How is chlorophyll made in plants? I've been studying photosynthesis and something made me think about how chlorophyll itself is made in plants.How exactly is it formed ?

In the first phase of chlorophyll biosynthesis, the amino acid glutamic acid is converted to 5-aminolevulinic acid (ALA) . This reaction is unusual in that it involves a covalent intermediate in which the glutamic acid is attached to a transfer RNA molecule. This is one of a very small number of examples in biochemistry in which a tRNA is utilized in a process other than protein synthesis. Two molecules of ALA are then condensed to form porphobilinogen (PBG), which ultimately form the pyrrole rings in chlorophyll. The next phase is the assembly of a porphyrin structure from four molecules of PBG. This phase consists of six distinct enzymatic steps, ending with the product protoporphyrin IX.
All the biosynthesis steps up to this point are the same for the synthesis of both chlorophyll and heme. But here the pathway branches, and the fate of the molecule depends on which metal is inserted into the center of the porphyrin. If magnesium is inserted by an enzyme called magnesium chelatase, then the additional steps needed to convert the molecule into chlorophyll take place; if iron is inserted, the species ultimately becomes heme.
The next phase of the chlorophyll biosynthetic pathway is the formation of the fifth ring (ring E) by cyclization of one of the propionic acid side chains to form protochlorophyllide. The pathway involves the reduction of one of the double bonds in ring D, using NADPH. This process is driven by light in angiosperms and is carried out by an enzyme called protochlorophyllide oxidoreductase (POR). Non-oxygen-evolving photosynthetic bacteria carry out this reaction without light, using a completely different set of enzymes. Cyanobacteria, algae, lower plants, and gymnosperms contain both the light-dependent POR pathway and the light-independent pathway. Seedlings of angiosperms grown in complete darkness lack chlorophyll, because the POR enzyme requires light. These etiolated plants very rapidly turn green when exposed to light. The final step in the chlorophyll biosynthetic pathway is the attachment of the phytol tail, which is catalyzed by an enzyme called chlorophyll synthetase (Malkin and Nyogi 2000).

Source: Plant Physiology Online: Chlorophyll Biosynthesis


What is the difference between chemical equilibrium and dynamic equilibrium? Is chemical equilibrium and dynamic equilibrium the same thing? Both point to the reversible reactions and no net changes in the concentrations of product and reactant.

Dynamic equilibrium from Wikipedia:

  In chemistry, and in physics, a dynamic equilibrium exists once a reversible reaction occurs. Substances transition between the reactants and products at equal rates, meaning there is no net change. Reactants and products are formed at such a rate that the concentration of neither changes.

Chemical equilibrium from Wikipedia:

  In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system. Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the forward and backward reactions are generally not zero, but equal. Thus, there are no net changes in the concentrations of the reactant(s) and product(s). Such a state is known as dynamic equilibrium.

Chemical equilibrium is a type of dynamic equilibrium, but not every dynamic equilibrium is a chemical equilibrium.

In a chemical equilibrium there is no change on the macroscopic scale. That means that if you look at the system it seems like nothing is happening, but at molecular scale there are reactions going on and the rate of forward reaction = rate of backward reaction.

There is another example of dynamic equilibrium (although not the best one, as it is not a closed system):
Imagine a bath into which water is flowing in at constant rate. Water is also flowing out of the bath at the bottom with the same rate.
This is an example of dynamic equilibrium, which has nothing to do with chemistry. If you look at the bath, the water level is constant, so on the macro level nothing is happening, but there is flow of water, so on micro level water from the tap is replacing water flowing out of the bath.


Salt Bridges/Porous Disks I am slightly confused about the concept of a salt bridge/porous disk. I understand that these are necessary in a galvanic cell since there is a charge buildup and the salt bridge and porous disks allow ions to flow in order to neutralize this charge and allow for the electricity to continue flowing.
What I am confused about is what ions are flowing. As an example I'm going to talk about a zinc-copper galvanic cell in which the zinc is oxidized at the anode and the copper is reduced at the cathode. What ions flow through the porous disk? Is it some of the $\ce{Zn^{2+}}$ ions and the anion from the copper solution, or is it separate ions? Is this the same in a salt bridge or is the salt used in the bridge the one that moves?
For example, if a salt bridge is made with $\ce{KCl},$ does the $\ce{K+}$ go to the cathode and the $\ce{Cl-}$ go to the anode or do the $\ce{Zn^{2+}}$ ions move through the salt bridge?
In other words, what ions flow through a salt bridge? Is it the ions from the separate solutions in the two half cells or the ions inside the salt bridge? What about a porous disk that has no ions inside of it?

It depends on the geometry/construction of the bridge. In a typical glass-tube bridge, some small amount of the ions from the cell may eventually make their way all the way across the salt bridge, but over reasonable amounts of time, it's mostly the ions from the salt bridge that move into the half-cells (and the ions from the half-cells moving into the ends of the bridge). For a small filter paper strip being used as a bridge, the time it takes should be shorter, as there is much less solution in the salt bridge and the distance is likely shorter. Sometimes, glass salt bridges are filled with gels to limit diffusion further (at the expense of increasing the resistance of the bridge).

The goal of using a salt bridge and separating the half-cells is to prevent the solutions of each from directly mixing. If significant quantities of the ions from the half-cells are crossing the bridge, then they can react directly, bypassing the electrode connections. e.g. if we have a simple galvanic cell of $\ce{Zn∣Zn^2+∥Cu^2+∣Cu}$ and $\ce{Cu^2+}$ makes it across the separator to the zinc size, it can be directly reduced onto the zinc electrode, instead of by the electrons flowing through the electrode connections. Having a single beaker with both $\ce{Cu^2+}$ and $\ce{Zn^2+}$ would thus make a poor galvanic cell as most of the reagents would be wasted reacting directly, rather than doing work in the outside circuit.

With a simple porous separator, there are no other ions than the ones in the half-cells, so naturally, these are the ions that flow across, with the goal simply to prevent bulk mixing of the two solutions.


Why is propanediol soluble in water? I'm a crafter who likes to make her own body oils and butters but I have no background in chemistry. Now, I just read something about propanediol (which I'm considering for modifying the feel of my anhydrous product) and on one site it says that in appearance it is a clear colourless  oil but then it also mentions it's solubility is in water.

Confused.

So, I want to know how it can be an oil and be soluble in water, but more importantly, I want to know if this means it's not soluble in oil and hence I can't use if for my anhydrous product?

This question requires some additional specification before it can be answered.

Propanediol is ambiguous as a chemical name. There is 1,2-propanediol and 1,3-propanediol, and these are different compounds (namely, structural isomers).

Though either way, if you check references, you will see that both compounds are miscible with water, so the solubility is infinite.

See: 1,2-propanediol and 1,3-propanediol.

The term "oil" simply refers to the fact that it is a viscous liquid at room temperature.

Unfortunately, generalized solubility is hard to determine. 1,2-propanediol appears to be miscible in several different polar solvents, so you might be able to dissolve in something anhydrous.


Electron affinities of the chalcogens and halogens Here are the electron affinities of the 16th and 17th groups.

The general trend for electron affinity down the group is that it decreases because of the increase in atomic radius.The exception of $\ce{Cl}>\ce{F}$, I can understand is because fluorine has a high electron density and it is unfavourable to add more electrons as it would only increase the electron -electron repulsion.

The same reasoning can be applied to $\ce{S}>\ce{O}$. But oxygen's electron affinity is lesser than all the elements down group 16. Why is this so? And how is it so different from group 17 or why doesn't group 17 have $\ce{F}$ being less than the rest?.

I tried to apply the reasoning that the electron-electron repulsion in oxygen is so much that it is less favourable for $\ce{O}$ than any other element in group 16 to gain an electron. But in that case $\ce{F}$ is an even smaller atom hence with even more repulsion. So why are these two groups so different?

The amount of screening is the same in the two groups but the effective positive charge that the incoming electron feels is stronger in Group 7.

For instance, oxygen has the electronic structure of
$$\ce{1s^{2} 2s^{2} 2p_{x}^{2} 2p_{y}^{1} 2p_{z}^{1}}$$ and 8 protons. The effective positive charge that pulls the incoming electron (which approaches the second level) is $8 - 2 = 6$.

In fluorine we have,
$$\ce{1s^{2} 2s^{2} 2p_{x}^{2} 2p_{y}^{2} 2p_{z}^{1}}$$ and 9 protons. The effective positive charge is $9 - 2 = 7$.

Therefore, the net pull from Group 7 is stronger than Group 6.


Density of cyclic hydrocarbons Why is the density of cyclic hydrocarbons more than that of their acyclic counterparts?

For ex:

Cyclopentane: 751 kg/m³
Pentane: 626 kg/m³

Cyclohexane:779 kg/m³
Hexane:655 kg/m³

In fact, as you can see above, cyclopentane is even denser than hexane.

My reasoning so far is:

Attractions in hydrocarbons depend only on London forces, which increases with surface area.
An alkane will have more surface area for bonding as compared to a cycloalkane (The inside of the ring cant be used), so it will have more forces, and hence will be packed closer (be denser).

But this contradicts the facts.

A simple explanation might be that the ratio of carbon to hydrogen is higher in the cyclic hydrocarbons, and carbon is a heavier element than hydrogen.


Is iron(III)nitrate changed and if so how? We have the solution:

$\ce{Fe^3+(aq) + SCN-(aq) <=>Fe(SCN)^2+}$

When we add more $\ce {SCN-}$ to the solution above, the solution produces more product.

My question is, how does it affect $\ce {Fe^3+}$? Do the concentration of iron(III)nitrate change or something?

If a stoichiometric amount of $\ce{SCN-}$ is added then the reaction is mostly

$$\ce{Fe^3+(aq) + SCN-(aq) <=>Fe(SCN)^2+(aq)}$$

If an excess of of $\ce{SCN-}$ is added then the reaction is mostly

$$\ce{Fe^3+(aq) + 2SCN-(aq) <=>[excess SCN-] Fe(SCN)2^+(aq)}$$


Can aluminium form blue-colored solution in liquid ammonia? I am a high school student and in one class, my inorganic chemistry teacher told group 1 elements with the exception of lithium and group 2 elements with  exception of Be form blue-colored solutions in liquid ammonia (Mg requires an electrolytic process). What are the conditions for a metal to form a solution in liquid ammonia, and which metals (other than what I have mentioned) form blue-colored solution in liquid ammonia? Can aluminium also form such solution?

The SE answers quoted in the comments provide a good broad answer but with respect to the question of aluminium – aluminium will form a blue solution in liquid ammonia following electrolytic reduction of $\ce{AlI3}$ in liquid ammonia according to J. Am. Chem. Soc. 1950, 72 (11), 5178–5180.
Europium and ytterbium have also been reported to dissolve in liquid ammonia in J. Phys. Chem. 1956, 60 (11), 1590–1591.


How much of Sodium dichloro-s-triazinetrione dihydrate should I use to replace bleach? I am trying to determine how much sodium dichloro-s-triazinetrione dihydrate powder I should add to water to create a dilution that can be injected into my water supply to react with the Hydrogen Sulfide and remove it.

I have a system in my home that uses sodium hypochlorite (bleach) and water. There is a 30 gallon water barrel that we mix 1 gallon of bleach (5.25% Sodium hypochlorite) with 29 gallons of water. I believe that means the resulting mixture is 0.16666% Sodium hypochlorite. That is injected using a small pump into the water supply where it mixes in a tank and oxidizes the Hydrogen Sulfide.

