Rusting in every other bar I have noticed that, when a structure with bars (such as a gate or a grate) develops rust, this sometimes happens differently across the bars. In particular, it seems that the bars are alternatively clean and covered in rust.
The photos below (taken by me) illustrate this effect (I have highlighted the rusted bars with red arrows). In both cases, from inspection, all bars appeared to be made from the same material and covered in the same kind of protective paint (the rusted parts were exposed to the elements).
My theory is that perhaps the rusting process activates some sort of "circuit" between bars, such that one bar protects the next one from rusting.
Is this possible?

Rebar for concrete is frequently stored in heaps, and rusts. It actually develops a stronger bond if it is not too rusty, because the rebar surface is rougher and better able to adhere to the hardened cement. It's interesting to see a pile of old rebar with a fresh delivery on top - silvery, not a trace of rust - on top of a pile of rebar that looks almost disgustingly rusted. And then you think, yeah, but the stuff on the bottom will have a better bond.
The key word in the observation about the fences is "sometimes".
I'm going to guess that the maker of the fence in the top picture also had a pile (or two) of iron bars, and some were rusty from being stored long or outdoors, or were a teensy bit different from some other pile of iron bars. This fence maker, knowing how his fence weathers, has to make it as uniform as possible, so he takes one bar from one pile, then a bar from the other pile... He prepares them identically, primes them identically, paints them identically... But then nature takes over and the most corrodable bar develops a spot, which spreads until it is all corroded-looking, before the more corrosion-resistant bar has even gotten started!
And then you come along and take a picture just before the other bars begin to corrode! This experiment is in the initial stages, not yet at equilibrium. Check back on the fence every year or two - I believe things will even out in time.

I would like to propose an alternative to those dismissing the advent of possible complex electrochemistry behind the phenomenon and asserting instead perhaps just poor planning and/or design decisions.
In this regard, an interesting 2011 article, 'An Fe3O4–FeO–Fe@C composite and its application as anode for lithium-ion battery', discussed here, to quote:

X-ray diffraction, Raman spectrometry, scanning electron microscopy with energy dispersive spectroscopy, transmission electron microscopy and galvanostatic cell cycling have been used to characterize the structure and electrochemical performance of the as-prepared Fe3O4-FeO-Fe@C composite. The electrode shows a stable and reversible capacity of over 600mAhg−1 at a current of 53mAg−1 for up to 60 cycles, which could be ascribed to the unique double-layer core-shell and embedded structures.

Also, per another reference discussing the same electrode, to quote:

Here, we report the synthesis, characterization and evaluation of the electrochemical properties of pristine Fe3O4 nanoparticles synthesized from a base catalyzed method mechanically mixed with electrospun Polyacrylonitrile (PAN) derived carbon fibers as simple composite material. The Fe3O4/C electrodes are characterized by galvanostatic cycling experiments using currents as high as 2000 mA g−1, revealing high capacity values and capacity retention, together with a very good capacity recovery during the rate capability experiment.

The picture appears to display a red pigment, possibly Fe3O4 created at elevated temperatures. Other bars, resemble Fe/FeO.
So speculatively in the current case, an alternate proposition that if the iron metal fence was exposed to say a brush fire, heating the iron and also depositing fine carbon, this may have resulted in a series of Fe-FeO@C and Fe3O4@C electrodes naturally developing over time (aka, yes, some possible complex electrochemistry).


In the nonideal solution, why can't one of the solute be positive deviation and the other be negative deviation? In the nonideal solution, why can't one of the solute be positive deviation and the other be negative deviation?
Let's say A and B makes a nonideal solution.
The moleculat force between A and B means A-B.
Let's say A-A > A-B > B-B.
Then A would evaporate more than in the pure state, and B would evaporate less than in the pure state.

For multiple component systems we use activity coefficient $\gamma$ in modified Roult's law as:
$$K_i=\gamma \frac{P_i^\star}P $$

and $\gamma$ is defined in the terms of $f$ : fugacity or fictitious pressure for a real gas.

You should read further about "Vapour Liquid Equilibrium" for multiple components.

Maybe:  http://www.journalrepository.org/media/journals/BJAST_5/2012/Dec/1355741742-Olaleke312012BJAST1789.pdf. You can see there is no reason why it can't have deviation in multiple directions.


Why are sigma bonds stronger than pi bonds? Why are sigma bonds stronger than pi bonds?

The reason behind this is the orientation of the overlapped orbitals. Sigma bonds result from head-on(co-axial) overlapping while pi bonds are outcome of lateral(para-axial) overlapping. Here is a pictorial representation of ethene(sp2 hybridized C atoms) :

The greater the extent of overlapping, the higher the probability of finding the valence  electrons in between the nuclei and hence the bond will be stronger & shorter.

In MOT, this can be explained using Overlap Integral. This is how Atkins depicts it :

In simple terms, after forming a sigma-bond (a pre-requisite for pi-bonds), the two atoms get locked along the inter-nuclear axis. As a result, the orbitals available for pi-bonding can only partially overlap, thus forming a weaker bond.

Unsaturated compounds (due to the existence of a double bond) are more reactive than saturated compounds, because the electron density at the internuclear axis is zero in a pi bond (in contrast to the sigma bond).


Why do branched chain compounds have lower boiling points than the corresponding straight chain isomers? The branched chain compounds have lower boiling points than the corresponding straight chain isomers. For example,

  $\ce{CH_3CH_2CH_2CH_2CH_3}$ -
  No branching-Pentane (n-pentane) ($\mathrm{b.p.}=309~\mathrm{K}$)
  $\ce{CH_3CH(CH_3)CH_2CH_3}$ -
  One branching-2-Methylbutane (Iso-pentane) ($\mathrm{b.p.}=301~\mathrm{K}$)
  $\ce{C(CH_3)_4}$ - Two branches-2,2-Dimethylpropane (Neo-pentane) ($\mathrm{b.p.}=282.5\mathrm{K~}$)

  This is due to the fact that branching of the chain makes the molecule more compact and thereby decreases the surface area. Therefore, the intermolecular attractive forces which depend upon the surface area, also become small in magnitude on account of branching. Consequently, the boiling points of the branched chain alkanes are less than the straight chain isomers.

The above extract from my book, mentions clearly that branching makes the molecule more compact and thereby decreases the surface area. Even for isomeric alkyl halides, the boiling points decrease with branching. The reason is said to be because of the decrease in surface area, same as explained before. How does decrease in the surface area make the intermolecular forces small in magnitude?

The simple explanation is that weak intermolecular forces (the forces that make something condense to a liquid when things are cold enough) depend on the surface area (as well as many other things). But in the case of relatively similar non-polar isomers (where weak intermolecular forces are the dominant forces), the larger surface area will lead to the large force and hence the highest boilling point.

2,2 dimethyl propane is a compact, almost spherical molecule. Pentane is long and "floppier" so will experience more forces between molecules.

The underlying, but slightly simplified, explanation of this is that intermolecular forces (often called van der Waal's forces) depend on attractions caused by quantum fluctuations in the surface electrons of the molecule. These lead to short-lived dipole moments that can also induce dipoles in neighboring molecules to which the original dipole is attracted. The larger the surface area, the more opportunity for such dipoles to exist and therefore a stronger force.

The London / van der Waals forces are weak forces but their effect depends both on the magnitude of the forces themselves and the distance over which they act. Linear molecules can align and lie parallel to each other, giving small distances of separation. This makes attraction produced by the same forces/moments greater.


What roles do neutrons play in an atom? An atom typically consists of electrons, protons, and neutrons. Electrons are negatively charged and participate in bonding to stabilize the atom.

Conversely, protons are positively charged and balance the charge of the atom. In addition, their positive charge attracts the negatively charged electrons.

What role do neutrons play in an atom?

Neutrons bind with protons and one another in the nucleus through the strong force, effectively moderating the repulsive forces between the protons and stabilizing the nucleus.$^{[1]}$

$\ce{^2He}$ (2 protons, 0 neutrons) is extremely unstable, though according to theoretical calculations would be much more stable if the strong force were 2% stronger. Its instability is due to spin–spin interactions in the strong force, and the Pauli exclusion principle, which forces the two protons to have anti-aligned spins and gives the $\ce{^2He}$ nucleus a negative binding energy. $\ce{^3He}$ (2 protons, 1 neutron), on the other hand, is stable, and is also the only stable isotope other than $\ce{^1H}$ with more protons than neutrons.$^{[2]}$

$^{[1]}$ Wikipedia, Neutron, Beta Decay and the Stability of the Nucleus
$^{[2]}$ Wikipedia, Isotopes of Helium

Neutrons play very important role in an atom. They provide stability to the atom and also prevent protons from repelling one another.


Why Cl⁻ can't act as bidentate ligand? $\ce{Cl-}$ is a monodentate ligand. It has four lone pairs then why it can't donate two lone pairs to central atom and act as a bidentate ligand?

A bidentate ligand is one that forms bonds to the metal center through two different atoms, not through two lone pairs. While chloride does have more than one pair of valence electrons, it only has one atom. Thus it cannot be bidentate.

If the chloride shares a second pair of electrons with the metal center, there would still only be one bond site between the ligand and the metal:

$$\ce{[M+]-Cl <-> [M]=Cl+}$$

A bidentate ligand needs two lone pairs on two separate atoms. Consider ethylenediamine (en), one of the classic bidentate ligands: $\ce{H2NCH2CH2NH2}$. Each nitrogen atom has a lone pair that can be used to form a bond to the metal center, creating two bonds from two different sites on the ligand. See the structure of the tris(ethylenediamine)cobalt(III) ion from tris(ethylenediamine)cobalt(III) chloride $(\ce{Co(en)3Cl3})$.

Update:
As Nicolau points out in a comment, chloride can bridge two metal centers. This is not the denticity phenomenon, which is one ligand with two bonding sites to the same metal. This is one ligand with bonds to two metal centers. See the structure of niobium pentachloride (really $\ce{Nb2Cl10}$ or better $\ce{(NbCl5)2}$.

Chlorine ion is monodentate.For bidentate ligand it should have atleast two bonding sites.In chlorine it has only one donatable lone pair.So it can't act as bidentate ligand.


