more of the product, the reactants must have a fixed ratio.

The law of definite proportions is not about the amounts present, but about the change in those amounts. I can set up a reaction with 1000 times more elemental oxygen than hydrogen, but they will still react in a 1:2 ratio if the reaction yields water.


Using Gelatin as Rocket Fuel? I'm doing a DIY project on if it's possible for gelatin to be used as rocket fuel. This is just standard gelatin in a standard rocket. I know that gelatin is made from animal fat and that it has lots of sugar in it. So is it possible?

Powdered sugar is often used in chemistry demonstrations, in the presence of oxygen it is highly combustable (there have been significant explosions and fires at sugar processing plants in the past due to this).

The sugar is high in energy, and the fine powdered variety has a greater surface area with which to interact with oxygen, producing the burst of energy needed. The sugar by itself however, is probably not enough to get any propellant effect, you'd need to add some kind of oxygen source. A quick google suggests mixtures of powdered sugar and potassium nitrate (link).

Dry gelatine (the powdered version) is essentially just sugar plus, as you noted, animal extracts or vegetarian friendly alternatives, so the same should apply, though using actual sugar is probably a lot cheaper.

Edit:  As pointed out by @MaxW in the comments below, and as eluded to in my opening comment, igniting a fuel in the presence of an oxygen source can be highly explosive. Be sure to take necessary safety precautions (work on a small scale, away from buildings. Ignite from a distance, and be sure any spectators are also out of harms way etc.) and also ensure that you're not breaking any laws in your area.


What is a pure substance? I found as I was preparing to teach "Introduction to thermodynamics" that there are varying definitions or examples of pure substances. For instance the text book I have to use "Thermodynamics an engineering approach, Cengel and Boles" defines a pure substance as one that has the same chemical composition throughout. Air in the gaseous form is cited as an example. Air in its liquefied form is NOT since it seperates into it's various components.

The Dummies book on Chemistry suggests that sugar can be considered a pure substance as it has the same chemical composition throughout. And water can be considered a pure substance for the same reason.

So would a homogeneous mixture of sugar and water be considered a pure substance? After all, a homogeneous mixture has "identical properties" throughout it's phases.

Yet another definition of a pure substance suggests that any material made of the same kind of atoms is a pure substance. So individual elements in the periodic table and stuff made "purely" of them are pure substances. Other stuff is just homogeneous or heterogeneous mixtures.

So are pure substances (irrespective of whether we view them as elements, compounds or mixtures) just very difficult to break down into their individual components?

Or would the definition of a pure substance change depending on who you speak to (chemists, phycisists, thermodynamicists)?

"Thermodynamics an engineering approach, Cengel and Boles" defines a pure substance as one that has the same chemical composition throughout.

This is the correct definition of a pure substance. However, air, regardless of phase, is not a pure substance.

All matter is categorized as either a "pure" substance or a mixture. The word "pure" in front of "substance" is unnecessary, since the definition of substance implies purity. A substance is a sample that is chemically uniform in composition. In other words, all particles of that sample are the same chemically (whether they are atoms or molecules or ionic lattices or what have you). Substances cannot be separated into simpler components by any physical process (i.e. without breaking chemical bonds). Mixtures are samples that contain two or more substances. All mixtures can be physically separated, although the means to do so may be neither easy nor practical.

Air is a mixture of nitrogen ($\ce{N2}$), oxygen ($\ce{O2}$), carbon dioxide ($\ce{CO2}$), argon ($\ce{Ar}$), water vapor, and traces of other gases. Each of these gases separately is a substance. Air can be easily (but not in the comfort of your own home) separated into its constituent parts. Cooling air will cause the constituent components to condense out of it at progressively lower temperature: first water vapor, and then $\ce{CO2}$ (195 K), followed by $\ce{Kr}$ (120 K), $\ce{O2}$ and $\ce{CH4}$ (90 K), $\ce{Ar}$ (87 K), $\ce{N2}$, $\ce{Ne}$ (27 K), $\ce{H2}$ (20 K), and $\ce{He}$ (4 K).

As each substance condenses out of air it can be physically separated from the remaining gas, reducing the complexity of the mixture. Each substance, if returned to the gaseous state separate from the original mixture is physically and chemically distinct from the original mixture (or whatever is left of it). Such physical separations do not need to be easy to make air a mixture, they need to be possible.

A pure substance does not exhibit this behavior. When pure water is cooled, it undergoes a phase change to ice, which is physically separable from liquid water. However, the solid and liquid samples are still water. You have not reduced the complexity of the sample, just changed its form. If you allow the separated solid water to melt, it will be physically and chemically indistinguishable from the original water sample. Pure water cannot be separated into other chemically distinct substances without performing a chemical reaction, which means breaking and forming bonds between atoms.