I want to stop buying bottles of bleach which is expensive and contains additives other than sodium hypochlorite. I would like to switch to sodium dichloro-s-triazinetrione dihydrate which I already have an abundance of in the form of Spa Sanitizer. The Spa Sanitizer label indicates it is 99.99% pure sodium dichloro-s-triazinetrione dihydrate.

So how much of the sodium dichloro-s-triazinetrione dihydrate powder do I need to mix with 29 gallons of water to create a solution similar to the one created by the bleach? (0.16666% sodium hypochlorite)

Or, maybe what I should be asking is what is the amount of free chlorine in the 0.16666% sodium hypochlorite solution and how much sodium dichloro-s-triazinetrione dihydrate powder do I need to mix with 29 gallons of water to achieve the same amount of free chlorine? Since I believe the free chlorine in sodium hypochlorite and sodium dichloro-s-triazinetrione dihydrate are not actually the same.

According to a relevant reference (Ref.1) Following reactions happen when sodium hypochlorite and sodium dichloro-s-triazinetrione dissolve in water, respectively:

$$\ce{NaOCl + H2O -> HOCl + NaOH} \tag {1}$$

$$\ce{NaCl2(NCO)3 + 2 H2O <=> 2 HOCl + NaH2(NCO)3} \tag {1}$$

Hence it is safe to assume that you need only $\pu{0.5 mol}$ of $\ce{NaCl2(NCO)3}$ (sodium dichloro-s-triazinetrione) or in your case $\ce{NaCl2(NCO)3.2H2O}$ (sodium dichloro-s-triazinetrione dihydrate), in place of each $\pu{1.0 mol}$ of $\ce{NaOCl}$ used. Let's see how much of $\ce{NaOCl}$ in $\pu{1.0 US gal}$ of $5.25\%$ bleach solution:

$$5.25\%(w/v) = \\
 \frac{\pu{5.25 g} \;\ce{NaOCl}}{\pu{100 mL} \;\text{solution}}\times \frac{\pu{1.0 mol} \;\ce{NaOCl}}{\pu{74.44 g} \;\ce{NaOCl}}\times \frac{\pu{1000 mL} \;\text{solution}}{\pu{1.0 L} \;\text{solution}}\times \frac{\pu{3.79 L} \;\text{solution}}{\pu{1.00 US gal} \;\text{solution}}\\
 = \pu{2.673 mol/US gal}$$

As states above, you need only half of this amounts in solid $\ce{NaCl2(NCO)3.2H2O}$. You can calculate the amount of $\ce{NaCl2(NCO)3.2H2O}$ in grams as follows:

$$\pu{1 US gal} \: \text{of } 5.25\%(w/v)  = \\
\pu{2.673 mol}  \; \text{of } \ce{NaOCl} \times \frac{\pu{0.5 mol} \; \ce{NaCl2(NCO)3.2H2O}}{\pu{1.0 mol} \; \ce{NaOCl}} \times \frac{\pu{255.98 g} \;\ce{NaCl2(NCO)3.2H2O}}{\pu{1.0 mol} \;\ce{NaCl2(NCO)3.2H2O}} \\
= \pu{342.1 g}  \;\ce{NaCl2(NCO)3.2H2O}$$

Thus, you can dissolve $\pu{342.1 g}$ of $\ce{NaCl2(NCO)3.2H2O}$ in your 30-gal (US) water barrel to get the same effect of 1-US gal of 5.25% bleach solution. Keep in mind that rate of chlorine release from $\ce{NaCl2(NCO)3.2H2O}$  is slower than that of bleach solution (Ref.1).

Late Edit:

This is about comment made by OP on my post. In that comment, OP states that the cost for dichloro-s-triazinetrione dihydrate (Di-Chlor) and beach are almost same. But I do not agree with that. Just for curiosity, I checked the prices for two chemicals: You can buy $\pu{50 lbs}$ bucket of Di-Chlor granules [online] from Pool Supply World for $\pu{132.99 USD}$; and You can buy $\pu{121 fl \: OZ}$ container of 6% bleach solution from Target for $\pu{4.49 USD}$ (let's assume its about $\pu{1 US\:Gal}$ of 5.25% bleach solution). According to my calculations above, $\pu{50 lbs}$ bucket of Di-Chlor granules equal to:

$$\pu{50.0 lbs} \; \ce{NaCl2(NCO)3.2H2O} \times \frac {\pu{454 g}}{\pu{1.0 lbs}}  \times \frac{\pu{1.0 US\:Gal}  \;\text{5.25% }\ce{NaClO}}{\pu{342.1 g}  \;\ce{NaCl2(NCO)3.2H2O}}\\
 \approx  \pu{66.5 US\:Gal} \text{ of 5.25% }\ce{NaClO} \text{ solution} $$

As a consequence, the cost for $\pu{50 lbs}$ bucket of Di-Chlor granules is $\pu{132.99 USD}$ while the cost for $\pu{66 US\:Gal}$ of 5.25% bleach solution  containers is $\pu{4.49 USD} \times 66  \approx \pu{296.34 USD}$ (more than twice the cost for Di-Chlor granules needed).

References:

Thomas Clasen, Paul Edmondson, "Sodium dichloroisocyanurate (NaDCC) tablets as an alternative to sodium hypochlorite for the routine treatment of drinking water at the household level," International Journal of Hygiene and Environmental Health 2006, 209(2), 173-181 (https://doi.org/10.1016/j.ijheh.2005.11.004).


When can I use concentrations instead of activities in equilibrium calculations? To calculate an equilibrium constant, it is said that you should actually use activities $a_i$ instead of concentrations $c_i$. But it is also said that within a certain range, you can use concentrations instead. This is also what is usually done because it's easier.

So what is this range? When can I safely use concentrations and when should I definitely switch to using activities?

To calculate any equilibrum constant we need to be sure of the fugacity of each species. Note that fugacities are defined for every common state of matter (gas, liquid and solid), although they are most commonly used for gases. Look at the end of this answer for more details.

The activity of the coumpound $i$ written $a_i$ is defined as:

$$a_i=\exp\left(\frac{\mu_i -\mu_i^{\circ}}{RT}\right)$$

where $\mu_i$ is the chemical potential, which depends on the fugacity.

However this expression can be hard to use if you have not a strong idea of what you are doing because the are a lot of different ways to calculate things, depending on the formulas and laws or theories used.

What you must remember all the time

Activity can also be defined as $$a_i=\gamma_i \cdot \frac{X_i}{X_i^\circ}$$

where $X_i$ can be a concentration, a partial pressure, a molar fraction, a molality, and so on, and where $\gamma_i$ is the activity coefficient. Be careful; this coefficient depends on what $X_i$ is.

The role of $\gamma_i$ is to correct for the interactions between each constituent of the mixture.

If you are in the ideal regime, which means that the amount of your solute is almost negligible with respect to the entire mixture, then $\gamma_i \approx 1$. Only then can you use $X_i$ in place of $a_i$.

In general, the determination of this coefficient is not quite simple, especially for ions: see the example here for hydrogen ion.

Ionic solutions

Electrolyte solutions (e.g. a solution of $\ce{NaCl}$) can be modelled by Debye–Hückel theory. There are different versions of this model depending on the precision you want. One of the most used is given in the paper about hydrogen ion:

$$\log(\gamma_i)=-\frac{A_m z_i^2\sqrt{I_m}}{1+\sqrt{I_m}}$$

where $I_m$ is the molar ionic strength, given by:

$$I_m=\frac{1}{2}\sum_i z_i^2m_i$$

where $m_i$ is the molality of $i$.

To answer the other part of your question, whether your system is behaving "ideally" or not ultimately depends on the precision you want.

For example if you have a reaction constant expressed as:

$$K=\frac{a_\ce{A}^2 a_\ce{B}^3}{a_\ce{C}}=\frac{\gamma_\ce{A}^2 \gamma_\ce{B}^3}{\gamma_\ce{C}} \cdot \frac{(c_\ce{A}/c^\circ)^2 (c_\ce{B}/c^\circ)^3}{c_\ce{C}/c^\circ}=\frac{\gamma_\ce{A}^2 \gamma_\ce{B}^3}{\gamma_\ce{C}} K^\circ$$

which corresponds to the reaction $\ce{C -> 2A + 3B}$, you will have a large error if you assume you are under ideal conditions when you aren't.

For example if you assume ideality in calculating the $\mathrm{pH}$ of $\pu{5 mol/L}$ of $\ce{HCl}$ solution, like I did here, the answer will not be the same as if you were to consider the activity of the ion, like this.

So at least for strong acid, if the concentration is much than $\pu{1 mol/L}$, you must use activity coefficients. Depending on your problem you'll need to think by yourself and do calculations with and without assuming ideality. If you are unsure as to whether your solution can be considered as being ideal, it's good to do a check (and feel free to use a computer in hard cases).

I would add, as an example, if one of your compounds is 10% of the solution in mass, the solution must no longer be considered as ideal. But, as said earlier, the limit of what is ideal is fixed by you, depending on the precision you want.


How do biochemists seperate nucleotides from DNA to add to the PCR How do chemists separate single nucleotides from DNA strands?  How do they add those nucleotides to e.g. a PCR reaction in order to replicate DNA?

I ask this because in my microbiology class, my teacher didn't know the exact answer.

Do you cleave off nucleotides by hydrolysis?  Do you somehow build nucleotides?

You just buy the nucleoside triphosphates from your favorite supplier, they're commercially available.

As far as I know the NTPs you can buy are still produced biologically and not synthesized (see this example from Sigma Aldrich, which states a microbial source). So you grow a large amount of bacteria, lyse the cells to get at the contents and then digest the DNA or RNA with an appropriate enzyme that cuts it into single nucleotides. To get the triphosphate from that you have to phosphorylate the monophosphates again with the appropriate kinase.


Activity of Polonium in half-time in Becquerels I got an exercise but I don't know the exact answer.
In the catalog note of the reference source of Polonium-210 radiation with a half-life of 138 days, it was stated that one day the activity was $\pu{2 \times 10^6 MBq}$. Calculate activity after 558 days.
After my calculations the final answer was ${126468.14}$, but my professor has told me that my answer should be about 1,7 times higher.