Prove that a Change of Volume will not Change the Value of the Equilibrium Constant According to my book, a decrease in volume will result in an increase in pressure for an equilibrium mixture, which will in turn cause its equilibrium to shift to the side with the least number of gas molecules, given the reaction involves gases (and according to Le Chatelier's Principle).

It also says that Volume and Pressure changes will cause the position of equilibrium to shift, but the equilibrium constant remains unchanged...

If the equilibrium constant (K) depends on the concentrations of reactants and products...

and according to the molarity formula, Volume is inversely proportional with respect to concentration...

... then why does the equilbrium constant remain unchanged?

So far I have this, but I don't know what it means...

In your workyou have changed the "equilibrium constant" which is a fixed while keeping the number of moles constant of all species in both instances. Rather its the opposite

Consider the reaction;$$\ce{AB <=> A + B }$$ with equilibrium constant $K_c$ at temperature $T$ and volume $V_0$

Initially let $n(AB) = x_0$ , $n(A) = y_0$ , $n(B) = z_0$ where $n()$ refers to the number of moles.

then $$K_c =  {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}}$$

Now the volume is reduuced from $\ce{V_0\bond{->}V}$
Hence all the concentrations increase.
Then the reaction quotient at the moment the volume is changed;$$Q_c =  {\frac{{\frac {x_0}{V}}{\frac {y_0}{V}}}{{\frac {z_0}{V}}}}$$

Now $Q_c > K_c$ hence equilibrium shifts to the right reducing the number of moles of $A$ and $B$ increasing the number of moles of $AB$.Due to this effect $Q_c$ will gradually reduce and at acertain point will becaome equal tp $K_c$ with different number of moles.

$$K_c =  {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}} = {\frac{{\frac {x}{V}}{\frac {y}{V}}}{{\frac {z}{V}}}} $$

The only parameter remaining constant(under constant temperature) is the Equilibrium constant is a certain parameter(i.e. Volume here) is changed, other parameters will vary keeping Equilibrium constant a constant.

Le Chatelier talks only about how the and on which side will the equilibrium position shift. Not about the change in equilibrium constant. Let me make two points. :-

Change in volume affecting the equilibrium constant:- $\Delta$n = (sum of stoichiometric coefficient of products)-(sum of stoichiometric coefficient of products). E.G.

$$\ce{aA + bB <=> cC + dD}$$

here $\Delta$n= (c+d)-(a+b)

now,

$$K_c=\frac{[C]^c [D]^d}{[A]^a[B]^b}$$

since $$[A]=\frac{moles_a}{volume(V)}$$

similarly for all .
 $$K_c=K_x*V^{\Delta n}$$

So if

$\Delta n= 0$ then it doesn't affect

$\Delta n >0$ the it is directly proportional

$\Delta n <0$ the it is inversely proportional

E.G.

$$\ce{H_2 + I_2 <=> 2HI}   \Delta n= 0$$

$$\ce{N_2 + 3H_2 <=> 2NH_3}  \Delta n= (-2)$$

$$\ce{ PCl_3 <=> PCl_5 + Cl_2}  \Delta n= (1)$$

Le Chatelier  says that if volume is increased then number of moles per volume decreases then the equilibrium will shift towards the side with more moles which is evident from the previous explanation.

If you still find a doubt then let me know I will like to give another try with more explanation


Alkali and alkali earth metal carbonates and bicarbonates decomposition Which of the following can not decompose on heating to give $\ce{CO_2}$?

$\ce{Li_2CO_3}$
$\ce{Na_2CO_3}$
$\ce{KHCO_3}$
$\ce{BaCO_3}$

What I have read till now: Lithium carbonate is not so stable to heat and bicarbonates of alkali metals decompose at approximately $\pu{300 ^\circ C}$. The stability of alkaline earth metals increases down the group and they decompose on red heating.
This only leaves us with one option as correct (2nd), but what is the reaction of decomposition of $\ce{Na_2CO_3}$?

From your question, the heating temperature is unclear. See these links for sodium and lithium carbonates, the hydrocarbonate or barium.

They are all also mentioned on Wikipedia.

In other words, they all seem to decompose at some point, so what is meant by heating?

The reaction for sodium carbonate decompositon at melting point should be:

$$\ce{Na2CO3 -> Na2O + CO2}$$

Na2CO3 is more thermally stable. So there is no effect of heating on it. Hence there is no reaction and carbon dioxide is not released.


Conductometric titration curve when KCl is gradually added to AgCl solution The situation given is that we are performing conductometric titration where KCl is gradually added to AgCl solution and we have to plot the variation of conductivity with the volume of aq. KCl added. I am new to conductometric titration and I have covered the cases of titration of acid v/s bases but I have never come across a case like this.

My thought was that since AgCl is a weak electrolyte it's initial conductivity would be low. And now we add KCl to it which is a strong electrolyte so the conductivity should increase. I also think that common ion effect would come into play and the dissociation of AgCl would be pushed back further but I'm still unsure about how an end point would be obtained here. Can someone guide me? Any help would be appreciated.

I'm a bit puzzled by the phrase "AgCl solution," since there wouldn't be a solution but rather colloidal size particles of AgCl in the aqueous solution. I don't think you could have a bottle of colloidal size particles of AgCl since once dried the particles would stick together in a lump.

For AgCl, a tiny amount would dissolve hence there would be conductivity slightly better than pure water.

$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$

However AgCl also form a complex with chloride:

$$\ce{AgCl(s) + Cl-(aq) <=> AgCl2-(aq)}$$

So as you add KCl the curve would show one slope as KCl builds up enough concentration to dissolve the AgCl dissolves, then a second slope as the AgCl dissolves, then a third slope after all the AgCl has dissolved.

For ecah change the curve won't look like two straight lines intersecting at a point but more like a parabola. So you'd need to extend linear portion of each and find the end point at the intersection of the linear extensions.

While yes, for AgCl, a small amount could dissolve per the reactions

$\ce{AgCl(s) <=> Ag+(aq)+Cl−(aq)}$

$\ce{AgCl(s) + Cl−(aq) <=> AgCl2−(aq)}$

However, I believe the presence of the AgCl as a suspension would have an impact on conductivity per this source, as well, where AgCl particles replace Al2O3–H2O per the abstract below:

  In this study, Al2O3–H2O nanofluids were synthesized using sodium dodecylbenzenesulfonate (SDBS) dispersant agent by ultra-sonication method. Different amounts of SDBS i.e. 0.1, 0.2, 0.3, 0.6, 1 and 1.5 wt.% were tested to stabilize the prepared nanofluids. The stability of nanofluids was verified using optical microscope, transmission electron microscope and Zeta potential. After selecting the suitable amount of dispersant, nanofluids with different volume fractions of Al2O3 were prepared. Zeta potential measurement of nanofluids with low alumina and intermediate fractions showed good dispersion of Al2O3 nanoparticles in water, but nanofluids with high mass fraction were easier to aggregate. The stabilized nanofluids were subjected for measuring of rheological behavior and electrical conductivity. The electrical conductivity was correlated to the thermal conductivity according to Wiedemann–Franz law. The results revealed that the nanofluid containing 1% SDBS was the most stable one and settling was observed for the fluid contained 0.75 vol.% of Al2O3 nanoparticles which gave higher viscosity. The rheological measurements indicated that the viscosity of nanofluids decreased firstly with increasing shear rate (shear thinning behavior). Addition of nanoparticles into the base liquid enhanced the electrical conductivity up to 0.2 vol.% of Al2O3 nano-particles after which it decreased.


Why aren't all bonds in cyclobutadiene not equal? Cyclobutadiene is like benzene in that it has alternating single and double bonds in a ring. However, it's bonds are not the same length, the double bonds being shorter than single bonds. The molecule is rectangular, not square.

How would this be explained?
I have one plausible explanation that I tried to give...that is as follows

Cyclobutadiene is anti aromatic with 4n = 4 pi electrons, moreover it has huge strain in the molecule. To avoid such strain the molecule goes out of plane to compensate, but thus loses its ability to resonate.

Thus resonance structures like ones in benzene cannot be drawn for cyclobutadiene.

Is this correct? Please correct me if I am wrong..and some more data about this would be appreciated

Strictly speaking cyclobutadiene is not antiaromatic but non aromatic, ie the same energy 4A + 4B is attained by applying the Hückel LCAO, as for two isolated double bonds, 2 times 2A + 2B.

Given that cyclobutadiene can only be isolated as non delocalized system when in matrixes and a very low T points to strain as a reason for instability of both forms, the delocalized one remaining hypothetical.

Edit for the readers not familiar with the terms: A system obeying all the requisites (cyclic, pi orbitals on each atoms) is antiaromatic when the hypothetical delocalization destabilizes it as compared to the localized system. If no energy difference arises, the system is called non aromatic.

Beware there are plenty of pages discussing non aromaticity in a trivial way, e.g. aliphatic or linear compounds. Cyclobutadiene is non aromatic within the Hückel picture. Also it seems worth to point out again that antiaromatic refers anyway to an hypothetical structure, not to an actual delocalization. A system which cannot avoid it by distorting would be unstable tout-court. Cyclobutadiene, albeit being non aromatic, is on the same line. Indeed you can find antithetical assertions about it, such as it being "the archetype of antiaromatics“ along with “it reacts as diradical“ as if the two things would be straight forward correlated. Normally a system escape antiaromaticity. Being cyclobutadiene non aromatic results in a distorted rectangular molecule, but only when it cannot escape this anyway unstable situation. See also Resonance structure of cyclobutadiene?. Although the situation of cyclobutadiene is again erroneously described as an antiaromatic case, it gives the answer. Also worth reading is the answer by Georg.


Why is 11-oxabicyclo(4.4.1)undeca-1,3,5,7,9-pentaene aromatic? I have been told by my instructors that for a compound to be aromatic, it must be planar. In this given molecule the oxygen seems to be out of the plane of the conjugated system: how can it still be aromatic, then?

The aromatic system should be planar. The oxygen is not part of the aromatic system.
Please note, of course, that your drawing does not capture the 3-dimensional geometry of the system. You would still have to analyze the carbon skeleton to see if it satisfied the planarity constraint enough to be aromatic.


