Acetaldehyde with few drops of concentrated sulphuric acid Acetaldehyde on treatment with a few drops of concentrated $\ce{H2SO4}$ gives?

Attempt:
First, I formed an enol intermediate in acidic medium $\ce{CH2=CHOH}$. Then, I performed electrophilic addition of another acetaldehyde molecule on this enol intermediate to get $\ce{CH3CH(OH)CH2CHO}$.
But answer given in my book is:

Please let me know the error in my attempt.

You're not wrong.

Your mechanism describes the classic acid catalyzed synthesis of 3-hydroxybutanal by dimerisation of acetaldehyde (Borodin 1869, Wurtz 1872).

Image: Whoopie23, Wikipedia

However, under different conditions acetaldehyde forms different addion and condensation products. E.g. with concentrated sulphuric acid at low temperatures metaldehyde is formed whereas at room temperature paraldehyde (the answer given in your book) is formed.


Controled burn colored fire and safety I am a martial artist who uses fire weapons in my shows. I am wondering how I can get the flames to be different colors.

I've done some research about it, and have found that the salts of some metals can be used for the intended purpose (for instance, copper sulphate burns green).

However, I use kerosene as my fuel and am wondering if it is advisable to mix the chemicals with it (kerosene) without burning myself. Also, I would like to know how to obtain the chemicals and what amounts should be added to a given volume of fuel.

I've done some research about it, and have found that the salts of some metals can be used for the intended purpose (for instance, copper sulphate burns green).

You're right! Various melts salts are known to colourize a flame. In the lab and/or for demonstration purposes, this is typically achieved by holding a magnesium oxide stick with a bit of the salt attached into the flame of a Bunsen Burner. There's a wikipedia page on Flame tests that shows some possible effects.

    I use kerosene as my fuel[…]

Here's a problem. The salts you could use soluble in water, but insoluble in a solvent as unpolar as kerosene! Consequently, you might want to give a flammable, but more polar solvent a try.

Is it conceivable to change the flammable solvent used in your performances? I reckon that you might achieve the desired effect using ethanol or methanol. With respect to the toxicity of both solvents, I'd definitely try ethanol first!


Does the emf of an electrode potential change over time? In an electrode potential with a salt bridge and high resistant voltmeter, does the emf change over time as the reactions occur?

Yes. Remember that the standard electrode potentials are for 298K at 1 atm of pressure and 1N concentration of reagents. As the concentration of ions to be reduced at the cathode decreases the emf will fall as well and is reflected in the Nernst equation.$$E_{cell} = E^\circ_{cell} - \frac{RT}{nF}ln(Q)$$


Reaction between lye and water? I am planning on making some homemade soap soon and have read online that breathing in the fume generated from the reaction between lye and water is not recommended because it is "hazardous."

I am a little bit confused because I thought the reaction between lye and water is simply a highly exothermic dissolution reaction and therefore the "fumes" would consist of only steam?

                           NaOH + H2O -> OH- + Na+ + H2O + HEAT

If this fume is corrosive, does that mean there is some gaseous hydroxide as well generated from the heat? Why else would this fume be dangerous?

The fumes will contain trace amounts of NaOH. The fumes are microscopic water droplets, and these droplets will still be caustic, as they are essentially a solution of sodium hydroxide.

There is no real chemical reaction going on here, but rather a dissociation of the sodium and hydroxide ions, which is an exothermic process. Solid sodium hydroxide is 'hazardous', the resulting solution is 'hazardous' and the fumes above the dissolving lye are 'hazardous'.


Are monosubstituted cyclopropanes achiral or chiral? A classmate an I were discussing the chirality of this molecule which is a monosubsituted cyclopropane. I argue this is an achiral molecule because that carbon which is bonded to bromine cannot be a stereocenter. If you try to determine a point of difference between the other two "substituents," there is none. But if you draw a mirror image, it appears to be non-superimposable which is a trademark of chiral molecules. Is my reasoning on the right track, or is there something unique about the chiral or achiral nature of cyclopropanes such as these.

Here is a 3-D conformer from PubChem

As you can clearly see, a plane of symmetry can be sent along the black line perpendicular to the plane of the screen. Hence, the molecule is achiral. If you take a mirror image, you can ultimately super-impose it again on the parent form

Here is an illustrative 3D image(courtesy of andselisk) which clearly shows the plane of symmetry in pink and ring plane in blue.


Synthesis of 1-naphthaleneacetic acid: Methods I am trying to synthesizing 1-naphthaleneacetic acid from naphthalene and monochloroacetic acid. I found some patents and papers, and I have tried to synthesis but the yield is low. I used iron powder and $\ce{KBr}$ as catalyst. Papers and some patents mentioned about the yield is 70% but I achieved only 35%. I tried different iron compounds such as $\ce{FeCl3, Fe2O3}$ with $\ce{KBr}$ but the maximum yield gained is 35%.
Could you please help me for this synthesis method for high yield?

If you have read the reference 1 carefully, you would realized that the amounts of $\ce{KBr}$ and $\ce{Fe2O3}$, as well as reaction temperature and time is very important to the reaction yield. That's the reason I asked you to list your procedures so that we can analyzed them and suggest some changes. Well, as reference 1 listed, follow the following procedure exactly and see the results:

Naphthalene with 0.0010% iron $(\pu{57.6 g},\ \pu{0.449 mol})$, chloroacetic acid $(\pu{14.1 g},\ \pu{0.149 mol})$, ferric oxide $(\pu{87.6 mg},\ \pu{0.549 mmol})$, and potassium bromide $(\pu{420 mg},\ \pu{3.53 mmol})$ were placed in a long-necked round-bottom flask fitted with a rubber stopper carrying a thermometer and an air-cooled tube about a $\pu{1 m}$ long, and the mixture was boiled gently on a sand bath for twenty hours. The best yield was obtained when the heating was controlled according to a temperature-time curve in which the temperature of reactants attained $200^\circ$ after ten hours and $218^\circ$ after twenty hours. After the reaction, a greater part of the unreacted naphthalene $(\pu{43 g})$ was recovered by distillation. The residue was extracted with hot sodium hydroxide solution, cooled, and filtered. Acidification of the filtrate with hydrochloric acid yieldeed the brownish precipitate, $\pu{19.4 g}$ (70%), m.p. $108-113^\circ$ ....

It is noted that the melting point can be raised to $132^\circ$ by repeated crystallization. First crystallization the crude product with 250-fold distilled water has yielded $\pu{9.5 g}$ of product with m.p. $124-126^\circ$. This yield is 34% based on  chloroacetic acid used, which is the limiting reagent and has reacted completely, and 45% based on naphthalene reacted.
The condition described above is the optimum conditions regarding to temperature, time, increased amount of $\ce{KBr}$, and the amount ratio of naphthalene to chloroacetic acid. Also note that:

In the absence of $\ce{KBr}$ and $\ce{Fe2O3}$, no naphthaleneacetic acid was formed.
In the absence of $\ce{KBr}$ but the presence of $\ce{Fe2O3}$, only 3% of naphthaleneacetic acid was formed.
In the absence of $\ce{Fe2O3}$ but the presence of $\ce{KBr}$, only 4% of naphthaleneacetic acid was formed.
$\ce{KBr}$ is the most effective additive for the reaction. The promoting action of several other metallic halides such as $\ce{NaCl},$ $\ce{KCl},$ $\ce{KI},$ hydrated $\ce{AlCl3},$ hydrated $\ce{AlBr3},$ hydrated $\ce{FeBr3},$ and $\ce{HgBr2}$ when mixed with $\ce{Fe2O3}$ in place of $\ce{KBr}$ was also observed but they are less effective than $\ce{KBr}$.

References:

Yoshiro Ogata and Jiro Ishiguro, "Preparation of $\alpha$-Naphthaleneacetic Acid by the Condensation of Naphthalene with Chloroacetic Acid," J. Am. Chem. Soc. 1950, 72(9), 4302 (https://doi.org/10.1021/ja01165a533).


How to correctly define anode and cathode? I am very much confused: in a simple voltaic cell ($\ce{Zn}$, $\ce{Cu}$ in $\ce{H2SO4}$), $\ce{Cu}$ is positively charged and is the anode, where in a Galvanic cell ($\ce{Zn}$ in $\ce{ZnSO4}$ and $\ce{Cu}$ in $\ce{CuSO4}$),
$\ce{Cu}$ is positively charged but is the cathode. How can I define it, where both are electric or voltaic cell?

Cathode is the electrode where the Reduction reaction takes place.

Anode is the electrode where the Oxidation reaction takes place


Pyrolysis of canola oil This is a topic that I cannot find much information about. I would like to know what compounds are in the products formed from pyrolysis of canola oil. I found out by accident that when trying to ignite some used canola oil in an aluminum can without a cover, the oil had become a mix of black solids and black liquid. I have not tried heating the oil in a container that has no exposure to oxygen. I could probably get better results by limiting all exposure to oxygen. I recently tried this experiment again and the oil is slowly turning black and looking like crude oil. I did find one article on this subject that says there are alkenes in the oil. Does anyone have more information on this?

Basically canola oil is obtained from a cultivated form of Rapeseed. Canola oil is just what rapeseed oil is called in North America.  From a marketing perspective it was thought that having the term "rape" in the product name would not help sales.

If you look for articles on rapeseed oil pyrolysis you'll find a lot of information since rapeseed (canola) oil is used in biodiesel production.

This article states that the major products produced from pyrolysis in a reactor include

gases: $\ce{CO2, CO, CH4}$ and $\ce{H2S}$
oils: oleic acid, 1H-indole, 2,3,5-trimethoxy toluene, toluene, (Z)-9-octadecanamide, psoralene, phenol and phenol derivatives

I didn't know what "psoralene" was, so here is a picture and a link


I am trying to picture how electrons move around in atomic orbitals Are they thought to continuously pop in and out of existence at various points inside the orbital defined by probabilities or do they follow definite paths that are made fuzzy by the Heisenberg uncertainty principle?
I sometimes find myself as thinking of the electron cloud as a gas with varying charge density corresponding to the shape of the atomic orbital.

I think you miss an important notion to get a better understanding of the behavior of the electrons: wave-particle duality. I believe that a better visualization takes into account that you have to deal even with a wave. From what you say it seems that your picture is shifted only to the particle-side,  keep meditating for a half an hour trying to visualize the electron as a wave. For me is a very interesting mind experiment. Then move to the orbital concept.

At Lund University researches took a picture of a moving electron. This image maybe could be helpful for your meditation:

Lund University. "Electron Gets Film Debut In First-ever Video Of Its Kind." ScienceDaily. ScienceDaily, 25 February 2008. www.sciencedaily.com/releases/2008/02/080222095358.htm.


Why do amides require much harsher conditions for hydrolysis than esters? Amides require much harsher conditions to hydrolyse than its ester homologue. An explanation given is that the orbitals holding the lone pair on the nitrogen overlaps with the C=O π-bond to give conjugation, thus introducing a "partial" π-bond between nitrogen and the carbonyl carbon. Consequently, this makes the bond harder to break.
However, for the ester homologue, doesn't this conjugation also happen? An explanation given is poorer overlapping of the orbitals. However, how is that true?
In addition, such explanation is given to show that C=O is shorter in ester than it is in amide. However, I still don't get why this is the case.

