Percentage of a conformer in vacuum I just started with Molecular dynamics simulations. My background is computer science but I'm learning about scientific computing and thus my jump into MD.

Having said that, I was asked to calculate the percentage of each conformer in vacuum at a certain temperature of the Methylcyclohexane molecule. Assuming that entropies and molar volumes are identical.

Can someone point me out to some website or something where I can see how to calculate this?

Page 11699 of this paper has one approach to what you want: basically, you compute the relative Gibbs free energies of the possible stable conformers, assume they can interconvert, and use a Boltzmann distribution to work out what the percentages are.

  First, we selected a reference structure $R$ for a given species (neutral or protonated).

  Next, for every structure $M$ other than the reference structure $R$ we determined the
  equilibrium constant $K_M$ $$K_M = \frac{[M]}{[R]}$$ from the difference in Gibbs free energies for $M$ and $R$. The fraction of $M$ in the equilibrated sample is given by:

  $$x_M = K_M/\left(1 + K_1 + K_2 + ...\right)$$

  where the sum in the denominator goes through all structures for a
  given species. The fraction of $R$ in the sample is:

  $$x_R = 1/\left(1 + K_1 + K_2 + ...\right)$$

They don't include it in the paper, but the way to calculate an equilibrium constant $K$ from a difference in Gibbs free energy $\Delta G$ is:

$$ \Delta G = RT \ln K $$

where $R$ is the molar gas constant, and $T$ is the temperature (in Kelvin).

The tricky part is then working out what your stable conformers are, generally using some form of either guesswork or conformation sampling algorithm. They use a genetic algorithm of some sort in the paper I linked to, but given that methylcyclohexane isn't that complex a molecule, you might want to just take a look at an energy diagram and discussion of conformers of cyclohexane for some hints.

If you're getting deep into molecular dynamics, you could also use some sort of algorithm designed to use dynamic simulation to sample a wide area on a potential surface, like metadynamics or umbrella sampling, to find your conformers, and probably calculate your free energies at the same time.

You can use a static method to calculate the free energies instead, though -- the method you use to do this will depend on what sort of accuracy you're after, and may vary from a classical forcefield technique through Density Functional Theory to coupled-cluster theory (in ascending order of computational expense).


What is hyperconjugation? What is hyperconjugation and how do you know if there is hyperconjugation in a compound? Which orbitals are involved in hyperconjugation?

In the ethyl cation, one can write the following resonance structures:

The structure on the right can be actually written 3 times since there are 3 hydrogens available. Note also that the "ethylenic-like" structure on the right involves a $\mathrm{sp^3-p}$ bond (the $\ce{C-H}$ bond is $\mathrm{sp^3}$ hybridized and the cationic carbon has a $\mathrm{p}$ orbital), so it is not a true pi bond ($\mathrm{p-p}$) like that found in ethylene.  The resonance structure on the right is meant to show the "hyperconjugation" effect of the $\ce{C-H}$ bond ($\mathrm{sp^3-s}$) with the $\mathrm{p}$ orbital located on the adjacent cationic carbon.  The drawing is a shorthand way of saying that when the $\ce{C-H}$ bond is rotated into the same plane as the $\mathrm{p}$ orbital on the cationic carbon, there can be a stabilizing overlap - and it is called "hyperconjugation." In the 2-propyl cation, there are 6 of these hyperconjugative resonance structures (there are 6 $\ce{C-H}$ bonds on the 2 methyl groups adjacent to the cationic carbon and in the t-butyl cation there are 9 such resonance structures. This can be used to explain why tertiary carbonium ions are more stable than secondary carbonium ions, which in turn are more stable than primary carbonium ions.


Why does 2,4 DNP give orange colour crystals? Why does 2,4 DNP give orange crystals on reacting with solution of aldehydic or ketonic group? Why does this test give orange crystals? Is this related to the electron emissions?

This is just a question from a curious student after 2 hours of lab work.

Generally, you will notice that compounds containing chromophore and auxochrome groups have a colour. Chromophore groups are groups that are responsible for colour of the molecule once they are “attached” to it: benzene is colourless, but nitrobenzene is pale yellow. Apart from a chromophore group there are groups called auxochrome groups which alter this colour and so the intensity of absorption ( example: nitrobenzene is pale yellow, but p-nitroaniline is a bright yellow to red, the auxochrome enhanced the colour ). Among the chromophore structures, there are groups such as nitro, azo, nitrile, thioketone, conjugated double bonds ( like in retinal, carotene ). Some auxochrome structures are: the aldehyde group, the methyl mercaptan group, hidroxyl group, amine group. To sum it up, functional groups such as these will account for the colour.
Now, onto analysing DNP. This compound has 2 nitro groups and a hydrazo group ( 2,4-dinitrophenylhydrazine ). From the start, we have a colour: red ( if it is pure ). When it reacts with an aldehyde or a ketone in a condensation reaction, a double bond forms between the carbon on which oxygen was attached and the unsubstituted nitrogen of DNP. The formation of this double bond will alter the colour, shifting it to an orange. It also depends on the carbonylic compound you use. If the carbonylic compound contained conjugated double bonds ( like propenaldehyde ), the condensation product will contain conjugated double bonds which are chromophoric ( so in this case I can only guess what the resulting colour would be  because we have more chromophoric groups ). So, DNP is red in its purest form. Its colour is the result of the existing functional groups. A reaction with an aldehyde or ketones will shift its colour to orange.


mechanism in bi cyclic compounds In both cases ring expansion is happening. What is Carbon tri chloride purpose?

" What is Carbon tri chloride purpose?" is not answered in the link provided but the answer starts with a carbene attack on furan .

Trichloromethane in presence of a base undergoes 1,1-elimination (or an alpha elimination) because both groups are removed from the same carbon.The electron-withdrawing effect of the halogens makes the hydrogen acidic enough to be removed by a strong base such as hydroxide or alkoxide ion.

A halide ion then acts as a leaving group from the conjugate base, producing the
 carbene.1

Carbenes find a pair of electrons  to complete their valence shell of electrons.carbenes attack a lone pair, a C=C double bond (electron-rich or -poor), or even a C–H bond.In its reactions ,

It is an electrophilic reagent and is known to give ring expansion of aromatic rings.
Electrophillic attack takes place on pyrrole where the intermediate is most stable.

This leeds to normal Reimer–Tiemann reaction product pyrrole 2-aldehyde.

Carbene attack of a C=C double bond leads to an ring expansion leading to 3-chloropyridine.

The reaction of bicyclic compounf acts on similar lines of dichloro carbene attack.2

References

1 ORGANIC CHEMISTRY ,JOSEPH M. HORNBACK ,UNIVERSITY OF DENVER

2 MARCH’S ADVANCED ORGANIC CHEMISTRY REACTIONS, MECHANISMS,AND STRUCTURE ,SIXTH EDITION


Home made dispersion/solution with detectable >0.1μm particles Sorry for the possibly wrong terminology. I'm not a chemist and I'm not a native English speaker.
Please edit the question as needed. Thank you.

Is it possible to buy or prepare a dispersion with the following properties?

the particles should be >0.1μm in size, for example 0.11μm - 0.2μm
there must be a way to detect them - even in a small amount (colour, taste, chemical reaction with something else, ...)
it mustn't be a "glue"; it must be possible to flush them completely away with water or something else readily available
it mustn't be a toxic or corrosive substance
the dispersion (or components to prepare it) should be readily available (drugstore, grocery etc.)
the total price shouldn't be higher than $20 (I know this difficult with prices different in every country - this is just a guideline)

Why do I need it?
There's a very slight chance that a Sawyer Micro Squeeze Water Filter might have been exposed to a below zero temperature and might have been damaged by frozen residual water.
The manufacturer says there's no way to test the filter functionality in that case and a new filter should be bought instead. I wonder if this is really needed or there's a way to test the filter at home.

Try to get some clay and shake it vigorously with water. Let it settle for a several hours. Test your filter with the supernatant water and collect the filtrate in a very clean glass tumbler. Colloids have an interesting property of scattering light. In a dark room, try to shine light (ordinary flashlight might work or perhaps an ordinary pointer used in presentations/offices). Be careful, never ever look at the pointer directly. Look at the filtrate at $90^o$ (right angles to the light beam), if you see a light beam travelling in water, it means that colloidal particles are seeping through the filtrate and the filter is damaged.

Read more about the Tyndall Effect https://www.sciencesource.com/archive/Tyndall-Effect-SS2423500.html or on Wikipedia before doing anything.


Calculate Ethanol content after heating for N minutes Does there exist a formula I can use to calculate alcohol content of a solution after heating it for some time?

I have the following use-case:
I am making a pecan pie. I have a thick sugary solution. I am not sure how much the makeup of the solution matters. Just assume its more viscous than water. I add 75 ml of 40%v  alcohol and I heat it at a temperature high enough to cause boiling for 20 minutes. How much alcohol remains at the end of 20 mins.

Obviously there are a lot of variables here that would make it impossible to give a precise answer. I would be interested in an answer that solves an "easier" version of this problem, for example: having 1 liter of water, adding to it 100ml of 100%v alcohol (Ethanol), boiled for 100 minutes, how much alcohol is left?

How quickly does alcohol boil off? Is it dependent on is concentration within a solution? is there a formula for that?

For a pure substance, to be able to calculate how much has boiled off after $N$ minutes you will need the enthalpy of vaporization, $L$ (in kJ/kg) and the heat input, $Q$ (in kJ) which will give you:
$$\tag{1} m=\frac{\epsilon \, Q}{L}$$
where $m$ is the mass that has boiled off and $\epsilon$ is the fraction of the heat input that actually goes into the liquid. $1-\epsilon$ would be the fraction of heat lost to the surroundings.

The very crude story

An extremely crude calculation would be to take the mixture enthalpy of vaporization and assume that only ethanol vaporizes such that the fraction of ethanol can be directly related to the total mass loss. Note that this calculation will give you a lower bound for the time needed to get rid of the ethanol

We can calculate the mixture enthalpy of vaporization by averaging the enthalpies of vaporization of water and ethanol:
$$\tag{2} L_{mixture}=x_{\ce{EtOH}} L_{\ce{EtOH}}+x_{\ce{H2O}}L_{\ce{H2O}} $$

Now we need to figure out the heat input. A precise calculation is impossible here, but we can use some reasonable estimates. According to an estimate on this (dutch) website when we are cooking we use about $164$ l/h of methane. Assuming that that is for cooking with 2 pots this translates to a gas flow rate of $\dot{V}=1.4$ l/min for a single pot. With the heat of combustion of methane, $\Delta H_c$ and the density $\rho$ we can then calculate the amount of heat released per minute:
$$\tag{3} \dot{Q}=\rho \dot{V} \Delta H_c $$
plugging in the values for methane we find $\approx 50$kJ/min

The enthalpy of vaporization of the mixture will constantly change, because the composition of the mixture constantly changes over time during the boiling process. Therefore, instead of simply using Eq. 3 in Eq. 1 to find $\dot{m}$ and multiplying with $N$ we will need to do an integration:
$$\tag{4} m_{boil-off}=\int_0^N \dot{m}\; dt $$
where $\dot{m}=\frac{\epsilon \, \dot{Q}}{L(t)}$. Now this is a bit problematic because $L(t)$ depends on $m_{boil-off}$, but it can be solved numerically (I feel like this should also be possible analytically, but can't figure it out right now, anybody?!).