Well, first of all, I should point out that your answer and the would be the answer ($1.7$ times your answer) are both wrong by the fact that after 552 days (4 half-lives), the activity should be $\displaystyle\frac{\pu{2E6 MBq}}{16} = \pu{1.25E5 MBq}.$ Since this is smaller than both your answer and the would be answer, they both would be incorrect. Accordingly, correct answer would be smaller than $\pu{1.25E5 MBq},$ because the sample would be further decays for another six days. Let’s do the proper calculations:
The half-life $\left(t_\frac{1}{2}\right)$ of a radioactive element is defined as
$$t_\frac{1}{2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}.\label{eqn:1}\tag{1}$$
The $t_\frac{1}{2} = \pu{138 d}$ in the equation \eqref{eqn:1} for your case. Thus from the equation \eqref{eqn:1}, you can calculate for $\lambda:$
$$\lambda = \frac{\ln 2}{t_\frac{1}{2}} = \frac{0.693}{\pu{138 d}} = \pu{5.02E-3 d-1}. \tag{2}$$
Also, the radioactive decay is according to the equation:
$$A_t = A_0\mathrm e^{-\lambda t}.\label{eqn:3}\tag{3}$$
where $A_0$ is initial activity and $A_t$ is the activity of the same same sample after time $t$ (Note: $t_\frac{1}{2}$ and $t$ should have same units). Thus, applying the $\lambda$ value on the equation \eqref{eqn:3}:
$$A_t = \pu{2E6 MBq}\times\exp\left(-\pu{5.02E-3 d-1}\times\pu{558 d}\right) = \pu{1.21E5 MBq}.$$
Note: I disregard the correct significant figures because I believe the initial activity should be given in three significant figures.
For fun, you can also checked the validity of the answer using my initial remark. You can use the activity of the sample after four consecutive $t_\frac{1}{2}$s (after $138 \times 4 = \pu{552 d})$ as its initial activity and find the activity after another six days using the equation \eqref{eqn:3}:
$$A_t = \pu{1.25E5 MBq}\times\exp\left(-\pu{5.02E-3 d-1}\times\pu{6 d}\right) = \pu{1.21E5 MBq}.$$


Which substance has the higher oxidation strength? I have to decide which one has a higher oxidation strength $\ce{ClO-}$ or $\ce{Cl2}$

$\ce{ClO- + H2O + 2e- -> Cl- + 2OH-}$

$\ce{Cl2 + 2e- -> 2Cl-}$

Will this reaction occur?

$\ce{ClO- + H2O + 2Cl- -> Cl- + 2OH- + Cl2}$

In my opinion not, because otherwise the $\ce{ClO-}$ is reduced to $\ce{Cl-}$. If the reaction works, the $\ce{Cl-}$ could react as reduction agent again. Therefore if I would add $\ce{Br2}$ to $\ce{ClO-}$:

$\ce{2 ClO- + 2 H2O + 4Br- -> \mathbf{2 Cl- }  + 4OH- + 2Br2 }$

$\ce{ClO- + H2O + 2Br- + \mathbf{2 Cl- } -> \mathbf{Cl- } + 2OH- + Br2  + \mathbf{Cl2}}$

So therefore chlorine has a higher oxidation strength than $\ce{ClO-}$?

Look right here ... .

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/electpot.html

Hypochlorite ion (+0.90 V) has a lower reduction potential in basic solution than elemental chlorine (+1.36 V), so the hypochlorite ion is less strongly oxidizing and more stable in basic solution (which is what you write the reaction for).

Household "chlorine" bleach, which really uses hypochlorite, is made alkaline as one way to stabilize the hypochlorite ion and prevent evolution of chlorine gas.


What is heterogeneous catalysis in gas phase? Give examples of heterogeneous catalysis (i) in gas phase and (ii) in liquid phase.

Now, I came to know that heterogeneous catalysis involves different phases between the catalyst used and the reactants. That way, either the catalyst can be in solid phase where the reactants can be in gas/liquid phase. But in the above question, what does "in gas phase" mean? Is it about the gas phase of the catalyst or the gas phase of the reactants?

Also, it would be of great help if you could provide me with examples of the above question.

"Gas phase" means that the reactants are gases
The basic answer is that heterogeneous catalysis "in the gas phase" just means that the materials being reacted are gases. They catalysts can be liquids or solids (more common).
A simple, but very, very important example is the reaction used to convert ammonia into nitric acid (vital for fertiliser production, important for many organic chemistry building blocks and explosives). In this reaction gaseous mixtures of air and ammonia are passed over platinum gauze and, initially, produce NO (nitric oxide) which is further oxidised homogeneously to dinitrogen tetroxide which ultimately reacts with water to give nitric acid. A good summary of the reaction and its history is given in this Johnson Matthey article.
So gas-phase reactants + solid catalyst -> product. Gas phase heterogeneous catalysis.


Why is the solid product washed with ice-cold solvent during in the last stages of recrystallisation? Why is the solid product washed with ice-cold solvent during in the last stages of recrystallisation? Is this to remove soluble impurities? (for example) Well, what if the impurities were also insoluble in the cold solvent?

Recrystallisation is weird in the sense that we have to make some assumptions in order for the process to fulfill the 'ideal' purification of the solid product that we're trying to get to.

For example, we have to assume that all soluble impurities are washed off during suction filtration, and that no impurities manage to creep up within the molecules of the crystals when they form. Would I be correct in saying this?

Possible sources of impurities in the recrystallization process can be described in terms of phenomena occuring during coprecipitation and which are the basis of gravimetry:

inclusion (impurity occupies positions in the crystal lattice);
occlusion (impurity is physically captured within the crystal as it has been growing);
adsorption (impurity is adsorbed on the surface layer of the crystal);
postprecipitation (impurity forms secondary solid phase once recrystallization of the target stopped/slowed down).

Washing crystals with a fluid only addresses the last two phenomena, adsorbance and postprecipitation. Postprecipitation is rarely a problem as it occurs on a large time scale or when a really poor solvent has been chosen.

Properties of the washing liquid vastly depend on the properties of the compound being recrystallized. In the simplest case it's the same solvent (as in your example), but it can also be an electrolyte, hydrolysis suppressor or another solvent (e.g. distilled water) altogether.

As for the temperature of the washing liquid, it's actually better to use many small portions of the hot solvent as filtration of the hot solutions happens much faster, surface adsorption decreases with the temperature and removal of mother solution from the surface is faster. However, if solubility of the recrystallized product is significantly increasing with the temperature (and this is typically the case), then one has to sacrifice the efficiency of the washing process for the better yield and use cold solvent instead.

Impurities, insoluble in the cold solvent, are practically not an issue. First, recrystallization starts from a homogeneous solution. Second, before crystals are washed, they are separated from the mother solution so that only minor amount of it is left on the crystals' surface, and if there were any impurities, they don't get a chance to nucleate as their concentration in washing solvent is going to be negligible.

As for your perception of the recrystallization as of a "weird" process, I'd say it's probably more complicated than one would expect from looking at the beaker with some crystalline mess. In fact, underlying processes are described with more or less appropriate models, and once a person
 gets experienced enough, it all starts to make much more sense.


What is the meaning of 'anion water' in coordination chemistry? In Mechanisms of Inorganic Reactions by Fred Basolo and Ralph G. Pearson, a passage discusses the history of coordination chemistry. Ammoniates had been assumed to be analogous to hydrates. They assess the comparision as,

  This analogy is essentially correct for in hydrates, except for interstitial and anion water, the water is coordinated to the central metal ion. [emphasis added]

What is the meaning of anion water, as used in this context?

I’m pretty sure that ‘anion water’ is a term they use to describe water which coordinates the anions by hydrogen bonding rather than having anything to do with the cations. This is again to be separated from interstitial water, whose job it is to stabilise either of the other water types by hydrogen bonding without interacting with cations or anions at all. (Or maybe, whose job it is to fill voids in the crystal structure).

Ammonia in coordination compounds typically does what water does if it is not interstitial or hydrogen-bonding to anions, thus the distinction and the mention.


How is bond angle of a molecule measured? Simply, how are (average?) bond angles for simple (or complex) molecules such as methane or ethene measured (determined)?
I'm guessing that X Ray crystallography can be used to determine bond angles for, well, materials that have a regular crystalline form. However, is there some other technique for determining bond angles in non crystalline  molecules?
I have researched on Chemistry SE and on the wider web and I can't find this question addressed - which I find strange as it seems to me an elementary question. Apologies if this question has already been addressed in Chem SE - I just wasn't able to find it using my search terms. I can find many resources about predicting bond angle and none on measuring them.

The most obvious method is single crystal xray crystallography, of course, and for simple molecules, IR and microwave spectroscopy in the gas phase can be used.
For many complex (e.g. bioorganic) molecules, this fails, but there are two (afaik) last hopes to help solve/prove structures if they don´t crystallise and the usual NMR spectroscopy with added computational chemistry is not enough.

Single molecule xray diffraction, a very new approach that can only just be done on the one most brilliant xray "laser" source in the world, the XFEL in Hamburg. It cost 1.2 billion euros to build, and this is the "killer" application it is supposed to prove feasible.

RDC (residual dipolar coupling) measurements with NMR, a very fancy method where the interesting compound is trapped in an oriented (sort of "stretched") gel. If the molecules of the compound have a preferred orientation in that gel (they usually do, because very few biomolecules are spherical), you observe dipolar splittings in the spectra, which would normally be averaged out due to the free tumbling of the molecules. These couplings can be used to calculate relations between distances and angles between the magnetic field and the direct connecting line between the atoms involved.

No 2 of course relies heavily on additional input from conventional (spectroscopic) methods and computer simulations.


Organic problem: acidity of benzene substituents Ortho nitro benzoic acid is more acidic than para benzoic acid since intermolecular H bonding comes into play in the latter compound. But why is that among para nitro phenol and ortho nitro phenol, para compound is more acidic than that of ortho? There is intermolecular H bonding in para nitro phenol too, right?

This isn't the answer you were looking for probably, but this is the answer you should take to heart as a chemist.

I want us to look at the difference in free energy changes between the $\mathrm{p}K_\mathrm{a}$'s in question:

\begin{array}{|c|c|}
\hline
\text{Compound} & \mathrm{p}K_\mathrm{a} \\ \hline
\text{2-nitrophenol} & 7.23 \\
\text{4-nitrophenol} & 7.15 \\ \hline
\text{2-nitrobenzoic acid} & 2.16 \\
\text{4-nitrobenzoic acid} & 3.44 \\ \hline
\end{array}

$$K_\mathrm{eq} = \frac{K_\mathrm{a}}{\ce{[H2O]}}$$

$$\Delta \Delta G = -RT\ln K_{\mathrm a1} + RT \ln\ce{[H2O]} - (-RT\ln K_{\mathrm a2} + RT \ln\ce{[H2O]}) = -RT\Delta\ln K_{\mathrm a}$$
$$=-2.303 RT\Delta \log K_{\mathrm a} = 2.303 RT \Delta \mathrm{p}K_\mathrm{a}$$

For $298\,\mathrm{K}$, we can compute the following $\Delta \Delta G$ values:
\begin{array}{|c|c|}
\hline
\text{Series} & |\Delta \Delta G| \\ \hline
x\text{-nitrophenol} & 456\ \mathrm{J}\,\mathrm{mol}^{-1} \\ \hline
x\text{-nitrobenzoic acid} & 7303\ \mathrm{J}\,\mathrm{mol}^{-1} \\ \hline
\end{array}

For reference, the standard entropy of solvation for acetate is $-192\ \mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}$, which at $298\ \mathrm{K}$ is equivalent to $-57\,216\ \mathrm{J}\,\mathrm{mol}^{-1}$.

Now, personally, I don't have a great handle on how the solvation of the conjugate bases are going to be relative to the benzoic acids or phenols. But you can see that solvation effects can play a huge role (and these are aqueous dissociation constants).

My personal opinion is that it's not particularly useful to understand where such a small difference comes from, particularly for the phenol series. We're talking about one-tenth of an order of magnitude. Perhaps, as you say, the benzoic acids are different because of intermolecular hydrogen bonding. I'm a bit suspicious since I don't see this acting in a differential way between the two species.

Nevertheless, the main point still stands. Acid-base equilibria in aqueous medium is a non-trivial process to understand because water is a complicated solvent. You can try to rationalize this, but you may well be wrong without very cleverly designed experiments.

And the fact of the matter is that rationalizing the difference in acidity is not that useful. Maybe if you were using linear free energy relationships, but in that case, you know there is a difference, and that's more important.

Again, as I said at the beginning, this is probably not the answer you wanted, but I think you should be trying to understand the limits of what is useful/practical to know.


