Why do we use the word phase and state of matter so interchangeably?

From the wikipedia page for phase (matter)

a phase is a region of space (a thermodynamic system), throughout which all physical properties of a material are essentially uniform [...] The term phase is sometimes used as a synonym for state of matter, but there can be several immiscible phases of the same state of matter.
A simple description is that a phase is a region of material that is chemically uniform, physically distinct, and (often) mechanically separable. In a system consisting of ice and water in a glass jar, the ice cubes are one phase, the water is a second phase, and the humid air is a third phase over the ice and water. The glass of the jar is another separate phase.

In the above system you can see that the ice and the glass jar are the same state of matter, but are two distinct phases. It's the same with oil and water. So, people who are being precise don't use the two terms interchangeably.


This is very poorly worded question, and really not just an equlibrium question. This is actually a solubility product question. Consider the given reaction:

$$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)} \tag{1}$$

If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, since the equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$, we can assume that all $\ce{Fe^2+}$ are reacted because it is the limiting reagent and would give $\pu{0.06 M}\times V$ amount of solid $\ce{FeS}$ (where $V$ is the volume of $\ce{Fe^2+}$).

However, the author of the question ignore that $\ce{FeS}$ would gives solubility product of $\ce{Fe^2+}$ and $\ce{S^2-}$.

Since most sulfides of transition metals are insoluble in water, following equilibrium would establish after the original reaction:

$$\ce{FeS (s)  <=> F^2+ (aq) + S^2- (aq)}\text{;} \,\,\, K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]=4.9 \times 10^{-18} \tag{2}$$

Here, $K_\mathrm{sp}$ has not be provided, so I found it in here at $\pu{298K}$ ($\pu{25 ^\circ C}$). The concentration of $\ce{Fe^2+}$ at this equilibrium is given as $A \times 10^{-17}$. We can also calculate the concentration of $\ce{S^2-}$ from equation $(1)$ that $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ (since equal volumes has made final volume dobuled and hence concentration of remaining ions are halves). Since this value is much larger than the $[\ce{S^2-}]$ generated by dissociation of $\ce{FeS}$ ($\approx \sqrt{4.9 \times 10^{-18}}$), we can assume $\pu{0.07 M$ is the same concentration of $\ce{S^2-}$ at this equilibrium. Accordingly:

$$K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]= A \times 10^{-17} \times 0.07 = 4.9 \times 10^{-18}$$

$$\therefore \; A = \frac{4.9 \times 10^{-18}}{1.0 \times 10^{-17} \times 0.07}  = 7.0 $$

Note that this final answer is highly depends on the $K_\mathrm{sp}$ value. Thus, the deviation of this answer from your given answer is based on what value should have been given at the first place.

Suppose we ignore the solubility product as author may be suggesting: Then, once the original reaction is completed, significant amount of $\ce{S^2-}$ would remain unreacted since it is in excess (compared to $[\ce{Fe^2+}]$). That amount would be $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ as indicated above. Since at this point, $[\ce{Fe^2+}]$ is $A \times 10^{-17}$, the equilibrium constant is set up as follows as OP has done:

$$K = \frac{1}{[\ce{Fe^2+}][\ce{S^2-}]}=\frac{1}{[A \times 10^{-17}][0.07]}= 1.6 \times 10^{17}$$

$$\therefore \; A = \frac{1}{1.0 \times 0.07 \times 10^{-17}\times 1.6 \times 10^{17}}= 8.93$$

Although, this setup gives the expected answer, it is not the correct way to solve this (at least to my knowledge).


Plotting ΔG vs Temp for Electrochemical Reaction

My question is, theoretically, what should happen to the voltage as I heat the cell, assuming that the concentrations of the solutions were off?

This question is tough to answer because I don't know what you mean by "off".  So let me ask (and answer) a few related questions that I think will help you.

What should happen to voltage as a function of temperature in an electrochemical cell

The electrical potential of electrochemical cells varies with temperature, even when the cell is at standard conditions.  A 1985 NIST document explains the situation well.  Only the standard hydrogen electrode has a standard potential that does not vary with temperature.  (The SHE potential is defined as 0 volts at every temperature.)  For other cells, usually the assumption of a linear variation with temperature is usually very good.  That is,

$${\cal E}_T^\circ = {\cal E}_{298}^\circ  + (T-298.15)\cdot \frac{d{\cal E}_{298}^\circ}{dT}$$

Note that this variation is completely unrelated to the Nernst equation.  The same reference gives values for $\frac{d{\cal E}_{298}^\circ}{dT}$ for zinc(II)/zinc metal of 0.119 milliVolts per Kelvin and for tin(II)/tin metal -0.32 milliVolts per Kelvin.

Don't be confused by the $T$ in the Nernst equation

The Nernst equation is

$$ {\cal E}= {\cal E}^\circ + \frac{RT}{nF}\ln Q$$

It does contain a $T$ in there, but don't be mislead into thinking that the $T$ you see is the sole means by which a cell potential ${\cal E}$ can vary with temperature.  ${\cal E}^\circ$ also varies with temperature (see above).

  According to the Nernst equation, it seems that the voltage or potential of the cell should not be affected by a change in temperature if the concentrations are equal because ln(0.1/0.1)=0.

Thus, hopefully my answer makes clear that this statement is wrong, because ${\cal E}^\circ$ varies with temperature.


How to obtain pure aniline from a mixture of phenol and aniline?

First dissolve phenol and aniline in diethyl ether and add excess HCl solution .
Then aniline will react with HCl and dissolve in water layer.
Easily remove organic layer.

Take the water layer and add excess NaOH to it and aniline will be precipitated.


What is the actual mechanism of Fenton reaction?

The Fenton reaction

The oxidation of organic substrates by iron(II) and hydrogen peroxide ($\ce{H2O2}$ ) is called the Fenton Reaction. It was first described by Henry John Horstman Fenton in 1894 who first observed the oxidation of tartaric acid by $\ce{H2O2}$ in the presence of $\ce{Fe^{2+}}$ ions [1]. After more than 100 years, the mechanism of the reaction is still unsettled. The recent review by Dunford (2002) stated that:

  The mechanism of reaction of hexaquo iron(II) with hydrogen peroxide has been unresolved for 70 years. Most scientists, perhaps by default, have accepted the free radical chain mechanism of Barb et al. (1957)[3]. However an earlier proposal involved formation of the ferryl ion, $\ce{FeO2+}$ [4]. Recent work has favored a mechanism involving $\ce{FeO2+}$ and $\ce{FeOFe^{5+}}$ species. Similarly there are differences of opinion on the mechanism of reaction of iron(III), both hexaquo and chelated, with hydrogen peroxide. These differences have fostered a recent burst of activity, with claims on one hand that hydroxyl radicals play a key role, and on the other, that there is a non-free radical mechanism. In contrast, the mechanism of reaction of the heme-containing peroxidase and catalase enzymes with hydrogen peroxide, orders of magnitude faster than reactions of iron(II)/(III), now appears to be well established.

According to the free radical chain mechanism [3], the formation of radicals $\ce{HO^.}$ and $\ce{HO2^.}$ have been proposed. Although, this mechanism has some flaws, it has still been used to explain result in current research as recently as 2014 [5]. The proposed mechanism was depicted in Figure 1, which is followed by several other well received mechanisms for the reaction discussed in reviews sighted.

Mechanism of the redox reaction involved in the Fenton’s reagent (Figure 1)[5]:

Basic reactions and intermediates involved in the classic Fenton and the metal centered Fenton reactions (Figure 2)[6]:

Proposed non-radical mechanism for the Fenton reaction (Figure 3)[6]:

Possible reaction pathways for the Fenton reaction in absence of organic substrates (Figure 4)[6]:

Requirements of the reaction:

$\mathrm{pH}$ should be adjusted to 3-5: if the $\mathrm{pH}$ is too high the iron precipitate in $\ce{Fe(OH)3}$ and will decompose the $\ce{H2O2}$ to oxygen. Basically, the optimal $\mathrm{pH}$ occurs between $3$ and $6$. It's really important to pay attention to the double $\mathrm{pH}$ drop due to the addition of $\ce{FeSO4}$ ($\ce{Fe^{2+}}$ source) and $\ce{H2O2}$. Indeed, $\ce{FeSO4}$ catalyst which contains residual $\ce{H2SO4}$ and the $\ce{H2O2}$ addition is responsible for the fragmentation of organic material into organic acids. Thus, the iron catalyst should be added as a solution of $\ce{FeSO4}$ with slow addition of $\ce{H2O2}$, you may able to control the suddern variation of $\mathrm{pH}$ and the temperature during the reaction. It's better to complete the reaction step by step with a continuous adjustment of these factors.

References:

[1] LXXIII.—Oxidation of tartaric acid in presence of iron: H. J. H. Fenton, J. Chem. Soc., Trans., 1894, 65, 899–910 (DOI: 10.1039/CT8946500899).

[2] Oxidations of iron(II)/(III) by hydrogen peroxide: from aquo to enzyme: H.Brian Dunford, Coordination Chemistry Reviews, 2002, 233–234, 311-318 (https://doi.org/10.1016/S0010-8545(02)00024-3).

[3] Reactions of ferrous and ferric ions with hydrogen peroxide. Part I.—The ferrous ion reaction: W. G. Barb,  J. H. Baxendale,  P. George,  and  K. R. Hargrave, Trans. Faraday Soc., 1951, 47, 462-500 (http://dx.doi.org/10.1039/TF9514700462).

[4] Ferryl ion, a compound of tetravalent iron: William C. Bray, and M. H. Gorin, Journal of American Chemical Society, 1932, 54(5), 2124-2125 (DOI: 10.1021/ja01344a505).

[5] Design Polysaccharides of Marine Origin: Chemical Modifications to Reach Advanced Versatile Compounds: Nathalie Chopin, Xavier Guillory, Pierre Weiss, Jean Le Bideau, and Sylvia Colliec-Jouault, Current Organic Chemistry, 2014, 18(7), 867-895 (DOI: 10.2174/138527281807140515152334).

[6] Fenton reaction - Controversy concerning the chemistry: Krzysztof Barbusinski, Ecological Chemistry and Engineering S, 2009, 16(3), 347-358.


What is the nomenclature of this alkane?

The nomenclature rule that you have missed is given in Section P-46 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) (explaining that compound acyclic substituents consisting of a principal chain and one or more acyclic substituents can be named in two ways which can be found in Subsection P-29.2) and in Subsection P-29.3.2.2, which clarifies that the preferred names for simple substituent prefixes derived from saturated parent hydrides with a single free valence that is not at the end of the longest chain are formed by using alkanyl substituents according to Method (2) of Subsection P-29.2:

  (2) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ are added to the name of the parent hydride with elision of the terminal letter ‘e’, if present, when followed immediately by the letter ‘y’. The locants for the atoms of free valences are as low as is consistent with any established numbering of the parent hydride and, except for mononuclear parent hydrides or the suffix ‘ylidyne’, the locant ‘1’ must be cited. This method is used to generate names of ‘alkanyl-type substituent groups’ that are simple substituent groups with free valences at positions other than ‘1’.

For the substituent that is mentioned in the question, the preferred alkanyl-type prefix according to Method (2) is ‘octan-4-yl’.

Note that, according to current IUPAC nomenclature, the prefix ‘isopropyl’ is only retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’, which is also formed according to Method (2).