Your example question (with $\epsilon=0.5$) leads to the following curve for the fraction of ethanol in your mixture:

As you can see the ethanol is gone after about 7 minutes. To do this calculation for your pecan pie mixture you will need to know the heats of vaporization of all your ingredients and the mass of each.

The more precise story

The more precise story is that not only the ethanol, but also part of the water will vaporize therefore slowing down the vaporization rate of ethanol. If you are 'lucky' you could assume that the vapor above your pan is at equilibrium with the boiling liquid. In that case you can use the theory available for distillation columns and something called the Ponchon-Savarit Method (the complete diagram for water-ethanol can be found here), which I will not recap here, because it is basically a book chapters worth of undergrad chemical engineering.

Note that the equilibrium vapor assumption is probably not a very good one due to air flows over the pot which keep pushing the system out of equilibrium. For the evaporation of ethanol this is good news by the way, because it will speed up the process a bit. Also note that with a true equilibrium you would run into the azeotrope of ethanol-water mixtures.


Does the amine group participate in resonance in aniline? When we draw resonance structures of aniline, 8 electrons participate in resonance, making it antiaromatic. But in our school textbook, it is said to be aromatic. Does the lone pair of $\ce{NH2}$ participate or not? If it doesn't, then it would make sense.

The lone pair of nitrogen is also part of the extended π-system; the molecule itself is still aromatic. As I already explained in the question about fulvene, aromaticity is nothing you can determine based on simple rules, especially not counting electrons.
Structural indicators like bond lengths and angles give a good idea how well established resonance is. Additionally you can draw a qualitative molecular orbitals diagram and look if it is similar to benzene. Eventually the determination of aromaticity is an experimental one, for example with NMR spectroscopy.

From the crystal structure (M. Fukuyo, K. Hirotsu, and T. Higuchi, Acta Cryst. 1982, B38, 640-643. DOI: 10.1107/S056774088200363X) we know that the carbon-carbon bond lengths are all similar and between 136 and 141 pm. The NMR spectrum clearly shows the protons have aromatic shift, see chemicalbook.com.

The molecular orbitals of aniline (calculated at DF-BP86-D3BJ/def2-SVP; Gaussian 09 D.01) show that the nitrogen is part of the π-system. They are also very similar to benzene. Here are the orbitals of the π-system in decreasing energy. Virtual (unoccupied) orbitals are red/yellow, and occupied orbitals are blue/orange.


Calculating internal energy change of a reaction We are asked to find the relation between $\Delta H$ and $\Delta U$, and also, the sign of the internal energy change of reaction:
$$\ce{CO(g) +\dfrac{1}{2}O2(g)   ->  CO2(g)}$$
This was part (3) of a question, and this one explicitly mentioned: "reaction happens in an isolated container". The other parts had "closed container".
Approach: For ideal gases, we know that $U=nCvT$. So, we have:
$U_{rectants}:(1*5R/2*T) +(1/2*5R/2*T)=\dfrac{15}{4}RT$
$U_{products}:(1*3R*T)=3RT$.
Thus $\Delta U=U_{products}-U_{rectants}=\dfrac{3}{4}RT>0.$
However, the answer given is $\Delta U$ is $0$.
And the answer also claims that $\Delta H$<$\Delta U$ , which is expected from the relationship $\Delta H$=$\Delta U$+$RT\Delta n$ is not true.
Now I do not know exactly how do we account for "isolated system" in the formula for internal energy.
What exactly does isolated system refer to in this context? And how exactly is it affecting the internal energy?
A possible approach might have to do with the FLOT. I (guess) that isolated system means that there is no heat exchange with the surroundings, so the expression reduces to $dU+P_{ext}dV=0$. Does "isolated system" somehow also imply that $P_{ext}dV=0$?

An isolated system is one that can exchange no mass, heat, or work with the surroundings.  So, from the first law, the change in internal energy of an isolated system in any process within the system is $\Delta U=0$.
Now for my analysis of this problem in greater detail to determine the enthalpy change.  I'm taking as the initial state of the system one mole of CO and 1/2 mole of O2 at 298 K and 1 bar in a rigid adiabatic container, and the final state as 1 mole of reaction product CO2 in this same container at whatever temperature and pressure result when the reaction occurs spontaneously in this same adiabatic rigid container.
My game plan is to examine the internal energy change for the process implemented in 3 successive steps, based on Hess' law to determine the overall internal energy change:

Reactants reacting at constant pressure and removing whatever heat is necessary to hold the temperature constant at 298 K.  So the final state in step 1 is one mole of CO2 product at 298 K and 1 bar, at 2/3 the original volume (since 3/2 moles goes to 1 mole).

Expanding the 1 mole of CO2 isothermally from 2/3 the original volume back up to the full original volume, while decreasing the pressure from 1 bar to 2/3 bar.  So the final state of step 2 is 1 mole of CO2 at 298 K and 2/3 bar pressure.

Adding heat to the 1 mole of CO2 at constant volume to increase its temperature to whatever value is necessary for the overall change in the internal energy for combination of the 3 steps to be equal to zero.

Analysis of internal energy change for step 1:
The process described for step 1 corresponds to the conditions for the standard heat of the reaction CO + 1/2 O2--> CO2 at 298 K and 1 bar, $\Delta H^0=-283000\ Joules$.  So the change in internal energy for this step is the standard change in internal energy $$\Delta U^0=\Delta H^0-RT^0\Delta n=\Delta H^0+\frac{RT^0}{2}=-283000-(8.314)(298)(-1/2)=-282000\ Joules$$
Analysis of internal energy change for step 2:
Step 2 involves an isothermal change for an ideal gas (without chemical reaction), so the internal energy change in step 2 for the system in this step is zero.
Analysis off internal energy change for step 3:
Step 3 involves the constant volume heating of one mole of the ideal gas O2 from $T^0=298 K$ to the final temperature T, so the change in internal energy from this step is $$(\Delta U)_3=C_v(T-T^0)$$ where $C_v$ is the molar heat capacity of CO2.
Overall combined internal energy change
It follows from all this that the overall internal energy change for this combined 3 step process is $$\Delta U=\Delta H^0+\frac{RT^0}{2}+C_v(T-T^0)$$But we know that this overall internal energy change must be equal to zero.  This enables us to calculate the temperature rise of the system in the adiabatic rigid container:  $$(T-T^0)=-\frac{\left(\Delta H^0+\frac{RT^0}{2}\right)}{C_v}$$The corresponding enthalpy change for the combined process is $$\Delta H=\Delta H^0+C_p(T-T^0)$$After some algebraic manipulation, this reduces to $$\Delta H=-(\gamma-1)\left(\Delta H^0+\frac{C_pT^0}{2}\right)$$Substituting literature values for the molar heat capacities and heat capacity ratio into this equation then yields the enthalpy change for the overall process:  $$\Delta H=+83000\ Joules$$ This indicates that answers B and D are both correct.
So, for this problem, we find that $$\Delta H>\Delta U=0$$
The main error you made in analyzing this (difficult) problem was failing to account for the heat of reaction (where the internal energy and enthalpy can change at constant temperature).


Does lower activation energy definitely mean higher rate? Does lower activation energy definitely mean higher rate? For example the decomposition of $\ce{H2O2}$. Can we say it has the highest rate with the catalyst making the reaction's activation energy the lowest?

The lower the energy of the intermediate, the faster the transition from the reactants to the products, at a given temperature.


Has IUPAC been inaccurate in their 1994 definition of kinetic stationary state? Summary of IUPAC definition
$\def\d{\mathrm{d}}$In IUPAC recommendations from 1994[1, 1166–1167], the authors discuss the process
$$\ce{A <-->[$k_1$][$k_{-1}$] X\\ X + C ->[$k_2$] D}$$
A kinetic stationary state or steady state is defined very reasonably:
$$\frac{\d\ce{[X]}}{\d t}=0.\tag2$$
They go on to emphasise that condition $(2)$ is not equivalent to $\ce{[X]}$ being constant, 'not even approximately'; simply that the change in $\ce{[X]}$ is small compared to changes in concentrations of $\ce{[A]}$ and $\ce{[D]}$ because $\ce{[X]}$ itself is small. If $\ce{[X]}$ were constant in a stationary state, they claim it would lead to the following contradiction.

Say reactant $\ce{C}$ is in excess. Assume that $\ce{[X]}$ is constant. Then the law of mass action $v = k_2\ce{[X][C]}$ would reduce to some $v = k'\ce{[C]} \overset{\mathrm{excess}}{\approx} k\in\Bbb{R}$. In other words, production of $\ce{D}$ could continue at a constant rate even after $\ce{A}$
has run out.

The document concludes that $\ce{[X]}$ cannot possibly be constant in steady-state conditions.
How I would resolve the apparent contradiction
I do not see a way around $\ce{[X]}$ being constant. Indeed, for any reasonable function[a] $\ce{[X]}:t'\to \Bbb{R^{0+}}$ equation $(2)$ implies
$$\int_0^t \frac{\d\ce{[X]}(t')}{\d (t')}\d(t')= \ce{[X]}(t) = r \in \Bbb{R^{0+}}\tag3$$
where $t'$ was simply a dummy variable.

Resolution to contradiction: If $\ce{C}$ is in a large excess, $\ce{[X]}$ is small, and $\ce{A}$ has run out, it is simply not possible for steady-state conditions to hold. In other words, it is our stationary state approximation $(2)$ which fails to hold for $\ce{X}$. So we can easily keep condition (3) for a steady state.
More accurately, if $\ce{A}$ has run out, the only possible stationary state would have been $\ce{[X]} = 0$. This is because, at a stationary state,
$$\ce{[X]} = \frac{k_1\ce{[A]}}{k_{-1} + k_2\ce{[C]}} \overset{\ce{[A]} = 0}{=} 0.$$
If $\ce{[X]} = 0$, it is unfeasible for the reaction to continue at non-zero constant velocity since $v = k'\ce{[C]} = 0$ (where $k' = k_2\ce{[X]}$).

Question

Is / is not the concentration of $\ce{X}$ constant, approximately or otherwise, in a kinetic stationary state (steady state)?

IUPAC states $\ce{[X](steady\ state)}$ is not even approximately constant because it would lead to contradictions.
I claim that $\ce{[X](steady\ state)}$ is constant, and contradictions surface when a steady state itself is impossible given certain constraints (or the constraints themselves are contradictory).

[a] Among other things that it is diffentiable everywhere in its domain (i.e., reaction rate is defined for every time $t'$).
[1] Muller, P. Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994). Pure and Applied Chemistry 2009, 66 (5), 1077–1184. DOI: 10.1351/pac199466051077. pages 1166–1167

IUPAC should say they are defining the "steady state approximation" rather than the steady state.
There are two approximations in this technique.
First, that final reaction step is not an equilibrium (reverse reaction rate is zero).
Second, that the intermediate is so unstable that its concentration (and therefore the rate of change of the concentration) is approximately zero.
Compare the IUPAC definition to the following 1954 U.S. National Research Council explanation:

STEADY STATE APPROXIMATION
The most generally applicable of these simple methods is the so-called "steady-state approximation." A steady state may be defined as a condition in which the rates of change of the concentrations of the several intermediates are very small compared to the rates of change of the concentrations of the reactants and products. This condition is realizable whenever the ratio of the concentrations of the intermediates to the concentrations of the reactants is very much less than unity. When this condition is not attained, the method is not applicable; however, it should not then be necessary since the (larger) concentrations of the intermediates could be measured by experimental means. ... The steady-state approximation consists in setting the rates of change of each of the intermediates equal to zero and in solving simultaneously the resulting algebraic equations.

