Mistake in finding bond angles in VSEPR theory Find the bond angles as indicated, in the compounds given below, according to VSEPR theory-

I got: $109.5^\circ$(6),$109.5^\circ$(7),$109.5^\circ$(8) for the compound (d). But the given answer says it is wrong. But I don't know why. Everything else I got them right.

Looking at compound D, one of the first things we notice is the carbonyl group. The carbon atom in a carbonyl is $\ce{sp^2}$ hybridized, so angle 6 involves an $\ce{sp^2}$ hybridized carbon. In such cases the $\ce{H-C-O}$ bond angle is ~ 120 degrees.

Just to elaborate a bit more:

If you have four objects arranged in space about a central carbon atom (the $\ce{sp^3}$ hybridization case), the four bonds from carbon are going to arrange themselves spatially so that the electron - electron repulsions between the four bonds are minimized.  This is achieved by arranging the bonds such that they point towards the vertices of a tetrahedron.  In a case like $\ce{CH_4}$, where the 4 atoms bound to carbon are identical, the C-H bonds adopt the perfect tetrahedral angle of 109.5 degrees. In the case of $\ce{CH_2Cl_2}$ where the four bonds are not equivalent, the bond angles will differ somewhat from 109.5 degrees (can you guess which angle will be greater then 109.5 degrees and which angle smaller) - but still, the tetrahedral angle is a reasonable first approximation.
In the case of an $\ce{sp^2}$ hybridized carbon, the 3 sigma bonds will arrange themselves in a plane with an approximate angle of 120 degrees between the bonds.  This arrangement again minimizes the electron - electron repulsion between the 3 sigma bonds in this geometry.
Finally, what geometry will minimize electron - electron repulsion between the two sigma bonds in the $\ce{sp}$ hybridized case?

(mouse over for answer)

   Yes, the two bonds will be in a plane 180 degrees from one another, just like the $\ce{H-C-C}$ bond angle in acetylene.

Take home message:

$\ce{sp^3}$ hybridized carbon has angles ~109 degrees
$\ce{sp^2}$ ~ 120 degrees
and $\ce{sp}$ ~180 degrees.


Is it correct to term boron trifluoride a "catalyst" in electrophilic aromatic substitution? In Friedel Craft Alkylation of aromatic compounds using $\ce{BF3}$ :

$\ce{BF3}$ acts as  a lewis acid

It is used as a catalyst

Amount of $\ce{BF3}$ used in the reaction is almost equal to that of amount of benzene used.

$\ce{BF3}$ acts as a nucleophile.

I am sure about options $1$ and $3$ being correct and $4$ th being incorrect but I am hesitant about option $2$.
Catalyst is a substance that lowers the activation energy and enhances the rate of the reaction, in the end remaining unconsumed.
Now, what Lewis Acids do is that they make halogens stronger electrophiles by forming  complexes with them. Taking this further we can say that yes they serve as "catalysts" in the reaction because otherwise the reaction would have been lot slower.
However, the given answer is $1,3$. Is my reasoning correct and the answer wrong? Is it functioning as a catalyst in the reaction?

On the formal definition, I think $\ce{BF3}$ may fail the remains unconsumed part. It will form $\ce{[BF3X]-}$, which would need to revert to $\ce{BF3 + X^-}$ in order to complete the catalytic cycle. I don't think the $K_\mathrm{d}$ of this complex is particularly high, so this is not likely to happen, hence it won't be a proper catalyst. This is also why you need a stoichiometric ammount of $\ce{BF3}$ for the reaction to proceed to completion.


Determination of Cell constant I read that cell constant is measured indirectly using a solution with known conductivity(like KCl).
But then how was the conductivity of that solution measured? (Is it by some theoretical consideration or another accurate experiment?)

Conductivity measurements were done in late 1800s by Friedrich Kohlrausch, who painstakingly measured the conductivities of solutions so accurately that they are still valid today. Most of his work is in German was published in relatively unknown journals. In his time, conductivity theory was not advanced enough to predict the conductivity of simple salt solutions in water, so there was no a priori theoretical prediction of conductivity, from the first principles. I don't think it is doable today after 100 years (may well be wrong(.
However, the accurate measurement of resistance was well established in physics. See below.
A paper in the J. Am. Chem. Soc. 1933, 55, 5, 1780 describes what Kohlrausch and his meticulous colleagues they did. Note the level of details which is remarkably missing from papers today.

After Kohlrausch had developed his method of measuring the electrical
conductance of solutions by means of an alternating current bridge, he
undertook with the aid of Holborn and Diesselhorst’ the determination
of the specific conductance of seven standard reference solutions in
absolute units in order that the results might be available to himself
and to other experimenters for the calibration of conductance cells.
For this purpose they used two cells whose cross sectional areas, a,
and length, l, were determined mechanically. The cell constant,
l/a, could then be computed in absolute units (cm$^{-1}$) from the measured dimensions.
The cell constants of a considerable number of other cells with
sealed-in electrodes of the usual type were determined electrically by
comparison with the primary cells by measurements of resistance when
filled with a common solution. These cells, whose cell constants were
thus determined indirectly, were then used for the determination of
the specific conductance of the seven standard reference solutions at
approximately 0, 9, 18, 27 and 36 C. The specific conductance of each
of these solutions for every degree between 0 and 36 C was obtained by
interpolation. No actual measurements were made at either 20 or at
25’. The water-bath used to fix the temperature of the cells was not
provided with an automatic thermostatic control. The majority of the
data on conductance in the literature are based on the values ascribed
to these standard reference solutions by Kohlrausch, especially the 1
N and 0.1 N potassium chloride solutions.
In the original paper by Kohlrausch, Holborn and Diesselhorst it is
stated that the normal potassium chloride solution was prepared by
dissolving 74.555 g. of potassium chloride weighed in air and making
up to 1 liter at 18 C. The 0.1, 0.02, and 0.01 normal solutions were
prepared by dilution by volume. The book “Das Leitvermogen der
Elektrolyte” by Kohlrausch and Holborn, which was published in the
same year as the paper referred to, contains alternative directions
for the preparation of these solutions by weight. This book has
probably been used by other experimenters more than the original paper
because it is more generally available.


Why do we have the concentrations powered to the coefficients in the equilibrium expression? $$\ce{A -> 2B} \qquad K_\mathrm{c} = \frac{[\ce{B}]^2}{[\ce{A}]}$$

Why do we have a power that is the same as the coefficient?
Assuming $K_\mathrm{c} = 1$, shouldn't the quantities of the mixture be $2B = A$ (in other words, the amount of $\ce{A}$ in the mixture equals two times the amount of $\ce{B}$ in the mixture)!? So why do we have $[\ce{A}] = [\ce{B}]^2$ rather than $[\ce{A}]=2[\ce{B}]$?

This is due to the way the equilibrium constant is defined.

A chemical equilibrium is a stable state at some point in the reaction (characterized by the extent of reaction $\xi$). Because it is a stable state the free enthalpy $G$ has to have a local minimum at exactly that place.
Therefore we can say
$$
\left( \frac{\partial G}{\partial \xi} \right)_{p,T} = 0
$$
On the other hand the following equation applies:
$$
\Delta_rG_m = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum_i \mu_i \cdot \nu_i
$$
With the free, molar reaction enthalpy $\Delta_rG_m$, the chemical potential of the component $i$ $\mu_i$ and it's reaction-coefficient $\nu_i$ (To understand this consider this general reaction equation: $\ce{\nu_1A + \nu_2 B <=> \nu_3 C}$ - note that the coefficients for the educts are negative by definition).

Therefore we can say that the following equation applies for the equilibrium state:
$$
\Delta_rG_m = 0
$$

we can rewrite above formula with the aid of the activity-dependency (basically a concentration-dependency) of the chemical potential ($\mu_i = \mu_i^* - RT\cdot\ln(a_i)$):
$$
\begin{split}
\Delta_rG_m &= \sum_i \mu_i \cdot \nu_i\\
&= \sum_i \nu_i\mu_i^* - RT \cdot \sum_i \nu_i\ln(a_i)
\end{split}
$$
Because $a \cdot \ln(b) = \ln(b^a)$ this is the same as
$$
\Delta_rG_m = \sum_i \nu_i\mu_i^* - RT \cdot \sum_i \ln(a_i^{\nu_i})
$$

Now we define the standard, free, reaction enthalpy $\Delta_rG_m^* = \sum_i \nu_i\mu_i^*$ and because we found out (in the beginning) that for the equilibrium state $\Delta_rG_m$ has to be zero we can write:
$$
\Delta_rG_m^* = -RT \cdot \sum_i \ln(a_i^{\nu_i})
$$
Because $ \ln(a) + \ln(b) = \ln(a\cdot b)$ we can rewrite this as
$$
\Delta_rG_m^* = -RT \cdot \ln\left(\prod_ia_i^{\nu_i}\right)
$$

The last step is now to define the equilibrium constant $K$ as
$$
K = \prod_ia_i^{\nu_i}
$$

And that is how the coefficients in the recation equation end up in the exponent when calculating the equilibrium constant.


Rate law of the reaction of propane with oxygen In class, I have a demonstration where I light a bubble filled with propane. As I add oxygen, the bubbles combust faster and more quickly. I have already balanced out the chemical reaction like so:

$$\ce{C3H8 + 5 O2 → 3 CO2 + 4 H2O + heat + light}$$

I have reasoned that the reaction must go faster because the limiting reagent is oxygen, and since the air is only about 20% oxygen adding more oxygen to the bubble will speed the reaction process. I have figured out the kinetics, the oxidation, the enthalpy diagram, etc.

However, the only thing I don't know how to do in this demonstration is how to find the rate law. How would I find the rate law and equation of a combustion reaction? Wouldn't it just be second order since it's a combustion reaction? I have no idea where to even begin or why it is so.

The oxidation of hydrocarbons ($\ce{RH}$) such as this one proceed via a chain mechanism. The initiation step is

$$\ce{RH + O2 -> HO2^. + R^.}$$

and the propagation

$$\ce{ R^.+O2 -> RO2^. \\ RO2^. +RH -> ROOH +R^.      }$$

But it does not stop there as the hydroperoxide decomposes $\ce{ ROOH -> RO^. + OH^.}$ and there are other reactions such as

$$\begin{align}
\ce{ R^. + O2 -> RO^. + O^. \\ RH + OH^. -> H2O + R^. \\ RCH2OO^. + O2 -> RCH2O + O3  \\ RH + O3 -> RO^. +HO2^.}\end{align}$$

and chain branching $\ce{ \quad O^. + RH -> OH^. + R^.}$.

The exact mechanism depends on temperature and pressure, and in a plot of $T$ vs $p$ there are regions of slow reactions, explosion and hot and cold flames. (An original paper is by Newitt & Thorne, J. Chem. Soc 1937, p1656). Thus you can appreciate that sorting out the mechanism is no simple task and will need much specialised equipment.


Intramolecular Claisen-Condensation My task was to do an intramolecular Claisen condensation with this molecule.

I already draw the product to the reaction, but I'm not sure, if I did it right. Could anyone help me out or confirm the solution? There should be only one product since there is symmetry.
Here is the way how I got the product in my question:


If A is a stronger acid than B then is B necessarily a stronger base than A? It seems like this must be true. So if it is could someone please prove this?

In general this is not true.

For instance, water is a stronger acid than ethane and is also a stronger base than ethane.


What is the reason for the blue color of an ammoniated electron? I am quoting a line from a book regarding reactivity of alkali metals with liquid ammonia.