The majority of matter that we interact with in our everyday lives is comprised of mixtures.

We encounter a few pure substances in the kitchen. Table sugar is a single compound (sucrose). So is table salt (sodium chloride), baking soda (sodium bicarbonate), and cream of tartar (potassium bitartrate). Water is a pure substance, so long as it has been purified. Most water that we encounter has other substances dissolved in it.

Gold (as long as it is 24K) is a pure elemental substance. So is diamond (carbon), graphite (also carbon), and the neon in neon lights. All minerals are pure substances, including asbestos. Glass is a pure substance. Most precious stones are pure substances.


How do surfactants spread in the case of multiple materials? For example, consider a pot of water with a drop of oil floating on it, with surfactants present in the water.

Will the surfactants in the water spread evenly across all three interfaces (water/metal, water/air, and water/oil)?

Are there material-specific surfactants, such that each of the three interfaces might attract different surfactants?

If so, are there also surfactants that would tend to spread across all three interfaces evenly?

A surfactant is a material that has two regions, one that tends to be attracted to its environment, and another that is repelled by the environment. The amount to which they will be attracted to one interface or another depends upon whatever would minimize the total free energy of the system. Thus, unless the interactions for water/oil, water/air, and water/surface were equivalent, you should see that the surfactant preferentially moves to the interface where it will have the biggest impact on the free energy.


Reduced-pressure (vacuum) evaporation/drying of 15 liters of aqueous solution I have 15 liters of aqueous solution inside an aluminum stock pot which rests upon a hot plate. I want to reduce the boiling point from 150°C to 100°C using reduced pressure.

What is the most economical way efficiently to dry the solute? Economical means does not require specialty equipment (i.e., laboratory vacuum pumps, expensive cold traps, etc.)?

I own a water aspirator pump (attaches to sink, 27.4 in Hg maximum vacuum, 6 liters/minute) and a 2-stage electric vacuum pump (15-micron, 3CFM) that I use for degassing epoxies. I don't know whether either of these will be of any use here.

The aspirator pump would be unlikely to be damaged by the distillate (and can be cheaply repaired or replaced), but the electric vacuum pump likely will require a cold-trap to prevent the vapors from destroying it.

The aspirator pump would seem the way to go, but you must factor in the cost of water in your location. This Nalgene aspirator uses ~6.5 liter/minute. A recirculating aspirator pump might be the best compromise for a production environment, where efficiency is important, and it might be feasible to save and recycle the distillate.


Which metals with high density do not react with mercury? I would like to know the highest density material between 10-17 g/mol range that is less than mercury and does not react with mercury.

Also would I like to know the highest density metal that is in solid form naturally.

Which ones are easily available and are highly dense?

Here is a link to a chart of all the elements in the Periodic table and their densities.  Iridium at $22.65~\mathrm{g/cm^3}$ is the densest of all the elements with Osmium a close second at $22.61~\mathrm{g/cm^3}$.  Sigma-Aldrich sells both of these metals (Iridium \$126.50 for 0.5 gram; Osmium \$195 for 1 gram).

  the highest density material between 10 17 range that is less than
  mercury and does not react with mercury

Mercury has a density of $13.534~\mathrm{g/cm^3}$.  It sounds like you're asking for something with a density right below mercury that doesn't react with mercury. Rhodium at $12.45~\mathrm{g/cm^3}$ is probably the most readily available element with a density less than mercury.  As to rhodium's reaction with mercury, an alloy is formed at 175 C.  If you're working at room temperature, especially for shorter periods of time (like hours) you'd probably be OK, but it really depends on your specific application.


Do eta-4 benzene ligands participate in Diels Alder or other cycloadditions? While refreshing myself on denticity to answer another question, I started to read about hapticity, which is the coordination of a ligand through multiple continuous atoms. A famous case is ferrocene (bis[$\eta^5$-cyclopentadienyl]iron, $\ce{Cp2Fe}$), where all five carbon atoms in each cyclopentadienyl ligand (Cp) are coordinated to the iron center.

Benzene can also be a polyhaptic ligand, forming $\eta^4$ or $\eta^6$ complexes depending on the oxidation state of the metal. In particular, bis(benzene)ruthenium complexes can undergo redox shuttling, where one benzene ligand switches between $\eta^4$ and $\eta^6$.