Overall, if the situation was truly, not just approximately, steady state, yes the concentration of the intermediate would be constant, but the point is that the "stead state approximation" is useful even when the concentration of the intermediate, though always near zero in absolute terms, decreases to say half of its original concentration over the course of an experiment (as the starting material, A, decreases to half its original concentration).
Additionally, a very small $\ce{[X]}$ does not purely-mathematically imply a small rate of change of $\ce{[X]}$, because $\ce{[X]}$ could rapidly oscillate within a small, near-zero, range, resulting in brief instances of large rate of change.  Instead, $\ce{[X]}$ being small over a given period of time really just places a limit on how long of a time the rate of change can exceed a certain value.  For example, if $\ce{[X]} < 0.001\ \mathrm{M}$,
$|\mathrm{d}[\ce{X}]/\mathrm{d}t|$ must not exceed $1\ \mathrm{M/s}$ for more than $0.001\ \mathrm{s}$.


Most negative and most positive value for proton chemical shifts What are the most negative and the most positive values for proton chemical shifts recorded till present?

I don't know if the following two examples present the "largest" upfield and downfield proton-nmr chemical shifts, but I suspect they're in the running.  The dihydropyrene dianion example has 16 pi electrons

delocalized around the periphery of the pyrene frame. It fits the 4n rule with n=4, so it is antiaromatic.  The [16]-annulene dianion has 18 pi electrons and fits the 4n+2 rule with n=4, so it is aromatic.  Note how the direction of the ring current reverses between antiaromatic (paramagnetic current) and aromatic (diamagnetic current) systems.  More intesting proton chemical shifts can be found in this compilation.

Edit

Guilty of thinking organic.  If we're including inorganic proton shifts, then how about IrHCl2(PMe(t-Bu)2)2 which has a chemical shift of -50.5!


Ranking gases by value of the van der Waals equation-of-state constant 'a'? How can I find out which gas has the highest value of the van der Waals constant $a$?

Example set of gases in ascending order of $a$: $\ce{NO2}$, $\ce{CO2}$, $\ce{H2}$, $\ce{H2O}$ and $\ce{CCl4}$.

You look up the values of "a" in a table of constants for the van der Waals equation.

\begin{array}{lrr}
   \text{gas}                  &
    a / \mathrm{L^2bar\,mol^{-2}} &
    b / \mathrm{L\,mol^{-1}}      \\\hline
   \text{Hydrogen}             &    0.2476  &         0.02661 \\
   \text{Carbon dioxide}       &    3.640  &         0.04267 \\
   \text{Nitrogen dioxide}     &    5.354   &         0.04424 \\
   \text{Water}             &    5.536  &         0.03049 \\
   \text{Carbon tetrachloride} &  19.7483  &         0.1281  \\
\end{array}


p-d sigma bond - Why not possible? Why is a head on (sigma) bond between a p-orbital and a d-orbital not possible? Is it because extent of overlap becomes too little? Can you please illustrate?

Your premise is incorrect. $\unicode[Times]{x3c3}_\mathrm{p-d}$ are very much possible. Consider any hexahalidometal complex such as $\ce{[FeF6]^3-}$. Since fluoride is essentially unhybridised as iron would be, too, we can separate the atomic orbitals into irreducible representations of the $O_\mathrm{h}$ point group. We find that iron’s d-orbitals transform as $\mathrm{t_{2g} + e_g}$ and the 18 fluoride p-orbitals as $\mathrm{a_{1g} + e_g + 2t_{1u} + t_{1g} + t_{2g} + t_{2u}}$. If we separate out those 12 p-orbitals that are perpendicular to the iron-fluoride axis, the ones that remain are $\mathrm{a_{1g} + e_g + t_{1u}}$. You may immediately realise that $\mathrm{e_g}$ is an irreducible representation of both and thus that the fluorides’ p-orbitals can and will mix with the iron’s $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$.

Professor Klüfers included the calculation results of said ion in his internet scriptum of his coordination chemistry course. The ones of relevance are:

You may be slightly surprised at the shape since you cannot recognise the typical $\mathrm{d}_{z^2}$ shape. But as Brandhorst, Grunenberg and Tamm explain, these two orbital shapes are mathematically equivalent descriptions for the $\mathrm{e_g}$ orbitals.[1]

The molecular orbitals shown are both of antibonding type (formally $\mathrm{e_g^*}$). The corresponding bonding orbitals are much lower in energy and centred on the fluoride ligands. Both of the depicted ones are occupied only by a single electron in α spin.

References

[1] K. Brandhorst, J. Grunenberg, M. Tamm J. Chem. Educ. 2008, 85, 1692. DOI: 10.1021/ed085p1692.


Why does HCN boil at a higher temperature than NH3? The boiling point of ammonia is −33 °C while that of $\ce{HCN}$ is 25 °C. In a recent AP (Advanced Placement) Chemistry test, a free response question asked why this is the case. Can someone shine light on this?

Based on Jan's answer:

Although the $\ce{C-H}$ bond does not usually exhibit good hydrogen bond, $\ce{H-CN}$ is a special case in which the bond is polar enough to provide better hydrogen bonding than $\ce{NH3}$. I agree with Jan on this point.
However, Jan's explanation is for the polarity of the $\ce{H-CN}$ bond is based on $\mathrm pK_\mathrm a$ (lower $\mathrm pK_\mathrm a$ suggests higher acidity and good hydrogen bonding). As one user pointed out, $\mathrm pK_\mathrm a$ is only defined in aqueous solution, so stating pK$_a$ as an evidence is not suitable for the system I have which is a pure $\ce{HCN}$ liquid. In such system, $\ce{HCN}$ does not significantly dissociate. A suggestion was made on using proton affinity as the evidence for the acidity of $\ce{HCN}$.

I perceive that the answers (reasons and evidence) are mostly targeted at the chemical process of dissociation of $\ce{HCN}$. Should this question be addressed more in terms of a physical process? I mean there is little dissociation of pure-liquid $\ce{HCN}$.

What I have been thinking about is a MO picture in which the cyano group withdraws electron density from the hydrogen atom, in a similar way to how acetate group withdraws electron from the hydrogen in acetic acid. What is your thought on this?

The enthalpy of vaporization of $\ce{HCN}$ is higher than for $\ce{NH3}$, which suggests that $\ce{HCN}$ molecules interact more strongly than $\ce{NH3}$ molecules. $\ce{C-H}$ bonds are not usually considered good hydrogen bond donors, but $\ce{HCN}$ is unusual. For example $\ce{HCN}$ has a $\mathrm pK_\mathrm a$ value of 9.2, indicating that the $\ce{CN}$ group is electron withdrawing and that it is a reasonably good hydrogen (bond) donor. This is likely due to the electronegativity of nitrogen, and also the high "s-content" of the sp-hybridized $\ce{CH}$ bond, which keeps the electron pair close to the nucleus.


What's the reason that alkanolamines change their color at storage? What's the reason that alkanolamines (e.g. monoethanolamine, diethanolamine, triethanolamine) of technical purity change their color (from colorless over yellowish to red-brown) at storage?
Is it because of oxidation, or UV-light?

Oxidative degradation by oxygen certainly occurs and has been studied 1, 2. I can't access the papers so I have no information on possible mechanisms. I've seen no mention of UV degradation, but that doesn't mean it does not happen.


Selectivity in aldol condensation between pivaldehyde and acetone In a crossed aldol condensation (condensation between aldehyde and a ketone), the primary products are such, that the carbonyl group of the ketone remains intact, and the $\alpha$-hydrogen of ketone participates, i.e. gets removed by the base to form the resonance stabilised nucleophilic enolate which later attacks the aldehyde. These are formed together with the self-condensation products, but the reverse products (aldehyde-derived enolate + ketone) are rarely formed.

I had a look at the mechanism, and also how a base catalyses condensation by favouring enolate formation.

  Why is the ketone involved in enolate formation and not aldehyde in
  crossed/mixed aldol-condensation reactions?

Ketones have two alkyl groups attached to the carbonyl moiety, which are electron-donating by hyperconjugation. That is, the overlap of $\ce{C-C}$ and $\ce{C-H}$ $\sigma$-bonds with the $\pi$-type molecular orbitals of the carbonyl increases the electron density around the carbon of the carbonyl group, thus making it less electrophilic.

Alternatively, I think it's also equally (if not more) valid to consider the reactivity from the perspective of the frontier orbitals. Alkyl substituents tend to slightly lower the energy of the HOMO of the carbonyl (by comparison to hydrogens), but concomitantly they raise the energy of the $\pi^{*}$ LUMO of the carbonyl, into which the electrons of the enolate would flow, resulting in lower electrophilicity (since the energy of the tetrahedral intermediate and transition state resulting from attack by the enolate are going to be correspondingly higher).


The IUPAC name of sodium nitroprusside, and back-bonding in it In the standard test for the sulfide ion, sodium nitroprusside $(\ce{Na^_2[Fe(CN)5(NO)]})$ is used. The IUPAC name for this compound as stated by Wikipedia is sodium pentacyanidonitrosylferrate(II)

I have three queries related to this:

Does the nitrosyl ligand donate a pair of electrons (through back bonding) to form $\ce{(NO)^+}$ or is $\ce{(NO)^+}$ the ligand?
In both the cases I guess the name should be sodium pentacyanidonitrosoniumferrate(II) instead.
Now the product formed after the sulfide test is $\ce{Na^_4[Fe(CN)5(NO)S]}$ whose IUPAC name is sodium pentacyanidonitrosylsulfidoferrate(III). But the nitrosyl ligand is neutral in this case. So why in this case there is no back bonding?

Nitrosyl is the textbook example for a non-innocent ligand — that is, one that can appear in different oxidation states with different numbers of donated electrons and can make determining the bonding situation rather complicated. It can appear as the nitrosyl cation $\ce{NO+}$ (isosteric to $\ce{CO}$ or $\ce{CN-}$), the nitrous oxide radical $\ce{NO^{.}}$ or the ‘hyponitrite’ anion $\ce{NO-}$ (isosteric to either $\ce{{}^1O2}$ or $\ce{{}^3O2}$). I’m not going to start talking about possible donated electron counts yet.

Nomenclature

For nomenclature purposes, any form of $\ce{NO}$ bound to a metal is always treated as a neutral three electron donor and is always termed nitrosyl. This is because IUPAC wants rules that can be applied blindly without having to perform complicated experiments to determine how exactly the ligand is bound. And we shall see in a second that it is very non-trivial.

Considering that and checking the $\ce{[Fe(CN)5NO]^2-}$ fragment, we take away 5 $\ce{CN-}$ leaving us with $\ce{[Fe(...)NO]^3+}$ which indicates that the correct IUPAC name for sodium nitroprusside is sodium pentacyanidonitrosylferrate(III). (Not (II).)

The product formed would be, by the same approach, sodium pentacyanidonitrosylsulfidoferrate(III). But we’ll come back to that.

But what is the actual ligand?

As I already stated, nitrosyl is a non-innocent ligand. It can be $\ce{NO+}$ and allow significant π-backbonding (metal to ligand) as exemplified in the tricarbonylnitrosylferrate($\mathrm{-I}$) system $\ce{[Fe(CO)3NO]-}$. By formality, this is considered a neutral donor giving iron a formal oxidation state of $\mathrm{-I}$. But in reality the ligand is $\ce{NO+}$ meaning that we are dealing with iron($\mathrm{-II}$) — analogous and isoelectronic to the tetracarbonylferrate($\mathrm{-II}$) complex.