The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution

Now my question is:

How does the independent electron get excited (without transitional levels)?

Why do salts of particular alkaline metal always have a fixed flame color if transitions of electrons can be from any energy level.
(I am asking this question based on my knowledge of school where we only studied transitions among hydrogen like species)

Both of your questions could benefit from more background, so please forgive me if I understood something wrong. However, I'll try to answer as best I can:

The excess electron in liquid ammonia is not really independent, otherwise, as you stated correctly, the electron would not absorb a certain wavelength but act as a black radiator/absorber like in a plasma (for example the sun). However, how the electron is bound and how the transitional states are formed is not completely understood yet. In this paper, I found the following sentence: "The excess electron in liquid ammonia ("ammoniated electron") is commonly viewed as a cavity electron in which the s-type wave function fills the interstitial void between 6 and 9 ammonia molecules." And then they go on to present a different hypothesis...
In a flame, the spectrum is recorded in the gas phase and formed not by the salt but by the individual atoms (even if you put the salt into the flame). Thus the spectrum depends only on the elements contained in the salt, not on the type of salt. This is different, if the spectroscopy is performed in the liquid phase of alkali metal halides. There the spectrum depends on the mole ratio of the anions and cations.


Values for heat of reaction example I am currently studying the textbook Mass Spectrometry, third edition, by Jürgen H. Gross. Chapter 2.4.3 Bond Dissociation Energies and Heats of Formation says the following:

Energetics of $\ce{H}^\bullet$ loss from $\ce{CH_4^{+ \bullet}}$ The minimum energy needed to form a $\ce{CH_3^+}$ ion and a hydrogen radical from the methane molecular ion can be estimated from the heat of reaction, $\ce{\Delta H_r}$, of this process. According to Fig 2.6, $\ce{\Delta H_r} = \ce{AE_{(CH_3^+)}} - \ce{IE_{(CH_4)}}$. In order to calculate the missing $\ce{AE_{(CH_3^+)}}$ we use the tabulated values of $\ce{\Delta H_{f(H^\bullet)}} = 218.0 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_3^+)}} = 1093 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_4)}} = -74.9 \ \text{kJ mol}^{-1}$, and $\ce{IE_{(CH_4)}} = 12.6 \ \text{eV} = 1216 \ \text{kJ mol}^{-1}$. First, the heat of formation of the methane molecular ion is determined based on the experimental value of $\ce{IE_{(CH_4)}}$:
$$\ce{\Delta H}_{f(CH_4^{+\bullet})} = \ce{\Delta H_{f(CH_4)}} + \ce{IE_{(CH_4)}} \tag{2.14}$$
$$\ce{\Delta H}_{f(CH_4^{+\bullet})} = -74.9 \ \text{kJ mol}^{-1} + 1216 \ \text{kJ mol}^{-1} = 1141.1 \ \text{kJ mol}^{-1}$$
Then, the heat of formation of the products is calculated from:
$$\ce{\Delta H_{f(prod)}} = \ce{\Delta H_{f(CH_3^+)}} + \ce{\Delta H_{f(H^\bullet)}}  \tag{2.15}$$
$$\ce{\Delta H_{f(prod)}} = 1093 \ \text{kJ mol}^{-1} + 218 \ \text{kJ mol}^{-1} = 1311 \ \text{kJ mol}^{-1}$$
Now, the heat of reaction is obtained from the difference
$$\ce{\Delta H_r} = \ce{\Delta H_{f(prod)}} - \ce{\Delta H}_{f(CH_4^{+\bullet})} \tag{2.16}$$
$$\ce{\Delta H_r} = 1311 \ \text{kJ mol}^{-1} - 1141.1 \ \text{kJ mol}^{-1} = 169.9 \ \text{kJ mol}^{-1}$$
The value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{(CH_3^+)}} = 14.35 \ \text{eV}$, which is in good agreement with published values of about $14.3 \ \text{eV}$ (Fig. 2.7).

Note: The amount of energy needed to be transferred to the neutral $\ce{M}$ to allow for the detection of the fragment ion $m_1^+$ is called the appearance energy ($AE$) of that fragment ion.

Taking the values from Fig. 2.7 and using this calculator, we get that $1311 \ \text{kJ mol}^{-1} = 13.588 \ \text{eV}$ and $1386 \ \text{kJ mol}^{-1} = 14.365 \ \text{eV}$.
Unless I have made a mistake or am misunderstanding something, these values do not agree with what the author has presented. It seems that, if we take the values from Fig 2.7, the value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 13.588 \ \text{eV}$, which implies that the published value should be around $14.365 \ \text{eV}$.
So has the author made a small mistake here? Or am I misunderstanding something?
I would greatly appreciate it if people would please take the time to review this.

The exercise in question is a prediction of the appearance energy of the methyl cation fragment $\ce{CH3+}$. The authors predict that value by estimating the enthalpy for the reaction of methane ionization followed by loss of a hydrogen atom. The enthalpy change estimation is based on published values for heats of formation and bond dissociation energies that have been determined experimentally.

The values presented, however, are rounded to the nearest integer or tenth, and the authors continue to round values during the explanation. For example, 12.6 eV is converted to the closest integer value of 1216 kJ/mol. All values in Fig 2.7 are presented as integers.

There seem to be two possible explanations for the final value being given as 14.35 rather than your calculated value of 14.365. Most likely, the authors have chosen to round to the nearest multiple of 0.05 rather than the nearest tenth or hundredth. Alternatively, they could have used unrounded values throughout the calculation rather than the rounded numbers presented in the step-by-step description. Either way, the difference is insignificant relative to the (im)precision with which the value can be predicted or experimentally determined.


Why is the pKa of the thiol group in N-acetylcysteine higher then the pKa of the thiol group in cysteine? I understand why cysteine has a $\mathrm pK_{\mathrm a}$ of 8.3. But the $\mathrm pK_{\mathrm a}$ of thiol in N-acetylcysteine is 9.27. Since there is an acetyl group attached to the nitrogen, doesn't this lower the $\mathrm pK_{\mathrm a}$ via the inductive effect?

Shown below are the relevant ionization events of cysteine and N-acetylcysteine. The question is asking about the second ionization of each, the deprotonation of the thiol groups, but we see a similar difference in pKas for the carboxylic acid groups as well. The difference in each case is an ammonium group vs. an acylated nitrogen. It's true that the acetyl group is an electron withdrawing group, but it turns out that the cationic ammonium group is even more inductively electron withdrawing. For both the first and second ionizations, the difference in electron withdrawing power is enough to lower the pKa by >1 pKa unit.

iad22agp's explanation is a good way of rationalizing this. Looking at the first ionization, cysteine is cationic while N-acetyl cysteine is neutral. Since the rest of the structures are similar, we would assume that it's easier to take a positive charge away from a positively charged compound than a neutral compound. Similarly looking at the second ionization, it's easier to take a positive charge away from a neutral compound than a compound that's already negatively charged.

pKa refs: cysteine, N-acetylcysteine


Reaction of 1-(1-chloroethyl)-1-methylcyclobutane under E1 Conditions What is the major product of the following reaction?
$\hspace{40 mm}$
$\hspace{20 mm}$

It looks to me that the $\ce{Cl}$ group leaves then there's an alkyl shift to where the $\ce{Cl}$ group leaves in order to form a tertiary carbocation. Then $\ce{EtOH}$ attacks the beta hydrogen and forms a double bond to give product B:
$\hspace{35 mm}$
However, the answer I came up with was incorrect. Any ideas why?

You are correct that an alkyl shift will take place but there are two possible options for which group moves. Your option is possible but there is another option which involves a ring expansion to generate product A.

This mechanism is favoured because it relieves the ring strain in the four membered ring. Since the reaction is run under reversible conditions, the thermodynamic product (with the least ring strain) will dominate.


What [if anything] is special about the boiling process of purifying water (excluding temperature)? I know that water boils at $100~^{\circ}\rm C$. I also know that the standard way to purify water is to boil it for roughly a minute. What I'm curious about is, what properties of the boiling process purify the water?

Is it purely a temperature thing? Are we saying "roughly a minute" so that the temperature actually increases beyond $100~^{\circ}\rm C$, thus reaching some actual temperature needed to kill bacterial/viral stuff? Or, does the "rolling" of the boil actually add a physical benefit to the purification process?

There nothing special about boiling that is doing anything.  For example, you can put a beaker of water under vacuum and get the water to boil at room temperature, but that won't kill any bacteria or render any viruses impotent.

It's purely a heat thing.  You want to destroy any pathogens that may cause illness.  Sufficient heat will denature proteins in a virus and kill bacteria, etc.  Boiling simply gives you a visual cue that you've reached a very high temperature (so you don't need a thermometer) and the ~1 minute is just to ensure that you sustain a high temperature long enough to destroy any bacteria or viruses.


Temperature dependence of molarity and molality My textbook says

  Molarity is dependent on temperature, whereas molality is not.

I'm aware of the definitions of both molarity and molality, but consider this scenario:

A solution of $\ce{NaCl}$ in $\ce{H2O}$ is heated. Eventually, all the water will escape and I will be left with only $\ce{NaCl}$ crystals. At the end, there is no solvent to even measure the concentration of what was once a solution. In this case, how does the argument "molality is independent of temperature" still hold true?

Recall that molarity $c$ is

$$c=\frac{n_\text{solute}}{V_\text{solution}}$$

whereas for molality $b$ is

$$b=\frac{n_\text{solute}}{m_\text{solvent}}$$

While in the extreme case you mentioned, both quantities are not defined because all the solvent is either boiled away (or for heat-sensitive species, the solvent can even decompose under heating), molarity depends on volume, and the volume of the solution depends on temperature due to expansion/contraction.


Friedel-Crafts Alkylation and hybridization Why must the alkyl halide be sp3 hybridized with respect to the carbon connected to the halogen for the alkylation to occur?

Why, for example, can't we use acetylide with a bromine substituted for hydrogen on one side in conjunction with aluminum tribromide to alkylate benzene?

If anything this seems to be more favorable as an sp carbocation is likely less stable than an sp3 carbocation.

If anything this seems to be more favorable as an sp carbocation is likely less stable than an sp3 carbocation.

That's the problem, sp2 and sp hybridized carbocations are less stable than sp3. Those carbocations are difficult to form and will likely undergo rearrangement faster than attack of the aromatic ring.

Modern organic synthesis has much better ways of making sp-sp2 and sp2-sp2 carbon-carbon bonds by palladium catalyzed cross-coupling.


Best consumption-safe liquid for caffeine solubility? The question is simple:  What human-consumption safe liquid has the best caffeine solubility?

It's 200mg/10mL of water.  That's a lot of water for one dose!

I found some ramblings on a fourm stating it's "low" solubility in propylene glycol and glycerine (glycol) but no sources or exact numbers, so that doesn't mean much.

In boiling water, it seems much higher, 66g/100mL, which is more of the range I am hoping.  So what happens if I boil a liter of water, dump 100g into it, and then let it cool?

I am hoping for something in the range of 200mg in 1mL-3mL, but as low as possible, and relatively easy to acquire.

All help very much appreciated.

Safety Warning

  Caffeine is a toxic substance. When following the advice below, always keep in mind the factors that will help you kill yourself if you are not careful! For more details you can start out on Wikipedia.

Supersaturation

Your idea of preparing a supersaturated solution is not that bad, but it might be that you have some nucleation sites in the bottle where you want to store it. Furthermore, it is probably not that easy to transfer the supersaturated solution from the container where you prepared it to the vial.

