Does direction matter in breaking a bond? Im not sure if this question belongs here or in another community. I was wondering if the direction that we apply a force in order to break a bond matters and how it affects strength. Say  we have one compound e.g NaCl. Compared to breaking the bond between the Na ion and Cl ion in a tensile manner, would it require a different force to "tear" it in a shearing manner.

A possible real-world example of directional force use may be apparent from the commercial application of so-called magnetizers (see discussion here).
I once experimented with a small inexpensive one and it appears to accelerate select reactions.
A commercial example for reducing the amount of chlorinating agent required in public swimming pools in England demonstrated success to reduce scale build-up, moderate pH and increase available chlorine in a pool, employing magnetic field treatments here.
Some discussion details on how magnetic fields may potentially impact chemical reactions per comments posted previously on StackExchange.
An important point concerns radical based reactions, which per chlorine-based chemistry in sunlight treated swimming pools, is likely extensive and may be potentially impacted. This could particularly alter pool chemistry especially surrounding the fate of the chlorine radical, where an alteration of spins along with a possible decrease in the reverse recombination reaction to Cl2 via the normal reactions:
$\ce{Cl2 + hv -> •Cl + •Cl}$
$\ce{ •Cl + •Cl -> Cl2}$
where the above reaction is impaired due to spin changes by magnetic fields, resulting in more chlorine radicals. The latter can slowly react with water, for example, importantly producing the powerful hydroxyl radical:
$\ce{•Cl + H2O <=> HCl + •OH}$
Source: See Table l in "Photochemistry of HCl and Other Minor Constituents in the Atmosphere of Venus" by Prinn, R.G. as an example.
More detail on mechanics is provided by this source, to quote:

Radical pairs in chemical reactions are known to respond to magnetic fields. This involves pairs of transient radicals being created simultaneously, such that the two electron spins, one on each radical, are correlated. Their chemical fate is largely dictated by weak magnetic interactions, so that you get radicals with unpaired electron spins either antiparallel (singlet state) or parallel (a triplet state).
Even though the interactions of electron spins with an applied field is very weak, there is enough time to affect the balance between singlet and triplet radical pair states and therefore to change the yields of the products formed from them,...


Path to take when labeling R/S I'm trying to find the R/S for the carbon marked with the asterisk. As I branched out to from the asterisk carbon to the red carbons, the methyl group has the 3rd priority (with the hidden hydrogen 4th). Here comes the confusion. In order to find the first point of difference for the two carbons, for one of the carbon, I can simply move outwards to the next green carbon (C, C, H). But for the other carbon, there are two paths I can take, one which gives (C, H, H) and the other (C, C, H). Now, which path should I take?

Assuming I take the path that gives me the (C, C, H) carbon. This gives me a tie. Again, I have two paths to take (indicated by the green arrow). Which path should I take? If I choose to go 'down', the two arrow clash at a single carbon.

I have consulted numerous textbooks but none of them seem to even surface this as an issue. In general, if there are multiply paths, which path should one take?

Extension of question: In a scenario that there are two carbon, C1 and C2 that 'ties', and it so happens that each of the two carbons has 'two paths' to take: C1 with path 1 and path 2, and C2 with path 3 and path 4. What happens if the priority of taking the different paths is not clear. Eg. Priority when taking path 1 is higher than the path taken by path 3, but path 2 has a lower priority then when taking path 4?

The compound you are presenting is a natural product, compactin. The numbers of the carbon atoms are arbitrary. The digraph of C9 shows how the configuration is determined. Sphere 2 contains the atoms C8, C10, C14 and H. Clearly, hydrogen has the lowest priority. Sphere 3 has the locants C10{15,5,H}, C8{7,(7),H} and C14{H,H,H}. In a one-to-one comparison of the locants, C10 and C8 have priority over C14 because C>H. So C14 has the next lowest priority. C(7), which is a duplicate atom of atomic number 6, is attached to three atoms in sphere 4 of atomic number zero. Zero is as low as you can go. Therefore, the priorities are C10>C8>C14>H. C9 has the (S)-configuration.


Carbon dating: What to compare the new value against? Formula for calculating age by using carbon-14 (source: HowStuffWorks)
$$
t = \frac{\ln (N_\mathrm f/N_\mathrm o)}{-0.693}t_{1/2}
$$
The page doesn't explain what the $-0.693$ represents. What is $-0.693$?
Secondly, it says that $N_\mathrm o$ is the original level of $\ce{C-14}$ and $N_\mathrm f$ is the measured value. How do you get the original value? Is it the same for all living organisms?

Wikipedia says that the age of an object (e.g. a fossil) can be calculated by comparing the amount of $\ce{C-14}$ in the fossil and the amount of $\ce{C-14}$ in a sample of living tissue. The half-life of $\ce{C-14}$ ($\pu{5,730 years}$) can be used to find out how much time has passed since the death of the organism because you should be able to calculate how much time has passed by measuring how much $\ce{C-14}$ has decayed. But how can you calculate how much has decayed if you don't know how much was there in the first place?

Large elements of my answer is drawn from my own notes that I use for teaching General Chemistry II.

What is −0.693?

Radioactive decay processes follow first-order kinetics.  A first-order reaction is one where the rate depends only on the concentration of one of the reactants raised to the first power.  So consider the following reaction:

$$\mathrm{A} \longrightarrow \mathrm{products}$$

The rate can be expressed as the rate of change in the reactant concentration with a change in time such that

$$\mathrm{rate} = -\dfrac{\Delta[\mathrm{A}]}{\Delta t}$$

This simply means that [A] is being consumed (i.e. converted to products) as the reaction proceeds with time.  The corresponding rate law can be written as

$$\mathrm{rate} = k[\mathrm{A}]$$

where $k$ is a rate constant.  We can set these two expressions equal to each other to give

$$-\dfrac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]$$

At this point we must apply a bit of calculus.  Let us write the previous equation in differential form to give

$$-\dfrac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]~~~\mathrm{or}
~~~\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -kdt$$

Integrate over $t=0$ to $t = t$ to give

$$\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -k\int_0^t dt$$

which results in

$$\ln[\mathrm{A}]_t - \ln[\mathrm{A}]_0 = -kt$$

or

$$ \ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt$$

where $[\mathrm{A}]_0$ is an initial concentration and $[\mathrm{A}]_t$ is the concentration at some time, $t$.

Now that we have our first-order integrated rate law, we can now discuss half-life.  Half-life (denoted as $t_{\frac{1}{2}}$) is simply the time required for some reactant concentration to drop to half its original value (where the original value is simply [A]$_0$.  So

$$t = t_{\frac{1}{2}}~~~\mathrm{when}~~~[\mathrm{A}]_t = \dfrac{1}{2}[\mathrm{A}]_0$$

Given this relationship, we can rearrange our first-order integrated rate law to solve for $t$ such that

$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0} = -kt~~~\longrightarrow~~~
t = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t} $$

Therefore, we can make some simple substitutions into our first-order integrated rate law at $t = t_{\frac{1}{2}}$ to give

$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{\frac{1}{2}[\mathrm{A}]_0} $$

We can then simplify this equation such that

$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln2\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_0} ~~~\longrightarrow~~~
t_{\frac{1}{2}} = \dfrac{1}{k}\ln (2)
~~~\longrightarrow~~~
t_{\frac{1}{2}} = \dfrac{0.693}{k}$$

So to answer your question, the 0.693 value comes from the $\ln(2)$ term.  As ringo pointed out below (and I will include here to be thorough), you can now solve your first-order integrated rate law for $t$ and substitute in your half-life equation such that

$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt
~~~\longrightarrow~~~
\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}{-k}=t
~~~\longrightarrow~~~
\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}
{
-\frac{0.693}{t_{\frac{1}{2}}}
}
=t
$$

or more cleanly...

$$t = \dfrac{\left(t_{\frac{1}{2}}\right)
\left(\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}\right)}
{-0.693}
$$

How do you get $N_0$?

You could get $N_0$ by examining the activity of carbon-14 in a relatively recent object or plant.  Carbon-14 in a living plant is not expected to have decayed at all (since the half-life is over 5000 years!).  Therefore you could use its activity for $N_0$.  Of course this procedure relies on a few assumptions but carbon-14 dating is an approximation to begin with.


Can F2/HF be separated using acid/base solutions in series? I am trying to verify the results of EPA method 26A (http://www.epa.gov/ttnemc01/promgate/m-26a.pdf). The Method claims to be able to separate and quantify a mixed halogen/halide gas stream. Although I believe the chemistry holds true for Cl₂/HCl and Br₂/HBr, I'm not so sure about F₂/HF. Despite the Method not mentioning F₂, it commonly produces data for both HF and F₂. I am dubious that the F₂ results are real and am trying to develop a chemistry argument to support that position, if possible.

In summary, the Method takes a mixed gas stream (F₂/HF) and passes it through a 0.1 N H₂SO₄ solution, then a 0.1 NaOH solution; sodium thiosulfate is later added to the alkaline potion. The theory is that the HF is captured in the acid solution, while the F₂ passes through unreacted. In the alkaline solution, the F₂ reacts to form HF and the hypohalous acid (HOF, I assume). The thiosulfate converts the HOF to another HF. When the acid and alkaline solutions are analyzed for fluoride ion, the quantity is assumed to be the HF and F₂ in the original sample, respectively.

Considering that F₂ is so reactive, I would think that any F₂ in the gas stream would react with the water in the acid solution and that theoretically no F₂ would make it to the alkaline solution. Is this a reasonable assumption?

If any F₂ did get to the alkaline solution, I am not sure that the conditions are favorable for the creation of HOF; F₂ reacting with water to produce HF seems more probable. Unless the acid solution is ignored by the F₂, I don't think that a true separation is taking place and any fluoride detected in the alkaline solution would represent blow through or some other error. Does this seem like a reasonable conclusion?

Thanks in advance for any input!

The answer is no, because as you remark $\ce{F2}$ will react with water and produce $\ce{HF}$.  So it is not the case that $\ce{F2}$ passes through unreacted.

I'm puzzled by this sentence:

  Despite the Method not mentioning F₂, it commonly produces data for both HF and F₂

Can you clarify what you mean by "it produces data"?


Why are reactions of acids with metals exothermic? I realised the other day that I'm not sure why the reactions of strong acids with metals are exothermic. Any reaction is exothermic if there's more energy released from bonds being made than is used in breaking bonds, of course (using the word 'bonds' in a sufficiently loose sense).

In these reactions, metallic bonds are broken; covalent bonds are formed in hydrogen, but the hydrogen is in a gaseous state, so its molecules must be less attracted to each other than the hydrogen ions were to each other in the liquid. I'd vaguely guessed that the latter fact would make this an endothermic reaction (it works for sodium carbonate, after all) but evidently not!

Is there a rigorous way of calculating the total enthalpy change, or an easy explanation that I'm missing?

I guess I can calculate the enthalpy change of every one of the state changes involved, but it's not obvious that that's always going to come out negative (which I gather it does) so I feel like there's probably a less mathematical way of explaining this observation.

Yes there is a rigorous way of calculating the enthalpy change.

First write down an example reaction.  It might not be as easy as you think.  What species are really formed when zinc chloride (for example) dissolves in water?

The simple version below...

$$\ce{Zn(s) + 2 H+(aq) -> Zn^2+(aq) + H2(g)}$$

...won't do, solvated zinc ions in hydrochloric acid are present in a variety of tetrahedral chloro-aqua complexes.

So maybe something like

$$\ce{Zn(s) + 2 H+(aq) + 2 Cl-(aq) + 2H2O(aq) -> ZnCl2(H2O)2(aq) + H2(g)}$$

This at least shows which bonds are being broken and which are being formed:

Zn-Zn bonds (metallic) are broken
H-H bonds are being formed
Zn-Cl and Zn-aqua (actually zinc-oxygen) bonds are being formed

To figure out the enthalpy change, you can figure out the approximate enthalpy of formation of each bond type.  H-H bonds are pretty "strong", i.e. the bond enthalpy is relatively high at 436 kJ/mol.  So when those bonds are formed, that enthalpy is released.  Zinc-water (i.e. zinc-oxygen) and zinc-chlorine bonds are also pretty strong at about 284 and 229 kJ/mol, respectively.  Zinc-zinc bonds enthalpies are only ~29 kJ/mol.