Dimethyl acetamide (DMA) is less susceptible to hydrolysis than ethyl acetate (EA). This statement is true for aqueous acid or base. The infrared (IR) frequency for the carbonyl group of DMA is 1662 cm-1 while EA is 1742 cm-1. The lower the frequency, the lower the energy and the more, relatively speaking, single bond character in the carbonyl group. Thus, the carbonyl group of DMA has less double bond character than that of EA. Because nitrogen can sustain a positive charge better than oxygen, resonance is more pronounced in DMA than in EA. Because resonance is stronger in DMA, its carbonyl group is less susceptible to nucleophilic attack. The carbonyl bond length of EA would be expected to be shorter than that for DMA because the more double bond character, the shorter the bond. It is no different than the carbon-carbon bond length in the series ethane, ethylene and acetylene.


How can one make graphene cables? You might have seen in the short 2 minute video here: billionsinchange.com/solutions/limitless-energy that the Billionaire Founder (net worth $4 billion) Manoj Bhargava talks about creating cables from graphene to conduct heat from the underground sections of the earth to the surface.

But as per the properties of graphene, it is only a good conductor if the number of layers is one or very few. Once you start stacking it up, it is just normal graphite and it's conductivity gets reduced.

What's proposed in the video is a cable made from graphene. My question is how is it possible to make a graphene rope without stacking up graphene layers (and thus ending up with graphite) or without converting it into nanotubes?

I would also like to know why stacking up graphene reduces it's conductivity? Why don't the layers behave independent of each other?

According to Jesus de La Fuente, CEO of Graphenea:

  Graphene is, basically, a single atomic layer of graphite; an abundant mineral which is an allotrope of carbon that is made up of very tightly bonded carbon atoms organised into a hexagonal lattice.

He also states that there are three commonly-known naturally occuring allotropes of carbon - Diamond, Graphite, and amorphous carbon. Graphene's structure is similar to that of Graphite, but Graphene itself is much different because it has a $sp^2$ hybridization, compared to $sp^3$ hybridization of carbon in the form of Diamond, and Graphene has a thickness of just 0.345Nm (nanometers). This means that it does not form tetrahedrals like common carbon atoms, making Graphene similar but special compared to other forms of carbon.

Also stating Jesus de La Fuente in Graphite-Graphene Comparison:

  Graphene is fundamentally one single layer of graphite; a layer of $sp^2$ bonded carbon atoms arranged in a honeycomb (hexagonal) lattice. However, graphene offers some impressive properties that exceed those of graphite as it is isolated from its ‘mother material’.

To elaborate, Graphene has special properties that set it apart. There are similarities: stacking layers of Graphene may be similar to creating Graphite; creating a rope of Graphene may be similar to creating Crabon Nano-tubes; both have $sp^2$ hybridizations. However, Graphene has special properties that make it stronger when stacked because of how it is extracted and isolated.

As for the conductivity, Jesus de La Fuente states that it's electronic properties is what sets it apart (See Graphite-Graphene Comparison). Fundamentally, Graphene is a zero-overlap semi-metal, with both holes and electrons as charge carriers, that has very high electrical conductivity by itself. Graphite carbon allotropes have 4 electrons in it's valence shell available for bonding, however, Graphene has 3 electrons that are bonded with other Graphene atoms in the 2D Plane (a.k.a. other Graphene atoms next to it on the same one-atom thick sheet). This makes a single sheet very conductive, because the remaining electron that isn't bonded with other Graphenes in the same sheet are in the 3D Plane (above and below the sheet), causing what is called, pi electrons. However, when Graphene sheets are stacked together, their pi electrons bond with each other, thus reducing the electric conductivity of the Graphene.


Treating styrene in the atmosphere I was actually inspired to ask this question in certain circumstances. In the outskirts of my city, there happened to be a styrene gas leak early morning today which killed thirteen and hospitalized many people.

Since styrene is an aromatic compound, it is suspected to be carcinogenic in nature but I think it is inert towards other atmospheric gases as well as many compounds in non-laboratory conditions. The temperature here is 31 °C, so it can stay for a large time unaffected at the same area without much spreading. I think so because it has a high molecular weight(104 g/mol).

My question is: "is there a way to treat this styrene in the atmosphere without causing much more toxicity in that area?" (For information regarding its toxicity, refer to this safety data sheet). Of course, I understand that styrene is broken down within 1-2 days in the atmosphere naturally but is there any other way to do it quicker?

Spraying another compound to neutralize styrene (which is difficult) or whatever it was is probably not helpful, as it takes too long to spread and other potentially harmful materials are introduced, as one would need to apply the substance in excess.

Having readily installed outlets for water curtains at the plant might be helpful. The quickest emergency measure would have been if every household in the area was equipped with sufficient amounts of respirators and suitable absorbers ("filters") and training the people how to use them correctly (and teaching them that the absorbers have a limited capacity) might give enough time to escape. This would be a good and rather cheap provision for the future. Otherwise ventilation to dilute the materials is the only thing one can do, I'm afraid.


Discovery of elements How did the first chemists know that a sample contained so and so elements?
If they tried to find through reactions,they should know what elements make up the other reactant(s), and how it reacts beforehand (which is absurd as elements weren't discovered & studied before that)

The notion of chemical element in modern sense arised at the end of 19 century.

Before this, the term "element" was obviously too fuzzy to discuss here, as well as methods and proofs involved.

After this, the optical spectroscopy was method of choice to find new elements and prove their existence,
because of distinct and reproducible spectra.
See the history of helium, for example.

Of course, besides of spectroscopy, chemists proved the finding of new elements by separation of individual substances.
For example, metals are expected to form oxides and salts with (at least roughly) expected properties and mutual transformations.
In particular, salts, even in microamounts, may be recrystallized to form visually distinct crystals of individual substance.
If such a substance possessed properties not seen before, this was indication/proof of the presence of new element.


Is coupled cluster variational for two electrons? I know that coupled cluster (CC) is not variational for the general case. However, if we only have two electrons with one nucleus, CCSD should be exact for this system like full configuration interaction (CI). Since full CI is variational, can we argue that CCSD is also variational for this special case?

No. The reason for this is not to be found in the excitations, but in the evaluation of the method, i.e. the working equations.
$$%Introducing some shortcuts
\require{cancel}
\newcommand{\op}[1]{\mathrm{#1}}  %\op{H}
\newcommand{\bracket}[2]{\left\langle#1\middle|#2\right\rangle}
\newcommand{\bra}[1]{\left\langle#1\right|}
\newcommand{\ket}[1]{\left|#1\right\rangle}
\newcommand{\order}[1]{^{(#1)}}         %E_n\order{1}
\newcommand{\overlap}[3]{\mathcal{#1}_{#2}\order{#3}} %\overlap{S}{m}{1}
\newcommand{\integral}[3]{\mathcal{#1}_{#2,#3}} %\integral{V}{i}{j}
$$

The variational principle

In short: a trial wave function always has a higher energy than the exact wave function.

Assume we know the exact solutions to the Schrödinger equation, which form a complete basis and there is one wave function $\Psi_0$ which leads to the lowest energy $E_0$.
\begin{align}
\op{H}\Psi_i &= E_i\Psi_i &
\{i \in \mathbb{N}, i\geq 0\}\tag{1}\label{schrodinger}
\end{align}

We choose the solutions to be orthonormal (boundary conditions).
$$\bracket{\Psi_i}{\Psi_j} = \delta_{ij} =
\begin{cases}1, & i=j \\ 0, & i\neq j \end{cases}
\tag2\label{orthonorm}$$

The trial wave function $\Phi$ can be expressed as a linear combination of the complete set of wave functions.
$$\Phi = \sum_i^\infty a_i\Psi_i\tag3\label{expansion}$$

The expectation energy of the trial wave function is given through
$$W = \frac{\bra{\Phi}\op{H}\ket{\Phi}}{\bracket{\Phi}{\Phi}}.
\tag4\label{expen}$$

With $\eqref{expansion}$, $\eqref{orthonorm}$, and $\eqref{schrodinger}$ we can expand and simplify.
\begin{align}
W &=
  \frac{
    \sum_i^\infty\sum_j^\infty a_i a_j \bra{\Psi_i}\op{H}\ket{\Psi_j}
  }{
    \sum_i^\infty\sum_j^\infty a_i a_j \bracket{\Psi_i}{\Psi_j}
  }\\
  &=
  \frac{
    \sum_i^\infty\sum_j^\infty a_i a_j E_j\bracket{\Psi_i}{\Psi_j}
  }{
    \sum_i^\infty\sum_j^\infty a_i a_j \bracket{\Psi_i}{\Psi_j}
  }\\
  &=
  \frac{
    \sum_i^\infty a_i^2 E_i\bracket{\Psi_i}{\Psi_j} +
    \sum_{j\neq i}^\infty a_i a_j E_j\bracket{\Psi_i}{\Psi_j}
  }{
    \sum_i^\infty a_i^2 \bracket{\Psi_i}{\Psi_j} +
    \sum_{j\neq i}^\infty a_i a_j \bracket{\Psi_i}{\Psi_j}
  }\\
  &=
  \frac{
    \sum_i^\infty a_i^2 E_i
  }{
    \sum_i^\infty a_i^2
  } & | -E_0\\
W - E_0 &=
  \frac{
    \sum_i^\infty a_i^2 (E_i - E_0)
  }{
    \sum_i^\infty a_i^2
  } \\
W - E_0 &=
  \frac{
    \sum_{i>0}^\infty a_i^2 (E_i - E_0)
  }{
    \sum_{i>0}^\infty a_i^2
  } + \frac{a_0^2 (E_0 - E_0)}{a_0^2}\tag{4'}\label{notnegative}\\
W - E_0 &\geq 0\tag5\label{varprin}
\end{align}

In $\eqref{notnegative}$ the last term is zero. Since $E_i > E_0$ (by definition) and $a_i^2$ is always greater or equal zero, the expectation value of the trial wave function must always have a higher energy than the exact solution of the Schrödinger equation, i.e. $\eqref{varprin}$.

Configuration Interaction

The CI wave function is set up as a linear combination of determinants from a reference calculation. This reference calculation can in principle be any other method the only requirement is that it also forms a complete (orthonormal) basis.
When Hartree-Fock is chosen, we know that we obtain a complete set and that this method itself is variational. The CI trial wave function can then be expressed as a linear combination of solutions of the HF wave function.
$$\Psi^\mathrm{CI} = \sum_i^\infty a_i\Phi_i =
a_0\Phi^\mathrm{HF} + \sum_{i>0}^\infty a_i\Phi_i$$
An alternative expression of the above is using an "excitation operator" $\op{T}$ to generate the higher energy determinants from the reference.
$$\Psi^\mathrm{CI} = (1 + \op{T})\Phi^\mathrm{HF}$$

The energy of the CI trial wave function can be evaluated with the Lagrange multiplier method, which requires, that the wave function is normalised (adding zero).
$$L = \bra{\Psi^\mathrm{CI}}\op{H}\ket{\Psi^\mathrm{CI}}
     - \lambda\left[\bracket{\Psi^\mathrm{CI}}{\Psi^\mathrm{CI}}-1\right]$$

Since the boundary conditions did not change, the variational principle still holds.