In the $\eta^4$ complex on the right, the top benzene ligand is now bent out of planarity, implying a break of aromaticity. This compound looks like a good dienophile for a Diels-Alder reaction. The article linked in the Wikipedia page on this switching phenomenon is published in Angewandte Chemie, International Edition, which is behind a paywall for me.

  Does coordination of benzene to a metal in the $\eta^4$ mode activate it toward cycloadditions? Because of paywalls, I would appreciate summaries of articles in addition to links to them. My attempts to find articles using Google Scholar led me to articles about the construction of benzenes by metal mediated cycloadditions, and not to reactions of benzene itself.

Coordinated benzene as a dienophile

It would be logical to suggest that an η4-benzene ligand could behave as a dienophile. The reason why benzene is characteristically unreactive, both as a dienophile and as a diene, is because of its aromaticity; if aromaticity is lost, one would expect it to become more reactive.

However, to the best of my knowledge (and SciFinder-fu), such reactivity does not seem to have been achieved. :(

In the rest of this answer I'll try to summarise some of the research that is (somewhat) related to what you were asking about.

Benzene as a dienophile - in general

Benzene itself (and some substituted benzenes) can act as a dienophile under some conditions:

with 1,2,4,5-tetrazines which act as electron-poor dienes (i.e. inverse demand Diels–Alder).1 Presumably, the reaction is favoured due to the loss of $\ce{N2}$ gas.

Under high pressure.2

Intramolecularly, when constrained to be in close proximity with a reactive diene.3

With incredibly reactive dienophiles (cyclobutadienes).4

(This list is probably not exhaustive.)

Coordinated benzene, but as a diene

In ref 4, Inagaki et al. also summarised the ways in which benzene has been activated towards Diels–Alder reactions:

  Currently, there are only a limited number of options to activate benzene sufficiently, including coordination of a transition metal complex to benzene, naphthalene with small resonance energy and the use of benzene derivatives with a sterically induced strained geometry [...] (emphasis mine)

The only reference given for the bolded sentence is the same article that Klaus has linked in his answer.5 It describes the reaction of η2-benzene as a diene, with a hypothesis very similar to what you have written in the question:

  In the proper coordination environment, both osmium and rhenium can form stable complexes with arenes in which only two of the six carbons are coordinated. Once complexed in a dihapto fashion, the uncoordinated portion of the arene more closely resembles a conjugated diene. We therefore hypothesized that the complexation should increase the propensity of arenes to undergo cycloaddition reactions.

In this reaction scheme taken from the paper, the carbons numbered 1 through 4 are acting as the diene component in a Diels–Alder reaction with N-methylmaleimide:

References

Seitz, G.; Hoferichter, R.; Mohr, R. Benzol und linear anellierte Arene als Dienophile in der Diels-Alder-Reaktion mit inversem Elektronenbedarf. Angew. Chem. 1987, 99 (4), 345–346. DOI: 10.1002/ange.19870990413.
Jarre, W.; Bieniek, D.; Korte, F. Benzene as Dienophile in the Diels-Alder Reaction. Angew. Chem. Int. Ed. 1975, 14 (3), 181–182. DOI: 10.1002/anie.197501812.
Saito, K.; Omura, Y.; Maekawa, E.; Gassman, P. G. The Addition of 1,3-Diphenylisobenzofuran to Cyclooctatetraene and to Dimethyl Cyclooctatetraene-1,2-dicarboxylate. The Benzene Ring as a Dienophile in an Intramolecular Diels–Alder Reaction. Bull. Chem. Soc. Jpn. 1990, 63 (2), 395–400. DOI: 10.1246/bcsj.63.395.
Inagaki, Y.; Nakamoto, M.; Sekiguchi, A. A Diels–Alder super diene breaking benzene into C2H2 and C4H4 units. Nat. Commun. 2014, 5. DOI: 10.1038/ncomms4018.
Chordia, M. D.; Smith, P. L.; Meiere, S. H.; Sabat, M.; Harman, W. D. A Facile Diels−Alder Reaction with Benzene: Synthesis of the Bicyclo[2.2.2]octene Skeleton Promoted by Rhenium. J. Am. Chem. Soc. 2001, 123 (43), 10756–10757. DOI: 10.1021/ja011689q.


Does the term 'silver nitrate' express all of its constituent chemical elements? My professor gave my class a set of practice questions for the mid-term. One of those questions asks for the molar mass of a number of chemicals, such as lithium bromide and silver nitrate. We haven't studied those chemicals, so there's little reason for the professor to expect us to have memorized the elements that constitute those chemicals. In the case of lithium bromide, the term informs us of its constitution. Whereas,'silver nitrate' informs us that the chemical $ \mathrm{AgNO_3}$, includes silver and nitrogen, but I don't see anything in the term that would inform the reader that the chemical includes $ \mathrm{O_3}$.

