Elements other than carbon that can form many different compounds by bonding to themselves? My textbook says the following:

  Unique among the elements, carbon can bond to itself to form extremely strong two-dimensional sheets, as it does in graphite, as well as buckyballs and nanotubes.

Is carbon the only element that can do this?

If not, then what are the other elements can also do this? Is there a term to describe such elements?

What is the chemical characteristic that allows this to occur?

I would greatly appreciate it if people could please take the time to clarify this.

Is carbon the only element that can do this?

No, carbon is not the only element with such characteristics.

  If not, then what are the other elements can also do this?

There is a whole number of elements such as silicon, arsenic, germanium.

  Is there a term to describe such elements?

At least I'm unaware of such a term, which might be furnished by our far wiser community.

  What is the chemical characteristic that allows this to occur?

Catenation.

More information:

According to the Molecular Orbital Theory, the condition for a compound to exist is that it should have more electrons in the bonding orbitals than in the anti-bonding orbitals. So, as long as you have the bonding orbitals filled more, you can have pretty anything, more than just chains of atoms.

Thus, the existence of a compound also depends on the precise conditions in which the compound is kept, for example sodium forms different types of chlorides under different conditions and that as pointed out by Poutnik in the comments, $\ce{He2^1+}$ and a ton of others are discovered and still more awaiting discovery.

So most of this has been answered, but the main elements that can bond to themselves are called diatomic atoms. This includes hydrogen, nitrogen, fluorine, oxygen, iodine, chlorine, and bromine. There are other elements that can do this as well, however these main elements occur naturally in their diatomic state as gases.


Bond dissociation energy among halogens Among the halogens, why is it that the fluorine has the lowest bond dissociation enthalpy, considering the fact that fluorine is the smallest and the internuclear distance between the fluorine molecule is least and $2p-2p$ overlap should also very very strong ,the bond dissociation enthalpy should have been very high but it is lowest among the halogens. Why is it so??

The strength of the bond is due to 2 things : Difference in electronegativity and atomic radius of the elements .Halogens except fluorine , have a very big atomic radius so less bond energy.And they have 0 difference in electronegativity so even less bond energy.They both "want" to get the bonding pair of electrons closer to them , so because they cant both atoms are "unhappy".


Electron orbital diagram of vanadium I am learning about electron configurations and I came across the "exceptions" of $\ce{Cr}$ and $\ce{Cu}$ where electrons actually half fill the $\mathrm d$ orbitals instead of the expected $\mathrm s$ orbitals. I thought maybe vanadium would do the same, but my textbook gives the answer of an electron configuration exactly following the Aufbau principle.

I've attached a picture of my drawing for the textbook answer and what I had thought might happen — could someone explain why vanadium is not an "exception" like chromium or copper?

Electrons always fill in the lowest energy configuration possible.

Cr and Cu, as well as Cu and Ag, are exceptions in the "typical" filling order.  In the case of Cr and Cu, they are stabilized by having 2 half filled orbitals, which maximizes exchange energy and minimizes electron repulsion.  In their case, the energy to promote an s electron to the d orbitals is compensated for by the effects of exchange energy and no repulsion.  This effect is called "the stability of half filled subshells", or something to that effect, in most textbooks.

The energy cost to promote 2 electrons from the s subshell, as would be the case for Vanadium, is too much to be compensated for by the exchange energy gained and the repulsion lost, so the lower energy configuration is to have two electrons in the s subshell and 3 in the d.

Here's a nice website that shows the electron configuration for all elements: with http://www.ptable.com/#Orbital

Its very simple. You cant skip orpitals. Since 4s is smaller than 3d you will have to fill 4s first and than 3d. But u cant excite both the electrons from 4s cause it would take a very high amount of energy and even if u did the 4s becomes vacant half filling the 3d orbital and thats would be highly unstable. So the only possible stable configuration would be 4s2 3d3.
And in Cr and Cu the excites 1 electrons because by doing that both 4s and 3d orbital gains half filled i.e the stable configuration. So in vanadium if u excite 1 electron the d orbital will still not be half filled. But 4s orbital will be half filled. Since fully filled orbital are more stable than half filled the vanadium will have 2 electrons in 4s orbital.


What exactly is happening when sodium, potassium, or other alkali metal explodes in water? There are lots of videos on YouTube showing sodium, potassium, etc. exploding when dropped into water (this, for example).

I understand that when an alkali metal is exposed to water, a violent exothermic reaction occurs where a hydroxide and hydrogen gas are produced, but why and how does the sample of metal end up detonating and fragmenting?  Physically speaking, how can a block of metal seemingly blast apart from the inside when the reaction occurs on the surface of the sample?

The Wikipedia article on alkali metals explains this, but I still don't seem to understand how this would result in an explosion.  Diagrams would be very helpful.

The key of the question is how does the release of hydrogen gas and energy result in an explosion?  Is there thermal runaway, and is the metal physically destabilized in some way during the reaction?

Until recently the answer was unknown, but a short time ago it was discovered that the reaction is in fact a Coulombic explosion. The rapid exchange of electrons between the sodium and the water causes the surface of the sodium droplet to become positively charged, and the ions repel each other. This behaves very like a negative surface tension, and the surface of the droplet increases in area rapidly forming a spiny, porcupine like shape as fingers of the molten metal are shot into the liquid at astonishing speed. The larger the surface area gets the faster the reaction occurs, leading to a runaway effect. The study was published in Nature Chem. (Ref.1).
References:

Philip E. Mason, Frank Uhlig, Václav Vaněk, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, "Coulomb explosion during the early stages of the reaction of alkali metals with water," Nature Chemistry 2015, 7, 250–254 (https://doi.org/10.1038/nchem.2161).

The reaction between alkali metals and water produces hydrogen, which will rise up and mix with the oxygen in the air above. If the reaction generates enough heat, it can ignite this mixture - hydrogen and oxygen will detonate when mixed in the correct ratio. This explosion will either directly spray out the metal (which by then will probably be molten) or force it down into the water where it will produce even more hydrogen gas and steam, propelling it back up and outwards.


Order for acidic strength of hydrides across a period I have learnt that in a group , say group $15$, the acidic strength increases on going down the group as

$NH_3 < PH_3<AsH_3<SbH_3<BiH_3$

This is due to decrease in the thermal stability on going down the group. But we can also reason it as follows : on going down the group , the electronegativity of the central atom decreases , and the bond between the central atom and hydrogen becomes less polar , so the bond strength also decreases hence it becomes easier to lose a hydrogen atom thus increasing the acidity.

Now if we compare the hydrides of a period, that is

$PH_3 , HCl , H_2S$ , according to the above logic , the order should’ve been $PH_3>H_2S>HCl$ because the electronegativity increases across a period. However , the correct order seems to be the reverse. Could you please explain why ? Is there a flaw in my reasoning?

The reasoning you gave for the increasing acidity of hydrides of group $15$ down the group is not the best one. You claim that less polar bonds are easy to break which is absolutely wrong. Non-polar covalent bonds are very difficult to break (think of beaking  $\ce{C-H , N-H , O-H }$ bond, which is easier to break ?)The logic can be modified as follows.

Actually the basicity of those hydrides decrease down the group, which in turn results in the increase of acidity of them. The reason for decrease in their basicity down the group is the increase in inability to donate their lone pair to accept the proton. In $\ce{NH3}$, the lone pairs recide in the $sp^3$ hybrid orbitals, whereas for other hydrides like  $\ce{PH3 , AsH3 , BiH3}$ the lone pairs recide almost in pure $s$ oritals (i.e. the hydrogen atoms are bonded to almost pure $p$ orbitals). So, as the $s$ character of the orbital where the lone pair is residing increases, it becomes closer and tightly bound to the nucleus, and thus difficult to protonate. Thus the basicity of group  $15$ hydrides decreases down the group, alternatively, the acidity increases.

Coming to the case of  $\ce{PH3 , H2S}$ and  $\ce{HCl}$, two factors govern the acidity of them. One is the  ease of bond-dissociation, and other is the solvation of the medium. When you consider the gas-phase acidity, i.e. no solvation is present, the bond-energy will only be the determining factor and bond energy is given as,  $\ce{P-H < S-H < Cl-H}$ (easily justifiable by  decreasing size of central atom), which tells us in gas phase the acidity of  $\ce{PH3}$ should be the highest. But normally when we talk about acidity, we talk about $pK_a$ in water. Whenever there is a medium, it stabilises the dissociated species via solvation (in water , this is hydration). As. the size of the conjugate base decreases across the period, the extent of hydration energy released will dominate over the bond-dissociation energy. Chloride is the smallest ion among them having higher electronic charge density, and $\Delta H_{hyd} \propto \frac{1}{r} $, It is highest for hydrolysis of  $\ce{HCl}$, making it a stronger acid among three.

@aditi I would like to suggest another dimension to look at it...acidity means the ability to evolve H+...now it's well know that EVERYTHING in this universe wants stability...now if ph3 evolves H+ it leaves Ph2-...now observe that phosphorus being more electronegative draws the electron cloud with hydrogen towards itself...causing high charge density on it in addition to the already present negative charge...but not being VERY electronegative it's unable to hold it...this thing is known as inductive effect you shall study in organic...but on the other hand look at cl-...no additional charge due to inductive effect as no hydrogen is present...plus it's far more electronegative than phosphorus...so it will accommodate negative charge far better...now you might be thinking how electronegative atoms accomodate negative charge better than comparitively electropositive ones...well it's cause greater effective nuclear charge causes them to attract electrons towards themselves...They LOVE charge...but remember...this is the case in a solvent...IN CASE WE ARE DISCUSSING WHAT IS HAPPENING IN VAPOUR STATE...acidity of PH3 is very much due to bond length...AS H-O bond is very strong hence in protic solvents (in this case water) will remove h from compounds...in that case cl- is more stable and hence HCl is most acidic...but otherwise h-cl bond is weak due to smaller size of cl and hence cl doesn't kick away h while PH3 and H2S do comparitively more.


How do I calculate the pressure of a known liquid in a sealed container heated above boiling point? If I have water in a sealed container heated to say 150 degrees, how do I determine the amount of pressure being generated in the container? What about for other liquids? I have searched extensively and cannot figure this out.

I was looking for a formula of sorts, but for example reasons let's say a 5000 ml container with 4500 ml of water in it, with the rest of the space air, heated to 150 degrees Celsius.

Note: for convenience, "gas" refers to any gas at a temperature beyond its boiling point; "vapor" refers to any gas evaporated from its liquid state, also implying that the liquid itself is below its boiling point, such as the 150 degrees water in your example.

Dalton's law of partial pressure would come in handy here. The law states that the partial pressure generated by each type of gas particles sum up to the total pressure in a sealed container. The partial pressure of "gasses" can directly be calculated using the ideal gas law. Of course the partial pressure of "gasses" is 0Pa if there is no gas but only vapor in your container.