Acidic Medium

I've come across this PDF that mentions something interesting:

  Solubility in water is increased by adding dilute acid (e.g. HCl or citric acid).

It would be very easy to prepare an acidic solution (I'd aim at a pH of about 3 or so, in the general vicinity of energy drinks and Coca Cola, beverages known to contain large amounts of caffeine) from some citric acid¹ and dissolve caffeine in there. Maybe you could even conduct a study about the solubility of caffeine at different pH values!

Water/Ethanol Mixture

There is one dissertation (PDF) that touches on the solubility of caffeine in a mixture of water and ethanol. I wouldn't trust it too much, since it was written using MS Word and contains several formatting errors (it says $K_{ow}$ for example, instead of $K_\text{ow}$). The graphic containing the data can be found on page 52, but for your convenience it is reproduced below:

(Seriously, the figure caption is useless, and the plot is so bad I'm tempted to redo it just for this answer.)

Never mind the linear relationship that is indicated by the line, the diamonds are the important part here. As you can see, there seems to be a solubility maximum at around $x=0.5$, where $x$ is the volume fraction of ethanol in the water-ethanol mixture. So that's a whopping 50 vol.-% of ethanol, about equivalent to a cask-strength whisky!

However, it is probably not a good idea to use this knowledge to its fullest, because that will be quite a lot of ethanol, and as already mentioned in the comments it is probably not very wise to combine the consumption of caffeine with ethanol.

Does it have to be dissolved?

There are also caffeine tablets and pills that you can take. They come with the additional guarantee that some FDA-like agency has looked into the production of these products, so you can assume they are safe to ingest. Also, they come with instructions and more than enough warning labels so you can educate yourself further.

Conclusion

If you really want a caffeine solution, it is probably your safest and best bet to create an acidic solution where you can dissolve above-normal (i.e. pH=7) amounts of caffeine.

¹ Though please do not ask here how to calculate the amount you need, that would be considered a homework question and get deleted pretty quickly...


Predict the products of the controlled oxidation of 2-methylpentan-2-ol Predict the products of the controlled oxidation of 2-methylpentan-2-ol.

From what I know, since 2-methylpentan-2-ol is a tertiary alcohol, there should be no reaction. However, my textbook states that the product includes 2-methylpentanone.

Edit: In response to the comments, my textbook neither specifies the isomer nor mentions any given conditions. I hate Nelson

Tertiary alcohols are resistant to oxidation in neutral or alkaline solution, but are readily oxidized by acidified oxidizing agents.

In your case two products will be acetone and propionic acid.

http://chemistry.tutorcircle.com/organic-chemistry/tertiary-alcohol.html


Why do atoms need 8 electrons to stabilize? As the title says. I have surfed all of the net but could never find the answer to this question. Why do atoms need 8 electrons to stabilize? I mean why not 7 or 5 or 10 electrons? Why specifically 8? And what does stabilization of atoms even mean? Are the atoms going to burst without 8? Why do they need 8 electrons?

This is a question which is not often discussed in chemistry and I need an answer which really makes any sense. Not just `it is because the atoms need to fill their octet.

The valence orbitals of atoms are composed of suborbitals (s and p)  there is 1 s suborbital which is spherical and can hold 2 electrons (one with up spin and one with down spin). There are 3 p suborbitals which are dumbbell shaped (look like two balloons tied together at the ends) and align along the x,y&z axies and hold a total of 6 electrons (2 per axis, 3 with up spin and 3 with down spin).  Since an atom is energetically satisfied when all of the electrons are paired and 2+6 = 8, an atom must have 8 electrons in its valence shell to pair all of the electrons.

The octet rule does not apply to all atoms though, for transition metals and later elements there is a d suborbital which can hold an addition 10 electrons, meaning that now an atom may have 18 electrons in its valence shell to be satisfied.  This allows for materials such as metal complexes and the sulfate poly-atomic-ion to exist despite having more than eight electrons in the valence shell.


Frequency region and "uncoupled oscillators" in vibrational spectroscopy I am currently studying the textbook Infrared and Raman Spectroscopy by Larkin. In the first chapter of the textbook, the author presents the following table and accompanying explanation:

The vibrational spectrum may be divided into typical regions shown in Fig. 1.1. These regions can be roughly divided as follows:

$\ce{X-H}$ stretch (str) highest frequencies (3700-2500 cm$^{-1}$)
$\ce{X#Y}$ stretch and cumulated double bonds $\ce{X=Y=Z}$ asymmetric stretch (2500-2000cm$^{-1}$)
$\ce{X=Y}$ stretch (2000-1500cm$^{-1}$)
$\ce{X-H}$ deformation (def) (1500-1000cm$^{-1}$)
$\ce{X-Y}$ stretch (1300-600 cm$^{-1}$)

The above represents vibrations as simple, uncoupled oscillators (with the exception of the cumulated double bonds). The actual vibrations of molecules are often complex and as we will see later, typically involved coupled vibrations.

I have two questions:

What do the $\ce{X}$, $\ce{Y}$, and $\ce{H}$ values mean in the above explanation?
Can someone please elaborate, with some detail (including mathematics), on this concept of "uncoupled oscillators", and its relevance in the above context? To be specific: What specifically is an "uncoupled oscillator", how is it described mathematically, and what is its relevance in the above context?

I appreciate it if someone would please take the time to clarify these points. Thank you.

$\ce{H}$ is just hydrogen and $\ce{X}$, $\ce{Y}$, and $\ce{Z}$ are just other elements, in this case likely carbon, nitrogen, or oxygen. They give an example in the table of $\ce{-X=Y}$ where $\ce{-C=N}$ matches this template of two (generally) different elements double bonded together.
They use coupled here in a slightly different way than most literature. I would may switch to delocalized/localized rather than coupled/uncoupled. The modes they mention are mostly localized to a single bond (e.g. an $\ce{-X-H}$ stretch) whereas in general vibrational modes are delocalized/coupled over the whole molecule (e.g. an aromatic $\ce{CH}$ stretch where all the $\ce{C-H}$ bonds of an aromatic ring move in unison).


Chemistry Calculation Dilutions of Acid I was presented with the following problem in my lecture and I am confused as to what to do.

  You are required to prepare $\pu{250 cm3}$ of $\pu{0.100 mol dm-3}$ HCl by diluting $\pu{0.600 mol dm-3}$ HCl with water.

  Calculate the volume of $\pu{0.600 mol dm-3}$ HCl that must be diluted with water.

This was my attempt.

Moles of the $\pu{250 cm3}$ of $\pu{0.100 mol dm-3}$ HCl (target acid)

= $\frac{250}{1000} \times 0.100 = 0.0250$ moles.

Hence

$\frac{\pu{mol}}{\mathrm{conc}} = \mathrm{vol}_{\pu{cm3}} = \frac{0.0250}{0.600} \times 1000 = 41.66... \approx \pu{42.00 cm3}$

I'm not sure if the second step is right.

If I am right (hopefully) can someone please explain why.

Many thanks.

The clue is the moles in the dilute solution is the same as the moles of the concentrated stock solution - it has to be constant.

The method I have used is (therefore) correct.


when do we indicate the "+ or -" sign in uncertainty how can we know when to indicate "+ or -" sign in uncertainty calculations because I noticed some problems has been solved by this sign and some has not.

You either use both "+" and "-", $\pm$, or neither in which case it is implied that you mean $\pm$.

Usually uncertainty is expressed in a way that is symmetric about the mean, such as $123\pm4$, meaning to a certain degree of confidence the true value is between 119 and 127.

However, you could write $123^{+5}_{-3}$ to indicate the best estimate of the true value is 123 and to a certain degree of confidence the true value is between 120 and 128.

For example, if you were highly certain that an object is a cube, and measure its edge as 100 $\pm 20$, then it might be better to express the volume of the cube as

$1.0^{+0.7}_{-0.5} \times 10^6$


Determining the order of molar conductivity This was a question which came in a objective type examination.

Which of the following have a $\lambda^\infty$ (molar conductivity at infinite dilution) larger than $\ce{KCl}$?
A. $\ce{CH3COOH}$
B. $\ce{HC}$
C. $\ce{NaOH}$
D. None of the above

I was not provided any experimental data to arrive at the answer.I know that $\ce{KCl}$ is a strong electrolyte and so my guess was that it must have greater molar conductivity than at least $\ce{CH_3COOH}$. But the answer key gives all A, B, and C. Is there a definite rule here?

I would agree with the answer key. The trick here is it's limiting molar conductivity, the molar conductivity at infinite dilution. For a neutral electrolyte compound, I was taught the notation $\Lambda ^0$, and $\lambda ^0$ was reserved for individual ions. (The $0$ superscript represents zero concentration, equivalent to $\infty$ dilution.)

Anyway, limiting molar conductivity is an interesting property because it does not represent a physically possibly scenario (a solute conducting electric charge when no solute is present). It can be extrapolated by the Kolhrausch Law, where it $\Lambda^0$ is the y-intercept of molar conductivity $\Lambda$ vs $\sqrt C$, concentration (note this relationship is only valid for strong electrolytes).

The degree of ionization of a weak electrolyte depends on the concentration, and tends to unity (100% dissociation) in the limit of concentration approaching zero. Limiting molar conductivity $\Lambda^0$ is defined (only) for this exact limiting scenario.

So when we consider limiting molar conductivity, we are doing so in a (abstract) condition where both strong and weak electrolyte fully dissociate. Thus for $\Lambda^0$ it makes no difference whether the electrolytes being compared are "strong" or "weak" at finite concentrations (for we are comparing them at "zero" concentration, i.e. infinite dilution). So we should have no problem accepting that a "weak" electrolyte might have a higher limiting molar conductivity value than a "strong" one, as in this case.

Why might $AcOH$ (organic chemist abbreviation), $HCl$, and $NaOH$ have higher $\Lambda^0$ values than $KCl$? Well, the $H^+$ ion is better at transporting its charge through an aqueous medium than $K^+$ because it can "water hop" via the
Grotthuss mechanism. This makes acidic compounds very effective electrolyte conductors. $OH^-$ is also an extremely effective charge carrier by an analogous "deprotonation-chain" mechanism. Because they use water itself to "tunnel"* through the bulk solution, protons and hydroxide have extremely high ion mobility in aqueous solutions and thus provide abnormally high molar conductivity.

So in general acids and bases would be expected to out-conduct a pH-neutral salt ($KCl$), and in the limit of zero concentration/infinite dilution, even "weak" acids and bases will have higher limiting molar conductivity values.

*Not in the quantum sense.


A and B irreducible representations? In character tables, I know that A and B are one-dimensional irreducible representations, but what is the difference them?

And why do some character tables list them as A1/A2, B1/B2 and others have Au/Ag, Bu/Bg?

The traditional labelling of irreducible representations is the following:

A, when rotation around the principal axis is symmetrical
B, when rotation around the principal axis is asymmetrical

You can see, for example, that $C_n$ groups of even order have both A and B representations, while those of odd order cannot feature B representations.

As others have commented, there are additional rules… E for doubly degenerate representations, T for triply degenerate, u and g for parity under inversion (odd and even, respectively), greek capital letters Σ, Π, Δ for infinite groups (C∞v and D∞h).

Finally, if there are several irreducible representations that are not distinguished under those conventions, they are numbered.


How many atoms can be reduced in a single redox reaction? I am doing high school chemistry and I am busy working with redox reactions.

The explanation in my text book, of how a redox reaction works, left me a bit confused as to how many atoms can be reduced/oxidized in a single reaction equation.