Why is the voltage produced by a voltaic cell affected by temperature as described by the Nernst Equation? I understand why changing the concentrations of the electrolytes in a voltaic cell affects the voltage. Given a zinc-copper cell for example, if I were to increase the concentration of the zinc sulphate electrolyte, the equilibrium between zinc ions and zinc metal would shift towards the side of the zinc metal. This would cause the reduction potential for zinc ions to increase, decreasing the overall cell potential. Please correct me if I’m wrong as this is how I understand it!

However, I don’t quite understand why temperature affects the voltage the way it does. According to the Nernst equation, if the reaction quotient is greater than one (meaning that the zinc sulphate concentration is greater than the copper sulphate concentration in this example), increasing temperature decreases voltage. Conversely, if the reaction quotient is less than one, increasing temperature increases voltage.

I am familiar with Le Chatlier's principle, but if I were to apply it to this scenario, given that the redox reaction taking place is exothermic, it would suggest that increasing temperature would always decrease the voltage produced since it would cause a shift towards the reactant side. This, of course, does not correspond with the relationship predicted by the Nernst equation.

I have looked all over the internet but I could not find any answer to this seemingly straightforward question. Any help is greatly appreciated!

Consider that even when exist a "driving force" to move ions to electrodes, it is necesary that particles stay not static, they need some capacity to move to change direction and take way to electrodes, and what gives this capacity to move, Energy, and in any particle system this energy distribution follows Maxwell-Bolzmann and this means that temperature counts.. e^-(1/kT)


What do positive, negative and zero overlap of atomic orbitals mean? I am confused about positive, negative and zero overlaps.

Do they represent the extent of overlap or are they related to the bonding and anti-bonding orbitals as described in Molecular Orbital Theory?

The terms in the lecture slides you refer to describe the situation in a certain, very specific subset of molecular orbital theory: the idea of linear combining single atomic orbitals to form bonding and antibonding orbitals. While one can arrive at the same bonding/antibonding orbitals with elaborate molecular orbital theory through the localisation of orbitals approach, the simplified view these lecture slides (and many others) present is not consistent with molecular orbital theory. (However, a subset of the result can be obtained by a proper symmetry analysis.)

Let’s consider a rather simple case such as $\ce{HCN}$. It is obvious that the carbon atom’s atomic orbitals will in some way interact with the atomic orbitals of both hydrogen and nitrogen to form bonds. In the method presented in the lecture slides, you would look at certain pairs of orbitals and then determine what type of interaction would be between. I will only present a few representative examples. For consistency, I will define the molecule axis to be my coordinate system’s $z$ axis.

The $2\mathrm p_z$ orbital of carbon and the $2\mathrm p_z$ orbital of nitrogen.

These two p-type orbitals are aligned in the same way. Without taking their phases into account, we can see that the $-z$ lobe of the nitrogen orbital and the $+z$ lobe of the carbon orbital can interact. Depending on whether the phases match or do not match, we would arrive at a bonding and an antibonding orbital, respectively. The resulting bonding orbitals would be a $\sigma$ and $\sigma^*$ orbital.
The $2\mathrm p_x$ orbital of nitrogen and the $2\mathrm p_x$ orbital of carbon.

These orbitals are parallel and still well aligned. However, at any one time both lobes of one will interact with both lobes of the other. There is a nodal plane that completely contains the bond axis. If the $+x$ lobe of both orbitals feature the same phase, the interaction is bonding. Otherwise it is again antibonding. The bonding orbitals are $\pi$ and $\pi^*$, respectively.
The $2\mathrm p_x$ orbital of carbon and the $1\mathrm s$ orbital of hydrogen.

Now, we are trying to mix an orbital that only has one phase (the s orbital) and one that has two phases (the p orbital). This would be okay if only one of the two phases of the p orbital were pointing towards the s orbital; however, because the p orbital is perpendicular to the bond axis, both its phases would have to interact at the same time. Thus, we have one interaction on the $+x$ half of the coordinate system which is bonding and at the same time an interaction on the $-x$ half which is antibonding. Mathematically, the bonding and antibonding contributions cancel out perfectly. Thus, there is no interaction between the two orbitals. This is zero overlap.
The $2\mathrm p_x$ orbital of carbon and the $2\mathrm p_y$ orbital of nitrogen.

Now, we are attempting to mix two different p orbitals. However, they are not oriented in the same direction. The $+x$ direction of one orbital and the $+y$ direction of the other have the same phase while the $-y$ and $-x$ directions have the other. If we start at $+x$ and go around in a circle, we will start in an area of no interaction (the $\mathrm p_y$ orbital has a nodal plane in the $xz$ plane). Then, turning towards $+y$ we reach an area of bonding overlap. After crossing $45°$, the bonding overlap decreases until we reach zero again at $+y$. From there to $-x$, we experience an increasing and decreasing antibonding interaction. We then rinse and repeat what we just saw for the remaining $180°$.

It would probably not be impossible to consider this interaction a curly bond that rotates around the central axis by $90°$ between the two atoms. However, traditional analysis would note that at the same time there is something like an antibonding second part displaced by $90°$ to this. Therefore, the analysis would consider these two orbitals nonbonding with zero overlap.

In a proper MO or LCAO approach, molecular orbitals will stretch across the entire molecule. Thus, the analysis of whether an orbital is bonding or antibonding with respect to a certain bond reduces itself to the question of ‘does this orbital feature a nodal plane along the bond axis or does it not?’. From the $+,-,0$ picture we had above, we arrive at a black and white picture. This is because nonbonding interactions are typically separated out before the molecular orbitals are even calculated by a symmetry analysis. If you do that, you would realise that the $2\mathrm px$ orbital and the $1\mathrm s$ mentioned in case three are representative of two different irreducible representations. (Strictly speaking, $2\mathrm p_x$ and $2\mathrm p_y$ both belong to the same degenerate irreducible representation.) Only orbitals that belong to the same irreducible representation can be combined in bonding or antibonding orbitals.

The atoms combine by colliding with each other. But what does this mean on atomic level. This situation refers to the process in which the two atoms comes so close to each other that they penetrate each other’s orbital and form a new hybridized orbital where the bonding pair of electrons reside.  This hybridized orbital has lower energy than the atomic orbital and hence are stable. It is in the minimum energy state. This partial penetration of orbital is known as orbital overlap.

The extent of overlap depends on the two participating atoms, their size and the valence electrons. In general, greater the overlap, stronger is the bond formed between the two atoms. Thus, according to the orbital overlap concept, atoms combine together by overlapping their orbital and thus forming a lower energy state where their valence electrons with opposite spin, pair up to form covalent bond.

The importance of orbital overlap was emphasized by Linus Pauling while explaining the molecular bond angles observed through experimentation and is the basis for the concept of orbital hybridization.

Directional Properties of bond

The molecular bond angles were explained through the directional properties of bond. The molecule of hydrogen is formed by overlap of 1s orbital in head on collision.

Overlapping of Atomic Orbital

When two atoms come in contact with each other to form a bond, their overlap can be positive, negative or even zero depending upon the phase and sign of the two interacting orbital.

Positive Overlapping of Atomic Orbital – When the phase of two interacting orbital is same, then the overlap is positive and in this case, the bond is formed. The phase of the two interacting orbital (+ or -) comes from the sign of orbital wave function and is not related to the charge in any sense.

Negative Overlapping of Atomic Orbital – When the phase of two interacting atomic orbital is opposite, then the overlap is negative and in this case, the bond is not formed.

Zero Overlapping of Atomic Orbital – When the orientation of two interacting atomic orbital is such that there is no overlapping of the orbital, that is known as zero overlapping.


Which will be the major monochlorination product? Which of these compounds represents the major monochlorination isomer formed in the following reaction?

According to selectivity theory, the answer should be (c). However, the given answer is (b).

c is the least likely.

Alkane reactivity order is tertiary > secondary > primary > methyl reference here

I would be interested to see an explanation of why b predominates over a


Shouldn't the pH at the equivalence point always be 7? I learned in class that the equivalence point in an acid-base titration is reached when the solution contains an equal amount of substance of $\ce{OH-}$ and $\ce{H+}$ ions. However, in a weak acid and strong alkali titration, the pH at the equivalence point (when I added the right amount of acid and alkali to make it a neutral solution) is 9. Also, in the titration of a strong acid and weak alkali, the pH of the solution when I have added the correct amounts is 5.

My question:

If I add 1 mole of ethanoic acid to 1 mole of sodium hydroxide, I should get a neutral solution and the pH should be 7. However, why is the pH 9? This does not make sense to me as a pH 9 suggest that the concentration of $\ce{OH-}$ is higher than the concentration of $\ce{H+}$. But how is that, I reacted the correct amounts?

Two kinds of reactions have to be considered:

Neutralisation of the weak acid with the strong base
Secondary reactions from the products

The neutralisation is fairly straightforward:
$$\ce{HA + NaOH -> NaA + H2O} $$

However, and here comes number two from the list: It doesn't just stop there because we still have unreacted acid in the solution and lots of water. So what will happen? The conjugated base of the acid can also react (as a base):
$$\ce{A- + H2O <=> OH- + HA}\; ,$$

thus raising the pH of the solution. This depends on the equilibrium constant $\mathrm{K_b}$ of the conjugated base.

So although you have reacted exactly equal amounts of a (strong) base with the (weak) acid, you also have to consider the (weak) conjugate base of the acid, which also swims around in your titration soup.

You have confused neutrality point, inflection point and equivalence point. The former is where there are equal numbers of protonated and deprotonated solvent species ($\mathrm{pH}~7$ in water at $\pu{25 ^\circ C}$). The inflection point is where the second derivative of the titration curve (amount of base added plotted against $\mathrm{pH}$) is a maximum (maximum buffering) and the third the point where the amount of analyte and titrant are equivalent. If you take acetic acid and start adding base to it even a small increment of added base will produce a large change in $\mathrm{pH}$. The base isn't reacting with the $\ce{HOAc}$, it is reacting with the solvent. The happens until the inflection point $\mathrm{pH}~4.76$ for acetic acid, is approached in which region the base is reacting with the $\ce{HOAc}$:
$$\ce{HOAc + OH- -> OAc- + H2O}$$

and a small change in $\mathrm{pH}$ requires much more base. Eventually, (about $\mathrm{pH}~4.76,$ that is two $\mathrm{pH}$ units above the inflection point) nearly all the $\ce{HOAc}$ is reacted and additional base just goes to deprotonate water, increasing the $\ce{OH-}$ ion concentration and $\mathrm{pH}$ at a rate much faster than near the inflection point. Thus the fact that you wind up at a $\mathrm{pH}$ near $9$ is simply a consequence of the fact that you are well past the inflection point of your acid which is, because the inflection point is at $4.76$, considered a weak acid. Once all the acid has been deprotonated, additional additions of base go to deprotonate water and the pH rises.