Now the partial pressure of the vapor of liquids. In a sealed container, if left to attain dynamic equilibrium (which is the steady state in this case) the partial pressure of the vapor of any liquid would be equal to its equilibrium vapor pressure, which is a function of temperature only. Though not excessively accurate, the Antoine equation can be used to estimate this:
$\log P = A- {B \over C+T}$
where P is the equilibrium vapor pressure, A, B and C are substance-specific constants and T is the thermodynamic temperature of the substance. The three substance-specific constants can be checked up on the internet, or provided by other sources, whereas conducting experiments can be a last resort (as it is a very tedious experiment that you need the whole curve of the function to approximate the constants).

If the liquid in question is a mixture of more than one liquid substance, then you'd need a third law - Raoult's law to determine the partial vapor pressure of each component of the mixture:
$p_i = p_i^\star x_i$
where $p_i$ is the partial vapor pressure of component $i$ of the mixture, $p_i^\star$ is the equilibrium vapor pressure of a pure sample of component $i$(in itself, not in a mixture), and $x_i$ is the mole fraction of $i$ in the liquid.
At dynamic equilibrium, combining Dalton's law of partial pressure and Raoult's law yields the expression for total vapor pressure:
$p_{total} = \sum p_i^\star x_i$

There are various empirically derived equations which are used to predict pressure given a temperature. As already mentioned the Antoine equation is for vapor pressure but can be used (although extrapolation should never be preferred to interpolation, and most empirical equations list the range of temperatures they are applied to (and derived from). These equations all require experimentally determined parameters and are essentially "curve fitting". (but see discussion on Van der Waals equation of state) Note that the ideal gas law rarely has accuracy to +/- 1% and occasionally as bad as 5% error...if I recall, the fit for CO2 near STP is in the "not so good" category for the IGL. Wikipedia discusses several alternatives to the IGL and The Handbook of Chemistry and Physics has tables for some of them (at least it used to, I don't have access to the most recent editions), as does Perry's Handbook of Chemical Engineering.


What is the name of the formula ΔE = q + w? What is the name of this formula? Google could not provide me with an answer.

$$\Delta E=q+w$$

This is a statement of the first law of thermodynamics. The first law says that the energy of an isolated system is conserved. An isolated system is one in which neither mass nor energy is exchanged with the surroundings. This equation shows the two means by which a closed system (meaning a system in which mass is conserved but energy is not) can exchange energy with its surroundings - by heat, or by work.

$$
\Delta E=q+w
$$

All energy transfers can be described as either heat flow or as work done on or by a system, and therefore the total amount of heat energy that goes in or out ($q$), plus the total amount of work energy that goes in or out ($w$), must be equal to the total internal energy change of the system ($\Delta E$).

This equation is particularly important in chemical thermodynamics because we have no way of measuring the total internal energy of a real system directly - all that we can measure is the amount of heat and work that goes in or out of the system, and therefore the change in internal energy.

This is the first law of thermodynaics.

$$
\Delta E=q+w
$$

Delta E donates change of internal energy

q - Amount of energy supplied

w - work done by the gas


What is the mechanism for this reaction? The closest reaction I can find is Lossen rearrangement but in that hydroxyamic acid is first converted to its derivative but how will that be possible here?

Moreover, even if I make an amine in this step I cannot proceed further, how will the amine and alcohol react in the next step? Any hints would be appreciated.

In your question, you have mentioned:

The closest reaction I can find is Lossen rearrangement but in that hydroxyamic acid is first converted to its derivative but how will that be possible here?

You have recognized the reaction correctly but the Lossen rearrangement is the conversion of a hydroxamate ester to an isocyanate (only the first step of given mechanism). Although the original Lossen rearrangement needed hydroxamate ester $\left(\ce{R(C=O)NH-O(C=O)R'}\right)$ to rearrange,  recent research has shown that the rearrangement is possible with free hydroxamic acids $\left(\ce{R(C=O)NH-OH}\right)$ as well (Ref.1):

Without using activating agents, a variety of free aromatichydroxamic acids could be rearranged to aromatic amines in the presence of base alone:

Source of the image: Ref.1
According to $\bf{(a)}$, free aromatichydroxamic acid $\left(\bf{1}; \ce{R(C=O)NH-OH}\right)$ react with a base to give  the corresponding carboxylate $\left(\ce{R(C=O)-O-}\right)$ as intermediate and hydroxyamine $\left(\ce{NH2-OH}\right)$ byproduct. The intermediate $\left(\ce{R(C=O)-O-}\right)$ reacts with another aromatichydroxamic acid $(\bf{1})$ to give necessary hydroxamate ester $\bf{A}$, which undergoes Lossen rearrangement in basic conditions to give the isocyanate $\bf{B}$. Mechanism for aromatic amine from isocyanate is depicted in $\bf{(b)}$.
For instance, the base-mediated rearrangement of free 4-methylphenylhydroxamic acid $\bf{(1a)}$ to  4-methylaniline $\bf{(2a)}$ is given as follows:

A mixture of N-hydroxy-4-methylbenzamide $\bf{(1a)}$ $(\pu{0.363 g}, \  \pu{2.4 mmol}),$ $\ce{K2CO3}$ $(\pu{0.332 g},$  $\pu{2.4 mmol})$, and DMSO $(\pu{2 mL})$ was heated to $\pu{90 ^\circ C}$ and stirred at that temperature for $\pu{2 h}$. The mixture was cooled to rt, and then treated with $\pu{2 M} \ \ce{ HCl}$ $(ca. \pu{3 mL})$. After the mixture became the clear solution, $\pu{2 M} \ \ce{NaOH}$ $(ca. \pu{3 mL})$ was added and extracted with $\ce{Et2O}$ $(\pu{15 mL} \times 3)$. The combined organic layers were dried over anhydrous $\ce{Na2SO4}$, filtered and evaporated under reduced pressure. The residue was purified by silica gel column chromatography $(\ce{hexane/Et2O}, 1:1)$ to yield the pure 4-methylaniline $(\pu{0.253 g}, \ 98\%)$ as a white crystalline solid.

The effects of the amount of base, reaction time, and presence of water are summarized in following diagram (Ref.1):

Thus, as shown in above diagram, presence of water may have direct effect on the conversion of the substrate to corresponding amine (even in DMSO). Further, Ref.1 clearly shows that this reaction requires polar aprotic solvent to proceed through Lossen rearrangement to give thiocyanate intermediate, which eventually converted to corresponding amine in the presence of aqueous base. For instance, if $\bf{(1a)}$ is reacted with $\ce{K2CO3}$ in methanol instead of in DMSO or DMF at for $\pu{90 ^\circ C}$ for $\pu{2 h}$, the yield was found to be only $10\%$ while it was $98\%$ in DMSO and $92\%$ in DMF under the same conditions. The authors reported $86\%$ of the starting compound has been recovered unreacted (none of it recovered in DMSO or DMF).
Note that the first reaction given in the question is in aqueous medium. Thus, according to the results found in Ref.1, the resulting product in the given reaction would be a trace of corresponding amine $\left(\ce{R-NH2}\right)$ and most of substrate would be recovered unreacted. Consequently, there won't be a reactive isocyanate $\left(\ce{R-N=C=O}\right)$, thus the reaction would not proceed to give the anticipated product in methanol as indicated.
Suppose this first reaction has been done in $\ce{OH-/DMSO}$ in place of $\ce{OH-/H2O}$. That would give you the anticipated isocyanate intermediate $\left(\ce{R-N=C=O}\right)$. It is well known that isocyanates reacts very differently in the presence of water, alcohol, and amine. In the presence of water, it would give you corresponding amine with evolution of $\ce{CO2}$. However, in the presence of alcohol or amine, it would give you corresponding carbamate or urea derivative, respectively$:

Since the given second reaction is in methanol (an alcohol), the product would be a methyl carbamate as shown in the scheme above (Ref.2). For example, polyurethane is prepared from diisocyantes:

References:

Yujiro Hoshino, Moriaki Okuno, Eri Kawamura, Kiyoshi Hondab and Seiichi Inoue, "Base-mediated rearrangement of free aromatichydroxamic acids (ArCO–NHOH) to anilines," Chem. Commun. 2009, (17), 2281-2283 (DOI: https://doi.org/10.1039/B822806J).
Greet Raspoet, Minh Tho Nguyen, Michelle McGarraghy, and Anthony Frank Hegarty, "The Alcoholysis Reaction of Isocyanates Giving Urethanes:  Evidence for a Multimolecular Mechanism," J. Org. Chem. 1998, 63(20), 6878–6885 (DOI: https://doi.org/10.1021/jo9806411).

Mechanistically, the N-H is deprotonated to form an amide ion. In basic medium, the hydroxyl group may be slightly deprotonated as well, but the pKa are not drastically different and this is not mechanistically useful. You can envision an isocyanate forming upon rearrangement, which is then subject to nucleophilic attack by methanol. Graphically:

It is also worth nothing that hydroxide ion is the base and the leaving group from the hydrodyxamic acid is also hydroxide. This is likely by design and not accident to prevent unnecessary by-products from existing.
Does this help? :)


Why is strontium(II) ion bigger than krypton atom? $\ce{Sr^2+}$ is exactly the same as $\ce{Kr}$, in terms of electrons and orbitals.
The only difference between the two, is that $\ce{Sr^2+}$ has a couple of extra protons in the nucleus (and probably a couple of extra neutrons too, but these don't influence ionic/atomic size).
Considering the two extra protons in the nucleus of $\ce{Sr^2+}$, and the overall 2+ charge on the ion,
shouldn't this draw the valence electrons in closer to the nucleus in $\ce{Sr^2+}$ in comparison to $\ce{Kr}$?
(Krypton size= 88 pm, while Sr2+ is 132 pm.)

Sr2+ is exactly the same as Kr, in terms of electrons and orbitals.

I think the appropriate word for this relation is isoelectronic.
According to Ptable and LumenLearning, $\ce{Kr}$ has an atomic radius of 88 pm, while $\ce{Sr^2+}$ has an atomic radius of 132 pm. Although it is true that $\ce{Kr}$ and $\ce{Sr^2+}$ have the same electron orbital configuration and $\ce{Sr^2+}$ has two more protons than $\ce{Kr}$, $\ce{Sr^2+}$'s greater nuclear attraction is likely countered by the nature of metal ions being more willingly to lose electrons, which implies the electrons are more loosely held by the nucleus than the noble gas $\ce{Kr}$.
Another reason I conjecture is that $\ce{Sr^2+}$ has unfilled 5s orbitals to readily accept promoted electrons, while $\ce{Kr}$'s electrons have less tendency to fill the 5s orbitals because the property of being a noble gas.


Why protons combined with electrons will release heat? During water electrolysis, the electrochemical reaction on the cathode side can be expressed as protons combined with electrons to form hydrogen. I believe this electrochemical reaction is a heat release process, but I don't know the reason.

Why protons combined with electrons will release heat?

It will be better if you could provide some references.