Add it all up, and you get a lot of enthalpy being released!


Effect of redefining Avogadro's constant and kilogram on molar mass Avogadro's constant $N_\mathrm A$ is defined as the number of constituent particles (usually atoms or molecules) contained in the amount of substance given by one mole.

The value of Avogadro's constant $N_\mathrm A$ is going to be set to exactly $6.02214076\times 10^{23}\ \mathrm{mol}^{-1}$ in 2019.

I might add that the kilogram is also going to be redefined in 2019 based on Planck's constant.

Will these changes result in the molar mass of a substance no longer being exactly equal to the molecular mass expressed in grams?

Will these changes result in the molar mass of a substance no longer being exactly equal to the molecular mass expressed in grams?

Yes. The ratio between the molar mass and the relative molecular mass is called the molar mass constant, and it used to be 1 g/mol exactly. Now that Avogadro's constant is set to a fixed value, it's value has to be determined experimentally.

NIST quotes its value as 0.999 999 999 65(30) x 10-3 kg mol-1 (CODATA 2018).

The relative molecular mass of the carbon-12 isotope is still exactly 12 and the atomic mass unit u (or dalton Da) is still defined as 1/12 of the mass of a carbon-12 atom, so the relationship between relative molecular mass and molecular mass still holds. However, its molar mass is no longer exactly 12 g/mol.

As @MaxW says in a comment, these changes are so minute that they have no practical implications outside of metrology (which was one of the goals of the International Bureau of Weights and Measures when they suggested the change).


Why do acids donate protons? I would have thought the answer to this question would be simple, but after searching this site I was unable to find any answers.

I know Brønsted-Lowry acids are defined in terms of proton donation, but what is it that makes the acids donate protons?

I guess the same would apply to Lewis acids, but in terms of electron pair acceptance.

Using the following generic reaction, where an acid donates $\ce{H+}$, i.e. a proton:

$$\ce{HX + B <=> X- + HB+}$$

If a lone pair of electrons on $\ce{B}$ has a higher energy than than than the resultant lone pair of electrons on $\ce{X-}$, then the $\ce{H+}$ on $\ce{HX}$ will be transferred to $\ce{B}$ to form $\ce{HB+}$ in order to stabilize the high energy lone pair on $\ce{B}$ in the form of a bond.

So, there are two things that influence whether an acid will transfer a proton:

The energy of the lone pair on the base
The energy of the lone pair on the resultant the conjugate base

If the conjugate base of the acid is more able to stablize a lone pair than the basic starting material is, an acid-base reaction will occur, and a majority of the basic reactant will end up being protonated.


Which is a better nucleophile: hydroxide anion or amide anion? I know that nucleophilicity order depends on what type of solvent we take, polar protic or polar aprotic and also on the basicity of nucleophile.

  In DMF, which is a better nucleophile $\ce{NH2-}$ or $\ce{OH^-?}$

Since the solvent is aprotic polar, it doesn't have any donor hydrogen to form  hydrogen bonds with nucleophile. So, the nucleophilicity should depend on which among them is more basic. Now, since $\ce{N}$ is less electronegative than $\ce{O}$, it's lone pair is more readily available than that of $\ce{OH-}$.

So, according to me, $\ce{NH2-}$ should be more nucleophilic than $\ce{OH-}$, but the answer given is the opposite, that is, $\ce{OH-}$ is more nucleophilic than $\ce{NH2-}$. How is my reasoning incorrect?

And also what are the factors we should keep in mind while deciding nucleophilicity?

I was not able to find data on this nucleophilicity of $\ce {NH2^-}$ and $\ce {OH^-}$ in dimethylformamide, $\ce {Me2NCHO}$. However, I have a plausible explanation as to why the answer may not be what you expected.

Firstly, it is important to remind ourselves that nucleophilicity is always relative to the electrophile that we are concerned with. However, it is also possible to discuss nucleophilicity generally, i.e. not with respect to any particular electrophile. This is the case now.

The fact that the solvent is polar and aprotic suggest that hydrogen bonds cannot be made with the $\ce {N}$ and the $\ce {O}$ of the nucleophiles. However, it does not restrict us to form hydrogen bonds between the lone pair on the carbonyl oxygen with the partially positively-charged $\ce {H}$ atoms on the ions. This is not at all unlikely considering that the strong delocalisation of the lone pair on the $\ce {N}$ in amides, gives a significant resonance structure with this carbonyl oxygen bearing a formal negative charge.

$\ce {NH2^-}$ has two such $\ce {H}$s and $\ce {OH^-}$ has only one such hydrogen. Thus, perhaps, we may still expect greater extent of solvation for the former. Relative to $\ce {OH^-}$, the greater amount of solvation of $\ce {NH2^-}$ increases the activation energy required to initiate the subsitution reaction since solvation can be said to bring down the energy of the reactant in the reaction energy profile. Thus, the rate of substitution with $\ce {OH^-}$ would be faster than $\ce {NH2^-}$ in this polar aprotic solvent.

Alternatively, the answer provided could very well be wrong. That is a possibility that cannot be discounted...


How do you find the pH of a solution with a known pKa and Molarity? How do you find the pH of a solution with a known pKa and Molarity?

Do you know the formula:

$$K_\text{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$

?

You will need to find a concentration first, however, on the equation.

$$\ce{HA <=> H+ + A-}$$

And solve for the dissociated hydrogen ion, say $x$.

$$K_\text{a} = \frac{[x][x]}{[\ce{HA}]-x} = \frac{x^2}{[\ce{HA}]-x}$$

The logic is: if $x$ of $\ce{HA}$ dissociated, then we would get $x$ of $\ce{H}$, $x$ of $\ce{A}$ and $[\ce{HA}]-x$ left of $[\ce{HA}]$.

Does this make sense?


pKa vs range in choice of an indicator When choosing the ideal indicator during a titration, there are two considerations to keep in mind:

The pH = pKa of the equivalence point
The indicator has a small range (<±1) over which it can change color

Which factor is more important when deciding over two indicators to see which is best?

The pH = pKa of the equivalence point.
it is the best factor to decided during the titration .
because if the range of the indicator is less , then we can't find the correct volume used in titration.this factor should be considered but not most affective one.


Activation energy and degrees of freedom According to the Arrhenius equation the rate of a reaction increases as at higher temperature a greater percentage of molecules have translational kinetic energy greater than or equal to activation energy Wikipedia.
I was recently introduced in molecular potential surfaces and I read that some reactions can proceed when excess kinetic energy is stored into vibrationals degrees of freedom than translational. But according to Wiki it seems that only translational kinetic energy matters. First, I want to understand if the factor:
$$e^{\frac{-E_a}{RT}}$$
takes into account only the percentage that have the necessary kinetic energy in translational degrees of freedom or in general the percentage with at least this energy irrespective if this energy is stored in a combination of various degrees of freedom? Also for two reactants $\ce{A}$ and $\ce{B}$ does it calculate the percentage of "pairs" $\ce{A}$, $\ce{B}$ that have this amount of energy or just the percentage of the population with at least this energy?
Secondly, if some reactions speed up depending in which type of degrees of freedom (vibrational or translational) are excited, does this mean that the Arrhenius equation even for an elementary bimolecular reaction holds only if the reactants have only translational degrees of freedom (like collision of two atoms $\ce{A}$ and $\ce{B}$ to form $\ce{A-B})?$

This depends upon the theory used to model the reaction; there are two major ones.
Collision Theory
If your molecules may be assumed to be spherical particles with no rotational or vibrational DOF, then collision theory appropriately describes your system. In this case, only translational energy is accounted for. The rate expression may then be derived from the Maxwell-Boltzmann kinetic theory of gases, where the energy of each particle is $\frac{1}{2}mv^{2}$.
I. Filot has a decent explanation of this.
Transition State Theory
Unfortunately, most reaction systems are much more complicated; the reactants experience not only translational, but also rotational and vibrational DOF. This must be accounted for in the rate expression. Again, I. Filot has a good explanation for this.
A general derivation includes the following:
Consider a bimolecular reaction:
$$A+B \rightarrow C$$
We assume there exists a transition state (TST), which defines the highest point along a minimum energy path on the potential energy surface. Moreover, we assume that our reactants are quasi-equilibrated with this TST; the rate of the reaction depends upon the frequency of reactants overcoming this barrier $(k_{f})$. Then:
$$A+B \leftrightarrow C^{\ddagger} \rightarrow C$$
Assuming this reaction occurs in the gas phase under ideal conditions, we can approximate their fugacity using the partial pressure of each reactant. Let us also assume that the reference partial pressures are unity. We derive an equilibrium constant between the TST and the reactants:
$$K^{\ddagger}=\frac{P_{C^{\ddagger}}}{P_{A}P_{B}} = exp(-\Delta G^{\ddagger}/RT) = exp(-\Delta H^{\ddagger}/RT)exp(\Delta S^{\ddagger}/R))$$
Our rate expression becomes:
$$r = K^{\ddagger}k_{f} P_{A}P_{B} = [(k_{f}) exp(\Delta S^{\ddagger}/R))exp(-\Delta H^{\ddagger}/RT)]P_{A}P_{B} = kP_{A}P_{B}$$
Our rate constant is:
$$k = (k_{f}) exp(\Delta S^{\ddagger}/R))exp(-\Delta H^{\ddagger}/RT) = (A)exp(-\Delta H^{\ddagger}/RT) \equiv (A)exp(-E_{a}/RT)$$
Where the partition functions defining the enthalpy and entropy include translational, rotational and vibrational DOF.
Therefore, to answer your questions:
"First, I want to understand if the factor $e^{-E_{a}/RT}$ takes into account only the percentage that have the necessary kinetic energy in translational degrees of freedom or in general the percentage with at least this energy irrespective if this energy is stored in a combination of various degrees of freedom?
No, it takes into account all DOF. In transition state theory, both the activation energy and pre-exponential factor consider the translational, rotational and vibrational DOF of the transition-state irrespective of the DOF providing the energy (This is directly tied to the canonical partition function, which is used in defining the enthalpy and entropy. It only cares about the total microstate energy and not from which DOF it comes from).
"Secondly, if some reactions speed up depending in which type of degrees of freedom (vibrational or translational) are excited, does this mean that the Arrhenius equation even for an elementary bimolecular reaction holds only if the reactants have only translational degrees of freedom?"
Reactions do not speed up depending on which type of DOF are excited. However, certain DOF speed up the reaction much more than other DOF. For example, based on the derivation I have provided, the pre-exponential factor scales exponentially with the transition-state entropy.
$$S_{total} = S_{translations}+S_{rotations}+S_{vibrations}$$
The DOF with the largest contribution to the entropy are translations. In theory, we may increase the rate of the reaction by increasing the vibrational entropy of our transition state. But, we may increase it to a much larger extent by increasing it's translational entropy.


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

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

Thank you.

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

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

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

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

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


Why does pKa of a acid-base indicator equal to the pH when the equivalence point is reached? To be honest, I just don't get it.
I had this question while reading a Chemguide article, on this topic:

I think my confusion is caused by the way I look at pKa of acid-base indicator. Should you think of it more as a reaction quotient. Because technically, at equilibrium, the two concentration aren't the same so cannot be cancelled out.

Look at the following equation:

  $$\mathrm{pH} = \mathrm pK_{\mathrm a} + \log\frac{[\ce{A^-}]}{[\ce{HA}]}$$

At the half equivalence point, say we have 10 moles of $\ce{WA},$ and so there will be 5 moles of $\ce{SB}$ as the name suggests ("half equivalence"). So we know that the 10 moles will be neutralized to 5 moles of $\ce{WA}$ by the $\ce{SB}.$ If we do this then the $\ce{CB}$ is also now equal to the $\ce{WA}$ since it will increase by 5 moles.

$$\ce{[WA]~ =~ [CB]}.$$

Thus,

\begin{align}\mathrm{pH} &= \mathrm pK_{\mathrm a} + \log 1\\  &=\mathrm pK_{\mathrm a}\;.\end{align}

So this is why the the $\mathrm{pH} = \mathrm pK_{\mathrm a}$ at the Half-Equivalence Point.