The pH at the equivalence point is the pH at which the solution is neutral. Finding this pH requires that we compute the charge of each species in the solution and totalling them up as a function of pH. We then seek the pH value that zeroes the sum. The species are $\ce{Na+}$, $\ce{OH-}$, $\ce{AcOH}$ and $\ce{AcO-}$. The charge on a liter of water is $Q_w(pH) = 10^{-pH} - 10^{(pH - pK_w)}$ which is the sum of the hydronium and hydroxyl ion charges. For the acetate its a bit trickier. We start with the Henderson Hasselbach equation writing it in the form $r_j = 10^{(pH - pK_j)}$ This is the ratio of the concentration (activity, really) of species which have lost j protons to the that of the species which has lost $j-1$. In the case of acetic acid there is only one proton to lose and so there is only one $r$ but in the case of polyprotic acids there is, of course, 1 $r$ for each proton. Now it shouldn't take you long to figure out that if the ratios are $r_1, r_2, r_3...$ the fraction of undissociated acid molecules is $f_0 = 1/(1 + r_1 + r_1r_2 + r_1r_2r_3)$ As r_1 is the ratio of singly deprotonated to undissociated its clear the the fraction of singly deprotonated is $f1 = r_1f_0$. At this point we have all we need to calculated the charge. Assuming 1 mole $\ce{NaOH}$ and 1 mole $\ce{AcOH}$ in 1 L of water the total charge would be $+1 + 1*Qw(pH) -1*f_1(pH)$ So the question now turns to how we find the value of pH which brings this expression to 0. Clearly the fastest way to a solution is to have and Excel spreadsheet compute it and either grope for the answer (pH 9 gives + $4.75·10^-5$ and pH 9.5 gives $4.75·10^-5$ so clearly the answer lies between those 2 pH values. This suggests a bisecting root finder as a path to a solution and that works very well. Excel has the Solver which does the iterations for you and takes you directly to the solution $pH = 9.38$. Note that the solution for 1 mole each in 10 liters of solution is $pH = 9.38$ which makes it clear that the water is playing a significant role here. Not all the $\ce{OH-}$ are going to deprotonate acid. Many are going to shift the waters charge balance.


Real boiling temperature I've been thaught that water at 1 atm water boils at 100 °C (the so called “normal boiling point”).

However, I came across a more rigorous description of the boiling process saying that in order to fulfill thermodynamic equilibrium between liquid and vapour the pressure inside the bubble must be higher than that outside and accordingly the liquid results superheated (to be at the same temperature of the vapour).

I also understand the nucleation process, but here for sake of simplicity I assume that the bubble is already present.

So my question is why we say that water boils at 100 °C (at 1 atm)?

Simply speaking: The rising bubbles in your pot shrink wherever it is too cold (< 100 °C), and grow whereever it is too hot (> 100 °C), equillibrating the temperature, by exchanging some heat of evaporation. If you keep your thermometer at a distance above the heat source, it will show (mostly, see below) exactly 100 °C, or whatever is the boiling point at the given outside pressure.

Very small bubbles have to have a measurably increased inside pressure, but while the water is boiling, the bubbles grow, and so this difference vanishes. You can calculate it with the surface tension of water (0.06 N/m at 100 °C) using the https://en.wikipedia.org/wiki/Kelvin_equation, and it is negligible for bubbles of millimeter size.

Of course with increasing distance to the water surface, the temperature rises  because of the pressure increase of ~ 0.1 bar/m. Slightly less, because water at 100 °C is less dense.

The boiling point is not identical to the temperature at which a material boils. Unless, of course, you define it that way. I don't know what your technical background it, so I'll mention a couple of basics which you can easily find out more about using widely available on-line resources.

The first point I'll make is that significant figures matter. The correct technical interpretation of your claim about the BP of water at 1 atm is that you meant 1.0 ± 0.5 atm. This is, of course an enormous range of atmospheric pressures. So, while the technical interpretation is clear, you probably didn't intend it. (See Wikipedia for both a discussion of significant figures and for one on atmospheric pressure (range).) The statement that water (assumed to be substantially pure and free of trace contaminants) boils at 100 °C at 1 atm is correct. The statement that the boiling point of water at 1 atm is 100 °C is also correct. Hence, to the accuracy of the values provided, there is no discrepancy.

As an aside, I am impressed that you understand the nucleation of vapor bubbles; I don't. Also, I've experienced (unfortunately) the explosive boiling of a cup of water I placed in a microwave oven. I don't recall its purity. After the heating cycle was complete, I opened the door and apparently the vibrations from that was enough to allow enough of the superheated (liquid) water to vaporize to blow most of the water out of the cup. Not an experience I recommend.

The point here is that while the equilibrium vapor pressure = gas phase (overhead) pressure is the criteria for the definition of boiling point, it just isn't true that water always boils at that temperature (and pressure).

There are two effects that are generally ignored when discussing (thermodynamic) boiling points: surface tension and isotopic composition. Not all water has the same isotopic composition, this fact is used in the study of a variety of historical, geological, and biological data. Surface tension can also be significantly affected by small trace contamination which are often quite difficult to remove. Other important factors are the height of the water column (assuming heating is from below), mixing of the liquid (rarely will you see isothermal samples of water, on a macroscopic scale), and even the composition of the gas phase (it is the Free Energy which determines the equilibrium; N2, O2, Ar, and CO2 all have different energies of interaction with H2O.) Surface tension is important in bubble formation. Small bubbles are especially sensitive to its value. As surface curvature flattens (radius gets bigger), surface tension becomes much less of a factor in the growth of bubbles. Vienna Standard Mean Ocean Water is the 'standard' water typically used in discussions of water's BP. (It is a pure water (not salt water) with a specified isotopic composition.) Depending on which temperature scale is used, at 1.00 atm water's BP is 99.99 °C.

The question of when and where bubbles nucleate and how fast they grow depends on convection, heating rate, water column pressure, the liquid's vessel's interior surface (both roughness and composition), and the purity of the water, both inorganic and organic contaminants, as well as dissolved gasses and particulates present. You are correct that in order to form a small bubble (here, small means on the order of 10 to 100 microns) surface energy (surface tension) requires a temperature and pressure higher than the equilibrium over the (flat) surface. Once bubbles attain sizes of 100 microns or so, the decrease in surface energy as the bubbles continue to grow is negligible. (Growth scales with surface area which scales at r², which is why large bubbles form preferentially to the same volume of small bubbles.) You can look at this as work, in the thermodynamic sense, as long as you realize that thermodynamics was long established before the atomic theory of matter (and before quantum mechanics). The two views, looking at the thermodynamics (energy and entropy) and looking at the molecular forces involved, are separate and complimentary perspectives. It's simplest not to mix them. From the simple equilibrium thermodynamic point of view, the process of boiling violates the assumption of "near equilibrium". That is, we need to be very careful about using thermodynamics in the discussion of bubble formation. All this is background to my last statement on the subject. Water boils near enough to 100 °C at 1 atm, given the typical ranges of composition, air pressure, temperature calibration, heating, vessel, etc.etc., to make it a piece of useful information.


Is 6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol a valid name? I recently study organic chemistry and I read a book by Sana Jamshaid. I came across this compound and I cannot understand why it named 6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol:

I understand that the longest chain is 8, hence octa (but if we go up on the 4th longest can be 9) but I don't understand the 4-(prop-2’-enyl). Are there any explanations to why the compound named in such way? The double bond also sits on the 7th and not the 6th.

The suggested name
“6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol”
is not correct; the closest valid name is
6-ethenyl-4-(prop-2-en-1-yl)oct-6-en-3-ol.
However, this name does not describe the compound that is shown in the picture. Note that the parent chain actually is non-7-en-3-ol instead of oct-6-en-3-ol, and the longest side chain actually is but-3-en-1-yl instead of prop-2-en-1-yl.
Therefore, the correct name for the compound that is shown in the picture is
(7E)-4-(but-3-en-1-yl)-6-ethenylnon-7-en-3-ol.

I also agree with @Loong, the suggested name "6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol" seems incorrect. The correct name for this compound should be (7E)-4-(but-3-en-1-yl)-6-ethenylnon-7-en-3-ol, as @Loong mentioned.
In this compound, the longest carbon chain has 10 carbon atoms, but we select the carbon chain with 9 carbon atoms as it has a functional group (-OH) attached. So, parent chain will be non- as there are 9 carbons.

Now, we number the chain from left to right, as the main functional group is always assigned the least possible locant.

So, we see that we get -OH at locant 3, -CH2CH2CH=CH2 at locant 4, -CH=CH2 at locant 6 and double bond at locant 7. So these substituent/functional groups can be named accordingly:

3-ol
(but-3-en-1-yl) (as carbon atom attached to parent chain (alkyl) is always numbered 1)
ethenyl (vinyl group)
7-ene (NOTE: the ending -e will be dropped while writing the complete IUPAC name)

Also, since it is a trans- (w.r.t double bond at locant 7), we put (7E)- in the beginning of the name. (Z is put for cis- )
So, now, the IUPAC name should be (7E)-4-(but-3-en-1-yl)-6-ethenylnon-7-en-3-ol
Here are a few resources:
NCERT Class XI Organic Chemistry - some basic principles and techniques
Systematic IUPAC nomenclature - Aditya Vardhan V
cis- and trans- nomenclature in organic compounds


How to calculate the concentration H3O+ in a solution with pH=6.99? What is the correct way to calculate the concentration $\ce{H3O+}$ in a solution with $\ce{pH}=6.99$?