Coupled Cluster

Because of the inherent complexity of the CC ansatz, one key element to evaluate to CC energy is missing: the requirement of the normalised CC trial wave function.
We generate the trial wave function with an exponential approach (rather than a linear) expanded into a Taylor series.
\begin{align}
\Psi^\mathrm{CC}
&= \mathrm{e}^{\op{T}}\Phi^\mathrm{HF}\\
&= \left[\op{1} + \op{T}_1 + \left(\op{T}_2 + \frac12\op{T}_1^2\right)
   + \left(\op{T}_3 + \op{T}_2\op{T}_1 + \frac16\op{T}_1^3\right)
   + \cdots \right] \Phi^\mathrm{HF}\\
&= \Phi^\mathrm{HF} + \left[\op{T}_1 + \left(\op{T}_2 + \frac12\op{T}_1^2\right)
   + \left(\op{T}_3 + \op{T}_2\op{T}_1 + \frac16\op{T}_1^3\right)
   + \cdots \right] \Phi^\mathrm{HF}\\
\end{align}

Or to shorten the journey
$$\Psi^\mathrm{CC} = \Phi^\mathrm{HF} + \sum_i^\infty t_i \Phi_i.$$

One can easily see what kind of mess it would become requiring the wave function to be normalised $\bracket{\Psi^\mathrm{CC}}{\Psi^\mathrm{CC}}=1$, so the Lagrange ansatz won't work. Instead we choose the wave function to be orthonormal to the reference, which in the Hartree-Fock (complete orthonormal basis) takes away a lot of work.
\begin{align}
\bracket{\Phi^\mathrm{HF}}{\Psi^\mathrm{CC}}&= 1&
\implies\quad \bracket{\Phi^\mathrm{HF}}{\Phi_i} &= 0
\end{align}

We can write the CC-Schrödinger equation as
$$\op{H}\mathrm{e}^{\op{T}}\Phi^\mathrm{HF} =
  E^\mathrm{CC}\mathrm{e}^{\op{T}}\Phi^\mathrm{HF}$$
and evaluate the energy as
\begin{align}
\bra{\Phi^\mathrm{HF}}\op{H}\mathrm{e}^{\op{T}}\ket{\Phi^\mathrm{HF}}
  &= E^\mathrm{CC}
     \bracket{\Phi^\mathrm{HF}}{\mathrm{e}^{\op{T}}\Phi^\mathrm{HF}}\\
E^\mathrm{CC}
  &= \bra{\Phi^\mathrm{HF}}\op{H}\mathrm{e}^{\op{T}}\ket{\Phi^\mathrm{HF}}\\
E^\mathrm{CC}
  &= E^\mathrm{HF} +
  \bra{\Phi^\mathrm{HF}}\op{H}
    \ket{\left[\op{T}_1 + \left(\op{T}_2 +
     \frac12\op{T}_1^2\right)
     + \left(\op{T}_3 + \op{T}_2\op{T}_1 + \frac16\op{T}_1^3\right)
     + \cdots \right]\Phi^\mathrm{HF}}.\\
\end{align}

From this we can see that the evaluation of the expectation value of the energy is significantly different from what is required for the variational principle $\eqref{expen}$.

Without going further into details, the dropping of the requirement of a normalised trial wave function leads to not obeying the variational principle.

Additional considerations

It is often said that Full-CC is equivalent to Full-CI, which is a convenient half truth. It is extremely important that this gets more true with a better description of the reference wave function, i.e. with larger basis sets. Full-CI is exact at the complete basis set limit (for the electronic Schrödinger equation). This is not necessarily true for Full-CC.

Since we are in computational chemistry hardly ever concerned about absolute energies, the non-variational nature of the method is by far outperformed by the "infinite" approach generating the "excited" determinants. Relative energies don't suffer significantly (compared with the size consistency and extensivity) under the non-variational approach.


NaCl dissociation not spontaneous? It's a well-known fact that the dissociation of $\text{NaCl}$ can be represented by the series of equations $$ \begin{eqnarray*} \text{Na} &\rightarrow& \text{Na}^+ + \text{e}^- \qquad \Delta H = 495 \, \text{kJ} \\ \text{e}^- + \text{Cl} &\rightarrow& \text{Cl}^- \qquad \qquad \Delta H = -348 \, \text{kJ} \end {eqnarray*} $$

Adding gives $ \text{Na} + \text {Cl} \rightarrow \text{Na}^+ + \text{Cl}^- $ with $ \Delta H = 147 \, \text{kJ} $, which is positive so it is not spontaneous. What explains this?

You might be confusing $\ce{\Delta G}$ and $\ce{\Delta H}$.

$$\ce{\Delta G = \Delta H - T \Delta S}$$

$\ce{\Delta G}$, free energy, measures the "usefulness" of a system. If $\ce{\Delta G \gt 0}$, then the reaction is not thermodynamically favored - it's NOT SPONTANEOUS.

$\ce{\Delta H}$ being positive means that the reaction is endothermic, or makes the surroundings cooler. While this would CONTRIBUTE to a positive value of $\ce{\Delta G}$, it DOES NOT GUARANTEE it.

In the case of your example above, I would assume the $\ce{T \Delta S}$ factor would make $\ce{\Delta G \lt 0}$ if that reaction really is SPONTANEOUS (I don't have an entropy table with me).


Effects of evaporating the solvent of a supersaturated solution What happens to the solute if you heat up a supersaturated solution? Since a supersaturated solution is already more than saturated with solute and unstable, my theory is that as you heat the supersaturated solution, the solubility limit will have been increased enough to allow all of the solute to dissolve. So technically, the solute does not precipitate out if you heat it enough.

Two caveats:

If you sufficiently disturb a supersaturated solution, the solute will spontaneously come out solution.  So you would need to heat it without sufficiently disturbing it.

The solubilities of some solutes decrease with temperature.  So let's assume your solute's solubility increases with temperature (as is the case for most, but not all, solids dissolved in a liquid).

So if we allow for the above then, as you heat the solution, you will eventually reach at point at which it is no longer supersaturated.
Indeed, the reverse of this is how supersaturated solutions are typically prepared:  A solution is prepared at a high temperature with a concentration of dissolved solid just below the saturation point (you don't want any undissolved solid left, since that will act as a nucleation site and thus prevent supersaturation).  Then it is carefully cooled.  So, by heating it sufficiently, you would simply be moving it back towards the starting point.
Finally, someone might argue that, if the solution is supersaturated near the boiling point, it might not be possible to heat it up sufficiently to get below the saturation concentration.  I suppose that's possible in theory, but I don't know how you would prepare such a system, since it couldn't be created using the typical cooling process.


How does VB theory explain the Si-O-Si bond angles in SiO2? Silicon dioxide has a huge variety of structures. Most of them are built up from connected $\ce{SiO4}$ units — the $\ce{O–Si–O}$ angle is $109.5°$, accordingly. The VB/hybridization approach to this tetrahedron would consequently assign $4\text{ }\mathrm{sp^3}$  orbitals to the silicon atom.
Now, the interesting part about silicon dioxide is the flexibility of its $\ce{Si–O–Si}$ angles, i.e. the bond angle of the bridging oxygen atoms; they vary from $100$ to $180$ degrees (the latter occurs in hexagonal tridymite, see e.g. here), depending on the polymorph. In vitreous samples, these angles show a distribution centered around $145$ degrees or so (source). In common forms of crystalline $\ce{SiO2}$, the angle also is close to $145$ degrees (e.g. this source). How is this possible? Since the oxygen bonds to two silicon atoms, one would expect a $\mathrm{sp^3}$ hybridization, with two orbitals filled with lone pairs — hence, a $104°$ angle like in water, or a slightly bigger one (since silicon is less electronegative than hydrogen — thus, the $\ce{Si–O}$ bond is more polar than the H–O bond and the angle bigger). But never an angle as large as $170°$: that would almost point at an $\mathrm{sp}$ hybridization of the oxygen atoms. How is such an angle energetically more convenient than the typical bent one?

Tl;dr version: the O-Si bonds utilize sp hybrid orbitals so that “lone pairs” are in nearly pure p orbitals and, rather than being true lone pairs, engage in some sort of $\pi$ bonding, the exact nature of which is unknown.  Other hypotheses are that the bond is highly ionic or that it is favored due to oxygen-oxygen repulsion.
Long version: First, let me point out a slight inaccuracy in the posted question:

Since the oxygen bonds to two silicon atoms, one would expect a sp3 hybridization, with two orbitals filled with lone pairs —

This is an incorrect interpretation of VB theory. Although we might approximate a molecule such as $\ce{H2O}$ as having an sp3-hybridized O atom, quantitative approaches to VB theory account for the fact that the contribution of s and p atomic orbitals to localized orbitals differs if the group differs. That is, lone pairs and bond pairs can have (and usually do have) different contributions from the s and p orbitals, ie different hybridization. A commonly cited example of this is $\ce{SH2}$, where the S-H bond orbitals are nearly entirely comprised of p orbitals, so the lone pairs are necessarily essentially sp hybrids.
Furthermore, one quantitative approach calculates the p orbital contribution based on the bond angle (Coulson's Theorem). In silicate, the Si-O-Si bond angle varies between ~140$^\circ$ and 180$^\circ$, depending on polymorph. We can calculate that the hybridization of these bond orbitals thus varies from sp (180 degree angle) to sp1.3 (140 degree angle), which is actually a relatively small variation in the s orbital fraction (43-50%). The lone pairs, therefore, are nearly pure p (93-100%) orbitals.
Although it is known that the $\alpha$-quartz polymorph (bond angle 144 degrees) is the most stable at low temperature, the observation of many polymorphs suggests that there is not much of an energetic penalty for the variation in bond angle within this range.
For simplicity, let’s consider the extreme case of the 180 degree bond angle, so that the O-Si bonds are formed from sp orbitals on O, and the lone pairs are in pure p orbitals. Both VB theory and MO theory indicate that this arrangement is only energetically favorable relative to a bent conformation if either the p orbitals are empty (as in $\ce{BeH2}$ for example) or if the electrons in the p orbitals are involved in $\pi$ bonding, for example the carbon atom of $\ce{CO2}$ or $\ce{HCN}$.
COnsistent with this idea, researchers have argued since at least as far back as the 1960's that the “lone pairs” on oxygen in silicate are in fact not lone pairs, but are involved in $\pi$ bonding to the silicon. This is supported by the short Si-O bond lengths and high bond strength which suggest a bond order greater than 1.
Historically, this interaction has been proposed to involve empty d orbitals on silicon, but more recently there has been a shift away from d orbital involvement, for similar reasons that d orbital involvement in phosphorus bonding has generally been disregarded. Interaction with empty $\sigma^*$ antibonding orbitals on Si, however, also does not appear satisfactory, as that would weaken the Si-O bonds (assuming that all Si-O bonds are equivalent, so all would have increased bond order from $\pi$ bonding, but decreased from addition of density to $\sigma^*$). Essentially, there would be a significant contribution of a resonance structure in which each Si has two Si-O bonds and one Si=O bond, similar to a carbonate.
Nonetheless, the idea that a large bond angle is favored due to some sort of $\pi$ bonding seems to be the prevalent view. Two other hypotheses, however, are still supported by some.
Other hypotheses
Ionic interactions
One alternative explanation is that there is very little difference in bond orbitals and lone pairs, which could result from a strongly ionic structure of $\ce{O^2-}$ anions and $\ce{Si^4+}$ cations[1]. On the oxygen, all of the electrons then are lone pairs, and the “bond angle” (which isn’t between any true bonds) is simply optimized to minimize repulsion between negatively charged oxygen atoms and between the positively charged silicon atoms. This would result in a tetrahedral arrangement of the negative charges around Si (as is observed) and a linear arrangement of the two Si atoms around each oxygen (the occasionally observed 180 degree angle). Since there aren’t any bonds, VB theory doesn’t really apply, and we consider the oxyanion as a lone atom with completely filled valence shell.
The problem with this argument is that free $\ce{Si^4+}$ is not observed under any circumstances, and most computational approaches suggest a high degree of covalency in the Si-O bond.
Oxygen-oxygen repulsion
A third hypothesis is that VB theory does not apply here because the minimization of energy of the Si-O-Si bond angle is not the only factor. Specifically, oxygen-oxygen repulsion favors a larger angle so as to increase the distance between silicon atoms and, therefore, the oxygens attached to adjacent silicon atoms.
Even at the small scale of a single molecule of $\ce{(HO)3Si-O-Si(OH)3}$, it has been argued that oxygen-oxygen repulsion is sufficient to cause a large Si-O-Si bond angle without any need for $\pi$ bonding[2]. The observed angle is a compromise between repulsion, which favors a 180 degree angle, and bond orbital energy, which favors a bent structure with bond angle closer to 110 degrees.
A problem with this hypothesis is that the Si-O-Si bond angle in disiloxane ($\ce{H3Si-O-SiH3}$) is also observed to be 144 degrees, even though there should be no oxygen atom repulsion effects and general steric effects should be much less than in silicates.
[1]Gillespie, RJ and Johnson, SA (1997) Inorg Chem 36:3031-3039. doi:10.1021/ic961381d
[2]Noritake F (2019) J Comput Chem Jpn 5:1. doi:10.2477/jccjie.2018-0016