Therefore, the complete name for the compound that is given in the question is 6-(octan-4-yl)-5-(propan-2-yl)dodecane.


It's stronger because of a dipole moment. oxygen is more electronegative than carbon, so there is a stronger attraction between C and O than a C=C bond.


Why does transition state loses its ability to vibrate? (Transition State Theory)

Only some of the statements you quote are true, certainly a transition state cannot be isolated  since it lasts for less than a picosecond. In fact there is hardly any direct measurement of transition states, such as a spectrum or time profile or any structure. It is not that they are too fleeting to measure but that they occur very infrequently in any reaction so their population is always vanishingly small. This is a real experimental challenge.

What we do know is that the transition state is just like any other molecule with bonds vibrating/rotating  normally except for the bond(s) that are involved in reaction. They stretch and break then form new ones as the transition state is crossed. This pathway is the lowest point on the trajectory from reactants to products and is usually drawn at the top of the reaction energy profile as a 'saddle point'. (When at the saddle point looking at right angles to the reaction path the energy increases, looking forwards or backwards along the reaction path the energy decreases).  As the new bond has lots of energy it is initially vibrationally hot and rapidly loses energy either to solvent or into other vibrational modes in the product molecule or both.

In a complex molecule you can imagine the reaction starting as the species approach one another. The energy starts to increase as the atoms that are going to react  approach one another, but collisions with solvent can randomly knock them off course, or further on course for a while, or vibrations in the molecule may randomly add or remove some energy and so hinder or accelerate the reaction. One can then imagine the species  behaving somewhat as if they are diffusing along the reaction path, sometimes going forwards but sometimes being hindered. At any point, including the saddle point, they can diffuse around a bit and either return to reactants or proceed to products.

Even in simple atom-diatom reactions in the gas phase, such as $\ce{H + Cl_2}$ vibrations in the diatomic molecule  can inhibit reaction even when there is enough total energy to react; sometimes the energy is just in the wrong place and cannot aid the reaction. This can be the case because the diatomic is both rotating and vibrating and reaction will depend on the angle which it has to the line of approach ( atom to diatom) at the separation at which reaction starts.


In the wet medium of an ocean, how does a hard shell form?

Carbon dioxide from the atmosphere (or from decaying matter in the ocean) reacts with water to form carbonic acid:

$$ \ce{ CO2 + H2O -> H_2CO_3 }$$

and this reacts with calcium ions to form calcium carbonate:

$$ \ce{H2CO3 + Ca^{2+} -> CaCO3 + 2H+ }$$

The solubility of calcium carbonate is about $13~\mathrm{mg \over L}$, so if the concentration of calcium carbonate is greater than this the excess will precipitate out as solid calcium carbonate.

Shell-forming organisms actively absorb calcium from the water around them, so they are able to increase the concentration of calcium carbonate to above $13~\mathrm{mg\over L}$ and precipitate the excess to form their shells. They can get the carbon dioxide from the water around them or from their own metabolism. The actual details of shell formation is far more complex than this as it's controlled by processes with the cells of the organism rather than just being uncontrolled precipitation. I'm not sure how well the details are understood even today.

Incidentally, this is why shelly fauna are not keen on the acidification of the oceans that results from increased atmospheric carbon dioxide. The solubility of calcium carbonate is strongly $\mathrm{pH}$-dependent and rises sharply as the water gets more acid. Given that the shell is usually intended to stop other animals eating you, having your shell dissolve is generally not good for life expectancy.


What is the reason for phase separation?

The change in Gibbs energy has to be negative for a process to happen (at constant pressure in the absence of work). Gibbs energy is a combination of entropy and enthalpy.

  I do not understand why this phenomenon occurs as a counter example to the natural laws of diffusion along the concentration gradient.

Diffusion is driven by entropy changes, specifically the entropy of mixing (in an ideal solution, the enthalpy of mixing is zero, so the entropy term is the only consideration). The entropy goes up when a solute moves from high to low concentration.

Phase separation occurs when the enthalpy term dominates. If substances A and B are initially mixed and then undergo phase separation, some change occurred (change of temperature, added third component, etc.) so that now phase separation more exothermic (lower energy for [A to interact with A and B interact with B] than [A interacting with B]). Under these circumstances, phase separation can happen.


Why the energies of orbitals in the same subshell decrease with increase of atomic number?

Short answer: more protons in the nucleus means higher nuclear charge which means stronger attraction between the electron and the nucleus.

Long answer: The energy of the orbital is driven by the Hamiltonian operator $\hat{H} = \hat{T} + \hat{V}$ where $\hat{V}$ is the potential energy. For a simple 1-electron atom/ion, the potential energy is the electrostatic attraction between the positively charged nucleus $Q$ and the negative electron $q$: $V = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r}$. Lithium has 3 protons in the nucleus versus the 1 proton for hydrogen, so $Q$ is three times higher and the potential energy of the 1s orbital in $\ce{Li^{2+}}$ is roughly three times more negative than the 1s orbital in $\ce{H}$ (remember $q$ is negative).

If you have more than one electron, the potential energy becomes more complicated because you have to include the electrostatic repulsion from all the other electrons as well, but the general trend still holds, all else being equal.


What is the IUPAC nomenclature for the cyclic mono-alkene ions formed after proton or hydride abstraction?

Anions

In general nomenclature, anions may be named by using functional class nomenclature; however, the resulting names are not the preferred IUPAC names. The corresponding subsection of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-72.2.1 Functional class nomenclature

  Functional class nomenclature can be used, in general nomenclature, to describe anionic compounds. An anion that can be considered as derived formally by adding an electron to a radical may also be named by adding the class name ‘anion’ as a separate word to the name of the substituent group. The names are formed by using the names of corresponding radicals (not necessarily the name of substituent groups) and the class name ‘anion’ as a separate word. The multiplying prefixes ‘di’, ‘tri’, etc., are added to the class name to denote multiple anions. This type of nomenclature is limited to anions having anionic centers in the same structure. Systematic names (see P-72.2.2) are preferred IUPAC names.

The functional class names for the anions that are given in the question are:

cycloprop-2-en-1-yl anion
cyclobut-2-en-1-yl anion
cyclopent-2-en-1-yl anion
cyclohex-2-en-1-yl anion

The preferred systematic names for anions that can be derived by deprotonation are formed according to P-72.2.2.1:

  P-72.2.2.1 Anions derived from parent hydrides and their derivatives

  An anion derived formally by the removal of one or more hydrons from any position of a neutral parent hydride is preferably named by using the suffix ‘-ide’, with elision of the final letter ‘e’ of the parent hydride, if any. Numerical prefixes ‘di’, ‘tri’, etc. are used to denote multiplicity; locants identify positions of the negative charges.

Therefore, the preferred IUPAC names for the anions that are given in the question are:

cycloprop-2-en-1-ide
cyclobut-2-en-1-ide
cyclopent-2-en-1-ide
cyclohex-2-en-1-ide

Cations

In general nomenclature, also cations may be named by using functional class nomenclature; but again, the resulting names are not the preferred IUPAC names.

  P-73.2.1 Functional class names

  Cationic compounds that can be considered as being derived formally by removal of electrons from the corresponding radical may be named by adding the class name ‘cation’ as a separate word after the name of the radical. Polycations are indicated by adding the multiplying prefixes ‘di’, ‘tri’, etc., as appropriate, to the class name. Systematic names formed by using the suffix ‘ylium’ are preferred IUPAC names (see P-73.2.2). When the charge is not localized, the structure is enclosed in square brackets.

The functional class names for the cations that are given in the question are:

cycloprop-2-en-1-yl cation
cyclobut-2-en-1-yl cation
cyclopent-2-en-1-yl cation
cyclohex-2-en-1-yl cation

The preferred systematic names for cations that can be derived by removal of a hydride ion are formed according to P-73.2.2.1.2:

  P-73.2.2.1.2 General method

  According to the general method, cations formally derived by the removal of one hydride ion, $\ce{H-}$, from any position of a parent hydride are named by adding the suffix ‘-ylium’ to the name of the parent hydride, with elision of the final 'e' in the name of the parent hydride, if present. Di- and polycations formally derived by the removal of two or more hydride ions from the parent hydride are named by using the suffix ‘ylium’ and the multiplying prefixes ‘bis’, ‘tris’, etc. (…)

Therefore, the preferred IUPAC names for the cations that are given in the question are:

cycloprop-2-en-1-ylium
cyclobut-2-en-1-ylium
cyclopent-2-en-1-ylium
cyclohex-2-en-1-ylium


Atomic mass number and mass number in atomic symbol

The IUPAC preference for A in $\ce{^A_ZX}$ is mass number. The term would also apply to a molecular fragment, for instance in a mass spec. So depending on the context atomic mass number or molecular mass number might be used to be more specific.

The problem with you arbitrarily defining A in $\ce{^A_ZX}$ as atomic mass number is that various references might just use mass number. So depending on context you'll need to be able to decide which possibility is the correct one. Of course in spite of IUPAC definitions there are no doubt multiple variations in the past and current literature.


Serious introduction to the chemistry of explosives recommendation

To be clear, that is absolutely not what I'm after. I want an academically oriented introduction, not some "recipe" book.

I'm glad to read that!

I'm aware of Chemistry of Energetic Materials (Eds. George A. Olah and David R. Squire, Academic Press, 1991) which seem to cover

  The Structural Investigation of Energetic Materials.
  Studies of Initial Dissociation Processes in 1,3,3-Trinitroazetidine By Photofragmentation Translational Spectroscopy.
  Studies of Molecular Dissociation By Means of Ultrafast Absorption and Emission Spectroscopy and Picosecond X-Ray Diffraction.
  Computer-Aided Design of Monopropellants.
  Polycyclic Amine Chemistry.
  Metallacarboranes of the Lanthanides and Alkaline-Earth Metals: Potential High-Energy Fuel Additives.
  Methods for Preparing Energetic Nitro-Compounds: Nitration with Superacidic Systems, Nitronium Salts, And Related Complexes.

On a side note, if you are interested in current research on high-energy materials free of nitro groups, have a look at the work of Thomas M. Klapötke.


Difference between lab-grade and food-grade purity?

So-called food-grade stuff is meant to be consumed by humans. It's important that every single compound of a foodstuff is food-grade or at least GRAS. In the European Union the law giver has defined the state when a compound is good for consumption as "save".

Lab-grade chemicals often contain toxic materials, so when something is called lab-grade it is not save per definition.

For example: $\ce{NaOH}$ is widely used in preparation of foodstuff as well as in the lab. Now it's a fact that lab-grade (standard) $\ce{NaOH}$ (Sigma Aldrich) contains a certain amount of heavy metals. On the other hand, food-grade $\ce{NaOH}$ will not contain any heavy metals (usually food-grade $\ce{NaOH}$ will have a purity of more than $99\%$ while standard $\ce{NaOH}$ pellets are sold with purities less than $97\%$). So using lab-grade $\ce{NaOH}$ in foodstuff is a crime (at least in the EU).

So my advice is: Do not use lab-grade materials for any food-related applications.


How to calculate the molecular weight for a volatile substance introduced into a Dumas bulb?

There are a couple of things wrong.

First, the mass of the volatile gas is not $0.8384\ \mathrm g$. It's the difference between $12.4528\ \mathrm g$ and $12.0468\ \mathrm g$.

