text:
Why is it easier to add bromine to styrene than cinnamic acid?
Why is it easy to add bromine to styrene and why is it so difficult to add bromine to cinnamic acid? It is obvious that the $\ce{COOH}$ group of cinnamic acid affects the reaction quite a lot, but I just can't figure out why. I am guessing that the electronegativity and inductive effect of the carboxylic acid plays a key role. Any help would be greatly appreciated!

Welcome to ChemSE. We like to keep questions neat and to the point without any distractions like "Greetings", so I edited your question accordingly. I hope to see your question answered soon. Good luck!

Maybe consider doing some research into alpha beta unsaturated carbonyl compounds.

Right now I seriously have the same question as you do @christofer!


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What is the IUPAC name of K4[Fe(CN)6]?

The IUPAC name of $\ce{K4[Fe(CN)6]}$ is:
A. Potassium hexacyanido ferrate(II)
B. Potassium hexacyanido ferrate(4-)
C. Tetrapotassium hexacyanoferrate
D. All are correct

Actually I had been going through the IUPAC exercise of my book when I encountered just this question.
I know that the correct answer is option A but in the answer key the answer is given as option D.
Can this really happen ? Can a compound have more than one IUPAC name?
Our teacher told that whichever name can be used to deduce the structure of the coordination compound can be termed as the IUPAC name of that compound but I am not at all convinced with this fact.

There shouldn't be a space between cyanido and ferrate AFAIK. So...

All three would appear accepted by IUPAC, though A is preferred generally.
Quoting from the red book (the IUPAC guide for nomenclature of inorganic compounds):
IR.5.4.2.2:

It is possible to provide information on the proportions of the constituents in names by using one of two other devices: the charge number, which designates ionic charge, and the oxidation number, which designates oxidation state. In nomenclature, the use of the charge number is preferred as the determination of the oxidation number is sometimes ambiguous and subjective. It is advisable to use oxidation numbers only when there is no uncertainty about their assignment.
  The charge number is a number whose magnitude is the ionic charge. It is written in parentheses immediately after the name of an ion, without a space. The charge is written in arabic numerals, followed by the sign of the charge. Note that unity is always indicated, unlike in superscript charge designations (which are used in formulae). No charge number is used after the name of a neutral species.
The oxidation number (see Sections IR-4.6.1 and IR-9.1.2.8) of an element is indicated by a Roman numeral placed in parentheses immediately following the name (modified by the ending ‘ate’ if necessary) of the element to which it refers. The oxidation number may be positive, negative or zero (represented by the numeral 0). An oxidation number is always non-negative unless the minus sign is explicitly used (the positive sign is never used). Non-integral oxidation numbers are not used for nomenclature purposes.

Potassium hexacyanidoferrate(II) and potassium hexacyanidoferrate(4–) are specifically given as examples based on this rule, making A and B both valid IUPAC names.
Section IR.5.2.1:

The proportions of the constituents, be they monoatomic or polyatomic, may be indicated in generalized stoichiometric names by multiplicative prefixes, as was the case for the constituents of binary

Tetrapotassium hexacyanidoferrate is given as an example based on this rule, making C also a valid IUPAC name.

As an aside, the red book specifically makes provision for the fact that, at present, there are multiple allowed IUPAC names:

A planned future project on inorganic PINs will need to face the problem of choice between the equally valid nomenclature systems currently in use.

This differs from organic nomenclature when one preferred IUPAC name is given (known as the PIN), even where multiple valid names may exist (and are often preferred in everyday use for historical reasons).

IUPAC does allow multiple names to compounds? I always thought that IUPAC Nomenclature was unambiguous.

@PrittBalagopal IUPAC nomenclature is unambiguous unidirectionally. A single valid IUPAC name will only point to a single substance but multiple different IUPAC names may point to the same substance.

@Jan Oh in that way huh? I always thought every compound had only one unique name.

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Equilibrium involving water and solid component
$$\ce{A + B <=> C(s) + H2O(l)}$$
Why when $\ce{H2O}$ is liquid and $\ce{C}$ is solid, the equilibrium constant of the reaction is
$$K_\mathrm{eq} = \frac{1}{[\ce{A}][\ce{B}]} \quad ?$$

I remember this being described as a "convention", as if to acknowledge that the role if a solid component of an equilibrium is ill -  defined. Would appreciate hearing from others on the subject.

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What's the mechanism of water catalyzing a reaction of hydrogen sulfide gas and sulfur(IV) oxide to yield sulfur?
The reaction
$$\ce{2 H2S(g) + SO2(g) -> 2 H2O(l) + 3 S(s)}$$
goes in presence of a catalyst, usually $\ce{Al2O3}$ or $\ce{TiO2}$, but also just in presence of moist or water (YouTube video)
I don't understand the catalytic mechanism - why doesn't the reaction go without water and how exactly does presence of water help?

Yes, hydrogen sulfide and sulfur dioxide reacts with each other in the presence of moisture. The reaction has been studied previously and was noted that the reaction did not proceed if there was no water vapor(dry gases). The mechanism of  reaction was determined and was adsorption. The gases adsorped on the surface of water and collided with each other to deposit sulfur and water. The formation of water further catalysed the reaction. This is a excerpt from the paper which extensively studies the mechanism of the reaction.$\ce{^{[1]}}$

The results of experiment 3 are best explained by assuming that both
  the  $\ce{H2S}$ and $\ce{SO2}$ are adsorbed and that reaction takes
  place only between these  adsorbed molecules.

There is another paper which extensively discuss the mechanism of reaction. It says that both the gases react with a reaction to form a solution. This is called Wackenroeder solution which is a mixture of sulfur based acid(polythionic acids). These acids dissociate in aqueous solution and form and water. Have a good read at the paper:$\ce{^{[2]}}$

Volynskii proposed a mechanism for the reaction of $\ce{H2S}$ and
  $\ce{SO2}$ in aqueous solutions. The formation of sulfur follows the
  steps below.
$$\ce{H2S + SO2 -> HSSO2H }$$ $$\ce{HSSO2H -> S(OH)2 + S }$$
  $$\ce{S(OH)2 + H2S -> 2H2O + 2S (or~S2) }$$
The overall equation for the three steps is reaction 2. The formation
  of polythionic acids results from the interaction of ionized $\ce{SO2}$ and
  monatomic or polyatomic sulfur, i.e.,
$$\ce{SO2 + H2O -> HSO3- + H+ }$$
  $$\ce{HSO3- + S -> HS2O3- }$$
  $$\ce{S_{x} + 2HSO3- -> S_{(x +1)}O6- + H2S }$$ $$\ce{S_{x} + 2HS2O3- ->S_{(x +2)}O6- + H2S }$$
  $$\ce{S_{x} + HSO3- + HS2O3- -> S_{(x +3)}O6- + H2S }$$
where, x can range from 1 to 7. Tiwari believed that the reaction of
  $\ce{H2S}$ and $\ce{SO2}$ in dilute sulfuric acid (below 10 wt %)
  follows the scheme from (7) to (9). His kinetic data show that the
  reaction is first order with respect to both $\ce{H2S}$ and
  $\ce{SO2}$, consistent with the scheme because he assumed reaction 7
  to be the rate-controlling

Reference
$\ce{^{[1]}}$ http://pubs.acs.org/doi/abs/10.1021/j150272a005
$\ce{^{[2]}}$ http://pubs.acs.org/doi/abs/10.1021/ie020275i


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How to find out which organic molecules would be good electron acceptors/donors on a GaAs semiconductor surface
I'm a physicist in way over my head.
I have a list of organic molecules to investigate, all common aldehydes and amines. My supervisor wishes to know which of these molecules would be good electron donors and which would be good electron acceptors.
The idea is to use some of these molecules to dope some GaAs nanowires, in order to be able to tweak the Fermi level at the edge states. What are the properties I should be looking for? I am guessing that compounds with very electronegative elemental constituents will be good electron acceptors and vice versa for those with electropositive constituents but other than that I am unsure where to find the information I need. Could anybody point me in a helpful direction?
Some one else mentioned electron affinities and ionisation potential but I have no idea where to look for these things in the literature. My chemical understanding is limited to what I know about physics and advanced high school chemistry.