What happens when KCl, K₂Cr₂O₇, and concentrated H₂SO₄ reacts? What happens when $\ce{KCl}$, $\ce{K2Cr2O7}$, and concentrated $\ce{H2SO4}$ reacts?

I have no idea how to approach these questions. I'd love if anybody can help me see the logic behind this.

Have a look at the properties of the compounds involved.

Dichromate ($\ce{Cr(VI), Cr2O7^{2-}}$) is a strong oxidant and is reduced to $\ce{Cr(III)}$ if it finds a suitable partner (which then is oxidized).

Here, a potential partner in a redox reaction is chloride: $\ce{2 Cl- + 2 e^{-} -> Cl2}$.


It seems to be a bug with optimization in internal coordinates. As I already mentioned in comments, ORCA couldn't converge at geometry relaxation step when optimization is done in internal coordinates. Switching to optimization in Cartesian coordinates (coordsys cartesian) or old internal ones (coordsys redundant_old) in the %geom section of the input, or using the ! COPT simple input as OP has done, solves the issue.

P.S. OP might report it as a bug at ORCA forum so that developers can take a closer look at this case.


Iron-platinum refining If I have an iron-platinum alloy, can I put this into a solution of $\ce{HCl}$ to dissolve the iron, leaving a $\ce{Pt}$ precipitate?

I know that Fe dissolves in $\ce{HCl}$ to form $\ce{FeCl2}$ and $\ce{H2}$, and I know that $\ce{Pt}$ does not dissolve in $\ce{HCl}$.

But will the $\ce{Fe-Pt}$ bonding affect the Fe dissolution?

It depends on the concentration of Pt in your alloy.

If the concentration of Pt is to high the HCl will not or not sufficently attack the alloy.

In this case you could try to melt your alloy with some additional Fe, to lower the Pt concentration and then hit it with HCl again.

You could also try to boil it in HCl to increase the reactivity of the acid (do it in a fumehood or outdoors).


What exactly is the parasitic white layer that forms after iron parts are treated with ammonia? I have little to do with chemistry and my only background is the inorganic chemistry I learned at school when I was 13-14 years old.

I need to have a basic understanding of the nitrating process (for an automation engineering application, more precisely an automated furnace used for nitrating iron or steels parts with the purpose of improving their mechanical and chemical properties).

A few things are not so clear for me:

What is the chemical formula of the iron lattice mixed with nitrogen atoms (see the picture)?
What is the unwanted white layer that I understand forms on the surface of the metal part treated with dissociated ammonia?
What is this dissociated ammonia?

Gas nitriding

Steels are nitrided in ammonia gas at 900 to 1050 F ; It forms a very hard , very thin ( < 0.01 ") hard layer of iron nitride . An addition of aluminum alloy to the steel ( 0.5 to 1.0 % ) enhances the nitriding. Nitriding is applied typically to cutting tools like drill bits and wearing surfaces. At higher temperatures "carbo-nitriding " is done where C and N are diffused into the steel surface. The "white layer" is something seen in a metallographic sample at high magnification -100 x and higher. It is undesirable and apparently unidentified as the ASM  handbooks refer to it as only "white layer" ; they give procedures to remove it if necessary . The white layer should be thin , even compared to a 0.005 " thick layer of nitride. Plain ammonia or dissociated ammonia may be used with slightly different procedures.


Can neutralization be reversed? So far, maybe it's just that my course isn't advanced enough, but I have never seen anything called 'neutralization constant'.

Any such reaction involving an acid and a base will be complete no matter the strength of either reactant.

Buffer works the same way. If the reaction is reversible then the buffet is not effective.

The reverse reaction of neutralization is called hydrolysis ( X salt + H2O = acid + base )


Is there a difference between potassium manganate and permanganate? Is potassium manganate the same as potassium permanganate?

Also, I have seen potassium manganate written as potassium manganate (VII) or potassium manganate (VI).

What do the roman numerals stand for?

No, they are somewhat different. $\ce{K2MnO4}$ is potassium manganate and $\ce{KMnO4}$ is potassium permanganate. Notice the two potassium for each $\ce{MnO4}$.

The Roman numerals mark the oxydation state. So, manganate will have a (VI) and permanganate will have a (VII).


Why is water ignored in the ionic equilibrium of HF? While trying to understand the solution of a problem given in my textbook, I realized I'm having some difficulty with the solution. The problem is as follows:

  The ionization constant of $\ce{HF}$ is $3.2 \times 10^{-4}$. Calculate the degree of dissociation of $\ce{HF}$ in its $\pu{0.02 M}$ solution. Calculate the concentration of all species present $\ce{H3O+}$, $\ce{F-}$ and $\ce{HF}$ in the solution and its PF.

In the solution of this problem, the equation is given as
$$\ce{HF + H2O <=> H3O+ + F-}$$

The concentration at the time of equilibrium are given as:

\begin{align}
\ce{[HF]}   &= 0.02 - 0.02x, &
\ce{[H3O+]} &= 0.02x, &
\ce{[F- ]}  &= 0.02x
\end{align}

I have the following questions:

Why are we not adding the contribution of water to the $\ce{H3O+}$ ions?
Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?

Why are we not adding the contribution of water to the $\ce{H3O+}$ ions?

Write down the dissociation constant equation with and without $\ce{H2O}$.
What changes?
Can you find a formulation of the dissociation constant where it seems natural not to include $\ce{H2O}$?
Read up on "activity".

  Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?

Write down and solve the equation with $0.02 x$ and with $x$ only. What is the difference?


Calculating the viscosity of water using particles There are two particles with radius 0.212 μm. The particles are in a thin liquid film and the microscope can only detect movement in the $x$- and $y$-axis. The temperature is 25 °C.
$$
\begin{array}{c|rrr}
\hline
t/s & 30 & 60 & 90 & 120 \\
\hline
x_1 & 4.4 & 10.7 & 11.0 & 12.4\\
y_1 & -3.1 & -5.2 & -10.8 & -9.3\\
x_2 & 3.9 & 2.0 & 8.3 & 5.8 \\
y_2 & 3.5 & 6.4 & 14.1 & 12.0 \\
\hline
\end{array}
$$
Use this information to estimate the viscosity of water.

I know that I should use the equations:
$$x^2 = 2Dt \tag{1}$$
$$D = \frac{kT}{6πηa} \tag{2}$$
Where if I plot $x^2$ against $t,$ I can calculate the viscosity from the slope $\displaystyle\frac{2kT}{6πηa}.$ However, what I don't understand is how to calculate $x^2.$
I tried taking the average of the two particles in direction $x$ and squaring the result, but that doesn't give me the right answer. I also tried taking the average of $x^2$ of the two particles, but once again I get the wrong answer.
How should I calculate $x^2?$

The basic idea here is that the spherical particles, each of radius a, are being bounced around via Brownian motion. So the two relevant equations are
$$r(t)^2 = 2Dt \tag{1}$$
$$D = \frac{kT}{6πηa} \tag{2}$$
Note that the first equation uses r(t) rather than x, to avoid needless confusion with the x and y cartesian coordinates of the spherical particles.
Now consider particle 1. At t = 0 s, it is at the defined location 0,0, where all distances are in $\mu m$. At four subsequent times, the x and y coordinates are as given in the OP's table. The following figure shows particle 1's location at t = 0s and at t = 30 s.

At any given time,
$$r_1(t)^2 = x_1(t)^2 + y_1(t)^2 \tag{3}$$
To estimate D, an ordinary least squares fit is performed on the tabulated data for particle 1. The result is shown in the next figure:

Note that the origin is used: at t = 0 s, then, by definition, $x_1(0) = 0 \space µm$ and $y_1(0) = 0 \space µm$. So $D = 1.1486 \times 10^{-12} m^2 s^{-1}$. Rounding off to two significant figures will be performed at the end.
Solving equation (2) for $\eta $ yields:
$$\eta = \frac{kT}{6πDa} \tag{4}$$
where T = 298.15 K, $a = 2.12 \times 10^{-7} m$ and $k = 1.38064852 \times 10^{-23} J K^{-1} $. The result for $\eta $ is
$$\eta = 8.97 \times 10^{-4} Nt \space m^{-2} \space s \tag{5}$$
where the fact that $1 J = 1 Nt \times 1 m$ has been used. N.B. 1 Joule equals 1 Newton times 1 meter.
Other units for $\eta $: Since 1 Pascal of pressure is 1 Newton of force per square meter, i.e., $1 Pa = 1 Nt / m^2$, and since 1 centipoise (symbol: cP) is 1 milliPascal times 1 s, i.e., $1 cP = 1 mPa \space s$, then
$$\eta = 8.97 \times 10^{-4} Pa \space s = 0.897 \space mPa \space s = 0.897 cP\tag{6}$$
Finally, rounding to two significant figures, $\eta = $0.90 cP. The viscosity of water at 25 °C is reported to be 0.91 cP.


What is the reason why protons and electrons do not collide? can someone give me an intuitive picture of why electrons don't collide with protons?

I know that electrons move in a sort of cloud, which is our 'orbital', and that they mainly behave like electromagnetic wave because of their very tiny mass. However, how comes they don't collide with the protons? I seriously didn't find a good explanation of these neither by professors I talked to nor by books, so a thousand points from me if you are able to get me into this.

Keep in mind that I am not looking for values, equations, or quantitative explanation of this, for that I have read enough, but for a picture. Think about the subatomic particles as you see them in your mind and describe their relationship.