  Vocabulary:

  SB : Strong Base. (eg. $\ce{NaOH}$)
  CB : Conjugate Base. (eg. $\ce{Ac^{-}}$)
  WA : Weak Acid. (eg. $\ce{HAc}$)

NOTE: We can never be 100% accurate here, this is just used to derive that $\mathrm{pH} = \mathrm pK_{\mathrm a}$


How does one test for lead in dry paint? What analytical methodologies are capable of identifying lead in peeling paint? I am a hobby chemist owner of a Clinical Research Investigative Site.  Please tell me if this site is only for professional chemists.
If so, I will cancel my registration.

This site is not specialized and reserved to professional chemists. On the contrary. A lot of teenagers are using it every day.
Now if you want to identify lead in a dry paint, you may burn it in the flame of a Bunsen burner, and save the ashes. They contain a mixture $\ce{Pb + PbO + Pb3O4}$. Then dissolve the ashes in nitric acid (concentrated $\ce{HNO3}$ diluted with its own volume of water). Filtrate the insoluble part. Dilute the filtrate still more, by adding an equal volume of water. This gives you a lead nitrate solution, with which you carry out the following tests. Add some $\pu{1 M}$ $\ce{HCl}$ to a part of the obtained solution : it should produce a white precipitate of $\ce{PbCl2}$, if the solution contains some lead ions. Or add some $\ce{KI}$ solution : You would obtain a yellow precipitate of $\ce{PbI2}$, if the solution contains lead. The reactions are :
$$\ce{Pb(NO3)2 + 2 HCl -> PbCl2(s) + 2 HNO3}$$
$$\ce{Pb(NO3)2 + 2 KI -> PbI2(s) + 2 KNO3}$$
A further proof is obtained by heating the solution containing the yellow deposit of $\ce{PbI2}$, because this substance gets redissolved at $\pu{100°C}$. And after slow cooling, $\ce{PbI2}$ will precipitate back and it looks like wonderful golden plates, floating in the liquid. It is amazing. If these precipitate do not occur, there is no lead in your paint.


What style clamp holder is this? I cannot for the life of me find this style of clamp holder with beaker base:

I can find beaker bases on sites like Fisher Scientific/etc., and flimsy alligator clamp soldering stands on Mouser/etc., but not this specific alligator (or other wire holding) clamp with sliding connector, on solid rod, attached to a beaker base.

Does anyone know what those sliding clamp holders are called, or where I can get this as a kit? I use this for e-chem and love it but need more.

Edit, 15-Jan This is an older version of my actual setup, which I would typically have on top of a magnetic stirring plate:

I would then be taking 2 of the 3 probes and passing them through a number of other poured solutions. My current clamping setup is great because I can simply release the counter electrode and just move my working and reference to different baths; really, I just need more of what I already have. As these experiments are limited in scope and do not require it, I do not make provisions for salt bridges, nitrogen purging, hydrogen gas aeration, etc.

Cyclic voltammetry (CV) is often performed in an enclosed vessel with the purge gas (argon/nitrogen) constantly bubbling through to prevent the solution from aeration and unnecessary side processes, mainly oxidation by the atmospheric oxygen. You'd rarely see alligator clamps in any CV apparatus also because they are not suited for holding moderately thick round objects (such as electrodes) in place, they can damage the inlet and outlet piping (e.g. the one for the purge gas, or chemicals if one studies kinetics), plus they make the experiments less reproducible since the position of the electrodes is changed each time you change the beaker.

Typically a teflon lid with 4 (or more) holes is used, where all 3 electrodes and the purging line are adjusted in place with rubber rings or plugs so that they are all firmly attached to the lid. This is very convenient since you don't have to manually adjust the apparatus every time the solution needs to be changed. All that's required is take the lid off, rinse the vessel and electrodes, fill the new portion in and put the lid back in place.

Alternatively, one can use a 3-neck round bottom flask with the electrodes attached through the rubber stoppers in the side necks. The downside is problematic cleaning, so this setup is more appropriate for the continuous measurements when one doesn't have to switch solutions too often. This is the simplest setup; a more sophisticated one can as well look like this:

  Image source

Either way, I would discourage from using alligator clips for CV. Even though they might work for simple measurements, in my opinion there are too many downsides.


Are all anomers diastereomers? I am told that all anomers are diastereomers. I think this is BS. If there is a carbohydrate with only one chiral center, then changing the configuration of the anomeric carbon gives us enantiomers.

Or is it impossible to have carbohydrates with only one chiral center?

The only saccharides I know of that have only a single chiral centre are the trioses L- and D-glyceraldehyde, but, at least according to the Gold Book definition, anomers are specifically diastereomers of cyclic forms of sugars. Trioses are too short for a cyclic form to normally arise and therefore can't anomerize.


Why are there different values for enthalpy of combustion, depending on the calculation method? Why are there different values for enthalpy of combustion, depending on the calculation method?
Take for example the combustion of ethanol:
$$\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O (l)}$$

If I calculate this using enthalpy of formation:

\begin{array}{c|c}
\text{Substance} & \Delta H^\circ /\pu{(kJ/mol)}  \\
\hline
\ce{H2O(l)}    & -285.8 \\
\ce{CO2(g)}    & -393.5 \\
\ce{C2H5OH(l)} & -277.0
\end{array}

\begin{align}
  \Delta H &=
    \left[2\cdot(\pu{-393.5 kJ/mol}) + 3\cdot(\pu{-285.8 kJ/mol})\right]
    -\left[\pu{-277.0 kJ/mol} +  3\cdot(\pu{0 kJ/mol})\right]\\
  &= \pu{-1367.4 kJ/mol}
\end{align}

If I calculate this using bond energy:

\begin{array}{c|c}
\text{Bonds} & \Delta H^\circ /\pu{(kJ/mol)}  \\
\hline
\ce{C-C}      &  347 \\
\ce{C-H}      &  414 \\
\ce{O-H}      &  464 \\
\ce{C-O}      &  351 \\
\ce{C=O(CO2)} &  803 \\
\ce{O=O}      &  498
\end{array}

\begin{align}
  \Delta H &=
    \pu{347 kJ/mol} + \pu{351kJ/mol} + \pu{464 kJ/mol}
    + 5\cdot(\pu{414 kJ/mol}) \\
  &\quad
    + 3\cdot(\pu{498 kJ/mol})
    + 4\cdot(\pu{-803 kJ/mol}) + 6\cdot(\pu{-464 kJ/mol})\\
  &= \pu{-1270 kJ/mol}
\end{align}

Why is there such a big difference between them?

Average bond enthalpies (note the word "average") are calculated in a different way from formation enthalpies. Formation enthalpies are well-defined and precise, meaning that two different people could measure them the same way and get the same answer. Average bond enthalpies are averages over many different types of bonds.

For example, there is only one way to make $\ce{CH4}$, but there are many different molecules with $\ce{C-H}$ bonds. The enthalpy of dissociation of a $\ce{C-H}$ bond in $\ce{CH4}$ is different from that in $\ce{C2H6}$, and in $\ce{C6H6}$, and really in any other molecule you can think of. To find average bond enthalpies, a selection of a large number of those bonds in many different molecules is taken, and the enthalpy of dissociation for each is measured. Then, the average (mean) is recorded. If you look at some different charts of bond enthalpies, you will notice two things: First, there are some numbers marked "exact" - $\ce{H-H}$, $\ce{Cl-Cl}$, $\ce{F-F}$, for example. This is because there is only one molecule with each of those types of bonds. The other thing you will notice is that depending on which chart you are using, many of the numbers will be slightly different. This is because there is no universal, unchanging standard describing which molecules are used to determine each bond - it depends upon what the people making the chart decided to use.

Because of this difference, when making predictions, average bond enthalpies are less accurate than formation enthalpies. In practice, they are used for quick estimations, or as a last resort when formation enthalpies are not available.

When I taught chemistry, I would have my students measure reaction enthalpies in lab, then calculate them using formation enthalpy tables and average bond dissociation enthalpy tables. The goal was to show them that average bond enthalpies can be reasonably close for rough estimates, but that formation enthalpies are much better in practice. It also showed them that accurately measuring reaction enthalpy in the lab is difficult!


experimentally determine oxidation state of Iron compound I need to make several Iron compounds but I am having a bit of trouble finding a decent way to determine the oxidation state of Iron.
I'm starting with a Iron(II) compound that I try to oxidize using several methods. I can use raman to determine the compound but that requires me to create the compound I want to make and check that spectrum against my experimental sample. This is quite a bit of work and I was hoping someone could point me towards a better solution.
I can make an EDS but since my compounds usually contain carbon or oxygen as counter ion so EDS isn't very accurate. I don't (currently) have access to an XPS and I don't have unlimited access to a mass spectrometer (I have to travel quite a bit to get to one).

Can someone help me find a better way to determine the oxidation state??

In all the (physical) experimental literature I've read, at least for biomolecules (usually enzymes), the technique of choice is Mossbauer spectroscopy. Here is a good starting point.


Number of two electron integrals I was trying to write a C++ code following the instruction given in this website. The problem is in the two electron integral file, only 228 entries for 7 molecular orbitals are provided. But my code says it should be 406 (please see below)

int NFI=0; //NFI=number of four electron integrals
for (int i=0; i<NAO; i++){
 for (int j=0; j<=i; j++){
  for (int k=0; k<=i; k++){
   for (int l=0; l<=k; l++){
    int ij=i*(i+1)*0.5+j;
    int kl=k*(k+1)*0.5+l;
     if (kl<=ij){NFI += 1;}
   }
  }
 }
}
printf("%d\n",NFI);

I'm using 4 criteria as can be seen in this code snippet. Can anyone explain me why the number of unique integral is 228, not 406?

As per suggestion, I compared my index with the index provided, here they are

1 1 1 1                   1     1     1     1
2 1 1 1                   2     1     1     1
2 1 2 1                   2     2     1     1
2 2 1 1                   2     1     2     1
2 2 2 1                   2     2     2     1
2 2 2 2                   2     2     2     2
3 1 1 1                   3     3     1     1
3 1 2 1                   4     4     1     1
3 1 2 2                   5     5     1     1

Left 4 numbers are generated from my code and right 4 numbers are given in the input file. (3,1,1,1) is missing in the data file provided (I checked the whole file), but it fulfills the criteria of $ij \geq kl$.

Update:: I think that 228 integral was a mistake, I found another project file from Dr. Crawford which listed 406 unique integrals for the same system ($H_2O$ with 7 molecular orbitals). In case if anyone wants to work on that project, they can use this data file.