Number of resonance structures of nitrobenzene I was studying about the resonance and here's my doubt:

How many resonating structures are there in nitrobenzene? Is it 5 (as drawn) or is it 4 because first and the last structures are same or is it 3 because second and fourth structures are also same?

In short, do we count fifth structure when we take the number of resonating structures?

We never really need to know the number of resonance structures. The whole concept is qualitative, so we can't draw any quantitative conclusions from it. But if you want to do it anyway, then of course 1 and 5 are the same, while 2 and 4 are not (they have positive charges on different atoms).

Also, both oxygens are in fact equivalent, so you might want to add the structures with their roles reversed.


Choice of lower bound of curve for PV work I am told that the area under the PV curve gives the magnitude of work done by a gas.
Let $V_i$ be the initial volume, $V_f$ be the final volume, and the pressure as a function of volume be given by $P=f(V)$. To my understanding, the work done is given by the area bounded by $x=V_i$, $x=V_f$, $P=F(v)$ and $y=0$.

Proceeding in the above stated manner, the work done comes out to be the sum of the areas of green and yellow region, but the answer stated neglects the yellow region.
Origin of Question : JEE MAIN
Here's the answer according to the official answer key released by the testing agency:

The error comes from reading the graph: the crossing of the axes corresponds to the point (2,2) and not (0,0)


What type of organic compounds are used in OLED technology? I know that there are organic compounds that exhibit electroluminescence, and are those types of materials difficult to synthesize?

The short answer is a variety of pi-conjugated materials.

A huge variety of compounds, both small molecules, oligomers, polymers, and organometallic species have been used. Key requirements for the emissive material is a high efficiency emission process, either fluorescence or phosphorescence.

Some of the first compounds were simply conjugated aromatic molecules and dyes. Martin Pope at NYU observed luminescence from anthracene (e.g. Wikipedia) in the 60s.

A wide variety of conjugated dyes have been tried with varying success.

In general, there are 2-3 different types of compounds used in an OLED:

Hole-transport materials: Easily oxidized and conduct positive charges (holes).
Electron-transport materials: Easily reduced and conduct negative charges (electrons).
Emissive materials: Holes and electrons recombine and emit light.

In many cases, these are made in a sandwich device: ITO anode with hole transport layer on top, emissive layer, then electron transport layer, then cathode. (I'm leaving out various interface agents, "blocking layers," adhesion layers, etc.)

You're asking about the emissive component. Basically, you want materials that will efficiently emit (fluoresce or phosphoresce) at different wavelengths / colors.

The biggest real breakthrough, IMHO was from $\ce{Alq3}$ which is an aluminum organometallic compound with remarkably bright emission. But emits green, so for displays, you need other things.

I've been out of the field for a while, but here are some examples. Chemical companies like lumtec sell various components.

Polyfluorenes: different copolymers have tailored emission colors. Usually the "bridge" carbon is substituted with two alkyl substituents for solubility and processability (and to avoid aggregation).

  In OLEDs, polyfluorenes are desirable because they are the only family of conjugated polymers that can emit colors spanning the entire visible range with high efficiency and low operating voltage. Furthermore, polyfluorenes are relatively soluble in most solvents, making them ideal for general applications[...]

Various vinyl carbazoles: Similar reasons to the polyfluorenes.
Phosphorescent metal complexes (e.g., iridium): If you work out statistics, you get a lot of electron-hole combinations that are triplets, so phosphorescent emission can be very efficient, particularly with transition metals with a high spin-orbit coupling.

I think most leading emissive materials still largely fall into those categories. Small molecule purely organic targets exist, but aren't as efficient.

A bigger question is usually how you intend to deposit the compounds. Vapor deposition (i.e., evaporation or sputtering) can be done with almost anything that sublimates, but if you want to do drop-casting, spin-coating, or inkjet printing, you'll want soluble compounds.


Simple soda–lime decarboxylation This is my homework question I tried it hard the question is as follow:

I attempted it as follow:

According to my understanding as acid base reaction are fastest the carbanion will gain a $\ce{H+}$. From the water formed and will the product as I made, but the answer is completely different:

Please guide me where I am wrong.

You are correct in the first step - the decarboxylation gives the anion of the chloroalkene.

It is important to remember that such soda lime reactions are run at high temperature and CaO is a powerful drying agent, so there will be no water present to quench the anion. What happens next would be obvious if you were dealing with an alkane; there is a leaving group beta to an anion so there would be an elimination with loss of Cl-. This is what happens; Cl- leaves to give the triple bond (again remember that this is a high temperature reaction so plenty of energy for forming unstable species). This is clearly unstable within a [6] ring so it dimerises by a 2+2 process to give the product drawn.


Can two metals combine to form a compound? what I mean is can a metal react with another metal to form a compound? non-metals react with each other and form compounds with covalent bonds can the same happen with metals.

In short, yes they can. What exactly forms however is quite complicated.
Certain rules have been created to estimate structures but in general it's not as easiy as predicting simple inorganic compounds.

It's part of a topic that we call intermetallic compounds. Sometimes also called alloys, although this term should not be used. We destinguish four different type of compounds that might arise from two metals.

Intermetallic Compound: This has a clear stoichiometry (Sodium thallide)
Intermetallic Phase: There is a certain range
Solid Solution: Two metals can be mixed in all ratios
Alloys: A compound that has more than one of these phases present

Why these compounds form however is often attributed to certain abilities like radii, sometimes electronegativity and something we call valence electron concentration.

In general we seperate the periodic table into four groups:

Group I + II
Group III - XI
Group XII - XIV (parts)
The rest (if metallic)

Now we can consider each combination of two, either from the same group or from different groups. For example group I + group I will often yield simply a solid solution are both compounds simply mix. Group I + say group XIII however have quite some difference in electronegativity. Here we get stronly polar or perhaps even more like ionic bonds, as compounds like NaTl form.

In between stuff like Laves-phases can form, compounds that mostly form deltahedra only, or Hume-Rothery-Phases like brass. At some point you get a lot of possible phases and some means to explain why they form and not something else depending on size, electronegativity, valence electron concentration, etc.

I had a few lectures on these topics and every new lecture would basically introduce a new set of compounds with new rules. You could relate them back to this simple scheme I gave you above but it would always have it's own limitations and set of rules.


Understanding group theory easily and quickly Please can you explain me the group theory especially the reducible and irreducible representation and class and subclass and how to tell they are in the same group or not. I know it is a tedious task but help will be very much appreciated.

The figures below show you how to navigate your way round point group tables. The irreducible representations (irreps) are shown as the row of characters. A reducible representation is a collection of the irreps that can be reduced to a number of irreps. Some reminders about definitions, important but tedious. The web page molecule-viewer.com gives many example structures that you can manipulate and to which you can add planes and axes.
some definitions :
A symmetry operation is an operation that moves the molecule about a point into a new orientation indistinguishable to its original one.
A symmetry element is a line point or plane to which the symmetry operation is applied.
(This can be confusing as operations and elements tend to get merged as for example $\ce{C_2}$ can be an element, a 2-fold axis or operation, rotation by 180 degrees)
The first thing to determine is the principal axis, this is the axis with highest rotational symmetry (choose one if there are two or more that are equal). Make this the z axis and choose x and y on your molecule. Make sure that the x and y correspond to the same order of axes as used by the set of point group tables you are using. (There is no standard way). Do this by checking with a simple molecule such as water with $\ce{C_{2v}} $ point group.
There are five types of symmetry operations (1) The identity always present and means 'do nothing' (2) rotations about a symmetry axis $\ce{C_1, C_2, C_3...}$ where subscript 2 means 180 degree rotation, 3 by 120 degree and so on. (3) mirror planes $\ce{\sigma}$ (three types, parallel to principal axis, called vertical, $\ce{\sigma_v}$, perpendicular to principal axis, called horizontal, $\ce{\sigma_h}$ and dihedral $\ce{\sigma_d}$ which bisect two 2 fold axes) Note that mirror planes are defined by their relationship to rotation axes. (4) Centre of inversion i,(5) Rotation reflection $\ce{S_1, S_2, S_3...}$
The figure shows the principal axis and one 2 fold horizontal axis. The point group is $\ce{D_{5h}}$.(from www.molecule-viewer.com)

Common groups are $\ce{C_n}$ and $\ce{D_n}$ types, $\ce{T_d}$ tetrahedral,and $\ce{O_h}$ octahedral. The C groups have rotation axes, and NO perpendicular 2 fold axis, the D groups have a rotation axes but do have a perpendicular 2 fold rotation axes.

The table below shows the Mulliken labels.

The table shows the symmetry elements for several point groups.It is usually only necessary to identify as few symmetry elements to work out the point group.

How to find irreps from reducible representations.
There are several ways to do this, most books give a formula but this can be tricky as its easy to make little sign errors so I always use a tabular method as explained in the figure below taken from my lecture notes. The reduced representation in $\ce{C_{2v}}$ is shown at the top. You also need this point group show below

To make the table to work out the irreps an extra row is added at the top of the point group table to hold the reducible representation and then the entries in the table are calculated as shown below. The values in the rows are added up and divided by the order of the group. The answer must be a whole number or zero, if not a mistake has been made. (the order is the total number of symmetry operations (columns in table) each first multiplied by number in its class, 4 for $\ce{C_{2v}}$ but 6 for $\ce{C_{3v}}$)
The method is quite quick as you once you have one value worked out others in each column are either the same or multiplied by an integer or zero depending on the point group. Don't forget that not all the characters are 1 or -1, in some tables they can be 2 or 3 etc, so you must always multiply by these values.


Oxidation of 1,4,5,8-tetrahydronaphthalene by meta-chloroperoxybenzoic acid I came across this question in Himanshu Pandey. I can't understand why the m-CPBA attacks the central pi bond and not the lateral ones. The answer given is C.

As the following figure indicates, the terminal oxygen in a peracid (the oxygen involved in the epoxidation) has a partial positive charge ; therefore this oxygen is electrophilic.

(image source)

The central double bond in your compound is tetrasubsituted, the lateral bonds are only disubstituted with alkyl substituents (remember that alkyl groups are electron donating).  Consequently the central double bond is the most electron rich and will be the preferred site of attack for the electrophilic oxygen.


Calculating the amount of oxygen in a glass using candle burning time I would like to calculate the amount of oxygen or concentration of oxygen in a glass or conical flask using candle burning time.
I am thinking of a simple way to calculate this based on the fact that the atmosphere contains $21\,\%$ oxygen. But I am not sure if this would give correct values.
Here is what I am thinking about:
I could do a candle and glass experiment and count the time for burning of candle.
First I will take a glass with atmosphere air and then keep it inverted on burning candle and count the time for burning.
Next I will take a glass with oxygen-enriched air (with unknown concentration) and then keep it inverted on burning candle and count the time for burning.
Here is a simple formula using cross multiplication, I could use
$$x = \frac{t_\mathrm{OEA}}{t_\mathrm{air}}\cdot 21\,\%,$$
where $x$ is the concentration of oxygen-enriched air, $t_\mathrm{OEA}$ is the time taken to burn in oxygen-enriched air, $t_\mathrm{air}$ is the time taken to burn in atmospheric air.
Is this possible or is there a way to do this without having to buy oxygen sensors?