If the process of uniting protons and electrons is assumed to occur isothermically in a vessel of constant volume, then the heat released will be equal to the change in internal energy of the system:

$$q_v=(\Delta U)_v$$

Now if the mixture were to behave as an ideal gas, the change in internal energy would be zero and therefore so would the heat. But we can expect a mixture of oppositely charged species such as protons and electrons to behave as an extremely non-ideal mixture, consistently with the large Coulombic potential between the particles. Since $\Delta U$ for the process of uniting protons and electrons (and thereby also minimizing repulsions) is negative, the process is then expected to be exothermic ($q<0$).

One way to think of it is the electrical potential energy of the separated charges is released (as heat energy, eventually) when the charges unite. First the energy will be released as photons, and these photons will be absorbed by nearby molecules, causing them to increase in kinetic energy, which on the macro scale, is heat.


Single electrode potential measurement My textbook says that in a galvanic cell, it's not possible to measure single electrode potential independently. Instead, a Standard Hydrogen Electrode is used as the system is under equilibrium. Can someone explain this concept more clearly?

Electrodes transport electrons produced from one half-cell to another, thus producing an electric charge. You need to have a reference value to compare it to in order to obtain the relative potential. Relative to the Standard Hydrogen Electrode (SHE), that is.

The SHE has a potential that is set arbitrarily to zero giving you a basis with which to compare the potentials from the electrode you're testing.

For further reading, I feel this article on the Standard Hydrogen electrode is pretty descriptive.

Potential is a relative measurement. You will need to set a reference for measuring any potential.

As in this case, Hydrogen electrode is set as the reference.

All other potentials are measured against Standard Hydrogen electrode to get their potentials under standard conditions.


Tap water pH adjustment using citric acid I've lowered tap water pH from 7 to 6 using citric acid. After couple of hours pH raised again to 7.

What is the reason for this raise? Citric acid overcomes waters buffering  feature then raise the $\ce{H+}$ concentration in water. Why pH raise again?

The pH was measured using a pH meter.

Maybe citric acid have had a negligible effect on the $\mathrm{pH}$. Everybody knows that in contact with air, pure water is carbonated by $\ce{CO2}$ from the atmosphere. And the $\mathrm{pH}$ may go down to $5.5$. Later on, the water may loose its $\ce{CO2}$, and the $\mathrm{pH}$ goes back to $7$.

Oxygenated tap water is rich in transition metals including Fe and Mn ions. Citric is a source of H+ and a good chelate and can drive a redox reaction in the presence of oxygen and H+ proceeding as follows:
$$\ce{4 Fe^2+/Mn^2+ + O2 + 2 H+ --> 4 Fe^3+/Mn^3+ + 2 OH-}$$
There is also a likely equilibrium reactions that can be effective in recycling ions (to continue the reaction, as occurs in natural waters) in the presence of citrate acting a chelate for, say, ferric:
$$\ce{Fe^2+ + Mn^3+ <=> Fe^3+ + Mn^2+}$$
$$\ce{Cu+ + Fe^3+ <=> Fe^2+ + Cu^2+}$$
Note: The above are active charge equilibrium reactions, referred to as a metal redox couple, and support the role of transition metals in their lower valence states engaging in notable radical creation formation via Fenton and Fenton-type (non-iron case) reactions. Further, in the presence of lab or sunlight, some further recycling of metal ions is possible.
Reference: See my comments and sources cited here.
The reaction with oxygen is electrochemical in nature with an inception period and proceeds with time.
Note: $\ce{H+}$ is consumed, so the $\mathrm{pH}$ would be expected to rise.
Here is also an electrochemistry reference, in particular, Table 2, where the first listed half-cell reaction corresponds to the above after adding two $\ce{H+}$ to both sides (creating water as a product). However, as this reaction can produce basic salts, as I have discussed previously, I believe it is more informative. Further, I do prefer my rendition, as per my prior link supplied above, it could also display a possible radical chemistry underpinning to the underlying half-cell reaction mechanics.


What is more acidic: D3O+ in D2O or H3O+ in H2O and why? What is more acidic:

$\ce{D3O+}$ in $\ce{D2O}$

or

$\ce{H3O+}$ in $\ce{H2O}$

and why?

I think it's $\ce{D3O+}$ in $\ce{D2O}$ as I saw somewhere that this property is used in mechanistic studies (the inverse isotope effect), but I need a proper explanation.

EDIT: I found a statement about this in Clayden:  Water $\ce{H2O}$ is a better solvating agent for $\ce{H_3O^+}$ than $\ce{D2O}$ is for $\ce{D_3O^+}$ because O–H bonds are longer than O–D bonds hence $\ce{D_3O^+}$ in $\ce{D2O}$ is stronger than $\ce{H_3O^+}$ in $\ce{H2O}$. That is the origin of the inverse solvent isotope effect.

This is in contrast with all the answers given yet, so I am pretty confused.

For the reasons explained in New point of view on the meaning and on the values of $K_\mathrm{a}(\ce{H3O+, H2O})$ and $K_\mathrm{b}(\ce{H2O, OH-})$ pairs in water Analyst, February 1998, Vol. 123 (409–410), the $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in $\ce{H2O}$ and the $\mathrm{p}K_\mathrm{a}$ of $\ce{D3O+, D2O}$ are undefined.

The entire point of the above reference is that

$\ce{H3O+ + H2O <=> H2O + H3O+}$

(which would correspond to an equilibrium constant of 1) is not a genuine thermodynamic process because the products and reactants are the same.

$\ce{D3O+ + D2O <=> D2O + D3O+}$

would also correspond to an equilibrium constant of 1

So when Clayden and the OP write

  $\ce{D3O+}$ in $\ce{D2O}$ is stronger than $\ce{H3O+}$ in $\ce{H2O}$

it is wrong for the above reason.

Two genuine thermodynamic equilibriums are

$\ce{2H2O <=> H3O+ + HO-}$ and $\ce{2D2O <=> D3O+ + DO-}$

Experimentally, the self-dissociation constants of $\ce{H2O}$ to $\ce{H3O+}$ and $\ce{OH-}$ and $\ce{D2O}$ to $\ce{D3O+}$ and $\ce{OD-}$ can be measured as in The Ionization Constant of Deuterium Oxide from 5 to 50 [degrees] J. Phys. Chem., 1966, 70, pp 3820–3824 and it is found that $\ce{H2O}$ is about 8 times more dissociated (equilibrium constant is 8 times greater).

But using the above data to say $\ce{D3O+}$ is stronger is misleading, because this corresponds to a reaction with $\ce{OD-}$, not $\ce{D2O}$.  $\ce{D3O+}$ simply has a lower concentration in heavy water than $\ce{H3O+}$ has in light water.

As for why the $\ce{D2O}$ is less dissociated than $\ce{H2O}$, The ionization constant of heavy water ($\ce{D2O}$) in the temperature range 298 to 523 K Canadian Journal of Chemistry, 1976, 54(22): 3553-3558 breaks the differences down in to enthalpy and entropy components, which both favor ionization of $\ce{H2O}$ and states that $\ce{D2O}$ is a more structured liquid than $\ce{H2O}$. Not only the bonds of each product and reactant molecule need to be considered, but also the intermolecular forces: the number and strength of intermolecular hydrogen bonds for each species.  See Quantum Differences between Heavy and Light Water Physical Review Letters 101, 065502 for recent (2008) experimental data.

Numerous references characterized $\ce{D2O}$ as "more structured" than $\ce{H2O}$, meaning more hydrogen bonds, and a more narrow distribution of hydrogen bond lengths and angles.  According to Effect of Ions on the Structure of Water: Structure Making and Breaking Chem. Rev. 2009, 109, 1346–1370 "It is indeed generally agreed that heavy water, $\ce{D2O}$, is more strongly hydrogen bonded (structured) than light water, $\ce{H2O}$."  My explanation would therefore be that there is a greater penalty for placing ions in $\ce{D2O}$ than $\ce{H2O}$ as far as disruption of a hydrogen bonding network.

Also the equilibrium constant for

$\ce{H2O + H2DO+ <=> HDO + H3O+}$

can be measured and it is 0.96 according to Isotopic Fractionation of Hydrogen between Water and the Aqueous Hydrogen Ion J. Phys. Chem., 1964, 68 (4), pp 744–751

Explanation of Normal/Inverse Solvent Isotope Effect

For a kinetic normal/inverse solvent isotope effect
there will be a reactant and transition state.  If (for example) there is a single solvent exchangeable proton that is the same group in the reactant and transition state, for example, $\ce{ROH}$ in the reactant and $\ce{R'OH}$ in the transition state, switching solvents from $\ce{H2O}$ to $\ce{D2O}$ will either favor the reactant or the transition state relative to each other (considering the respective $\ce{OH}$ bond strengths as well and intermolecular hydrogen bonds to solvent). IF $\ce{D2O}$ favors the reactant relative to the transition state (activation energy is increased), this is a "normal kinetic solvent isotope effect".  Oppositely, if $\ce{D2O}$ favors the transition state relative to the reactant this is an "inverse kinetic solvent isotope effect." More complex scenarios involving more exchangeable sites can of course occur.

Similarly there can be equilibrium normal/inverse solvent isotope effect, if there is an equilibrium reaction and then it is reactant vs. product (rather than reactant vs. transition state) that matters.

Well, if you are interested in organic chemistry, you might be aware of the fact that the hyperconjugation effect of -CH3 is greater than -CD3. The reason for this is that the atoms in the molecules keep vibrating and the bonds of H being slightly weaker than D are easily delocalised. (due to heavier weight of D than H, the vibrations are weaker and thus the bonds are stable). The same is the case with water molecules where the bonds between O and H are weaker than O and D bond. Hence H2O is slightly acidic than D2O.


Disposal of Silver Halides dissolved in Ammonium Hydroxide I was reading the Lab Manual for Chemistry prepared by the NCERT . While going through the the test for Chloride ions (pg 88) I came across the following Hazard Warning :

The solution obtained after
dissolving AgCl/AgBr/AgI
precipitate in ammonium
hydroxide should be
acidified with 2M HNO3
and should be discarded
quickly to avoid serious
explosion.

How does it cause a serious explosion?

In the test for Cl- the chloride ions are converted into AgCl by the following reaction,
Cl- + AgNO3
→
NO3- + AgCl
which is then dissolved in ammonium hydroxide by the formation of diammine silver chloride by the following reaction:
AgCl + 2NH4OH
→
[Ag(NH3)2]Cl + 2H2O
On standing, the diammine silver complex subsequently breaks down to Ag3N.
The standard free energy of Ag3N is about +315 kJ/mol . Hence it is unstable and decomposes explosively to metallic silver and nitrogen gas .
2Ag3N → 6Ag + N2
So the diammine silver complex has to be destroyed after use so that it does not break down into Ag3N.
One way of destroying the complex is by adding nitric acid.
[Ag(NH3)2]Cl + 3HNO3
→
AgNO3 + 2NH4NO3 + H2O
So in all reactions involving the formation of diammine silver complex, the complex must be destroyed using acids after the purpose.
The complex is formed when AgCl, AgBr(sparingly soluble)and AgI(the complex will be formed only in trace amounts) are dissolved in NH4OH . The comlex is also formed in the well-known Tollen's test. Tollen's reagent is [Ag(NH3)2]OH. Ammoniacal Silver Nitrate is also used in Fontana–Masson Stain and also in silver mirroring(such as inside insulated vacuum flask).
The Wikipedia page on Silver Nitride
A reported accident

Actually, the mechanism of $\ce{Ag3N}$ creation (and also the associated products, Silver amide $\ce{AgNH2}$, and the imide $\ce{Ag2NH}$), have not been definitely accounted for, either by accident investigation reports or per a recent review of the literature, on possible paths to explosive Silver nitride.
I have ideas of a wider chemistry spectrum based answer which accounts for the apparent role involving $\ce{Ag2O}$.
To begin, I cite the more common background description, per a government source, to quote:

Silver nitride is an explosive chemical compound with symbol Ag3N. It is a black, metallic-looking solid which is formed when silver oxide or silver nitrate is dissolved in concentrated solutions of ammonia, causing formation of a silver-amide or imide complex which subsequently breaks down to Ag3N.