-----------------------------------------------------------------------
      PSI4: An Open-Source Ab Initio Electronic Structure Package
                          PSI 4.0.0-alpha Driver
T. D. Crawford, C. D. Sherrill, E. F. Valeev, J. T. Fermann, R. A. King,
M. L. Leininger, S. T. Brown, C. L. Janssen, E. T. Seidl, J. P. Kenny,
and W. D. Allen, J. Comput. Chem. 28, 1610-1616 (2007)

                     Additional Contributions by
Francesco Evangelista, Andrew Simmonett, Justin Turney, Jeremiah Wilke
-----------------------------------------------------------------------

Using LocalCommunicator (Number of processes = 1)

Memory level set to 256.000 MB

./plugin_aointegrals.so loaded.
Calling plugin plugin_aointegrals.so.
    Molecular point group: c2v

Geometry (in Angstrom), charge = 0, multiplicity = 1:

   Center              X                  Y                   Z
------------   -----------------  -----------------  -----------------
       O          0.000000000000     0.000000000000    -0.071151380605
       H          0.000000000000     0.757939245855     0.564612021746
       H          0.000000000000    -0.757939245855     0.564612021746

Number of unique atoms: 2

Atoms equivalency:
   unique atom 0: 0
   unique atom 1: 1 2

Nuclear repulsion energy:       8.90770810

## Overlap ##

             1            2            3            4            5            6            7

1    1.0000000    0.2367039    0.0000000    0.0000000   -0.0000000    0.0500137    0.0500137
2    0.2367039    1.0000000    0.0000000    0.0000000   -0.0000000    0.4539953    0.4539953
3    0.0000000    0.0000000    1.0000000    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000    1.0000000    0.0000000    0.2927386   -0.2927386
5   -0.0000000   -0.0000000    0.0000000    0.0000000    1.0000000    0.2455507    0.2455507
6    0.0500137    0.4539953    0.0000000    0.2927386    0.2455507    1.0000000    0.2510021
7    0.0500137    0.4539953    0.0000000   -0.2927386    0.2455507    0.2510021    1.0000000

## Kinetic ##

             1            2            3            4            5            6            7

1   29.0031999   -0.1680109    0.0000000    0.0000000   -0.0000000   -0.0045400   -0.0045400
2   -0.1680109    0.8081280    0.0000000    0.0000000   -0.0000000    0.1137475    0.1137475
3    0.0000000    0.0000000    2.5287312    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000    2.5287312    0.0000000    0.1993355   -0.1993355
5   -0.0000000   -0.0000000    0.0000000    0.0000000    2.5287312    0.1672036    0.1672036
6   -0.0045400    0.1137475    0.0000000    0.1993355    0.1672036    0.7600319    0.0083249
7   -0.0045400    0.1137475    0.0000000   -0.1993355    0.1672036    0.0083249    0.7600319

## Potential ##

             1            2            3            4            5            6            7

1  -61.6882822   -7.4363118    0.0000000    0.0000000   -0.0186797   -1.6150635   -1.6150635
2   -7.4363118  -10.1101908    0.0000000    0.0000000   -0.2221598   -3.6569582   -3.6569582
3    0.0000000    0.0000000   -9.9595668    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000  -10.0957534    0.0000000   -2.0901916    2.0901916
5   -0.0186797   -0.2221598    0.0000000    0.0000000  -10.0553869   -1.8259929   -1.8259929
6   -1.6150635   -3.6569582    0.0000000   -2.0901916   -1.8259929   -5.7165220   -1.5685885
7   -1.6150635   -3.6569582    0.0000000    2.0901916   -1.8259929   -1.5685885   -5.7165220

## One Electron Ints ##

             1            2            3            4            5            6            7

1  -32.6850823   -7.6043227    0.0000000    0.0000000   -0.0186797   -1.6196036   -1.6196036
2   -7.6043227   -9.3020628    0.0000000    0.0000000   -0.2221598   -3.5432106   -3.5432106
3    0.0000000    0.0000000   -7.4308356    0.0000000    0.0000000    0.0000000    0.0000000
4    0.0000000    0.0000000    0.0000000   -7.5670222    0.0000000   -1.8908561    1.8908561
5   -0.0186797   -0.2221598    0.0000000    0.0000000   -7.5266557   -1.6587893   -1.6587893
6   -1.6196036   -3.5432106    0.0000000   -1.8908561   -1.6587893   -4.9564901   -1.5602636
7   -1.6196036   -3.5432106    0.0000000    1.8908561   -1.6587893   -1.5602636   -4.9564901