I think you will be better off measuring the volume change after complete burning.
A common demonstration  was to place a candle in a dish of shallow water and light it, then put a clear cup over it and see how much water was sucked into the bottom as the oxygen is consumed. utilizing the same effect in a more controlled way perhaps would be a straightforward way to get a rough oxygen concentration.
When calculating it remember that 2 oxygen molecules are replaced by a single co2 molecule and you will want your water deep enough so that when the heated air expands it doesn't escape out the bottom and account for temperature differences either by doing the math or waiting for the temp to be the same.
Fully burning steel wool and measuring the gain in weight due to the oxygen getting incorporated into solid iron oxides might also work.


How would Dalton's law be affected when there are two ideal gases in a container at different temperatures? How would Dalton's law be affected when there are two ideal gases in a
  container at different temperatures?

Let the gas with higher temperature be gas A and the gas with lower temperature be gas B. Then heat will be transferred from gas A to gas B due to which kinetic energy of the molecules of gas A will decrease and kinetic energy of molecules of gas B will increase. Hence, molecules of gas B should make more collisions to the walls of the container because of the increased kinetic energy and hence the total pressure exerted by gas B should increase and using the same argument, pressure exerted by gas A should decrease.

Now, we can no longer use P(A) + P(B)= P(T)

; P(A) is the partial pressure of gas A,

P(B) is the partial pressure of gas B

and P(T) is the total pressure exerted by the mixture of gas A and gas B

Am I correct with this logic?

  A glass bulb of volume 400 ml is connected to another bulb of volume
  200 mL by means of a tube of negligible volume. The bulbs contain dry
  air and are both at a common temperature and pressure of 293 K and
  1.00 atm. The larger bulb is immersed in steam at 373 K ; the smaller, in melting ice at 273 K. Find the final common pressure.

In the above problem, we cannot simply use PV= nRT to calculate final pressure because we don't know the final common temperature. So, we have to individually calculate the partial pressure of both the gases and then sum them up to get the total pressure. But according to me, as stated above, that should not be correct.

Is my reasoning correct or I am missing out on something? Also, how should I solve using my logic and please tell if there are other ways to go about solving this problem.

Determine total number of moles of air in bulbs:
$$n=\frac{PV}{RT}=\frac{(1)(0.4+0.2)}{(0.082)(293)}=0.025$$

Consider final state of system.  Let $P_F$ be the final equilibrium pressure, $n_1$ be the final number of moles in the larger bulb, and $n_2$ be the final number of moles in the smaller bulb:
$$n_1=\frac{(P_F)(0.4)}{(0.082)(373)}=0.0131P_F$$
$$n_2=\frac{(P_F)(0.2)}{(0.082)(273)}=0.0089P_F$$
$$n=n_1+n_2=0.022P_F=0.025$$
$$P_F=1.14\tag{atmospheres}$$


How to convert Gaussian input file .gjf to SMILES? Is there a way to convert Gaussian input files (.gjf) to SMILES-codes (.smi)?

I'm aware of Open Babel (http://openbabel.org/docs/2.3.0/FileFormats/Overview.html), but this lists .gjf as a "write-only" format. Is there, nevertheless, a way (which ideally can be integrated into a Python application) to convert .gjf to .smi?

Edit:

(as mentioned in the comments there could be two different formats in .gjf?!)
Some example to clarify:

...

Winmostar

0 1
C
N 1 R2
C 2 R3 1 A3
C 3 R4 2 A4 1 D4
C 3 R5 2 A5 1 D5
C 5 R6 3 A6 2 D6
C 4 R7 3 A7 2 D7
C 6 R8 5 A8 3 D8
H 5 R9 3 A9 2 D9
N 1 R10 2 A10 3 D10
H 8 R11 6 A11 5 D11
C 10 R12 1 A12 2 D12
...
R2 1.380897451
R3 1.347679561
R4 1.373428060
R5 1.389879183
R6 1.397784043
R7 1.383546211
R8 1.399306496
R9 1.102440160
R10 1.375509032
...
A3 108.5501889
A4 108.1923111
A5 131.7222115
A6 117.5631443
A7 123.5229334
A8 121.6972504
A9 120.8892161
A10 105.3917963
...
D4 3.7044863
D5 -177.0963408
D6 -178.8876730
D7 178.9809960
D8 0.1218056
D9 0.5014806
D10 -7.3468625
...

It is not aiming for code golf, but the following offers the conversion to intermediate .xyz which may be processed further (cf. vide infra):

import fnmatch
import os
import sys

gjf_register = []

for file in os.listdir("."):
    # collect all .gjf:
    if fnmatch.fnmatch(file, "*.gjf"):
        gjf_register.append(file)
    gjf_register.sort()

for entry in gjf_register:
    # recover only the atom block:
    per_file_register = []
    with open(entry, mode="r") as source:
        for line in source:
            per_file_register.append(line)
    del per_file_register[:5]  # clip the first lines

    # the other mandatory entry in the .xyz file:
    number_of_atoms = len(per_file_register) - 1

    # write an intermediate .xyz file:
    intermediate_xyz = str(entry)[:-4] + str(".xyz")
    print("File {} written.".format(intermediate_xyz))

    with open(intermediate_xyz, mode="w") as newfile:
        newfile.write(str(number_of_atoms))
        newfile.write(str("\n"))

        newfile.write(str(intermediate_xyz))
        newfile.write(str("\n"))

        for entry in per_file_register[:-1]:
            newfile.write(entry)

sys.exit(0)

The conversion into .smi may then be delegated to obabel, e.g. by

obabel *.xyz -osmi -m

if there are multiple files to consider.

Proof:  The following .cif

data_EFAHAW
_cell_length_a 8.1819
_cell_length_b 10.4285
_cell_length_c 9.2965
_cell_angle_alpha 90
_cell_angle_beta 99.962
_cell_angle_gamma 90
_symmetry_space_group_name_H-M 'P 21/c'

loop_
_symmetry_equiv_pos_site_id
_symmetry_equiv_pos_as_xyz
1 x,y,z
2 -x,1/2+y,1/2-z
3 -x,-y,-z
4 x,1/2-y,1/2+z

loop_
_atom_site_label
_atom_site_type_symbol
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
O3 O 0.19290 0.35524 0.46580
C9 C 0.16780 0.24532 0.42690
O2 O 0.05757 0.21719 0.30246
C7 C 0.05180 0.07991 0.27798
C1 C 0.16866 0.02605 0.40441
C6 C 0.23660 0.12584 0.49162
C3 C 0.35070 0.10279 0.61788
C5 C 0.39278 -0.02130 0.65283
C2 C 0.32245 -0.12294 0.56458
C4 C 0.20840 -0.10008 0.43826

was converted by

babel -icif EFAHAW.cif -ogjf test.gjf

into

#Put Keywords Here, check Charge and Multiplicity.

 EFAHAW

0  15
O           0.82917         3.70462         4.26502
C           0.68636         2.55832         3.90884
O          -0.01540         2.26497         2.76942
C          -0.02324         0.83334         2.54528
C           0.72957         0.27166         3.70291
C           1.14519         1.31232         4.50144
C           1.87569         1.07195         5.65752
C           2.16377        10.20637         5.97753
C           1.73027         9.14642         5.16948
C           1.00028         9.38482         4.01285

The above script converted it into .xyz

10
test.xyz
O           0.82917         3.70462         4.26502
C           0.68636         2.55832         3.90884
O          -0.01540         2.26497         2.76942
C          -0.02324         0.83334         2.54528
C           0.72957         0.27166         3.70291
C           1.14519         1.31232         4.50144
C           1.87569         1.07195         5.65752
C           2.16377        10.20637         5.97753
C           1.73027         9.14642         5.16948
C           1.00028         9.38482         4.01285

and already mentioned babel -ixyz test.xyz -osmi test.smiles offered the SMILES string

O=C1O[C][C]=C1[C].[C][C][C] test.xyz

Due to the definition of asymmetric unit and unit cell, other original input formats than .cif may work even better (i.e., may yield molecules which are not fragmented as here), e.g. «native» .mol2 or .sdf.


Comb and hair, is that an example for ionic boding? As i understand, when brushing hair, some electron transfer from hair to the comb. Thus, make the comb have different charged from hair and they attract each other.
It's look likely what i have learn from school about ionic bond. One electron from Na is transfer to F and so we have NaF . Isn't it ???

No, not at all. It is an example for triboelectric charging. A mechanical example for this type of electrostatic generators would be a Van de Graaff generator.

But all this has nothing to do with ionic bonds.


Why does water splash if you put it into a hot pan? I had kept my tea pan on the stove (At high)for about 30seconds . I was making tea and getting the water to put in the tea pan. There was nothing in it before that .After that , I quickly poured some water in it and boom.
The pan kind of like jumped out of the stove. The water from the pan instantly became steam and some droplets of water touched my kitchen ceiling.
The scene looked like a bomb exploding. As a single droplet of water just touched the pan , the pan just went of control.
Also , you hear a very noisy sound if you do it.
What could be reason for this little amount doing such a reaction ?
Yes, I do not advise kids to do without parents or some elders support.

When you pour water on a very hot plate, much hotter than $100°$C, the water cannot touch the plate. It boils immediately, even before touching the plate. This makes a sudden and huge production of vapor, which takes much more room than a liquid : this throws the rest of the water anywhere in the vicinity with a great speed. It looks like an explosion. Some water may even be thrown to the ceiling.
Another famous experiment can be done directly on the electric plate, without a pan. If you pour maybe one milliliter water on a red hot plate, which has a small cavity in its center, the water floats above the plate in the cavity without touching it. For a couple of seconds, the vapor produced is not strong enough to eject the drop, which stays calmly in levitation. This effect is called Leidenfrost effect, or calefaction.


Are all molecular structures symmetric? Are all molecular structure symmetric either in relation to a plane within itself or in relation to other molecules? Are there any completely asymmetrical structures when looking at molecular geometry? Are there any studies into the reason for symmetry?

It is not quite clear what do you mean by "in relation to a plane within itself or in relation to other molecules", but I think the answer is no, not all molecules are symmetric. Many of them belong to what is known as $\mathrm{C}_1$ point group, or, in other words they are are totally asymmetric. For more information on molecular symmetry and point groups read this Wiki article.


Can hydrogen gas reduce copper(II) ion? I have a solution of copper(II) sulfate, and I would love to precipitate the copper metal. From looking at the standard reduction potential table, it appears that I can do this by bubbling hydrogen gas through the solution. However, the potential is not very large, so I am unsure of how well this approach would work. I was also wondering if sodium metabisulfite could be used for this purpose. I know I can precipitate copper carbonate with sodium carbonate and reduce with carbon, but this will take more time, and probably result in a large loss of product. I would also like to avoid precipitating the metal with another metal, because I am demonstrating how metals can be isolated so it would be silly to use a metal that is even harder to reduce.

Copper reduced from solution is readily oxidized by air.

When I cared, I made a moderately fine copper powder from citrate. Threat copper salt solution with slight molar excess of sodium (potassium) citrate solution (can be made on site from corresponding hydroxide and easily available citric acid), shake and let the solution stay for a while. It immediately turns deep blue, but the precipitate may appear several minutes later. Filter off the precipitate and then heat on a sand bath at ~200-300 Celsium in a vial plugged with cotton wool. I didn't check if evolved gases are toxic, so use a fume hood or good ventilation. The citrate goes through several changes in color, but eventually produces metallic copper powder. Cool it with wool in. The powder seems to be mildly (inconsistently) pirophoric, so store in a small air-tight container.

Hydrogen on it's own certainly can reduce copper, but the process is neither fast, nor pragmatic.