So, of interest is the cited reaction:
$\ce{Ag2O(s) + 4 NH3 + H2O ⇌ 2 [Ag(NH3)2]OH}$
where I would expect any Silver oxide formed from the reverse reaction of dissolving Ag2O in ammonia would then likely create a high surface area (with enhanced reactivity) $\ce{Ag2O}$. It should also importantly be noted that the reaction, itself, is reversible (see as a source, "Second year college chemistry" by William Henry Chapin, page 255), to quote:

As might be expected, the silver-ammonium complex dissociates slightly into its constituents as indicated by the equation

$\ce{[Ag(NH3)2]+ ⇌ Ag+ + 2 NH3}$

This is a reversible reaction, very much like the ionization of a very weak acid or base.

So, once any Ag2O is formed (by say moving the equilibrium to the left from a possible loss of water on standing in an open vessel or upon addition of dry alcohol), a radiation source could further accelerate the process (see, for example, Ag2O as a New Visible-Light Photocatalyst: Self-Stability and High Photocatalytic Activity). To quote from the abstract:

Ag2O is unstable under visible-light irradiation and decomposes into metallic Ag during the photocatalytic decomposition of organic substances. However, after partial in situ formation of Ag on the surface of Ag2O, the Ag2O-Ag composite can work as a stable and efficient visible-light photocatalyst.

As such, the introduction of appropriate light, or other irradiation paths (like X-rays commonly associated with radical formation) may be instrumental. In fact, here is an interesting pertinent radiation study: The X-ray activated reduction of silver (I) solutions as a method for nanoparticles manufacturing.  To quote from a passage discussing an associated radiation treatment:

The next step was aimed at eliminating glucose-containing ammonia solutions. The reactions taking place in these solutions resulted sometimes in the formation of precipitates with unstable composition. These precipitates contained considerable amounts of silver nitride Ag3N, which formed in some ammonia solutions of the silver salts and exhibited explosive properties in a dry state [15]. The results from analysis of the deposits obtained for three selected types of solutions are presented in Table 1.

As the irradiation of glucose is a path to radicals, my proposed mechanics incorporates an ostensible promotive radical pathway for the allude to creation of $\ce{Ag3N, Ag2NH and AgNH2}$, which also appears to be accelerated in the presence of $\ce{OH- and Ag2O}$ in the presence of ammonia.
The proposed reaction path is:
$\ce{NH3 + H2O ⇌ NH4+ + OH-}$
$\ce{NH4+ ⇌ NH3 + H+}$
$\ce{[Ag(NH3)2]+ ⇌ Ag+ + 2 NH3}$  (per above)
In the further presence of light:
$\ce{Ag+ + hv -> •Ag + h+}$
Source: See Photodecomposition and Luminescence of Silver Halides
$\ce{OH- + h+ -> •OH + hv}$  (same source as above)
$\ce{OH- (aq) + hv -> •OH + e- (aq)}$
As a reference, see: Flash photolysis in the vacuum ultraviolet region of sulfate, carbonate, and hydroxyl ions in aqueous solutions
$\ce{NH3 + •OH -> •NH2 + H2O}$
From a related work: Kinetics and Mechanism of the Reaction of •NH2 with O2 in Aqueous Solutions
$\ce{NH3  + •H <=> •NH2 + H2}$
As noted in this work.
$\ce{•NH2 + e- -> NH2-}$
$\ce{Ag+ +  NH2- <=> AgNH2}$
$\ce{AgNH2 -> Ag2NH + NH3}$
$\ce{AgNH2  + Ag2NH -> Ag3N + NH3}$
Note, my suggested pathway is largely promoted from the recognition of Ag2O as a recent (article was 2011) visible-light photocatalyst with the possible introduction of contributing radical based reactions.
In summary, I am suggesting chemically diverse mechanisms, involving both light and radical based reactions, as a more likely explanation.


My text seems to say that the high(est) energy orbitals are not valence orbitals - everything else says otherwise. What's true? A paragraph in my text reads

  Bonding involves the valence orbitals almost exclusively because these orbitals have the appropriate energies to interact strongly. Examine the electron energy-level diagram of a fluorine atom. The core $1 \mathrm{s}$ orbitals on a fluorine atom are at much lower energies than any of the $n = 2$ orbitals and do not participate effectively in bonding with other atoms. On the other hand, orbitals with $n > 2$ lie at considerably higher energy than the $n = 2$ orbitals. These orbitals can interact, but the resulting orbitals are too unstable to form strong covalent bonds. Therefore, the only orbitals of fluorine that form chemical bonds are the valence orbitals, those with $n = 2$. Typically the highest energy occupied orbitals are the ones used to bond with other atoms.

In contrast this simple paragraph from SparkNotes expresses what I had believed to be the case

  The outermost orbital shell of an atom is called its valence shell, and the electrons in the valence shell are valence electrons. Valence electrons are the highest energy electrons in an atom and are therefore the most reactive.

I had thought that the outermost shell, the valence shell, the most energetic shell, and the shell involved in molecular bonds were all the same thing. However, the quoted paragraph of my text says otherwise. Where have I gone wrong?

The problem is the words used to describe energy. I am not native english speaker so I am not the best person to answer your question but I'll try.

As smaller $n$, smaller energy. Problems arises when the words higher lower are used to compare numeric values (as those of the energy). They must say greater or smaller.

Without accurate words all of us intuitively agree in that:
"2 is higher that 1", and we should say that 2 is greater than 1 (2>1)

When negative numbers are involved it is not so clear. For some authors "-2 is higher than -1" and "-2 is higher than 1", for others "-2 is lower than -1" and "-2 is lower than 1". In the later cases the correct word is "smaller" (-2<-1).

In this context energies are negatives because of the definition of the reference states. So the energy of an $1s$ orbital is smaller than the energy of a $2s$ orbital.

I have some discrepancies about the meaning and usage of orbital energies but I adopted the common usage to avoid extra confusions.


Do metal ions of 4d and 5d series always form low spin complex? I've recently come across a source where it stated that all 4d and 5d metals form low spin complexes irrespective of the strength of ligand.

I'm not sure of the authenticity of the source and couldn't figure out the reason for this phenomenon, thereby enquiring about it.

I feel that the splitting energy is higher than pairing energy in this case, but I am not able to justify it properly.

While the vast majority of the compounds and complexes of the heavier transition metals are low spin (where that concept has any meaning) $\ce{PdF2}$ is one exception, which is unusual not only this way but also having $\ce{Pd^2+}$ in an octahedral environment (see e.g. Wikipedia). Thus your source is not quite correct.

As for the reason why 2nd and 3rd row transition metals are more likely to form low spin complexes than the lighter elements, the reason is given in the answer linked above in the comments. In fact, while the question may be different, the answer is almost a duplicate.

You can simply remember that CFSE of 4d and 5d series is far more than that of 3d series. Therefore, However strong the ligand, The pairing energy will always be lesser than the CFSE. So mostly, all the complexes of these two series are inner orbital complexes.
And yes the reason for high CFSE is diffused state of 4d and 5d orbitals.
Order of CFSE: 5d>4d>3d.


Thermodynamic stability of benzene derivatives Q1. How can we comment on the thermodynamic stability of the following benzene derivatives?

Q2. What does thermodynamic stability mean, in general? What does it mean to compare the thermodynamic stability of more than compound?

Q3. How do we compare the thermodynamic stability of benzene derivatives? Is there a specific rule, or way to go about it? Would MOT (frost circles, to be specific) help?

I looked at the extent of resonance, and other such stabilizing phenomena/interactions. However, that only tells the extent of stabilisation of the compound due to those phenomena, not as a whole.

Edit: I wish to predict the order of stability using knowledge of chemistry, and not check enthalpy of formation data and then establish the order.

Wikipedia defines thermal stability as .

  Thermodynamic stability occurs when a system is in its lowest energy state, or chemical equilibrium with its environment. This may be a dynamic equilibrium, where individual atoms or molecules change form, but their overall number in a particular form is conserved.

For the particular problem in hand the lowest energy state definition is more relevant for us .

Enthalpy of formation can be used for judging the energy states.

A  negative enthalpy change implies that the product has released heat thus lowering the amount of energy present in the product

This implies.

  More negative the enthalpy of formation, more stable the compound.

Enthalpies of formation

Phenol
$\ce{ -> -165.0 kJ/mol}$.
Hydroquinone
$\ce{-> -371.1 ± 1.3 kJ/mol}$
P-Nitrophenol
$\ce{-> -207.1 ± 1.1 kJ/mol}$

All data has been taken from http://webbook.nist.gov , for solid state.

The order can be calculated now as

Hydroquinone $>$ paranitrophenol $>$ phenol

Thermodynamic stability of compounds can be determined by obviously enthalpy of formation ($\Delta H_{_\mathrm f}$) of individual compounds. The enthalpy of formation will be lesser if the compound is formed from its constituent elements enjoys some greater stability.

Among the above given compounds, considering any extra stability of the compounds, their thermodynamic stability can be guessed.

For phenol, when 1 mole of it is formed there can be slightly weak intermolecular Hydrogen bonding($\ce{C6H5-OH...OH-C6H5)}$. The hydrogen bonding is slightly weak because the lone pair of oxygen atom is delocalised with the benzene ring thus creating a $\delta$+ charge on the $\ce O$ atom, due to which the dipole dipole interaction between $\ce H$ of one $\ce{OH}$ and $\ce O$ of other phenol molecule is weakened. Thus, phenol enjoys only weak intermolecular hydrogen bonding.

For 4-nitrophenol, there is stronger intermolecular hydrogen bonding, which can exist between two phenol molecules as:

Due to enhanced no. of hydrogen bonds, the compound enjoys more stability than ordinary phenol and therefore its enthalpy of formation is more than phenol.

For hydroquinone, the number of hydrogen bonding is even more. One molecule can have a hydrogen bond with four other molecules through four hydrogen bonds as shown:

Thus the molecule of hydroquinone enjoys much more stability due to large number of hydrogen bonding, making its thermodynamic stability even more than p-nitro phenol.