Two-electron Integrals

( 0  0 |  0  0) =    4.785065404705502
( 1  0 |  0  0) =    0.741380351973408
( 1  1 |  0  0) =    1.118946866342470
( 1  0 |  1  0) =    0.136873385354388
( 1  1 |  1  0) =    0.256633394730974
( 1  1 |  1  1) =    0.817206321526058
( 2  0 |  0  0) =    0.000000000000000
( 3  0 |  0  0) =    0.000000000000000
( 4  0 |  0  0) =   -0.000000000000000
( 2  1 |  0  0) =    0.000000000000000
( 3  1 |  0  0) =    0.000000000000000
( 4  1 |  0  0) =   -0.000000000000000
( 2  0 |  1  0) =    0.000000000000000
( 3  0 |  1  0) =    0.000000000000000
( 4  0 |  1  0) =   -0.000000000000000
( 2  1 |  1  0) =    0.000000000000000
( 3  1 |  1  0) =    0.000000000000000
( 4  1 |  1  0) =   -0.000000000000000
( 2  0 |  1  1) =    0.000000000000000
( 3  0 |  1  1) =    0.000000000000000
( 4  0 |  1  1) =   -0.000000000000000
( 2  1 |  1  1) =    0.000000000000000
( 3  1 |  1  1) =    0.000000000000000
( 4  1 |  1  1) =    0.000000000000000
( 2  2 |  0  0) =    1.115813812152427
( 3  2 |  0  0) =    0.000000000000000
( 3  3 |  0  0) =    1.115813812152427
( 4  2 |  0  0) =    0.000000000000000
( 4  3 |  0  0) =    0.000000000000000
( 4  4 |  0  0) =    1.115813812152427
( 2  0 |  2  0) =    0.024477412258099
( 3  0 |  2  0) =    0.000000000000000
( 3  0 |  3  0) =    0.024477412258099
( 4  0 |  2  0) =    0.000000000000000
( 4  0 |  3  0) =    0.000000000000000
( 4  0 |  4  0) =    0.024477412258099
( 2  2 |  1  0) =    0.256683985810103
( 3  2 |  1  0) =    0.000000000000000
( 3  3 |  1  0) =    0.256683985810103
( 4  2 |  1  0) =    0.000000000000000
( 4  3 |  1  0) =    0.000000000000000
( 4  4 |  1  0) =    0.256683985810103
( 2  1 |  2  0) =    0.037808607416361
( 3  0 |  2  1) =    0.000000000000000
( 4  0 |  2  1) =    0.000000000000000
( 3  1 |  2  0) =    0.000000000000000
( 3  1 |  3  0) =    0.037808607416361
( 4  0 |  3  1) =    0.000000000000000
( 4  1 |  2  0) =    0.000000000000000
( 4  1 |  3  0) =    0.000000000000000
( 4  1 |  4  0) =    0.037808607416361
( 2  2 |  1  1) =    0.817022605320914
( 3  2 |  1  1) =    0.000000000000000
( 3  3 |  1  1) =    0.817022605320914
( 4  2 |  1  1) =    0.000000000000000
( 4  3 |  1  1) =    0.000000000000000
( 4  4 |  1  1) =    0.817022605320914
( 2  1 |  2  1) =    0.180518392104632
( 3  1 |  2  1) =    0.000000000000000
( 3  1 |  3  1) =    0.180518392104632
( 4  1 |  2  1) =    0.000000000000000
( 4  1 |  3  1) =    0.000000000000000
( 4  1 |  4  1) =    0.180518392104632
( 2  2 |  2  0) =    0.000000000000000
( 3  2 |  2  0) =    0.000000000000000
( 3  3 |  2  0) =    0.000000000000000
( 4  2 |  2  0) =    0.000000000000000
( 4  3 |  2  0) =    0.000000000000000
( 4  4 |  2  0) =    0.000000000000000
( 3  0 |  2  2) =    0.000000000000000
( 3  2 |  3  0) =    0.000000000000000
( 3  3 |  3  0) =    0.000000000000000
( 4  2 |  3  0) =    0.000000000000000
( 4  3 |  3  0) =    0.000000000000000
( 4  4 |  3  0) =    0.000000000000000
( 4  0 |  2  2) =   -0.000000000000000
( 4  0 |  3  2) =    0.000000000000000
( 4  0 |  3  3) =   -0.000000000000000
( 4  2 |  4  0) =    0.000000000000000
( 4  3 |  4  0) =    0.000000000000000
( 4  4 |  4  0) =   -0.000000000000000
( 2  2 |  2  1) =    0.000000000000000
( 3  2 |  2  1) =    0.000000000000000
( 3  3 |  2  1) =    0.000000000000000
( 4  2 |  2  1) =    0.000000000000000
( 4  3 |  2  1) =    0.000000000000000
( 4  4 |  2  1) =    0.000000000000000
( 3  1 |  2  2) =    0.000000000000000
( 3  2 |  3  1) =    0.000000000000000
( 3  3 |  3  1) =    0.000000000000000
( 4  2 |  3  1) =    0.000000000000000
( 4  3 |  3  1) =    0.000000000000000
( 4  4 |  3  1) =    0.000000000000000
( 4  1 |  2  2) =    0.000000000000000
( 4  1 |  3  2) =    0.000000000000000
( 4  1 |  3  3) =    0.000000000000000
( 4  2 |  4  1) =    0.000000000000000
( 4  3 |  4  1) =    0.000000000000000
( 4  4 |  4  1) =    0.000000000000000
( 2  2 |  2  2) =    0.880159093375045
( 3  2 |  2  2) =    0.000000000000000
( 3  2 |  3  2) =    0.047444445118384
( 3  3 |  2  2) =    0.785270203138277
( 3  3 |  3  2) =    0.000000000000000
( 3  3 |  3  3) =    0.880159093375045
( 4  2 |  2  2) =    0.000000000000000
( 4  2 |  3  2) =    0.000000000000000
( 4  2 |  3  3) =    0.000000000000000
( 4  2 |  4  2) =    0.047444445118384
( 4  3 |  2  2) =    0.000000000000000
( 4  3 |  3  2) =    0.000000000000000
( 4  3 |  3  3) =    0.000000000000000
( 4  3 |  4  2) =    0.000000000000000
( 4  3 |  4  3) =    0.047444445118384
( 4  4 |  2  2) =    0.785270203138277
( 4  4 |  3  2) =    0.000000000000000
( 4  4 |  3  3) =    0.785270203138277
( 4  4 |  4  2) =    0.000000000000000
( 4  4 |  4  3) =    0.000000000000000
( 4  4 |  4  4) =    0.880159093375045
( 5  0 |  0  0) =    0.159028953813045
( 5  1 |  0  0) =    0.381788700431517
( 5  0 |  1  0) =    0.029107255214247
( 5  1 |  1  0) =    0.088639046604491
( 5  0 |  1  1) =    0.054208207239172
( 5  1 |  1  1) =    0.311911628346485
( 5  2 |  0  0) =    0.000000000000000
( 5  3 |  0  0) =    0.208975170450133
( 5  4 |  0  0) =    0.175289465612119
( 5  0 |  2  0) =    0.000000000000000
( 5  0 |  3  0) =    0.001028596856434
( 5  0 |  4  0) =    0.000862792421254
( 5  2 |  1  0) =    0.000000000000000
( 5  3 |  1  0) =    0.049151047350638
( 5  4 |  1  0) =    0.041228155506758
( 5  1 |  2  0) =    0.000000000000000
( 5  1 |  3  0) =    0.005091750233326
( 5  1 |  4  0) =    0.004270986718219
( 5  0 |  2  1) =    0.000000000000000
( 5  0 |  3  1) =    0.001923027401919
( 5  0 |  4  1) =    0.001613045439387
( 5  2 |  1  1) =    0.000000000000000
( 5  3 |  1  1) =    0.184701361441018
( 5  4 |  1  1) =    0.154928467698285
( 5  1 |  2  1) =    0.000000000000000
( 5  1 |  3  1) =    0.038683534029509
( 5  1 |  4  1) =    0.032447950603007
( 5  2 |  2  0) =    0.009336058122476
( 5  2 |  3  0) =    0.000000000000000
( 5  2 |  4  0) =    0.000000000000000
( 5  3 |  2  0) =    0.000000000000000
( 5  3 |  3  0) =    0.011813922734513
( 5  3 |  4  0) =    0.002078445792233
( 5  4 |  2  0) =    0.000000000000000
( 5  4 |  3  0) =    0.002078445792233
( 5  4 |  4  0) =    0.011079469320272
( 5  0 |  2  2) =    0.054104219459936
( 5  0 |  3  2) =    0.000000000000000
( 5  0 |  3  3) =    0.054291571211162
( 5  0 |  4  2) =    0.000000000000000
( 5  0 |  4  3) =    0.000157151628507
( 5  0 |  4  4) =    0.054236039067604
( 5  2 |  2  1) =    0.054125156914812
( 5  2 |  3  1) =    0.000000000000000
( 5  2 |  4  1) =    0.000000000000000
( 5  3 |  2  1) =    0.000000000000000
( 5  3 |  3  1) =    0.075855681518665
( 5  3 |  4  1) =    0.018227677656995
( 5  4 |  2  1) =    0.000000000000000
( 5  4 |  3  1) =    0.018227677656995
( 5  4 |  4  1) =    0.069414628237947
( 5  1 |  2  2) =    0.306655599347507
( 5  1 |  3  2) =    0.000000000000000
( 5  1 |  3  3) =    0.315738293535417
( 5  1 |  4  2) =    0.000000000000000
( 5  1 |  4  3) =    0.007618611374202
( 5  1 |  4  4) =    0.313046130402530
( 5  2 |  2  2) =    0.000000000000000
( 5  2 |  3  2) =    0.008064485483452
( 5  2 |  3  3) =    0.000000000000000
( 5  2 |  4  2) =    0.006764532589140
( 5  2 |  4  3) =    0.000000000000000
( 5  2 |  4  4) =    0.000000000000000
( 5  3 |  2  2) =    0.176295759644369
( 5  3 |  3  2) =    0.000000000000000
( 5  3 |  3  3) =    0.197638924560966
( 5  3 |  4  2) =    0.000000000000000
( 5  3 |  4  3) =    0.011138225736852
( 5  3 |  4  4) =    0.179964436324773
( 5  4 |  2  2) =    0.147877804962089
( 5  4 |  3  2) =    0.000000000000000
( 5  4 |  3  3) =    0.152251498109802
( 5  4 |  4  2) =    0.000000000000000
( 5  4 |  4  3) =    0.011733162163856
( 5  4 |  4  4) =    0.164484175321776
( 5  5 |  0  0) =    0.517125360977304
( 5  0 |  5  0) =    0.006270522616106
( 5  5 |  1  0) =    0.122013556886528
( 5  1 |  5  0) =    0.019208705902818
( 5  5 |  1  1) =    0.491873027324534
( 5  1 |  5  1) =    0.146642827474200
( 5  5 |  2  0) =    0.000000000000000
( 5  5 |  3  0) =    0.009188483135799
( 5  5 |  4  0) =    0.007707347696814
( 5  2 |  5  0) =    0.000000000000000
( 5  3 |  5  0) =    0.011128525245716
( 5  4 |  5  0) =    0.009334665162233
( 5  5 |  2  1) =    0.000000000000000
( 5  5 |  3  1) =    0.095810960370799
( 5  5 |  4  1) =    0.080366734512074
( 5  2 |  5  1) =    0.000000000000000
( 5  3 |  5  1) =    0.105474819021288
( 5  4 |  5  1) =    0.088472829676139
( 5  5 |  2  2) =    0.471913115835107
( 5  5 |  3  2) =    0.000000000000000
( 5  5 |  3  3) =    0.506704669305872
( 5  5 |  4  2) =    0.000000000000000
( 5  5 |  4  3) =    0.029183336960891
( 5  5 |  4  4) =    0.496392251464638
( 5  2 |  5  2) =    0.021048752318672
( 5  3 |  5  2) =    0.000000000000000
( 5  3 |  5  3) =    0.096048971720199
( 5  4 |  5  2) =    0.000000000000000
( 5  4 |  5  3) =    0.062910576176908
( 5  4 |  5  4) =    0.073818472547383
( 5  5 |  5  0) =    0.026686622886860
( 5  5 |  5  1) =    0.289333860384054
( 5  5 |  5  2) =    0.000000000000000
( 5  5 |  5  3) =    0.237637271677700
( 5  5 |  5  4) =    0.199331385983227
( 5  5 |  5  5) =    0.774605943919898
( 6  0 |  0  0) =    0.159028953813045
( 6  1 |  0  0) =    0.381788700431517
( 6  0 |  1  0) =    0.029107255214247
( 6  1 |  1  0) =    0.088639046604491
( 6  0 |  1  1) =    0.054208207239172
( 6  1 |  1  1) =    0.311911628346485
( 6  2 |  0  0) =    0.000000000000000
( 6  3 |  0  0) =   -0.208975170450133
( 6  4 |  0  0) =    0.175289465612119
( 6  0 |  2  0) =    0.000000000000000
( 6  0 |  3  0) =   -0.001028596856434
( 6  0 |  4  0) =    0.000862792421254
( 6  2 |  1  0) =    0.000000000000000
( 6  3 |  1  0) =   -0.049151047350638
( 6  4 |  1  0) =    0.041228155506758
( 6  1 |  2  0) =    0.000000000000000
( 6  1 |  3  0) =   -0.005091750233326
( 6  1 |  4  0) =    0.004270986718219
( 6  0 |  2  1) =    0.000000000000000
( 6  0 |  3  1) =   -0.001923027401919
( 6  0 |  4  1) =    0.001613045439387
( 6  2 |  1  1) =    0.000000000000000
( 6  3 |  1  1) =   -0.184701361441018
( 6  4 |  1  1) =    0.154928467698285
( 6  1 |  2  1) =    0.000000000000000
( 6  1 |  3  1) =   -0.038683534029509
( 6  1 |  4  1) =    0.032447950603007
( 6  2 |  2  0) =    0.009336058122476
( 6  2 |  3  0) =    0.000000000000000
( 6  2 |  4  0) =    0.000000000000000
( 6  3 |  2  0) =    0.000000000000000
( 6  3 |  3  0) =    0.011813922734513
( 6  3 |  4  0) =   -0.002078445792233
( 6  4 |  2  0) =    0.000000000000000
( 6  4 |  3  0) =   -0.002078445792233
( 6  4 |  4  0) =    0.011079469320272
( 6  0 |  2  2) =    0.054104219459936
( 6  0 |  3  2) =    0.000000000000000
( 6  0 |  3  3) =    0.054291571211162
( 6  0 |  4  2) =    0.000000000000000
( 6  0 |  4  3) =   -0.000157151628507
( 6  0 |  4  4) =    0.054236039067604
( 6  2 |  2  1) =    0.054125156914812
( 6  2 |  3  1) =    0.000000000000000
( 6  2 |  4  1) =    0.000000000000000
( 6  3 |  2  1) =    0.000000000000000
( 6  3 |  3  1) =    0.075855681518665
( 6  3 |  4  1) =   -0.018227677656995
( 6  4 |  2  1) =    0.000000000000000
( 6  4 |  3  1) =   -0.018227677656995
( 6  4 |  4  1) =    0.069414628237947
( 6  1 |  2  2) =    0.306655599347507
( 6  1 |  3  2) =    0.000000000000000
( 6  1 |  3  3) =    0.315738293535417
( 6  1 |  4  2) =    0.000000000000000
( 6  1 |  4  3) =   -0.007618611374202
( 6  1 |  4  4) =    0.313046130402530
( 6  2 |  2  2) =    0.000000000000000
( 6  2 |  3  2) =   -0.008064485483452
( 6  2 |  3  3) =    0.000000000000000
( 6  2 |  4  2) =    0.006764532589140
( 6  2 |  4  3) =    0.000000000000000
( 6  2 |  4  4) =    0.000000000000000
( 6  3 |  2  2) =   -0.176295759644369
( 6  3 |  3  2) =    0.000000000000000
( 6  3 |  3  3) =   -0.197638924560966
( 6  3 |  4  2) =    0.000000000000000
( 6  3 |  4  3) =    0.011138225736852
( 6  3 |  4  4) =   -0.179964436324773
( 6  4 |  2  2) =    0.147877804962089
( 6  4 |  3  2) =    0.000000000000000
( 6  4 |  3  3) =    0.152251498109802
( 6  4 |  4  2) =    0.000000000000000
( 6  4 |  4  3) =   -0.011733162163856
( 6  4 |  4  4) =    0.164484175321776
( 6  5 |  0  0) =    0.154983826095369
( 6  0 |  5  0) =    0.006178562861131
( 6  5 |  1  0) =    0.036324481156051
( 6  1 |  5  0) =    0.018637075964466
( 6  0 |  5  1) =    0.018637075964466
( 6  5 |  1  1) =    0.138899129503383
( 6  1 |  5  1) =    0.120424410513661
( 6  5 |  2  0) =    0.000000000000000
( 6  5 |  3  0) =    0.000000000000000
( 6  5 |  4  0) =    0.002735070642527
( 6  2 |  5  0) =    0.000000000000000
( 6  3 |  5  0) =   -0.009851358272803
( 6  4 |  5  0) =    0.009063088333113
( 6  0 |  5  2) =    0.000000000000000
( 6  0 |  5  3) =    0.009851358272803
( 6  0 |  5  4) =    0.009063088333113
( 6  5 |  2  1) =    0.000000000000000
( 6  5 |  3  1) =    0.000000000000000
( 6  5 |  4  1) =    0.024748716021254
( 6  2 |  5  1) =    0.000000000000000
( 6  3 |  5  1) =   -0.059473510026833
( 6  4 |  5  1) =    0.070735331310826
( 6  1 |  5  2) =    0.000000000000000
( 6  1 |  5  3) =    0.059473510026833
( 6  1 |  5  4) =    0.070735331310826
( 6  5 |  2  2) =    0.135540657848790
( 6  5 |  3  2) =    0.000000000000000
( 6  5 |  3  3) =    0.138097677983239
( 6  5 |  4  2) =    0.000000000000000
( 6  5 |  4  3) =    0.000000000000000
( 6  5 |  4  4) =    0.142908608378693
( 6  2 |  5  2) =    0.016764912164332
( 6  2 |  5  3) =    0.000000000000000
( 6  2 |  5  4) =    0.000000000000000
( 6  3 |  5  2) =    0.000000000000000
( 6  3 |  5  3) =   -0.013036508898613
( 6  3 |  5  4) =   -0.034131389240033
( 6  4 |  5  2) =    0.000000000000000
( 6  4 |  5  3) =    0.034131389240033
( 6  4 |  5  4) =    0.057682230376614
( 6  5 |  5  0) =    0.007800931718220
( 6  0 |  5  5) =    0.025610632874342
( 6  5 |  5  1) =    0.065308066611221
( 6  1 |  5  5) =    0.196819899854417
( 6  5 |  5  2) =    0.000000000000000
( 6  5 |  5  3) =    0.040452941573134
( 6  5 |  5  4) =    0.044287849234219
( 6  2 |  5  5) =    0.000000000000000
( 6  3 |  5  5) =   -0.085489074048849
( 6  4 |  5  5) =    0.126882792323867
( 6  5 |  5  5) =    0.133977082851046
( 6  6 |  0  0) =    0.517125360977304
( 6  0 |  6  0) =    0.006270522616106
( 6  6 |  1  0) =    0.122013556886528
( 6  1 |  6  0) =    0.019208705902818
( 6  6 |  1  1) =    0.491873027324534
( 6  1 |  6  1) =    0.146642827474200
( 6  6 |  2  0) =    0.000000000000000
( 6  6 |  3  0) =   -0.009188483135799
( 6  6 |  4  0) =    0.007707347696814
( 6  2 |  6  0) =    0.000000000000000
( 6  3 |  6  0) =   -0.011128525245716
( 6  4 |  6  0) =    0.009334665162233
( 6  6 |  2  1) =    0.000000000000000
( 6  6 |  3  1) =   -0.095810960370799
( 6  6 |  4  1) =    0.080366734512074
( 6  2 |  6  1) =    0.000000000000000
( 6  3 |  6  1) =   -0.105474819021288
( 6  4 |  6  1) =    0.088472829676139
( 6  6 |  2  2) =    0.471913115835107
( 6  6 |  3  2) =    0.000000000000000
( 6  6 |  3  3) =    0.506704669305872
( 6  6 |  4  2) =    0.000000000000000
( 6  6 |  4  3) =   -0.029183336960891
( 6  6 |  4  4) =    0.496392251464638
( 6  2 |  6  2) =    0.021048752318672
( 6  3 |  6  2) =    0.000000000000000
( 6  3 |  6  3) =    0.096048971720199
( 6  4 |  6  2) =    0.000000000000000
( 6  4 |  6  3) =   -0.062910576176908
( 6  4 |  6  4) =    0.073818472547383
( 6  6 |  5  0) =    0.025610632874342
( 6  5 |  6  0) =    0.007800931718220
( 6  6 |  5  1) =    0.196819899854417
( 6  5 |  6  1) =    0.065308066611221
( 6  6 |  5  2) =    0.000000000000000
( 6  6 |  5  3) =    0.085489074048849
( 6  6 |  5  4) =    0.126882792323867
( 6  5 |  6  2) =    0.000000000000000
( 6  5 |  6  3) =   -0.040452941573134
( 6  5 |  6  4) =    0.044287849234219
( 6  6 |  5  5) =    0.342541413419711
( 6  5 |  6  5) =    0.035634211559360
( 6  6 |  6  0) =    0.026686622886860
( 6  6 |  6  1) =    0.289333860384054
( 6  6 |  6  2) =    0.000000000000000
( 6  6 |  6  3) =   -0.237637271677700
( 6  6 |  6  4) =    0.199331385983227
( 6  6 |  6  5) =    0.133977082851046
( 6  6 |  6  6) =    0.774605943919898

There are 406 unique integrals


Ammonia decomposition on platinum surface I was given a problem I have to solve, unfortunately I cant wrap my head about it. I've searched the page for answers but they doesnt seem to fulfill my needs, so i'm asking for little guidance.