Is there anything in the term 'silver nitrate' that would inform a reader that the chemical includes $ \mathrm{O_3}$?

More generally, is there a set of rules that relates the chemical constituents of chemical substances to the linguistic constituents of the terms that designate them?

You should know that nitrate is a polyatomic ion with the molecular formula $\ce{NO3^-}$. For example:

Sulfate is a polyatomic ion with the molecular formula $\ce{SO4^{2-}}$;
Sulfite is a polyatomic ion with the molecular formula $\ce{SO3^{2-}}$;

Nitrate is a polyatomic ion with the molecular formula $\ce{NO3^-}$;
 Nitrite is a polyatomic ion with the molecular formula $\ce{NO2^{-}}$

At first glance, the nomenclature of the polyatomic negative ions seems difficult.
When the name of the ion  ends in either -ite or -ate:

The -ite ending indicates a low oxidation state.
The -ate ending indicates a high oxidation state.

I agree with you that at the beginning, it seems difficult. But by practicing, you find this quite easy.


How to interpret the instruction of the dilution of hydrogen peroxide? I am required to prepare $\pu{120cm3}$ of hydrogen peroxide to each of one of my students tomorrow. Some material is given to me on the method of preparing the solutions. Within the method of preparing $\ce{H2O2}$, it says:

  Prepare from freshly purchased $100$ volume ($30\%~\text{w/w}$, $\pu{8.3 mol/dm3}$) or $20$ volume ($6\%~\text{w/w}$, $\pu{1.7 mol/dm3}$) $\ce{H2O2}$. Dilute $\pu{200 cm3}$ of $100$ volume solution to $\pu{1 dm3}$ of $20$ volume solution, then dilute $\pu{59 cm3}$ of $20$ volume solution to $\pu{1 dm3}$ of $\pu{0.1 mol/dm3}$ solution.

Can someone please explain the quoted part?

Take $\pu{200 mL}$ of the $30\%$ solution, add water to a final volume of $1$ liter. Then you have a $6\%$ solution. Next, take $\pu{59 mL}$ of the $6\%$ solution and add water until it is $1$ liter.

After the first dilution it's not exactly $6\%$, because the density of $30\%$ $\ce{H2O2}$ is $\pu{1.135 g/cm3}$. It's more like $6.6\%$, but the instructions are referring to it as $6\%$. If you want exactly $6\%$ you should dilute by weight instead of volume.


Chemical formula of a compound crystallized in a cubic lattice Question

  In a cubic lattice of $\ce{XYZ},$ $\ce{X}$ atoms are present at all corners except one corner which is occupied by $\ce{Y}$ atoms. $\ce{Z}$ atoms are present at face centres. What is the formula of compound?

Answer

   $\ce{X2YZ24}$

My solution

Let the number of atoms in a unit cell be $N.$ Then

$$
\begin{align}
N(\ce{X}) &= 7 × \frac{1}{8} = \frac{7}{8}\\
N(\ce{Y}) &= 1 × \frac{1}{8} = \frac{1}{8}\\
N(\ce{Z}) &= 6 × \frac{1}{2} = 3
\end{align}
$$

Therefore, the ratio would be

$$N(\ce{X}):N(\ce{Y}):N(\ce{Z}) = \frac{7}{8}:\frac{1}{8}:3,$$

which is the same thing as $7:1:24.$ So, the formula is $\ce{X7YZ24}.$

What am I doing wrong?

I would like to say your answer is right, but the question is wrong.

Remember that a lattice is "infinitely" repeated units in 3 dimensional space so you should be able to expand the unit cell in x, y and z axes, indefinitely.

Now imagine your cubic lattice with an $\ce{Y}$ atom in just 1 (out of 8) corners.

Expanding the unit cell in the x-axis will necessarily duplicate the $\ce{Y}$ atom in the x-axis because all expanded unit cells must be identical in composition. Similarly, expanding the unit cell in the y-axis will duplicate the $\ce{Y}$ atom in the y-axis, and expanding the unit cell in the z-axis will duplicate the $\ce{Y}$ atom in the z-axis.

In the end you will find the having one corner as $\ce{Y}$ atom will necessarily have ALL corners as $\ce{Y}$ atoms. In other words, $\ce{X}$ IS $\ce{Y}$, and the formula is necessarily $\ce{XZ_3}$ or $\ce{YZ_3}$

Unfortunately, I don'k think the designer of the question saw his fatal contradiction because this very question is seen in the (mock) Joint Entrance Examination (JEE) in India.