Thus the stability order will be $$\text{hydroquinone > 4-nitrophenol > phenol}$$


How does density of alkali metals increase down the group? Size of alkali metals increases down the group, so volume also shows increment, and since volume is inversely proportional to density, how does density also increase down the group?

$Density=mass/volume$

1) If mass is increasing and volume is decreasing, then density (mass/volume) will increase.

2)If mass is decreasing and volume is increasing simultaneously, then the density (mass/volume) will decrease.

*3)If both mass and volume are increasing, then we need to check which one of them is increasing at a faster rate(since both are contradictory factors)

a) If mass is increasing at a faster rate than volume, then density will increase.

b) If volume is increasing at a faster rate than mass (i.e denominator in mass/volume is increasing making the overall fraction smaller), then density decreases.

You can apply same logic when both volume and mass are decreasing simultaneously.

Generally, we see that in alkali metals the rate of increase of mass is greater than rate of increase of volume, therefore the density increases down the group.

Also, note that there is no definite reason for why mass is increasing at a faster rate than volume. So asking "why" does it happen isn't a great question.

It is the total differential of the d = m/v equation, not the equation itself, that gives you the effect of increasing row number on density. d = m/v tells you nothing about what happens to m and v as you increase the row number of the atom. Don't confuse an equation with the change in the variables in that equation as a function of something else (in this case, At. No.). There's two important effects in answering your question. First, mass increases as you increase At. No.,but it for every 1 unit increase in charge (1 proton and 1 electron), the mass increases by more than 1. (the mass of an electron is less than 1/1800th of the mass of a proton, so its mass can be ignored, as can the volume of the nucleus - its volume is negligible) As at no. increases the  element, on average, requires an increasing number of neutrons to keep the atom together, and since a neutron is almost exactly the same mass as a proton, mass increase faster than at. no. Check out this graph https://en.wikipedia.org/wiki/Isotope#Nuclear_properties_and_stability and note that almost all of the stable isotopes are above the neutron/proton = 1 = Z/N line, meaning there is more than 1 neutron to each proton (on average). Note also the curve in the distribution of stable elements. So, mass increases faster than at. no. - what about volume? It turns out that the valence electron determines an element's atomic (covalent) radius. It turns out that all the electrons are held closer in towards the positively charged nucleus as the charge on the nucleus increase (as at. no. increases). There's a couple of reasons for that. One is that the repulsion between electrons as you add an electron to the atom isn't as strong as the attraction between all the electrons and the additional proton. Another way to explain it is that volume increases as the 3rd power of radius,which can be thought of to mean that a little increase in radius gives the electrons a lot more room, so that they're not going to push that valence electron too far out from the nucleus. Another way to explain this is the fact that the inner electrons are not 100% efficient at shielding outer electrons from the nuclear charge (pretty obvious, right? I mean if they were 100% efficient then you'd expect the outer electrons to wander away...) So, each added inner electron doesn't "cancel" each added protons charge as felt by the valence electron, that is it doesn't fully, 100% cancel the attraction. All this means size increases slower than mass (in a given period). Note that going across a row requires a more detailed (quantum mechanical/orbital) analysis because the different orbitals (s, p, d, f, etc.) have different shielding efficiencies. Since you're considering only down the alkali metals, this doesn't have to enter in to your considerations.


Can all water soluble ionic compounds conduct electricity? I'm taking a first year university chemistry course, and I am reading a lot of contradicting things. Some people say that ionic compounds that are soluble in water are always electrically conductive, others say that it maybe conductive.  Which one is it?

I have this problem to think about:

  Suppose that an unknown chemical compound exhibits the following properties:

  it is crystalline but shows no electric conductivity in the solid state
  it melts at $\mathrm{300^\circ C}$ and decomposes at $\mathrm{400^\circ C}$
  the compound is soluble in water, the solution shows no electrical conductivity.

  What kind of chemical bond would you expect in the
  given compound? Try to describe this type of bond in detail.

Everything screams "ionic" (high melting point, solid at room temperature, crystalline structure, non-conducting in solid form) besides the fact that it doesn't conduct in water.

DavePhD is right! Your material is an organic compound.

In order to be soluble in water, it should have some polar substituents. I'd however exclude carboxylic acids or phenolic $\ce{OH}$ (except maybe in the proximity of a carbonyl group) since these will partly dissociate and yield to minor conductivity.

If the melting point wouldn't be that high, inositol would be a candidate.

The nucleobases thymine and uracil do show melting points in the 300 °C range and are both soluble in water.

I wouldn't say "always". But yes, and by default, ionic compounds are usually going to conduct electricity. I would study electrolytes before trying to decide these things, if you know electrolytes, the answer would be obvious to you


What happens with enol and ketone What process happens between an enol and a ketone, so that the ketone will enolate and become acidic?

Please, explain what happens with them in this process?

Because the enol and ketone aren't stable, equilibrium happens; therefore the enol has a chance of losing a hydrogen ion and because of that, the enol becomes an enolate.

Also keep in mind that an acid is a molecule that can donate a proton.
The enolate is a base and it's gathering/accepting a proton from the hydrogen.

I assume this is what you meant when you asked the question.


Why does Vapor Pressure decrease on addition of solute? Consider a vessel with a volatile liquid in it. Let's say it has a vapor pressure $P^0$ when it is pure.
Now a non-volatile solute is added to it. Its vapor pressure decreases as $P^0\chi_{solvent}$.
Why does this happen at a molecular level?

The reason that I have come across says that less surface area is available for evaporation and hence this effect. But this does not make sense to me as we can have liquid in container of all shapes and sizes and will still get the same Vapor Pressure.

Some users have suggested me a link to a similar question. The answers in the link are remarkable but the explanation I was hoping for was more likely to involve forces (as in Physics) and maybe a sort of "free body diagram" explaining how these forces are acting(words rather than actual diagram will be fine too) rather than thermodynamic picture.

For solvent at temperature $T$, surface area $A$ and  having the molar fraction $x$, the rate of molecules leaving liquid is:
$$\frac {\mathrm{d}n}{\mathrm{d}t}=-C_1 \cdot A \cdot x \cdot f(T)$$
Lower molar fraction means lower surface density and lower evaporation rate per surface area.
Factor $C_1$ includes pure solvent molecular surface density and $f(T)$ includes probability of a molecule leaving liquid.
For vapour of the partial pressure $p$, the rate of molecules coming to liquid from vapour is:
$$\frac {\mathrm{d}n}{\mathrm{d}t}=C_2 \cdot A  \cdot p \cdot  g(T) $$
The $g(T)$ involves temperature dependent relation of pressure and collision rate with liquid surface. Higher temperature means higher kinetic energy and less collisions needed for the same pressure.
For equilibrium, the net rate must be zero, therefore:
$$p= \frac{C_1 \cdot x \cdot f(T)}{C_2 \cdot  g(T)}$$
Therefore the equilibrium pressure is proportional to the molar fraction.
It is quite analogical to the steady state of water level(=pressure), when it is simultaneously being filled up(=evaporation) and drained(=condensation). When you dilute a non-volatile solute, you decrease the filling up (= surface concentration of molecules is lower), so draining (=condensation), brings the water level(=pressure) to lower value to match draining rate with the filling.

If we add non volatile solute to solvent for example water,  we decrease the tendency of water to evaporate in gas phase because solute particle obstructs the evaporation of water in gas phase therefore vapour pressure decrease when  we add non volatile solute to solvent like water.  This image is from ncert:


Name for a spin state with a multiplicity of 13 If I want to name a spin state with 12 unpaired electrons,
what is the correct name?
Tredecim is the latin name for thirteen, but tredecimet sounds odd.

I have seen multiplicity 13 referred to as simply a 13-let. This #-let nomenclature is sometimes used for high multiplicities (e.g. higher than 10).

I would go for the Greek form and call it trisdecatet. Or should it be with a k, trisdekatet?


Why can't oxalate ion donate two pairs of electrons from the two double-bonded oxygen atoms So, I read that an oxalate ion is a bidentate ligand but since I saw that it has 4 oxygen atoms I was confused as to why it isn't a tetradentate ligand instead.

I searched it on Google and found this:
Oxalate ion dentiticity - why is it bidentate and not tetradentate?

Now, in the link it's implied that the oxalate ion can donate the 2 pairs of electrons either from the 2 oxygen atoms having negative formal charge or from 1 "double-bonded" O atom and 1 negatively charged O atom.

But why can't the two pairs of electrons be donated from Both the "double-bonded" O atoms?
 Why isn't this "third" bonding mode possible?

EDIT:  From the comment below, I infer that because of resonance, there is no double-bonded or negatively charged O atom in the Oxalate ion; rather all of the O atoms have a partial negative formal charge and double-bond character, is that correct?

So, can electrons be donated from any O atom in the ion?

(And so did the answer in the provided link just ignore this for the sake of simplicity?)

Oxalate is a planar molecule. The four oxygen atoms share the pi electron density through resonance, as noted in comments, and thus are all identical. Imagine those four oxygen as four corners of a table. Now you are a Lewis acid, perhaps a transition metal cation, and you walk up to the table. How many of the four corners ligate you? Perhaps you imagine the planar ligand dropping on top of you, so that the four corners are around you. For effective overlap with all four corners of the table, you'd need to be a pretty big cation. In almost all cases you'd achieve better orbital overlap, and a more energetically favorable coordination complex, if you just cozy up to two of the donor oxygens.


What is the link between electrode surface areas and EMF? I was wondering if someone would help me with my question.
I was completing a question which asked about the effect of increased surface area of a platinum electrode on the EMF of the overall cell.

This is the cell: Pt | H2O2(aq), O2(g) || IO (aq), I2(aq) |Pt

When I answered, I thought the EMF would increase due to the surface area of the electrode being able to now accept more electrons but I found out the answer is that the EMF is unaffected by any change in surface area, why is this?

Thanks for any help anyone can provide!

If by EMF you mean the ideal, thermodynamically determined, voltage then the answer is that increasing the surface area does not affect voltage/EMF. If by EMF/Voltage, the questions means operational voltage then it will increase because increasing the surface area of electrodes decreases the overpotential losses (Tafel equation) by essentially decreasing the resistance due to the kinetics of the reaction thus increasing operational/real voltage of a cell.

When electrode surface area increases,internal resistance decreases.But emf does not change as it is a characteristic of a cell.Emf only depend upon nature of electrodes and electrolyte.


Stability of carbocations We know that if a structure has all equivalent resonance structures, then it will be highly stabilised. So then why benzylic carbocation is more stable than other cyclic allylic carbocations, though allylic carbocation has all equivalent resonance structures?

More the number of resonance structures (alias mesomers) more is the stability of the carbocation so benzylic carbocation with 4 mesomers is more stable than 2 structures of the allylic carbocation.


If alloys are homogeneous mixtures, why can't we separate their components? An alloy is a material composed of two or more metals or a metal and a nonmetal. And, they are usually formed by heating the elements to their melting points, and then cooling them, so that the components mix. Now, why doesn't this works backwards i.e. if we heat the alloy again to melting point of their constituents, and they should separate?