On the other side of the spectrum we have the complex formed in one nitrate detection method: $\ce{[Fe(H2O)5NO]^2+}$ — pentaaquanitrosyliron(II) according to IUPAC. Approaching this complex is difficult, however most evidence points towards some structure between $\ce{[Fe^{II}NO^{.}]^2+}$ and $\ce{[Fe^{III}NO- ]^2+}$. The $\ce{Fe-N-O}$ unit is still linear (VSEPR would predict $120^\circ$ for  a hyponitrite). This can be explained by considering the $\ce{NO-}$ (or $\ce{NO^{.}}$) in a triplet state much like $\ce{^3O2}$ (or a corresponding dublet state in case of $\ce{NO^{.}}$). There is a slight tendency for the unpaired electrons in iron and $\ce{NO}$ to couple; this is usually understood as antiferromagnetic coupling and results in an overall observed spin of $+\frac{3}{2}$.

In the $\ce{NO+}$ case, the bonding picture is rather clear: $\ce{NO+}$ behaves exactly like $\ce{CO}$. We have a σ donating (forwards) bond and significant π back bonding. The other case is rather hazy except for the σ bonding, which is of course ligand to metal. As I wrote, the spin coupling is considered antiferromagnetic coupling and not actual bonding usually, and if anything, it would be a covalent type of bond (each fragment donating one electron). Thus it would be wrong to speak about back-bonding — especially since metals in high oxidation states usually don’t back-bond significantly.

Removing the five cyanido ligands in nitroprusside leaves us with a $+3$ charge as shown above. Thus we are either dealing with $\ce{[Fe^{II}NO+]}$ or $\ce{[Fe^{III}NO^{.}]}$. The extremely short $\ce{N#O}$ distance points strongly towards an $\ce{NO+}$ cation. $\ce{Fe^{II}}$ is again not good at back-bonding and no π forward bonding is to be expected due to the mismatched energy of the π electrons of $\ce{NO}$. Therefore, the $\ce{NO}$ group is rather labile and can be displaced at ambient temperature and pH. This is the mode of action of nitroprusside as a drug: It releases $\ce{NO}$.

And what about the sulfide test?

I didn’t dig into it deeply, but it seems like you are mistaken in the product formed. What I found on the web[1] implies (I couldn’t read the actual article, only the abstract) that sulfur does not coordinate additionally to the iron centre (there would not be much space anyway) but rather attacs the $\ce{N#O}$ triple bond nucleophilicly on the nitrogen atom. Therefore, the final complex is better written as $\ce{[Fe(CN)5(SNO{-}\kappa N)]^4-}$ and better named pentacyanidothionitrito-κN-ferrate(III).

References

[1] S. L. Quiroga, A. E. Almaraz, V. T. Amorebieta, L. L. Perissinotti, J. A. Olabe, Chem. Eur. J. 2011, 17, 4145. DOI: 10.1002/chem.201002322

Background from Prof. Klüfers’ internet scriptum to his coordination chemistry II course.


Final amount of Hg present after $\beta$ decay I have been trying to solve an exercise on nuclear chemistry, but I was not sure whether I should ask this on physics or chemistry but I think that if chemistry had an exercise based on this then chemistry experts might be having the way to overcome this problem. So here is what my nuclear chemistry question says:

  One gram atom of $ _{79}^{198}\mathrm {Au} $ (having a half life of 65
  hours) decays by $\beta$-emission to produce stable nuclide of Hg.
  How much Hg will be present after 260 hours?

The thing which is making me confused is $\beta$ decay. I think I have a formula for this question but that $\beta$ thing is creating a problem. I would like to make that equation public. Here it is:

$$N_t=N_0\cdot\exp(-\lambda\cdot t)$$

where,

$\lambda$ is the radioactive decay constant
$N_0$ is initial number of nuclides present
$N_t$ is the co. of parent nuclides present at time $t$

$$\ce{^198Au -> ^198Hg + e^{-}  ~~~~~~~~ \beta~ decay}$$

260 hours corresponds to 4 half-lifes (4 x 65 = 260).  After each half-life, one-half of the $\ce{^198Au}$ will have been transformed into $\ce{^198Hg}$.  After 4 half-lifes ((1/2)^4 = 1/16) 1/16 of the gold will remain and 15/16 of the gold will have been transformed into mercury. Therefor, after 260 hours 15/16 gram atom of mercury will be present.  15/16 gram atom of mercury corresponds to (15/16 x 198 = 185.63 gm) 185.63 gm of mercury.

Using your equation we (of course) get the same result.
$$\ce{N_t=N_o * e^{-kt}~~(I)}$$ where k is the rate constant and t the elapsed time.  For radioactive decay (a first order reaction)
$$\ce{k=\frac{ln(2)}{t_{1/2}}~~(2)}$$ where $\ce{t_{1/2}}$ is the decay half-life.  Substituting into equation (I) we find
$$\ce{\frac{N_t}{N_o}= e^{-\frac{ln(2)}{t_{1/2}}*4}=1/16}$$


Why can't the energy of an electron exceed 0 eV? Today we were learning about atomic structure, our teacher introduced the mathematics of the 'Niels Bohr Model of Atom', and calculated the value of total energy $\frac{-13.6z^2}{n^2} \pu{eV}$. He explained that when the value of $n$ tends to infinity, the value of total energy tends to $0$. Mathematically I understood this.
I couldn't understand, or comprehend, how or why the energy of an electron can't exceed $\pu{0 eV}$? What is the reasoning behind it; why is it not possible and what is the physical significance of it.
For the sake of the argument, what would it look like if the total energy of an electron were to exceed $\pu{0 eV}$?

The value of the energy in the Bohr model is zero when the quantum number is infinity  because that is the limiting value of the Coulombic potential at large distances, and because the electron is assumed bound to the nucleus (the atom is stable), which constrains the value of the total energy.
The energy of a stationary hydrogen-like atom is described by the equation
$$ U(Z,n) = R_\mathrm{E} \frac{Z^2}{n^2} $$
where $R_\mathrm{E} = \pu{-13.6 eV}$. It can equally well be written as
$$ U(Z,n) = \frac{-Zk_e e^2}{2r_n} \tag{1}$$
where $r_n$ is the allowed orbit radius in the Bohr model, which illustrates how the energy becomes more positive (goes to zero) as the average distance between the nucleus and the electron increases.
The Bohr model assumes the universe consists of just one electron and nucleus. It is implicit in the model that the electron is bound (the electron and nucleus can never get far enough apart to escape their mutual attraction). According to the virial theorem the total energy in the above equations is one half of the Coulombic potential energy and the negative of the kinetic energy. That means the expectation values of the potential and kinetic energy for a given quantum number are fixed, and that the value of the total energy is bounded by the value of the potential energy (see Eq. 1):
$$U(Z,n)=\frac{\left<U_{\mathrm{Coulomb}}\right>}{2}$$
The zero limiting value of the Coulombic potential can be interpreted as indicating that the interaction between the particles vanishes at large distances. On the other hand, the zero limiting value of the total energy can be interpreted as indicating that the reference frame of the atom is at rest. If the atom is assumed to have kinetic energy than this can be represented as a non-zero  offset of the total energy. In that case the energy converges to this threshold as the particles move farther apart.
In addition, in practice external fields might interact with the atom, contributing to the potential. It is possible for instance to superimpose a uniform linear potential on the Coulombic potential. The solutions (wavefunctions) for the new perturbed atom are not identical to those of the original isolated atom. Suffice it to say that if you now pull the particles apart far enough  they will eventually be free to follow trajectories that are no longer bound.
The following figures illustrate (roughly) the energies of an atom (for Z=1) before and after superimposing an external linear potential, plotted against the quantum number or the orbit radius in the original Bohr model. In the case of particles starting at rest, what determines whether particles will remain bound or fall apart is relative values of the potential. It is the slope of the potential that determines forces that drive particles together or apart. In the last figure, the (slight) downward slope at large r as the radius increases indicates that particles at rest starting at this internuclear distance are no longer bound.
Hydrogenic atom in Coulombic potential...

... and perturbed by linear potential

An electron can also be excited by a photon into the continuum of non-quantized states with energy greater than zero, but then the energy is no longer described by the Bohr model. The excess energy (greater than the ionization energy) amounts to the kinetic energy of the dissociated (nonbound) state (which is alluded to in the other answer).
You might also want to read these other posts.


Why isn't pure ortho hydrogen possible to obtain? In school, I've learned that we can only get an ortho hydrogen content upto 75%. Why is that we can obtain a pure sample of para hydrogen, but not ortho hydrogen? What prevents us from isolating ortho hydrogen?

Ortho and para hydrogen differ in the alignment of their nuclear spins. The two hydrogen atoms each have a nuclear spin of 1/2 which results in a total nuclear spin of 0 or 1. The states with zero spin are called para and those with 1 ortho. The combination of the nuclear spins also results in the formation of a new spin wave function. There are 3 possibilities to get a nuclear spin of one with an even wave function and there is one possibilitie to get a spin of zero with an odd wave function. The overall wave function of the hydrogen molecule is given by the product of the electronic, vibrational, rotational and spin wave functions and should we antisymmetric with respect to exchange of the fermionic hydrogen nuclei. Because the electronic ground state of molecular hydrogen is even and the vibrational wave function of a diatomic molecule is always even, the wave function of para hydrogen has to be combined with the even rotational states and ortho with the odd rotational states. The absolute ground state of molecular hydrogen is thus the rotational state with zero rotation and is therefore para. The next state, with one quantum of rotation, is ortho.

You can only convert ortho and para hydrogen into each other using collisions or chemical reactions. To make pure para hydrogen, people use a catalyst at very low temperature. Because of the low temperature, all produced hydrogen will eventually end up in the absolute ground state, that is, para hydrogen. You cannot make ortho hydrogen this way because it does not represent the absolute ground state of the system. The 75% comes from the degeneracy of the Boltzmann factor: at low temperature (with no reaction catalyst) there are three times more ortho molecules with a spin of 1 than para molecules of zero spin.

Although you cannot make pure ortho hydrogen, you can study it by using narrow lasers that only excite odd rotational states.


How can pKas differ across similar protons of the same atom? For example, citric acid has three acidic protons, all of which are carboxylic acids. Despite being part of the same functional groups, they all have very different pKas. Why (and how?) is this?

Here is a picture of citric acid.

As you noted, there are 3 carboxylic acid groups in citric acid, so there are 3 acidic protons. A chemist looking at this unionized molecule would say that the carboxylic acid group drawn on the right side of the molecule is equivalent to (the same as) the carboxylic acid group drawn on the left.  Therefore, these 2 equivalent carboxylic acid groups should have the same pKa.  The carboxylic acid group drawn in the middle of the diagram is different from the other two (it is not the same as the other two) and should therefore have a different $\ce{pKa}$.  In chemistry, groups are either the same or different.  If they are the same they must be indistinguishable in all aspects, including physical properties like $\ce{pKa}$

That said, we might expect when we treat citric acid with more and more base to see only 2 different $\ce{pKa}$'s, but that is not the case.  In actuality when we treat citric acid with base one of the two different (right and left or middle) carboxylic acid groups will ionize.  Once that happens the ionized citric acid molecule contains a negative charge.  It will be harder to ionize the molecule again and remove the second proton and place a second negative charge on the molecule (bringing like charges together is destabilizing and costs energy).  The same argument tells us that removing the third and final proton will be even more difficult because we wind up with 3 negative charges on the molecule.  As a consequence of this increasing difficulty to remove protons, we observe 3 different $\ce{pKa}$'s.