Look for the literature entering "organic donor and acceptors" "P and N-type organic semiconductors" for instance. If both you and your supervisor are new to these things, "organic electronics" is somehow an established field of research. You will find plenty of reviews. As it is quite intrinsic to their nature, carbon compounds tends to be of P type. Triphenylamibes and carbazoles are known since long time to be holes transporting materials. Conjugated polymers are relatively newer. . Although purely carbon based, fullerenes are the most outstanding molecules on the N side. Just start search.

I realise you cannot choose the molecule. :(

You probably need to look for tables of redox potentials such as in the CRC handbook of Chemistry & Physics. The free energy of the reaction is $\Delta G = E_{donor}-E_{acceptor}$. So GaAs could be donor or acceptor. Look for values, say, for a known donor and compare other molecules with this. Also look at Marcus theory of electron transfer as too large a $-\Delta G$ can lead to slow electron transfer reaction, as a rule of thumb  -$1$eV is ok


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Symmetry and NMR linewidths of quadrupolar nuclei
The linewidth of peaks when taking NMR spectra of quadrupolar nuclei (11B or 14N for example) is often very broad, making their interpretation challenging.
Recently, when acquiring 11B-NMR spectra of some boronic esters I'd made, the NMR technician pointed out that the linewidth is much improved in highly symmetrical environments (which my boronic esters were not) – she proved this by running a sample of $\ce{LiBH4}$, but couldn't really offer an explanation of why this was the case, other than to point out that its a relaxation effect.
Claridge[1] confirms this point about symmetry:

14N linewidth of $\ce{N(Me)4+}$ is less than 1 Hz whereas that for $\ce{NMe3}$ is nearer to 80 Hz.

Arguably, both $\ce{N(Me)4+}$ and $\ce{NMe3}$ are fairly symmetrical, being Td and pseudo-C3v respectively, which leads me to think that what the molecule needs is some kind of spherical symmetry, which only Td (and Oh, arguably less useful in synthetic organic chemistry) can provide.
Some further reading through Claridge[1] reveals that this is something known as quadrupolar relaxation, with Reich[2] confirming, though not really expanding beyond an equation describing the relaxation process and a comment that there is "friction" between the nucleus and the surrounding electron.
All of the explanations I've found have focused a lot on explaining this mathematically (Levitt describes it quite nicely, but slightly beyond my understanding). Is there an intuitive explanation for the relaxation taking place in quadrupolar nuclei, and why high symmetry environments lead to narrow peak shape?

[1]: Claridge, T. D. W. High Resolution NMR Techniques In Organic Chemistry; Elsevier:Amsterdam, 2016
[2]: https://www.chem.wisc.edu/areas/reich/nmr/08-tech-01-relax.htm, accessed 20/08/2017.

A large electric field gradient at the nucleus facilitates quadrupolar relaxation, in a highly symmetric environment the field gradient is (near) zero and quadrupolar relaxation becomes much slower.

I know that you asked for an intuitive explanation, but I'm afraid that if one wants to go beyond ron's comment, the technical aspects are somewhat necessary. It isn't incredibly important to understand the maths, but I have included it as I want to enable interested readers to go further, and it provides a basis for the last section, so I make no apology for it. I have also assumed some basic knowledge of NMR theory.1

Linewidths and relaxation
Following a pulse sequence, the spectrometer detects transverse magnetisation (i.e. magnetisation in the xy-plane). The signal that is detected is generally of the form $s(t) \propto \exp(\mathrm i \Omega t)\exp(t/T_2)$, where $T_2$ is the transverse or spin-spin relaxation time.
If one takes the Fourier transform of $s(t)$, one gets the spectrum $S(\Omega)$, and it can be shown that the widths of the Lorentzian peaks thus obtained are proportional to $1/T_2$. To be precise, the linewidth at half-height $\Delta \nu_{1/2}$ is given by
$$\Delta \nu_{1/2} = \frac{1}{\pi T_2}$$
Therefore, fast transverse relaxation (which corresponds to small $T_2$) leads to broad lines, and slow relaxation leads to narrow lines.

Quadrupoles and the electric field gradient
For quadrupolar nuclei ($I > 1/2$), a major source of relaxation arises due to interaction of the nuclear electric quadrupole moment with the electric field gradient at the nucleus.
An introduction to the above two concepts is as follows: a quadrupole moment reflects an asymmetry in the charge distribution of the nucleus, and can be thought of as two opposing dipoles placed next to each other. The electric field gradient refers to the derivative of the electric field $E$, which is itself the derivative of the electric potential $V$ (so, the field gradient is the second derivative of the potential). These derivatives are evaluated at the spatial position of the nucleus.
The electric field gradient therefore most generally has nine components, since two derivatives, each with respect to 3 axes, are taken. These are represented by $q_{ij}$ (where $i,j$ are one of $x,y,z$):
$$q_{ij} \equiv \left.\frac{\partial^2V}{\partial i\,\partial j}\right|_\text{at nucleus}$$
However, by a suitable transformation of coordinates, the cross derivatives $q_{ij}$ ($i \neq j$) can be set to zero (this essentially involves diagonalising the matrix $\mathbf{q}$). The remaining quantities $q_{xx}$, $q_{yy}$, $q_{zz}$ are referred to as the principal values, and without loss of generality we can adopt the convention $|q_{zz}| \geq |q_{yy}| \geq |q_{xx}|$. Since the electric potential obeys Laplace's equation $\nabla^2 V = 0$, this also means that $q_{xx} + q_{yy} + q_{zz}$ is necessarily equal to 0 (this is often referred to as "traceless", since the trace of the matrix $\mathbf{q}$ is 0). Therefore, there are only two independent parameters. These parameters are chosen to be $q_{zz}$, and $\eta$, a quantity called the biaxiality and defined as
$$\eta = \frac{q_{yy} - q_{xx}}{q_{zz}}.$$