Why can primary cells not be recharged effectively? Primary cells cannot be recharged effectively because their reaction products are not in contact with the electrodes. True or false?

I got it correct just by rote-learning the theory so the answer is true. But when I tried to understand it properly I'm getting confused between a galvanic cell and a primary cell. They both work on the same principles (chemical energy is converted to electrical), but they are not the same thing? Since in galvanic cells are where there are 2 beakers with 2 electrodes, 1 in each, their products are always in contact with electrodes, opposed to primary cells where it is not modelled in the same way (i.e. just a battery where the products can move away from the electrodes)?

Galvanic/electrolyic refers to how the cell is being run. If you have an electrochemical cell, and you connect a load to it, and it produces a spontaneous current, it's a galvanic cell. If instead you connect a voltage source, and the current from the voltage source induces a reaction, it's an electrolytic cell. There's no requirement that a galvanic cell be made with two beakers and a salt bridge, that's just a very common setup for introductory chemistry experiments using the Daniell cell.

The defining characteristic of a secondary battery is that the net "electrolytic" reaction is the reverse of the net "galvanic" reaction. So a secondary battery is a galvanic cell when it's discharging and an electrolytic cell when it's charging. In a primary battery, whatever reaction happens during charge doesn't return the battery to its initial state. A common reason for this is that the galvanic reaction produces insoluble products, so the concentration is really low by the electrode. Then when you reverse the current and run it as an electrolytic cell, something else reacts instead, like water electrolyzing.


On Wavefunctions of Hydrogen atom In the context of the wave functions of electron in a hydrogen atom, I just want to get clear regarding the terms probability and probability density of an finding an electron. To be precise, we know that the value of $|\Psi|^2$ at a point gives the probability density of the electron at the point. Now, does that imply that if the value of $|\Psi|^2$ is maximum at a point, the electron is most likely to be found at that point?

Reason for my doubt: The question given by my instructor goes:

  Given that the normalized wave fuction of the electron in $\mathrm{3d}_{z^2}$ orbital of an Hydrogen atom is:
  $$
\Psi = N \sigma^2 \mathrm{e}^{-\frac{\sigma}{3}}(3\cos^2\theta-1) \, ,
$$
  where $\sigma={\frac{r}{a_o}}$

  Now where is the electron most likely to be found?

The solution given goes like:
If $P$ represents probability of finding the electron and $\mathrm{d}V$ the volume element then,
$$
\frac{\mathrm{d}P}{\mathrm{d}V} = |\Psi|^2
\implies
\mathrm{d}P=|\Psi|^2 \mathrm{d}V$$
Also since,
$$\mathrm{d}V = r^2 \sin\theta \mathrm{d}r \mathrm{d}\theta \mathrm{d}\phi$$
We get,
\begin{align}
\mathrm{d}P &= |\Psi|^2 r^2 \sin\theta \mathrm{d}r \mathrm{d}\theta \mathrm{d}\phi \\
&= N_1 \sigma^4 \mathrm{e}^{-\frac{2\sigma}{3}}{(3\cos^2\theta-1)^2} \sigma^2 \sin\theta \mathrm{d}\sigma \mathrm{d}\theta \mathrm{d}\phi \\
& = N_1 [ \sigma^6 \mathrm{e}^{-\frac{2\sigma}{3}} \mathrm{d}\sigma ][ {(3\cos^2\theta-1)^2} \sin\theta \mathrm{d}\theta ][ \mathrm{d}\phi]
\end{align}

He later maximized the $[\sigma^6 \mathrm{e}^{-\frac{2\sigma}{3}}]$ and $[{(3\cos^2\theta-1)^2} \sin{\theta}]$ terms to get the values of ${\sigma}$ i.e. hence $r$ and ${\theta}$ at which the electron is most likely to be found. Infact, he stressed on the maximization of ${\theta}$ part which came out to be $0$. And hence, a conclusion made that for an electron in $\mathrm{3d}_{z^2}$ orbital, the electron is most likely to be found in the XY plane.

I find all this stuff confused and fishy. Can someone explain me this solution or prove it wrong via a right way of solving this question.

Now, does that imply that if the value of $|\Psi|^2$ is maximum at a
  point, the electron is most likely to be found at that point?

It certainly does imply something similar to what OP is saying, but not quite. As always in quantum theory one has to be very careful with his or her wordings. Namely, we are talking about a continuous random variable here, so the only probability which makes sense is that of it falling within a particular range of values, i.e. probability of finding an electron in some infinitesimal element of volume $\mathrm{d}V$ centered at a particular location $r$ and not on this single point. Still, if you divide the whole space into volume elements of an identical (infinitesimally small) size, the bigger the value of $|\Psi|^2$ is, the bigger is the above mentioned probability.

Now to the actual findings. So, for an electron occupying the $\mathrm{3d}_{z^2}$ orbital, the probability of finding it in an infinitesimal element of volume $\mathrm{d}V$ centered at a particular location $r$ is given as $|\Psi|^2 \mathrm{d}V$, where $|\Psi|^2$ is the probability density. We analyze the probability density
$$
|\Psi|^2
=
N^2 \sigma^4 \mathrm{e}^{-2\sigma/3} (3\cos^2\theta-1)^2 \, ,
$$
or, more precisely only its angular $(3\cos^2\theta-1)^2$ part. So, when does this expression have its maximum value? The global maxima are at $\theta = 0$ and $\theta = \pi$, i.e. alongside $z$-axis, plus there are local maxima at $\theta = \pi/2$ and $\theta = 3\pi/2$, i.e. in a plane perpendicular to $z$-axis. 1

From the discussion with OP, it seems it must be noted that terms "probability density" and "probability" are defined with respect to a particular random event. Above we analyzed the probability density of finding an electron in an infinitesimal element of volume $\mathrm{d}V$ centered at a point $r$. The radial distribution function arises when considering a probability of a different event, namely, the probability of finding an electron between two concentric spheres of radii $r$ and $r + \mathrm{d}r$. We start from the probability of finding an electron in infinitesimal volume element $\mathrm{d}V$ at a point $r$
$$
  |\Psi|^2 \mathrm{d}V
  =
  \underbrace{R_{nl}^2(r) |Y_l^m(\theta, \phi)|^2}_{|\Psi|^2}
  \underbrace{r^2 \mathrm{d}r \sin\theta \mathrm{d}\theta \mathrm{d}\phi}_{\mathrm{d}V} \, ,
$$
and integrate out $\theta$ and $\phi$ dependences
$$
  \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi}
  |\Psi|^2 \mathrm{d}V
  =
  \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi}
  R_{nl}^2(r) |Y_l^m(\theta, \phi)|^2
  r^2 \mathrm{d}r \sin\theta \mathrm{d}\theta \mathrm{d}\phi \, .
$$
And since the spherical harmonics $Y_l^m$ are normalized (to unity)
$$
  \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi}
  |Y_l^m(\theta, \phi)|^2 \sin\theta \mathrm{d}\theta \mathrm{d}\phi
  =
  1 \, ,
$$
for the probability of finding an electron between two concentric spheres of radii $r$ and $r + \mathrm{d}r$ we simply get
$$
  R_{nl}^2(r) r^2 \mathrm{d}r \, .
$$
The corresponding probability density is, of course, $R_{nl}^2(r) r^2$. Such analysis is different from what we did above for the angular $\theta$ dependence: we simply ignored the radial and the other angular dependence instead of integrating them out. This ignorance makes our analysis a bit rough, but not worthless: at least we were able to identify some general trends in the angular $\theta$ dependence.

1) See Wolfram Alpha for details.


reaction between Iron(ii) and Vanadium(v) When we reduce Vanadium (V) using iron(II), why are we adding 85% Phosphoric acid?
After the above reduction, if we are going to titrate the mixture with Potassium permanganate, why are we adding dipotassiumpersulfate to the mixture before the titratioin?

  Here's the procedure:
  To 5 ml of vanadium (V) solution, add 2 ml of 6 M sulfuric acid, 1 ml of phosphoric acid and 0.20 g of hydrated ferrous sulfate to reduce vanadium (V). Then swirl the solution and allow it to stand for 3 minutes. Then add 0.23 g of dipotassiumpersulfate and let to stand for 5 minutes. Stir the contents vigorously and titrate with 0.02 standard potassium permanganate.

I think I realize (a part of) the answer, concerning the conditions of a redox titration of vanadium(V), for which a metavanadate, such as $\ce{NH4VO3}$ is typically used as starting material. Under acidic conditions, i.e. in the presence of 85% $\ce{H3PO4}$, the colourless $\ce{VO2+}$ is formed.

$\ce{VO3- + 2 H+ -> VO2+ + H2O} $

Again under acidic conditions, the latter species is reduced by Fe(II) to the blue $\ce{VO^{2+}}$.

$\ce{VO2+ + Fe^{2+} + 2 H+ -> VO^{2+} + Fe^{3+} + H2O} $

Is is conceivable that the peroxodisulfate is used to oxidize the excess of Fe(II), which otherwise would be oxidized by the permanganate?


What really is an equation of state (EOS) and how are they developed? Are EOSs merely correlations that mathematically represent (in a better or worse way) experimental data or do they have a theoretical background? How are they developed (suppose one wishes to develop his/her own EOS -- how would he/she start and which script should he/she follow?).

They are merely brief and approximated ways to describe observed relations among state parameters. They do not need to have a theoretical background, but many times they have. It helps to create them and provides some reliability. The theory used to guide the construction of a EOS is anyone that turns to be useful.

how would he/she start and which script should he/she follow?

It is a very broad question. In thermodynamics there are few state parameters used normally. Thermodynamic theory is a macroscopic theory of somewhat general applicability, but EOS can be very specific, and this specificity can not be provided by thermodynamics, so, it turns to be a wide search.
In cases where simple general relations exist (mostly already discovered), it can be found just by fitting a data to a function.


Total spin for lithium atom Why does the ground state of a Li atom have a total spin of S = 1/2? I get that the total z component of the spin has to be either Ms = 1/2 or Ms = -1/2 because of the Pauli exclusion principle, but that could also correspond to a total spin of S = 3/2, so why is it always S = 1/2?

Probably the easier way to explain it is to rely on the fact that if Li indeed had a ground state with $S = 3/2$, this would necessarily comprise four levels with $M_S = +3/2, +1/2, -1/2, -3/2$. This sort of splitting can be observed under a magnetic field, for example (the Zeeman effect).

Note that it's not possible for the Pauli principle to forbid $M_S = +3/2$ and $-3/2$ while simultaneously allowing $M_S = +1/2$ and $-1/2$. The Pauli principle doesn't mean that "this state can exist, but the electrons can't reach that state". It means that "this state simply cannot exist": because any state that violates the Pauli principle is equal to zero, which isn't a valid state. So it's not as if the $M_S = \pm 3/2$ states are there, and that the electrons simply choose to go into the $M_S = \pm 1/2$ states. The $\pm 3/2$ states are nonexistent. And consequently, because a $S = 3/2$ state necessitates the existence of $M_S = \pm 3/2$ states, it cannot exist either.