Can only one atom (species) be reduced/oxidized in a single equation or can there be more than one reduced/oxidized in the same chemical equation?

All of the reaction equations I have come across so far have just one atom reduced and one atom oxidized, for example:

$$ \ce {Mg(s) + 2HCl(aq) -> H_2(g) + MgCl_2(aq)} $$

In this reaction only Mg is oxidized and only H is reduced. Is there a reaction in which more than one species of atom it reduced or oxidized?

I'm not sure if I understand your question correctly, but consider the reaction

$$\ce{CH3-SH + 3 O2 -> SO2 + CO2 + 2 H2O}$$

where methanethiol is oxidized with oxygen to sulfur dioxide, carbon dioxide, and water.

Thereby the oxidation states of sulfur and carbon change from $-2$ to $+4$ and the oxidation state of oxygen from $\pm0$ to $-2$. So sulfur and carbon atoms are both oxidized while oxygen is reduced.

I hope this answers your question.


How can nucleophilicity be estimated accurately? Given the following: $\ce{CN-,OH-,SH-,I-}$. Determine  the relative order of nucleophilicity in a protic solvent.

I know that, more the nucleophile is solvated by hydrogen bonding in a protic solvent, the lesser will be its nucleophilicity. Also, solvation depends upon the size of the anionic nucleophile or its charge density (smaller the size greater the solvation).

Again larger the size, greater will be the polarizability of the anionic nucleophile and thus stronger will be its nucleophilicity.

All these indicate that the larger the size, the greater the nucleophilicity will be, and so the relative order of nucleophilicity should be $\ce{I-}>\ce{SH-}>\ce{CN-}>\ce{OH-}$. But the answer is $\ce{SH-}>\ce{CN-}>\ce{I-}>\ce{OH-}$. So where have I gone wrong? How can I determine nucleophilicity accurately? Is there any other factor on which nucleophilicity depends?

It has been the lifetime research project of Professor i.R. Mayr of the University of Munich (LMU) to determine reactivity scales of nucleophiles. They key points are that any nucleophilic addition’s rate constant can be derived by the equation

$$\lg k_\pu{20^\circ C} = s_\mathrm{N}(N + E)\tag{1}$$

where $E$ is an electrophilicity parameter, $N$ is a nucleophilicity parameter and $s_\mathrm{N}$ is a solvent and nucleophile-dependent additional nucleophilicity factor. A multitude of nucleophiles and electrophiles have been entered into this scheme over the years and a few posters showing nucleophilicity scales have been created (see the webpage).

Mayr’s research could show that nucleophilicity depends on the solvent and on the nucleophile, that it is a kinetic parameter and that it can only be estimated to a certain extent. Simple parameters such as size or even the charge per volume ratio are never enough to explain a nucleophile’s relative nucleophilicity unless all other parameters remain the same.

So your discussion which seems to base itself on size alone is bound to fail.

I wholeheartedly invite you to scrutinise the database data (also linked on the webpage) to attempt to find truly general rules. If you do, do not hesitate to submit them to the academic journal of your choice.


What is the @ symbol used for in material science? I came across the phrase "Silicone Rubber/Paraffin@Silicon Dioxide" also written as $\ce{SR/pa@SiO2}$, utilising the @ symbol.

How can I understand what this means?

The at symbol @ universally means "at the site"  or "at the rate of".
It is routinely used in supramolecular chemistry to denote guest@host relationship because

it follows the original designation of denoting the binding site;
its shape resembles a guest (letter "a") entrapped within a host shell (circle sign "○") providing a visual cue for an intuitive image (e.g. see endohedral fullerene).

In the context of the paper by Guo et al. [1] you are referring to, paraffin is encapsulated inside $\ce{SiO2}$ shell (obtained in situ), and these silicon dioxide particles were then embedded in silicone rubber matrix.
Here, in the notation SR/Pa@SiO2 the forward slash "/" implies composite material (phase separator), and the at sign "@" refers to host-guest interactions as mentioned before.
Here is a brief infographics I roughly created with Inkscape to illustrate the concept:

An obligatory reference quotation [2, p. 203]:

3.6.17 At Symbol (@)
The at symbol ( @ ) is most commonly used in email addresses. Its modern use is primarily in accounting (where it means “at the rate of”) and should not be used in scientific writing as a substitute for “at.”
Two legitimate uses of the at symbol are

In chemical formulae, the @ is used to denote trapped atoms or molecules. For instance, $\ce{La@C60}$ means lanthanum inside a fullerene cage.
In genetics, an at symbol after a gene symbol indicates that it is part of a gene cluster.

References

Guo, Y.; Yang, W.; Jiang, Z.; He, F.; Zhang, K.; He, R.; Wu, J.; Fan, J. Silicone Rubber/Paraffin@silicon Dioxide Form-Stable Phase Change Materials with Thermal Energy Storage and Enhanced Mechanical Property. Solar Energy Materials and Solar Cells 2019, 196, 16–24. DOI: 10.1016/j.solmat.2019.03.034.
The Manual of Scientific Style: A Guide for Authors, Editors, and Researchers, 1st ed.; Rabinowitz, H., Vogel, S., Eds.; Elsevier/Academic Press: Amsterdam; Burlington, MA, 2009. ISBN 978-0-12-373980-3.


Why does rock salt change colour when it's heated We heated rock salt for a chemistry lab and it changed from white crystally rock to brown rock and ash-like materiak on the bottom of the test tube. Also there was a sizzling\popping noise. Why did it change colour and why was there a noise? What caused this?

"Rock salt" contains many impurities, in addition to its main ingredient, sodium chloride. Some trace chemicals in rock salt, such as lithium chloride or sodium carbonate, hold onto water, which might boil out, sizzling and popping, and if your specimen got hot enough, carbonates could decompose.

Even if you had used "table salt", rather than rock salt, chemicals are intentionally added, such as sodium ferrocyanide (to prevent caking), which has a yellowish color and potassium iodide (as a food supplement).

Had you heated pure sodium chloride, there should not have been a color change.


What is non-hückel double bond of Aromaticity? Chemistry of Aromaticity. What's non hückel double bond ?
What are the difference between hückel and non-hückel double bond ?
[ Here, in figure the green one is non hückel double bond ]

In the given molecule Pyrene , the π  bond shown in green is in cross conjugation (shown in green in figure 1). This bond( figure 2) is excluded from interaction with rest of conjugated system. Hence , the central $\ce{C=C}$ isn't hosted by the largest p orbital loop, so it's not counted towards Hückel's rule for aromaticity.

The outer periphery (shown in blue in figure 3), is a loop of $\ce{14 π}$ electrons , therefore aromatic. This may be possible hückel double bond you are referring to.

The π bond in cross-conjugation (shown in green) maybe non-hückel double bond.


How did Williard Gibbs come up with the Gibbs equation? The equation I am referring to is

$$\Delta G = \Delta H - T\Delta S$$

I understand that this Gibbs free energy determines if a process is spontaneous or not. My question is how did Williard Gibbs, presumably, come up with this equation? Did he do a bunch of experiments and guessed the answer, or did he define this thing called Gibbs free energy because of well motivated theory?

He derived this formula by fully appreciating the fact that entropy tends to a maximum.

As the first paragraph of the relevant paper says,

  It is an inference naturally suggested by the general increase
  of entropy which accompanies the changes occurring in any
  isolated material system that when the entropy of the system
  has reached a maximum, the system will be in a state of equilibrium.
  Although this principle has by no means escaped
  the attention of physicists, its importance does not appear to
  have been duly appreciated. Little has been done to develop
  the principle as a foundation for the general theory of thermodynamic
  equilibrium.

That is to say, he saw it as obvious that entropy increases until it cannot increase anymore and this makes a for a good definition of equilibrium. Gibbs offers a second definition of equilibrium, however, by saying,

  For the equilibrium of any isolated system it is necessary and
  sufficient that in all possible variations of the state of the system
  which do not alter its entropy, the variation of its energy shall
  either vanish or be positive.

The new move, however, is to construct some kind of function which makes these two definitions simply the same definition stated in a different way. That is, Gibbs has introduced two conditions:

$$(\delta S)_{\epsilon} \le 0$$
$$(\delta\epsilon)_{S} \ge 0$$

In words, at equilibrium a system's entropy (for a constant amount of energy) can only change by decreasing, and at equilibrium a system's energy (for a constant entropy) can only increase. Or simply, entropy is a maximum and energy is a minimum.

Then, because these two properties are opposites in the sense that one tends to a maximum and the other a minimum, it is a reasonable thing to subtract them and say that when this subtraction is equal to zero, we are at equilibrium. This is exactly what Gibbs did. He literally just multiplied the entropy by temperature without justifying it ($TS$ is common on thermodynamics though so he may have thought it obvious), and wrote down:

$$\psi=\epsilon-TS$$

where in this case $\psi$ is what we now call the Gibb's free energy. With this definition, the two conditions for equilibrium above are turned into one condition which is,

$$(\delta\psi)_T \ge 0$$

That is to say, at a constant temperature physical systems tend to the minimum of a free energy surface.

The reasoning Gibbs' has walked through thus far is sufficient to derive the equation as you have written it if we say that $\epsilon=\epsilon(U,P,V)$. That is, the energy of the system is determined fully by the internal energy (imagine microscopic motions) and the pressure on the substance and volume of the substance. Then we can simply write (again he just guesses a proportionality of one and uses units to figure out the rest),

$$\zeta=U+PV-TS=H-TS$$

where $\zeta$ is again the Gibb's free energy in a more explicit form. He takes the next couple of pages to take the total differential and then simple integration would give the familiar form but this is just a mathematical exercise not worth going through here.

A more formal, but not much more complicated, derivation of Gibbs' energy from the same condition that entropy increases can be found here under free energy of reactions. In this case two conditions for the entropy are used. Namely, the internal and external entropies. That's worth looking at as it contains the same principle which Gibbs used.


How does cooling increase the selectivity of a reaction producing the kinetic vs the thermodynamic product? I know that cooling gives a lower ratio of molecules with sufficient energy to overcome the activation barrier, but I'm not sure how this selects for a kinetic or thermodynamic product. I also know that the kinetic product is favoured if there is not enough energy supplied to reach equilibrium, and the opposite for the thermodynamic product. Can cooling select for either product in different circumstances?

First, there is no rule that "the" kinetic product will be different than the thermodynamic product.  It depends on the particular reaction you are interested in.

Thus, I think you are really asking "in reactions where the kinetic product is different than the thermodynamic product, what does cooling select for?"

Let's consider two reactions, the first with a reactant R to a thermodynamic product T and a second, separate reaction where R goes to the kinetic product K.  The rate constants for these reactions are $k_1$ and $k_2$, respectively.

$$\ce{R ->[k_1] T}$$
$$\ce{R ->[k_2] K}$$

Let's suppose both reactions follow classical transition state theory, which is a very good approximation to many everyday reactions and is a great starting place even for the weird cases that do vary.

TST says (if you assume that the $\kappa$ transmission coefficient is 1):

$k_1 = \frac{k_b T}{h} e^{-\Delta G^{\ddagger}_1/RT
}$

$k_2 = \frac{k_b T}{h} e^{-\Delta G^{\ddagger}_2/RT
}$

The "kinetic" product is the one with the higher rate constant; this is the definition of kinetic.  Doing some algebra,

$$\ln{\frac{k_2}{k_1}} = -\frac{1}{RT}\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$$

...we see that for $k_2$ to be higher than $k_1$, i.e. for $\ln{\frac{k_2}{k_1}}$ to be greater than 0, then $\Delta G^{\ddagger}_2$ must be smaller than $\Delta G^{\ddagger}_1$ (note the negative sign).