Quadrupolar relaxation2
The interaction of the quadrupole moment with the electric field gradient contributes to the energy of the nucleus (exactly analogous to how an electric dipole moment interacts with an electric field).3 This interaction is modulated by molecular tumbling. That is to say, as the molecule rotates in solution, the magnitude of this interaction changes as well.4
If this interaction varies at the frequency corresponding to the transition between two spin states, then it is possible to induce this transition, thereby leading to spin relaxation. (More strictly speaking, it has to contain a component that oscillates at the transition frequency.) In this aspect, it is analogous to the more well-known case of how magnetic dipolar interactions between two nuclei can lead to relaxation, which is described quite thoroughly in Keeler's book.1 For spin-1/2 nuclei in solution-state NMR, dipolar interactions are the primary method of relaxation since these nuclei do not have quadrupole moments. For quadrupolar nuclei, though, quadrupolar relaxation is the primary method of relaxation, simply because this interaction term tends to have a large magnitude.
For our purposes, it is sufficient to use a derived result, which you seem to have already come across. I know that I am glossing over a lot, but as far as I can tell, the derivation of this is very involved:5
$$\frac{1}{T_1} = \frac{1}{T_2} = \frac{3\pi^2}{10}\frac{(2I+3)}{I^2(2I-1)} \left(\frac{e^2q_{zz}Q}{h}\right)^2 \left(1 + \frac{1}{3}\eta^2 \right) \tau_\mathrm{c}$$
Definitions:

$T_1$ and $T_2$ are the spin-lattice (longitudinal) and spin-spin (transverse) relaxation times respectively; these only account for relaxation via the above mechanism and not for other possible mechanisms
$I$ is the spin of the nucleus
$e$ is the elementary charge, $\pu{1.602 \times 10^-19 C}$
$q_{zz}$ is the main component of the electric field gradient, as explained above
$Q$ is the quadrupole moment of the nucleus
$h$ is Planck's constant, $\pu{6.626 \times 10^-34 J s}$
$\eta$ is the biaxiality, defined above
$\tau_\mathrm{c}$ is the correlation time, which is a constant that essentially measures the rate of molecular tumbling (see books in ref 1 for more information).

Please, no more maths
OK - the only remaining maths is to analyse the equation above and identify the factors which contribute to relaxation, and hence, linewidths. For the actual question, the relevant portion is $e^2q_{zz}Q/h$, which is often written as $\chi$ and termed the nuclear quadrupole coupling constant. Some representative values of $\chi$ are given in ref 2.
In particular, the value of $q_{zz}$ is mainly determined by the distribution of electrons close to the nucleus, and this provides an obvious link to the molecular geometry of the quadrupolar nucleus. For nuclei in tetrahedral, octahedral, cubic, or spherical environments, $q_{zz}$ is, by symmetry, equal to zero. Therefore, quadrupolar relaxation by the above mechanism does not take place, $T_2$ is long, and linewidths are small. One sees very narrow linewidths for the central atoms in, for example, $\ce{Cl- (aq)}$ (neglecting solvation effects, spherical symmetry); $\ce{NMe4+}$ and $\ce{SO4^2-}$ (tetrahedral), and $\ce{[Co(CN)6]^3-}$ (octahedral). On the other hand, if such symmetry is lacking, then $q_{zz} \neq 0$, there is usually fast quadrupolar relaxation by the above mechanism, and linewidths are large.
$q_{zz}$ is directly tied to the distribution of electrons and not the distribution of bonds. Although there is nearly always a correlation, there are some interesting cases where this is not true; in $\ce{Et2N-NO2}$, for example, the $\ce{NO2}$ nitrogen has an extremely sharp peak, despite the local molecular symmetry not intuitively suggesting that it would be the case.2 (The $\ce{Et2N}$ nitrogen, as expected, has a very broad peak.)
Since we went the full distance to obtain the above expression, we can also extract a bit more information from it. The rate of relaxation is also proportional to the nuclear quadrupole moment $Q$; some quadrupole moments are tabulated in ref 2. Some nuclei which have small quadrupole moments (namely deuterium, lithium-6, and cesium-133) have smaller linewidths and are much more amenable to NMR studies. Also, the factor of $I^2$ in the denominator also means that higher spins tend to give sharper lines.

A bonus: coupling to quadrupolar nuclei
$\ce{^13C}$ NMR experiments are often run with proton decoupling - this means that when the carbon signal is collected, the protons are irradiated with a radiofrequency field. This has the effect of inducing rapid transitions between the two spin states of the protons, and each carbon-13 nucleus effectively "sees" an average of the two spin states, causing the C–H couplings to be lost.
Since quadrupolar relaxation also induces rapid transitions between spin states of quadrupolar nuclei, it leads to what is effectively a "self-decoupling" of all quadrupolar nuclei. However, the couplings can still be seen when the quadrupolar nuclei are in highly symmetrical environments: for example, the $\ce{^19F}$ spectrum of $\ce{[BrF6]+}$ is shown here.6 Both $\ce{^79Br}$ and $\ce{^81Br}$ have roughly equal abundance (~50%) and a nuclear spin of $3/2$, so we expect to see two 1:1:1:1 quartets of equal intensity, one arising from $\ce{[^79BrF6]+}$ and one from $\ce{[^81BrF6]+}$.

A few other examples are presented on Hans Reich's page. Of course, these can be understood from the discussion above about how molecular symmetry affects quadrupolar relaxation.
In fact, because deuterium has a small quadrupole moment and undergoes slow quadrupolar relaxation, coupling to deuterium can often be seen. All organic chemists should recall that $\ce{CDCl3}$ appears as a 1:1:1 triplet - this is why!

Notes and references

For an exposition of basic NMR theory see either Keeler, Understanding NMR Spectroscopy, 2nd ed. (Wiley) or Hore, Nuclear Magnetic Resonance, 2nd ed. (OUP).
A necessarily mathematical, but still fairly accessible, introduction to the topic is provided in: Gerothanassis, I. P.; Tsanaktsidis, C. G. Nuclear electric quadrupole relaxation. Concepts Magn. Reson. 1996, 8 (1), 63–74. DOI: 10.1002/(SICI)1099-0534(1996)8:1<63::AID-CMR5>3.0.CO;2-N.
The full Hamiltonian is provided in Appendix A.7 of Levitt, Spin Dynamics, 2nd ed. (Wiley).
Abragam (ref 5) writes that this is best understood in terms of "a fluctuating electric field gradient acting on the quadrupole moment of the nucleus", and refers to: Bloembergen, N.; Purcell, E. M.; Pound, R. V. Relaxation Effects in Nuclear Magnetic Resonance Absorption. Phys. Rev. 1948, 73 (7), 679–712. DOI: 10.1103/PhysRev.73.679. (Non-paywall version from Harvard available here.)
However, as far as I can tell, Bloembergen et al. simply describe the case for dipolar relaxation of spin-1/2 nuclei, and then write that "the interaction of the electric quadrupole moment of the deuteron with a fluctuating inhomogeneous electric field can bring about thermal relaxation. We omit the analysis of this process, which parallels closely the treatment of dipole-dipole interaction".
This equation is given without explanation in Gunther, NMR Spectroscopy, 3rd ed. (Wiley). An explanation can be found in Abragam, The Principles of Nuclear Magnetism (OUP), including the conditions under which this equation holds true. Quadrupolar relaxation is dealt with on p 313, but going through the earlier sections is necessary to understand it. (I didn't do that.)
Gillespie, R. J.; Schrobilgen, G. J. Hexafluorobromine(VII) cation, $\ce{BrF6+}$. Preparation of hexafluorobromine(1+) hexafluoroarsenate(1-) and hexafluorobromine(1+)-fluorodecafluorodinabinonate(1-) and characterization by fluorine-19 nuclear magnetic resonance and Raman spectroscopy. Inorg. Chem. 1974, 13 (5), 1230–1235. DOI: 10.1021/ic50135a043.