However, since that line of argument doesn't seem to fully convince you, you may want to go through the "exercise" of proving it. Unless I'm terribly wrong, this should be a completely foolproof (albeit terribly tedious) method. You can construct the full wavefunction for the Li atom as a Slater determinant (without loss of generality we choose $\alpha$ for the third spin):

$$\Psi = \frac{1}{\sqrt{6}} \begin{vmatrix}
1s(1)\alpha(1) & 1s(1)\beta(1) & 2s(1)\alpha(1) \\
1s(2)\alpha(2) & 1s(2)\beta(2) & 2s(2)\alpha(2) \\
1s(3)\alpha(3) & 1s(2)\beta(3) & 2s(2)\alpha(3) \\
\end{vmatrix}$$

so "all" we need to do is to apply $\hat{S}^2$ to this wavefunction and see what eigenvalue it returns. We know that

$$\hat{S}^2|\Psi\rangle = S(S+1)|\Psi\rangle$$

and so if $S = 1/2$ then the eigenvalue will be $3/4$; and if $S = 3/2$ the eigenvalue will be $15/4$. The $\hat{S}^2$ operator can be expanded as

$$\begin{align}
\hat{S}^2 &= (\hat{S}_1 + \hat{S}_2 + \hat{S}_3)^2 \\
&= \hat{S}_1^2 + \hat{S}_2^2 + \hat{S}_3^2 + 2(\hat{S}_1 \cdot \hat{S}_2) + 2(\hat{S}_1 \cdot \hat{S}_3) + 2(\hat{S}_2 \cdot \hat{S}_3) \\
&= \hat{S}_1^2 + \hat{S}_2^2 + \hat{S}_3^2 + 2(\hat{S}_{1x} \hat{S}_{2x} + \hat{S}_{1y}\hat{S}_{2y} + \hat{S}_{1z}\hat{S}_{2z}) \\
&\quad + 2(\hat{S}_{1x} \hat{S}_{3x} + \hat{S}_{1y}\hat{S}_{3y} + \hat{S}_{1z}\hat{S}_{3z}) + 2(\hat{S}_{2x} \hat{S}_{3x} + \hat{S}_{2y}\hat{S}_{3y} + \hat{S}_{2z}\hat{S}_{3z})
\end{align}$$

We already know what the individual operators $\hat{S}_i^2$, $\hat{S}_{ix}$, $\hat{S}_{iy}$, and $\hat{S}_{iz}$ do to the basis states $|\alpha(i)\rangle$ and $|\beta(i)\rangle$. Bearing in mind that operators don't act on different degrees of freedom, you can work out what this monster operator does to the Li wavefunction.

If you look online, there are examples of how to perform this calculation for a two-spin system (e.g. "prove that the singlet and triplet wavefunctions are eigenstates of $\hat{S}^2$"). The extension to a three-spin system is not difficult, but is extremely tedious, which is why I gave up on writing it out in full. But I am fairly certain that at the end of the day you will find the eigenvalue of $3/4$.


Which has higher priority (Cahn-Ingold-Prelog)? Which group gets higher priority and why?

$\ce{R-CH_2OCH_3}$

$\ce{R-CH=O}$

As mentioned by Yomen , the double bond counts for two atoms with a single bond between them . They are called ghost atoms . This makes aldehyde group at higher priority than the ether .

Wikedia has a very nice explanation on it :

  If an atom A is double-bonded to an atom B, A is treated as being singly bonded to two atoms: B and a ghost atom that has the same atomic number as B but is not attached to anything except A. In turn, when B is replaced with a list of attached atoms, A itself is excluded in accordance with the general principle of not doubling back along a bond that has just been followed, but a ghost atom for A is included so that the double bond is properly represented from both ends.

A triple bond is handled in a similar manner .

Wikipedia link : Cahn–Ingold–Prelog priority rules


Is there open source software for SEM EDX data analysis? This can't be unique to me and would apply to anybody who has to use equipment in other departments has had departments move etc.

What open source packages exist to read and potentially manipulate SEM EDX (electron dispersive x-ray elemental analysis) data after the data has been collected?

(In my case I have data from an Oxford Instruments EDX but there are many more manufacturers.)

You could use HyperSpy library for Python.

Some examples of what you can do with it are available here.


Resonance hybrid of nitrosobenzene In nitrosobenzene, there are two ways that the $\ce{-N=O}$ group can show mesomeric effect.

  It can show positive mesomeric effect by becoming $\ce{=N+=O}$
  It can show negative mesomeric effect by becoming $\ce{=N-O-}$

  So will the electron density on the benzene ring increase or decrease?

I think it should decrease because in the first case we have a $\delta^+$ charge on nitrogen and a $\delta^-$ charge on carbon, and in the other case we have a $\delta^+$ charge on carbon and a $\delta ^-$ charge on oxygen.Is this reasoning correct? Or will there even be such an decrease?

Your reasoning is not quite correct. Since electron density on $\ce{N=O}$ bond is more around $\ce{O}$ than $\ce{N}$ (electronegativity of $\ce{O}$ is greater than that of $\ce{N}$), $\ce{N}$ should have a slight $\delta+$ charge, which demands electrons from adjacent $\pi$-system. As a result, the electron density on the phenyl nucleus of nitrozobenzene is less compared to that of benzene. This phenomena is shown as a ring deactivating factor. However, this fact may be hard to prove practically because of the instability of nitrozobenzene.

Yet, in computational studies, one set of descriptors of electron structure of aromatic compounds may come from the pEDA/sEDA approach (Ref.1). The pEDA and sEDA descriptors are defined as populations of the $\pi$- and $\sigma$-orbital electrons, respectively, in a given planar molecule or its planar part (Ref.1 & Ref.2):

The abstract of Ref.2 states that:

  An application of quantum chemical modeling allowed us to investigate a substituent effect on a $\sigma$ and $\pi$ electron structure of a ring and the nitro group in a series of meta- and para-$\ce{X}$-substituted nitrobenzene derivatives ($\ce{X =}$ $\ce{NMe2}$, $\ce{NHMe}$, $\ce{NH2}$, $\ce{OH}$, $\ce{OMe}$, $\ce{Me}$, $\ce{H}$, $\ce{F}$, $\ce{Cl}$, $\ce{CF3}$, $\ce{CN}$, $\ce{CHO}$, $\ce{COMe}$, $\ce{CONH2}$, $\ce{COOH}$, $\ce{NO2}$, and $\ce{NO}$). The obtained pEDA and sEDA parameters (the $\pi$- and $\sigma$-electron structure characteristics of a given planar fragment of the system obtained by the summation of $\pi$- and $\sigma$-orbital occupancies, respectively) of the $\ce{NO2}$ group and the benzene ring allowed us to reveal the impact of the substituents on their mutual relations as well as to analyze them from the viewpoint of substituent characteristics. The decisive factor for dependence of pEDA on sEDA of the ring is electronegativity of the atom linking the substituent with the ring; in subgroups an increase of sEDA is associated with a decrease of pEDA. The obtained mutual relation between pEDA($\ce{NO2}$) and pEDA(ring) characteristics documents strong resonance interactions for electron-donating substituents in the para position. The observed substituent effect on the $\sigma$-electron structure of the nitro group, sEDA($\ce{NO2}$), is significantly greater (∼1.6 times) for meta derivatives than for the para ones.

Although, there is no general correlation between these two contributions, pEDA(ring) and sEDA(ring), to the description of the electronic structure of the transmitting ring, it's worth comparing effect of $\ce{NO}$ substituent with the known effect of $\ce{NO2}$. It is clearly the population of the $\pi$-electrons are lower in nitrobenzene when $\ce{NO}$ group substituted in para- or meta-position of it compared to that of second $\ce{NO2}$ group substituted in either position (see $\color{green}{\text{green circle}}$ of the plot). Both $\ce{NO}$ and $\ce{NO2}$ groupd have shown $\pi$-electon withdrawing effects on the ring compared to nitrobenzene itself ($\ce{X = H}$; $\color{blue}{\text{blue circle}}$). Therefore, it is safe to say that the electron density on the phenyl ring decreases in nitrozobenzene compared to that of benzene. This conclusion is supported by the positive Hammett substituent constant ($\sigma$-value) listed for $\ce{NO2}$ substituent in benzoic acid: $\sigma_{p-\ce{NO}} = 0.91$ and $\sigma_{m-\ce{NO}} = 0.91$. By comparison, $\sigma_{p-\ce{NO2}} = 0.78$ and $\sigma_{m-\ce{NO2}} = 0.71$ (The authors referred Ref.3 for these $\sigma$-values).

References:

Halina Szatylowicz, Anna Jezuita, Tadeusz M. Krygowski, “On the relations between aromaticity and substituent effect,”  Structural Chemistry 2019, 30, 1529–1548 (https://doi.org/10.1007/s11224-019-01360-7).
Halina Szatylowicz , Anna Jezuita, Krzysztof Ejsmont, Tadeusz M Krygowski, “Substituent Effect on the σ- And π-Electron Structure of the Nitro Group and the Ring in Meta- And Para-Substituted Nitrobenzenes,”  J. Phys. Chem. A 2017, 121(27), 5196-5203 (https://doi.org/10.1021/acs.jpca.7b03418).
Corwin  Hansch, A. Leo, R. W. Taft, “A survey of Hammett substituent constants and resonance and field parameters,”  Chem. Rev. 1991, 91(2), 165–195 (https://doi.org/10.1021/cr00002a004).


What Are Examples of Reducing Sugar in Mailliard Reaction? I'm interested in food science, and I've been reading about Maillard reaction.

All of the articles / videos about Maillard reactions that I found online mention that reducing sugar and amino acids are required for the reaction to occur, but never talks about what kind of reducing sugars are present while searing a steak, roasting a chicken, etc.

When we are cooking a steak with just olive oil, for example, what kind of reducing sugars are present? I initially thought glycogen and ribose in the meat might be acting as reducing sugars, but I learned that most meat we buy don't have glycogen, and there is only a very small amount of ribose present in the meat. Similarly, what kind of reducing sugar / amino acid is present when baking bread?

Lastly, my understanding is that table sugar we buy from stores is sucrose, which is a non-reducing sugar. Does this mean that adding sugar while cooking meat wouldn't help if your goal is to start a Maillard reaction? I know that sucrose does break down to glucose and fructose, which are reducing sugars, when heat is added, but does that break-down occur fast enough for it to be meaningful?

Any input would be greatly appreciated!

Quoting from medium.com

  If it wasn’t for Maillard, we would not have juicy seared steak, deep fried chicken wings, nor grilled sausages. All these browning, aromas and flavor bursts are a result of simple sugars reacting together with proteins. However, Maillard is not the only reaction occurring between these different molecules in food.

This is one of the chemical reaction that whether you have an interest in chemistry or not, we all carry out on a regular, maybe even in daily basis.The Maillard Reaction${^1}$ causes two types of browning of foods: enzymatic & non-enzymatic browning. Non-enzymatic browning can again be split in two types: the Maillard reaction and caramelization${^2}$.

The Proteins contained in meat and other comestible goods are usually rich in the amino acids arginine and lysine. The side chain func- tional groups of Arg and Lys react with reducing carbohydrates such as glucose or lactose to form Amadori reaction products${^3}$.

Maillard reaction chemistry

The Maillard reaction is a chemical reaction between an amino acid and a reducing sugar, usually requiring the addition of heat.The Maillard reaction is a ‘non-enzymatic browning' reaction involving reduced sugars with compounds possessing free amino groups. A reactive sugar, such as glucose(a reducing sugar) can react with amino groups in amino acids, peptides, and proteins as well as with other molecules that contain free amino groups${^4}$.

Why a Reducing sugar?