The effect of temperature is still in the equation:

$$\ln{\frac{k_2}{k_1}} = -\frac{1}{RT}\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$$

The assumption of TST is that $\Delta G^{\ddagger}$ is not a function of temperature.  As temperature is lowered, then the $RT$ divisor in the equation gets smaller, which means that the difference $\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$ affects the ratio of rate constants more.  Thus, under the assumptions we have made, cooling will favor the kinetic product more. As temperature is raised, eventually $RT$ becomes much greater than $\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$, so selectivity for the kinetic product vanishes because $\ln{\frac{k_2}{k_1}} \rightarrow 0$, i.e. $\frac{k_2}{k_1} \rightarrow 1$.

You might ask, how can the ratio go to one; doesn't that mean that we have assumed the equilibrium constant between T and K is also 1?  The answer is no, we haven't assumed that, but we have made a number of other assumptions that mean we we won't be able to predict the net rates of T and K formation.  The most important one is that we have neglected the reverse reactions $\ce{T -> R}$ and $\ce{K -> R}$; introducing them would give the same qualitative effect of cooling the temperature but would complicate the math considerably (but would let us wind up with net rates of T and K instead of the gross single-directional rates we have calcualted here.  Second we have assumed that $\kappa = 1$, which might not always be true.


What is the noble gas configuration of copper? Copper has an electron configuration of $\ce{[Ar] 3d^10 4s^1}$. Now sometimes the noble state is written as $\ce{[Ar] 3d^10 4s^1}$ or as $\ce{[Ar] 4s^2 3d^9}$.

Now the first noble state seems to be the same as his normal configuration and the latter seems to have equal electrons but divided in another way.

Now copper has two oxidation numbers $+1$ and $+2$ and as copper is d-element of the $4$ period it follows the $18$-electrons rule. But probably the $18$ electrons aren't created to get the krypton noble state. So how does the noble state configuration of copper work?

$\ce{[Ar] 4s^1 3d^10}$  is the correct configuration for copper. Why?

In simplest terms, some elements do not follow the Aufbau principle, so there are some alternate ways that electrons can arrange themselves that give these elements better stability.

Copper is a transition element and transition elements like either half filled orbitals or fully-filled orbitals. This makes them more stable.

For example:

Using the Aufbau principle, you would write the following electron configurations:

$\ce{Cr = [Ar] 4s^2 3d^4}$
$\ce{Cu = [Ar] 4s^2 3d^9}$

However, the actual electron configurations are:

$\ce{Cr = [Ar] 4s^1 3d^5}$
$\ce{Cu = [Ar] 4s^1 3d^10}$


Can the change in internal energy be nonzero if temperature is constant? I have learnt that internal energy, $U$, is a state function and it only depends on temperature... So if $\Delta T = 0$ then $\Delta U = 0$.

However when I was studying exothermic and endothermic reactions it was written in my textbook that at a constant temperature and pressure, $\Delta U$ is negative for exothermic reactions. How is that possible? Shouldn't $\Delta U$ be zero since the temperature is constant?

TL;DR: Do not just memorise thermodynamics equations! And if you have an issue with the equations $\Delta U = 0$ or $\Delta H = 0$ for an isothermal process, read on...

The first problem
You said that an exothermic reaction corresponds to $\Delta U < 0$. This is not true. It is defined by $\Delta H < 0$.
However, we should still discuss this issue because $H$ is also a state function and in some cases $H$ is dependent only on temperature.

How to study thermodynamics
This kind of issue in thermodynamics frequently crops up here, and it is a very common mistake amongst students to indiscriminately use equations that they have learnt.
So, you have an equation that says $\Delta U = 0$ for an isothermal process, i.e. one at constant $T$. The important question here is not "what is the equation" or "what is the answer"! Instead, you should be asking yourself "how do I derive this result", and the answer will naturally follow. This should really apply to everything you do - how can you expect to apply a formula that you do not actually understand?
Furthermore, if you are able to actually understand how an equation comes about, then you don't need to memorise it - you can derive it. For example, what's the point of memorising $w = -nRT\ln(V_2/V_1)$, along with the conditions, if you can simply obtain the equation from the very definition of work $đw = -\int p \,\mathrm{d}V$ by substituting in $p = nRT/V$ and integrating? What's more, the fact that you actually substituted in the ideal gas law should tell you something about the conditions that accompany the equations - for one, it's only applicable to ideal gases. What about when you reach the equation $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$ and happily memorise it, only to find out later that there are actually three more equations that look exactly like it? Are you going to memorise all four? No - you need to know where they come from.

Internal energy
In general, the internal energy $U$ of a substance is a function of at least two variables, for example $T$ and $p$. However, it can be shown using statistical mechanics that for an ideal gas,
$$U = n\left(\frac{3NRT}{2} + U_0\right)$$
Here, $N$ is the number of atoms in one molecule of the gas (for example, $N = 2$ for $\ce{H2}$, and $N = 3$ for $\ce{H2O}$). $U_0$ is the molar internal energy at absolute zero (this contains terms such as electronic energy), and crucially, this depends on the exact identity of the gas. This result is one example of the equipartition theorem in action (although the theorem itself is more powerful).
We do not need to concern ourselves with its derivation now. The point is that: this equation can only be applied to a fixed amount of a fixed ideal gas at a time. For example: if I have one mole of $\ce{H2}$ at $298\mathrm{~K}$, it is going to have a different internal energy from one mole of $\ce{Ar}$ at $298\mathrm{~K}$, even though the amount of substance and the temperature are the same.
What does this mean?
Firstly, it means that for anything that is not an ideal gas, you cannot assume that $U = U(T)$ and therefore you cannot assume that $\Delta T = 0$ implies $\Delta U = 0$. Let's say I compress $10\mathrm{~g}$ of water by reversibly increasing the pressure from $1\mathrm{~atm}$ to $10\mathrm{~atm}$, at a constant temperature of $323\mathrm{~K}$. Is the change in internal energy equal to zero? No, because water is not an ideal gas.
Secondly, it means that if the amount of ideal gas changes in any way, $U$ is going to change. Let's say I have a balloon filled with $1\mathrm{~mol}$ of argon gas, and I add a further $2\mathrm{~mol}$ argon gas into the balloon, all at a fixed temperature of $298\mathrm{~K}$. Is the change in internal energy equal to zero? No, in fact it will triple because you are changing $n$ from $1\mathrm{~mol}$ to $3\mathrm{~mol}$!
Lastly, it means that if the chemical composition is changing in any way, you are automatically not allowed to say that $\Delta T = 0$ implies $\Delta U = 0$. And this is the case even if all reactants and products are ideal gases. Why not? Let's look at this reaction, and let's assume that everything there behaves as an ideal gas, and let's say that the reaction vessel is kept at a constant temperature of $300\mathrm{~K}$.
$$\ce{H2 (g) + Cl2 (g) -> 2HCl (g)}$$
Let's write an expression for the internal energy of the reactants. On the left-hand side, we have:
$$U_\mathrm{reactants} = \left[\frac{3}{2}RT + U_0(\ce{H2})\right] + \left[\frac{3}{2}RT + U_0(\ce{Cl2})\right]$$
On the right-hand side, we have:
$$U_\mathrm{products} = 3RT + 2U_0(\ce{HCl})$$
So, is $\Delta U$ equal to zero? No; even though the multiples of $RT$ cancel out with each other, the values of $U_0$ for the reacting species are different, which makes
$$\Delta U = 2U_0(\ce{HCl}) - U_0(\ce{H2}) - U_0(\ce{Cl2}) \neq 0.$$

Enthalpy
Let's go back to our (closed) system where there is only one ideal gas and it's not undergoing any chemical reaction. Within this system, we can safely say that if $\Delta T = 0$, then $\Delta U = 0$. So:
$$\begin{align}
\Delta H &= \Delta (U + pV) \\
&= \Delta U + \Delta (pV) \\
&= \Delta (pV) \qquad \\
&= \Delta (nRT) \\
&= nR \Delta T \\
&= 0
\end{align}$$
We haven't made any assumptions here apart from the ideality of the gas (the fact that $n$ is constant came from the fact that it is a closed system). So, for an ideal gas, $\Delta T = 0$ implies $\Delta H = 0$.
However, we did use the fact that $\Delta U = 0$. Therefore, what I said about the cases above where $\Delta U \neq 0$ also applies here.
Regarding the terms endothermic and exothermic, they are used to describe chemical reactions. And for chemical reactions, there is absolutely no way you can say that constant temperature implies $\Delta U = 0$ or $\Delta H = 0$. That's why we can actually use the terms endothermic and exothermic. If every single chemical reaction had $\Delta H = 0$, life would be rather plain!


In a saturated solution, why is there ongoing dissolution and crystallisation? I was studying equilibrium, and had this question..

In a saturated solution with sugar dissolved, why is there a constant process of dissolving and crystallising? What I mean by this is, why don't the already dissolved sugar just stay that way, and the excess undissolved sugar stay that way? As I thought this would seem easier, and neater (?), wouldn't it??

Is it because the undissolved sugar is still in water, and that means it will want to dissolve, but since the solvent has reached maximum amount of solutes it has to kick some sugar out and hence there is a process of crystalisation??

I am sorry for a rather stupid question... But any explanation is much, much appreciated...

Because equilibrium is dynamic not static

There are many things in the world of chemistry which could be simpler but are not. Equilibrium is one of them.

There are few chemical processes where equilibrium is equivalent to the situation where everything just stops. Equilibrium normally occurs when the rates of processes going forward are the same as the rates going backward, not when things stop happening. In this case the rate of dissolution is the same as the rate of crystallisation.

On a molecular level things are very dynamic. Molecules in the solution are hitting crystals all the time. Molecules in the crystal are vibrating and being hit by molecules of the solvent: sometimes the molecules in the crystal have enough energy to escape from the crystal and go into solution. Equilibrium is the point where those two reactions have equal rates, not where all the activity stops.

We know this is true in real solutions from several experiments. We can, for example, label the molecules in the crystal with radioactive versions of one of their atoms (eg carbon 14). Even if we start with a saturated non-radioactive solution and add radioactive crystals of the same compound, the radioactivity will spread to the solution (see this question Proof of Dynamic Nature of Equilibrium).

The key thing is that all equilibria in chemistry are dynamic. The major difference in many cases is how dynamic. Some happen very slowly, others very quickly. But the position of the equilibrium only has a weak relationship to the speed of the equilibrium.


Do perfect gases react when placed in a container whose volume can vary? Suppose that $n$ moles of four generic perfect gases are made to react in a container, whose volume can vary, at a constant temperature and pressure, according  to the formula
$$\ce{A(g) + B(g)<=>C(g) + D(g)}$$
and suppose also the the equilibrium constant $K_c$ is known.

Now, if temperature and pressure are constant, then so will the density of the gases be. But if the density is constant, then the concentration of the gases will be constant as well, regardless of the number of particles present in the container (this is more or less the same argument that my chemistry textbook makes to justify the exclusion of concentration of solids and liquids from the expression of the equilibrium constant).

But if the concentrations are constant, then at the outset of the reaction, when there are $n$ moles of each gas, the reaction is already at equilibrium, because the reaction quotient $Q=K_c$.

My question is the following: is there something wrong about this argument? If not, how will the number of moles of each gas vary with time?

You have yourself all wrapped around the axle with odd conditions.

  Suppose that $n$ moles of four generic perfect gases...