Attempt 1.

  pH<7, therefore there are only $\ce{H3O+}$ particles in the solution. $[\ce{H3O+}] = 10^{-\ce{pH}} = 10^{-6.99} = 1.02 \cdot 10^{-7}$

Attempt 2.

  We have $[\ce{H3O+}] = 10^{-\ce{pH}} = 10^{-6.99} = 1.02 \cdot 10^{-7}$ and $[\ce{OH-}] = 10^{-\ce{pOH}} = 10^{-7.01} = 9.77 \cdot 10^{-8}$.   Because of $\ce{H3O+ + OH- -> 2 H2O}$ we are left with $[\ce{H3O+}] = 1.02 \cdot 10^{-7}- 9.77 \cdot 10^{-8} = 4.6 \cdot 10^{-9}$

When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking very large. So I wonder what the correct way is?

If you take a sample of pure water, there will be few hydroxide and hydronium ions. Of course, they can combine to form water and yes they do combine but there will be few water molecules which break/combine to form the ions again. Hence, there exists a dynamic equilibrium between concentration of ions and water molecules.

$\textrm{pH}$ by definition is the negative logarithm of hydronium ion concentration.

$$\textrm{pH} = -\log [\ce{H^+}] = -\log [\ce{H3O^+}]$$

You can obtain the concentration of H+ ions by substituting the value of pH in the following formula,

$$[\ce{H3O^+}] = 10^{\mathrm{-pH}}.$$

Your attempt 2 is flawed because your assumption that all the ions combine to form water molecules is incorrect. There will always be some concentrations of the ions and all of them needn't combine to produce water molecules. Your attempt 1 is correct.

It appears like you are not aware of the concept of equilibrium and self ionization of water, I have picked few good materials which you might(should) want to refer to,

Chemical Equilibrium

Self Ionization of water

The concept of chemical equilibrium is very important and you will come across it frequently in chemistry, so you must learn it. Also, self-ionization of water along with chemical equilibrium are central concepts for learning acids and bases.


Is the total entropy change of all isothermal processes 0? We define  total entropy as: $$\Delta S_\mathrm{total}=S_\mathrm{system}+S_\mathrm{surroundings}$$

I know that for an isothermal AND reversible expansion: $S_\mathrm{total}= 0$

But what about an isothermal expansion at a constant pressure? Does $S_\mathrm{total}$ also equal 0? If not, then would the equation look something like this:
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$

where $q_\mathrm{surroundings}$ is the heat lost or gained by the surroundings by a process and $q_\mathrm{rev}$ is the heat lost or gained by the system IF that process was to become reversible.

Is the total entropy change of all isothermal processes 0?

No. Only for reversible processes.
However your equation is correct.
For an irreversible process in which an ideal gas expands isothermally against a constant external pressure
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$
becomes
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{w}{T}$$
where $w=-p \Delta V$, since $q_\mathrm{surroundings}=-q_\mathrm{system}$ and for the isothermal process the first law stipulates that $w=-q_\mathrm{system}$.
For a free expansion $p_\textrm{ext}=0$ and no work is done. No heat is exchanged and the entropy of the surroundings does not change. The change in entropy of the system then drives the process.

Isothermal expansion at constant pressure is NOT a reversible process and hence ∆S(universe) is always greater than zero.
I.e. $$\Delta S_\mathrm{univ}=\frac{Q_\mathrm{actual}}{T}-\frac{\int PdV}{T}$$

$$i.e. ~ \Delta S_\mathrm{univ} >= 0 $$

Equality holds only for reversible processes.

\n

Only reversible processes are iso-entropic.


How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond? For example, which is the right way to name this compound:

Is it 2-bromo-4-methylenehexane or 4-bromo-2-ethylpent-1-ene?

Does the larger number of carbon in a chain take priority, as in the former, or does the carbon-carbon double bond take priority, as in the latter?

The IUPAC recommendations on this matter have been changed. In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed:

  (…), the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second criterion.

For example, a preferred IUPAC name is 3-methylidenehexane (not 2-ethylpent-1-ene) since the longer chain now supersedes a shorter unsaturated chain. (This example is given in the Blue Book.)

Besides, the prefix ‘methylene’ is no longer recommended in IUPAC nomenclature to designate the $\ce{=CH2}$ group. The preferred prefix is ‘methylidene’.

Therefore, the correct name for your example is 2-bromo-4-methylidenehexane.

From wikipedia:

Identification of the parent hydrocarbon chain. This chain must obey the following rules, in order of precedence:

It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.

So the alkene takes priority over the alkane and 4-bromo-2-ethylpent-1-ene is the correct name.


How to manage complex equilibria when studying a mecanism? Consider a complex multistep reaction mechanism with several intermediates. There are multiple possible pathways as well as equilibria in solution. One wants to study the reaction mechanism using computational chemistry.

It is easy to calculate the $\Delta G$ and $\Delta G^\ddagger$ of individual reactions, but how do you take in consideration all the possible equilibria?

For example, consider an aldol reaction in presence of water (taken from Wikipedia):

Acetone can be either as a ketone, a protonated ketone, an enol or an enolate. This means that there are 16 possible pairs (although only 4 which make sense). These will yield different intermediates, which can dehydrate in different ways (ex. by protonation of the hydroxyde first or not, through a ketone or an enol or an enolate...)

In this case, many possibilities could be eliminated as too high in energy just by general knowledge of chemistry, but if you don't know enough about the system, how do you tackle this problem? How do you handle all the equilibria possible?

If you know what the final equilibrium products are or may be, and you know how much mass you start with, either as elements or molecules, you can calculate the Gibbs energy of the products as a function of (unknown) concentrations (assuming constant temperature and pressure) and then minimize the Gibbs energy with respect to those concentrations. You have to constrain the solution to conserve all of the elements that you began with, and constrain it to have all concentrations be non-negative.


Catalytic poison in Pt catalysed manufacture of H2SO4 I came across a question that goes as:

  In manufacturing of sulfuric acid in presence of platinum catalyst, which metal impurity acts as a catalytic poison?

  Fe
  Cr
  Cu
  V

I searched everywhere I could but did not find anything that talks about this.
I read about $\ce{As2O3}$ being a catalytic poison in this reaction and recommendations for using the Contact process using $\ce{V2O5}$ instead of this reaction.

I can understand that this is more of a fact based question but reasons or explanation (if any) should please be included.

$\ce{Fe}$ and $\ce{V}$ are used in the contact process.
$\ce{Cu}$ is used in the preparation of chlorine gas.
Thus $\ce{Cr}$ can be used as a catalyst poison.


Are there methods for iron content determination in spinach that do not require thiocyanate? I am looking for a way to measure the iron content of spinach for a chemistry experiment. Unfortunately I am not sure that thiocyanate will be available and I'm therefore trying to find an alternate method. The most common one seem to be a titration with potassium permanganate. The issue with this method is that other much more abundant metals in spinach would also react with the permanganate.
I had the idea to ash the spinach, dissolve the ashes in water. Then use a displacement reaction since iron is the least reactive metal in the chemical makeup of spinach. I am not sure exactly how but I thought it might be possible to use a compound to precipitate the more reactive ions out of solutions.
Do any of you have some ideas regarding this experiment?

There are iron testing kits in water which are available at an affordable price. I have not used them but read more about them here Testing iron in well water. You can test your spinach extracts for a yes no type answer if iron is present in significant quantities or not. Just keep in mind that PopEye's spinach is really not that rich in iron. Too much iron is a liver toxin.
I am afraid to share the bad news for you that with the resources you have, you cannot reliably determine iron content in spinach. Micronutrients or minerals cannot be determined by classical methods because they are present in a complex matrix and in very small quantities (parts per million range). They need instrumental analysis (absorption by molecular complexes, atomic emission or atomic absorption). I saw a couple of papers from ResearchGate which talk about the titration of spinach extract with permanganate. Nothing can be more wrong than this procedure in analytical chemistry. They are basically titrating all the organic stuff with it. Those methods are horribly wrong. Someone on SE Chemistry posted a query on how to determine iron in spinach by titration Iron in spinach...that is all incorrect. Yes, you will generate some numbers with those burette readings but those numbers have no significance.
Regarding your ashing method, that will not work either, because you need at several kgs of spinach to get a small amount (milligram) of iron. What should be the ashing temperature? How would you isolate iron from the ashes? Selective precipitation to get rid of other minerals? Nope, that is a bigger headache. For that you would need more reagents.
If this is a school project, make sure you have the resources after searching the literature. If this is an home experiment- classical methods will not work.

As you are looking for inventive ways to possibly measure Fe presence for a school presentation, may I suggest a crude (but likely interesting) visual aid in possibly determining relative iron concentration.
One idea is that Fe ions in the presence of acidic H2O2 will engage in radical production. The latter reaction in the presence of suitable light over time could result in a cyclic reaction system (for example, photo-Fenton). The latter is a conceivable path to magnify even small transition metals presence.
The latter radicals (especially, the powerful hydroxyl radical) can breakdown organic matter, including, for example, a strongly colored dye over time, resulting in eventual de-colorization. The relative periodic time evolution for the latter could be a manner to simply display/document apparent differences in even small relative iron (also copper, manganese,..) concentrations.
Here is a citation from the literature as an example, Effect of iron salt on the color removal of water containing the azo-dye reactive blue 69 using photo-assisted Fe(II)/H2O2 and Fe(III)/H2O2 systems.
So, spinach grown in iron-rich soil, reduced to ash added to acidic H2O2 (plus perhaps more acid added 1st to the spinach,  further lowering pH) in the presence of a photo-sensitive dye subject to photolysis, should display more rapid de-colorization (as captured by periodic snapshots) than one grown in iron-poor soil.
You may experiment with various ways of numerically quantifying the tentative iron content, as deemed viable given limited resources (including funds which alternatively, could purchase an iron testing kit, available here, for example, at $99, or less expensive water testing strips here).
Note: I am assuming one does not have the time to actually grow the same variety of spinach. Otherwise, I would simply speculate that iron-richer spinach may possess different coloration (and perhaps other discernable quantifiable attributes) for the purpose of a simple relative classification as to iron content.
This would conceptually be based, relatedly to above, on naturally occurring photosynthesis. To quote from a supporting source:

Iron, one of the structural elements of organic components that play an essential role in photosynthesis and nitrogen assimilation of plants, is available at extremely low concentrations in large parts of the Southern Ocean's surface waters. We tested the hypothesis that photosynthesis is the primary target of iron stress in phytoplankton living in this specific environment, resulting in a reduced carbohydrate production.