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How can nucleophilic substitution happen at the sp²-hybridised carbon of 4-chloropyridine?
Why can substitution of the C–Cl bond occur at the sp2-hybridized carbon in the following reaction (taken from Clayden et al., Organic Chemistry (2nd ed.), p 700)?
Ordinarily, SN1- and SN2-type reactions should not occur at sp2 centres, as described here: Why do SN1 and SN2 reaction not occur at sp2 centres?. What is the mechanism of this substitution?

The mechanism is indeed a nucleophilic substitution but it does not follow the bimolecular substitution paradigm ($\mathrm{S_N2}$) or the unimolecular one ($\mathrm{S_N1}$) that one learns about in basic organic chemistry classes — and it also does not follow the general electrophilic aromatic substitution that is also often presented in introductory courses.
$\mathrm{S_N2}$ can be ruled out because a rear-side attack would require an attack through the ring — a path which is blocked by the nitrogen atom (and the other bits of the ring).
$\mathrm{S_N1}$ is highly unlikely because we would need a phenyl cation — a highly unstable intermediate.
The aromatic ring of 4-chloropyridine can be understood as electron-deficient since pyridine’s nitrogen exercises a strong $-I$ and $-M$ effect. You can draw the $-M$ effect by pushing the electrons of the $\ce{C=N-C}$ double bond to nitrogen to give $\ce{\overset{+}{C}-\overset{-}{N}-C}$ — and this cation can be moved around the ring, most notably also to the 4-position where chlorine is located. See the figure below.

Figure 1: $-M$ effect of the pyridine nitrogen.
The nucleophilic attack is that of the amino nitrogen’s lone pair into this π system; the attack flips electrons so that the pyridine nitrogen has an additional lone pair, two $\ce{C=C}$ bonds are adjacent to this and a tetrahedral carbon bonded to both $\ce{Cl}$ and $\ce{\overset{+}{N}H2R}$ completes the ring.
This zwitterion can break down by reverse electron flipping which displace the chloride to give the skeleton of the reaction product. Additionally removing a proton from the amino nitrogen gives the neutral product. Considering that we have an acid (2-aminobenzoic acid) in the reaction mixture, I have drawn the pyridine nitrogen (the more basic one) to be protonated throughout the mechanism. This assists in the nucleophilic addition step by further reducing the ring’s electron density. See figure 2 for the mechanism.

Figure 2: Mechanism of the nucleophilic aromatic substitution reaction. Nitrogen being protonated due to the acidic effect of 2-aminobenzoic acid facilitates the reaction.
This mechanism is called nucleophilic aromatic substitution $\mathrm{S_NAr}$ and can only occur on very electron-deficient aromatic compounds: typically pyridines or pyrimidines with additional electron-withdrawing potential leaving groups such as chlorine.

Hi Jan, I was wondering if a N atom would exert an electron-withdrawing effect as strong as a nitro group (-NO2). Is that the case here? Also, how would we be able to deduce the strength of electron-withdrawing effect (-I or -M) of a given group?

@JonathanSmith The pyridine nitrogen is as strong or even stronger than the nitro nitrogen. We can’t really deduce the effects without chemical intuition.

@JonathanSmith: To follow-up on Jan, have one equivalent of anthranilic acid react with 1 equiv. of 4-chloropyridine and 1 equiv. of p-chloronitrobenzene. Analyze the rxn. mixture for remaining electrophiles or how much of each product is formed.

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text:
Products formed when solid CuCN is dissolved in aqueous KCN solution
Copper cyanide forms a complex coordination compound, when dissolved in a solution containing excess ($\ce{CN^-}$) ions.
My question is whether it forms $\ce{[Cu(CN)_4]^{3-}}$ only or both $\ce{[Cu(CN)_3]^{2-}}$ and $\ce{[Cu(CN)_4]^{3-}}$ complexes?
Wikipedia says that both complexes are formed, but I am not sure if its a reliable source.

No, both $\ce{[Cu(CN)3]^2-}$  and $\ce{[Cu(CN)4]^3-}$ species are formed. According to Wikipedia article of copper cyanide:

... but rapidly dissolves in solutions containing $\ce{CN−}$ to form
  $\ce{[Cu(CN)3]^2-}$ and $\ce{[Cu(CN)4]^3-}$, which exhibit trigonal
  planar and tetrahedral coordination geometry, respectively. These
  complexes contrast with those of silver and gold cyanides, which form
  $\ce{[M(CN)2]−}$ ions in solution.The coordination polymer
  $\ce{KCu(CN)2}$ contains $\ce{[Cu(CN)2]−}$ units, which link together
  forming helical anionic chains.

See this paper for more details.


text:
What atom is always paramagnetic and why?

Regardless of its electron configuration, it must always be paramagnetic when it's a single, neutrally charged atom:
  (A) Carbon (B)
  Nitrogen (C) Oxygen (D) Neon (E) Argon

At first, I eliminated D and E. Then, I tried to find the answer by checking the electron confiurations of the atoms; I saw the phrase "regardless of its electron configuration", but I didn't know what else I can do. Then, I found out that all three atoms (A, B, and C) have some unpaired electrons …
I checked charts with paramagnetic elements filled in; three first pictures on google said different things (check here), but after all the info I have found on the net I understood that C (oxygen) is the most likely answer. I don't understand why … I checked a few videos that said that if an atom has 1 or more unpaired electrons it is paramagnetic; then, A, B, and C are all paramagnetic. A few sites said we need to look not at the elements, but at the compounds (not oxygen, but $\ce{O2}$); the task said, anyway, we are looking at the single, neutrally charged atom.
Please tell if it is oxygen and if yes why it is paramagnetic but other elements are not.

Answer is nitrogen, because it has an odd number of electrons, and therefore there must be at least one unpaired electron. For carbon and oxygen which have an even number of electrons, it's always possible to pair them up somehow (not in the ground state, though; as your research indicates, all of C, N, and O are paramagnetic in the ground state). I don't like the question's phrasing, but that's what it's going for.