This is a really odd way to specify the initial concentrations of four gases. Say I have 5 moles total. There are four gases, so I have four unknowns and only two equations (total moles and equilibrium equation) which is unsolvable.

This is also ambiguous. It is $n$ moles total, or $n$ moles for each gas?

  Now, if temperature and pressure are constant, then so will the density of the gases be. But if the density is constant, then the concentration of the gases will be constant as well, ...

Since you're holding temperature and pressure constant, and given the reaction:
$$\ce{A(g) + B(g)<=>C(g) + D(g)}$$
then the density will always be the same. Even if you have all A and B which are totally converted to C and D, the density of the gas mixture will be the same (assuming ideal gas behavior which you stipulated).

If at a constant volume and the reaction had been:
$$\ce{2A(g) + B(g)<=>C(g) + D(g)}$$
then the density would change as the equilibrium shifted.

But since you stipulated constant temperature and pressure then the volume would also have to change as $\ce{2A(g) + B(g)}$ was converted to $\ce{C(g) + D(g)}$ to maintain constant pressure, so the overall density of the gas remains constant.

  But if the concentrations are constant, then at the outset of the reaction, when there are n moles of each gas, the reaction is already at equilibrium, because the reaction quotient $Q=K_c$.

Let's break this down. If the concentrations are constant, then the reaction is already at equilibrium.

It does not necessarily follow that when there are $n$ moles of each gas then the reaction is at equilibrium. That would only be true if $K_c = 1$, but $K_c$ has not been specified.

  If not, how will the number of moles of each gas vary with time?

Ok, let's assume $n$ moles of each of the four gases. You'd also need to provide the rate constants for the forward and reverse reactions to calculate concentrations over time. Without the rate constants it isn't clear if the equilibrium will be established in seconds or years.


Why does dry boiling discolor steel cookware? I have a stainless steel kettle that I let boil dry. That is, I left it on a hot coil burner with nothing in it, and noticed a few minutes later that the bottom was red hot and smoking. To make matters worse, I panicked and quenched it with cool water.

Both the inner and outer surface were left discolored with various shades of yellow and purple. This seems to be a known consequence of overheating, but I can't find an explanation of what actually happened.

I understand why heat (especially combined with quenching) would warp the metal, but why would it cause discoloration (that persists after the metal has cooled)?
After hunting around, I saw several suggestions that boiling white vinegar could remove the discoloration, and to my suprise, it worked pretty well. Why does this work?

Stainless steel have many different compositions. If you heat it up the surface undergoes oxidation. Diffusion at higher temperatures plays also its roll - shifts surface composition of the kettle. The various shades of yellow and purple are metal oxides. The vinegar removes the oxides, thus removes the coloration. Vinegar should remove the oxides even without boiling. It will just take more time.


Why isn't cyclobutane planar? In short, cyclobutane adopts a certain, non-planar, configuration in order to reduce ring strain caused by the fact that carbons are $\mathrm{sp^3}$ hybridised. That much is understandable.

In an exercise it was suggested to propose another kind of hybrisation for the carbons in order to justify a (theoretical) planar configuration. I thought that carbons could be $\mathrm{sp}$ hybridised. This way, $\mathrm{p}$ orbitals could form the $\ce{C-C}$ bonds and $\mathrm{sp}$ orbitals could form the $\ce{C-H}$ bonds.

I checked the answer and I was right. However, I couldn't think of something that could justify why this particular structure is not possible. That exactly is my question.

TL;DR Planar structure is definitely possible, and more so in solid state as a statistically averaged geometry in space and time (in gaseous state ring is bent by $35^\circ$). Cyclobutane can be approximately attributed with $\mathrm{sp^4}$/$\mathrm{sp^5}$ hybridization, and the strain caused by non-bonded $\ce{H...H}$ repulsions ("eclipsing") can be reduced to minimum by introducing ring substituents:

substituing $\ce{H}$ with $\ce{OH}$ (octahydroxycyclobutane, planar);
substituing $\ce{C}$ with $\ce{O}$ (oxetane, dihedral angle $10^\circ$).

Single-crystal diffraction studies of octahydroxycyclobutane, an $\ce{OH}$-substituted derivative, in late 1960s showed that unit cell symmetry implies a planar ring; further it's been supported by possible H-bonds formation [1]. A hypothesis that a statistically disordered nonplanar ring may exist has been rejected as it possessed worse agreement with experimental data.

20 years later crystal structure of cyclobutane at low temperature ($\pu{117 K}$) has also been elucidated [2]. From [2] and references therein:

  The gas-phase equilibrium conformation, as derived from electron diffraction and spectroscopic experiments, is nonplanar with a dihedral angle of about $35^\circ$ and an inversion barrier of only $\pu{6 kJ mol^-1}$. The puckering of the molecule is observed in Raman-spectra, and the corresponding absorption band
  is still present in the solid at $\pu{113 K}$. Theoretical calculations using ab initio methods also result in a bent conformation as the energetic minimum.

  As one would expect from the shallow potential of the ring inversion, there are various planar and bent derivatives, and it does not seem possible to predict the ring conformation from the substitution pattern.

In contrast with octahydroxycyclobutane, a statistically disordered model was considered more reliable. Although no particular close $\ce{H-H}$ contacts were observed, an averaged ring is considered planar:

Most importantly, electron density map suggests rather increased $\mathrm{p}$-character in hybridisation (like $\mathrm{sp^5}$ in case of cyclopropane) with higher $\pi$-character of $\ce{C-C}$ bonds in comparison with the one expected for expected for alkanes:

  The residual electron density shows significant maxima in positions between the carbon atoms. These maxima are displaced tangentially by $\pu{0.10(3) A}$ to the outside of the $\ce{C-C}$ bond vectors of the time- and space-averaged molecule, indicating a "banana bond" type electronic arrangement which is frequently found in highly strained small ring systems. In cyclopropane the covalent bonding maxima are displaced by $\pu{0.2 A}$ from the $\ce{C-C}$
  vectors of the carbon triangle. Since in cyclobutane there is reduced ring strain our observation is in good agreement with the cyclopropane findings.

This can be considered as an experimental confirmation of predicted earlier  hybridization of cyclobutane ($\psi(\ce{C-C}) = \mathrm{sp^{4.28}}$, $\psi(\ce{C-H}) = \mathrm{sp^{2.19}}$) [3]:

  The difference between the bond overlaps of the planar and non-planar models is very small. [...] This clearly indicates that other interactions (even if small), must be dominant in causing non-planarity of the molecule. Non-bonded repulsions make a dominant contribution in determining conformations in acyclic and larger cyclic systems (e. g. ethane, cyclohexane), and they favour staggered arrangements of $\ce{CH}$-bonds. In a planar model of cyclobutane all hydrogen atoms are in eclipsed positions, this is a cause of additional strain and is energetically less favourable. A relieve is obtained by bending the skeleton of the molecule, since the $\ce{CH}$ bonds are then approaching a staggered conformation.

Cyclobutane, just as cyclopropane, is a strained cycle with the bent "banana" bonds with intermediate character between $\sigma$ and $\pi$, which allows for mimicking the behavior of double-bond compounds, such as decolorization of bromine water [4, p. 195], even though protons in cyclobutane are less acidic  than in cyclopropane:

  Four-membered rings also exhibit angle strain, but much less than three-membered rings, and for that reason are less easily opened. Cyclobutane is more resistant than cyclopropane to bromination, and although it can be hydrogenated to butane, more strenuous conditions are required.

Bibliography

Bock, C. M. J. Am. Chem. Soc. 1968, 90 (11), 2748–2751. DOI 10.1021/ja01013a002.
Stein, A.; Lehmann, C. W.; Luger, P. J. Am. Chem. Soc. 1992, 114 (20), 7684–7687. DOI 10.1021/ja00046a012.
Maksić, Z.; Klasinc, L.; Randić, M. Theoretical Chemistry Accounts: Theory, Computation, and Modeling (Theoretica Chimica Acta) 1966, 4 (3), 273–286.
Smith, M. March’s advanced organic chemistry: reactions, mechanisms, and structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013.


What will happen if I put this chemical into my water tank? There is an 8000 liter water tank on top of my house which provides water for the house.

Bacteria, fungi, algae, etc often start developing in the tank, giving unpleasant smells, etc.

This chemical seems to be chlorine based. According to the bottle, the composition is:

  Sodium hypochlorite 5% max.

  Sodium hydroxide 1% max.

  Amine oxide 1% max.

I'm thinking of maybe putting 50 ml into the tank about once a week.

Nobody is drinking or cooking with the water. It's used for stuff like laundry, showers, washing dishes, etc.

I know it seems like a strange idea, but I live in the countryside in a developing country and it's not easy to find stuff here.

The water is ground water that we pump up there with our own pump.

What's gonna happen? Should I decrease or increase the amount? Or drop the idea entirely? (And maybe do something else?)

It's very hot here, maybe 40 celsius in the shade. This tank is in the baking sunlight.

The product which you have mentioned in your post may have other chemicals like fragrance etc... which may make the water not suitable for washing dishes.

You can purchase sodium hypochlorite (bleach) which is sold separately as powder or solution and add to the water tank. The concentration added to water tank is crucial.

You can use unscented / fragrance free laundry bleach solution also. This will be easier to find.
If you get solution (6% solution of bleach in water) then you can add 40mL to the 8000L full tank first time and mix.

Then, say after a week when your tank has 2400 Liters full (using 800L per day ); you can add 28-30mL again and then fill up tank to full (8000L). (Approximate ratio is 1-1.5mL per 200 Liters of water).

Preventing algal growth:

To prevent the growth of algae it is always recommended to prevent light exposure of stored water. Dark colored/black tanks do well. You can cover the manholes/ crevices through which light may enter the tank.
Additionally, warm water aids in algal growth, so you can try to cover the tank with some tarpaulin etc... to prevent heating.
Regular cleaning is important. Algae already growing on the inner wall of the tank is difficult to to deal with. Proper regular cleaning will help to remove such growth.


How to determine which of two similar compounds has the larger dipole moment? I am solving some problems involving the order of dipole moments of some compounds. I could solve most of them but the following:

Which has greater dipole moment?

$\ce{CH3F}, \ce{CH_3Cl}$

Since fluorine is more electronegative my assumption was that $\ce{CH3F}$ should have a larger dipole moment.
On the other hand since the angle between the dipole moments is less in 3-nitro phenol than in 4-nitrophenol, my assumption was that 3-nitro phenol has a greater dipole moment. But this contradicts the answer given.

The dipole is the vector sum of each atoms's partial charge times its distance from the origin. Lacking this information individual bond dipole values can be used

In the first case it is that of $\ce{CH3}$ vs. the halogen, and then which halogen has the greater electronegativity. Looking up some bond values it seems that a $\ce{C-F}$ bond has the slightly larger dipole ($1.6$ vs $1.5$ Debye ) so the difference is small and not easy to predict as the C-F bond is shorter than the C-Cl.

In the second molecule, geometry can be used as the two groups are the same on each molecule but positioned differently. The para molecule will have the smaller dipole vs. the meta if both groups act in the same direction, either both to withdraw or add electron density with respect to the ring, as then they act to cancel one another to some extent.

If they act in a different direction then the para molecule will have the larger dipole vs the meta. Looking for dipole moments it seems that phenol has a dipole of $1.45$D pointing into the ring and nitrophenol $3.95$D out of the ring so act in different directions and so the para should have the larger dipole.