On the suggested coloration argument relating to iron content, also a confirming reference, to quote:

Leafy vegetables featuring a deep, dark green color, such as spinach and collard greens, are usually high in iron.


How can we replace volume in ideal gas equation with V - nb? We know that ideal gas equation is $PV= nRT$ where P is the pressure of the ideal gas and V is the volume of the ideal gas. Thus, we can write :
$$ P_{\textrm{ideal}}V_{\textrm{ideal}} = nRT$$
Now from volume correction term of van der Waals' equation we know that
$$ V_{\textrm{real}} = V_{\textrm{ideal}} - nb $$ where $ V_{\textrm{real}} $ is the volume of the real gas and $ V_{\textrm{ideal}} $ is the volume of the ideal gas. Clearly, $ V_{\textrm{ideal}}$ and $ V_{\textrm{real}} $ are different.
Then how can we write :
$$P_{\textrm{ideal}} (V_{\textrm{ideal}} - nb) = nRT$$
How can we replace $ V_{\textrm{ideal}}$ with $ V_{\textrm{real}}$?

The actual volume occupied by the gas is less than the measured volume(volume of the container).
In reality gaseous particles are not point objects and don't occupy zero space, which is against one of the postulates of the Kinetic Theory of Gases.
Thus while dealing with real gases we have to account for the volume which has been occupied by the gaseous particles. We do so by assuming the volume of gaseous particles to be $nb$.
Now, the ideal volume(actual volume) occupied by the gas is the measured volume(volume of container) minus the volume occupied by the gaseous particles.
Thus, $V_{ideal} = V_{real} - nb$ and not the other way around.
Note, $V_{ideal} =V_{actual}$ and $V_{real} =V_{measured}$.
The mistake you are making is formula you are using and thus
$P_{ideal}(V_{ideal}-nb) ≠ nRT$, but
$P_{ideal}(V_{real}-nb) = nRT$

There is no definite answer to this question. It depends on you. It is only a question of precision required. To give you an example, the ideal volume of $1$ mole of an ideal gas is usually taken as
$\pu{22.4 L}$ at $\pu{0°C}$ and $\pu{1 atm}$. Here are the observed molar volumes $\pu{V_o}$ of a couple of gases at $\pu{0°C}$ and $\pu{1 atm}$ :
$\ce{V_o(N2) = 22.402 L}$
$\ce{V_o(O2) = 22.393 L}$
$\ce{V_o(H2) = 22.430 L}$
$\ce{V_o(CO2) = 22.262 L}$
If you are satisfied with the usual value ($\pu{22.4 L}$), forget about the van der Wals coefficients at usual pressures and temperatures. If you want to be more precise, do use these coefficients. Of course at much higher pressures, like about $\pu{100 atm}$, it makes sense to use van der Wals formula.


Woodward Fieser rules for dienes - non conjugated double bond as exocyclic? When looking for exocyclic double bonds in a diene, should a double bond not connected to the conjugated system be considered?

For example, in this compound, should the marked double bond considered as an exocyclic double bond?

Woodward's rules are an empirical method for predicting the maximum absorption of a molecule based on the functional groups present (that is, conjugated functional groups which possess a distinctive maximum absorbance such as a diene or an unsaturated carbonyl).

  Should the marked double bond considered as an exocyclic double bond?

Exocyclic is defined as being outside of the ring system (or in more complex systems, outside of the ring system being considered), the bond highlighted is indeed inside the ring system and is as such endocyclic.

If we start with the alkene at the bottom left, then it is in conjugation with an endocyclic alkene (to the immediate right) and an exocyclic alkene (in the ring to the right hand side). These double bonds are all in conjugation with one another.

The bond you've pointed at isn't in conjugation with the 'triene'

The Woodward rules are specifically related to the effect of groups in conjugation, a simple test of which is to see if you can push curly arrows between the pi bonds. If you try this for your molecule you'll quickly see that the bond you've highlighted is too distant from the conjugated system to have any involvement.

If you find a double bond outside a ring (as in the 4 ring system here), it will be considered as exocyclic. So, the pointed double bond is exocyclic to the adjoining cyclopentane ring.


What exactly is temperature? I've read at many places that temperature is the average kinetic energy of particles present in an object. I just don't intuitively get how kinetic energy is connected with temperature. And how is heat connected with temperature then? Then what exactly is temperature? All of the descriptions given online are very confusing.

Heat is the transfer of energy to or from the body in forms other than matter flow or work (organized energy transfer, such as pushing).

Temperature is only a well-defined property for a collective body (you wouldn't be able to tell me the temperature of a single atom, for example). Like you said, it's the property of matter describing the amount of kinetic energy of the particles in the body. As to why this is, I'd ask: what happens at absolute 0?

   At absolute 0, heat has transferred out of the system so much so that you cannot lower the energy of the system any more.

(Note for the knowledgeable reader: there is indeed still a quantum mechanical phenomena -- zero point energy-- that prevents some energy from leaving the molecule, but that's a conversation for another time).

From a thermodynamic definition, temperature is the description of how the internal energy changes with entropy for a closed (no matter flows in or out) system of constant volume:

$T=\left(\frac{\partial{U}}{\partial{S}}\right)_{N,V}$

By increasing the entropy of the system by a fixed amount, the temperature of the system tells me by how much the internal energy will increase. Now, this is not a very useful form, as you cannot directly increase the entropy of a body (you must add energy and then let the entropy indirectly increase). It's much more useful to consider the inverse temperature:

$\frac{1}{T}=\left(\frac{\partial{S}}{\partial{U}}\right)_{N,V}$

At absolute zero, everything is in the lowest energy state. Any small transfer of energy to the system will result in a large increase in entropy. But this was only a small change, so the internal energy doesn't increase by much. Compare to a system at room temperature, where I must transfer much more energy to the system to achieve the same magnitude of increase in entropy.

Temperature is the average kinetic energy of the particles making up a system. That's it, and it is correct. Any other definition, and there are many of this page, are either equivalent or incorrect. What's the problem?


Why Zn has highest ionisation enthalpy in 3d series? Zn which has the highest ionisation enthalpy in 3d series.The reason given in my textbook is:

  The value of zinc is higher because it represent ionisation from 4s level.

This is not correct because for every 3d element ionisation is from 4s orbital. The explanation I found for this is from this site is:

https://www.chemguide.co.uk/atoms/properties/ies.html

  Increase in nuclear charge outweigh the electron electron repulsion of 4s electrons rendering Zn an higher ionisation enthalpy than Cu.

This seems like a probable explanation but if Zn has higher ionisation enthalpy than Cu due to its increased nuclear charge.Why can't Zn have smaller atomic radius than Cu due to its increased nuclear charge which can decrease the size of the atom. But experimentally it has been found that Zn is bigger in size than Cu.

So why does Zn has the highest ionisation enthalpy in 3d series?

TEXT SOURCE:NCERT textbook (chapter:d and f block)

There is band structure formed in the metallic lattice  in the crystals of d and f block elements due to partially filled orbitals. But as you move along the periodic table to 'Zn' , you will find that the orbitals are completely filled hence there is no possibility for inter-orbital mixing. Another thing worth noting is that due to the filled d-orbital, zn and copper behaves more like s-block rather than d-block elements.

"The filled d-shells in these elements contribute little to interatomic interactions, which are dominated by the s-electrons through metallic bonds. Unlike metals with incomplete d-shells, metallic bonds in copper are lacking a covalent character and are relatively weak. "

An alternate explanation can be on the metallic lattice:
Copper has a face centered cubic structure and zinc has a hexagonal close packed structure

reference : wikipedia

In the Zinc atom, the levels $4s$ and $3d$ are completely occupied. This increases the stability of the atom. And electrons need more energy to be removed.


Para effects on substituents in aromatic compounds I wanted to know how to compare the following compounds on basis of acidic strength.

I have to find the acidic strength of following compounds with $\mathrm{X}=\ce{CH3, \ OCH3,}$ and $\ce{NO2}$.

For ease of explaining, denote the first compunds as (A), the second as (B), and the third as (C)

Due to steric inhibition of Resonance (SIR) effect, compound (A) will be the most acidic, as its anion will be the least stable.

Now for ordering the other two, I know that there are two things affecting acidic character:

Bond strength; the weaker the bond, the higher its acidic strength
Charge on hydrogen; the more positive charge their is on the hydrogen, the more its acidic strength.

So now if we take $\mathrm{X}=\ce{CH3}$, we should have that $\mathrm{(A)>(C)>(B)}$; as $\ce{CH3}$ is an electron donating group and it would make the hydrogen less positive, and inductive effect is distance dependent. This answer is incorrect as per my textbook, which says $\mathrm{(A)>(B)>(C)}$.

If we take $\mathrm{X}=\ce{OCH3}$ which is an electron withdrawing group, we should have $\mathrm{(A)>(B)>(C)}$. This answer is correct.

If we take $\mathrm{X}=\ce{NO2}$, which is also an electron withdrawing, we should have $\mathrm{(A)>(B)>(C)}$. But my textbook says the answer is $\mathrm{(A)>(C)>(B)}$.

I wanted to know where am I going wrong, and if there is some other effect I am not taking into account.

Note: my earlier question was closed due to the similar question: Ortho-effect in substituted aromatic acids and bases.
But that question only deals with the ortho effects of substituents, not the para.

At meta position the M effect is not considered. The reason behind is that during resonance of a six membered cyclic compound charges never reaches at meta position that why it is so try to proof it by drawing resonance structure and if you didn't get us then tell me in the comments below.


Why won't water freeze if you put ice in it, despite ice being frozen? Just now I got some room temperature water and added some ice to cool it, and it just came to me, like the scientist I wish I was ... why doesn't frozen (freezing point) ice added to higher temperature liquid compound force the water molecules to solidify?

Why won't water freeze if you put ice in it?

It will, even at room temperature. You just need a big enough, cold enough ice cube.