Second, the volume is incorrect. $11.6144\ \mathrm g + 999.97\ \mathrm{kg/m^3} \times 0.339012\ \mathrm{m^3} \ne 350.6264\ \mathrm g$


According to Textbook of metabolism and metabolic disorders (1964):

  The ribosides derived from purines have the suffix -osine; those from pyrimidines, the suffix -idine. The corresponding deoxyribosides, with the exception of thymidine, do not have any such simple designation. Occasionally, instead of, for example, guanine-deoxyriboside, the term deoxy-guanosine is used.

Beyond the more common adenosine and guanosine, there are also:

inosine  and

xanthosine


Why does mixing two solutions decrease the concentration of the solutes?

Comment turned to answer:

Because it has been mixed.

Each substance is now in doubled volume, compared to volume before mixing.

If you had whiskey and if you mixed it with the same volume of soda, it wouldn't have the same strength as before, would it ?


Separating a powdered mixture of Zinc Oxide and Silver Oxide

Your second suggest that you also want to save the silver oxides apart from zinc oxide. Sodium hydroxide won't help because it also reacts with silver oxides. So, you need a weak acid to selectively react with zinc oxide but neither silver oxide or silver. Hydrobromic acid will work because it reacts with zinc oxide to form zinc bromide which is highly soluble in water. leaving silver oxide and silver untouched. Silver bromide is also expected to form but hydrobromic acid is not strong enough to do that. Instead, if an alkali bromide salt was used, silver bromide would definitely form. From wikipedia article of zinc bromide:

  $\ce{ZnBr2 · 2H2O}$ is prepared by treating zinc oxide or zinc metal
  with hydrobromic acid.

  $$\ce{ZnO + 2 HBr + H2O → ZnBr2 · 2H2O}$$

  The dihydrate can be dehydrated by passing it over hot $\ce{CO2}$


Can bridged and bridgehead positions be chlorinated using free radical mechanism?

The "bridged position" is just another ring, when viewed differently.

Suppose you look at the molecule a little differently (rotate it around a bit), then you could have a five-membered main ring, with a two membered "bridge", as opposed to what we have now (6 membered main ring with 1 membered bridge).

It's definitely possible to have a free radical halogen attack on that position, and with two stereoisomers at the positions which you have identified, it becomes total 3 monochlorinated derivatives.

(in case you're wondering how to check if the two stereoisomers are not identical, you could try to superimpose them, or go from one form to the other without breaking any bonds. In that case, you'll find that they're indeed different. You could even check for a plane/center of symmetry)

--EDIT--

My answer is wrong, as @Abcd and @AkshatJoshi pointed out, because I counted stereoisomers along with the required structural isomers.

I read about attacks at Bridgehead positions and there's NO place that mentions the existence of Bridgehead free radicals. Thus, there will only be 2 structural isomers, but 3 isomers overall.

So, I believe the answer ought to be 2.


Why is the orbital angular momentum of a pi electron along the axis of two atoms' molecule one?

Let the molecular axis be the $z$ axis, A and B the two atoms forming the bond. The $2p_z$ AO has atomic quantum number $m_l = 0$ and so has $L_z =0$. The MOs formed from
2pz AOs are therefore $\sigma$ MOs, bonding and antibonding.
The $p_x$ AO is a linear combination of $m =+ 1$ and $m = -1$ AOs and has $|m| =1$.

$2p_x \equiv \frac{(2p_1 + 2p_{-1}) } {\sqrt 2}$

$2p_y \equiv \frac{(2p_1 - 2p_{-1}) } {i\sqrt 2}$

Therefore the MOs made from the $2p_x$ and $2p_y$ AOs have $|m| = 1$  and are $\pi$ MOs.


Relationship between solubility and enthalpy of solution

For equilibria such as $\ce{[C6H5COOH]_{(s)} <=> [C6H5COOH]_{(aq)} }$.  The equilibrium constant $K_{eq}$ can be written in the very simple form:

$$K_{eq}\equiv \ce{[C6H5COOH]_{(aq)}} \equiv \text{solubility}\,  (sol)$$

That's more than likely where the equation comes from. Then you get

$$\dfrac{d \ln Sol}{dT}=\dfrac{\Delta H}{RT^2}$$

when you separate the variables and integrate with respect to both sides you get:

$$\int_{\ln sol_1}^{\ln sol_2}d(\ln sol) = \int_{T_1}^{T_{2}}\dfrac{\Delta H}{RT^2}\, dT$$

$$\ln sol_2 - \ln sol_1 = \dfrac{\Delta H}{R}\Bigg{(}\dfrac{1}{T_1}- \dfrac{1}{T_2}\Bigg{)}$$

Which gives the useful result

$$\ln \dfrac{sol_2}{sol_1}= \dfrac{\Delta H}{R}\Bigg{(}\dfrac{1}{T_1}- \dfrac{1}{T_2}\Bigg{)}$$


The inductive effect is reflective of the comparison of electronegativity between an -R group and carbon. There is an -I effect because the nitrogen is more electronegative than carbon, so it will inductively withdraw electrons. I will also note that the inductive effect is measured relative to the -R group being hydrogen.


Order of priority in hybridization of d-orbitals

The rotational frame of reference which defined which axis is the z-axis is changed such that $d_{z^2}$ and $d_{x^2-y^2}$ are hybridized first is a convention / practice.

In theory (meaning mathematically), you should be able to get equivalent geometry if you used the other d orbitals for hybridzation.

Basically the donut thing is weird enough we try to fold it in so we can deal with the more normal cloverleafs as leftovers.


I can't find reaction enthalpy ∆H for water formation reaction

Here is the tally of bonds:

One $\ce{H-H}$ bond broken
One half $\ce{O=O}$ double bond broken
Two $\ce{O-H}$ bonds made

Breaking bonds costs (positive sign), making bonds gains (negative sign). The result will be an estimate of turning reactants to products in the gas phase. It is an estimate because the strength of an $\ce{O-H}$ bond depends on what else is attached to the oxygen. For H-H and O=O, this is no problem because there is only one molecule, respectively, with that kind of bond.

$$\Delta H = \pu{436 kJ/mol} + \pu{249.5 kJ/mol} - \pu{928 kJ/mol} =  \pu{-242.5 kJ/mol}$$

We got the right sign and are not too far off in magnitude.

Why is the physical state important?

In solids and liquids, there are intermolecular interactions that contribute to the enthalpy. This is why the enthalpy of sublimation and boiling is positive - it takes energy to break the intermolecular forces. Here are the values for enthalpy of formation for water in the two states at room temperature:

$$\Delta H_\mathrm{f}(\ce{H2O(l)}) = \pu{-285.8 kJ/mol}$$

$$\Delta H_\mathrm{f}(\ce{H2O(g)}) = \pu{-241.8 kJ/mol}$$

And here is the enthalpy of vaporization

$$\Delta H_\mathrm{vap}(\ce{H2O(g)}) = \pu{43.99 kJ/mol}$$


Determining the product which is not possible in the reaction

@Pete
B1 is same as D1

But, D1 & D2 are not same.

@Aniline
I see some problem in the options while it was framed/typed/published

I have shown the 6 possible stereoisomers, which have have labelled independently as A, B, C, D, E & F

For the sake of clarity have generated their names with some software [not from my own] to be assured that these are actually different or same compounds.

Showing the mechanism, how there is a possibility of getting SIX by a free radical mechanism:

This one a redraw of the options A, B, C & D with their names:

The relationship of D1 & D2 as per question doesn't work to the the same:

In my opinion, there are six stereoisomers possible, but the correct option does not exist, in the question[options given, in fact].

Hope it throws some more light on your query.


How does initial rate of reaction imply rate of reaction at any time?

First things first

When we talk about the rate of a reaction, we mean the rate at which the products appear. This means that it is also equal to the disappearance of the reactants. So, when we quantify the rate in terms of reactants, they must be in opposing magnitude. $$\text{rate }= \frac {\Delta [\text{product}]}{\Delta t}=-\frac{\Delta [\text{reactant}]}{\Delta t}$$

From A to B (and C)

Suppose that I am running the reaction $\ce{A ->B +C}$ in two separate vessels, flask 1 and flask 2, each with a different concentration of $\ce{A}$. We'll say that flask 1 is more concentrated. If you agree that the initial rate is given by $\text{rate}_\mathrm{i} = k[\ce{A}]_\mathrm{i}$, then it follows that for each the initial rate is $\text{rate}_{i,fl1} = k[\ce{A}]_{i, fl1}$ and $\text{rate}_{i,fl2} = k[\ce{A}]_{i, fl2},$ respectively.

If we were to add something to flask 2, would it change the reaction rate? As long as we didn't add any $\ce{A},$ the rate would be the same. The rate is dependent on $\ce{A}$ but nothing else. So let's add a little bit of $\ce{B}$ and a little bit oc $\ce{C}$. We'll add the same amount of both. Let's call that amount $x$ and set it equal to the difference in concentrations in $\ce{A}$ in the two flasks. $$x=[\ce{A}]_{i,f1}-[\ce{A}]_{i.f2}$$

Now let's let the reaction in flask 1 proceed until the concentration of $\ce{A}$ is equal to the amount in flask 2. To distinguish the conditions now from the initial conditions, we'll call this time $t=1$ (arbitrary units). By the stoichiometry of the reaction, we know the amount of $\ce{A}$ and $\ce{A}$ will be $\Delta[\ce{A}] = [\ce{A}]_{i,f1}-[\ce{A}]_{i,f2}=x$ There is no perceivable difference between the two flasks, and the reaction rate is thus the same in flask 1 as in flask 2. Because we have chosen arbitrary concentrations, we can abstract it to any concentration. $$\text{rate}_{t=1,f1}=\text{rate}_{i,f2}=k[a\ce{A}]_{i,f2}=k[\ce{A}]$$

Finally, we want the instant rate. We do this by decreasing the time interval to an infinitesimal amount. $$\lim_{\Delta t \to 0}\frac{-\Delta[\ce{A}]}{\Delta t}=-\frac{\mathrm{d} [\ce{A}]}{\mathrm{d} t}=k[\ce{A}]$$


Why is the concentration halved at the intersection of Hg²⁺/Hg₂²⁺?

The total concentration of all ions is $c = \pu{0.10 mol L-1}$. The conservation equation of element mercury requires that
$$\left[\ce{Hg^2+}\right] = \left[\ce{Hg2^2+}\right] = \frac{c}{2}.$$

Generally, it's a convention to make the concentration of ions at the border equal. On the other hand, matter conservation implies that the concentration of each ion is $\frac{c}{2}$.


How can I get self-diffusion coefficient from MSD (mean sqaure displacement)

I would usually recommend the second option (fitting to the slope of the MSD at long times) over the first (dividing MSD by the time). If you compute $\langle \Delta r^2(t)\rangle/6t$, there is a residual effect of the velocity correlations at short time.

Suppose the velocity correlation function decays exponentially in time
$$
\langle \mathbf{v}(0)\cdot \mathbf{v}(t)\rangle = \langle |\mathbf{v}|^2\rangle \exp(-t/\tau) .
$$
This is related to the diffusion coefficient by
$$D= \frac{1}{3} \int_0^\infty \langle \mathbf{v}(0)\cdot \mathbf{v}(t)\rangle \, dt
=\frac{1}{3}\langle |\mathbf{v}|^2\rangle\tau .
$$
Since
$$
\Delta \mathbf{r}(t)=\mathbf{r}(t)-\mathbf{r}(0)=\int_0^t \mathbf{v}(t') \, dt',
$$
it follows that
$$
\frac{d}{dt} \langle | \Delta \mathbf{r}(t)|^2 \rangle =
 6D\, [1-\exp(-t/\tau)] ,
$$
and this converges towards the correct answer, $6D$, exponentially fast in the time $t$.
On the other hand
$$
\frac{1}{t} \langle | \Delta \mathbf{r}(t)|^2 \rangle =
 6D\, \left[ 1 - \frac{\tau}{t} + \frac{\tau}{t}\exp(-t/\tau)\right]  .
$$
This converges more slowly, proportional to $t^{-1}$.