(To add two vectors of length b and c at an angle $\theta$ the resultant c is found using the cosine rule $c^2 = a^2 + b^2 -2ab\cos(\theta)$ . If the dipoles point in the opposite direction make either a or b negative)


Why does the weaker C–H bond have a higher wavenumber than the C=O bond? My understanding is that a stronger bond has a higher wavenumber in IR spectrum. But why does the C–H vibration have a higher wavenumber than the C=O vibration? The latter is a double bond, so I think it should be a stronger bond than the C–H single bond.

A property of the harmonic oscillator is that the oscillation frequency, $\omega$, is dependent not only on $k$ (the spring constant) but also on the mass $m$ of the object:

$$\omega = \sqrt{\frac{k}{m}}$$

We can crudely model a chemical bond as a two-body harmonic oscillator, which largely obeys the same rule, except that the mass must be replaced with the reduced mass $\mu$, defined by:

$$\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}$$

where $m_1$ and $m_2$ are the masses of the two bodies. So, for a C–H bond, we have

$$\frac{1}{\mu_\ce{CH}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{1 u}} \implies \mu_\ce{CH} = \pu{0.923 u}$$

and for a C=O bond, we have

$$\frac{1}{\mu_\ce{CO}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{16 u}} \implies \mu_\ce{CO} = \pu{6.86 u}$$

From J. Chem. Phys. 1946, 14 (5), 305–320 the force constant $k$ for the C=O bond in formaldehyde is approximately $2.46$ times that for the C–H bond in methane (the units are in the old cgs system, so I am lazy to quote the actual values).

So, we can come up with a very rough theoretical estimate. The wavenumber used in IR spectroscopy, denoted by $\bar{\nu}$, is directly proportional to the (angular) frequency $\omega$.

$$\begin{align}
\frac{\bar{\nu}_\ce{CH}}{\bar{\nu}_\ce{CO}} &= \frac{\omega_\ce{CH} / 2\pi c}{\omega_\ce{CO} / 2\pi c } \\
&= \frac{\omega_\ce{CH}}{\omega_\ce{CO}} \\
&= \sqrt{\frac{k_\ce{CH}}{k_\ce{CO}} \cdot \frac{\mu_\ce{CO}}{\mu_\ce{CH}}} \\
&= \sqrt{\frac{1}{2.46} \cdot \frac{\pu{6.86 u}}{\pu{0.923 u}}} \\
&= 1.74
\end{align}$$

This is not too far from the experimental value of $\pu{2900 cm-1}/\pu{1730 cm-1} = 1.68$ (using values in the middle of the typical ranges for alkanes and aldehydes).


Hybridisation of terminal nitrogen in diazomethane I have a few questions about the terminal nitrogen (highlighted in red) in diazomethane, $\ce{CH2N2}$.

Is that nitrogen $\mathrm{sp}$ or $\mathrm{sp^2}$ hybridised?
What type of orbitals do the lone pairs on that nitrogen reside in?
Does the negative charge on nitrogen undergo resonance?

As usual with these small molecules, the bonding situation is a lot more complicated than expected. In this very molecule this is to the extend that not only the Lewis formalism, but also the conventional resonance formalism breaks down.
However, Jan's description is sound, and provides a good starting point for further analysis.
Unfortunately formulating a clear cut valence bond description is difficult, because it involves strongly coupled electrons. This is in part due to the high symmetry of the molecule, $C_\mathrm{2v}$, and the even higher local symmetry of the $\ce{CNN}$ moiety, $C_\infty$. Because of the nature of molecular orbital theory, i.e. being better at describing delocalisation, it is actually easier to understand.

Here are the molecular orbitals computed at the DF-M06L/def2-SVP level of theory:

And here is the same thing in numbers

 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-14.323 is N2-s=1.00
 Alpha occ 2 OE=-14.252 is N1-s=1.00
 Alpha occ 3 OE=-10.099 is C3-s=1.00
 Alpha occ 4 OE=-1.055 is N2-s=0.51 N1-s=0.29 N1-p=0.10
 Alpha occ 5 OE=-0.802 is N2-p=0.31 C3-s=0.31 N1-s=0.14 N2-s=0.12 C3-p=0.07
 Alpha occ 6 OE=-0.518 is C3-p=0.26 C3-s=0.24 H5-s=0.16 H4-s=0.16 N1-s=0.07 N2-s=0.06
 Alpha occ 7 OE=-0.487 is N2-p=0.45 C3-p=0.24 N1-p=0.14 H5-s=0.08 H4-s=0.08
 Alpha occ 8 OE=-0.450 is N2-p=0.61 N1-p=0.29 C3-p=0.09
 Alpha occ 9 OE=-0.407 is N1-s=0.44 N1-p=0.41 N2-p=0.08
 Alpha occ 10 OE=-0.366 is N1-p=0.28 C3-p=0.24 N2-p=0.16 H5-s=0.15 H4-s=0.15
 Alpha occ 11 OE=-0.206 is C3-p=0.57 N1-p=0.38
 Alpha vir 12 OE=-0.082 is N1-p=0.55 N2-p=0.32 H5-s=0.06 H4-s=0.06
 Alpha vir 13 OE=-0.005 is C3-p=0.33 N2-p=0.33 N1-p=0.32

From the visual inspection we can see that the terminal nitrogen has only one orbital that could be classified as a lone pair, i.e mo 9; the hybridisation according to this calculation is approximately sp. There are three bonding π orbitals, i.e. mo 7, 8, 10. And there is also one anti-bonding π orbital, i.e. mo 11. The sigma bond is formed by two bonding (mo 4, 5) and one anti-bonding orbital (mo 6).
This is no coincidence, but pretty much always the case for terminal atoms.

Due to symmetry constraints, the hybridisation pattern at terminal atoms is usually:

for the orbital forming the single bond $\mathrm{sp}^{\geq1}$
two orthogonal (π) lone pairs $2\times\mathrm{p}$
one (σ) lone pair $\mathrm{sp}^{\leq1}$

Localisation of the orbitals is not easy for this compound. I performed a natural bond orbital analysis, which transforms the (delocalised) canonical orbitals into hybridorbital like ones, but it results in a low occupied carbon lone pair orbital and a partially occupied anti-bonding nitrogen-nitrogen π orbital.

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   1. - 3. core orbitals (omitted)
   4. (1.97583) LP ( 1) N  1            s( 67.37%)p 0.48( 32.60%)d 0.00(  0.03%)
   5. (1.33940) LP ( 1) C  3            s(  0.00%)p 1.00( 99.92%)d 0.00(  0.08%)
   6. (1.99655) BD ( 1) N  1- N  2
               ( 43.82%)   0.6620* N  1 s( 33.02%)p 2.02( 66.72%)d 0.01(  0.26%)
               ( 56.18%)   0.7495* N  2 s( 46.00%)p 1.17( 53.92%)d 0.00(  0.08%)
   7. (1.99587) BD ( 2) N  1- N  2
               ( 51.93%)   0.7206* N  1 s(  0.00%)p 1.00( 99.63%)d 0.00(  0.37%)
               ( 48.07%)   0.6933* N  2 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
   8. (1.98025) BD ( 3) N  1- N  2
               ( 39.11%)   0.6254* N  1 s(  0.00%)p 1.00( 99.54%)d 0.00(  0.46%)
               ( 60.89%)   0.7803* N  2 s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)
   9. (1.99580) BD ( 1) N  2- C  3
               ( 65.65%)   0.8103* N  2 s( 53.76%)p 0.86( 46.18%)d 0.00(  0.05%)
               ( 34.35%)   0.5861* C  3 s( 30.11%)p 2.32( 69.75%)d 0.00(  0.14%)
  10. (1.95419) BD ( 1) C  3- H  4
               ( 60.92%)   0.7805* C  3 s( 34.95%)p 1.86( 65.03%)d 0.00(  0.02%)
               ( 39.08%)   0.6251* H  4 s( 99.91%)p 0.00(  0.09%)
  11. (1.95419) BD ( 1) C  3- H  5
               ( 60.92%)   0.7805* C  3 s( 34.95%)p 1.86( 65.03%)d 0.00(  0.02%)
               ( 39.08%)   0.6251* H  5 s( 99.91%)p 0.00(  0.09%)
 ---------------- non-Lewis ----------------------------------------------------
  12. (0.01076) BD*( 1) N  1- N  2
               ( 56.18%)   0.7495* N  1 s( 33.02%)p 2.02( 66.72%)d 0.01(  0.26%)
               ( 43.82%)  -0.6620* N  2 s( 46.00%)p 1.17( 53.92%)d 0.00(  0.08%)
  13. (0.64416) BD*( 2) N  1- N  2
               ( 48.07%)   0.6933* N  1 s(  0.00%)p 1.00( 99.63%)d 0.00(  0.37%)
               ( 51.93%)  -0.7206* N  2 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  14. (0.07754) BD*( 3) N  1- N  2
               ( 60.89%)   0.7803* N  1 s(  0.00%)p 1.00( 99.54%)d 0.00(  0.46%)
               ( 39.11%)  -0.6254* N  2 s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)

This is usually an indication of a strongly delocalised system such as a three-centre bond. Investigating the molecule with natural resonance theory gives the rather odd contribution of long range N1 - C3 bond.

 TOPO matrix for the leading resonance structure:

     Atom  1   2   3   4   5
     ---- --- --- --- --- ---
   1.  N   2   2   0   0   0
   2.  N   2   0   2   0   0
   3.  C   0   2   0   1   1
   4.  H   0   0   1   0   0
   5.  H   0   0   1   0   0

         Resonance
    RS   Weight(%)                  Added(Removed)
 ---------------------------------------------------------------------------
    1*     33.02
    2*(2)  32.31    N  1- N  2, ( N  2- C  3), ( N  1),  C  3
    3*(2)  28.81    N  1- C  3, ( N  2- C  3), ( N  1),  N  2
    minor contributions omitted
 ---------------------------------------------------------------------------
          100.00   * Total *                [* = reference structure]

According to NBO, the terminal nitrogen has an $\mathrm{sp}^{0.5}$ σ-type lone pair orbital, two $\mathrm{p}$ π-type bonding orbitals, and a $\mathrm{sp}^2$ σ-type bonding orbital.

Hybridisation is a mathematical tool to analyse a bonding situation. There is no correct way of applying it. It is also very important to understand an atom is never hybridised. (Even though organic chemists tend to use this jargon - even in textbooks.) The orbitals at an atom can be hybridised.


Why do complex ions not emit light from de-excitations of electrons? I understand that complex ions are coloured due to d-orbital splitting which results in electrons being able to absorb wavelengths of visible light and become excited to the higher energy state meaning the transmitted light is coloured. However I don't see why the electron would not simply de-excite and emit the same wavelength of light that it absorbed which would result in no visible difference. Is there some reason why the same wavelength is not emitted or perhaps the electron does not de-excite at all?

The fate of electronically excited states is not immediately obvious.

First of all the rate at which the electron emits a photon and falls down a level is dependent upon the energy difference between the two levels, it is only an expectation value of a macroscopic observable, on an individual atomic scale it is impossible to predict when the electron in the upper level will emit a photon, however multiple instances of the same measurement will tend towards the expectation value. The direct result of this is that there is a path difference between the emitted photon and the 'rest' of the photons that were not absorbed by the complex and the result is a different interference between photons in the transmitted light compared to the incident light and hence a new colour is observed.