Don't believe it? Add a few drops of water to an ice cube in an ice cube tray (which is the same as adding an ice cube to a few drop of water). Wait a few seconds, turn the tray upside down. No water will fall, presumably because it has frozen. Repeat, but immediately turn the tray upside down. You'll get a wet floor. You can conclude that, given enough time, the water will be frozen by the ice. To the whys and hows:

From the second law of thermodynamics we know that:

  An isolated system, if not already in its state of thermodynamic equilibrium, spontaneously evolves towards it.

And

  Heat cannot spontaneously flow from a colder location to a hotter location.

So if you put ice and water together in a cup, heat will flow from the water to the ice until they reach thermal equilibrium, which will be a temperature between the water's initial temperature and the ice's initial temperature. If the temperature of the ice was low enough or if there was enough ice for the water to keep giving away heat, then the water will freeze. That's considering our cup is an isolated system.

Why don't you usually see that happening? Well, regular ice cubes are probably around −5 °C, and there usually is more water than ice in a cup, so before the water has a chance to reach 0 °C all the ice has already melted. Besides that, the ice is receiving heat from the room, unless you're in a place where's below 0 °C. In that case, the water WILL eventually freeze, but mostly because of the environment.

Now if we have a look on values of sp heats of ice and water.

L(ice)=334 kJoule per kg

C(water)=4200 Joule per kg per degree Celsius

C(ice)=2100 Joule per kg per degree celsius

If we want 1 kg of water at 1 degree celsius to first reach zero degree Celsius and then be converted to ice the amount of heat released will be greater than 4200 joules

On the other hand,  If we want 1 kg of ice at minus one degree celsius to first reach zero degree Celsius and then be converted to water the amount of heat required will be around 2100 joules

Therefore in general if we want water to be converted to ice the ratio of ice to water should be more than 2 isto 1.
(Assuming temp of both are at same distance from 0degree celsius)


Dichloromethane cleaning Is it safe to drink out of a bottle that use to contain Dichloromethane?  What should I use to clean a bottle of Dichloromethane?  Is there any chemicals that I could use to dissolve the Dichloromethane and make it safe to drink out of?

More info:

So if I had a glass bottle containing >= 99.5% Dichloromethane with amylene as a stabilizer and left out overnight is that safe? The bottle was used in a chemistry department so it is high quality, I would assume. Made from the company Sigma-Aldrich and in a Glass bottle.

Safety regulations in every lab I have heard of strictly disallow mixing of containers of any sort that contain, have contained or will contain chemicals with containers of any sort that contain, have contained or will contain food, beverages or anything else intended for human consumption. You should never drink out of a bottle that had contained chemicals, you should never store food or drinks and chemicals together and you should not consume anything where there are chemicals. Your question has a premise that should never happen due to safety regulations.

With that gotten out of the way the answer is quick and dirty: It’s not the dichloromethane that is your prime point of concern. Dichloromethane boils at $40~\mathrm{^\circ C}$ and if you keep a bottle that contained it open and lying down overnight at room temperature, there are no traces of dichloromethane in harmful concentrations left. But unless you really go out of your way to buy $99.999~\%$ purity dichloromethane, it will contain impurities which generally do not evaporate as easily, i.e. which will stay in the bottle. Since no bottle of dichloromethane is designed for drinking out of it, there are no safety tests performed with these impurities or additives and thus you cannot assume it to be harmless even if dichloromethane is gone.

You should not read this answer to say that bottles of highest-purity dichloromethane be safe to drink out of. Never mix food or beverages and chemicals. Never, ever!

There are chemicals that kill and some that don't, learn to differentiate and how to read MSDSs.

Try not to mix random chemical bottles with food containers for a lot of reasons, mostly because it might kill someone (you) due to compounded mistakes.

As for this scenario I would say sure.

A typical residue on evaporation for this type of product is <0.0005% that is 5 parts per million. If you have an empty 1000ml bottle you might have 2ml of remaining product.  If you let the left over product in your empty bottle evaporate (bottle mouth down is better due to drainage and density of vapour) you would have 0.000005 x 2 = 0.000010g or 10ug or residue left. If you somehow were to dissolve in water and ingest all the residue and it was very toxic you would be getting 10ug total dose that is made of all the stuff that is non-volatile.

In any event the 2ml (near 2700mg) of product left in an empty bottle would be enough to kill half of a population of four 500g rats if NOT evaporated. By simple (not accurate) extrapolation a 70kg adult would need to drink 100g (75ml) of product to have a 50% chance of dying, instead we are talking about the residue from 2ml (2.7g) product with all the poisonous stuff gone.

You probably don't even need to rinse the bottle, though that is what I would do (using clean acetone if I was feeling paranoid, our bodies create acetone and know how to metabolise it in small quantities so don't fret) both solvents will evaporate to dryness (and be clean for almost any use), what is left may be less toxic than the residue from dish-washing liquid.

Also note it is used in some extraction processes in the food industry an produced in some way during the chlorination of municipal recycled drinking water.  It has been linked with lab rat cancer when administered in significant doses but not in humans working with it over long periods.

Please note that some chemicals are not this easy to avoid and may cause problems down the line even with low doses. I believe sugar and prescription drugs are a bigger concern than Dichloromethane residue.

Please down vote only if you have read the linked documents and have made an informed comment about Dichloromethane evaporation residue toxicity so people know where I have made errors, I will happily correct any brought to my attention.

EDIT:
Hmm, down votes without useful reasons.  If you look up the toxic doses for caffeine, ethyl alcohol, nicotine, dish washing liquid, dishwasher tablets, quinine (tonic water), cream of tartar, acetic acid and too many other 'acceptable' chemicals then you may find they are worse than dichloromethane and don't evaporate as well.

What gives, did you people not actually read this answer before down voting?

Is fear mongering so much fun that you have to remove your own judgement from your own actions. One day you may actually need to decide which of two chemical bottles you need to carry water across the desert to save your life, are you going to pick the right one or are you going to say, yuk, chemicals and walk to your death from dehydration.


Reason for vibration of molecules Why do molecules of a body exhibit vibrational motion? Is it due to interaction between various molecules of the body, or interatomic interactions between atoms in the same molecule? If it is because of any of them, how can these interaction cause the molecules to vibrate?

Here is a picture and an equation that may add to the comment LordStryker made.  Molecules vibrate because they have energy. First the equation, as $T \to 0$ (absolute zero), the first term will vanish, but the molecule will still have an energy given by $h\nu/2$.  So even at absolute zero a molecule will have some vibrational energy, this is called the "zero point energy".  The parabola (representing the hydrogen well, for example, that two hydrogen atoms will "fall into" when they bond and become a stable hydrogen molecule) gives a pictorial representation of why this must be - energy levels are quantized.  Even the lowest level is above the bottom of the parabola, so even in the lowest energy level at absolute zero, the atom or molecule will still have a non-zero energy which will cause it to vibrate.

$\displaystyle\varepsilon = \frac{h\nu}{\mathrm e^{h\nu/kT} - 1} + \frac{h\nu}{2}$

Taking a stab at theoretical physics here.

As long as there is some energy, the electrons and other sub atomic bits will move randomly. This randomness will result in tiny fluctuations in the distribution of energy /mass inside the atom resulting in vibrations that can be passed on to adjacent atoms.

The fluctuations will also result in fluctuations in charge distribution inside the atom, that can affect the state of adjacent atoms, again causing vibrations and rotations.

Before long, a feedback loop could in theory arise between adjacent atoms whereby the motions of a neighborhood of atoms may appear random to the macro observer, but be quite deterministic at the sub particle level.

I have no proof of this from an experimental standpoint, but from my understanding of the topic as an energy engineer, this makes sense to me. Cheers


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

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

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

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

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

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

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

I think the question wants you to consider the SN1 mechanism as well in your rankings. As Q and S each have a more electron rich ring than R, they generate a more stable carbocation on loss of the chloride ion than R. Therefore, the rate of SN1 has increased considerably and so a smaller proportion of the molecules go through the SN2 route. The carbocation forms on the carbon bonded to the ring therefore it is easily stabilised by the mesomeric effect.
The fact that the question doesn't provide a specific nucleophile makes me believe that they're probing electronic effects rather than steric hinderances.


Why does the Leidenfrost effect (seemingly) not apply to the case of molten NaCl when it is poured into water? On pouring molten sodium chloride into water you can see that when molten $\ce{NaCl}$ (table salt, i.e. sodium chloride) is poured into water ($\ce{H2O}$), the Leidenfrost effect appears to be nonexistent, whereas it clearly manifests itself in instances where other molten salts such as sodium tetraborate (1) (borax) and sodium carbonate (2) are poured into water.

Why does the Leidenfrost effect in molten NaCl last much shorter than that of the other salts, appearing to be nonexistent?

Input from paracetamol:

  "The Leidenfrost effect (physical phenomenon) does show up. See the "cushion" of water vapour around the blob of molten salt in the video? Though short-lived, it's pretty apparent. Now why molten table salt would react with water faster than molten borax or soda (as implied in your post) is an interesting question."

In the video, TheBackyardScientist offered an explanation for the huge explosive effect when 50 grams of molten NaCl is poured into water. He said that some water penetrated the blob of molten NaCl and blew it apart from the inside. A Leidenfrost effect on the outside of the blob is visible, but not the major effect.

The borax blob did not explode, so it must not have been penetrated by a significant amount of water. Borax is a tetrahydrate; when it melts and then solidifies, it goes thru a glassy stage which might be expected to be somewhat cohesive, perhaps resembling Silly Putty. In addition, it loses a lot of water on melting: an initial amount of 50 grams melts to only 26 grams. (Boric acid solidifies similarly.)

NaCl does not exhibit a glassy stage on cooling (note the crystalline surface on the patty from solidified NaCl). Perhaps 26 grams of NaCl would also not explode but merely show the expected Leidenfrost effect. As a 50-gram blob of NaCl cools, solid (strong) regions form over weak molten interior; the weak part of the blob splits open, allowing water into an interior hot enough to boil the water quickly and blast the blob skins apart. The hardened skins of the blob serve as a partial container - like a pipe bomb.

50 grams of washing soda (sodium carbonate decahydrate) will melt down to 18 grams; again, perhaps this is too little to give an explosion.