If you melt the alloy and then use electrolysis, or if you use a redox reaction, you should be able to separate the components.

For example, if you used a basic acid, let's say hydrochloric acid, on a zinc-copper alloy, I'd imagine the zinc would be removed from the alloy and taking the hydrogen's place, forming zinc chloride, leaving hydrogen gas and copper metal.

This is based off of simple redox reaction memories and no personal experience. However, if you're curious, try it out!

Once the alloy has been formed the atoms from the different metals will have shared there electrons with each other and come to an equilibrium. In this state the metal atoms have formed a complex structure which has a different reactivity or properties than each individual metal did in its original form .


Is alcohol Lewis base or Brønsted base? I know that Lewis' concept of acid base deals in terms of electrons and Brønsted in terms of $\ce{H+}$ and base which can provide electrons or $\ce{OH-}$. If I am wrong (partially or fully) please explain Lewis and Brønsted-Lowry concept along.

According to me it is a Lewis base because of the lone pair on oxygen of the alcohol and a Brønsted acid.

Alcohols can act as both.  Many reactions of alcohols are believed to start with their protonations or deprotonation. However, isolating salts where they act as bases is much harder, while alcoholates are quite easy both to make and isolate. Protonated alcohols are very active alkylating agents and if fitting $\beta$-hydrogen is present, are prone to beta-elimination.

Alcohols is a brønsted base because of the presence of unpaired electron over Oxygen atom which make them proton acceptor.


Do electron donating groups make a molecule nucleophilic? Do electron donating groups make a molecule nucleophilic?
One group, the alkyl group in this example, is electron donating in $\ce{CH3CH2O-}$. Will the molecule be nucleophlic?

Do electron donating groups make a molecule nucleophilic?

Sometimes.

Sure, electron donating groups can make a molecule more nucleophilic by virtue of increasing the negative charge density on the nucleophilic atom.

However, nucleophilicity is a kinetic property, and having a lot of electron donating groups attached can significantly affect the ability of the nucleophile to approach the electrophile. This reduces nucleophilicity. An example would be tert-butoxide. Yes, oxygen with a negative charge is generally nucleophilic, and yes, the tert-butyl group is an inductive donor, but tert-butoxide is not very nucleophilic because of steric hinderance.

  One group, the alkyl group in this example, is electron donating in $\ce{CH3CH2O-}$. Will the molecule be nucleophlic?

Maybe. Nucleophilicity also depends on the solvent. A charged ion such as ethoxide ion would be well-solvated in a polar solvent, such as water, and less well-solvated in a non-polar solvent, such as hexane.

Solvation impacts reaction kinetics, and in turn, impacts nucleophilicity. Consider the halides for example. In polar protic solvents, iodide ion is the strongest nucleophile. Protic solvents are not able to solvate iodide ion as well as they can solvate the fluoride ion, which has a much higher negative charge density. In polar aprotic solvents, however, fluoride ion is generally a stronger nucleophile than iodide ion because fluoride ion is not as strongly solvated and it also has a great negative charge density.

In sum, nucleophilicity is not an inherent property. It depends on a variety of factors - both external and internal to whatever molecule or ion that you are considering.

Yes electron donating group increases nucleophillic nature either by $+I$ (Inductive effect) or  by $+M$ (Mesomeric Effect). They do so by increasing the electron-density over nucleoplillic atom. In $CH_3CH_2O^-$ the alkyl group has its $+I$ effect which Increases nucleophillic nature.

Hyperconjugation also increases nucleophillic nature by $+H$ effect. According to Baker Nathan Hyperconjugation dominates Inductive effect. Therefore $CH_3-$ will Increase more nucleophicity than $CH_3CH_2-$ because of presence of 3$\alpha H$ rather than 2$\alpha$.

The general order of effects is :-
$+M > +H > +I$

Hope it Helps...!


What are the ways to make radioactive water drinkable again? Let's say survivors in the area contaminated by radioactive fallout need to somehow process the water from rivers, rain or pipes. How could they go about it? Is it even possible to 'remove radioactivity' from the water?

The water itself isn't radioactive, rather particles dissolved in it, because water is a quite good solvent. Therefore, filtration (e.g. with activated carbon) or precipitation can remove such particles. Also a ion exchanger can remove radioactive species. But this can take quite long, especially if the needed amount is large.


Finding mass of MnF3 is produced in a reaction The question is as follows:

  If $\pu{1.23 g}$ of $\ce{MnI2}$ reacts with $\pu{25.0 g}$ of $\ce{F2}$, what mass of $\ce{MnF3}$ is produced?

$$\ce{2 MnI2 + 13 F2 -> 2 MnF3 + 4 IF5}$$

I think this is a limiting reactant problem but I'm having difficulties with starting the question off. Do I find the mass of both $\ce{MnI2}$ $\pu{(308.74 g/mol)}$ and $\ce{F2}$ $\pu{(38g/mol)}$ and divide them both by their given masses $\pu{1.23 g}$ and $\pu{25.0 g}$ respectively?

Once I do that, do I find the limiting and excess reactant, take the limiting reactant and do stoichiometry to find the mass of $\ce{MnF3}$ ?

Here we have an already balanced equation given in the question-

$$\ce{2 MnI2 + 13 F2⟶ 2 MnF3 + 4 IF5}$$

We find the given number of moles of each reactant by dividing their given masses by their respective molar masses

Given mass of $\ce{MnI2}=\pu{1.23g}$
Molar mass of $\ce{MnI2}= \pu{308.74 g/mol}$

Amount of $\ce{MnI2} = \frac{1.23}{308.74}= \pu{0.0039mol}$

Given mass of $\ce{F2}=\pu{25g}$
Molar mass of $\ce{F2}= \pu{38 g/mol}$

Amount of $\ce{F2} = \frac{25}{38}= \pu{0.657mol}$

Now we find the mole ratio from the given information and compare the calculated ratio to the actual ratio.
If more than $\pu{13 moles}$ of $\ce{F2}$ are available per $\pu{2 moles}$ of $\ce{MnI2}$, the $\ce{F2}$ is in excess and $\ce{MnI2}$ is the limiting reactant. If less than $\pu{13 moles}$ of $\ce{F2}$ are available per $\pu{2 moles}$ of $\ce{MnI2}$, $\ce{F2}$ is the limiting reactant.

Let's say that all of the $\pu{0.657 moles}$ of $\ce{F2}$ were to be used up, there would need to be

$\frac{2}{13}\times0.657=\pu{0.101  moles} \text{ of } \ce{MnI2}$.

But there is only $\pu{0.0039 moles} \text{ of }\ce{ MnI2}$ available which makes it the limiting reactant.

If all of the $\pu{0.0039 moles} \text{ of }\ce{ MnI2}$ were used up, there would need to be

$0.0039\times \frac{13}{2} = \pu{0.0254 moles} \text{ of } \ce{F2}$.

Because there is an excess of $\ce{F2}$, the $\ce{ MnI2}$ amount is used to calculate the amount of the products in the reaction i.e. amount of $\ce{MnF3}$ produced.

From the equation given its easy to observe that the stoichiometric ratio of $\ce{MnI2}$ and $\ce{MnF3} = 1:1$

Thus amount of $\ce{MnF3}$ produced = $\pu{0.0039 mol}$
Molar mass of $\ce{MnF3} = \pu{111.94 g/mol}$

Mass of $\ce{MnF3} = 111.94 \times 0.00398 = \pu{0.445g}$

$F_2$ is present in great excess. Do the calculation with the manganese compounds.

$1.23$ g $MnI_2$ contains $1.23 g /308.74$ = $0.004$ mol $MnI_2$. The reaction produces the same number of moles of $MnF_3$. As the molar mass of $MnF_3$ is  $55 + 3·19 = 112$ g/mol, $0.004$ mol $MnF_3$ weighs $0.004·112$ g = $0.448$ g


Does brewing tea with hard water reduce the amount of bioavailable fluorine? I read that brewed tea contains a relatively large concentration of Fluorine.

Fluorine can have negative health effects such as Skeletal Fluorosis. EDIT: I understand some Fluorine is necessary and has benefits. This is just to inform why I have structured this hypothesis. I am interested in the chemistry not the health effects as that is a different topic.

Tap water and especially hard water contains lots of Calcium ions.

Would Calcium and Fluorine bond to form Calcium Fluoride? I'm obviously not a chemist, but I assume it would be a lower energy state than the soluble ion pair in the tea, correct? It forms Fluorapatite on tooth enamel.

This Calcium Fluoride SDS indicates that eating solid CaF2 straight won't immediately kill you, so I assume Calcium Fluoride does not decompose in HCl stomach acid, correct?

So with that logic would the following be true?

Mol bioavilable F-  =  Mol F- from brewed tea  -  2*(Mol of Ca2+ in hard water)

If that is the case, it brings up some other interesting questions, like if moderately hard municipal water actually has any Fluorine in solution or not...

EDIT: To select a best answer I need to know:

if Ca ions bond with F ions in solution to form CaF2
and to know the percent or time-rate solubility of CaF2 in stomach acid.

Thanks!

$\ce{CaF2}$ has a relatively low solubility in water, about 15 mg/L, with a $K_{sp}$ of about $4\times 10^{-11}$.  According to the paper you linked, the concentrations of fluoride in tea are around 5 mg/L, which corresponds to about $1.3\times 10^{-4}$ M. Thus, the concentration of $\ce{Ca^2+}$ required to precipitate the fluoride out is about 2.5 mM, which corresponds to 0.1 g/L. Hard water is defined as having a calcium concentration above about 0.6 mM, with very hard water being 1.8 mM calcium. Thus, it is unlikely that there would be high enough concentrations for $\ce{CaF2}$ to precipitate out of solution.

Because of the large solvation energy of fluoride ion, the molecules in solution are likely to be nearly completely ionized. That is, the fluoride anions and the calcium cation will each have their own solvation shell of water molecules. That means that they will not interfere with each other until the concentrations are high enough that solid $\ce{CaF2}$ begins to precipitate.

As long as the concentrations are below the solubility limit, the presence of calcium ions will therefore not affect the availability of the fluoride ions.

Furthermore, since HF is a weak acid, in the acidic environment of the stomach some of the fluoride ions will be protonated to HF, decreasing the concentration of free fluoride and increasing the apparent solubility of $\ce{CaF2}$.

Even I am not a chemist, but according to my textbook there is already a counter ion present in hard water to the $\ce{Ca^2+}$ and $\ce{Mg^2+},$ which are mostly bicarbonates (temporary hardness) or fluorides, chlorides and sulfates (permanent hardness). These counter-ions would by themselves prevent any appreciable amount of fluoride forming a bond with the $\ce{Ca^2+}.$

Also, even though $\ce{CaF2}$ probably won't decompose fully in stomach acid, it would decompose enough to cause the negative effects of fluorides to come into effect, according to my professor. My professor also says that this the reason $\ce{EDTA^4-}$ is used to complex with $\ce{Ca^2+}$ and $\ce{Mg^2+}$ in hard water: since coordinate compounds don't dissociate in water, the $\ce{Ca^2+}$ and $\ce{Mg^2+}$ are not able to make the water hard. The anions are removed using organic ion exchangers.