Looking at a log-log plot may be useful, as it may highlight changes between different regimes of behaviour (for example, at long times) that you would miss otherwise. This might be especially true of a polymer system like yours. But your option (3) "slope of log-log plot"  does not give you the diffusion coefficient! It just gives you an idea of whether the behaviour is diffusive (slope $=1$) or not (slope $\neq1$). And option (4) "slope of semi-log plot" does not give you $D$ either.


When to use iso and sec while naming compounds in alkyl halides?

iso- is used when the all carbons form a straight chain except one methyl at "2nd Carbon", like 2-methyl-R (where R refers to straight chain alkane)

And, sec- is used when any functional group is linked to a secondary carbon.


Is there a single MO that represents the π system of a Möbius aromatic system? If so, what does it look like?

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

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

So it goes.


I gotchu.

It's definitely a dilution question. Just plug it into the equation you have, C1V1=C2V2.

$C_1V_1=C_2V_2$

$(0.2\: M)(500\: mL)=(18\: M)(x\: mL)$

Solve for x:

$x=5.56\: mL$

Therefore, you want to use 5.56 mL of 18 M stock solution. To make your 500 mL of 0.2 M sulfuric acid, just take the 5.56 mL of the stock solution, then fill it up with water until you have 500 mL total solution.

The idea behind the equation, by the way, is to preserve the number of moles in the solution while changing the total volume of the solution. Here, we find out that (0.2 M)(0.5 L) = 0.1 mol of sulfuric acid is needed in the final solution. So, we find the amount of stock solution needed to provide that 0.1 mol of sulfuric acid, by dividing 0.1 mol by the concentration of the stock solution.


How can elements have the same number of electrons, but different electron configurations?

The reason is very complicated, but in short:
The energy gap between $\ce{3d}$ and $\ce{4s}$ is wider in $\ce{Cr^2+}$ than in $\ce{Ti}$ because $\ce{Cr^2+}$ has more protons.

When you promote an electron from $\ce{3d}$ to $\ce{4s}$, the electron gained the energy thanks to orbitals, but you also lost the energy thanks to repulsion (of electrons in the same orbital). Remember that electrons like to be in states of less energy, not more.

The energy gap between $\ce{3d}$ and $\ce{4s}$ is less in $\ce{Ti}$, so the electron, by promoting, lost more energy thanks to repulsion than it gained thanks to the orbital, so its configuration is $\ce{[Ar] 4s^2 3d^2}$.

In $\ce{Cr^2+}$, the enrgy gap widened, so the electron would gain more energy than its lost by promoting, making its configuration $\ce{[Ar] 3d^4}$.

See also chemguide.co.uk.


How do P-H bonds in phosphorus acids impart reducing properties?

The $\ce{P-H}$ bond imparts an oxidation state of -1 per bond to the acid. Upon oxidation, these are converted into $\ce{P-OH}$ bonds, which imparts a +1 oxidation state per bond. You can clearly see this in the reaction:
$$\ce{H3PO2 + 4AgNO3 + 2H2O -> 4Ag + H3PO4 + 4HNO3}$$
There are two $\ce{P-H}$ bonds in $\ce{H3PO2}$, both of which are converted into $\ce{P-OH}$ bonds in $\ce{H3PO4}$.

Moreover the $\ce{P-H}$ bond has an enthalpy of 322kJ, while the $\ce{P-O}$ bond has an enthalpy of 335kJ/mol. This slight increase in bond energies drives the reaction in the forward direction. Resonances may also be a factor that contributes as well.


Is a meta-sodium sulfonate group on a benzene ring electron withdrawing or donating?

As $\ce{SO3H}$ is a strongly electron withdrawing group, I'd say that $\ce{SO3Na}$ is basically the same, but to a slightly less extent. The reason for this is that being a group of highly electronegative atoms both groups have strong inductive electron withdrawing effects but they are both at donating electrons via conjugation (only possible from $\ce{S\bond{2}O}$ bond). This means that the sodium sulfonate group will decrease the electron density on the aromatic ring.


Molarity to Percent, Mole Fraction, Molality

I figured it was 6M per 1L of water

That is incorrect you have 6 mol per 1L of water. M(stands for mol/L) not for mol!

How to tackle this problem?
Check water density, is 1 $g/cm^{2}$. So yours is the density of the solution. Then understand what the problem is asking what is percent? What is mole fraction? What is molality? Are you able to answer these questions? If not go back to the text book.
Then is only a matter of simple algebra and units matching. And remember if you have weight and molecular weight you can calculate the number of mole!

So suppose you have a liter of this solution what's the weight of NaOH?

   molecular weight ($\frac{g}{mol}$) of NaOH $\times$ number of moles

Now what's the Overall weight?

   Overall weight=density $\times$ volume (remember to use the right units)

Percent what? In chemistry there are a lot of percent, you mean percentage by weight (wt%) I imagine. Try to figure out from your self:

  $ = \frac{NaOH weight}{Overall weight}\times100 $

Mole fraction require mole of both, we start from the weight, so whats the weight of water?

   you got the overall weight and the weight of NaOH

Now calculate the moles of water, and it is simple to find the mole fraction, and molality. Make these steps in the opposite way and you will find the answer to your last question.


From the Wikipedia article for phosphine:

  The low dipole moment and almost orthogonal bond angles lead to the
  conclusion that in PH3 the P-H bonds are almost entirely pσ(P) – sσ(H)
  and phosphorus 3s orbital contributes little to the bonding between
  phosphorus and hydrogen in this molecule. For this reason, the lone
  pair on phosphorus may be regarded as predominantly formed by the 3s
  orbital of phosphorus. ... This electronic structure leads to a lack
  of nucleophilicity and an ability to form only weak hydrogen bonds.

And, one might assume, likely also explains the lowered basicity of phosphine relative to ammonia.

Ammonia is classically $sp^3$-hybridized, which entails that the lone pair resides in an an $sp^3$-orbital. Based on empirical evidence (some of which is mentioned above), phosphine is less $sp^3$-hybridized, with the lone pair residing in an orbital that is more like an $s$-orbital than an $sp^3$-orbital. Thus phosphine's orbital has more s-character, which means the electron pair will be held closer to the nucleus and be less inclined to bond. This is an argument rationalizing phosphine's lower basicity than ammonia.

Countering this argument, one might argue that phosphine's lone pair orbital is a $3s$, which is inherently bigger than the $2s$-orbital ammonia uses for hybridization. So "after"* hybridization, ammonia's $sp^3$-orbital is certainly bigger than its original $2s$ orbital ($2s$ has more s-character than $sp^3$), however, it may not actually be bigger than phosphine's $3s$. Thus, perhaps ammonia's hybridized $sp^3$-orbital is not so different in size than phosphine's $3s$-orbital, I suppose we would need a quantum computation to know for sure.

At the end of the day, we are only trying to rationalize an observed phenomenon after the fact. In spite all of our human ideas of orbitals, hybridization, etc, nature does as it pleases. Perhaps our models can offer a correct explanation and someone can interpret it better than I did (which shouldn't be too difficult), but also for some phenomena we must realize that our models don't fit perfectly and there are exceptions to every trend.

*Quotes used because ammonia is inherently hybridized, the act of nitrogen's $\{s, p_x, p_y, p_z\}$ atomic orbitals hybridizing when the molecule forms is a human idea that helps explain empirical observation, but may not necessarily take place physically. Indeed, the very $\{s, p_x, p_y, p_z\}$ orbitals themselves are only exact solutions for the hydrogen atom (as treated by quantum mechanics), and are only approximations of the orbitals of all other elements. But let's not get bogged down in all that.


If molality is 1.88, then the change in boiling point is -0.96 degrees C, not the boiling point itself.


Look closely on that site. It's not 'valence electrons' but 'valence'. There is a difference between the valence electrons and valence.

Valence electrons are the number of free electrons in an atom that can take part in making the bond. There are 6 such electrons in the oxygen atom as you stated.

While the valence or valency is the number of bonds that can be made by the atom.

That's why valency of oxygen is 2


Why is the rate of nitration greater for thiophene than toluene?

I would suggest two reasons:

Thiophene is an electron-rich compound because of presence of two lone pairs on the sulphur atom. Hence, it attracts electrophiles readily.
Though its aromaticity is lesser than that of benzene, the delocalisation is among five atoms in thiophene whereas in benzene it is among six atoms. Hence electron density in thiophene is more.


How does a redox reaction between ClO- and Br- take place?

In the pH range between 10 and 14, chloride and hypobromide are formed (DOI):

$$\ce{OCl- +Br- ->  Cl- + OBr-}$$


Nitrobenzene reduction with (tin) Sn catalyst: Why is C-H bond cleavage preferred over O-H bond cleavage?

I don't buy that mechanism.  Like you noted, breaking a C-H bond in methanol seems unlikely, especially since the reaction is run in the presence of acid, so a ready source of protons is available.

Most of these reactions that involve a metal surface proceed through what is known as a single electron transfer (SET) mechanism.  Reductions of nitro compounds, be they catalytic hydrogenations or metal surface - acid catalyzed reductions, usually proceed through nitroso and hydroxylamine intermediates.  This link provides some discussion of the reaction and presents the following (more reasonable) mechanism where electrons are transferred from the metal surface to the various intermediates and the acid provides a ready source of protons.  These reactions are usually run in an organic solvent in order to bring the organic nitro compound into solution.


Which orbital in a subshell gets filled first?

The notation [-1,0,1] is given to address the sub shells.The energies of all these orbitals (unless you apply a magnetic field) are identical, therefore the filling is completely arbitrary. That means any of the shells can be filled first. But after 1 shell is filled the next electron goes to any one of the remaining shells. And the next goes to the other one. This is known as the Pauli's Exclusion law. Electrons with like spins are filled first and electrons with the opposite spin are filled next (note: 1 shell can share 2 electrons). This choice of spin is also complete arbitrary.


In which of the following compounds does Cl have the most positive partial charge?

The greater the difference in electronegativity the more the partial charges is true. Compare the electronegativities of the atoms bonded with chlorine and you can easily determine in which compound chlorine has the biggest partial charge. Chlorine can have a partial positive charge when it's bonded with a more electronegative element like oxygen or fluorine.


Assigning locants in halo alkanes/alkenes

Indeed, the first compound in your sketch is numbered incorrectly according to IUPAC. IUPAC rules for branched hydrocarbons say: "When series of locants containing the same number of terms are compared term by term, that series is "lowest" which contains the lowest number on the occasion of the first difference. This principle is applied irrespective of the nature of the substituents. " (emphasis mine). In your top case, this means that you have two numbering choices: either 2,2,4 or 2,4,4. The 2,2,4 choice has a lower locant at the first point of difference, so it is preferred. When you construct the name, you indicate the substituents in alphabetical order, so the full name of the top compound is 4-chloro-2,2-dimethylpentane.