They do collide.  But it's not like cars crashing and smashing each other up on the highway.  Their quantum mechanical wavefunctions overlap and the electrons do in face settle in with wavefunctions that are centered on the nucleus.  But the low mass of the electrons, the Heisenberg uncertainty principle (which has its greatest effect on low-mass particles), and the Pauli exclusion principle (electrons can't all pile up in the most stable quantum state)  cause the electron wavefunctions to get diffused out around the nucleus.  Instead of getting smashed up, the electrons basically surround snd embed the atomic nucleus.


What type of material is this and how is it corroding? So I was thinking about corrosion and I came across this structure(see image)

Do you think the inner metal/alloy that is corroding is steel?

Also, what do you think the outer layer (in white) is? perhaps the metal is just coated with paint to avoid corrosion?

Finally, what type of corrosion do you think its undergoing? General Corrosion, Galvanic corrosion, uniform corrosion and so on.

Though a picture is not really enough information, approximate location at least and any simple tests on the materials shown would help a lot.

Its structural and ugly so its steel.
Its mostly likely painted steel. I'd say the paint flaked off so it has general uniform corrosion. Why it looks black and not the regular red-brown is probably combination of oxides and sulphides formed as the corrosion products. I don't see a hint that its galvanised steel.
Steel in exterior uses  rarely suffers any corrosion other than  general uniform corrosion.


Why is the Chlorine missing an electron in Chlorine Monoxide? I was asked to draw the lewis-dot structure for ClO and I drew this:

   ..  ..
  :Cl-  O.
   ..  ..

The correct answer is apparently this:

   .   ..
  :Cl-  O:
   ..  ..

How would I know that? In the first example, all the formal charges are zero, but in the second, Oxygen has -1 and Chlorine has +1.

In fact the unpaired electron belongs neither to Cl nor to O, but to a molecular orbital spread over both atoms [but not equally: see below].  The Lewis structure is simply not a complete representation of what is going on.  (And for what it's worth wikipedia gives the same Lewis structure you drew: http://en.wikipedia.org/wiki/Chlorine_monoxide)

Edit: Actually it seems the highest occupied molecular orbital (where the unpaired electron is) is mostly localized on the oxygen atom.

MO Diagram (taken from here):

That is to say, your structure is more correct than the one in the answer key.  However the justification involves molecular orbital arguments outside the scope of a general chemistry course.

Another edit: I too like your argument about formal charges and would accept it as a justification that the unpaired electron belongs to O rather than to Cl.


Is there a single MO that represents the π system of a Möbius aromatic system? If so, what does it look like? I recently answered the question of whether twisted polyene chains were known by pointing towards Möbius aromatic system. While in conventional Hückel aromatic systems the lowest-lying π-orbital obviously has a nodal plane containing all the nuclei, this is not possible in a Möbius aromat since topologically these only have one side.

A number of Möbius aromatic systems have been both synthesised[1] and calculated;[2] for an X-ray determined structure of one, see the second structure in the figure below.[3]

One can easily see that the π chain would have green and white or plus phase and minus phase overlapping. This got me thinking and wondering: what would this system’s molecular orbitals look like? Specifically: is there a set of MOs that would correspond approximately to the π system of such a Möbius aromat? If not, how are the π bonds seen in an MO analysis? The only calculations I could find searching the literature were both very old and without nice pictures.

References:

[1]: M. Stępień, N. Sprutta, L. Latos-Grażyński, Angew. Chem. Int. Ed. 2011, 50, 4288–4340. DOI: 10.1002/anie.201003353.

[2]: H. S. Rzepa, Chem. Rev. 2005, 105, 3697−3715. DOI: 10.1021/cr030092l.

[3]: Y. Tanaka, S. Saito, S. Mori, N. Aratani, H. Shinokubo, N. Shibata, Y. Higuchi, Z. S. Yoon, K. S. Kim, S. B. Noh, J. K. Park, D. Kim, A. Osuka, Angew. Chem. Int. Ed. 2008, 47, 681–684. DOI: 10.1002/anie.200704407.

I suppose you know better than many that aromaticity is not about a single MO, but a whole set thereof. We may just solve the system by the simple Huckel method and see what they are like. Sure, it is terribly old-fashioned and won't give you nice pictures (unless you make some yourself), but I don't think we need them all that much. What we need is the qualitative description, and it will be there all right.

From a cursory glance, it seems to me that the eigenvalues of a Möbius system are given by the same Frost circles, only the polygon is pointing down with its side, rather than a vertex. This, BTW, explains the reversed stability rules ($4n$ instead of $4n+2$). The molecular orbitals will look kinda like those in usual aromatic systems, except that the lowest orbital will be doubly degenerate and indeed will have a node plane across the chain.

So it goes.


Why the products of combustion of Cu2S are Cu2O and SO2? $$\ce{Cu2S + 2 O2 -> 2 CuO + SO2}$$

Why are not these oxides $\ce{Cu2O}$, $\ce{SO}$ or $\ce{SO3}$?
Is it connected to the stability of these molecules?

The reaction gives different products when combusted at different temperature.

$$\ce{8Cu2S + 15O2 ->[\Delta] 6Cu2O + 4CuSO4 + 4SO2}$$

Copper(I) sulfide react with oxygen to produce copper(I) oxide, copper sulfate and sulfur dioxide. This reaction takes place at a temperature of 500-600°C.Here

$$\ce{2Cu2S + 3O2 ->[\Delta] 2Cu2O + 2SO2}$$

Copper(I) sulfide react with oxygen to produce copper(I) oxide and sulfur dioxide. This reaction takes place at a temperature of 1200-1300°C.Here

Sulfur monoxide is very unstable and it is unlikely to exist in Earth's atmosphere but it exist in interstellar spaces. Sulfur trioxide will not form from sulfur dioxide as it requires an oxidising agent.


Is there an all-natural, non-volatile and low viscosity oil? The goal is to thin a vegetable oil so that it spreads easily on top of water. It is not for consumption and the flash point of the mixture should be >=95° Celsius". Vegetable oils tend to have flash points greater than 225° Celsius so the flash point of any additives can be well below 95°, as long as special precautions would not be necessary in a manufacturing process.

Since my original post, I've discovered another reason for the low viscosity. It is required for the delivery system to work properly.

I am open to other means to achieve the goal.

If the goal is to have the oil cover the water surface, then you might need to add a natural surfactant, such as lecithin, to cause the oil to spread, rather than form globules. You can try egg-yolk or soy beans as a source of lecithin, or you can buy refined lecithin.

As for the smoke point, or maximum cooking temperature for the oil, many common vegetable oils could do: safflower, sunflower or refined canola (rape seed) oil all have a good margin of safety.

One issue might be that if the water is at a "roiling" boil, the turbulence will mix the oil into the body of water. For that reason, you'd need to keep the temperature below 100° C and prevent local hot-spots in the heat source from causing strong convection currents.


Why is Aluminium oxide amphoteric? I was reading my book, in which it mentioned that aluminium and gallium oxides are amphoteric and and indium and thalium oxide are basic in their properties.
But no explanation was given.
My main question is we know that mostly metal oxides are basic and non metal oxides are acidic. But we see some metal being amphoteric, what is the cause of this difference?
I tried to search online but found no reliable piece to read.

I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements
To start off, please go through my answer to this question: Why is chromate stable in basic medium and dichromate stable in acidic medium? to read about the preface to this method, including the potential concerns.
Essentially, the book talks about determining the type of reactivity of inorganic molecules by looking at their structures. While this may not reflect with 100% accuracy the actual reaction pathway, we can take it to be a pretty good approximation of what happens in reality. While the completely correct mechanism can only be found out experimentally, we can get a probable mechanism by employing an arrow-pushing approach, and that should at least mirror the actual pathway to a sufficient extent, if not entirely.
Now, if we look at the structures of Aluminium oxide and Gallium oxide ($\ce{Al2O3}$ and $\ce{Ga2O3}$), we can see that they possess a similar bridged dimer-kind of structure as shown below for aluminum oxide(the same structural skeleton is valid for gallium oxide as well, just by replacing aluminum with gallium below) :

This structure now possesses some unique properties. It contains an $\ce{Al-O-Al}$ bridge with an $\ce{Al=O}$ group on each side.Now,the $\ce{Al=O}$ group can be said to be structurally similar to the carbonyl group from organic chemistry.
Hence, like the carbonyl group, it can be said that there will be high partial positive charge on the $\ce{Al}$ atoms due to the electron-withdrawing effect of the $\ce{Al=O}$ groups. Hence, they can serve as the ideal electrophilic site for the attack of a nucleophile like $\ce{OH-}$ in basic medium, and lead to bond cleavage(and further reaction) in alkali as follows:

In a basic medium, the $\ce{AlO2H}$ moiety will also be deprotonated, and thereby you will end up with two moles of $\ce{AlO2^-}$ . If the base being used was $\ce{NaOH}$, then these products are better reported as $\ce{NaAlO2}$ for aluminum. In an analogous case, you would get $\ce{NaGaO2}$ if the we had started out with gallium oxide.
In excess of alkali, the reaction can proceed further, as further additions of hydroxide are still possible at the $\ce{Al=O}$ site as follows:

The reaction can either finish after the formation of the hydroxide, or a continued process may lead to the addition of yet another $\ce{OH^-}$ to form the complex ion $\ce{[Al(OH)4]^-}$ , which might be called the aluminate ion here.In another analogous case,we would have ended up with $\ce{[Ga(OH)4]^-}$ (called the gallate ion) if we had started out with gallium oxide.
Now, for the acidic medium, let us suppose that $\ce{Al2O3}$ reacts with $\ce{HCl}$. Here, the central oxygen of the $\ce{Al-O-Al}$ bridge represents a weak link that can be easily cleaved after protonation(because it has moieties that can function as good leaving groups on either side). Also,the counter-ion of the acid(that is, $\ce{Cl^-}$), can serve as a nucleophile to break down the dimer skeleton after protonation as follows:

The $\ce{AlO2H}$ can again be protonated to form an oxonium ion that can expel water and take up a $\ce{Cl^-}$ to form another mole of $\ce{AlOCl}$(shown in the next diagram).
Following that, the $\ce{AlOCl}$ can be engaged in a repetitive process of protonation, attack by chloride and removal of water to yield the final products as follows:

The products for these reactions can be verified for aluminium and gallium oxides from their Wikipedia pages respectively.
So we can see that one major reason for the amphoteric nature of oxides can be their bridged structures, which allows them to react with both acids and bases as illustrated above.
Now, coming to indium oxide, the formula of this compound as stated here is $\ce{In2O3}$, which bears a resemblance to the formula of $\ce{Al2O3}$ and $\ce{Ga2O3}$ which we stated above. So, I would expect this oxide to atleast have some amphoteric properties. This is indeed the case, as stated here for indium hydroxide(since hydroxides are simply hydrated forms of their oxides, we can expect a similar nature for indium oxide as well):

The result indicated conclusively that In(OH)3 is only slightly amphoteric, being much more basic than acidic.