The problem:

  When the decomposition of ammonia is carried out on platinum surface at 1000°C the
  hydrogen binds strongly on the surface. Prove the that the rate is
  equal to $\frac{dp(NH_3)}{dt}=-k\frac{p(NH_3)}{p(H_2)}$.

I dont understand why we have to divide with the partial pressure of hydrogen. In my understanding the rate is $\frac{dp(NH_3)}{dt}=-kp(NH_3)$. I know I have to use the fact that hydrogen binds strongly at the surface, but I'm lost. Can you please explain to me how should I derive this, or just hint me to the right direction.

I lack of good understanding on this topic, so sorry for this novice question.

What I've searched:

Why is decomposition of ammonia gas on quartz a 1st order reaction?

What is the molecularity of the RDS of a zero order complex reaction?

Thank you for your time.

Thanks to Poutnik's comment, I arrived to the answer and I would like to share it.

$$\ce{2NH_3 ->N_2 + 3H_2}$$
$$\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
$$\theta_{N_2}=\frac{K_{N_2}p(N_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
$$\theta_{H_2}=\frac{K_{H_2}p(H_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$

Since the hydrogen binds strongly on the surface, we can safely determine that $K_{H_2}>>1$, wich means the denominator $1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)\approx K_{H_2}p(H_2)$. So the surface coverage of ammonia can be written as $\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{K_{H_2}p(H_2)}$. By definition $v=-\frac{dp(NH_3)}{dt}=k_0\theta_{NH_3}=k_0\frac{K_{NH_3}p(NH_3)}{K_{H_2}p(H_2)}$, defining $k=k_0\frac{K_{NH_3}}{K_{H_2}}$ we arrive at the desired equation.

Feel free to edit.


Electrophilic Aromatic Substitution: Are protonated OH-groups ortho-/para- or meta-directing? In Electrophilic Aromatic Substitution ($\text{S}_{\text{E}}\text{Ar}$) reactions groups that can donate an electron pair into the aromatic ring, e.g. $\ce{-OH}$ or $\ce{-NH2}$, are ortho-/para-directing.

In the case of $\ce{-NH2}$-groups one has to be cautious because if the $\text{pH}$ value of the solution is too low the $\ce{-NH2}$-group will be present mostly in its protonated form, $\ce{-NH3^+}$, which lacks a free electron pair and has a meta-directing effect because of its strong $-I$ effect. This brings up the question: What is the situation like with protonated $\ce{-OH}$ groups, $\ce{-OH2^{+}}$? In contrast to $\ce{-NH3^+}$ this group still has a free electron pair which could, in principle, interact with the aromatic ring ($\Rightarrow$ ortho-/para-directing) but it should also have a very powerful $-I$ effect ($\Rightarrow$ meta-directing).

I'd say, if the electron pair can be effeciently delocalised into the aromatic ring then its ortho-/para-directing power should outweigh the meta-directing power of the inductive effect. But I wonder if there even is any appreciable degree of delocalisation present because the mesomeric structures don't look very favourable:

Thinking about it from a frontier orbital perspective I would say: The orbital of the $\ce{-OH2^{+}}$ group's free electron pair will be much lower in energy than that of the $\ce{-OH}$ group's free electron pair. Thus, the energy difference between free electron pair's orbital and the aromatic ring's $\pi$-system might be too large to facilitate a good interaction. Therefore, I'd expect a protonated $\ce{-OH}$ group to be meta-directing.

But what is the actual situation when phenols or phenol ethers undergo ($\text{S}_{\text{E}}\text{Ar}$) reactions in strongly acidic solutions? Does one get the ortho/para or the meta product?

Edit:

I looked up the $\mathrm{pK}_{\mathrm{a}}$ value for protonated phenol: It is approximately $-7$. This makes me wonder if there are any reasonable (real world) situations where the solution is acidic enough to protonate phenol (or phenol ethers) completely (even $\ce{HI}$ with its $\mathrm{pK}_{\mathrm{a}}$ of $-10$ shouldn't be enough for that, right). I'm not sure if even sulfonation or nitration reach such low $\text{pH}$ values.
But @Uncle Al's answer made me aware that I forgot about Lewis acids and I'd say that Lewis acids like $\ce{AlCl3}$ that coordinate strongly to oxygen might lead to a state that will be very similar to a $\ce{-OH2^{+}}$ group (please, correct me if I'm wrong and the coordination of Lewis acids is not strong enough to be compared to the $\ce{O-H}$-bond). If this is the case, then what about Friedel-Crafts reactions which make heavy use of Lewis acids: Would those Lewis acids interfere with the ortho-/para-directing properties of phenols? Does anyone have experience with the directional properties of a $\ce{-OH}$ group in the presence of a strong Lewis acid, maybe even in the context of Friedel-Crafts reactions?

O-protonated phenols direct in meta, according to an article on that topic. As it occurs, phenols can be, depending on acid, protonated on oxygen, but also on carbon, and even diprotonated (but not twice on the same atom), when most powerful mixtures of HF and SbF5 are used. O-protonated forms can be brominated, hydroxylated and alkilated. C-protonated forms don't posses such reactivity and are essentially Meisenheimer's complexes.


Is an ester or a ketone more acidic? Two organic chemistry textbooks I've read (by Bruice and by Tandon) say:

  Esters are more acidic than ketones, because the resonance between the two oxygen atoms gives less opportunity for the delocalization of the electron pair on the alpha carbon in esters contrary to what happens in ketones.

But, in esters, when the resonance happens between the two oxygen atoms, the oxygen atom gains a positive charge which is highly unstable, given that oxygen is highly electronegative. So, shouldn’t the order of acidity be the other way round?

So, shouldn’t the order be the other way round?

Yes, it is the other way around.  Ketones (pKa ~ 20) are more acidic than esters (pKa ~ 25).  The figure below shows the resonance structures for the enolate anion of a ketone and an ester.  Resonance structures II and IV stabilize the enolate anion in the ketone and ester respectively.  In the ester, there is an additional resonance structure, V. Through resonance structure V, the alkoxy group in the ester donates electrons to the carbonyl.  This movement of electron density towards the carbonyl will tend to destabilize the formation of any negative charge on the carbon attached on the other side of the carbonyl (electrostatic repulsion).  This reduces the likelihood of the formation of an enolate anion on that carbon. Therefore, in the ester, to whatever extent structure V contributes, structure IV will contribute somewhat less.  Hence, the ester enolate anion resonance structure IV plays a smaller role and the enolate anion is less stabilized in the ester compared to the ketone.  Consequently, the ester is less acidic than the ketone.


Latent Heat and Kinetic Energy The kinetic theory says that temperature is the measure of the average kinetic energy. That would mean then: The temperature rises with the rise in kinetic energy. We know that when matter changes state 'latent heat' helps in 'breaking the bonds' at the transition phase and that this 'extra heat' is not noticed on a thermometer. But energy can neither be made nor destroyed. Taking that into consideration latent heat has to convert into some form of energy. Since temperature is not rising, it's not kinetic energy it's getting converted to. What energy does it get converted to?

Note: I already know why the temperature does not change.
What I want to know is what the latent heat energy gets converted to.

Latent heat isn't heat, it is energy released or absorbed during a constant temperature phase change. The kinetic theory of gasses has little to say about phase changes. (Which by definition involve liquids or solids and may or may not involve gasses, most often "not".) Claiming a phase change releases latent heat is false, an error of comprehension (or articulation). If I slowly accelerate a tank of gas, its kinetic energy increases with no increase in temperature. Temperature is a measure of kinetic energy, not the only measure of kinetic energy. (Which is obvious since you know that temperature doesn't increase as most solids at their melting point do not increase in temperature as heat energy is added to them.) In the broadest (most general) sense, latent energy is the energy difference between the two phases. So where it goes (or where it comes from) is in converting one phase to the other. The assumption here is one phase is lower in potential energy than the other. You may know that entropy is defined in terms of heat and temperature. In classical thermodynamics energy and entropy are two fundamental concepts, they are on equal footing. In statistical thermodynamics, energy is still fundamental, while entropy can be derived from the ensemble of possible states that a system can be in. Whoops! Too much! In the case of ice melting, the intermolecular bonds (mostly hydrogen bonding) are 'broken' (which means there are a lot fewer of them and they are (on average) weaker). The energy that was potential energy of the bonds becomes "kinetic" energy as chemists understand the term in this context. In this context (as contrasted with the kinetic energy of a bullet or a ballistic electron), the kinetic energy includes all of the energy of translation, rotation and vibration: the rotation of a molecule doesn't change its speed, nor do the movements of the two hydrogen atoms with respect to the oxygen atom - their bond to the central O atom can be looked at as a (coil) spring and they can stretch and bend that spring. Faster and greater spring movement occurs during melting. (The rotation of the H atoms also occurs but the energy required to spin a H atom about its bond is very small as bond energies go, so is generally neglected with little loss of accuracy; this wouldn't be true for bonds between larger groups, where rotation around the bond could require quite a bit of energy). So, in the isothermal melting of ice, the place where the energy is going is in these intramolecular movements. (intra = inside). And in the isothermal freezing of water, the place where the energy is coming from is these same intramolecular motions. There is no one unique definition of kinetic energy and another for potential energy. In a group of atoms or molecules, the 'potential' energy is 'internal' and kinetic is 'external' and yet if one focuses in on a single molecule, some of its potential energy is due to its movement (or movements of its parts), so what is and what is not k.e. or p.e. may change with the context. In fact, the basic laws of energy conservation may be K.E. + P.E = constant or K.E. = P.E. (which is often expressed as K.E. - P.E. = 0. These two approaches are embodied in Hamiltonian and LaGrangian Mechanics which are the two related but complimentary ways modern physics is expressed. The LaGrangian is L = T-V (where T is k.e. and V is p.e.) and the Hamiltonian is H = T + V. These are used in both classical Physics and in quantum Physics.


Volcano plots: How is the data fitted? When studying about catalysts I normally come accross volcano plots like the one shown below:

What confuses me is how are the two lines fitted, how do you determine what points to include and what points not to include. For example I could have included the two points for Pt and Ir to obtain a slightly different shape of the volcano plot. Is there a guidline for this?

The most obvious approach for me would be to find a partition of the set of points into two subsets and two straight lines, one for each subset, such that the sum of the squared residuals over both partitions is minimized.