If unsaturations are present, then these have a higher priority than alkyl and halogen substituents. From the rules for unsaturated hydrocarbons: "The chain is so numbered as to give the lowest possible numbers to the double bonds." This rule is sufficient to impose the numbering you report.

As Loong correctly pointed out, the links provided above refer to the text of the now obsolete 1979 recommendations. Both the 1979 and 1993 recommendations are available on the same site, maintained by ACD Labs with IUPAC's permission, and are in accordance with what shown above.

Unfortunately, the text of the most recent (2013) recommendations is not freely accessible (that I know of). This is a link to the electronic version of the book, published by The Royal Society of Chemistry. The relevant section is P-14.3.5 in Chapter P-1 (General Principles, Rules, and Conventions): "The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set '2,3,5,8' is lower than '3,4,6,8' and '2,4,5,7'."


Negative compared to what?

Most chemical system have several unoccupied orbitals still in the negative energy range, meaning lower excited states are still bound systems (they don't fall just apart).


How does NMO reform Osmium tetroxide after reaction with an alkene?

N-methyl-morpholine-N-oxide (NMO) is used to reoxidize and regenerate osmium tetroxide ($\ce{OsO4}$) in what is known as the "Upjohn Dihydroxylation" procedure.

Mechanistically, the redox occurs when a lone pair of electrons on the reduced $\ce{Os(VI)O3}$ are used to form a bond to the morpholine N-oxygen; the electrons from the former $\ce{N-O}$ bond are, in turn, transferred to the morpholine nitrogen.  The $\ce{OsO3}$ gains an oxygen and is oxidized back to osmium (VIII) tetroxide and the NMO is reduced to N-methylmorpholine

image source

Osmium tetroxide is both toxic and expensive. Using the Upjohn procedure allows the reaction to be run with a catalytic amount of osmium teroxide and 1 equivalent of NMO.


Why does charcoal not produce smoke when heated?

You say that wood decomposes under heating and produces gases(1) which then burn. That's correct. It would be wrong, though, to assume that all combustible materials are like that. Indeed, many - arguably, most of them are. Wax, tar, and all kinds of plastic come to mind. But carbon is a different story.
See, carbon is very heat-resistant, about as much so as the most heat-resistant thing you know, only more. You heat it to $1000\;^\circ\rm C$, and nothing happens. Well, it may start burning provided there is air around, but short of that, it is nothing. You heat it a few hundred degrees more, it starts glowing red, then white, then steel melts, then you pass $2000\;^\circ\rm C$ and continue to push on, then everything else melts, by now your carbon glows so bright that you can't see a thing, but it is still intact.
Burning of carbon, unlike burning of many other materials, is a true heterophase reaction. You chip atoms off that piece one by one, hence no smoke is produced. Burning of wood or plastic is another story. Combustible gases are produced and fly away, and when some of them do not burn completely(2), they leave a solid residue (which, BTW, may or may not consist mostly of carbon). That's what we call smoke.

P.S. Thought I'd rather add two clarification notes:
(1) When heated, plastic turns into gases all but completely. Wood doesn't. Much of it remains in the form of charcoal, which is mostly carbon and burns like carbon, that is, producing no smoke.
(2) Incomplete combustion of carbon is also possible, and it results in carbon monoxide, which killed quite a number of people over the past few millenia. But it isn't smoke; it's a gas. You can't see it. You just see people dropping dead.
So it goes.


crystalline structure of Si wafers used for MEMS

The flats on Si wafers are not arbitrary.  The largest, or primary flat, is perpendicular to the <110> orientation in the plane of the wafer. If it exists, the secondary flat helps identify the wafer orientation and doping type (and can easily vary based on wafer maker and the company ordering the wafers). All modern wafers (anything 200mm or bigger, almost all 150mm) are (100) and boron doped, so the secondary flat isn't used.

The primary flat has a specific orientation since the chips need to be aligned along the <110> crystallographic orientations so the wafer can be diced up.  This usually uses a diamond scribe and careful breaking rather than sawing, because the saw is wider than a scribe line (so fewer chips/wafer) and dirtier .

So, you wafer has <001> pointing up, and <110> in plane towards the flat.


What is the difference between an ionic and molecular activity model? (Example: Diopside-Hedenbergite)

The activity of a chemical species is a key concept in thermodynamics of solutions. While the thermodynamic potential is the “true thermodynamic variable”, i.e. the one that plays a primary role in statistical mechanics, chemists do tend to like activity (which is univocally related to chemical potential) because it is close to another quantity which has clear physical meaning: concentration.

Now, relating the activity (or chemical potential) of a species to its concentration in a mixture is impossible to do in a general way. Models have been developed to provide approximations, however, the best known being the ideal solution, where activity equals molarity (concentration). Your “molecular mixing model” appears to be a model of ideal solid solution. However, one cannot expect this approximation to hold in case of strongly non-ideal solutions, and ionic solutions are prominent in that category. The “ionic mixing model” you describe thus appears to be another such approximation, more suited to these ions in a solid solution. I have never encountered it before, but I found a series of lecture slides where it seems to be very neatly introduced (see around slide 31).

While I cannot fully address the question, I hope this helps!


Belousov-Zhabotinsky reaction: questions about rate determining step, k and activation energy

I realise your homework was probably due a LONG time ago but I just stumbled across this.

In answer to (3) yes, the reaction is notoriously sensitive to a wide range of conditions and stirring is one of them...when the system is stirred faster (probably generating a vortex) oxygen is more readily available to react with radical species in the reaction - effectively these will react with oxygen rather than the bromide. As such the bromide in the system is above the threshold level for oscillations to occur - if you were to slow this down again then you'd see the oscillations resume.

Also I don't get why they'd make you use cerium sulfate...everyone knows ferroin is better.


Ionization of hydrogens in hypophosphorous acid

To begin with, removing another proton from anionic species is not very favorable. For example, the monohydrogen phosphate dianion is not very acidic $(\mathrm{p}K_\mathrm{a}\sim 12)$ despite it having an $\ce{OH}$ group.

In the case of $\ce{H3PO2},$ the second and third protons are bonded directly to $\ce{P},$ which is much less electronegative than $\ce{O},$ and consequently these P-H bonds are much less polarized than an $\ce{O-H}$ bond. So these protons are unable to ionize appreciably in aqueous solution.


Under what circumstances is working in the chemistry lab hazardous to health in the long term?

May I start with a disclaimer saying that whatever long term studies can be done, this will always be a subjective question. All I can do here is offer a few comparisons and examples to show that you shouldn't (in my opinion) be worried.

There answer to the question "Am I more likely to come into contact with carcinogenic/hazardous substances than a regular person?" is quite likely yes. At undergraduate level you will be using a range of strong, harmful, poisonous etc. substances that you could not get outside of a lab environment or at school.

However, the answer to the question "Am I more likely to suffer poor health as a result of working in a lab?" is more difficult.

As you correctly state in your question, there are always many safety procedures in place when you undertake experiments which have been carefully thought, require the scientists to have some minimum level of training in the techniques (perhaps learning with less dangerous substances) out and always err on the side of safe rather than sorry. I myself have been working with a new nanotube based material recently and though unlikely to be carcinogenic, all experiments were to be carried out using gloves in a fume cupboard with extraction as the material hasn't been fully classified or tested yet (again: "safe rather than sorry").

Of course, there are unfortunate cases (for example Marie Curie), but in this day and age the risk assessments are so extensive and always erring on the side of caution that the chance of something going terribly wrong are no worse, I feel, than something catastrophic happening in many other walks of life (e.g. mountaineering accident, plane crash, shark attack).

One perhaps reassuring and rather amusing case is that you are far less likely to succumb to terminal cancer if you work in the nuclear industry by virtue of the extensive risk assessments, precautions and health checks. There is so much safety in place at a reactor, that exposure is minimal (indeed, the background radiation from radon in Cornwall is higher than that next to a nuclear power plant!!) and the industry provides so many health checks for workers that and cancer (radiation induced or natural) is often picked up at a very early, curable stage. Earlier than if you simply went to a doctor when you began to feel ill as a regular person.

Therefore depending on the policy of your employer, you could in fact be at a lower risk of dying from cancer, and I see no need for you to worry for your health in the lab!!

As mentioned when I started, this is a subjective issue, but I really see no need for worry.

Best,

Jack


s, p, d, f and so on are the names given to the orbitals that hold the electrons in atoms.  These orbitals have different shapes (e.g. electron density distributions in space) and energies (e.g. 1s is lower energy than 2s which is lower energy than 3s; 2s is lower energy than 2p).

(image source)

So for example,

a hydrogen atom with one electron would be denoted as $\ce{1s^1}$ - it has one electron in its 1s orbital
a lithium atom with 3 electrons would be $\ce{1s^2~2s^1}$
fluorine has 9 electrons which would be $\ce{1s^2~2s^2~2p^5}$

Again, as we read from left to right the energy of the orbital increases and the superscript shows the number of electrons in the orbital. You can read more here.


The lower part of the Dutch wikipedia about the Gabrielsynthese indeed displays an example where this reaction is carried out to yield a secondary amine:

The original publication (DOI: 10.1021/jm9804376, J. Med. Chem., 1999, 42, 593-600) equally includes similar conversions with racemic secondary substrates like (R/S)-($\pm$)-1-methylpentyl bromide to yield (R/S)-($\pm$)-N-(1-methylpentyl)phthalimide which in turn was then converted into (R/S)-($\pm$)-(1-methylpentyl)amine hydrochloride.


What is the correct unit for the equilibrium constant?

Andselisk correctly identified the law of dilution and the name Ostwald is often connected with it.
$$K_\text{dissociation} = \frac{\alpha^2}{1-\alpha}\cdot c$$
However, the degree of dissociation is $\alpha$ and has "no" unit, i.e. dimensionless quantity. Therefore the unit for the equilibrium constant is that of a concentration, in SI that would be $\pu{mol m-3}$.

It is derived from the Law of mass action; the derivation can be found on Wikipedia and various other sources.

In principle there is no fixed unit for the (generalised) equilibrium constant, as it is simply defined as a product of the involved quantities according to the IUPAC gold book.

  Equilibrium Constant
  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always dimensionless, as it is defined differently (IUPAC gold book).

  Standard Equilibrium Constant $K$, $K^\circ$
  (Synonym: thermodynamic equilibrium constant)
  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{\mathcal{R}T}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $\mathcal{R}$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.


Odd/Even swap rule for chiral compounds in stereochemistry

While this rule is certainly not infallible, it is often used in organic compounds, especially for Fischer projections.

The odd/even rule implies that no change of configuration occurs if an even number of swaps are made at a single chiral center and change of configuration occurs if an odd number of swaps are made at a single chiral center.

Here, you have made 3 swaps at both ends which means you changed the configuration of 2 out of 4 chiral centers. Hence, you can make the inference that 2 chiral centers (C-2 and C-3) are the same, while another 2 (C-1 and C-4) are different. These are hence diastereomers.

I am assuming that you thought you made 3 + 3 = 6 swaps, so the compound should be identical. But the odd/even swap rule is only applicable to a single chiral center, hence the error.

Finally as pointed out already, this rule doesn't apply anywhere where the central atom doesn't have a tetrahedral geometry. As counterexamples, coordination complexes often have non-tetrahedral geometries:

octahedral: $\ce{[Co(NH3)6]^3+}$
square planar: $\ce{[PtCl4]^2-}$
dodecahedral: $\ce{[Ce(NO3)6]^2-}$

Note that these specific examples are optically inactive but by changing the ligands for different ones, it is possible to create optically active versions with the given geometries.
However, since organic chemistry mostly involves tetrahedral carbon as the chiral center, the rules certainly do apply there.