Thus I believe making tea in hard water won't reduce the fluoride concentration very much, and in any case drinking hard water is generally a bad idea because it is not pure enough to be drunk.


Moving water vapour between flasks I have a flask of silica gel connected to a flask that contains a bit of water, so when I heat the silica gel the water absorbed gets evaporated and I want it to move to the other flask, but it always condenses on the roof of the first flask, so is there any way to force the vapour to move flasks?

To prevent the condensation, you need to create the conditions under which the vapor does not become saturated. The way to do that is to heat up the whole transfer line (the roof of the first flask and all the tubes that follow) - the higher the temperature, the higher the saturated vapor pressure. Heated transfer lines are often employed in commercial equipment when there's need to get condensible gases from one setup (e.g. thermogravimetric) to the other (e.g. mass spectrometer).

In your flask, the water will always go from the drops of liquid water to the silica, and not in the opposite direction. If you want to remove the water from the silica, the only way to do it is to heat the hydrated silica in the air oder under a vacuum, and quickly close the hot silica when dry.


Is anode actually ever negative? Disclaimer: This question comes from a physics student.

I was studying electrostatics and I was wondering how to charge a conductor at home without any fancy furs etc. So an idea came to me: take a battery, touch for example the negative terminal with the conductor and then remove it. Since the negative battery terminal is at a negative potential and the conductor was previously grounded ($V = 0$) then as their potentials must become equal, it means that electrons will flow from the battery terminal to the conductor for a short period of time. Afterwards, I disconnect the conductor and the electrons remain 'trapped' in it hence the conductor is now negatively charged.

At least that was the idea. Now suppose that the battery is a galvanic cell ($Zn, Cu$). The question is whether there are free electrons floating inside the $Zn$ anode when it is not connected to the cathode?

The whole system (battery, wires) is electroneutral while the current is flowing e.g. Zinc atoms at anode dissolve and electrons stay at the anode while the $Zn^{2+}$ ions go to the solution, but in the same instant other electrons leave the anode.

When the electrodes are not connected, I'm not sure would the redox reactions happen. In my understanding, $Zn$ oxidizes because $Cu$ on the other side has greater electron affinity, but $Zn$ would have to be able to 'sense' $Cu$ on the other side somehow, otherwise the oxidation reaction happens anyways for a short time until it reaches some equilibrium point. This is what I think actually happens and that's why the anode would actually be negative until connected to the cathode.

So am I on the right track with this and ultimately, will my attempt to charge the conductor a bit succeed?

It is a very good questions from a student who seems to be thinking deeply. I have been thinking to write on this topic for the last 7-8 years. In short the poles of the batteries are electrostatically charged and one can actually "sense" this charge with the help of sensitive electroscopes which were known in the time of Volta.

The label anode or cathode is not defined with respect to the electrostatic sign as explained in the post by Maurice.

Think of a large parallel capacitor. Connect one terminal of the plate to one positive terminal and the other one the negative terminal. The capacitor plates would show a electrostatic charge.

By all means, using one single cell you would not be able to detect the electrostatic charge, like a charged comb. Only specialized electroscopes can sense that.

An anode is the place where oxidation occurs. In a galvanic cell, like the Daniell cell ($Zn - Cu$), the $Zn$ is oxidized at the Minus pole. $Zn$ is oxidized into $Zn^{2+}$ ion and the electrons are sent in the electric connexion going to the $Cu$ electrode. $Zn$ is the anode (negative sign) and $Cu$ is the cathode (positive sign)

If the same setup is used in electrolysis, the phenomena a going in the reverse sense. Electrons are obliged to flow from the copper plate to the zinc plate. $Zn^{2+}$ ions are reduced to metallic $Zn$ on the $Zn$ plate. The zinc plate is now working in the anode mode. But it is still the Minus pole, as, in electrolysis, cations (positive ions) are attracted by the negative pole. In the electrolysis, the $Cu$ plate is oxidized into $Cu^{2+}$ ions : Copper is the anode.

When comparing the same chemical equation working in a galvanic cell or in an electrolysis, $Zn$ is always the negative pole and $Cu$ is always the positive pole. The signs PLUS (+) or MINUS (-) can be printed on the plates. But Zn is an anode in a galvanic cell, and becomes cathode in an electrolysis system.

So you see that an anode is not always negative. The anode is the negative pole (Zn) in a galvanic cell. The anode is the positive pole (Cu) in electrolysis.


Does any element have nutritional (caloric) value? Is there any single element (e.g. possibly carbon) that can be eaten, which has a nutritional value such that it provides calories?

I found on Wikipedia (dietary element) that there are certain elements that our bodies require.  I suppose oxygen could count as such a nutritional element, although we breathe it instead of eat it.

Interesting question! You may get a better answer, but I believe the answer is no.

Calories are units of heat and represent energy available to the body. The form of fuels that can be (biochemically) "burned" by the body are hydrocarbons; i.e. molecules of carbon and water, like sugars. Carbon alone I do not believe can work.

The only other single element fuel is hydrogen, and again I do not think this is biologically available to the body. By the way, oxygen is required as the oxidizer in any combustion reaction, and we definitely need oxygen (from breathing) to convert the food we eat to energy. But oxygen is not a fuel: it is the oxidizer that is reduced during combustion.

Most metals consumed (albeit in only tiny amounts for select elements) will be directly converted to ions by the HCl in our stomach.

Then, as the question includes the word 'nutritional', commonly cited elements (in their ionic form) of particular import include: potassium/sodium (regulates fluid balance, muscle contractions, and nerve signals), iron (linked to hemoglobin and the transportation of oxygen in the blood), calcium/magnesium (maintains strong bones), iodine (required for creation of thyroid hormones and regulating the body's metabolism), chromium (important in the breakdown of fats and carbohydrates) and copper (instrumental in formation of red blood cells, maintenance of nerve cells and the immune system), to mention a few.

Of the mentioned elements, those having a caloric impact include chromium and iodine.

Note, elemental carbon, especially in the form of charcoal pills or activated carbon, should be avoided (not nutritional or even caloric), see discussion here.

Further, I would argue, that as the human body generates a chemical-based electric current (see this discussion), it is probably not wise to introduce a very good cathode (namely carbon). Unwanted solvated electrons and radicals created therefrom, most likely do not promote good health, and in ones' body, may supply an electrode to promote electrochemical (or battery) cell activity.


How many bonds can Hydrogen make? Recently I saw a video which told me that Hydrogen has a valency of 1, i.e, Hydrogen can only bond with 1 other atom. But since hydrogen wants a complete shell, it can have 2 covalent bonds with 2 electrons. Please explain.

Apart from some Group 13 weirdness, hydrogen can only make one bond. Covalent bonds require pairs of electrons and hydrogen can only have two electrons bound in one covalent bond.

Hydrogen forms 1 single bond when there is a s-s overlap which is the greatest overlap but it can also form 2 single bond where there are 2 s-p overlaps such as is
Even in compounds like $\ce{B2H6}$, H seems to make 2 bonds but it doesn't.

As you can see the two Hydrogen atoms in the centre make 2 bonds each with Boron.

The number of a bonds that hydrogen can make is dependent on the type of bonding.

It can form only a single ionic bond.

It can form two bonds when there is overlap involved (covalent bond)

It can form a single covalent bond.

If stability can be achieved it can form a lot of coordinate bonds, but due to it's small size and only a single proton I would think a max of 1 coordinated bond can be formed if we being with an ionised H-atom.


Why does the distance between molecules increase when the temperature is raised? I have learnt that when we heat ice-like substances it changes to water and when I asked my teacher she said  that the distance between molecules increases.

When I thought about it a bit more a question arose: Why does the distance between the molecule increase as we raise the temperature?

The answer "the distance between molecules increase" is incomplete if not plain wrong.

Temperature is an effect of energy present. Basically, it's an effect of little movements and vibrations of molecules and atoms due to their energy.

In an crystal, the energy of the molecules is so low, that they don't vibrate and move enough to break the structure. The more energy you put into the system, the more the molecules move. At one point, the movement is too much to keep the molecules in place, the crystal structure breaks apart, the ice melts.

A liquid (NOT WATER, IT IS A SPECIAL CASE) has lower density than the crystal because the molecules are moving around a lot and "need more space".

First you have to understand that molecules are randomly moving in matter.When you heat them,you essentially give them kinetic energy which increases their speed and hence the distance between molecules increases implying that interatomic forces decreases,leading to change in state from solid to liquid.The basic difference between solid,low,and gases are that molecules in gases have more kinetic energy than liquid which have more than solid.


Which is stronger, PET or PVC? I have been given the task of designing a new wheel rim (the metal part of the wheel) but instead of metal, it has to be made out of a polymer.

I'm thinking of using either PVC or PET but I don't know if either of those plastics will be able to support the weight of a car and am not sure how to find that out.

If anyone can think of a different polymer that I could use instead that would be better suited feel free to include it as an answer. This is only theoretical so cost/difficulty to manufacture is not a big hassle.

You need to define what you mean by "stronger".  Are you talking about resisting deflection (and thus rolling losses)?  Yield?  Break?  Impact?  Abrasion?  Lifespan?  Ability to withstand heating from its deflection as the vehicle rolls?  Also, any material can "withstand the weight of a car" if you use enough of it, do you know what sort of geometry you're talking about?

Lastly, the material properties of polymers depend a lot on the manufacturing processes and the details of the raw materials (chain lengths, branching, monomer ratios, etc) and additives.  A good example is PET.  Check out the difference in tensile strength between simple thermoformed PET and BoPET (Mylar), for example; they're vastly higher for BoPET.  And the reason is because it's elongated to near breaking at lower temperatures - thus orienting the polymer - and then given a heat treatment to relax it into its new shape.

I think you need to define your problem statement better.  Think about what properties you want, your geometry, and how you plan to manufacture it.  Also, what are the parameters you're trying to optimize?

Use PVC as it has a greater value of young's modulus:   490,000 psi

PET has a young's modulus of: approximately: approx 430,000 psi

Higher the young's modulus the greater the resistance to deformation.


Nucleation and Phase Equilibrium at the Melting Temperature I used to think that at the melting point, the solid and liquid phases are in equilibrium. That’s because the Gibbs free energy change for the transformation is 0, as is characteristic of equilibria.

However, I recently learnt about nucleation in phase transformations, and I’m very confused. My textbook tells me that below the melting temperature the Gibbs free energy change is not 0 and this energy can go into increasing nuclei surfaces and surmounting the nucleation activation energy barrier. It goes on to say that at the melting temperature, no such energy is liberated (since ∆G = 0) and the nucleation activation energy barrier cannot be achieved. Thus, no nucleation occurs at the melting temperature.

If no nucleation occurs, then how can the solid and liquid phases be in equilibrium? I know about supercooling and that the principle behind it is that it takes time for such an equilibrium to be achieved. However, if nucleation simply doesn’t occur at the melting temperature, then equilibrium will never be achieved...