And $\ce{X_2YZ_24}$ is not in the options even in the wrong question.


Mechanism for the ring contraction In this reaction I am unable to understand its mechanism. How the peroxide HCCL3 is causing the ring contraction. I know that the 8 member ring is unstable due to strain and hence it will try to contract. But I am not able to initiate the mechanism due to reagents present. Please can someone explain me how the contraction of 8 members ring to 2 five member ring is taking place?

As has already been said in the comments, this is a free radical-based reaction. The peroxide $\ce{R1-O-O-R2}$ acts as a radical initiator( where $\ce{R1}$ and $\ce{R2}$ represent alkyl groups). These peroxide radicals generate $\ce{^.CCl3}$ which attacks the double bond and completes the reaction as shown.

An important point to note is that $\ce{^.CCl3}$ is regenerated at the end of the reaction, so it can be described as a catalytic intermediate for this mechanism


I have to write a Python program relating with molecular weight. Any insight? Write a program that determines the molecular weight of a chemical compound based on the number of hydrogen, carbon, and oxygen atoms in the compound. You should use the following atomic weights:
  \begin{array}{|c|c|c|c|}
\hline
\text{Atom} & \text{Weight}~\pu{gm/mole} \\\hline
\ce{H} & 1.0079 \\\hline
\ce{C} & 12.011\\\hline
\ce{O} & 15.9994\\\hline
\end{array}

I don’t understand the relation between molecular weight and the number of $\ce{H}$, $\ce{C}$, and $\ce{O}$.

The professor has said "there is no need to be able to handle chemical compound names like $\ce{H2O}$ or $\ce{C2H5OH}$. (I plan to discuss how to handle these later in class) Instead, the user should just be asked to specify the number of hydrogen, carbon, and oxygen atoms in the compound.”

I need help to proceed.

It's pretty simple actually. The molecular weight of a compound is the sum of the atomic weights of all its atoms. The same numbers apply to a single atom as for a mole of that atom, but at the single-molecule level the unit of measure is the atomic mass unit or amu (aka the dalton or Da). The number of grams per mole of an element and the number of daltons that a single atom of the element weighs are intrinsically linked due to the definition of what a mole of something is (the amount of a pure chemical substance that weighs the same amount in grams as the mass of one atom or molecule of the substance in daltons).

So, methane (CH4), for instance, has a molecular weight of 1.0079 x 4 + 12.011 = 16.0426 Da. A mole of this gas would weigh 16.0426 grams (which is - interesting science fact - lighter by almost half than that of N2 or O2 gases making up most of the atmosphere, so a balloon of it will float). So, your Python script would simply prompt for the number of hydrogens (accepting integer values only), carbons (ditto) and oxygens (ditto), and plug them into the expression:

total_mass = num_hyd * 1.0079 + num_carb * 12.011 + num_oxy * 15.9994

... before returning total_mass as the answer.

Theoretically, the energy contained as enthalpy in the bonds between atoms contributes to a molecule's mass by general relativity and the first law of thermodynamics. In practice, the amount of energy inherent in even the least stable bonds (which produce the most net energy when they decompose to form more stable ones) is a rounding error when you run it through E=mc2, and so chemists don't bother. Nuclear physicists, who actually change the atoms themselves, do care, because the equation works in reverse for them; they get lots of energy from very small changes in the atomic masses of the reactants vs the products.


Dependence of EMF of a cell on stoichiometry In the reaction $\ce{Zn + 2Ag+ -> Zn^2+ + 2Ag}$ , my book says that the EMF is the difference between the standard reduction potential of $\ce{Ag}$ and $\ce{Zn}$.

Shouldn't it be the difference between twice the potential of $\ce{Ag}$ and $\ce{Zn}$ since two moles of $\ce{Ag}$ are getting reduced per mole of $\ce{Zn?}$

The electromotive force of a redox reaction is not dependent on the coefficients in the balanced equation as it is an intensive property (i.e. a property that does not depend on the size of the system). Thus, your reasoning is incorrect.

You can read more about electrochemistry here: https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Basics_of_Electrochemistry

Also, I am wondering if your textbook is actually correct because $\ce{Ag2+ + e- -> 2Ag}$ and $\ce{Ag+ + e- -> Ag}$ are definitely different. Thus, the standard reduction potential of silver (I) should not be used. Instead, the standard reduction potential of $\ce{Ag2+}$ should be used. Someone who is more knowledgeable in this area please correct me if I am wrong.