What is the hybridization of terminal fluorine atoms in molecules like boron trifluoride?

TL;DR: As a rule of thumb, terminal (heavy) atoms are almost always best described as having sp hybrid orbitals (at the most).

For more on this, I refer you to my answer on What is the hybridization of chlorine in vinyl chloride?

Here are the relevant visualisations for $\ce{BF3}$ at the DF=BP86/def2-SVP level of theory. First I'll start with the canonical orbitals obtained from a molecular orbital point of view. The molecule has $D_\mathrm{3h}$ symmetry, therefore some orbitals are degenerated.

With appropriate transformations, we can interpret these orbitals in a localised fashion, which will give us hybrid orbitals. In this view the σ bond orbitals, and the lone pairs are triply degenerated. Note that there is a significant amount of delocalisation of the fluorine lone pairs into the empty boron p orbital.

Here is the representative section from the natural bond orbital analysis:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
(core and symmetry equivalent orbitals skipped)
   5. (1.99345) LP ( 1) F  2            s( 63.79%)p 0.57( 36.19%)d 0.00(  0.01%)
   6. (1.94124) LP ( 2) F  2            s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
   7. (1.86896) LP ( 3) F  2            s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
  14. (1.99778) BD ( 1) B  1- F  2
               ( 17.29%)   0.4158* B  1 s( 33.25%)p 1.98( 65.98%)d 0.02(  0.76%)
               ( 82.71%)   0.9094* F  2 s( 36.24%)p 1.76( 63.62%)d 0.00(  0.14%)
 ---------------- non-Lewis ----------------------------------------------------
  17. (0.38551) LV ( 1) B  1            s(  0.00%)p 1.00(100.00%)d 0.00(  0.00%)
  18. (0.05374) BD*( 1) B  1- F  2
               ( 82.71%)   0.9094* B  1 s( 33.25%)p 1.98( 65.98%)d 0.02(  0.76%)
               ( 17.29%)  -0.4158* F  2 s( 36.24%)p 1.76( 63.62%)d 0.00(  0.14%)

Here you can clearly see the approximate two sp orbitals and the remaining p lone pair orbitals.

The above remains true for (almost) all terminal atoms, because of the local linear symmetry (only one bonding partner, negligible external field), although with lesser extend for heavier atoms (like bromine). This is because hybridisation in general becomes a less reliable description due to the increase in the s-p gap.


How to identify ions by their quantum number?

A pattern-based approach: Remember that the values for ${m_l}$ are centered about 0. This means for a d-orbital, where ${l=2}$, the values of ${m_l}$ = ${(-2, -1, 0, +1, +2)}$.

So, for ${l=2}$ and ${m_l=0}$ and ${m_s=+1/2}$, you should arrive at ${d^3}$. The orbitals each get an electron for all the positive spins first, then are filled completely with the negative spin ones.

That means you "visit" ${m_l}$ twice (in rising order: -2, -1, 0, +1, +2, repeat).

This means the first d-electron (↑ _ _ _ _) such as ${Sc}$ has the configuration ${m_l=-2, m_s=+1/2.}$

The 3rd electron (↑ ↑ ↑ _ _) would have ${m_l=0, m_s=+1/2}$.

A 5-electron configuration (↑ ↑ ↑ ↑ ↑) would have ${m_l=+2, m_s=+1/2}$.

Now on the right-hand side of the d-group, the 6th electron (↑↓ ↑ ↑ ↑ ↑) would have ${m_l=-2, m_s=-1/2}$, and then the magnetic quantum numbers rise to +2 again at ${d10}$.

Anyways, ${Mn^{+2}}$ is ${d^{3}}$ (5 to start, then subtract 2 because of its +2 charge) which matches the quantum numbers you specified.

It's not an 8 electron configuration because of the spin specified. That'd be (↑↓ ↑↓ ↑↓ ↑ ↑).


Well, $\ce{Ag2O}$ is just a basic oxide. As such, it would dissolve in suitable acids ($\ce{HNO3}$ would do), but I guess that's not quite what you want. Well, some metal oxides ($\ce{ZnO}$, for instance) are amphoteric and thus would dissolve in $\ce{KOH}$ as well, via formation of hydroxo complexes. Sadly, this is not the case with $\ce{Ag2O}$. Then we must resort to some specific complex which would be stable in alkaline media.

Indeed, there is such complex: $\ce{Ag(NH3)2+}$, the active component of Tollens' reagent, which is prepared by interaction of $\ce{Ag2O}$ with aqueous ammonia, like I said. If you are actually going to do it, read the page carefully, including the "Safety" section, or at least don't make that solution by gallons and don't leave it unattended for years. Also, note that it is normally made using freshly prepared $\ce{Ag2O}$; dry reagent off the shelf would likely dissolve as well, only much slower.


Single crystal x-ray diffraction experiment backwards: from CIF to genuine set of raw data without performing actual experiment

Structure determination by X-ray diffraction, regardless if powder diffraction, or single crystal diffraction analysis, is a spatial mapping of electron density.  After solving a crystallographic structure (phase problem), the subsequent structure refinement includes steps to complete the crystallographic model, and to minimise the differences in the Fourier maps mapping observed electron density ($F_O$), with electron density predicted by your current model ($F_C$).  This minimisation is done until this said difference is below a threshold value, and if the experimenter is satisfied with the result. For the later, this includes the complete and reasonable attribution of atoms (does this electron density belong to an atom of Carbon, or Nitrogen?), checking if bond distances and angles included in the model are consistent with the body of data determined in other experiments.  Some of the critera applied are interdependent to others.

If the crystallographic model is completed (in layman's terms, if there is a *.cif file), than it is easy to predict a theoretical powder diffraction pattern.  Programs like CCDC Mercury just ask about the wavelength to consider and offer graphical and numerical output like

Such a theoretrical diffractogram is then compared with an experimentally recorded one; for example to check the newly prepared sample's phase identity (polymorphism) and purity.

In PXRD, spatial arrangement and conformations of subunits in the crystallographic model are optimised (refined) until yielding a model that chemically and mathematically is reasonable. This is the reason you see in publications a difference trace below the the superposition of experimental and predicted diffraction pattern:

(source)

For single crystal diffraction, the generation of a 3D diffraction model computationally might be more costly than for PXRD, on the other hand, as said, programs like ShelX routinely compare experimentally recorded $F_O$ with predicted / calculated diffraction intensities $F_C$ derived from the current set of atoms (their coordinates, occupancy, etc.) of the model. It is done internally. ShelX' *.lst listings for example include a section like:

Hence I answer your question, if diffraction pattern may be simulated / faked with a "yes it is possible, provided there is a good data set about how the atoms / molecules are spatially arranged".  Initiatives like the recurrent CSP blind tests to predict crystal structures and databases like PCOD may provide a foundation for this.

Then how to spot an artifical and wrong diffraction model?  Checkcif / Platon are tools to check for consistency and integrity of such models against the raw data (hkl-files), structure factors. The recent installation of checksums into *.cif files (currently under watch as Platon errors PLAT012, 013, and 014), may be seen as one element of protection if few entries were changed after completing the work described above.  These bells probably remain silent if data were compromised very early on.


Which of the following compounds do not undego electrophilic aromatic substitution?

Wikipedia disagrees.  Selenophene does undergo EAS, and apparently faster than thiophene.

None of the proposed answers to the original question is correct.


Which salt is a surfactant and lowers the surface tension of the water?

Cetyltrimethylammonium bromide (CTAB) and its cheaper chloro homologue are just two examples of cationic detergents commercially available and frequently used e.g., for hair shampoos.  They are called cationic because beside lipophilic alkyl groups their head group is positively charged:

(source)
Note, there equally are anionic and nonionc detergents, too.
There are multiple approaches to demonstrate their effect to lower the surface tension of water.  Just a qualitative illustration is placing carefully a paper clip on water which will sink after the addition of a detergent:

(source)
If you are up to quantify the effect, the measurement of the contact angle between the liquid and the solid is a frequently used approach.  Again, the Journal of Chemical Education (2017JChemEduc230) published a lab class showcasing the relevance of surface tension e.g., when spraying leaves of lettuce.  A second nice demonstration I recall from this journal was to record the rising meniscus in pasteur pipettes (2016JChemEduc1647) in variation of the concentration of sodium dodecyl sulfate (a popular anionic surfactant).


Determining phase from intermolecular interactions and kinetic energy

Most likely gas.

In fact, the definition of an ideal gas is a gas in which its molecules have no intermolecular interactions (except elastic collisions).

As the strength of intermolecular attraction increases, the molecules will be brought closer together. Molecules in liquids are much more densely packed than molecules in gases.


Which compound should I avoid in deodorants to avoid "sticky, unwashable" feeling?

Regrettably, most deodorants work as antiperspirants,  by clogging the pores of apocrine glands. These include salts of aluminum and zirconium, which hydrolyze (break up and combine with water). One example is aluminum zirconium tetrachlorohydrex glycosine complex, which works by forming a sticky, colloidal gel with water (and sweat). It's like the tale of the scorpion and the frog -- its their nature to act that way.

A bad alternative is deodorants with antibacterial compounds such as triclosan, which don't block sweating, but kill some odor-causing bacteria. Some of these antibacterials are endocrine disruptors, and all promote a population of antibacterial-resistant germs on the skin.

Traditionally, personal deodorants were just masking scents, and in 16th century Europe, flowers were worn to cover the stench... personal and environmental.


Convert CIF file to SDF file using free Linux-compatible software other than Avogadro or OpenBabel?


How do I find the ground state term symbol for transition metal complexes?

To find the ground state term symbol, you should be using symmetry and group theory arguments, you shouldn't have to resort to searching Tanabe-Sugano diagrams to get the answer.

We'll start with octahedral complexes (the general idea can be extended quite easily to tetrahedral or square planar complexes). As requested in the question, I will only cover ground-state term symbols, but this procedure can also be extended easily to excited configurations.

In an octahedral complex, the d orbitals are split into a $\mathrm{t_{2g}}$ set and an $\mathrm{e_g}$ set. You will need the direct product table for the $O_\mathrm{h}$ point group, which you can find here (I will reproduce it here for convenience). You simply need to tack on the g subscripts on everything (since $\mathrm{g \otimes g = g}$, there will never be any ungerade states involved).

$$\begin{array}{c|ccccc} \hline
 & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\ \hline
\mathrm{A_1} & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\
\mathrm{A_2} & & \mathrm{A_1} & \mathrm{E} & \mathrm{T_2} & \mathrm{T_1} \\
\mathrm{E} & & & \mathrm{A_1 + [A_2] + E} & \mathrm{T_1 + T_2} & \mathrm{T_1 + T_2} \\
\mathrm{T_1} & & & & \mathrm{A_1 + E + [T_1] + T_2} & \mathrm{A_2 + E + T_1 + T_2} \\
\mathrm{T_2} & & & & & \mathrm{A_1 + E + [T_1] + T_2} \\ \hline
\end{array}$$

I also want to make the disclaimer that there is not necessarily a one-to-one correspondence between electronic configurations and term symbols. Therefore, the configuration $(\mathrm{d}_{xy})^2(\mathrm{d}_{z^2})^1$ does not necessarily lead to one specific term symbol. A term symbol may often be a linear combination of multiple electronic configurations, and vice versa. Nevertheless, I will use the traditional "orbital diagrams" to illustrate points as necessary.