Neon and argon are correctly eliminated, no questions asked. Henceforth, we do indeed need to pay attention to question semantics and the phrase ‘regardless of its electronic configuration’.
The key here is that following Hund’s rule will only give you one possible electronic configuration. For example for carbon, Hund’s rule will give $\mathrm{1s^2\, 2s^2\,2p}_x{}^1\, \mathrm{2p}_y{}^1 2\mathrm{p}_z{}^0$ or any other combination of two of the three equivalent p orbitals half-populated. However, in early organic chemistry courses you typically rapidly learn about the hybridisation approach which changes that electronic configuration to $\mathrm{1s^2\, 2s^1\,2p}_x{}^1\, \mathrm{2p}_y{}^1 2\mathrm{p}_z{}^1$ — four instead of two unpaired electrons. Likewise, it is possible to imagine an electronic configuration as follows: $\mathrm{1s^2\, 2s^2\,2p}_x{}^2\, \mathrm{2p}_y{}^0 2\mathrm{p}_z{}^0$. In this case, all electrons are paired and the atom is diamagnetic.
Performing this type of analysis with all the atoms at your disposal — carbon, nitrogen and oxygen — you will be able to succeed in pairing all electrons for carbon and oxygen but not for nitrogen since it has five valence electrons. Five, being an odd number, cannot be mapped pairwise. Thus, a nitrogen atom is always paramagnetic.


text:
Why is "Spontaneous" a part of 2nd law wording?
The way I understand the law, any process is implausible if the total entropy of the universe does not increase. Why is "Spontaneous" even a part of the wording, what does it mean?
Is it perhaps a part of the probabilistic theory of thermodynamics, in which it is not "impossible" for heat to flow from cold to hot, simply very very unlikely?

Spontaneous means that the process doesn't need an external source of energy - like light and so - to occur. Also, it gives a direction to the universe "process" because there isn't any external source of energy, so the universe progress in the direction of increasing entropy. But maybe I am misunderstanding what you say..

Please provide the statement of the second law that you're considering---there are many formulations of the second law, and (at least to me) it's not clear which one you're referring to.

The phrasing i am referring to is: "In a spontaneous process, the entropy of the Universe increases." Taken from Keeler and Wothers' "Why Chemical reactions happen" page 8, although i am sure i have seen it otherwhere as well.

In chemical thermodynamics we are interested in the amount of work that can be done by a reaction as it proceeds towards equilibrium and whether a reaction will spontaneously proceed towards products or, if not, how much work is needed to bring it about. Spontaneous is used because experimentally reactions and processes occur without us having to do anything to bring them about, and this needed an explanation.
The second law can be stated as$^1$  'Spontaneous changes are those which, if carried out under the proper conditions can be made to do work. If carried out reversibly they yield the maximum amount of work. In natural processes the maximum amount of work is never obtained.' This is also a general definition of equilibrium. A shorter definition of the second law is 'heat does not spontaneously flow from from a cold to a hot body'.
If there are differences in intensive properties then a spontaneous change is always that which eliminates the difference. Thus if there is a (perfect) piston with different pressures on either side, it will move to equalise the pressure. Similarly if there is a temperature difference between two bodies in contact with one another then 'heat flows' from the warmer to the colder. Similarly, a concentration difference in a solution will equalise. However, this cannot be used to predict the direction of a chemical reaction in which there may already be constant pressure, temperature or concentrations.
In a purely mechanical system, the direction of spontaneous change seems obvious and is one in which the potential energy is lowered to a minimum. It was thought at one time that the same would apply to chemical systems, i.e that the direction of spontaneous change was one that minimises the heat content of the system. But experiment shows that this is not the case, there are, for example, spontaneous endothermic reactions$^2$, so that another criteria is needed to predict the direction of a spontaneous reaction.  This means understanding why for example, a gas expands to fill a container even though no loss of energy occurs on doing this. The missing idea is the entropy$^3$.
It used to be thought, before the second law was properly understood, that the maxiumum work a chemical reaction could do was to be found by converting all the heat of reaction into work. This was not bourne out by experiment and it is the free energy $\Delta G$ that gives the maxmum amount of work. Gibbs was the first to understood that the maximum amount of work is given by the first law unless some heat is taken in or given out to the surroundings. Thus the work done may be greater than the heat of reaction.
The second law requires that there be an increase in entropy during any actual process. Thus $dS_{sys} + dS_{surr} \ge 0$ for the system and surroundings. The sum of the two entropy production processes is always a positive quantity, but which can be zero in the case of a reversible process.
(Your speculation about statistical mechanics is partly correct, it is possible for entropy to 'run backwards' if only a very small number of molecules is considered and then only for a very short time. In 'normal' lab chemistry the number of molecules involved is vast say, $10^{18}$ so deviations from the average are minute.)
$^1 $ E. B. Smith, 'Basic Chemical Thermodynamics'.
$^2$ e.g. $\ce{Ba(OH)2(8H2O)_{(s)} + NH4SCN_{(s)} +heat \rightarrow Ba(SCN)2 + 2NH3 + 10H2O}$
$^3 $ The second law is needed because the first law offers too much latitude compared to experimental observations; by the first law heat could spontaneously flow from a cold to a hot body or a broken glass could spontaneously reform itself.

Refer to Wikipedia. Please note that your understanding (If dS/dt isn't positive...) is tautological. That is (as far as we know), our Universe requires dS/dt to be positive (or perhaps I should say we've never seen otherwise). The 2nd Law is most useful for nearly perfectly isolated systems and systems near (or, arguably, at) equilibrium. "Spontaneous" refers to time-forward processes in systems which are not perturbed (by outside events/forces). Our laws of quantum mechanics do not distinguish between positive and negative time. Supposedly, any event which can happen, can "un-happen" (i.e. reverse itself). This can be contrasted with macroscopic processes which are never perfectly reversible. This is one way that "the arrow of time" is seen. This problem remains unsolved and is recognized as one of the major fundamental problems in physics. We can't go backwards in time. So, we're limited to what experimental evidence we have when we talk about the "direction" of time.
You should also understand that the 2nd Law has no one single written form. It is expressed in different ways for different purposes. By this I mean that the word "spontaneous" need not be part of its expression.
See
https://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Various_statements_of_the_law
https://en.wikipedia.org/wiki/Arrow_of_time#Arrows
(also Entropy_(arrow_of_time) and Entropy#Definitions_and_descriptions)
and maybe https://en.wikipedia.org/wiki/Equipartition_theorem
and of course Statistical_mechanics


text:
Remove only salt from salt water
I have a bottle of salt water I collected on a local beach. I want to see what is in it, apart from water and salt. If I boil the whole bottle and wait for it to evaporate, mostly salt will remain. I want to see everything except salt that is still in the pot after I boil the water. Is this possible with common household items and commonly available (and safe) chemicals?
Disclaimer: I know nothing about chemistry.

How do you know there is anything other than water and salt in the first place?

The water is often pretty dirty, supposedly because criminal organizations dump waste at night. I'm curious to see if there is anything visible to the naked eye after evaporating a bunch of jugs filled with water.

Is there anything visible now, before any evaporating?

I'm getting ready to go to the beach right now! I'll report later.

Sea water contains lots of salts not just sodium chloride. Are you interested in the others? or do you just want to remove the sodium chloride? And you are equating whether household items can replicate what analytical chemical labs with lots of dedicated equipment: that could be a stretch.

Evaporating won't give you much information, especially if there are traces of organic pollutants, therefore water samples are analysed as is.  Elemental analysis, photometry and chromato-mass spectrometry would be ideal tools to pinpoint the exact composition, but you need an analytical lab to do it for you (some will accept water samples and do analysis for free, in return you allow them to use the data).

Before boiling the water, just let the water sample sit over an extended period of time in a clear container and see what settles to the bottom.