The Leidenfrost effect should be expected for small blobs of anything. The explosive effect might be expected only for larger quantities, but including even materials cooling thru a glassy stage, if the amount were large enough.

Because, physically speaking, the salt water should never come in contact with the water, thanks to something called the Leidenfrost effect. This effect occurs when a liquid comes into contact with a substance significantly hotter than its boiling point and forms an insulating layer of vapour


The order of aromaticity of benzene, thiophene, pyrrole, furan, and pyridine I have been taught that the order of aromaticity is: benzene > thiophene > pyrrole > furan. However, I have been unable to deduce any logical explanation for that order. Also, I wish to compare pyridine's aromaticity with these, but my professor does not know about it.

According to me, all the rings are aromatic and satisfy the Huckel's rule of aromaticity with $6\pi$ resonating electrons and a planar ring. So, that point is useless for doing any comparison of aromaticity.

But, we note that $\ce{2p\pi-2p\pi}$ overlap is most favorable in conjugated systems. Hence, thiophene should be the least aromatic (since it has a $\ce{3p\pi-2p\pi}$ overlap instead).

I agree that benzene should be the most aromatic because both its resonating structures are equivalent and have no charge separation. Using exactly the same reasons, pyridine should be the second best aromatic ring (lesser than benzene because the N atom is an electron withdrawing group and is reducing the electron density in the ring)

That said, every resonating structure (except the uncharged one) for pyrrole and furan will create a formal positive charge on the hetero-atom. Since N is less electronegative than O, it will be slightly more stable than O with that positive charge. Hence, pyrrole will be more aromatic than furan.

Therefore, according to me, the aromaticity order should be: benzene > pyridine > pyrrole > furan > thiophene.  But it is obviously wrong.

Hence, my question is: what is the order of aromaticity of benzene, thiophene, pyrrole, furan, and pyridine. And why?

The main criteria of aromaticity is resonant or delocalization energy.
It is difference between energy contained in bonds if they were localized like in Kekule's formula for benzene and actual energy.
For benzene if is about 150 kJ/mol.
The more this energy the more stable molecule is.

The approaches for calculating this energy may differ.
As for β-values mentioned in paper by @orthocresol it looks like it is Dewar's resonance energy per one π-electron calculated based on modified Hückel's MO method as mentioned in GSE article.

There is a table in this article with the same β-values:

In English it is

Benzene                            0.065     Pentalene                     -0.018
Naphthalene                        0.055     Azulene                        0.023
Anthracene                         0.047     Fulvene                       -0.002
Tetracene                          0.042     Pyridine                       0.058
Cyclooctadecanonaene([18]annulene) 0.012     Pyrrole                        0.039
Cyclobutadiene                    -0.268     Furan                          0.007
Azete (Azacyclobutadiene)         -0.160     Thiophene                      0.032

All the three molecules- furan,pyrrole and thiophene are aromatic in nature. There is a quite simple explanation for why thiophene is more aromatic than furan. If you draw the resonating structures of thiophene and furan, they seem almost identical. This is until the point when the lone pair of oxygen is about to return to the oxygen atom. In furan, the lone pair comes back on the oxygen atom and oxygen simultaneously shifts the existing pi-bond pair to the adjacent carbon atom. However, in thiophene, when the lone pair comes back to the sulphur atom, thiophene does not need to break any existing pi-bonds as it has an empty d-orbital to store the lone pair in. This results in an extra resonance structure for thiophene, thus increasing the extent of resonance and by extention its aromaticity.


Is it possible to extract magnesium from epsom salt? Is it possible (at home) to extract magnesium from $\ce{MgSO4·7H2O}$, also known as Epsom salt (or, alternatively, from $\ce{MgSO4}$)?

Supposed you are trying to obtain magnesium metal from Epsom salt, $\ce{MgSO4.7 H2O}$:
Is is worth the effort?
Magnesium metal in the form of chips (particle size 4-30 mesh) in decent purity (99.98%) is available from suppliers such as Sigma-Aldrich for about 150 Euro per kilogram.
Can I use electrolysis to generate $\ce{Mg}$ metal?
Yes, but you wouldn't want to do that at home. An older industrial process uses electrolysis of molten magnesium chloride, $\ce{MgCl2}$ (mp. is around 700 °C). While $\ce{Mg}$ metal is formed at the cathode, chlorine gas is released at the graphite anode. Did I mention that this isn't meant to be done at home?
Can I convert $\ce{MgSO4}$ to something more usable to make $\ce{Mg}$ metal?
Yes, in principle. A group of Chinese researchers examined the conversion of  $\ce{MgSO4}$ to  magnesium oxide, $\ce{MgO}$ by heating in the presence of natural gas at temperatures around 500 °C and outlined their results in
Geochemical Journal, 2011, 45, 97-108.
The initial pressure by natural gas at room temperature was 3 MPa, but upon heating to the final temperature, the pressure in the reactor was as high as 12-20 MPa! This can't be done at home and it shouldn't be.
What if I would have $\ce{MgO}$? What could I do with it?
Magnesium oxide, $\ce{MgO}$, is indeed the most important starting material in the production of magnesium metal in the Pidgeon process. Here, magnesium oxide is heated in the presence of silicon in an evacuated reactor. Silicon is oxidized to silicon dioxide, magnesium oxide is reduced to magnesium metal. The latter boils off (bp. around 1100 °C) and is condensed. This isn't kitchen chemistry either.
In summary, you're better off buying some magnesium metal from a supplier of your choice!

Yes, it is possible. If you electrolyze MgSO4(aq), one of the products will be magnesium hydroxide. If you you add Hydrochloric acid to it you get magnesium chloride 2HCl(aq)+Mg(OH)₂(s)⟶MgCl₂(aq)+2H₂O(l), electrolyze that and you get magnesium and chlorine gas, the last step must be done outdoors
Here's a useful link: https://www.youtube.com/watch?v=6BThiJpbBJQ


Converting cyclic compounds to their corresponding aliphatic compounds Is there any reaction available to convert cyclic compounds to their corresponding aliphatic compounds?
eg. Converting cyclohexane to n hexane

The reaction you want to know about is "Cleavage reaction" in particular oxidative cleavage.

In this dihydroxylation, if the solution is warm or acidic or too concentrated, oxidative cleavage of the glycol may occur.

Then they can be reduced to aldehydes by reducing agents.Reduction of carboxylic acids to aldehydes is difficult because aldehydes are more reactive than carboxylic acids toward most reducing agents. Almost any reagent that reduces acids to aldehydes also reduces aldehydes to primary alcohols.

Lithium tri-tert-butoxyaluminum hydride, LiAlH(O-t-Bu)3 is a weaker reducing agent than lithium aluminum hydride. It reduces acid chlorides to aldehydes because acid chlorides are strongly activated toward nucleophilic addition of a hydride ion.

Under these conditions, the aldehyde reduces more slowly and can be isolated. Therefore, reduction of an acid to an aldehyde is a two-step process: Convert the acid to the acid chloride, then reduce using lithium tri-tert-butoxyaluminum hydride.

And now we can reduce it by the Wolff Kishner reduction.

Ethylene glycol, diethylene glycol, or another high-boiling solvent is used to facilitate the high temperature (140–200 °C) needed in the second step. I am assuming you can convert the cyclohexane to cyclohexene by any radical halogenation and then dehydrohalogenation.


Why is mercury liquid? Why is mercury the only metal that is liquid at room temperature?

Note: I am a high school student and would appreciate an answer, if possible, suited for my level of understanding.

Thank you.

There are a couple of reasons why Mercury is a liquid at STP, I will go through each of them with an explanation that won't be too complex for your understanding.

First of all understanding why most metals are metallic is a good place to start. This happens because of the sharing of valence electrons. The sharing of these electrons make metals malleable and ductile and thus not liquid. Generally the more the electrons shared means the higher the melting point and a harder metal.

OK, getting into the Mercury. Like others have probably mentioned above, Mercury doesn't like to share its electrons. It hangs onto them very tightly and thus Mercury-Mercury bonding is very weak and thus the compound melts and boils at very low temperatures. In other words, Mercury lacks the sea of electrons.

So then, why doesn't it like to share electrons? The $\ce{6s}$ electrons come very close to the nucleus and swing around it at near the speed of light. Due to this, relativistic effects occur. In other words, the $\ce{s}$ electrons behave as if they are more massive than other electrons. This lowers the energy of the electrons. To summarize, the electrons are buried deep in the atomic core and thus aren't able to participate in chemistry.

Overall, since you are a high school student, I would probably just say that Mercury doesn't readily bond well and that is due to the $\ce{6s}$ electrons being very attracted to the atomic core.

At room temp at atmospheric pressure mercury will not readily share valence electrons with other atoms. The mercury atoms won't interact with each other to form a structure or lattice to become a solid.


Do lone pairs contribute towards the pi electron count of heterocycles? I'm not able to understand how the following compounds are aromatic.

When should the lone pairs on heteroatoms be taken into consideration when counting the number of π electrons?

Only count the lone pairs/ pi-bonds/ groups which are participating in conjugation and ignore them in all other cases.

For example, in compound 2 (thiophene), there are two lone pairs on sulfur.

One lone pair (brown) is in a p-orbital, and hence participates in conjugation with the two π-bonds. The other lone pair (blue) is pointing outwards from the ring in an $\mathrm{sp^2}$ orbital. This lone pair is orthogonal, or perpendicular, to the π-system and hence cannot take part in conjugation. In total there are six π electrons, and the compound is therefore aromatic.

See also: Conjugated system on Wikipedia.

The Nitrogen atom is already an $\mathrm{sp^2}$ hybridized atom, so its lone pair does not take part in conjugation
The sulfur atom is $\mathrm{sp^3}$ hybridized, it has two lone pairs, so only one of its lone pair will take part in conjugation for it to be $\mathrm{sp^2}$ hybridized atom
$\ce{N}$ bonded to $\ce{H}$ is $\mathrm{sp^3}$ hybridized and has its lone pair taking part in conjugation, thus its lone pair counted as 2 pi electrons. the lone pair of the $\mathrm{sp^2}$ hybridized $\ce{N}$ atom is not pi electrons
Lone pair on $\ce{S}$: one pair is pi electrons, lone pair on $\ce{N}$: not pi electrons