$\mathrm{d^1}: {}^2T_{\mathrm{2g}}$

The ground state is a $\mathrm{(t_{2g})^1}$ configuration, which corresponds very simply to a $T_\mathrm{2g}$ state. The spin of a single electron is $S = 1/2$, corresponding to a multiplicity of $2S + 1 = 2$.

$\text{High-spin }\mathrm{d^6}: {}^5T_{\mathrm{2g}}$

The high-spin configuration of $\mathrm{(t_{2g})^4(e_g)^2}$ can be considered to be the sum of a $\mathrm{(t_{2g})^3(e_g)^2}$ $\mathrm{d^5}$ configuration (in black), and a $\mathrm{(t_{2g})^1}$ $\mathrm{d^1}$ configuration (in blue).

The five spin-up electrons are symmetrically distributed amongst the five d-orbitals, and as a consequence, these five electrons together have a spherically symmetric electron density. You can prove this quite easily by looking up the wavefunctions of the five d-orbitals, and evaluating the function:

$$\rho(\mathrm{d^5}) = |\mathrm{d}_{xy}|^2 + |\mathrm{d}_{xz}|^2 + |\mathrm{d}_{yz}|^2 + |\mathrm{d}_{x^2-y^2}|^2 + |\mathrm{d}_{z^2}|^2$$

In fact, it is probably much easier to add up the complex spherical harmonics (add up the square modulus of the five functions with $l = 2$). The radial part $R(r)$ can be factored out (and ignored) since it only depends on the quantum numbers $n$ and $l$ (and is therefore constant for all five 3d orbitals).

Whichever way you do it, you would find that the resulting electron density $\rho$ would only be dependent on $r$, and no longer on $\theta$ and $\phi$. Such spherically symmetric spatial components transform as the totally symmetric irreducible representation, which in our case is $\mathrm{A_{1g}}$. (This is always the case, but if you need further evidence, it is provided in the tables of descent in symmetry.) Since $\mathrm{A_{1g}} \otimes \Gamma = \Gamma$ (where $\Gamma$ can be any generic irrep), we can safely discard the $\mathrm{A_{1g}}$ component, and focus on the blue component...

...which we already found in the previous section to be $\mathrm{T_{2g}}$. Therefore, the spatial component of a high-spin $\mathrm{d^6}$ configuration is also $\mathrm{T_{2g}}$.

The spin, however, is different. This time, there are four unpaired electrons, such that $S = 2$ and $2S + 1 = 5$.

$\text{Low-spin }\mathrm{d^6}: {}^1\!A_\mathrm{1g}$

In fact, there is nearly no group theory here to be discussed. This is an example of a closed-shell configuration: all electrons are paired up and there is no unequal population of degenerate orbitals. Such configurations always lead to totally symmetric term symbols, i.e. a spatial part of $A_\mathrm{1g}$. Since there are no unpaired electrons, $S = 0$ and $2S + 1 = 1$.

$\mathrm{d^9}: {}^2\!E_\mathrm{g}$

This is a good point in time to introduce what is known as the hole formalism. The idea is that a $\mathrm{d^9}$ configuration is the same as a $\mathrm{d^{10}}$ configuration, minus one electron. A fully filled d subshell is spherically symmetric, meaning that the $\mathrm{d^{10}}$ configuration has a $A_{\mathrm{1g}}$ term symbol. Therefore, if the $\mathrm{d^{9}}$ configuration transforms as $\Gamma_{(9)}$, and the "missing" $\mathrm{d^{1}}$ transforms as $\Gamma_{(1)}$, then:

$$\Gamma_{(9)} \otimes \Gamma_{(1)} = \mathrm{A_{1g}}$$

There is a result in group theory that states that the totally symmetric irrep ($\mathrm{A_{1g}}$ in our case) only occurs along the main diagonal of the direct product table. In other words, the above equation can only be fulfilled if

$$\Gamma_{(9)} = \Gamma_{(1)}$$

Therefore, our task is simply to find what the term symbol of the "missing" electron is. The missing electron is known as a hole.

In our case, in order to minimise the energy of the $\mathrm{d^9}$ electron configuration, we would want to put the hole in the higher-energy $\mathrm{e_g}$ orbital. As a result, our task becomes to find the term symbol for a $\mathrm{(e_g)^1}$ hole configuration, which is simply $E_\mathrm{g}$.

The total spin of a $\mathrm{d^9}$ configuration is easily seen to be $S = 1/2$, leading to a doublet state again.

$\mathrm{d^2}: {}^3T_{\mathrm{1g}}$

The ground state is a $\mathrm{(t_{2g})^2}$ configuration. Clearly, the green configuration is more favourable than the red one (Hund's first rule). The question then becomes how you obtain the possible term symbols for a $\mathrm{(t_{2g})^2}$ configuration, and which of them correspond to the green configuration and which to the red. The different possibilities are given by the direct product

$$\mathrm{T_{2g} \otimes T_{2g} = A_{1g} + E_g + [T_{1g}] + T_{2g}}$$

The square brackets are used to indicate spatial states that are antisymmetrised, i.e. are antisymmetric under the permutation of two electrons. Due to the spin-statistics theorem, the total electronic wavefunction, which is a product of spin and spatial parts, must be antisymmetric. To satisfy this, you have two possible choices:

$$\begin{array}{cc|c} \hline
\text{Spatial} & \text{Spin} & \text{Overall} \\ \hline
\color{green}{\text{AS } (\mathrm{T_{1g}})} & \color{green}{\text{S (“triplet")}} & \color{green}{\text{AS}} \\
\color{red}{\text{S } (\mathrm{A_{1g}, E_g, T_{2g}})} & \color{red}{\text{AS (“singlet")}} & \color{red}{\text{AS}} \\ \hline
\end{array}$$

The antisymmetric spatial wavefunction is more stable than the symmetric spatial wavefunctions, due to exchange energy. Consequently, the ground-state term symbol has a $T_{\mathrm{1g}}$ component.

As can be seen from the above table, this necessitates that the spins of the electrons are parallel, i.e. $S = 1$ and $2S + 1 = 3$. We can therefore associate the $T_\mathrm{1g}$ state with the green electronic configuration, with two spin-up electrons in different orbitals.

$\mathrm{d^3}: {}^4\!A_\mathrm{2g}$

The $\mathrm{(t_{2g})^3}$ configuration here is simply a spherically symmetric $\mathrm{d^5}$ configuration, minus an $\mathrm{(e_g)^2}$ hole configuration. To find the symmetry of the two holes, we take

$$\mathrm{E_g \otimes E_g = A_{1g} + [A_{2g}] + E_g}$$

Since the two holes are in different orbitals, their spatial component has to be antisymmetrised. Therefore, it is really the $A_\mathrm{2g}$ state that we want here.

As for the multiplicity, $S = 3/2$ gives $2S + 1 = 4$.

Now that all the ground work has been done,

we can whizz through the remaining electronic configurations - how exciting! It's a good idea to try working these out yourself before looking at the answer, to see if you truly understand the concepts above. I'm also not going to comment on the spin multiplicities any more - hopefully you can look at the diagram, calculate $S$, and work out $2S + 1$ yourself.

$\text{High-spin }\mathrm{d^4}: {}^5\!E_\mathrm{g}$

Using the hole formalism, you simply subtract one $\mathrm{e_g}$ electron from a spherically symmetric $\mathrm{d^5}$ configuration, which straightforwardly gives the term symbol above.

$\text{Low-spin }\mathrm{d^4}: {}^3T_\mathrm{1g}$

The black $\mathrm{d^3}$ configuration has symmetry $\mathrm{A_{2g}}$, and the lone blue d-electron has symmetry $\mathrm{T_{2g}}$. The direct product gives the symmetry for the low-spin $\mathrm{d^4}$ case:

$$\mathrm{A_{1g} \otimes T_{2g} = T_{1g}}$$

$\text{High-spin }\mathrm{d^5}: {}^6\!A_\mathrm{1g}$

We already established earlier that this is spherically symmetric and therefore transforms as the totally symmetric irrep $\mathrm{A_{1g}}$.

$\text{Low-spin }\mathrm{d^5}: {}^2T_\mathrm{2g}$

This is best treated as a $\mathrm{t_{2g}}$ hole in a totally symmetric low-spin $\mathrm{d^6}$ configuration, which gives rise to a $T_\mathrm{2g}$ state.

$\text{High-spin }\mathrm{d^7}: {}^4T_\mathrm{1g}$

Treating this as a sum of a totally symmetric $\mathrm{d^5}$ configuration and a $\mathrm{d^2}$ configuration, we can tell that this has the same spatial symmetry as the original $\mathrm{d^2}$ configuration, i.e. it is a $T_\mathrm{1g}$ state.

$\text{Low-spin }\mathrm{d^7}: {}^2\!E_\mathrm{g}$

Noticing that this is the sum of the totally symmetric low-spin $\mathrm{d^6}$ and one $\mathrm{e_g}$ electron makes this easy.

$\mathrm{d^8}: {}^3\!A_\mathrm{2g}$

This is, again, a sum of the totally symmetric low-spin $\mathrm{d^6}$ configuration and the $\mathrm{d^3}$ configuration, which has $\mathrm{A_{2g}}$ symmetry.

For extra practice, why not try doing this using the hole formalism instead? You should reach the same answer!


Why are correlation consistent basis sets used with DFT calculations?

It is generally recommended not to use a cc basis set with a DFT method (and I guess conversely, a basis set aimed at DFT should not be used with a coupled cluster method).

This statement glosses over some specifics that might be important.

There is nothing technically wrong with using correlation-consistent or ANO basis sets with DFT, unless the basis is deficient to begin with (cc-pVDZ). That is, using cc-pVQZ to optimize the geometry of a small organic molecule will not lead to more "incorrect" results than 6-31G* unless linear dependencies arise in the basis due to over-completeness.
However, it is most certainly inefficient to use these basis sets for general usage (SCF-type methods) due to using general contraction rather than segmented contraction. Most modern integral algorithms still have a difficult time dealing with highly-contracted basis sets. This is one reason to use the TURBOMOLE/Ahlrichs/Karlsruhe Def2 and Frank Jensen's segmented polarization-consistent (pcseg-n) basis sets.
"Using a basis set aimed at DFT should not be used with a coupled cluster method" is more tricky. The TURBOMOLE basis sets are certainly used for MP2 and coupled cluster calculations, however they are more appropriate for "single" calculations (not CBS extrapolation), and only with the largest basis one can afford. They were designed to capture the total mean-field energy, which decays exponentially, not the correlation energy, which decays like ~$(L+1)^{-3}$, where $L$ is the maximum angular momentum being considered. The correlation-consistent and atomic natural orbital basis sets are better for coupled cluster because they were designed from the beginning to recover the correlation energy.
The use of Pople-style (STO-nG, 3-21G, 6-31G*, 6-311++G(d,p), etc.) basis sets would be incorrect with wavefunction-based methods, perhaps even for qualitative understanding depending on how "difficult" your system is. Like the TURBOMOLE basis sets, they are designed to capture the mean-field atomic energies, but many have constraints (identical s- and p-type exponents, the "SP" functions that are often seen) due to the computational resources available at the time of their creation. Many of the older basis sets (notably 6-31G and not 6-311G) were optimized to use Cartesian rather than pure d-type functions, leading to potential inconsistencies in results. It is also not as obvious how to systematically improve the quality of the basis sets, and none are larger than triple-$\zeta$. Ultimately, the virtual MO space created by these basis sets is not sufficient to recover the correlation energy, and the benefits of using coupled cluster are completely lost.