You could remove the bulk of the sodium chloride (and most other salts) by evaporation and crystallization. Most of the impurities you are looking for are going to stay in solution or suspension when you filter off the salts.
You could also use dialysis to remove the remaining salts: this requires a bag or length of thin-walled tubing made of a common polymer such as cellulose acetate. Tie off one end (the bottom), put the concentrate from the above step into the tubing, and tie off the top end. Then place it in a bucket of distilled water. The salts and other small molecules will diffuse out into the water, whereas the higher molecular weight materials like toxins will stay in the bag. You may need to change the water in the bucket a few times to get rid of all the salts.
You will still need some chemistry to figure out what substances remain in the bag. With seawater, it could be a very complex mix. However, looking at it with a fluorescent light might show something interesting. Good luck, and hope you get inspired to study chemistry!

Exactly what I wanted! :) I am inspired to study chemistry already, but right now I have a bit too much on my plate, especially with music software programming (I'm making my own DAW with most of my free time). Chemistry is one of the first things I want to learn when I have more time.

You could use ion exchange resin to remove the salt, I would use DOWEX-50 in the protonated form. I would wash it several times with water to remove the yellow water soluble impurity which is normally on brand new DOWEX-50- I would mix it with the water at the same time. I would then add to the water IRA-67 (weak base resin) in the free base form. I would have washed it first. This combination will remove the sodium chloride from water.
However if I wanted to see what was in the water, then this desalination treatment with resin would be counterproductive.
If I wanted to find out what organics were in the water, then I would extract with diethyl ether or ethyl acetate. The sodium chloride in the water will reduce the amount of the organic solvent which dissolves in the water. It will also sometimes increase the partioning of the organic solute into the organic phase. I would be then wanting to inject the organic extraxct into a gas chromatography machine after dilution with hexane and drying with sodium sulfate.
If I was to want to isolate metals from the water, then I would be looking to do solvent extraction. Depending on the metal I am interested in I would either use 30 % DEHPA in kerosene or 30 % aliquat 336 in toluene.
The DEHPA (Di- 2-ethylhexyl hydrogen phosphate) will extract many divalent metals from the water, depending on the pH it can extract calcium, strontium, barium and lanthanides. I would strip the metals using nitric acid.
The aliquat 336 is able to extract from sodium chloride solution things like copper, zinc, cadmium, iron and cobalt. I would want to strip (back extract) the metals from the aliquat phase into dilute nitric acid, then I would want to use ICPOES or AAS to measure them.
The general idea is that if you take a clean glass bottle and put in it 200 ml of the water sample, 20 ml of the organic phase. You then seal it and shake it. Then take a 15 ml sample of the organic phase to a smaller vial and add 5 ml of the strip solution. This has the potential to increase the concentration of the metals.
The problem is that if I was to use the DOWEX-50 resin on the orginal water sample then it would remove many of the metals which we want to measure using the solvent extraction pretreatment method.
In general if you want to measure things in salt water, I would try to devise methods of recovering the anaylte without needing to remove the common salt.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why does the keto tautomer of guanine have lower energy?

Looking at the keto and enol forms of guanine, it would seem as if the enol form is more stable, due to the fact that aromaticity is established with the enol form. However, according to a question on Chem.SE here: LINK there is a $\pu{8.7 kcal/mol}$ free energy difference between the enol and keto forms in aqueous solution.
Why is this? Shouldn't the enol form be much more stable due to aromatic stabilization?

Both structures are aromatic if you consider the keto tautomer’s mesomeric structure of $\ce{\overset{+}{C}-\overset{-}{O}}$.

The structure on the right is indeed a very tempting one, however, @jan is correct that both structures will be aromatic. If you look closer at possible resonance structures, it will become clear immediately.

Note that the only difference between these two structures is where the hydrogen is bonded. So, the question then becomes is there a 8.7 kcal/mol difference between the basicity of a nitrogen versus an oxygen?


text:
What chemicals are in anesthetic drugs
What chemicals are there in an anesthetic drug and how it reduces the pain of a person? Are the chemicals used in it safely, dangerous or both?

Asking what chemicals are used is fine, but in most countries they wouldn't be used if they were not safe when used properly. They could be unsafe if not used properly, but that is a far bigger question.

This is too broad - also the main component of the question (how they work) is not within the scope of chemistry.

There are so many different anesthetic drugs ranging from NSAID's, Opiates, and sedatives each group and even each drug working very differently.
For instance Opiates work in that they act as Neurotransmitters and attach to receptors that detect endogenous opiates (opiates created by the body, not sufficient to quell serious pain or to get high off of) in 3 parts of the brain, The Limbic System, The Brain Stem, and The Spinal Cord. The brain then releases dopamine causing the pleasure of opiates This blocking of neuro-receptors in the brain stem and spinal cord is rather complex and involves many levels of interaction including indirect affect by the stimulation of such hormones as serotonin and the complete blockage of some neural pathways.
However the function of such chemicals as lidocaine is very different as it simply blocks sodium from entering the nerve ending (only near site of injection -caine suffix means local anesthetic) thus blocking pain signals and nearly all other feelings from reaching the brain as that part of the nervous system is sort of frozen in place.
While Ibuprofen also works in the brain but on a much simpler count, by blocking the production of prostaglandins indirectly, (prostaglandins are released by the body and cause pain and swelling) Ibuprofen does this by limiting the enzyme Cyclooxygenase which converts arichidonic acid into Prostaglandins H2 which is then converted into Prostaglandins.
If you want something else you must be more specific.

One more: Hystorical narcotics such as chloroform or diethylether somehow disruption nerves communication by virtue of their lipofilicity, infiltrating membranes and sheaths. Not a detailed comment but gives an idea..


text:
How can I test fruits and vegetables for certain insecticides?
I live in Turkey and I don't feel safe eating the fruits and vegetables that I enjoy that are plentiful here.  After reading this article, I have been washing all fruits and vegetables with water and vinegar mixture, but I am not sure if this is sufficient.
As stated in said article fruits and vegetables exported to the EU (European Union) from Turkey have been returned to Turkey for having unacceptable levels of certain insecticides; particularly Chlorpyrifos (banned for causing hormonal problems in humans) and Methamidophos (hazardous carcinogenic chemical).
The article mentioned that the returned goods could not be traced after their return to Turkey and most likely were sold and consumed domestically.  Although Turkey has the same laws as the EU for banning and restricting these chemicals, these are obviously not enforced.
I need help in figuring out a practical method of testing for these chemicals.  In the EU, RASFF- the Rapid Alert System for Food and Feed is a key tool to ensure the flow of information across borders to swiftly react when risks to public health are detected in the food chain.  Perhaps the RASFF tools are practical and cost effective or maybe there is even a more practical detection technology or method.
I tried to find what specific tools and methods the RASFF uses to detect these pesticides but their website does not openly reveal these details.

I came across this article( https://www.insidescience.org/news/device-detect-pesticide-residue-produce-may-soon-be-available ).  If it can become available, ıt would allow the consumer to cost effectively test the produce before purchase.  No need to risk our health trustıng regulating agencıes or governments to ensure safe to consume food.  Databases of results can be self-posted online so consumers efficiently sort where to buy safe food.


text:
What solution is not in equilibrium?