Also, on the Wikipedia page for indium oxide it clearly states that this oxide is soluble in acids but not in alkali. This is pretty much expected because when you move down a group, the elements tend to become more metallic in nature, and hence their oxides tend to become more basic as well. Hence, an amphoteric oxide slightly tending towards the basic end seems reasonable for indium(III) oxide. The reaction between acid and $\ce{In2O3}$ should follow the general scheme that is illustrated above for $\ce{Al2O3}$ and $\ce{HCl}$, and should yield similar products. This indeed is the case as can be seen here
As for thallium, due to the inert pair effect, it generally forms an oxide in which thallium is in the +1 oxidation state, with the formula $\ce{Tl2O}$ as stated here. This doesn't fit in with the general formulation of $\ce{Al2O3}$ and $\ce{Ga2O3}$ that we discussed above. Hence, one would expect this oxide to simply become more basic than the other group members(as it is an oxide of the lower-most element of the group) and not show much signs of an amphoteric nature. This matches with what is stated here.
Before wrapping up the answer, I would like to re-iterate a point I made at the beginning of the answer:
All these pathways are simply possible mechanisms, and there is no way of knowing whether they are completely correct or not without performing experiments. However, they are highly probable propositions, and you can expect these pathways to represent the actual mechanisms to a good approximation


What is exactly the pH value of distilled water? This was asked by one of my students in my class. I know it is the pH of 7, but others say it can be 6.55 pH

keep in mind that water (distilled, deionized, or tap) is NOT "pure" (i.e., pH equal to 7). The moment it comes in contact with air, CO2 gas begins dissolving into it, forming carbonic acid. The actual pH, therefore, will often be slightly less than 7.


Do I really need to wear protective clothing while using isopropyl alcohol? I recently purchased a 500ml bottle of 99% isopropyl alcohol (aka isopropanol aka rubbing alcohol) for everyday use at home. The label for the bottle worryingly states:

  Wear protective gloves/protective clothing/eye protection/face protection

I'm surprised by this particular warning because I've always known of isopropanol being used for such everyday household appliances as cleaning and sterilising wounds, and have never known even the experts to wear or recommend wearing any sort of protective clothing while doing so.

The most I expect to handle the isopropanol is when pouring the solution from the 500ml bottle into smaller dropper bottles, and after that either using the 500ml bottle to wet cleaning cloths, or the dropper bottle pipettes to apply the solution directly to electronics.

Do I need to wear protective clothing for such applications? If not, at what volumes/for what applications would it be necessary to start wearing protective clothing?

You'd be surprised at what things you might not think of as "chemicals" can do.  One of my co-workers got a spritz of  oil in his eye while peeling an orange, and now he is afraid when I tell him that I eat orange rinds because they are rich in vitamin C.  The moral of the story: your eyes are so sensitive that at the very least, eye protection is a must with just about anything in the lab, even oils.  I second Jon Custer's suggestion of reading the MSDS.


Is this problem about acid-base titration wrong? I wouldn't post mere 'problems' here normally, but I will hopefully be starting to give chemistry education to some 10-11-12th graders and I want to make sure that it is not I that has a problem but rather the problem in the book itself. Here is the problem:

$100~\mathrm{mL}$ of $\ce{H2SO4}$ solution is titrated with $5\cdot 10^{-2}\ \mathrm{M}$ solution of $\ce{NaOH}$. Here are the information that are given in the titration graph:
Total volume of $\ce{NaOH}$ solution added: $0$ | pH: $x$
Total volume of $\ce{NaOH}$ solution added: $400$ | pH: $7$
Total volume of $\ce{NaOH}$ solution added: $600$ | pH: $12$
According to what has been provided above, what is the $\mathrm{pH}$ of the initial $\ce{H2SO4}$ solution ($x$) ? (Assume that $\ce{H2SO4}$ fully dissociates in water to give $\ce{2H+}$ and $\ce{SO4^2-}$ ions)
A) 1
B) 2
C) 3
D) 4
E) 5

My answer: $-\log (2\cdot 10^{-1})$
My steps:

The $\ce{H+}$ ions in the acid solution have been neutralized by $400\ \mathrm{mL}$ of $5\cdot 10^{-2}\ \mathrm{M}\ \ce{NaOH}$ solution. That means that the amount of $\ce{OH-}$ ions in $400~\mathrm{mL}$ of that $\ce{NaOH}$ solution must equal the amount of $\ce{H+}$ ions in the acid solution.

The amount of $\ce{OH-}$ ions in $400~\mathrm{mL}$ of $5\cdot 10^{-2}~\mathrm{M}\ \ce{NaOH}$ solution can be found by taking the product of the liter of solution and the molarity of the solution. And that gives us $0.4\times 5\cdot 10^{-2} = 2\cdot 10^{-2}\ \mathrm{mol}\ \ce{OH-}$ ions. (Yes, I didn't use the units, I don't have to, I explained with words what I was going to do)

Since we used $2\cdot 10^{-2}\ \mathrm{mol}\ \ce{OH-}$ ions to neutralize the acid solution, that should mean that there were $2\cdot 10^{-2}\ \mathrm{mol}\ \ce{H+}$ in the acid solution.

To find the pH of the acid solution, we need to find the concentration of $\ce{H+}$ ions in that solution. The volume of the solution is known, $100~\mathrm{mL}$. And the amount of $ce{H+}$ ions too is known, $2\cdot 10^{-2}\ \mathrm{mol}$. The concentration of $\ce{H+}$ ions in the acid solution should therefore be $2\cdot 10^{-2} / 0.1 = 2\cdot 10^{-1}\ \mathrm{M}$.

$-\log[\ce{H+}] = -\log (2\cdot 10^{-1}) =$ none of the choices above.

Your calculations are correct. I also did a second sanity check: Adding $200~\mathrm{ml}\ \ce{NaOH}$ to a neutral solution does indeed give a $\mathrm{pH}\ 12$ solution, so they also wanted you to assume that the neutral point is the equivalence point.

Which answer to select? Well, considering that:

$$-\log(2 \cdot 10^{-2}) \approx 1.699 \approx 2$$

I would probably go with $2$. Along with telling the pupils that this is a badly worded question and in their exam there will be clearly solveable ones.


Examples from everyday life for diastereoisomers ? Like the amino-acids and sugars that are enantiomers, can you give any examples of diastereoisomers ? where can we find them in everyday life ?

Nowadays, it's really easy to find information on the Internet or in books. Being resourceful and being able to search for stuff yourself will help you greatly.

Nevertheless...

Carbohydrates

Glucose and galactose, two of the most common monosaccharides found in the body, are diastereomers of each other:

Lactose, which is found in milk, is formed by a condensation reaction between glucose and galactose.

Eyesight

Retinal is a molecule found in the photoreceptor cells of eyes that normally exists in the 11-cis form. When it absorbs light, it undergoes an isomerisation to form the all-trans form:

This forms part of the biochemical pathway by which light is converted into an electrical signal in the nervous system.


Reactivity of group 13 elements with nitrogen Do all group 13 elements react with nitrogen to form nitrides of form $\ce{MN}?$
J.D. Lee's Inorganic chemistry book mentions the following:

Aluminium burns in dinitrogen at high temperatures forming $\ce{AlN}.$ The other elements do not react.

However, a reaction of boron with nitrogen is mentioned elsewhere:
$$\ce{2B + N2 -> 2BN}$$
I'm assuming these boron and aluminium do react. Do gallium, indium and thallium react with nitrogen as well?
Please post your answer mentioning a reliable reference or source.

In the case of boron, a reaction with nitrogen does occur under some eye-popping conditions:

Combustion of boron powder in nitrogen plasma at 5500 °C yields ultrafine boron nitride used for lubricants and toners.[1]

The characteristic molecular thermal energy $kT$ at this temperature is roughly half an electron volt, which is enough to at least partially break a significant fraction of the nitrogen-nitrogen bonds.  The boron is also vaporized at this temperature, and both of these factors facilitate the reaction.  Vaporizing the boron also favors forming the fine particles which are desired for the lubricant and toner applications noted above.
Most boron nitride, not requiring the ultrafine particles, is synthesized by metathesis reactions under more typical industrial conditions, which are also given in the WP article noted above.  These commonly involve boron oxide, which both becomes fluid at much lower temperatures than the element and can act as an acid towards basic nitrogen sources such as ammonia.
Cited Reference
1.
Paine, Robert T.; Narula, Chaitanya K. (1990). "Synthetic Routes to Boron Nitride". Chemical Reviews. 90: 73–91.


What will be the major product of the reaction between (2R,3R)-2,3-dimethyloxirane and triphenylphosphine? Problem

My Attempt

But answer is given B. Where did I go wrong?

First, note that with the mechanism you have drawn you should expect to get a mixture of cis and trans isomers. In the intermediate you drew, rotation about the C–C bond will be fast and will lead to loss of any stereochemical information that was previously there.

The only way to get exclusively trans-2-butene is for the reaction to occur in a single step without a discrete intermediate:

However, that's not the case. The mechanism of epoxide deoxygenation by triphenylphosphine isn't a concerted pathway, but is stepwise.

Triphenylphosphine opens the epoxide ring first;  because this epoxide is symmetrical, it doesn't matter which carbon you attack. This gets you to a charge-separated intermediate 1a (a "betaine"). These species tend to combine their positive and negative charges if they can, and here they certainly can, but first the C–C bond in the middle has to rotate to get the negatively charged oxygen and the positively charged phosphorus on the same side (1b). Note that this bond rotation brings the right-hand side methyl group to the front of the molecule.

Cyclisation to the oxaphosphetane 2 then occurs, and collapse of this to give triphenylphosphine oxide and cis-2-butene is concerted and stereospecific, exactly like the final step of a Wittig reaction.

See also the Barton–Kellogg reaction for a similar mechanism, just with sulfur instead of oxygen.


What is happening in this video of solvated electrons donated from sodium in ammonia? I just saw the Periodic Video Liquid Electrons - Periodic Table of Videos where sodium is added to liquid ammonia. The demonstration shows that even if electrons are solvated, if you have a high enough electron density, the substance will turn shiny and reflective and metallic. Sodium is dissolved in liquid ammonia and donates the electrons to the solution.

There are a number of things happening in the demonstration that I don't understand.

There is supposed to be a separation - once a high enough electron density is reached, the solution is supposed to separate into a high electron density and low electron density layer. Why the separation instead of a uniform concentration? And why would the high electron concentration layer float on top?

Also, at concentrations below the metallic appearance, the sodium in ammonia solution starts out very strongly blue colored. Why? What is it about adding sodium to liquid ammonia that instantly produces such a deep blue, almost black color?

It seems that the demonstration did not go quite as planned, so isopropyl alcohol was added, producing sodium isopropoxide. Why?

I'll try an answer to this question because I watched this video a while back and did a bit of reading on it at the time and I think I understand the big picture. The problem is that these solvated electrons are very complicated things, and do not lend themselves to the traditional ways that chemists would like to think about things. For that reason, there is quite a lot of literature coming from physicists which is very complicated but nonetheless valuable. Additionally, solvated electrons show up in more than just this rather unusual metal-liquid ammonia solution. For instance, a recent paper in Nature chemistry by Sieferman et. al. [1] showed that there are transient solvated electrons at the surface of water which can have major implications for natural electron transfer reactions. So, understanding these systems is truly quite important.

The Blue Color:

In order to get at the deep blue color which is seen in this sodium-ammonia solution, we must understand what it is that is absorbing (reddish) visible light and hence leading to the transmission of a lot of blue light.