Amine vs Amide Solubility I had a question regarding the solubility of amines and amides.
I was looking into the solubility of butanamide and n-Butylamine, and it turns out that whilst butylamine is miscible in water, the solubility for butanamide is 163g/L
This would imply that amines have a greater solubility than amides. My understanding is that
the carbonyl group would add two lone pairs, which act as hydrogen bond acceptors and thus should mean amides are more soluble. (more hydrogen bonds)
It'd be much appreciate if someone could point me in the right direction or let me know how my reasoning is flawed.
(I have viewed this link but the answer given isn't too helpful: Solubility of Amides)
Thanks in advance!
Sources for Solubility Values
https://en.wikipedia.org/wiki/N-Butylamine
https://chem.libretexts.org/Under_Construction/Walker/Chemicals/Substance%3AB/Butanamide

The mixing of two compounds is a process which requires consideration of three types of interactions: solute-solvent, solvent-solvent and solute-solute. You make a good argument for the solute-solvent interaction being stronger in mixtures of butanamide and water, compared to butylamine and water. However, as you say, this argument alone would predict a higher solubility for butanamide, which goes against the observed data. Also, in both cases the solvent-solvent interaction is the same (water with water), so this can't really be the source of an explanation.
By exclusion, we're guided towards considering the solute-solute interactions. At a first glance, both butylamine and butanamide display a combination of dipole-dipole and hydrogen bonding interactions, making it tricky to compare how their relative strengths change between compounds.
To make things easier, we can refer to additional experimental data. A nice way to gauge the strength of intermolecular interactions in a substance is to look at its melting and boiling points. According to Wikipedia, the melting and boiling points of butylamine are approximately -49 °C and 78 °C respectively. Meanwhile, for butanamide, the values are a whopping 115 °C and 216 °C! Clearly the mutual interactions between butanamide molecules are much stronger than those between butylamine molecules.
And there is our answer. You are correct that, compared to a butylamine molecule, a butanamide molecule should interact more strongly with water molecules. However, it turns out that butanamide molecules also interact very strongly with themselves, far more so than butylamine molecules. Taking everything into account, the effect of the solute-solute interactions is more prevalent; the water molecules have more difficulty keeping butanamide molecules apart, to the point that after a critical value of 163 grams of butanamide in a liter of aqueous solution (at 15 °C), the water molecules just can't stop the butanamide molecules from meeting and packing into a solid. Even if you add more butanamide, it doesn't visibly dissolve. Some butanamide molecules in the solid continuously manage to break loose from each other and get pulled into the aqueous solution, but they are replaced just as fast by molecules of butanamide already in solution which get too close and pack into the solid state.
You will find that this can be made into a very rough rule of thumb: when comparing the solubility of two roughly similar substances in a same solvent, if one of the substances has significantly higher melting and boiling points, they are likely to be less soluble.
Note that I have discussed only the influence of the enthalpic contributions to mixing. An entropic contribution also exists (favouring dissolution for both compounds), but in this case it can be safely neglected due to its comparatively small effect.


Reaction of phenol with Zinc dust I have seen this reaction many times:

  $\text {Phenol}+\text{Zn}\ce {->}\text{Benzene}+\text{ZnO} $

But what is the mechanism?! Tried a lot but couldn't cook up one...

Thanks for any help...

The net result is $\ce{C6H5OH + Zn -> ZnO + C6H5H}$, but under the reaction conditions (zinc DUST and strong heating (b.p. of phenol $= \pu{182 ^\circ C}$)), I suspect the reaction mechanism involves the zinc surface rather than individual $\ce{Zn}$ atoms or $\ce{Zn^{2+}}$ ions. And the zinc surface will be a complex oxide. I have read that the reaction is not clean and produces several byproducts, typical of high temperature reactions.

The acidity of phenol is not great enough to dissolve zinc (at least, not easily), but it could adsorb on the zinc/zinc oxide surface. First step: probably a phenolation of an oxygen to phenolate and hydroxide:

$\ce{Zn - Zn - ZnO - Zn - Zn- (surface) -> Zn - Zn - Zn (OH, OC6H5) - Zn - Zn}$.

At high temperature, the $\ce{H}$ on the hydroxyl could reach the phenyl group in a five-membered ring and go off as benzene ($\ce{C6H5H}$), leaving $\ce{Zn - Zn - ZnO - ZnO - Zn}$.

Magnesium and zinc have some similarities (ionic radius, oxide coat), but magnesium is significantly more reactive. I suspect that magnesium in phenol would react at lower temperatures and give magnesium phenolate and hydrogen: $\ce{Mg + 2 C6H5OH -> Mg(OC6H5)2 + H2}$, no benzene. In this case, the acidity of phenol would be enough to dissolve the magnesium; fine powder (pyrophoric) would not be needed.


Silver mirror and copper mirror There is a formation of silver mirror in the silver mirror test. why does the silver does not precipitate, it forms the mirror in the silver mirror test? Is there any test that copper can form mirror? Why silver can form silver mirror but copper cannot?

A copper mirror can be obtained by heating two or three grams of dry copper formate $\ce{Cu(HCOO)2}$ in a test tube with a Bunsen burner. For some reason, the copper atoms are deposited on the inner surface of the test tube. Try it! It is easy to do!


How to calculate the rate of reaction of Sodium and Chlorine $$\ce{2Na(s) + Cl2(g)   ->    2NaCl(s)}$$
How exactly do you calculate the reaction rate of the following equation?

Since reaction rate is usually calculated in moles per $\rm dm^3$ per second (M/s) and the reactants here are solid. I want to have a equation of $$R = K \times\ce{[Na]}^x\ce{[Cl]}^y$$

Rate laws are determined via experimental data. Reactant concentrations are held constant while others are modified in order to determine the effect of each reactant's concentration on a reaction. For this reason, one cannot determine the rate law from just a stoichiometric equation.
$$\ce{2Na + Cl2 -> 2NaCl}$$

We can propose reaction mechanisms to make an educated guess of the rate law:
\begin{align}
\ce{Na + Cl2 &-> NaCl + Cl}&&\text{(slow step)}\\
\ce{Na + Cl &-> NaCl}&&\text{(fast step)}\\
\end{align}

Using Hess' law, we can combine these to make sure it equals the original equation.
$$\require{cancel}
\begin{align}
\ce{  && Na + Cl2 &-> NaCl + \cancel{Cl}}\\
\ce{+ && Na + \cancel{Cl} &-> NaCl} \\\hline
\ce{  && 2Na + Cl2 &-> 2NaCl} \\
\end{align}$$

The coefficients of the slowest elementary step in reaction mechanism determine the exponents of the rate law. Hence:
$$\text{Rate}=k[\ce{Na}]^{1}[\ce{Cl2}]^{1}$$

Or just
$$\text{Rate}=k[\ce{Na}][\ce{Cl2}].$$


Why is the change in enthalpy computation for a change of state done like this? Suppose that a system whose initial state is Pressure = $P_1$, Volume = $V_1$; and it is taken through a process after which final pressure and volume are $P_2$ and $V_2$ respectively. Now change in enthalpy is defined as $\Delta H= \Delta U +P\Delta V+V\Delta P$The thing which I am confused with is which pressure and which volume do we hold constant for their respective parts in the calculation? Till now I had dealt only with problems involving the $P\Delta V$ part, and it was calculated as $P_2(V_2-V_1)$, same as irreversible work. But when I came across a problem in which Pressure also varied with volume, the second part needed to be taken as $V_1(P_2-P_1)$.Why do we take $P_2$ as constant in one case and $V_1$ in the other?One reason which I thought of was that in an irreversible process, The pressure is first changed by a finite amount and then the volume varies. Thus the pressure throughout the process was $P_2$. This can be a rationale for the $P\Delta V$ part, but I am not sure if it is reasonable to say that the volume at start of the process was $V_1$, which resulted in this method of calculation. So.. what is the reason?

You're confused because the equation is incorrect.  It should read $$\Delta H=\Delta U+\Delta (PV)$$If you insist on writing it the way you did, then P should be $\frac{P_1+P_2}{2}$ and V should be $\frac{V_1+V_2}{2}$.  That will give the same result as the first equation I wrote.


Are there any ionic amorphous solids? This question on NaCl crystalization actually got me wondering: are there any ionic amorphous solids? Like ionic crystals are crystalline materials of electrostatically-attracted ions, can ions form an amorphous phase? I can see no reason why not, but I cannot think of any example either…

The difficulty with creating an ionic bond glass is that creating true glasses depends on a combination of directional bonding between atoms (or molecules) and a rate of cooling that doesn't give those bonds time to move into optimal positions. The result is a bit like piling together sticky spheres that become locked before they can find an optimal stacking arrangement.

Pure ionic bonds in contrast are more like slippery spheres that are attracted towards each others' centers, but don't have any particular preferences about which parts of their surfaces make contact.

Given that, the most likely result by far from very rapid cooling of a molten ionic solid will be to produce micro or nanocrystalline ionic solids that may superficially resemble glasses, but which under X-ray or neutron diffraction crystallography would prove actually to have crystal structure at a very fine scale.

With that said, you nonetheless can come up with a recipe for trying to make a truly glassy ionic solid, especially since there's no such thing as a pure ionic bond. You would want to cool a very thin layer of molten salt at at an incredibly high rate, at least millions of degrees per second if I recall glassy metal cooling rates rightly. For ionic, you would also want the result to be pretty close to absolute zero, since whatever degree of "directional stickiness" you can get out of ionic solids is likely to be less even than metallic bonds.

Finally, if you really want to try to make such a thing as a serious project, you of course must check the literature in detail. My rule of thumb is that if you dig deep enough, there are almost no decently plausible ideas that you can think of in the physical sciences that someone has not at least written a paper on in some journal, and maybe even done some experiments.

Still, what you are asking is an intriguing idea for a real experiment, and it wasn't that many years ago that both metallic glasses and Penrose-tiled quasicrystals were new ideas.


How to take a sample of a hot salt solution without it precipitating in the cooler glassware due to the difference in solubilities at different temps? I'm doing an investigation to see the effect of temperature on the solubility product of strontium hydroxide.

I made a supersaturated solution left in the oven overnight at 85°C for equilibrium to be reached. Now, I want to titrate it against $\ce{HCl}$, but when I try to pipette a 25 mL sample of the (very hot) solution, it quickly crystallises in the pipette due to the large difference in solubility at 85°C compared to room temperature.

(NOTE: doing at lower temperature is not an option.... )

One idea I can think of is to keep the glassware in the oven overnight with the solution so that it's more or less at the same temperature. (Then, I'd have to handle with a cloth, of course). However, I do not have access to an oven that is big enough to fit a pipette.

Hence, the only option I have is to use a cylinder rather than a pipette, but that would increase my uncertainties by A LOT.

Can anyone think of anything else that I could do?
Would really appreciate any help.

Just use your room temperature volumetric pipette, and wash out the precipitate with water. The exact density at 85°C is going to be tricky anyway. The expansion coefficient of borosilicate you can find, and you need to measure or calculate (heat capacity of borosilicate, and your solution?!? IR thermometer?) the temperature of the solution.