You probably know that water can be supercooled down to well below -30°C. And it has probably been suggested (if not outright stated) that at those temperatures, the water is in a "metastable" liquid state and that "eventually" it will spontaneously solidify - even if "eventually" may require trillions of years. Since thermodynamics is empirical, you could wonder how that has been empirically observed...but let's not go there. You've been taught that ΔG is a measure of the spontaneity of a process at constant temperature and pressure and at equilibrium. (How a "process" happens when the system is at equilibrium (steady-state) is often ignored. It's fair to say that nothing "happens" at equilibrium, rather when we say "equilibrium process" we mean "a process occurring sufficiently close to equilibrium". But I think I'm getting off topic, again. In considering nucleation, it is vital you distinguish between homogeneous and heterogeneous nucleation. This is the crux of the issue. Water is almost never without contaminates which provide nucleation sites, even if those sites are only on the surface of the container holding the liquid (getting the water this pure is an enormously difficult feat, and how you create a liquid phase without a container I leave to your imagination (hint:rain)). So, when reading about nucleation, you need to distinguish between the situations where there is preexisting centers (contaminants) or whether the water is assumed to be absolutely pure and centers must form spontaneously. Perhaps to answer your question (it would have been most useful if you had copy/pasted the exact text from your book). Once centers (seeds?) exist, the continued growth is pretty much identical whether their origin was homo- or hetero-. Obviously there is no equilibrium between a dust particle (solid) and liquid water. So how can nuclei exist at the MP? Obvious, right? They're hetero-, no formation is required. (Here I avoid the issue of whether FP is (or is not) identical to MP). Wikipedia has a short article on Classical Nucleation Theory which may help. Consider a solid. At its surface, the atoms (or molecules) are different than in the bulk (for most solids, bulk properties aren't reached for top 2 or 3 (or more) layers) but just consider an ideal crystal. Obviously, the surface has fewer bonds and hence has higher energy (is less stable). We call that surface tension (or exactly equivalently, surface energy). The classical theory states it's always positive. This is the energy which needs to be "overcome" to get nuclei, and it's proportional to the surface area. You should note here that our discussion has gone down to the microscopic level, where classical thermodynamic quantities become less and less useful. In other words, the macroscopic classical world thermodynamics needs to be treated carefully here. Statistical thermodynamics is most suitable transition from the classical to the quantum world. These types of processes can be studied using either, but fundamentally the world behaves according to the rules of (statistical) quantum mechanics when you're concerned with atoms depositing on or leaving a surface. The perspective of Classical Thermodynamics is of limited use. (In this case, I think it is probably just as useful, if not more useful, than the quantum mechanical approach. But you're unlikely to get many "fundamental" insights using it.) I'll point out one other apparent flaw in the "activation energy" approach: if the energy barrier is symmetric, then at its peak there should be no preferential direction: the odds of the system moving towards the solid or towards the liquid should be exactly the same (i.e. 50:50), yet the Classical Nucleation Theory requires a fudge factor for this probability - it is an indication that there is more going on here than this approach can handle, that the nucleation activation energy is, at best an approximation. Meaning don't take it too seriously; unless it can be derived from the quantum mechanics, it has no fundamental basis. I don't know much about this area of chemistry. I don't know whether or not it can be derived from the QM, but given that the Classical Nucleation Theory article in wikipedia is speaking about radius as if it is continuous, then I have my doubts. In some systems you begin to see bulk properties at N=100 to N=1000 (N=number of particles), so you'd have to be above this threshold for "classical" behavior to emerge.


Is hybridization used in ab initio valence bond calculation? Many general chemistry textbooks introduced the concept "hybridization" to construct a symmetry-adapted VB-type wavefunction. In the textbooks, usually the minimal basis is used and without optimizing all resonance forms.

My question is, in the ab initio valence bond calculation with extended basis set and optimizing many resonance forms,

(i) do people still use hybridization to construct the wavefuntion?

(ii) if the answer to question (i) is yes, does the result differ from the textbook description? For instance, the lone pair electron in the $\ce{H2O}$ molecule? (is there any unitary equivalent result like orbital localization in MO-based method?)

I asked the question (ii), since NBO analysis tends to give non-equivalent composition for lone pairs in $\ce{H2O}$, one sp, the other p. I am wondering what is the answer in ab initio valence bond method.

(i) do people still use hybridization to construct the wavefuntion?

Yes and no. No: Only atomic orbitals are included in the base set. Yes: if the wavefunction is optimized, linear combinations of atomic orbitals are allowed, and we all know that one S-orbital + some P-orbitals give hybridized orbitals.

For $\ce{H2O}$, the lowest 3 molecular orbitals are generally these (see below):

At first, this result seems unlike the SP$^3$ bonds between hydrogen and carbon. There is a trick, however. It is called "localization of the orbitals", meaning that we generate linear combinations of molecular orbitals to generate new ones. Below, I'll give you an example in which we linear combinations of molecular orbitals $\varphi_2$ and $\varphi_3$ generate orbitals that look more like the SP$^3$ orbitals from text books.

(ii) if the answer to question (i) is yes, does the result differ from the textbook description? For instance, the lone pair electron in the $\ce{H2O}$ molecule? (is there any unitary equivalent result like orbital localization in MO-based method?)

As you may have noticed above, the results differ slightly, but I can elaborate.

Lone pairs are given by $\varphi_4$ and $\varphi_5$ below.

For $\varphi_4$, note that p$_z$ is along the z-axis, which is in the plane of the oxygen and hydrogen atoms: it is the 2-fold rotation axis.

For $\varphi_5$, note that p$_y$ is along the y-axis, which is perpendicular to the plane of the oxygen and hydrogen atoms: it is out of the plane.

For $\varphi_6$ and $\varphi_7$, note that they are unoccupied orbitals (i.e. lumo and lumo+1)

Now, orbital localization can be applied, linear combinations of $\varphi_4$ and $\varphi_5$ will give 2 lone pair orbitals, that stick slightly out of plane. I haven't tried drawing those yet, as it is not so easy to do so in this representation in 2D, so I'l try to switch to another representation...

As you see, those localized molecular orbitals closely resemble the SP$^3$ lone pair orbitals on oxygen, as predicted by valence bond theory. The only difference is the shape & probability of the small counter-part of the lobe; I included phase, not probability density here.

Does this answer your question?

I can answer your first question definitively from my experience in quantum chemistry. Yes, hybrid orbital basis sets are still used, albeit minimally. Typically, these basis sets are used in QM/MM in hopes to scale down the QM computation. There are severe limitations in this approach. You tend to fix the geometry to hybrid orbitals, as you can only overlap certain orbitals. It becomes "ball and stick" chemistry. For this reason, it is not popular. Better to use a traditional QM technique on a smaller region where you will have reactivity.

I can attempt to answer your second question from a comprehensive standpoint as I have never done ab-initio calculations myself using hybrid orbitals. The results should be similar for geometries of simple molecules and break down for increasingly complex molecules. This is because non-VB electrons can undergo orbital overlap.


Doubt on process to calculate pKa I have to solve an exercise that I've been trying for a long time without success.

Information that I have:

  pH @ certain temperature of an aqueous dissolution of an acid
  molar concentration of the acid

Unknown:

  approximate $\mathrm{p}K_\mathrm{a}$ (is not meant to be calculated with calculator I suppose that is the reason they say "approximate")

So far I tried to express the chemical equation and its two steps to finally get $\ce{H3O+}$ and an oxid.
I've calculated backwards the Hydronium concentration $[\ce{H3O+}]$ with the $\mathrm{pH}$ formula. $$ \mathrm{pH} = -\log([\ce{H3O+}])$$.
I've written down the formula to calculate $K_\mathrm{a}$ from where I expected to be able to calculate $\mathrm{p}K_\mathrm{a}$.

I don't know how to extract now the concentrations of every component on the $K_\mathrm{a}$ formula in order to be able to calculate that. Moreover, what am I suppose to do with the temperature?

I hope you can give me guidance in the process to solve such exercises.

Say your acid is monobasic: suppose its formula is $\ce{HA}$.
So, it dissociates according to following equation:
$$\ce{HA + H2O <=> H3O+ + A-}$$
Suppose its molar concentration (given) is $c$, and it dissociate $\alpha$ amount in water (assume total volume is $\pu{1.0 L}$) at given temperature at equilibrium. Thus, equilibrium concentrations of $\ce{HA, H3O+}$, and $\ce{A-}$ are $(c-\alpha)$, $\alpha$, and $\alpha$, respectively.

$$\therefore \; K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}=\frac{\alpha \cdot \alpha}{c-\alpha}$$

Since $\alpha = [\ce{H3O+}]$, $-\log \alpha = \mathrm{pH}$ (which is given) and you know $c$, which is also given, you can estimate the $\mathrm{p}K_\mathrm{a}$ (and hence $K_\mathrm{a}$ if you needed).

Late edition:

Take (-$\log$) on both side of above equation. Thus:

$$\mathrm{p}K_\mathrm{a} = -2\log \alpha + \log{(c-\alpha)}=2(\mathrm{pH})+ \log{(c-\alpha)}$$
Most of the time (in cases of $c\gt\gt\alpha$),  $\mathrm{p}K_\mathrm{a} \approx 2(\mathrm{pH})$

Thank you very much. Now I am gonna insert the values and show you what I get. I have no clue if it makes sense or not.

I start with an aquous dissolution of $\ce{H2SO3}$. I know it will dissociate in two steps, however I read in a book that the first dissociation is normally the strongest and almost every time you can easily ignore the second one, so I proceeded like it follows:

$$p\mathrm{H} = -\log([\ce{H3O+}]) = 1.1$$

This value was given. Then we know that the dissolution has a concentration of $0.5\,\mathrm{M}$. Thus:

$$\ce{H2SO3 + H2O -> H3O+ + HSO3-}$$
$$\ce{c - x -> x   +x}$$

$$K_a = \frac{[\ce{H3O+}][\ce{HSO3-}]}{[\ce{H2SO3}]}=\frac{x·x}{c-x}=\frac{x^2}{c-x}$$

I don't know how to determine this without a calculator. Is there an  approximation for this?

$$K_a = \frac{10^{-2.2}} {0.5-10^{-1.1}} = 0.015$$

From this, and using the expression by the last equation:

$$pK_a = -2 \log(x) + \log(c-x) = 2.2 + \log(0.5-10^{-1.1}) = 1.82$$

Does that make any sense to you?


How are orbitals displayed in computational Chemistry software? In typical software like Gaussian or GAMESS, the molecular orbitals (or even the basis functions) can be displayed as surfaces.

I understand that the surface is probably constructed from a series of points that give a fixed value when substituted into the wavefunction. However, how are these points calculated? Since typical software usually use GTOs, I’d like a formula if possible using those.

Thanks!

Only the program that you used to calculate the electronic structure knows all the idiosyncrasies of the computation to correctly evaluate the orbitals. Therefore, what most people do is to export the evaluation of the orbitals in a grid. Then, a plotting program performs an extraction of a polygonal mesh with a marching cubes (or similar) algorithm.