As it happens, at low concentrations of sodium (I'll address this in a bit), the solvated electrons exist in a bound state. This is not too dissimilar to ordinary electronic energy levels, except that there is not really a molecule to be found. Rather, the electron exists in a bound state which it creates for itself. That is, the electron polarizes the surrounding solvent such that it is contained in a bound state. Sometimes people refer to this electron as being a "polaron". One important difference between this and energy levels in a molecule is that electronic excitations generally take place from one bound state to another bound state. In the case of a solvated electron, however, the transitions are from a bound state to a continuum state.

Obviously one question that would be nice to answer is why is it such a deep blue solution? Well, this sort of question is always hard to address because the color of things is more or less an accident. The fact that it is blue just tells you that the electron is relatively weakly bound because it is excited by reddish light.

For more details on the bound-continuum transition, see Aulich's paper [2] which uses photo-electron emission to study solvated electrons as a function of concentration and photon energy. For a theoretical treatment of the bound states at dilute concentrations of metals, see Jortner's paper [3].

The Transition to a Bronze Color:

One reason that people have been very interested in this system is because of this transition that takes place as the concentration of metal increases. Let me first try to give some intuition as to why this transition takes place at all. First, going off of our model that the electrons polarize the solvent and exist in a low-energy, bound state, we should expect that as more electrons are added to the solution (i.e. as more sodium is added), these cavities which the electrons occupy will become closer and closer to each other. This means that electron-electron interactions become quite important. This means two things: first, we have to begin worrying about satisfying Pauli exclusion, and second, we have to be concerned that at some point the electrostatic repulsions will become larger than the binding energy of this electron.

These two points have been studied by some physicists. Alavi and Frenkel's paper [4] develops a model and then performs simulations on ideal fermions (I honestly don't know what makes a fermion ideal and don't understand this paper), demonstrates that the low-concentration bound states are very unstable as the concentration of electrons increases. Thus, some transition must take place.

The best paper on the theory of this transition can be found here [5]. The reason this transition is so interesting is that it is a "metal to non-metal" transition. Remember when you were first introduced to metals and the characteristic description is that they are a "sea of electrons"? That is, the electrons are free to move around, and this leads to high conductivity and other observable properties, one of which is that they absorb and reflect nearly all wavelengths of light, which leads to the silvery color of most metals.

Well, the exact same phenomenon is being observed here, but it's very peculiar because this is all happening in a liquid. For instance, Schroeder et. al. wrote a paper [6] which measures the conductivity of several metal-ammonia solutions as a function of concentration of the metal. I unfortunately couldn't add the relevant figure, but the conductivity basically increases linearly with concentration of the metal until it hits a plateau and does not increase any further.

This gradual increase explains what is going on when we see a separation into two phases. The explanation goes like this. There is only so much space which allows for the existence of the bound states we described above because these bound states occupy a cavity of relatively large volume in the solvent. As more metal is added, more electrons are free in the solution, but the solution is already saturated with these bound electrons. Thus, the electrostatic and exclusion effects become such that any additional electrons added can only exist in a metallic state. This is peculiar because this metallic state is in the liquid phase and is actually fairly dense (I read this somewhere but can't remember the reference... Will update later). If one continues adding electrons, they always become incorporated into the metallic state because the bound states are saturated. Eventually, enough electrons are present that the destabilizing effects due the presence of other electrons is large enough that no possible bound state can exist and the whole system becomes metallic. Hence the plateau in the conductivity plot I described above. At this point, I believe the sodium ceases to dissolve and just stays as plain old metal sodium.

I do not have a good explanation for why the metallic phase is specifically bronze colored as opposed to silvery. I suspect this is a very subtle feature and would certainly be temperature dependent. I haven't described the temperature dependence of all of this, but there quite a few papers that address only this point and it seems to be quite complicated.

I also don't know why the metallic phase forms on the top. I suspect this may just be an artefact of the experiment in that video. That is, it is entirely feasible that solid would sink to the bottom if it didn't just freeze to the side of the test tube. I could be wrong about that though.

References:

Siefermann, K. R., Liu, Y., Lugovoy, E., Link, O., Faubel, M., Buck, U., ... & Abel, B. (2010). Binding energies, lifetimes and implications of bulk and interface solvated electrons in water. Nature chemistry, 2(4), 274-279. DOI
Aulich, H., Baron, B., Delahay, P., & Lugo, R. (1973). Photoelectron emission by solvated electrons in liquid ammonia. The Journal of Chemical Physics, 58(10), 4439-4443. DOI
Jortner, J. (1959). Energy levels of bound electrons in liquid ammonia. The Journal of Chemical Physics, 30(3), 839-846. DOI
Alavi, A., & Frenkel, D. (1992). Grand‐canonical simulations of solvated ideal fermions. Evidence for phase separation. The Journal of chemical physics, 97(12), 9249-9257. DOI
Jortner, J., Cohen, M. H. (1976). Metal-nonmetal transition in metal-ammonia solution. Physical Review B, 13 (4), 1548-1568.
Schroeder, R. L., Thompson, J. C., & Oertel, P. L. (1969). Conduction in Concentrated Solutions of Several Metals in Liquid Ammonia. Physical Review, 178(1), 298. DOI
Thompson, J. C. (1968). Metal-nonmetal transition in metal-ammonia solutions. Reviews of modern physics, 40(4), 704.
Schroeder, R. L., Thompson, J. C., & Oertel, P. L. (1969). Conduction in Concentrated Solutions of Several Metals in Liquid Ammonia. Physical Review, 178(1), 298.


Where can I find a collection of density functional parameterizations? I want to implement a solution to the electronic Schroedinger equation using DFT. I know there are many, many functionals to choose from but don't know where to start. I'd like to start with a couple LDA and GGA functionals, is there a good place where I can look up some parameterization, or even the tabulated data itself to form my own parameterization?

Unless you are doing it just for fun and curiosity how the think works, you are better with some existing library. Most prominent is the libxc, offering the lowest level interface, i.e. given density (and possibly gradient) at series of points, it returns the energy. What is left for you is the construction of the grid and optimization of density, ie. you can go for the real Hohenberg-Kohn DFT, not just Kohn-Sham. For the usage examples (of the really nice and simple interface) see http://www.tddft.org/programs/octopus/wiki/index.php/Libxc:manual

And most importantly, you can find definitions of all the functionals in the list of implemented functionals


Is the enthalpy of a molecule determined by the energy of the electrons? I understand that the strength of the bond of a molecule is determined by the potential and kinetic energy of the electrons. I also understand that a reaction where the bond strength of the products is stronger than the bond strength of the reactant releases energy (is exothermic).

My textbook states that in a reaction where NaOH dissolves in water, the energy of the solid NaOH is greater than the energy of the dissolved ions in water. Is the "energy" of the reactants and products determined by the bond strength?

Also, my textbook states that enthalpy is "the amount of energy, in the form of either kinetic or potential energy, a substance has." I'm a bit confused about enthalpy. My textbook enthalpy changes:

Are these enthalpy changes determined by the bond strength also? And is the enthalpy of a certain molecule also from the bond strength?

I find that enthalpy is one of the concepts that students struggle with the most and I feel your pain in trying to understand it!

  I understand that the strength of the bond of a molecule is determined by the potential and kinetic energy of the electrons

The strength of a bond is best understood in terms of potential energy. A strong bond involves a deeper potential energy "well" that electrons and bonded atoms can exist in. The atoms involved in the bond (and the electrons too) will have kinetic energy and that can explain how the bond behaves over time, but the overall intrinsic strength of the bond is a reflection of how low the potential energy of the system is at the equilibrium bond length. We call this the bond dissociation energy and it directly tells us how strong the bond is.

https://en.wikipedia.org/wiki/Morse_potential

  My textbook states that in a reaction where NaOH dissolves in water, the energy of the solid NaOH is greater than the energy of the dissolved ions in water. Is the "energy" of the reactants and products determined by the bond strength?

Yes, this refers to the potential energy of the system that is decreasing when the stronger attractions are formed in the solution vs the weaker ones broken in the undissolved ionic lattice and bulk water. Essentially, you have to break weaker bonds/IMF's in the reactants then you form in the products. Please note that this is a general statement and includes all "bonds" (covalent, ionic or metallic) plus other types of attraction (IMF's, ion-dipole forces etc.) that exist before or after a process or reaction occurs.

  Also, my textbook states that enthalpy is "the amount of energy, in the form of either kinetic or potential energy, a substance has." I'm a bit confused about enthalpy. My textbook enthalpy changes:

The key part here is the potential energy. When this decreases (as in your example of NaOH dissolving), the potential energy turns into thermal energy. By itself, this doesn't change the enthalpy, as that is the sum of all internal energy at constant P. However, it will raise the temperature of the system and this makes it out of thermal equilibrium with the surroundings. Heat now flows from the system and this is when the enthalpy decreases as the system is losing energy. It is best to think of the enthalpy decreasing when the potential energy of the system decreases as the heat flow is what then tends to result. The reason this happens, as you are suggesting, is because stronger bonds are forming than existed before.

  Are these enthalpy changes determined by the bond strength also? And is the enthalpy of a certain molecule also from the bond strength?

Hopefully this last bit makes sense now, and the relationship between bond strength and the enthalpy of the system is a but less mysterious!!


Disease predicting software for small molecules Can anyone please help me in finding softwares for predicting diseases from small molecules?

Do you mean predicting health effects/toxicity of a given chemical? I haven't seen such software and I doubt it exists. Much easier problem (predict optimal synthetic route to compound X remains unsolved.

By hand your best way is to check toxicity of related compounds. A program that can predict health effects of compounds will revolutionize drug design (and cause job loss for 99% of RnD chemists in pharma).


Can the pyrophoric tendencies of a pyrophoric substance be suppressed in a pressurized environment? It's common knowledge that the pressure of an environment influences many properties of a substance, like the melting point, boiling point, or triple point. Pyrophoric substances ignite spontaneously at temperatures below $\pu{55^\circ C}$, like caesium for example.

I wish to ask if the tendency for pyrophoric substances to ignite is influenced at all by pressure? Obviously, these substances would not spontaneously ignite in a vacuum, but would their ignition temperature change and/or be dependent on the pressure of their environment? Is there a pressure apart from a vacuum that could cause the substance to lost its pyrophoric properties?

Depending on the pyrophoric substance, there can be differeing effects from changing air pressure. Strange as it may seem, sometimes combustion is slowed at very high pressure. For example, ethane, which reacts explosively with oxygen at STP, oxidizes slowly in a series of steps at ~100 atmospheres (~107 Pascals).

That said, it depends on the substances involved, pressure and presence of catalysts or other substances. Aluminum in a ball mill produces pyrophoric dust, but adding graphite during milling suppresses ignition.


Entropy as the driving force for osmosis How is entropy responsible for osmosis and is movement of solvent possible from its higher concentration to lower concentration?

By concentration, I will be referring to the concentration of solvent molecules below.

I understand that the entropy of the universe tends to increase. But I do not understand how this principle applies to osmosis. Basically, in osmosis the solvent moves from its higher concentration  to lower concentration. Here, it appears that the entropy of the lower concentration side is increasing and the entropy of the higher concentration of solvent side is decreasing. How can we claim that the net change is increase in entropy?