What are SN1' and SN2' reactions? I went through the problems on the pictures:

The answer of the problem 119 is $\mathrm{S_N1'}$, and that of 120 is $\mathrm{S_N2'}$. I actually can't figure out what $\mathrm{S_N1'}$ ($\mathrm{S_N1}$ prime) and $\mathrm{S_N2'}$ ($\mathrm{S_N2}$ prime) are.

Also, how does the answer change by changing the nucleophile?

The primed version of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ can only occur if there is a double bond in the vicinity of the leaving group as in your example. I have drawn both possibilities for the $\mathrm{S_N2}$ case in the scheme below.

Scheme 1: Comparison of the reaction products in an $\mathrm{S_N2}$ (top) and $\mathrm{S_N2'}$ (bottom) pathway.

The $\mathrm{S_N2}$ reaction is as you expect it to be. The $\mathrm{S_N2'}$ reaction uses the double bond as an electron relay system. Instead of the nucleophile (here: $\ce{OMe-}$) directly interacting with the $\unicode{x3c3}^*(\ce{C-Br})$ orbital, the π system interacts with the σ* orbital. The nucleophile then interacts with the π system’s LUMO (corresponding to the middle orbital of the allyl π system) to perform the attack. Since the nucleophilic attack and the leaving group are on different carbon atoms, relayed by the π system, this is not a direct $\mathrm{S_N2}$ but a derivative of it ($\mathrm{S_N2'}$).

The same logic can be applied to $\mathrm{S_N1'}$: here, the intermediate carbocation is not a localised one but an allyl cation.

The primed versions here are observed because not only the bimolecular attack on the highly substituted tertiary carbon ($\mathrm{S_N2}$) is unlikely — this carbon is tertiary which usually already implies no $\mathrm{S_N2}$ — but also the capturing of the carbocation under $\mathrm{S_N1}$ conditions is more likely to happen at the sterically much less hindered carbon atom.

As to why methanol predominantly attacks according to an $\mathrm{S_N1}$-type mechanism while methanolate predominantly follows an $\mathrm{S_N2}$-type mechanism I will refer you back to your textbook; the reasons are stated pretty often.


On the scale of stuff with a pH I am a confused (and somewhat anal retentive) non-chemist who is trying to understand what pH is.  I have come to understand it is a measure of acidity, but I have not yet figured out what that means to a precision that I find satisfying.

Please excuse me if these are dumb questions, but they will help me understand the definition:

  Does everything have a pH?  What is the scale that pH works upon?  Does a proton have a pH?  A quark?  Does the sun have a pH?  Does the Universe?

Did you not carry out some research?

Well, you can find information about it in Wikipedia, or even dictionaries like Merriam-Webster or Dictionary Online.

You will see that pH is calculated through:

$$pH = -\log(a_{\ce{H+}})$$

Or which can most of the time be stated as:

$$pH = -\log(\ce{[H+]})$$

$a_{\ce{H+}}$ is the hydrogen ion activity, the 'effective concentration' of hydrogen ions. Without active hydrogen ions, you don't have any pH.

You might find that the pH scale is from 0 to 14, but you can also have circumstances where you have negative pH and pH above 14. It depends on the extent of dissociation of an acid or base to produce active hydrogen ions, or hydroxide ions in the case of bases.

Since I mentioned hydroxide ions, it might also be worth nothing that:

$$14 = pH + pOH$$

And that basically, a pH of 1 is equivalent to a pOH of 13.


Why are there peaks in electronegativities in d-block elements? Looking at the Pauling electronegativities in the Periodic Table (below, from ChemWiki):

Asides from the overall trend of increasing electronegativity across and up the Periodic Table (towards fluorine), there seems to be a minor peak in the first 3 periods of the $d$-block elements, this peak becomes more pronounced as you go down the $d$-block, with the most noticable peaks in Periods 5 and 6 - seemingly the reverse of the overall trend. This is shown in the graph below (Image source):

In the ChemWiki page, there is also a reference to the Allred-Rochow Electronegativity Scale, which also has peaks occurring most prominently in Periods 4 and 6:

Image source

The Wikipedia page Electronegativity details several methods of calculation, which brings into question how well-defined the electronegativity values are; however, there still seems to be the peaks occurring, particularly in Periods 5 and 6, and to a lesser extent in Period 4.

What is the reason for these electrongativity peaks in the $d$-block?

You should know that closeness to filling a shell and atomic size are the reasons for the overall electronegativity trend. This also applies to the subshells, how close is an element to filling, or half-filling, a subshell? How large is the atom?

The atomic size trend for the d subshell is opposite to that of the overall trend. In period 4 the 3d subshell is deep in the core orbitals and the electron density is high, whereas in the lower periods the d orbital is more diffuse and can more easily accommodate more electrons, leading to a better energy payoff for completing the subshell. For the elements that take s electrons to fill the d subshell, this energy payoff is even greater from then filling the s subshell as they are very diffuse.


Why do some molecules show non-symmetric absorption and emission bands? I'm looking at a molecule of [Ru(bpy)3]2+ in water and the absorption and emission bands are not symmetrical. I want to know why the mirror image rule doesn't apply here. Is this because of the MLCT, is this something to do with the solvent? Are there more likely to be processes involving non-radiative decay?

Mirror image spectra are only observed in solution and then only if the ground and excited state potential energies have almost exactly the same shape. Thus in rigid molecules such as anthracene a good mirror image is observed but not in more flexible molecules such as stilbene ($\phi-C=C-\phi$). The spectra of each vibrational level is wide in solution due to solvent interactions altering energy levels v. slightly, this does not happen in the collisionless gas phase where lines are sharp. In solution the width of lines from different vibrations can overlap making a broad largely structureless spectrum.

There are different effects, a stokes shift due to solvent causing lowering of excited state energy shortly after absorption, a displacement in position of minimum energy, due to different bond lengths in each state and anharmonicity in one potential vs the other. Even slightly different Franck Condon factors between states can mean that a phosphorescence spectrum is similar, but different to its fluorescence spectrum, see figure below, when correction is made for detector sensitivity vs wavelength.

The figure below (not to scale) shows how ideal mirror image spectra are formed. Notice that displacement of potential energy is needed and that vibrational frequency in ground and excited state can be slightly different.


How much sulphuric acid can be obtained from 5 kg of sulfur How much sulfuric acid $(\ce{H2SO4})$ can be obtained from 5 kg of sulfur $\ce{S8}$?
answer: 15.312 kg

To  make 1 molecules of $\ce{H2SO4}$ we need 1 atoms of S.
Now coming to $\ce{S8}$. This gives us 8 atoms of S. so from this we can obtain 8 $\ce{H2SO4}$ molecules if we take 1 atom of $\ce{S8}$ and use it to make $\ce{H2SO4}$ molecules. So,

1 molecule of $\ce{S_8}$ $\ce{->}$ 8 molecules of $\ce{H2SO4}$
0.019 moles of $\ce{S_8}$   $\ce{->}$  $8 \times 0.019=0.15$ moles of $\ce{H2SO4}$

Formula mass of $\ce{H2SO4} = 1(2)+32+16(4) = 98$
we know the formula
for $$\text{Number of moles} = \frac{\text{Mass}}{\text{Formula Mass}}$$
So from this $\text{Mass}_\ce{H2SO4} = \text{Moles} \times \text{Formula Mass} = 0.15 \times 98=14.7$
Now you see that my answer doesn't match to the answer. What did I do wrong?

$\ce{S6}$, $\ce{S8}$, $\ce{S12}$ – does it make a difference or is it just a trick to make the question more complicated than it is?

What you know for sure is:

$M(\ce{S}) = 32.065\ \mathrm{g}\cdot \mathrm{mol}^{-1}$
$M(\ce{H2SO4}) = 98.079\ \mathrm{g}\cdot \mathrm{mol}^{-1}$

$\frac{5000\,\mathrm{g}}{32.065\,\mathrm{g}\cdot \mathrm{mol}^{-1}} = 155.93\,\mathrm{mol}$

$155.93\,\mathrm{mol} \cdot 98.097\,\mathrm{g}\cdot \mathrm{mol}^{-1} = 15.29\,\mathrm{kg}$, which is close enough to the answer.


What is it called when a benzene ring loses a hydrogen atom? Does not get substituted by another group it just gets removed  (so not aryl) what is it? Like when two benzene rings lose one H atom each we get arylene? Whats another name for it? I just need the name for when the atom is removed and NOT subtituted
Thank you

Edit: thats not the answer I asked for
I asked what it was called when it lost the hydrogen atom and you say its aryl but thats when a group is in place of the lost atom

The name of the benzene minus a hydrogen is called a phenyl group (C6H5). https://en.wikipedia.org/wiki/Phenyl_group Look this page up for more information. The question you've asked is very similar to the one that @Gaurang has commented, so just be careful when you ask questions on here. Hope this helps


Lewis Structure of SO2 What is the structure of SO2? I have seen two different ways the Lewis Structure is written:

The formal charges of the SO2 with the single bond and a double bond is larger than the SO2 with two double bonds. So I would assume that the one with two double bonds is the correct structure. But chemistry books I have looked at (Zumdahl Edition 5 and 7) says that it is the opposite.

Which is the correct Lewis Structure?

The Lewis structure most closely resembling reality consists of two resonance structures: the first one posted in the question and its mirror image. The reason is that the octet rule is observed this way (no hybridization of d-orbitals for main group chemistry necessary or possible) as well as the symmetry of the molecule: the two bonds are identical.


Feasibility of using ionic liquid fuels in an MHD type reactor First off I would like to say that my knowledge on chemistry, even general chemistry, is basic if not poor.

That said, I have some intuituions on this matter. I haven´t found any paper regarding this kind of system, and that is pretty much a killer for me.
Even with that I´d still like to ask the question.

Ionic liquids, liquid salts, are regarded as having some conductive capacities, being low to moderate conductors due to their ionic nature.
I know that salts have been used in MHD (magnetohydrodynamic) systems due to their high melting points -the ionic bond being particularly strong- and their conductive potential.

I have an specific fuel reaction in mind, An hydrazinium based propellant + LOX (liquid oxygen) as oxidizer,

[HO-C2H4-N2H4]+[NO3]- + 2.25 O2 = 2 CO2 + 1.5 N2 + 4.5 H2O

for example.
I read that ionic liquid salts do not need to be dissolved for the anion and the cation to conduct electricity (https://www.researchgate.net/post/How_is_the_electrical_conductivity_of_ionic_liquids_in_comparison_with_metals)
Outside of that I´d like to know if hydrainium based reactions do have some electric potential, or if by any reason wich I did not manage to grasp, it is simply out of the question to use them in such machines.

I would be very pleased to know if something is just worng or has been misunderstood.

Source for the reaction: http://cdn.intechweb.org/pdfs/20222.pdf, page 9 (455 in the original text).

Overal description of how a MHD machine works: https://en.wikipedia.org/wiki/Magnetohydrodynamic_generator

*Edit:
The majority of the volume is occupied by the reaction products. Using just conventional fuels like liquid hydrogen is just more practical.

Still, we could use a thermal/catalytical decomposition reaction, like 6 NH3OHNO3(aq) = 2 N2(g) + 2 NH4NO3(s or l) + 6 H2O(g) + 4 HNO3(g), where both a salt and water are produced.

If fuel is oxidized at high temperature, the conductivity of the fuel itself is irrelevant -- only the conductivity of the combustion products, which are ionized because of the high temperature and are therefor conductive. Brittanica states that temperatures greater than ~2,500 K are needed for sufficient ionization. Alkali metals can be used to increase ionization, or higher temperatures have been used with inert gases.

Note that the temperature of ions and electrons in the gas are not necessarily the same, due to the difference in mass and time to reach equilibrium. This impacts on conductivity and the rate of erosion of the MHD components.


Understanding the setup of a chemistry lab about redox In chemistry lab we were looking at oxidation and reduction reactions. To determine the relative activity of several metals we observed reactions between the metals and aqueous solutions of dissolved metals. We mixed metals like $\ce{Fe}$ with solution like $\ce{Cu(NO3)2}$ to see if a reaction occurred.