An example of a solution not in equilibrium
  (A) Chemical ph
  indicator
  (B) Acid/base buffer
  (C) Anhydrous solution
  (D)
  hypotonic solution
  (E) Supersaturated solution

I  eliminated A and B. Then, I chose E, which is a right answer. Nonetheless, I read then that theoretically any solution or system can be in equilibrium. So, why is the answer E but not C or D?

Supersaturated solution just can't be in equilibrium, otherwise it would not be supersaturated.

@IvanNeretin Isn't there an equilibrium established between the solid crystals and the solution? This is a dynamic process, is it not? I am unsure myself. Please clarify.

The situation when there is an equilibrium between the solid crystals and the solution is called saturated solution. The "super" part tells us quite explicitly that we are in a different situation.

@IvanNeretin Sorry got confused


text:
Reaction of Rubidium with vacuum system viewport
I have a vacuum system ($<1\times10^{-7}$ mbar) filled with Rubidium ($^{87}$Rb) vapour, at room temperature.
We also have a viewport near the Rubidium source, made of a stainless steel flange and a glass substrate, connected with a Kovar alloy.
We are seeing potential degradation of the seal, which we do not see on other parts of the system. The only difference is the Nickel and Cobalt based Kovar alloy, not present anywhere else.
What are the reaction rates of Nickel and/or Cobalt with Rubidium?

Please, check the mark of the glass and if it is suited for work with rubidium in this particular settings. In general, glasses are oxides that can potentially react with alkali metals in right circumstances. I see no reason for them to react with nickel/cobalt alloy. However, rubidium might react with oxide layer on the surfaces of this alloy and/or any dirt left on the seal (and glass, and any dirt left on the glass) and any passing air if the seal is... not good enough.

btw, thanks to ability of rubidium to form suboxides even tiniest amount of oxygen may bind a lot of rubidium.


text:
Are metal oxides thermodynamically more stable then fluorides?
I came across a statement that $\ce{PbO}$ is more stable then $\ce{PbF2}$. Now how can i justify that staement because fluorine being more electronegative and should form  stronger bond with metals than oxygen.

You put $\ce{PbO}$ on a shelf and it will stay there forever without any change. Ditto for $\ce{PbF2}$. Now what does "more stable" really mean?

Stability obiously mean thermodynamic stability. PbO is thermodynamically more stable then PbF2

This was not obvious at all, so you might want to edit your question to that effect. Also, do you have any numerical evidence to support your claim?

I read in a book n even i couldnt justify tht statement hence m askng are oxides more stable then fluorides?

O and F have different properties as ligands and relations between values of thermodynamic functions of their compounds are by no means simple.

Can there be any simple explanation based on periodic trends and properties of elements to justify their relative  stabilities.

"$\ce{PbO}$ is thermodynamically more stable then $\ce{PbF2}$" I don't think this statement has any meaning. You need to specify thermodynamically stable relative to what? The constituent elements?

According to the formula for lattice energy, it is proportional to the product of the charges on the ions, and inversely proportional to the ionic separation (we shall approximate it to be the sum of the ionic radii).  In this case, the product of the charges is higher for PbO, and the sum of the ionic radii is similar between the two compounds.  This results in the ionic bond in PbO being stronger than in PbF₂.  Thus PbO is more thermodynamically stable.
Note: this may not be correct, my friend and I just thought of this explanation when discussing this question.


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Removing aluminum oxide before galvanic current and potential measurements
I have an aluminum sample stored about 1 month ago. I am going to couple it for galvanic current and potential measurements. I think the Al sample grew an oxide film due to exposure to air. Would the aluminum oxide film can affect the measurements? Should I remove it first? How?

Aluminium electrode is going to corrode anyway, depending on the current it may even be thick layers of $\ce{Al2O3}$/$\ce{Al(OH)3}$ that peel off over time as the electrode is consumed. If aqueous electrolyte is used, also expect hydrogen evolution and corresponding pH change which may affect measured potential. If these are undesired processes, you probably want to use alloyed aluminium (say, with tin).

sure it's going to affect the test so you have to remove the oxide layer
Generally, we adopt the method of alkali corrosion to remove the oxide layer on the surface of the aluminum.The purpose of alkali corrosion is removing the oxide layer on the surface of the aluminum. (Note: Technological process of the method is as follows: under normal temperature, 20% nitric acid; the process parameters is about 0.5 minutes. Pre-degrease--degrease--rinsing--rinsing--alkali corrosion--rinsing--rinsing--bright dipping--rinsing--rinsing--pure rinsing. The purpose of bright dipping is to remove surface residual black ash after alkali corrosion.Temperature:50-60 degrees; time:0.5-1 minutes.The purpose of degreasing is to remove grease on the surface of the aluminum.The process parameters is 0.5-1 minutes. You can use acid to wash it

Can I use a dilute solution of HCl that has the same ph with the nitric acid?

Can mechanically polishing it using abrasive paper remove the oxide layer?

I prefer using the first method of HCl not the mechanical way

What concentration would do?

Use a concentration of 3%


text:
How can we get the toxicity level of a plant without using animals?
We need to know the answer for our research paper. We will appreciate it so much if you will answer our question.

Use people. Seriously though, this site is for asking specific questions about concepts of chemistry. You will need to take the short [tour] and read through the [help] to get a better understanding of how the site works. A good question will have some semblance of an answer, with a question or two about where you are getting stuck. We can't start your research paper for you :) Anyway, welcome to Chemistry.SE and best of luck!

I'm voting to close this question as off-topic because it is asking us to help write the OP’s research paper.

Short answer
You can't!
Longer answer

Toxicity is not defined for a plant (or animal), but for specific compounds which are either produced or accumulated in these "toxic organisms". The concentration of toxic compounds may vary during the vegetation cycle and can depend, particularly in the case of accumulation, on the growth conditions.

Toxicity of particular compounds is typically given against specific endpoints, such as daphnids, rats, etc.

You asked for "the toxicity level". Which one do you mean?
$\mathrm{LD_{50}}$, $\mathrm{LC_{50}}$, NOAEL/NOAEC, NOEL/NOEC, DNEL,...

If you cannot measure in vivo for whatever reasons, use established QSAR models for specific endpoints.


text:
What is it that is "symmetry-adapted" about symmetry-adapted perturbation theory?
I think this is a fairly straightforward question which I should know the answer to, but surprisingly cannot find a discussion of online.
Symmetry-adapted perturbation theory (SAPT) is a fairly common method to describe intermolecular interactions using physically meaningful terms (electrostatics, exchange, induction, dispersion), where the interaction is built up. That is, SAPT calculates the interactions directly as opposed to finding the interaction strength by subtraction.
So, quite simply, in SAPT, what does the "symmetry-adapted" specifically mean, and why is it necessary to specify this?
In other words, is there a non-symmetry-adapted perturbation theory which can be used for intermolecular interactions?
Finally, how, if at all, does this symmetry-adaptation relate to the kind we use when constructing molecular orbitals using group theory?