Also, is the reverse process i.e. movement of solvent molecules from  lower concentration side to their higher concentration side ever possible (= thermodynamically favorable) without the presence of external driving forces?

(a) The tendency for the solute undergoing diffusion to occupy as large a volume as possible is similar to that of a gas filling the volume available to it and in each case the driving cause is the increase in entropy. In any  distribution of particles there are more ways of distributing them in a larger volume than in a smaller one.

Recall that as the solvent enters the concentrated solution the solution volume increases and this continues until the chemical potential on either side of the semi-permeable membrane is equalised. The pressure is also increased as solution is pushed up into a column, the difference in pressure is the osmotic pressure.

(b) From the statements above you can see that reverse osmosis cannot occur without applying external energy.

Notes: The change in entropy of a gas with volume is $\displaystyle \frac{\partial S}{\partial V}=\frac{R}{V}$. In a dilute solution the mole fraction $x_s \sim 1/V$ where $V$ is the  volume of the solvent (or solution)  with one mole of solute. As $G-G^0=-T S = RT\ln(x_s)$ then $\displaystyle \frac{\partial S}{\partial V}=\frac{R}{V}$. Thus the change in entropy is the same for the dilution of a gas as for the solute in a dilute solution.


What makes elements radioactive? I know that certain elements like plutonium and certain isotopes of caesium are radioactive but what gives them this ability? If it's their electron/proton/neutron arrangement, what makes that arrangement give radioactive properties and how?

Radioactivity is not at all dependent on the atomic or molecular structure: it a purely nuclear phenomenon. The only thing that is relevant is the structure of the nucleus.

The nucleus contains only protons and neutrons and the distinguishing feature that makes nuclei radioactive is, crudely, the ratio of the two. Some arrangements are unstable for a variety of reasons a full understanding of which requires fairly advanced quantum theory. What you need to remember is that some ratios or arrangements of protons and neutrons are inherently unstable and spontaneously decay in a variety of ways to more stable nuclei. For example, 39K (which contains 19 protons and 20 neutrons) is stable but adding a neutron makes 40K which is slightly unstable and spontaneously decays in several possible ways. You can tell this isn't a simple phenomenon, though, as 41K, with two extra neutrons is stable again.

It is also important to note that radioactive decay is essentially uninfluenced by anything outside the nucleus. For any given nucleus there is a probability that it will decay but that event occurs randomly and we can't tell when it will happen (but the rate of decay is stable and not influenced by temperature, pressure or the chemical compounds the nuclei are present in).


Bonds in R-H3N+ group Source: Wikipedia

I am confused about the ammonium group in this molecule.

I can see that we have a nitrogen atom and 3 hydrogens. The group is attached to a carbon.

Nitrogen has atomic number 7 so my understanding is it can form up to 3 covalent bonds, and therefore cannot have covalent bonds with the carbon and all 3 hydrogens.

I can see a + on the nitrogen, suggesting it is an ion, and maybe some ionic bond?

Nitrogen along with it's 3 bonds has a lone pair. That's what comes into play. The H+ ion has no electrons of it's own and therefore when it forms bond with N, nitrogen coordinates it's lone pair to H+ completing it's duplet and at the same time it's octet and forms a coordinate bond also known as dative-bond.

You can find more about coordinate bond here


Negative oxidation states of Si According to List of oxidation states of the elements, silicon has a possible oxidation state of $-4$.

Now, I've been looking everywhere for a compound that contains Si(-IV), but I cannot find any definitive references. Perhaps one of the synthetic carbonyls of silicon has this OS? I've seen references to $\ce{[Mo(CO)_4]^{-4}}$ and $\ce{[W(CO)_4]^{-4}}$ in my research, but silicon never shows up.

I am also uncertain as to how these carbonyls come together, and whether or not they actually complex in the same way for metalloids as for the transition metals.

I am not sure if Silane ($\ce{SiH_4}$) can really be considered.

But there are silicides which silicon forms with strongly electropositive metals. In these compounds, silicon has a negative oxidation state.

For magnesium silicide - $\ce{Mg_2Si}$, the oxidation state of silicon would be -4.


Acidity order of xenon fluorides,oxides and oxofluorides The acidity order of xenon oxides,fluorides and oxofluorides is:
  $\ce{XeF6 > XeO2F4 > XeO3F2 > XeO4 > XeOF4 > XeF4 > XeO2F2 > XeF3 > XeO3}$.

How to justify this order?

My thoughts were that the acidity order of xenon fluorides, oxides and oxofluorides is based on its tendency to accept an oxide(as these are the most common reactions). As fluorine is more Electronegative than oxygen, we can safely say that $\ce{XeF6}$ would be the most(As $\ce{Xe}$ would be most electron deficient and  can accept an oxide easily). $\ce{XeO2F4}$ would be the second most and the last would be $\ce{XeO3}$, based on number Fluorine+Oxygen atoms, with priority to fluorine. However evidently, "the more number of fluorine atoms" logic doesn't apply to all of them. Looking at the structure of each molecule I'm unable to deduce any logical explanation for the above mentioned order.

Also, I know for a fact that $\ce{XeF6}$ corrodes glass and $\ce{XeF4}$ doesn't. Upto which xenon compound would corrode with glass?

You have to consider the oxidation state of the xenon.  Higher oxidation state makes the central atom a stronger Lewis acid in general, xenon is just one case.

So we should consider each oxidation state separately.  Look at the ordering when we do that:

+8:  $\ce{XeO2F4}>\ce{XeO3F2}>\ce{XeO4}$

+6:  $\ce{XeF6}>\ce{XeO2F2}>\ce{XeOF4}>\ce{XeO3}$

+4, +3: Only the fluorides are given, no oxygen vs fluorine comparison possible from the list.

The +8 and +6 orderings are consistent with the expected effect of oxygen versus fluorine within a given xenon oxidation state.


How to measure the color of honey? After doing some research, I found that honey's color could be a factor of climate. I am very interested in this subject and I'm a high school student so we don't have as much apparatus as you would find in a chemistry school. The color of the honey can be determined using wavelengths which means a spectrometer is required and it is not available in our school. It would be very helpful if any of you recommended me with a way from which I can carry out this experiment and measure the color of honey.

Thanks for your valuable help!

Could you use a reference color chart? Here is one originally made for skin colors:

Wikipedia has other examples.
Paint stores might give you a nicely printed color chart for free.

There are also online versions of color selectors, for example the one from Pantone, but then you have to be very careful about color calibrating your screen, or at least not changing it between comparisons.

A third option, more realistic if you are only measuring color along one scale, would be to make your own comparison material. Start with a light honey colored liquid and mix in a darker color till it matches your honey. The amount of dark color used gives you the darkness of the honey. You probably have all supplies available in the art department.

Edit:

MaxW found a much better solution, a special color scale for honey. It is called the Pfund scale:


Tautomer of acrolein Are $\ce{CH2=CH-CH=O}$ and $\ce{CH2=C=CH-O-H}$ tautomers?

There is only a shift of a hydrogen atom and a pi bond. So why are these two not tautomers?

The molecular geometry argument is pretty common-sensical, and is true to a certain extent. But I don't think it's the cleanest argument. If you look at a typical ketone and its enol, you will notice that the enol has a more constrained geometry than the ketone:

In the keto tautomer, the highlighted carbon is free to go wherever it likes, since it's part of an sp3 chain. But in the enol tautomer, by virtue of the 120° bond angles in the alkene, the highlighted carbon must be in the same plane as the double bond.

Furthermore, it's not possible to claim that the ketone and the enol have the same geometries in the carbon chain. Many of the bond lengths in the two tautomers will be different: after all one has a C–C (and C=O) bond and the other has C=C (and C–O). Unlike resonance structures, both of these tautomers are things which exist independently, and they have different bond lengths.

The safest way is to go back to the IUPAC definition. That's a bit boring, but when you read these, you can see the effort that has gone into coming up with a definition that is (as far as possible) consistent with edge cases such as these. Tautomerism is defined as (https://doi.org/10.1351/goldbook.T06252):

  Isomerism of the general form $\ce{G-X-Y=Z <=> X=Y-Z-G}$ where the isomers (called tautomers) are readily interconvertible.

Acrolein and propadienol do match the bonding pattern required for tautomerism, but don't pass the second criterion of being "readily interconvertible".

The reason is fairly simple. The process of tautomerisation involves protonating the oxygen, and deprotonating an α C–H (which way round this happens doesn't matter). To deprotonate a proton that is α to C=O, the C–H bond must be oriented parallel to the C=O π system, because the new C=C π bond is going to be made from that C=O π bond and the C-H σ bond. This cannot be the case in acrolein, because of the extended conjugation between the C=C and C=O bonds. Indeed, the α C–H bond in acrolein is perpendicular to the C=O π system.


Can you find [SO₃²⁻] with just the two acid dissociation constants Kₐ₁ and Kₐ₂? Sulfurous acid, $\ce{H2SO3}$, is a diprotic acid.
\begin{align}
K_{a1} &= 1.6 \cdot 10^{-2}\\
K_{a2} &= 6.4 \cdot 10^{-8}\\
\end{align}

Is it possible to find the concentration of $\ce{SO3^2-}$ from this?

I would think you would need to know the concentration of at least one thing. It is like trying to solve for 3 unknowns with 2 equations!

No, it is not possible.

For example, as total concentration of acid approaches zero, concentration of sulfate would also approach zero.


Reactivity of chlorobenzene and benzene in electrophilic substitutions As we know, chlorobenzene is a deactivating substituent and benzene has a stable structure. Now I'm confused.

My question is — considering electrophilic aromatic substitution is chlorobenzene or benzene more reactive?

considering electrophilic aromatic substitution is chlorobenzene or benzene more reactive?

Let's expand the question to include all of the halobenzenes.

All of the halogens ($\ce{F, Cl, Br, I}$) are more electronegative than hydrogen.  Therefore they will inductively remove electron density from a benzene ring. However, you can also draw resonance structures for the various halobenzenes where the halogen substituent is donating electron density to the ortho and para positions of the benzene ring through resonance.

Therefore, in the case of the halobenzenes these two effects are working in opposite directions.  In order to determine which effect controls the reaction, we need some data.

The following table compares the relative rates of electrophilic aromatic nitration of the halobenzenes to the rate for benzene itself. If we assign benzene a relative rate of 1, we see that all of the halobenzenes react slower than benzene itself. This tells us that the inductive effect is stronger than the resonance effect in the halobenzene series.
\begin{array} \hline
 \text{Relative rates of aromatic electrophilic nitration} \\ \hline
\end{array}
\begin{array}{|c|c|c|c|} \hline
\ce{Ar-X} & \text{Relative rate}  \\
\hline
\ce{Ar-H} & 1.0 \\ \hline
\ce{Ar-F} & 0.11 \\ \hline
\ce{Ar-Cl} & 0.02 \\ \hline
\ce{Ar-Br} & 0.06  \\ \hline
\ce{Ar-I} & 0.13 \\ \hline
\end{array}

(data source: see p. 26 here)

However, when the halobenzenes do undergo electrophilic aromatic substitution, they react preferentially at the ortho and para positions, because, as explained above, the resonance effect donates electron density to the ortho and para positions.  Therefore, these two positions are the least deactivated and react preferentially.