There is another fate of excited electrons, the Franck-Condon principle invokes the very plausible assumption that the nuclei do not respond immediately (at least on the time scale of electronic transitions) to the new equilibrium bond position that is determined by the new electronic state of the complex. This means the excited electronic state will most likely not be in the lowest vibrational state of the new equilibrium bond configuration, the splitting of vibrational energy levels is orders of magnitude lower than those of electronic transitions and hence the same law that governs electronic emission applies, the electron is much more likely to fall down these vibrational energy levels first, without emitting photons of visible light.

The electron will then be expected to, by the time it is occupying the lowest vibrational state of the excited electronic state, invoke a transition down to the ground state energy level. The energy of the excited state is described in a basic form by
$$H=E^*+v$$

where $H$ is the energy of the absorbed photon, $E^*$ is the energy of the excited state in the lowest vibrational mode and $v$ is the contribution from the initial vibrational state of the excited complex.


How can one distinguish an IR spectroscopy alcohol band on an organic molecule versus the alcohol bands from water in the sample? I'm reading this IR spectra and trying to determine if the large broad band at 3388 is alcohol from an intermediate or from water in the sample. The details of the reaction, conditions, and mechanism can be found below if it's helpful.

I attempted to run a double aldol condensation of p-anisaldehyde and acetone using 2 mL of 3 M NaOH, 0.3 mL of p-anisaldehyde, and 1.6 mL of 0.625 M acetone in ethanol. I let the reaction run for 30 minutes at room temperature and recrystallized the product.

I drew all of the steps I think the reaction mechanism could have stopped that have an alcohol here:

Thanks for any advice.

The main working rule of a good FTIR spectrum is that everything must be bone dry! The absorbance in IR is additive just like Beer's law, i.e. your spectra is the sum of water traces in your sample as well as your suspected alcohol. There is no simple way to deconvolute it. Dry your product thoroughly in a vacuum dessicator for at least a day and re-run the IR. If your big -OH band vanishes, it means your product was not dry. If it persists, your product likely has an hydroxide group.

KBr must be oven dried well before use as well.


Differentiating between Titration, Buffer, diluted buffer, Deionized H₂O I had a chemistry lab where we had to titrate three solutions with $\ce{HCl}$: a buffer, a diluted buffer and deionized water. My partner and I didn't label the charts we made, so now I have to figure out which one is which.

The way I see is that from the Henderson-Hasselbalch equation:
 $\mathrm{p}\ce{H} = \mathrm{p}K_a + \log\left(\ce{[A^{-}]/[HA]}\right)$,
 the diluted buffer should have a larger slope at the beginning then the original buffer solution.

My question is, is my logic correct, and how do I differentiate between the original buffer and the deionized water?

What is the expected pH of deionized water and what would the pH be after adding a few drops of acid? For both buffer solutions, there is an equivalence point, how would you use that to tell them apart? Usually deionized water is slightly acidic. I kind of mentioned that water doesn't have an equivalence point. Is there a graph without one?

What is the purpose of a buffer? A diluted buffer would have the same characteristic, but in a smaller amount. Water would be missing this characteristic. You should be able to figure out which graph is which now.


What does it mean to say that a strong base is only slightly soluble? I've often seen the hydroxides of Calcium, Strontium and Barium described as "strong bases that are only slightly soluble," or similar.  But strong bases dissociate completely in solution, so I would guess this means that such hydroxides dissolve only a small amount, but the dissolved portion dissociates completely.

But the problem with this interpretation is that these hydroxides are ionic compounds, and my understanding is that "dissolve" is essentially equivalent to "dissociate" for ionic compounds, i.e. ionic compounds dissolve by dissociating.

If there's no distinction between dissolving and dissociating for ionic compounds, what does the description "strong base that is only slightly soluble" mean?  How does it differ from the term "weak base?"

If there is a distinction, and it is possible for ionic compounds to dissolve without dissociating, how does such dissolution work?  I have a model for the distinction between dissolution and dissociation of covalent compounds: first the compound dissolves into separate molecules, then each molecule dissociates into two or more ions.  If ionic compounds can dissolve without dissociating, into what sort of particles do they dissolve? Formula units? Small clusters of ions?  Something else?

Solubility is different from dissolution, which is itself different from dissociation.

Solubility tells you the extent to which a substance dissolves. Dissolution is the kinetic process of solvation, when solvent molecules interact with solute molecules or ions. Dissociation is, from IUPAC,

  The separation of a molecular entity into two or more molecular entities (or any similar separation within a polyatomic molecular entity). Examples include unimolecular heterolysis and homolysis, and the separation of the constituents of an ion pair into free ions.
  The separation of the constituents of any aggregate of molecular entities.

Dissociation is the mode of solvation for ionic compounds.

I think your confusion stems from associating solubility and dissolution improperly. ChemGuide directly answers your question here:

  Some strong bases like calcium hydroxide aren't very soluble in water. That doesn't matter - what does dissolve is still 100% ionised into calcium ions and hydroxide ions. Calcium hydroxide still counts as a strong base because of that 100% ionisation.


Why does oxidized copper become green? Why does oxidized copper turn green? I am sorry if this question is too basic, but am really stumped here. It would be helpful if one could provide a research paper or website with your answer.

Copper oxide is not green, but green verdigris, basic copper carbonate (or acetate) forms on copper exposed to air and moisture.

Heating copper in air can form a near-transparent film of copper oxide, which can exhibit all the colors of the spectrum due to interference of light. The same effect cause a "rainbow" in oil slicks on water.


paramagnetism vs. dimagnetism this was a question for my chem class- and for some reason the answer was the last option. Why isn't it $\ce{Mn}$ which has the most unpaired electrons?

Which of the following species would you expect to be most
paramagnetic?
1.  $\ce{Mn}$
2.  $\ce{Mn^2+}$
3.  $\ce{Fe^3+}$
4. Both $\ce{Mn^2+}$ and $\ce{Fe^3+}$ are the same
and more paramagnetic than $\ce{Mn}$
5.  All have the same paramagnetism.

Para-magnetism: Para-magnetism refers to the electronic configuration of an atom having one or more unpaired electron. That means an atom can be called para-magnetic if it has one or more unpaired electron. Now if atom A and B has no unpaired electron they have same para-magnetism (in this case zero). If they both have one unpaired electron still they have same para-magnetism (Here it's one).
Now look at the electronic configuration of the above sated options::
$Mn :: [Ar]3d^54s^2$
$Mn^{2+} :: [Ar] 3d^5$
$Fe :: [Ar]3d^64s^2$
$Fe^{3+} :: [Ar] 3d^5$
So they all have five unpaired electron. That's why they have same paramagnetism.


Application of Nernst Equation in Potentiometric titration of Fe2+ with Ce4+ It is well Known that to find $E_{cell}$ during titration of  $Fe^{2+}$ with $Ce^{4+}$ we consider 3 domains:

Before Equivalance point ,  $E_+ = E_{Fe^{3+}/Fe^{2+}}^o+\frac{RT}{nF}log\frac{[Fe^{3+}]}{[Fe^{2+}]} \tag{1}$

At Equivalance point , $E_+=\frac{E_{Fe^{3+}/Fe^{2+}}+E_{Ce^{4+}/Ce^{3+}}}{2} \tag{2}$

After  Equivalance point ,$E_+ = E_{Ce^{4+}/Ce^{3+}}^o+\frac{RT}{nF}log\frac{[Ce^{4+}]}{[Ce^{3+}]} \tag{3}$

$E_+$ is the electrode potential of the positive electrode where the negative electrode is a fixed reference electrode.

In the book Quantitative Chemical Analysis by Daniel C Harris, Chapter 15 says that both equation 1 and 3 can be used either of the two domains and the reason why one is used instead of the other is due to the fact that the ratio of the concentration of a particular metal ion in different oxidation state is much easily found in that domain while the other ratio usually requires a complicted equilibrium equation to be solved to obtain the solution.

my questions:

Does this mean that if by some means I knew the ratio of concentrations of both metals I could substitute it in eqn 1 and eqn 3 and end up getting the same value for $E_+$.
If not does it only hold near the Equivalance point?
If one is True , The only explanation I have is that both $Fe^{2+}$ and $Ce ^{4+}$ are in equilibrium with the same electrode. But this still doesn't make sense as usually when 2 reactions are occurring at the same container the $\Delta G_{net}=\Delta G_{1}+\Delta G_{2} $ So in some sense shouldn't $ E_+ =E_{due-to-Fe}+E_{due-to-Ce}$ be the equation we require?

If I'm missing some essential concept it would be Great if someone could point it out and probably provide some links for me to look up.

It is an interesting question. Daniel Harris is revising his book with my former mentor. Hope he clarifies this section in the revised version. Your point number 1 is misleading. The reason is that before the titration, theoretically there is no Fe(III). So Nernst equation should not be used- electrode potential is infinite (log 0 is undefined). When you start the titration (Ce (IV) in the burette), the moment you add Ce(IV), it is immediately reduced by Fe(II) to Ce(III). You do not have the Ce(IV/III) couple in the solution. Before the endpoint, Fe(III/II) couple controls the potential of the indicator electrode.

Do you remember what that indicator electrode is made of? It is just a platinum wire and the reference electrode is usually silver-silver chloride electrode, all enclosed in a single unit. The platinum wire is oblivious to iron or cerium. All it "senses" is the charge. Imagine a single Pt electrode dipped in an iron solution and forget about the presence of cerium. Imagine that a metal is a source (sea) of electrons.

The moment you dip the electrode, instantaneous exchange of electrons takes place between the metal and the reducible or oxidizable ions. Electrons can be released by Fe(II) to form Fe(III) and similarly electrons can be gained by Fe(III) to form Fe(II). This exchange rate is dependent on the concentration ratios of Fe(II) and Fe(III).

As a result of this exchange, the Pt metal develops a charge. It is very surprising that you only need to remove a very very very small number of electrons to develop a measurable surface charge. Hence, there is no perceptible change in the concentration of Fe ions. If platinum electrode develops a charge, the interface (solution electrode boundary) develops an opposite charge until equilibrium is achieved.

Now given the fact that Pt electrode is charged i.e., if we connect Pt to another electrode, we can sense the potential difference between the Pt electrode and another electrode, which can be any other electrode. Now that you wish to have a stable. For that purpose you need an electrode whose potential is fixed i.e. a reference electrode.


Will the carboxylate groups of the citrate anion undergo hydrogen bonding? The citrate anion of trisodium citrate has three carboxylate groups. I am concerned with whether or not there will be hydrogen bonding in water, with hydroxyl groups of other ions or alcohols.

It has been suggested to me that the negative carboxylate oxygens won't hydrogen bond, due to resonance delocalisation of the charge. If it mattered whether or not the oxygen was partially charged, I would hypothesize that the oxygen is charged, because carbon has a lower electronegativity.
The oxygen should have two lone pairs, yet it has either been implied that they are not there, or for some reason that they will not interact in hydrogen bonding.

My question is: will the citrate anion hydrogen bond with other molecules or polymers with hydroxyl groups, in aqueous solution?

It has been made aware to me that there is an article that has done research to show that carboxylates are strong hydrogen bond acceptors in anhydrous solvents.

As the comments indicate, you don't need a formal charge on oxygen at all.  "Neutral" oxygen atoms are usually negatively charged anyway just from bond polarity, as in water.  That and the lone pairs on oxygen are quite enough to enable hydrogen bonds to protic hydrogen.


Calorimeter and bond energy Why does 1,3-Butadiene has different standard heat of formation value calculated from calorimeter than the one calculated from bond energies?

Its because of the resonance. buta-1,3-diene is resonance stabilized. The Heat of combustion actually will be less than calculated theoretically because some of energy released will be used to destabilize the molecule.