Sodium + Chloroform:
https://www.youtube.com/watch?v=du3ksrfnCcg

To prepare deuterochloroform I'd recommend the haloform reaction.


How does a double displacement reaction work?

Double displacement reactions involve ionic compounds, i.e. salts formed from positive cations and negative anions. Consider the following setup,
$$\ce{\color{blue}{A^{+}}\color{red}{B^{-}} + \color{blue}{C^{+}}\color{red}{D^{-}} -> \color{blue}{A^{+}}\color{red}{D^{-}} + \color{blue}{C^{+}}\color{red}{B^{-}}}$$
where the cations are indicated in blue and anions in red (and all charges are $\pm 1$ to reflect your question). All salts are comprised of cations and anions, hence $\ce{\color{blue}{AC}}$ wouldn't form since both are cations. $\ce{\color{blue}{C}\color{red}{B}}$ would form instead.


Yes, and no, and maybe.

Oftentimes the sign of the Laplacian at the bond critical point in terms of the quantum theory of atoms in molecules (QTAIM) is taken as a measurement for the covalency of a bond. That it is not that simple is quite logical from your question alone.

Just as in bonding in general, i.e. there is a spectrum from pure covalent (ideally, but not achievable) and pure ionic (ditto),* there cannot be a clear yes or no answer here, too. I think the case of $\ce{BCl3}$ versus $\ce{AlCl3}$ and respective dimers, which I have written about, demonstrates this feature and how QTAIM can help understand bonding. As with many things it certainly is hard to draw a line. In that sense it is probably a good criterion to judge the nature of a bond. Judging only based on the sign of the Laplacian, while being a rigorous and unambiguous way to look at things, might lead to the wrong conclusions though.

What does it matter anyway? In many cases such a strict separation is not helpful. It should be clear, that there are covalent and ionic contributions in every bond, and reactivity often does not care about that. You'll never obtain a complete picture, if you are looking at a single property.

* And then there is charge-shift bonding (What is charge shift bonding?), but I personally don't see any value in that.


Is the Aufbau principle violated in the diagram (b)?

Since the question specifically asks about the violation of Aufbau principle, your reasoning is correct.

Had it been based on the violation of Hund's rule then option (a) would be the answer.


Can natural gas appliances run on Hydrogen?

Pure hydrogen, as opposed to hydrocarbons, degrades metals.  $\ce {H_2}$ molecules are tiny, so they wedge themselves into the relatively large metallic crystal structures and form hydrides.  This is the entire reason nickel-metal hydride (NiMH) batteries work - the hydrogen is stored in an iron-hydride matrix. An appliance that might last 20 or 30 years burning natural gas might not make it past ten years burning $\ce{H_2}$, and a failure may take the form of a dangerous leak.


Why don't calcium chloride and sodium nitrate react?

It's because all the species are soluble in water and there is no driving force such as precipitation or gas evolution. You just have a mixture of ions.


Mixture of water, methanol and dichloromethane

This is a partial phase diagram. That seems to indicate that systems with roughly 50% methanol are miscible.

Source: http://shodhganga.inflibnet.ac.in/bitstream/10603/9995/11/11_chapter%206.pdf


How to order the reactivity of t-butyl iodide and neopentyl bromide for SN2

The order will be $\text{neopentyl bromide} > \text{t-butyl bromide}$. This is just an addition to @user377386's answer. From Solomons and Fryhle's Organic Chemistry, the relative rates of reaction (SN2) of different kinds of halide are (for the same halogen):
$$ \begin{array}{c|c}
 \textbf{Type of halide} & \textbf{Relative Rate} \\ \hline
 \text{Primary Halide} & 1 \\
 \text{Secondary Halide} & 0.03 \\
 \text{Neopentyl Halide} & 0.00001 \\
 \text{Tertiary Halide} & \approx 0 \\
\end{array} $$

You can easily see that neopentyl halides and tertiary halides are almost unreactive towards SN2 reaction, so it makes little sense to compare them. Neither of them is going to react by SN2 ever, they will instead react through SN1 mechanism. The slightly increased leaving group ability of iodine does not matter much. If you insist on comparing though, neopentyl halide will be better at SN2.


Since the gravitational force between two protons is negligible there must be another force holding the nucleus together. This is the strong nuclear force, which as the name suggests is extremely strong but it is also extremely short range and so it's effects are only felt on the scale of nuclei and baryons. As you can see in the graph, if two protons approach each other they repel each other due to electrostatic repulsion but once they get within about $\mathrm{3~fm}$ of each other the strong nuclear force begins to become significant and at about $\mathrm{2~fm}$ it starts to outweigh the electrostatic repulsion and bind the two protons together.

It is also interesting to note that at even shorter ranges the strong nuclear force is repulsive and so there is an equilibrium position, indicated on the graph, where the strong nuclear attraction and the electrostatic repulsion are equal in magnitude and so the net force on the protons is zero.

Note that the graph shown is for two protons and so in nuclei containing more than two protons the numbers will be different but the principle is the same.


In chemistry, some chemical species can still react with nitrogen so argon is usually favored, but, in the case of wines, none of the two should react whatsoever.

In both cases, however, the high density of argon is an advantage because it will always fall down to be in contact with the liquid in order to avoid contact with oxygen.

By contrast, nitrogen would just dilute oxygen in the remaining air and just slow down oxidation, because air is already composed of 80% nitrogen.


How do we know that water has [H+] = 10^-7?

Why is $[\ce{H+}] = [\ce{OH-}]$?

Because the water is electrically neutral.

In general, for all solutions, the sum of the concentration of each ion, multiplied by the ion's respective charge, must be zero.

  How do we know $[\ce{H+}] = 10^{-7}$?

This is found by experimentally by determining $K_\mathrm{w}$ and substituting, or by measuring the pH.

For details see: Bandura, A. V.; Lvov, S. N. The Ionization Constant of Water over Wide Ranges of Temperature and Density. J. Phys. Chem. Ref. Data 2006, 35 (1), 15–30. DOI: 10.1063/1.1928231.


Expression of the potential in the hydrogen atom while solving radial part of the wave function

The 'centrifugal' term appears when the coordinates are changed from $(x,y,z)$ to spherical $(r, \theta, \varphi)$. This is not to say that it is artificial, or a mathematical trick, but is a consequence of the fact that the electron has orbital angular momentum, and this is made clear by changing coordinate systems.
To summarize: the equation is $H\varphi(\pmb{r})=E\varphi(\pmb{r})$ where $\displaystyle H=-\frac{\hbar^2}{2\mu}\Delta +V(r)$. Changing to spherical coordinates produces,
$$\Delta =-\frac{1}{r}\frac{\partial ^2}{\partial r^2}+\frac{1}{r^2}\left(\frac{\partial ^2}{\partial \theta^2}+\frac{1}{\tan(\theta)}\frac{\partial}{\partial \theta}+\frac{1}{\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2} \right)$$
which means that eigenfunctions of $\varphi(\pmb{r})$ are variables in $r, \theta, \varphi$
The second term is almost the same as that for $\pmb{L^2}$ and substituting for this makes the Hamiltonian
$$\displaystyle H=-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial ^2}{\partial r^2}r+ \frac{1}{2\mu r^2}\pmb{L^2}+V(r)$$
As $\pmb{L^2}\varphi(\pmb{r})=l(l+1)\hbar^2\varphi(\pmb{r})$  then
$$\displaystyle H=-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial ^2}{\partial r^2}r + \frac{l(l+1)\hbar^2}{2\mu r^2}+V(r)$$
The potential is unchanged and still $V(r) = -e^2/r$ and the Hamiltonian is still a function of $r$ but now depends on variable $l$ thus the eigenvalues must depend on $l$.
The centrifugal term is so called because the corresponding force, the negative slope, always repels the electron from the centre and labelled a centrifugal potential or barrier. The term is always positive or zero and can change the shape of the potential the electron experiences and thus forms an effective potential written as  $\displaystyle V_{eff}=\frac{l(l+1)\hbar^2}{2\mu r^2}+V(r)$. When $l>0$ the centrifugal term dominates at small $r$ causing $V_{eff}$ to be repulsive. As larger $r$ the potential becomes attractive.


Reversible & irreversible adiabatic processes

For a given number of moles, once $P_1$ and $V_1$ are specified, that determines state 1 and $T_1$.  Once $P_2$ and $V_2$ are specified, that determines state 2 and $T_2$.  It doesn't matter what process is used to get from state 1 to state 2.  Only 2 intensive variables are required to specify the state of the system (assuming no phase change).  So $T_2=T_2"$.

Maybe you want to remove either the $P_2$ or the $V_2$ from the problem specification.


Acidity order checking for α-furoic acid and acetic acid

The conjugate base of 2-furoic acid is aromatic, but so is the acid.
'

The aromaticity does not contribute to the stability of the conjugate base because the negative charge from the carboxylate anion cannot delocalize into the ring. The major resonance contributors for the carboxylate anion are:

However, because of the oxygen atom, furan is an electron-rich arene, and one position that has higher than expected electron density is the 2-position:

Furan is therefor electron-donating, and electron-donating groups destabilize anions. Consider the following resonance contributor of 2-furoic acid and the close proximity of the formal negative charges:


If you use methanol as a solvent and add a suitable base, you should end up with the correct product, since any MeOH attacking the protecting group you want to keep will simply re-form the methyl ester, whereas the acetate will cleave to give the unprotected alcohol.

You could potentially also achieve this using sodium methoxide/methanol to speed the reaction up, but standard potassium carbonate/methanol is usually just fine.


I want to make electricity of aluminiun cans i.e. reverse the process of refinning aluminium. Any ideas?

You could "burn" the aluminum in the presence of oxygen to generate aluminum oxide.  For every two moles of aluminum consumed, a bit less than 1700 kJ of energy would be released.  You could use the energy to drive a turbine and generate electricity.


If the oxides are separate grains or compounds (a mixture, not a solid solution), you can find this with SEM microprobe (a geology technique).  For tricky new mixtures, may be better than powder x-ray diffraction (find small amounts of new compound, than resolve stoichiometry and do XRD on phase pure samples.  It will not give you the oxidation level though.

You could just do visual examination.  $\ce{PbO}$ is yellow and $\ce{PbO2}$ is black-brown.  $\ce{CuO}$ is black and $\ce{Cu2O}$ is red.  You can use a metallography microscope (high powered optical) to look at individual grains.  [May also be able to see the crystal habit.]

In general, people working with oxides know the expected oxidation state of the materials they have processed given the conditions.  E.g. cerium will always be $+4$, not $+2$ under atmospheric ceramic processing conditions.  Not always possible, but often the case.

If you could narrow down a little more what you are trying to do, might be easier to answer.  What substances, processing conditions and what info do you need on the end result (and why, really)?

ICP or AA are techniques helpful for trace metal analysis but you won't get oxidation state.

sorry, I can't give a more precise answer, but really your question is very broad.

If you can't discuss your project details on the net, consider to talk to the director of a research university chemistry department shared analysis lab.  I really don't think there is a single answer to your question and it will depend more on what you are looking at and how precise an answer you need, which technique is best.