You might think about using a second volumitric pipette with dest water, and wash everything into a beaker with known empty weight.

Have fun with calculating the uncertainties! ;-)

Afterthought:
Fill the RT volumetric pipette from the hot solution. Wait until cooling (the filling should get lower, because water has a larger expansion coeff. than glass) and precipitation is complete. Now completely fill the pipette, the additional amount of solution will be rather small, and you don't need to wait again for quillibration.

Now weigh and titrate the solution&precipitate (after washing out with a known volume of water). You know the density and molarity of cold strontium hydroxide solution, you know the density of solid hydroxide. I think that is sufficient information to calculate the concentration at 85°C.


Why apart from when building atoms in the first place, the 3d is the lower energy orbital? From chemguide.co.uk:

  If the phosphorus is going to form $\ce{PCl5}$ it has first to generate 5 unpaired electrons. It does this by promoting one of the electrons in the $\mathrm{3s}$ orbital to the next available higher energy orbital. Which higher energy orbital? It uses one of the $\mathrm{3d}$ orbitals. You might have expected it to use the $\mathrm{4s}$ orbital because this is the orbital that fills before the $\mathrm{3s}$ when atoms are being built from scratch. Not so! Apart from when you are building the atoms in the first place, the $\mathrm{3s}$ always counts as the lower energy orbital."

Could you please explain why this is so?

In P element, the energy of 3d is lower than 4s. Look at the link provided below. Only in K and Ca, 4s has lower energy than 3d.

http://www.cdeep.iitb.ac.in/nptel/Core%20Science/Engineering%20Chemistry%201/Slide/lect6/6_6.htm

Another reason could be because the symmetry of hybridized orbital with d orbital will be more preferred in bonding with that of s orbital, and the molecule obtain more energy compensation with sp3d over sp3s.


Can a magnesium atom combine with just one chlorine atom? If I have a magnesium atom and a chlorine atom why can't the magnesium atom combine with just one chlorine atom? Wouldn't the magnesium atom become more stable, than it was before combining, as it gets closer to octet after combining with one chlorine atom?

In solid state $\ce{MgCl2}$, can be approximately described assuming fully ionic bond, in which $\ce{Mg}$ atoms lose two electrons to form $\ce{Mg^2+}$, while a $\ce{Cl}$ atoms gain one electron to form $\ce{Cl-}$. To balance the two electrons released by $\ce{Mg}$, you need two chlorine atoms. The electrostatic force of attraction (aka the "ionic bond") binds them, and it has the formula $\ce{MgCl2}$.

Now, you wish to form $\ce{MgCl}$. To maintain charge neutrality, you'll want the $\ce{Mg}$ to lose only one electron. This means the valence shell of $\ce{Mg}$ would still have a single electron left in it. This is a very unstable configuration.

However, it is incorrect to say that $\ce{MgCl}$ cannot form. A research paper1 has explored the very $\ce{MgCl}$ molecule. Note that the conditions were very drastic, $\pu{500^\circ C}$ for pyrolysis and $\pu{2500^\circ C}$ for vaporization respectively. They ensured "the complete formation of $\ce{MgCl}$ molecular species, since $\ce{Cl}$ was the limiting reactant" as you had correctly identified yourself.

Another paper3 describes the formation of the $\ce{MgCl}$ molecule:

  Excited $\ce{MgCl}$ molecules were produced by two different techniques: (i) at Orsay, they were created by mixing Mg vapor with a flow of $\ce{He/Cl2}$ and excited in a “heated” Schuller-type discharge tube, and (ii) at Waterloo, they were generated by using a copper hollow cathode lamp loaded with $\ce{MgCl2}$ powder.

Note that these techniques are indeed quite "unnatural", in the sense that you might not find $\ce{MgCl}$ naturally, but you can certainly create it through artificial conditions. In fact, $\ce{MgCl}$ is fairly popular otherwise. Several papers2 have already discussed the emission spectrum of the $\ce{MgCl}$ molecule. Some4 also explore its laser cooling. Hence, by no means it is an unknown compound and in fact, can definitely be formed.

References:

Rina Lourena da S. Medeiros, Sidnei de Oliveira Souza, Rennan Geovanny Oliveira Araujo, Djalma Ribeiro da Silva and Tatiane de A. Maranhão, Chlorine determination via MgCl molecule in environmental samples
using high resolution continuum source graphite furnace molecular absorption
spectrometry, Talanta, http://dx.doi.org/10.1016/j.talanta.2017.08.026
Darji, A.B., Shah, N.R., Shah, P.M. et al. Pramana - J Phys (1985) 25: 571. https://doi.org/10.1007/BF02847235
Vervloet, M. Fourier Transform Spectroscopy 2001. https://doi.org/10.1006/jmsp.2002.8514
The Journal of Chemical Physics 143, 024302 (2015); https://doi.org/10.1063/1.4926389


Vibration motion at the imaginary frequency for TS of the Claisen rearrangment I understand that the vibration along the reaction coordinate is different for the transition state , as the motion of the atoms takes TS towards products/reactants.

However, what would be the vibrational motion corresponding to the imaginary frequency of the Claisen rearrangement?


Mechanism of formation of peak of mass spectrum of isobutane at m/z = 27 Here is the mass spectrum of isobutane:

(source: nist.gov)
Clearly the peak at $\mathrm{m/z} = 27$ corresponds to the species $\ce{C2H3+}$, the "protonated acetylene".
I am interested in possible formations for its formation from the molecular ion $\ce{t-BuH+}$.

Some of the first mass spectrometrists to see this ion were Stevenson and Hipple in 1942.1  They were puzzled by it.
Fifteen years later, Rosenstock and Melton explained the source of this ion in an exhaustive study.2  First, the molecular ion $\ce{C4H10+}$ dissociates into fragments via five possible pathways, each with its own activation energy.  See Table VIII in their paper, which I reproduce below.

None of these primary dissociation pathways gives rise to $\ce{C2H3+}$.  Instead, it arises from a variety of processes from a subset (the metastable subset) of these five primary transitions.  These processes are shown in Tables X and XI of their paper.

All of the reactions that give rise to $\ce{C2H3+}$ are probably happening to some extent, but the lowest-energy route there for isobutane (right column) appears to be via $\ce{C2H5+ -> C2H3+ + H2}$ (only 1.9 eV).  However, $\ce{C2H5+}$ itself requires more energy to form than $\ce{C3H7+}$ ($m/z=43$), so the route $\ce{C3H7+ -> C2H3+ + CH4}$ may also occur as well.
The general conclusions of their paper are worth quoting:

The basic concept of the quasi-equilibrium theory is
that mass spectra are formed by a series of competing
consecutive unimolecular decomposition of excited ions.
The results discussed above lend strong support to this
concept. It was found that of several possible reactions
of a given ion, the metastable transition was associated
with the reaction having the least activation energy.
The general rate equation appropriate for these systems
is the product of a frequency factor and a function of
the excitation energy $E$ and activation energy $\epsilon$. When
the value of $E$ is such that the rate constant is of the
order of $10^6~\mathrm{sec^{-1}}$, the rate constant is extremely
sensitive to the value of $\epsilon$. The result is that a second
reaction having an $\epsilon$ higher by more than a few tenths of
an eV will have a considerably smaller rate constant,
even if the frequency factor is larger by a factor of a
thousand. Thus, in general, the correlation can be
carried out in terms of activation energies alone.

References

D. P. Stevenson, J. A. Hipple, Jr., J. Am. Chem. Soc. 1942, 64 (7),  1588–1594.
H. M. Rosenstock, C. E. Melton, J. Chem. Phys. 1957, 26 (2), 314-322.


What is the difference between crystal structures of proteins, organic and inorganic materials? From the definition of crystal and the main differences between crystalline and amorphous material, it is known that crystal formation requires ordered bonding between atoms or molecules.

How can the protein with its irregular structure (when compared to organic and inorganic materials) make ordered bonding, since I often read about the crystal structure of the a protein and its ligand or inhibitor obtained by X-ray?

And is there any difference in crystal structure of an organic and inorganic material?

Imagine an infantry unit of soldiers marching in file. Each of them may be quite irregular, but together they form a repeating pattern. And that's exactly what happens with protein molecules in a crystal.

When we say that a protein is irregular, we mean it on a different level. Indeed, one molecule of our protein is irregular if we are talking about the amino acid residues in it, of which there are probably a hundred: here goes Ala, then Gly, then all of a sudden Leu - what will come next? You can never tell. But wait, there is another molecule of the same protein sitting nearby, and it looks exactly the same as this one: Ala, Gly, Leu and so on. And there is yet another next to it, also similar, and many more of them, and together they form an ordered crystal structure which makes it possible for us to diffract X-rays on it and measure the intensities and then do some hardcore math and voila, here is your structure.


Chloroantimonic vs fluoroantimonic acid Does chloroantimonic acid exists? What do we get when we mix hydrochloric acid and antimony pentachloride and is the mixture stronger or weaker than fluoriantimonic acid. How about bromo and iodoantimonic acids?

Chlroantimonic acid has been synthesised by dissolving antimony trioxide in concentrated hydrochloric acid followed by saturation with chlorine. The relevant information can be found here[1]:

Complex compounds are known to which the formulae $\ce{HSbCl6}$, $\ce{H2SbCl7 }$ and
$\ce{H3SbCl8}$ have been ascribed; these may be regarded as meta-, pyro- and
ortho-chlorantimonic acids [...] Neither the ortho- nor the
pyro-chlorantimonic acid appears to have been isolated, although salts
are known. Meta-chlorantimonic acid has been obtained by dissolving
antimony trioxide in concentrated hydrochloric acid and saturating the
solution with chlorine. The solution darkens at first, afterwards
changing to a bright greenish-yellow colour. Crystals can be obtained
by concentrating slightly over a water-bath, adding hydrochloric acid
and leaving over sulphuric acid, the temperature not
exceeing 0° C. The crystals are very hygroscopic; they are soluble in
cold water, alcohol, acetone and glacial acetic acid. The solution in
water undergoes hydrolysis, especially on warming, hydrated antimony
pentoxide separating out. It is prevented by
hydrochloric acid and to some extent by nitric acid. The solutions in
organic solvents are more stable. Various salts of chloroantimonic acid are known [...]

Bromoantimonic acid is prepared from crystallization from antimony bromide/bromine and sulfuric acid. From Here[2]:

Of the three possible acids only one has been obtained in the free
state. By crystallisation from a solution containing antimony
tri-bromide, bromine and either hydrobromic or sulphuric acid the
compound meta-bromoantimonic acid, $\ce{HSbBr6.3H2O}$, has been
prepared in the form of hygroscopic, irregular, six-sided, black
tablets. It decomposes readily, with evolution of bromine, leaving a
residue of antimony tribromide.

Pure iodoantimonic acid is not known although a mixed acid, $\ce{HSbBrI3}$ is known. Here[3]:

Salts of a complex antimony iodohydrobromic acid, $\ce{HSbBrI3}$, have
been obtained by triturating equimolecular proportions of antimony
triiodide and metallic bromide with a non-aqueous substance such as
acetic acid or xylene.