I'm wondering why is it not necessary to mix a metal like $\ce{Fe}$ into its own solution $\ce{Fe(NO3)2}$ (like you would do when setting up an electrochemical cell)? We mixed $\ce{Fe}$ with $\ce{Zn(NO3)2}$, $\ce{Cu(NO3)2}$, and $\ce{Ni(NO3)2}$.

I'm wondering why is it not necessary to mix a metal like Fe into its own solution $\ce{Fe(NO3)2}$ (like you would do when setting up an electrochemical cell)?

First remember what the purpose of your experiment was:

  To determine the relative activity of several metals we observed reactions between the metals and aqueous solutions of dissolved metals.

In an electrochemical cell setup, $\ce{Fe}$ would be in two-way equilibrium with $\ce{Fe^2+}$:

$$\ce{Fe + Fe^2+ <=> Fe + Fe^2+}$$

When power was drawn from the cell you then get:

$$\ce{Fe -> 2 e- + Fe^2+}$$

Where as with another salt such as copper which is less reactive, the copper will oxidize the iron in one direction with no reverse reaction.

$$\ce{Fe + Cu^2+ -> Cu + Fe^2+}$$

or with zinc which is more reactive no reaction will occur at all:

$$\require{cancel} \ce{Fe + Zn^2+ \cancel{\ce{<=>}} Zn + Fe^2+ ->Fe + Zn^2+}$$


What are the products of the reaction of 1-methoxycyclohexene with dilute hydrochloric acid? When 1-methoxycyclohexene is treated with dilute aqueous hydrochloric acid, methanol is produced along with another organic product. Identify this product. Explain why methoxycyclohexane is inert under these reaction conditions.

I don't understand why it isn't just a simple addition to the double bond. I get that there's conjugation but is this what makes things different? I don't get why methanol is the leaving group and therefore I cannot predict the other product. Also, it says that methoxycyclohexane (the same molecule without the double bond) is inert under these conditions. Why does it not undergo substitution? The oxygen could surely be protonated and then methanol would be a good leaving group which would give rise to substitution (probably SN1). Why is this not the case?

Vinyl ethers are "masked" carbonyls, therefore in acidic aqueous environment the following pathway takes place: alkene protonation assisted by the methoxy group followed by nucleophilic attack of water with hemiacetal formation. Subsequent intramolecular proton transfer to the methoxy group makes it a good leaving group and cyclohexanone (the product) is formed while methanol is released.


Enones and hydrogenation Are enones resistant to hydrogenation? My solutions manual thinks so.

I don't recall discussion regarding enones in my text. Nonetheless, could the resonance stabilization of the enone be preventing its hydrogenation? In that case, why is the alkene hydrogenated; both the carbonyl and the alkene are resonance-stabilized to some extent!

I read your question as asking why the C-C pi bond is reduced while the C-O pi bond is not. It appears to be possible to change the selectivity of hydrogenation based on the catalyst used (reference), but for simple palladium on carbon, reduction of alkenes (even conjugated to carbonyls) is much faster than an aldehyde or ketone. Coupled with complementary hydride reagents, which are selective for carbonyls, synthetic chemists can achieve the desired reduction without worrying about reaction of other reducible groups.


Formation of oximes by ketones and aldehydes Why do symmetric ketones form a single oxime and asymmetric aldehydes and ketones can form two isomeric oximes?

An oxime is formed by the reaction of hydroxylamine (NH2OH) with either an aldehyde or ketone. The oxime nitrogen has a double bond (sigma and pi) to the oxime carbon, a lone pair and a single (sigma) bond to the hydroxyl oxygen. If we use valence shell electron pair repulsion (VSEPR) theory to deal with the problem.

Then the oxime nitrogen will be trigonal planar.

If the two groups on a ketone are identical then the molecule will be the same which ever way around we put the lone pair and the hydroxyl. But if the two carbon groups are different then we can have two isomers.

Consider for a moment the 2-hydroxyoximes used as LIX extraction agents for copper. There is a paper from the 1970s on the separation of the isomers of LIX 63 (T.T.Tammi, Hydrometallurgy, 1977, Volume 2, Issue 4 (October), Pages 371-380) which you migh want to read.

To make it more clear we have diagrams here of the two isomers of LIX63.


Will octahedral splitting occur in tribromidotricyanidocobaltate(III)? Octahedral splitting occurs strong field ligands are coordinated with the central atom. What if we have both types of ligands attached to the central atom as in $\ce{[Co(CN)3(Br)3]^{3-}}$, knowing that $\ce{CN^-}$ is a strong field ligand while $\ce{Br^-}$ is a weak field ligand.

In reality it would perhaps boil down to experimental reasoning, but if it's an exam asking you this, I'd like you to remember a rule of thumb as you'd call it:

If the complex has a coordination number of $6$, with $3$ strong field ligands and $3$ weak field ligands, then consider all the ligands to be strong field ligands.
If the complex has a coordination number of $4$, with $2$ strong field ligands and $2$ weak field ligands, then consider all the ligands to be weak field ligands.


Comparing the pKb of piperazine and methylpiperazine In comparison to piperazine, the inductive effect will cause the $\mathrm{p}K_\mathrm{b}$ of methylpiperazine to be:

The answer given is that the $\mathrm{p}K_\mathrm{b}$ will be lower due to electron donation by the methyl substituent. I understand that the methyl group stabilizes through the inductive effect by releasing electron density. However, wouldn't this make the $\mathrm{p}K_\mathrm{b}$ higher (a weaker base) since it's more stable and less likely to react?

In terms of the logic employed by your answer key

Whether an inductive effect is stabilizing depends on the nature of the molecule and reaction you're thinking about.

You're right to think of the methyl group as inductively release electron density into the ring. However, this inductive release from the methyl group is actually increasing the electron density in the piperazine ring, which in turn makes the N lone pair higher in energy and more reactive, thus making it a stronger base. However!

What is actually the case in real life

I was able to find a detailed study of $\mathrm{p}K_\mathrm{a}$ values of protonated piperazines and methylpiperazines, and assuming $\mathrm{p}K_\mathrm{b} = 14 - \mathrm{p}K_\mathrm{a}$.

Then at 298 K...

$\mathrm{p}K_\mathrm{b}\text{(piperazine)} = 4.27$
$\mathrm{p}K_\mathrm{b}\text{(N-methylpiperazine)} = 4.86$
$\mathrm{p}K_\mathrm{b}\text{(N,N'-dimethylpiperazine)} = 5.62$
$\mathrm{p}K_\mathrm{b}\text{(2-methylpiperazine)} = 4.43$

This tells us that your answer key is actually wrong, and that in real life sterics may be playing a role in the basicity of the N atom. This data is in line with the fact that secondary bases tend to be slightly stronger than tertiary ones. For example, the

$\mathrm{p}K_\mathrm{b}(\text{diethylamine)} = 3.02$
$\mathrm{p}K_\mathrm{b}(\text{triethylamine)} = 3.25$


What reactions produce lots of Helium? I'm working on a pet project and I'll need some helium. I was wondering if there were any reactions that I could perform at home to produce helium cheaper than it would be to buy helium from someone else.

Anyone know of such reaction off the top of their head?

Helium doesn't form any compounds, being a noble gas. That is, you can't extract it from any reaction, because you don't have anything that it reacted with already. Also, it would've probably taken much more energy to form that helium compound than it would've to actually extracted it from the atmosphere.

Now, you can produce helium, technically. That's what an alpha particle is: A loose helium nucleus bouncing about. But mucking around with radioactive things is dangerous and can be quite expensive.

As a note, you could probably extract it yourself if you have a natural gas well that you can access directly and a way to separate it out with fractional distillation. Almost all the gasses will condense out before helium does, which means you'll end up with very pure helium left behind, with a minor impurity of neon. This process is dangerous, however, as you'll both be working with very cold temperatures and have liquid methane running amok.

You could probably get a tank from it from a chemistry supply store somewhere. (Party supply stores add oxygen to prevent people from suffocating off of their tanks. The impurity is not insignificant.)

TL;DR: No, unless you're willing to irradiate or freeze yourself.


Lewis dot structures of CO2 I know the general Lewis dot structure of carbon dioxide is the one where there are two double bonds connecting oxygens to carbon.

However the question is, does $\ce{CO2}$ have resonance structures?

We could have moved one of the bonds (pi bond in this case) to the other $\ce{O-C}$ bond, leaving us with a single bond and a triple bond forming $\ce{CO2}$.

It is almost always possible to draw resonance structures. In case of $\ce{CO2}$ you could imagine a resonance structure in which the carbon doesn't have an electron octet and a positive charge and one of the oxygen atoms carrying a negative charge.
You could also draw one in which the carbon is only connected via a single bound to each oxygen and carrying two positive charges while the two oxygens carry a negative charge each.

You might come up with more of those structures.

So the question you really wanna ask is whether these resonance structures have any real contribution to the actual bonding state of $\ce{CO2}$ and for that you will see that all of the described strucutures are highly unfavourable due to their energetic state. Therefore the normal Lewis-notation of $\ce{CO2}$ decribes the real bonding state pretty well as all other resonance structures don't contribute much.


How to prove at what pH the color of an indicator will change based on its pKa? The equivalence point of an acid-base titration can be determined by choosing the appropriate colored indicator. This indicator is a weak acid $\ce{HA}$. The undissociated acid has a different color than it's ion $\ce{A-}$. The $\mathrm{p}K_\mathrm{a}$ value of a colored indicator is given to be $\mathrm{p}K_\mathrm{a}=6$. At what value of the $\mathrm{pH}$  will the color change? Please explain your answer. (Assume that both colors are equally well recognizable).

I've had this question on an exam and I can't seem to figure out what exactly the answer could be.

I would intuitively say that the color change happens at $\mathrm{pH}=6$. If the $\mathrm{pH}$ is lower than 6 then there are less $\ce{A-}$ ions and more $\ce{HA}$ (acid-indicator) in the solution. Hence the solution will take on the color of the undissociated acid-indicator. If the pH is higher than 6 then the $\ce{HA}$ starts to dissociate more to $\ce{H3O+}$ hence there being a higher $\ce{A-}$ concentration and the solution takes the color of the indicator's conjugate base. But I'm not sure if my thought process is right or how to 'prove' it's right.

Thank you guys for your answers! I am fairly confident I have found the correct answer, or at least one of the possible correct answers. Here's it goes:

The indicator reacts with water according to the following reaction equation:
$$\ce{HA + H2O <=> H3O+ + A-}$$
Then using the Henderson-Hasselbach equation, I arrive at the following point:
 $$ K_s = \frac{[\ce{H3O+}]*[\ce{A-}]}{[\ce{HA}]} $$
Now, at the equivalence point, the concentration of the indicator an it's conjugate base are equal, hence:
 $$\require{cancel} K_s = \frac{[\ce{H3O+}]*\cancel{[\ce{A-}]}}{\cancel{[\ce{HA}]}} $$
$K_s = [\ce{H3O+}]$. After taking the negative log of both sides I arrive at the equation: $pK_s = pH$ (which was given to be 6).

Now, when the solution is acidic(pH lower than 6 and high [$\ce{H3O+}$] concentration), Le Chatelier sais then the balance will shift to the left of the top most equation. Favoring the indicator $\ce{HA}$ as opposed to it's conjugate base. Naturally, if the solution contains more indicator molecules, then it will also take on the indicators color.

 Analogue to the previous reasoning, when the pH reaches values above those of 6, Le Chatleier sais that the reaction will favor the formation of the indicators conjugate base [$\ce{A-}$](the right side) and the solution takes on the color accordingly.


Why don't electrons have the same spin even when in same orbit? Do electrons not have the same spin because that is just how it was found or does it contribute to another factor?

The Pauli Exclusion Principle says no two electrons can have the same quantum numbers, and there are roughly two ways to interpret this.

First, an 'orbital' may host two electrons so long as they have opposite spins.

Second, an 'orbital' may only host one electron, but there is a second orbital that is geometrically and locationally identical which hosts an electron of the opposite spin of the first orbital.

Why?  It is written into the fabric of the universe.  That is how the math works out.  Experiments have strongly supported Quantum Mechanics.