Freshly prepared solutions of glucose in water gradually change by the opening of the pyranose ring to the free aldehyde form. This reaction is a reversal of a hemiacetal formation reaction. Reducing sugars behave like glucose${^5}$.

Following are the steps involved.

Initial reaction between a reducing sugar and an amino group forms an unstable Schiff base${^6}$.

The Schiff base slowly rearranges to form Amadori products${^7}$.

Next, oxidative decomposition of Amadori products can lead to the formation of a wide range of reactive carbonyl and dicarbonyl compounds.The complex reactions are summarised in figure below${^7}$.

Different amino acids produce different amounts of browning.

The Maillard reaction is responsible for many colors and flavors in foodstuffs${^8}$.

  The Maillard reaction does not just make food taste better. It has a science involved in it. Understanding the key facts allows you to manipulate the process of cooking according to what your taste buds crave for. It is the difference between blindly following a recipe and being able to make a recipe work for your taste${^9}$.

UPDATE

  The Maillard reaction occurs most readily between temperatures 300°F-500°F. When the meat is cooked, the outside temperature becomes hotter than the inside, which triggers the reaction. Protein molecules in the meat are bonded in coils and when heat is applied, those bonds begin to break and the coils begin to unwind. Most of the water content in the muscle fibers then leak out. Reactive carbonyl groups of the sugars react with the nucleophilic amino acids, which produces glycosylamine and water. Pentose sugars react more than hexoses

The pentose sugars might be part of Nucleotide of DNA and RNA(In case no external reducing source is added).

That said, Reducing sugar free xylose, has been reported in onion, strawberries, prunes, apples, pears, grapes, juniper berries, barley malts, brewhouse worts, maple syrup, asparagus, the white and the yolk of eggs, corn, tomatoes, apricots, bamboo shoots, potatoes, beans, alfalfa, beer, and mangoes${ ^{10}}$.

Most of the carbohydrate of the carrot was found to be in the form of soluble reducing substance(You could try as additives for steak prepration)${^{11}}$.

References

${^1}$:https://www.compoundchem.com/2015/01/27/maillardreaction/

${^2}$:https://foodcrumbles.com/maillard-reaction-mechanism-hard-core-chemistry/

${^3}$:https://www.iris-biotech.de/global/knowledge-base/custom-synthesis/maillard-reaction-products

${^4}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx

${^5}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx

${^6}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx

${^7}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx

${^8}$:https://www.scienceofcooking.com/maillard_reaction.htm

${^9}$:https://medium.com/@vijinimallawaarachchi/maillard-reaction-the-science-of-browning-flavoring-and-aroma-1605133960ae

${^{10}}$:https://www.researchgate.net/publication/268009365_Free_Sugars_in_Fruits_and_Vegetables

${^{11}}:$http://www.jbc.org/content/46/3/537.full.pdf


How can melting point equal freezing point? I don't understand how the freezing point of a substance is the same temperature as the melting point of the same substance.

For example, if liquid water freezes at 0 °C how can ice also melts at 0 °C?

Because melting point and freezing point describe the same transition of matter, in this case from liquid to solid (freezing) or equivalently, from solid to liquid (melting).

What you may not realize is that while water is freezing or melting, its temperature is not changing!  It is stuck on $0\ \mathrm{^\circ C}$ during the entire melting or freezing process.  It is easier to see this for boiling points.  if you put a thermometer in water and heat it, the temperature will rise until it reaches $100\ \mathrm{^\circ C}$, and then it starts boiling.  And while it boils, it will stay at $100\ \mathrm{^\circ C}$!  All the way until the water has all boiled away.  Now if you could somehow trap the steam (gaseous water) and keep heating it, the steam could have a temperature higher than $100\ \mathrm{^\circ C}$.

So to sum this all up, when matter is transitioning from solid to liquid (melting) or liquid to solid (freezing), its temperature is fixed at the melting/freezing point, which is the same temperature.


Siliconization for Organic Compounds I was wondering if it was possible to create Silane from Methane using the process of siliconization to replace the Carbon with Silicon. For example Silicon Carbide is produced by replacing C-C bonds with a Si-C bond.

A key difference between carbon and silicon (and why you usually can't make an all-silicon version of an organic compound where all its carbons are replaced with silicon) is that long chains of silicon atoms are not stable. Another key difference is that a hydrogen attached directly to silicon, as in silane, is typically more reactive--say toward hydrolysis or oxidation--that one attached to a carbon atom. I do not know of any process to directly "silanize" a carbon compound, as stated in your question.


Does freezing a solution with water always cause the water to separate and form the ice lattice? I'm curious, I was trying to look into the affect of freezing a solution with water even when the solution is completely miscible. I came across something that detailed this regarding salt water and basically said when it gets low enough to freeze, the water forms it's tell-tale lattice basically pushing the salts out so the ice itself is pure water and less dense (considerably less dense because the salt solution allowed the density to increase all the way to the lowered freezing point).

I am curious, is this a general fact of the behaviour of any solution with water regardless of miscibility? Are there water solutions which break up the lattice while still allowing freezing, or does the affect of blocking the formation of that lattice completely block the solution from freezing altogether no matter what? Are there water solutions which will freeze solid even with the lattice without just separating the non-water compounds?

Note: I recognize amorphous ice can be created with wizardry, but my question is more in regards to the typical behaviour on earth.

The purification that occurs when an aqueous solution freezes is called Fractional Freezing and is based on the idea that impurities of a solvent (and this can be any solvent, not just water) have a much lower solubility in the solid phase of the solvent relative to the liquid phase.   Several techniques for performing fractional freezing for purification purposes can be found here.   As the solid is formed, the impurities are expelled into the liquid that hasn't yet solidified.  The purified crystals can then be removed by physical methods (e.g. filtration).  In the case of water, where ice is less dense than the liquid phase, the purified crystals float to the top of the impurity-laden liquid.  This is probably the reason why some people have observed a layer of impurities deep within a block of ice.  As mentioned above, the process is not limited to just water, and freeze crystallization is use to eliminate the purities of silicon wafers and has been known for many years as a way to make hard liquor.  This latter use of fractional freezing is close to the OPs question, which inquires about miscible substances (in this case, ethanol and water).  Despite their miscibility, ethanol and water can be separated by fractional freezing, just like they can be separated by fractional distillation.  Certain substances cannot be separated completely using distillation, and these substances are called azeotropes.  Water and ethanol make an azeotropic mixture that is approximately 95% water.  (So the moonshine you were thinking of making using the reference above will be limited to 190 proof.)  What I find interesting is the following figure,

which shows the freezing point of water/ethanol mixtures as a function of ethanol content.  At a point around 95% there is a significant dip in the freezing point.  At this point, called the eutectic point, the solution behaves as if it is a single component (much like an azeotrope) and will not likely be separable through 'simple' fractional freezing means.  Note I use the term simple because azeotropes can be separated using special distillation techniques.

So in summary: fractional freezing will work with any aqueous solution, regardless of the miscibility of the components, unless the component ratio is close the eutectic point.  We don't need to limit ourselves to aqueous solutions, as many species can be purified in this manner.  The purification mechanism is not so much a factor of the crystal lattice, but the relative solute solubilities in the solid and liquid phases of the solvent.  Finally, many water solutions will freeze solid even with impurities.  This and other types of separations work on equilibrium processes, and multiple iterations are necessary to "completely" remove impurities.  (or more correct, remove the impurities to a level that is acceptable for your application.)


Why are there exactly 5 types of two-dimensional lattices, and what distinguishes them? In the book Chemistry: The Central Science, they introduced the following 5 types of two-dimensional lattices:
The book said that the blue square represents the unit cell, the black dots are lattice points, and the vector a and b are lattice vectors. They don't say what the solid black lines represent. I feel as though they are redundant.
According to the illustration, hexagonal lattice is just a special case of rhombic lattice where γ = 120°. So why does a rhombic lattice have 2 types of unit cells (the blue and the green) whereas a hexagonal lattice only has 1 (and why do the solid black lines for these two lattices are different?). The same question also applies to rectangular lattice. Isn't it just a special case of oblique lattice where γ = 90°?
Why does it make sense to group lattices into 5 types, and what distinguishes them?

First, about the black lines. I think they are helpful for the hexagonal and the rhombic lattice. For the hexagonal system, it shows you the hexagons. One hexagon contains exactly the same space as one conventional unit cell. You just have move (translate by a lattice edge) certain areas as shown below.

For the rhombic lattice, it is nice to have the black lines when you are looking at the green cell. Without it, you would not understand why you would call it rhombic.

So why does a rhombic lattice have 2 types of unit cells (the blue and the green) whereas a hexagonal lattice only has 1 (and why do the solid black lines for these two lattices are different?).

The green cell is a centered cell, which makes the math easier because it has right angles. On the other hand, you suddenly have a translation by $½a + ½b$ on top of the lattice translations that you have the keep track of. In fact, the rhombic type is usually described as a centered rectangular lattice,  e.g. https://en.wikipedia.org/wiki/Bravais_lattice#In_2_dimensions.

The same question also applies to rectangular lattice. Isn't it just a special case of oblique lattice where γ = 90°?

Compared to the oblique lattice, all the other lattices are special cases. They aren't special cases because of some rare coincident, but because of the presence of point group symmetries, i.e. mirror planes and twofold, threefold, fourfold and sixfold rotations. A mirror plane will always allow you to choose a unit cell with a right angle. A fourfold axis will always allow you to choose a unit cell that is square, and a threefold or sixfold axis will always allow you to choose a unit cell that is of the hexagonal type. As you can see, the conventional unit cell is not a hexagon but a rhombus with a 60 degree angle. This makes the math easier.
I don't know why the illustrator chose to offset the black lines and the unit cell by half a unit cell length in both $a$ and $b$ for all lattice types (except the hexagonal one). Maybe it is for better visibility. In general, any point in the unit cell will have symmetry mates in other unit cells related by lattice translations.
If you want to see some examples of lattices that contain objects, check out this link: https://www.iucr.org/education/pamphlets/21/supplement
Here is the illustration for a lattice that contains a hexagonal axis. In this case, you have six identical objects in the unit cell, related by point group symmetry. They do no fit directly into the conventional unit cell, but it has the same volume as six objects (you can try to draw the black lines showing hexogons in your mind). The object is called the asymmetric unit, and it can have various shapes as long as it maps onto the unit cell filling the space without overlap.


How to compare acidity in the following aniline derivatives? Let the simplest one be a), ortho be b), meta be c) and para be d).
I have confusion between ortho and meta. How to compare their acidity?

Taking the case of ortho and meta in which you have confusion:

The acidity of aniline will increase if there is a electron donating group in the compound to stabilise the conjugate positive acid formed.

Since OMe is shows +R and -I effect, it is a net electron donating group as +R effect generally dominates over -I.So in case of ortho it is able to show electron donating nature due to +R effect which makes it more basic than the meta isomer which shows electron withdrawing effect due to only -I effect as +R effect will not affect from meta position.


Do you refer to a bonded atom as an "ion"? Take H2O, would you call the H a hydrogen ion? Or just a hydrogen atom?

Answer: Hydrogen atom.
Remember, the Hydrogen in this case is polar covalently bonded to the Oxygen. The Oxygen has not completely stripped the Hydrogen of it electron but it DOES have primary ownership.
Take a look at NaOH. In this case you would consider the Sodium portion of this to be an ion. Because the Oxygen has completely stripped the Na of its electron so the two are now held together by opposite charges. [Na+][OH-]
At least that's how i like to think about it.

All the best,
D