Carefully note that in this acid, antimony is in the +3 rather than +5 oxidation state.  Iodide ion, more easily oxidized than lighter halide ions, would accordingly be less stable in combination with antimony(V).
References

http://antimony.atomistry.com/chloroantimonic_acids.html

http://antimony.atomistry.com/antimony_pentabromide.html

http://antimony.atomistry.com/antimony_triiodide.html

$$\ce{Sb2O5 + 12HCl → 2H[SbCl6] + 5H2O}$$
http://www.easychem.org/en/subst-ref/?id=3722


How to fluorinate polyethylene? I seek to put a layer of fluorine on a rotationally molded cross-linked polyethylene (XPE) plastic gasoline tank. I sell tanks and they will be more eco-friendly if I can accomplish this.

I am an electrical engineer and my last experience with chemistry was Chem 200 lecture and lab at SDSU, a one semester intensive "all of chemistry for engineers" kind of course. I asked a chemistry professor at my alma mater about this, who said:

  Xenon difluoride is a powerful fluorinating agent, and is soluble
  (according to Wikipedia) in anhydrous methyl cyanide and several
  fluorinated solvents.  Offhand, this looks to me like it would do the
  job, but it's not a friendly compound (toxic, possibly corrosive).

The concept is already in place all around us: any time you buy a plastic jug with harsh chemicals in it, the jug has been fluorinated. So far, the majority of my learning is confirmation that yes, indeed, fluorinating polyethylene (PE) surfaces with $\ce{XeF2}$ leads to decreased gas permeability.

I've done a bit of Googling and bought a couple of great textbooks, Fluoropolymers 1; Synthesis and Fluoropolymers 2; Properties. Fluoropolymers 1 seems to have great stuff in a couple sections. Perusing Fluoropolymers 2, I haven't seen anything jump out at me yet.

There are two methods I have read about:

Involves an oven, steady high temperature, and plasma discharge.
"Surface Fluorination of Polymers Using Xenon Difluoride", or Gaseous $\ce{XeF2}$ and PE are placed in a "stainless steel reactor" and left alone.

Seeing the options, the 2nd seems the way to go, and this seems to be the epic equation concerning my quest quoted from Fluoropolymers 1 page 231, section 15.3:

  The comparative accessibility and desirable properties of $\ce{XeF2}$ enable
  its use for the synthesis of fluoroorganic compounds under “mild”
  conditions. As a fluorinating agent $\ce{XeF2}$ can be used to effect
  electrophilic fluorination of various organic compounds with good
  yield and it is a useful reagent for generating radical cations.
  ... In many cases XeF2 is an attractive alternative to the
  conventional Balz–Schiemann reaction for the preparation of
  fluorinated aromatic compounds, e.g., fluorobenzenes:

  $\ce{RC6H6 ->C[XeF2] Rc6H5F + Xe + HF ->C[XeF2] RC6H4F2 }$

  Xenon difluoride reacts with carbon–carbon single, double, and triple bonds
  giving addition fluoroorganic compounds, e.g.,

  $\ce{R-CH=CH2 ->C[XeF2] RCH2-CHF2 + Xe}$

  and

  $\ce{RCH2-CH3 ->C[XeF2] RCHF-CH3 + Xe + HF ->C[XeF2] RCF2-CH3 + Xe + HF}$

And this quote from page 236 gives me great hope:

  Another argument in favor of the $\ce{XeF2}$ method is the evidence that since 1995 it
  has been used by a manufacturing company that had no previous experience with
  direct fluorination or fluorine at all.

... for I'm in the same position now!

How do I actually perform a fluorination surface treatment my XPE products?

Where / how do I go about getting $\ce{XeF2}$? And if it is soluble in a "fluorinated solvent" like my prof said, is there a way I can flush this through the interior of my tanks in a controlled manner that would accomplish some degree of fluorination?

Edited to add 10-22-12: I called a few companies today. Everyone is pointing me to pay Fluoro-Seal to do this. They are the only company providing this service in the US. I'll need to ship my products to them, wait a couple weeks, and pay $15~45 each tank. I'd still like some kind of bath or wash to do this. This is why no-one making replacement tanks has them emissions legal... all this triples my cost.

Edited to add 12-30-13: Fluoro-Seal has given up on fluorinating rotationally molded PE. I'm pushing a plastics manufacturer to develop their existing acetel product to work with rotational molding. Their solution for injection and blow molding is in production and impervious to fuels (they have parts in nearly all automotive fuel systems in the US right now). So there's hope for a solution; however it seems impossible to fluorinate PE.

There is a simpler way to get fluorinated polyethylene.

Polyethylene is polymerized ethylene $\ce{CH2=CH2}$. As part of the polymerization, the double bond is no longer there. Polyethylene is essentially a very long saturated hydrocarbon:  $\ce{(-CH2-CH2 -)}_n.$

Another polymer already exists that replaces all the hydrogen atoms with fluorine atoms: poly(tetrafluoroethylene) (PTFE), made from tetrafluoroethylene $\ce{CF2=CF2}$. PTFE is a long saturated ferfluorocarbon: $\ce{(-CF2-CF2 -)}_n.$

The most well known trademarked brand of PTFE is Teflon. Get your coated with or made out of Teflon/PTFE. It will be easier, cheaper, and safer in the long run. If XeF2 will fluorinate polyethylene (which is reasonably inert), imagine what it could do to yyou.


Cleaning with sodium bicarbonate I have a small metal, most likely silver, cross. It is an old object and has turned dark. In order to clean it, my mother told me to boil it for a few minutes in water containing sodium bicarbonate.

Could someone explain what kind of reaction(s) occur(s)?

If originally made of silver, the dark layer on your jewlery is silver sulfide, $\ce{Ag2S}$.  The removal suggested by your mother is an electrochemical one, that will reduce the silver back into metallic, shiny $\ce{Ag}$, if in addition of hot water and bicarbonate, salt and an other metal acting as reducing component is in electrical contact with the silver piece are present.  Here, some foil of aluminum will serve well as sacrifical anode.

To quote UMass Amherst, presenting it as a lecture demonstration, the chemistry may be summed up to this equation

$$\ce{3 Ag2S(s) + 2 Al(s) + 3 H2O (l) –> 6 Ag(s) + 3 H2S (aq) + Al2O3(s)   }$$

The addition of salt (which is absent in the equation) serves to increase the electric conductivity of the solution, hence increasing the rate of this process.  Immerse the jewlery in the hot solution (it needn't be boiling), for example wrapped in the creased foil for some minutes; then rinse and wipe it with a cloth.


Crystal structure in function of pressure If you were to investigate crystal structure changes as a function of pressure, which metals would you study first? Last? Why?

Seems like homework, so I'll give some hints:
What do you expect to happen to a crystal under pressure?
What does the crystal structure have to do with the response to pressure?
Why would the crystal structure want to change under pressure?
What crystal structures are common for metals? (there aren't many)
Since there aren't many, what distinguishes their responses to pressure that might make the metal want to change crystal structure (a phase change).

That should get you going.


Number of chiral centers in penicillin V I have often seen that in many drugs chirality plays a role. So I was trying to analyze the structure of Penicillin V.

What is number of chiral centers in the structure of penicillin V? Is the nitrogen which is  a part of the four membered ring to be counted here because I have read that the nitrogen atom without pyramidal inversion is treated as chiral. So can the nitrogen at bridgehead position be treated as chiral? Because that makes the stereocenter count 4 while on pubchem website the count is mentioned  to be 3.

Yes, here Nitrogen has to be considered as a stereocenter. Also, you can assign R and S to it. The rule is that the lone pair has the least priority and then you can assign R and S as you do for other molecules.


How to identify the reduction in a given redox equation of iron chloride and chlorine? The equation shows a redox reaction between iron(II) chloride and chlorine gas.
  $$\ce{2FeCl2 + Cl2 -> 2FeCl3}$$
  Which equation describes the reduction process in this reaction?
  \begin{align}
\ce{2Cl- &-> Cl2 + 2e-}\tag{A}\\
\ce{Cl2 + 2e- &-> 2Cl- }\tag{B}\\
\ce{Fe^2+ &-> Fe^3+ + e-}\tag{C}\\
\ce{Fe^3+ + e- &-> Fe^2+}\tag{D}
\end{align}

I couldn't understand why the answers are not C instead of B.

The correct answer is B. A and D cannot be the answer.

A redox reaction consists of 2 parts, oxidation and reduction These two parts can be written up as half-equations where one half-equation shows oxidation and the other shows reduction.

It seems that you have mixed up the oxidation and reduction half-equations. Oxidation is losing electrons while reduction is gaining electrons. This can be remembered by the mnemonic:

The question is asking for the reduction half-equation. To do this, you must first recognise what is getting reduced (in other words, what is the oxidant). This can be done by looking at the oxidation states for each element before and after the reaction. Since it is getting reduced, it is gaining electrons and therefore its oxidation state should decrease after the reaction. So by looking at the oxidation states, it should be clear that chlorine is getting reduced as its oxidation state goes from $0$ to $-1$.

Therefore the answer is B

Note: The answer definitely can't be a or c since they are oxidation half-equations


Stability of Bohr Orbits In class we had been taught that Rutherford's model was unsuccessful because it failed to show that the orbits are stable because the electrons would lose energy because of electromagnetic radiation.

Bohr had suggested that electrons would move in orbits having well defined energy levels, but my question is that what would prevent these electrons from emitting radiation and collapsing into the nucleus.

Not every model is perfect. Rutherford's model suffered from the problem of electromagnetic radiation but answered important questions about the structure of atoms by showing the existence of a positively charged nucleus.

Bohr improved on this by realising that if electrons could only have certain energies then he could explain other things such as the spectra of hydrogen. As far as I am aware he did not attempt to explain why electrons are fixed in these orbits - this was only explained later by the advent of quantum mechanics. However, this does not mean that Bohr's model was useless since it accurately predicts useful things.

We do not have to know everything about why a model works in order for it to be useful. Einstein did not know why the speed of light was constant but he assumed that it was (based on all the available experimental evidence) and derived special relativity, which is an extremely useful theory. Nowadays we have quantum mechanics and general relativity which we know are incompatible but they are both exceptionally useful theories so we do not ignore them.


Why is DMSO more polar than other solvents, yet it dissolves ionic compounds worse? I'm working right now in the solvent importance in ionic character, and I noticed how a binary salt ($\ce{CuCl2}$) dissolved better in methanol than in dimethylsulfoxide, and in contrast, a coordination compound (Cu[salen]) dissolved better in DMSO that in the aforementioned solvent.

It doesn't make a lot of sense to me considering these properties I found:

DMSO:

Dipole moment: 3.96
Dielectric constant: 47

Methanol:

Dipole moment: 1.71
Dielectric constant: 33

I would have expected the binary salt to dissolve better in DMSO rather than Methanol, and the inverse in the coordination compound yet it didn't happen like that. Is there a intermolecular detail I'm not considering? An hydrophobic-hydrophilic portion in the structure? Thanks everyone.

Having large dipole moments and dielectric constants, as DMSO does, does not imply that it has the ability to solvate ionic compounds like $\ce{CuCl2}$.

Although both methanol and DMSO have oxygen atoms with their lone pair of electrons to help solvate cations, there is no corresponding functionality in DMSO to solvate a small, inorganic anion like chloride.  Methanol on the other hand can solvate chloride via hydrogen bonding in the same way that water does.

These properties are not necessarily applicable to a coordination compound, so there is no reason to expect methanol to act as a better solvent than DMSO in this case.


Reaction of concentrated nitric acid react with metallic lead? Here is my work:

Concentrated $\ce{HNO3}$ should first decompose into $\ce{NO2}$, $\ce{H2O}$ and $\ce{[O]}$. Nascent oxygen should combine with $\ce{Pb}$ to form $\ce{PbO}$. Now $\ce{PbO}$ should neutralize acid to form $\ce{Pb(NO3)2}$.

Therefore, the net products would be lead nitrate, $\ce{NO2}$ and $\ce{H2O}$.

Is this right?

Well, you were going along the right track.... just one correction. The products will be Pb(NO₃)₂, NO₂ and H₂O. Nitric acid is a very powerful oxidizing agent. It oxidized H₂ to H₂O.


Why does chocolate ice cream melt faster than others? The other day I was dipping out Neapolitan(strawberry, vanilla, & chocolate) ice cream and noticed that the chocolate ice cream tended to melt faster than that of the vanilla and strawberry.  I couldn't remember in which order I dipped the ice cream so I decided to do a controlled experiment with it the next time I dipped it out.

I took three bowls of the same type/shape and dipped out the strawberry first, then vanilla, and chocolate last.  Even dipping the chocolate ice cream last showed that the chocolate ice cream would melt faster than the strawberry or vanilla.

Neither the strawberry nor chocolate use artificial coloring, though the vanilla states that it does.  Is there a reason as to why chocolate ice cream melts faster?  Is it because of the darker coloring that attracts heat or perhaps the low melting temperature of chocolate in general?

This appears to be physics more than chemistry. A dark color means it emits less visible light but rather absorbes it with temperature rise. Have your next ice cream in a darker place as to visible and ir emisions to test this.


Why does ethylamine have two infra-red spectra? Are both correct? The one I have in my textbook is this:

Shouldn't the first trough at $3500$ to around $3000$ mean that it's an $\ce{O—H}$ group with a $\ce{C—H}$? And then the other trough at $1700$ be a $\ce{C=O}$? Why is this ethylamine $\ce{CH3CH2NH2}$?

The other one I found on Google:

Shouldn't the $\ce{NH2}$ group have double troughs? The only trough there is the $\ce{C—H}$ at $3000$. How is this ethylamine?

Can anyone explain those spectra for me? Are both correct? I'm very confused.

When looking at IR spectra of hydrogen-bonding groups, always check how IR spectra was recorded. It appears that NIST spectra is recorded using gas phase, while the first one used liquid film. In liquid film $\ce{NH2}$ groups form a network of hydrogen bonds, resulting in shape distortion and widening of $\ce{NH2}$-related bands. Same is true for carboxylic acids, alcohols, phenols, and many other group.  Comparison of water spectra in ice, liquid and gas is particularly enlightening.
If the spectrum is recorded in a solution, you also have overlapping spectrum of the solvent in your recording. If you look at water spectrum in liquid phase (nist webbook), you can see some wide peaks suspiciously similar to wide bands in the first spectrum you posted, so it might be that 'liquid film' is actually a water solution. In this case the situation is even worse, because in addition to ethyl amine spectrum you also have overlapping ethyl ammonium spectra.


Which has the largest bond angle between water, oxygen difluoride and dichlorine oxide? Which one out of $\ce{H2O}, \ce{Cl2O}, \&\ \ce{F2O}$ will have largest bond angle?

I think it should be $\ce{H2O}$ because oxygen is most electronegative in this case so electrons will be more towards oxygen hence increasing bond angle, but the answer is $\ce{Cl2O}$.  Why?

There is practically only one effect that determines the angle, but there is an underlying mechanism that you also should understand.

Ideally, a central atom such as oxygen would want to use its p-orbitals exclusively for bonding and have the s-orbital electrons sitting comfortably as a lone pair close to its nucleus. However, that requires a $90^\circ$ angle which can create rather strained molecules — remember that each atom is surrounded by an electron cloud with negative charge that repulses other atoms’ electron clouds. There are now two possibilities of how to relieve this steric strain:

enlargen the bond lengths, so that different substituents are further away from each other while keeping a bond angle as perfect as possible;
enlargen the bond angles so the substituents move further apart from each other while retaining a close distance to their central atom with a more perfect bond length.

In practice, almost all systems stabilise themselves by modifying the angles. Increasing bond lengths decreases bond strengths and thus energy gained much more strongly than tinkering around with angles, because the overlap will inevitably get significantly smaller in absolute value number no matter which way you look at it.

An average bond length can be thought of as the distance which is just large enough for those orbitals that interact to do so favourably while minimising nonfavourable interactions. For certain atom pairs, bond length can essentially be considered almost a constant.

All of this taken together means that the very small hydrogen atoms have a much easier time of crowding together closer to the $90^\circ$ angle than the larger fluorine or huge (in this context) chlorine atoms. Even though the latters’ bond lengths are longer in absolute values their larger radii makes the effective bond lengths shorter and thus makes them require more space. Since chlorine is the largest of the lot, the $\ce{OCl2}$ bond angle is largest.


What are similarities and differences among shells, orbitals, subshells, and energy levels? I think I know what these terms mean:

each row on the periodic table has a new shell of electrons,
a subshell is e.g. 1s, 2s, 2p, 3s, 3p, etc.,
and energy levels are the same as shells,
whilst orbitals are the same as subshells.

Is this correct or totally wrong?

I am using the chemistry textbook by Masterton/Hurley (Cengage) as an outline for this answer. It has a side note that reads "Our model for electronic structures is a pragmatic blend of theory and experiment". In the spirit of that quote, there is some experimental data mixed in with the concepts.

  Each row on the periodic table has a new shell of electrons

Yes, and you see that in the plot of atomic properties vs. atomic number. For example, atomic radii (image source)increase significantly as you go from noble gas (last element in a period) to group 1 metals (first element with the next shell occupied). The quantum number that defines the shell is the principal quantum number n.

  A subshell is e.g. 1s, 2s, 2p, 3s, 3p, etc.

Yes. The periodic table is organized into s-block and p-block (main group elements), d-block (transition metals) and f-block (lanthanides and actinides) according to which subshell is partially filled (or just filled). The quantum number that defines the subshell is the second quantum number, $\ell$.

  Are energy levels the same as shells?

This is true only for the hydrogen atom and other one-electron systems ($\ce{He+}, \ce{Li^{2+}}$) in the absence of a magnetic field:

$$ E \propto -\frac{Z^2}{n^2} $$

with $Z$ the charge of the nucleus and $n$ the principal quantum number.

For systems with more than one electron, distinct subshells in the same shell correspond to slightly different energy levels. One method to measure these energy levels is photoelectron spectroscopy, with some data on aluminium 2s and 2p shown below (image source).

In general, the energy levels depend on the atomic number (number of protons in the nucleus) and the electron configuration (how many electrons are present, and whether they are at ground state or in an excited state). Going from a hydrogen atom to a larger multi-electron atom, the energy levels of distinct shells can be roughly described as an electron bound to a nucleus surrounded by inner electrons with a combined effective charge $Z_\mathrm{eff}$ (smaller than the actual charge of the nucleus because the attraction by the nucleus is offset to a degree by the repulsion by the other electrons):

$$ E \propto -\frac{Z_\mathrm{eff}^{2}}{n^2} $$

For the innermost shells (n = 1), $Z_\mathrm{eff}$ is almost the same as the nuclear charge, whereas for the outermost shells, $Z_\mathrm{eff}$ can be much lower.

For example, comparing 1s electrons across the periodic table, the binding is tighter and tighter as the atomic number increases. On the other hand, the outermost shell (valence shell) in elements of the same group (e.g. Be, Mg, Ca, Sr, Ba, Ra) corresponds to similar energy levels, even though they differ in the principal quantum number. This explains why elements in a group have similar chemical behavior.

  Are orbitals the same as subshells?

No, there are seven different orbitals in the f-subshell, five different orbitals in the d-subshell and three different orbitals in the p-subshell (compare size of blocks in the periodic table). The three orbitals in the p-subshell, for example, are either designated (+1), (0), and (-1), or alternatively $p_x$, $p_y$, and $p_z$. For the s-subshell, there is a single orbital, so only in this special case is the subshell the same as an orbital. The quantum number defining the orbital in a subshell is the third quantum number $m_\ell$.

While the different orbitals of the p, d, and f subshells have distinct, sometimes intricate shapes, a filled subshell corresponds to a spherical distribution of electrons.

  How do these concepts help to understand chemical bonds?

We study electronic structure because it explains properties and reactivity of atoms and molecules. It might seem that atomic orbitals of hydrogen are of little use in understanding the electronic structure of larger atoms and of molecules. It turns out, however, that they can be used as building blocks for the more complicated systems, both qualitatively (talking about sigma and pi bonds and lone pairs) and quantitatively (computational methods use atomic orbitals as basis functions).

Surprisingly, the changes in electronic structure going from atoms to molecules are moderate, with spherical electron distribution around the atoms explaining most of the electron density observed, for example, in crystallographic studies. The small deviations from this are called electron deformation densities, and they give an experimental view of the subtle shifts in electron density upon bond formation. The picture below shows a contour diagram of the total electron density on the left (near the atoms, there would be many more contour lines, but they are not shown), and the deformation density on the right (lines: higher density than atomic density, dashed: lower density than atomic density, dot-dash: same density as atomic density).

(Image source)


How can you make magnesium carbonate? I am trying to start my own gym company that sells gym chalk however I have not been able to find a good process for making good magnesium carbonate. I have tried mixing equal amounts of magnesium sulfate and sodium bicarbonate into water and pouring it through filter paper but have not ended with a good batch.

The best way to make magnesium carbonate is to opt the laboratory route (mix magnesium chloride and sodium bicarbonate) as described in the Wikipedia article of magnesium carbonate.

$$\ce{MgCl2(aq) + 2NaHCO3(aq) → MgCO3(s) + 2NaCl(aq) + H2O(l) + CO2(g)}$$

The advantage of this process is that the byproduct do not interfere with the main product (sodium chloride is soluble in water, carbon dioxide bubbles out) leaving behind magnesium carbonate as precipitate.

Alternatively, magnesium carbonate can be prepared from magnesium carbonate by bubbling carbon dioxide in it. The advantage of this process is that there are no byproducts.

Note: If you use sodium carbonate, you will get a  precipitate of basic magnesium carbonate—a hydrated complex of magnesium carbonate and magnesium hydroxide—rather than magnesium carbonate itself is formed.

$$\ce{5MgCl2(aq) + 5Na2CO3(aq) + 5H2O(l) → Mg(OH)2·3MgCO3·3H2O(s) + Mg(HCO3)2(aq) + 10NaCl(aq)}$$

You can see that a basic salt is formed and also magnesium bicarbonate is formed which add to the difficulty of separating them and decomposing them to get magnesium carbonate and thus this process is not feasible.


Why do Hydrogen Halides(with the exception of HF) form mostly strong acids? Why do Hydrogen Halides(with the exception of HF) form mostly strong acids?
My guess is that the negatively charged Halide part of the molecule is attracted by the partially positive part of the water molecule. Thus it is separated more easily. Although because it has one more electron it is more stable once it has dissociated

I would like to correct a few points firstly that the halide part is not negatively charged. Also the electronegativity of the halide is the dominating effect in this case.

Hydrofluoric Acid as we know is not a very strong acid as compared to other hydrogen halide acids, had it been the case of electronegativity this would not have been the case.

The dominating factor here is the sizes of the atoms forming the molecule. As the size of halide increases, the bond strength of the H-X bond decreases, making it easier to break the Hydrogen halide bond and act as a stronger acid.

So the acidic strength comes out to be:
              HF < HCl < HBr < HI