The "symmetry" in symmetry-adapted perturbation theory refers to the anti-symmetry of the wave function with respect to electron exchange:
$$\Psi(\mathbf{r}_1,\mathbf{r}_2) = -\Psi(\mathbf{r}_2,\mathbf{r}_1) $$
The most usual way to enforce antisymmetry in electronic structure theory is to enforce orthogonality between spin-orbitals.  So SAPT is intermolecular perturbation theory that includes terms that depend on the overlap between orbitals in different molecules.
Intermolecular perturbation theory gives rise to concepts like eletrostatic interactions, polarization, and dispersion.  With SAPT you then get additional concepts like exchange repulsion (aka steric or Pauli repulsion).

Ah. That makes sense. It's kind of the only thing it could mean. Thanks for the concise clarification.


text:
Why doesn't NF₃ undergo hydrolysis?
I was reading Jan's answer to the following question-:
Why can NCl3 be hydrolyzed but CCl4 cannot?.
Since nitrogen is less electronegative than fluorine in $\ce{NF3}$, the electrophilic site is the Nitrogen atom. That means the water molecule should attack nitrogen atom in an $\mathrm{S_N2}$ manner. The stearic crowding around the atom is also less. Then why is the reaction not taking place? Is it only because fluoride ion is a poor leaving group due to its high basic strength?

I would assume that thermodynamic factors need consideration here too, NF3 is relatively stable and has a negative enthalpy of formation.

Bonds of other second period elements to fluorine are relatively strong.  So in addition to the other factors already mentioned, there is an inherent high activation energy.  Fluoroalkanes are not good substrates for nucleophilic substitution either, same reason.

Can anyone summarise everything into an answer with data and examples if possible?

@KartikeyaBadola They aren't related.

Nitrogen trifluoride is extremely inert to hydrolysis, attack by base or acids or oxidation because of the fact that there is no d-orbitals in Nitrogen trifluoride and fluorine is far more electronegative than nitrogen. However, in the presence of base or acids, nitrogen trifluoride can however undergo hydrolysis because there is now excess of nucleophile which can attack nitrogen trifluoride.
There is a old paper which discuss the hydrolysis of nitrogen trifluoride in presence of acids and base.

In the presence of aqueous base, however, slow hydrolysis occurs at
  100°C yielding nitrite and fluoride.
$$\ce{NF3 + 4 OH- ->[100℃] NO2-  + 3F- + 2H2O}$$

It further says:

This behavior differs sharply from that of nitrogen trichloride,
  which is known to give ammonia and hypochlorite under similar con-
  ditions. The latter products are readily explained in terms of
  nucleophilic attack directed at the chlorine atoms, a mechanism which
  appears reasonable in view of the fact that the electronegativities
  of $\ce{N}$ and $\ce{Cl}$ are very nearly the same and that the
  halogen may easily  expand its valence shell. Obviously, these
  considerations cannot be  applied to the nitrogen trifluoride molecule
  since fluorine is considerably more electronegative than nitrogen and
  it has no available  d orbitals. Although the nitrogen atom also has
  no free orbitals,  the relatively low electron density would at least
  offer less resistance to the approach of a nucleophile .

The proposed intermediate $\ce{HONF2}$ would be expected to be
  unstable with  respect to the loss of $\ce{HF}$, as is apparently the
  case with the unknown  perfluoro alcohols .
Two other reaction modes worth considering include the formation of
  an intermediate activated complex with water.

Find a paper that isn't outdated.

@Mithoron At least that paper tells us that $\ce{NF3}$ hydrolyses at $\pu{100 ^\circ C}$ in highly basic conditions. In reverse, that means hydrolysis is probably kinetically inhibited at room temperate. And it also states that it does indeed follow the expected $\mathrm{S_N2}$ mechanism with nucleophilic attack on nitrogen followed by displacement of the fluoride. (I reserve the right to be skeptical about the $\ce{HF}$ dissociation but it’s not entirely unlikely.)

@Mithoron Why? Did chemistry changed from the 40s ?

@permeakra Any explanation with "having d-orbitals" is outdated.

My book says NF3 undergoes hydrolysis through Sn1 mechanism under drastic conditions


text:
Why doesn't the N-Br bond dissociate quickly in NBS?
NBS is used to provide a constant, small amount of $\ce{Br2}$ which then dissociates to give $\ce{Br^.}$ free radical. Why doesn't the $\ce{N-Br}$ bond simply dissociate to give $\ce{Br}$? This dissociation should be just as fast or faster because the free radical formed would be stabilized by delocalization over the two oxygen atoms.

It does, and that's why it is used in the first place.

Actually, the sequence of reactions leading to a low but constant concentration of $\ce{Br^.}$ radicals is very simple:
$$\begin{align}\ce{R2N-Br &<<=> R2N^. + Br^.}\tag{1}\\[0.4em]
\ce{Br^. + Br-NR2 &<=> Br2 + R2N^.}\tag{2}\\[0.4em]
\ce{Br2 &<<=> 2 Br^.}\tag{3}\end{align}$$
The $\ce{N-Br}$ bond is very weak and can easily dissociate homolytically to give both bromine radicals and succinimide radicals. Bromine radicals can combine with other NBS molecules to form free bromine and another succinimide radical and bromine molecules can dissociate to give bromine radicals.
All of these steps are reversible and the equilibrium favours the non-radicals side. Whether your reaction needs a constant supply of bromine or a constant supply of radicals doesn’t matter as NBS can supply them.
Only if there is already some bromide in solution can a fourth reaction take place:
$$\ce{R2N-Br + Br- -> R2N- + Br2}\tag{4}$$
This is the only reaction that has a dedicated forward-direction. If bromide is present, bromine molecules can be generated much more rapidly than radicals, so the generation of radicals will have to come from bromine dissociation.
While it is true that the NBS radical is somewhat well stabilised, that does not make it that much easier for the $\ce{N-Br}$ bond to dissociate: radicals are still radicals and two-electron-two-centre bonds are still energetically favoured. The $\mathrm{S_N2}$ displacement is much better since an anion is equally well stabilised but has a  lower general energy.


text:
capacitive contribution voltammetry
I am working with a sheet of zinc metal where there is an oxide film on it (ZnO) in passivation conditions (pH=10). I have done voltammetry measurements for different scan rates with a 3 electrodes configuration, with zinc sheet as the working electrode. I would like to quantify the capacitive current. How should I process ? How to compensate the iR drop ? What is the crossing point in my curves ?
Thank you ]1


text:
Reactions of Grignard Reagents with amides
Can Grignard reagents add to amides? It appears from the synthetic pathway shown below that there has been some form of nucleophilic substitution at the carbonyl group.

Source: 49th IChO Preparatory Problems, Task 29

I don't see why not...There's even a sort of electron sink - the benzyl ester - for the nitrogen...

I'm a bit more puzzled why the amide is cleaved and not the carbamate...

Grignard can certainly react with tertiary amides (eg Weinreb amides) but I'm more than a little surprised than the ester isn't attacked first and that the chirality is retained. Is there any mention of  an additional reagent?

@Waylander looked it up (Org. Lett. 2004, 6, 1329) it's Grignard + TMEDA in THF

@Martin-マーチン the reference above also says that premixing TMEDA and the Grignard prior to reaction with the substrate minimises the extent to which cleavage of the Cbz group occurs


text:
What is the element with the greatest first ionization energy?

Identify the element with the greatest first ionization energy.
  (A) Ce
  (B) C
  (C) Cl
  (D) Ca
  (E) Cs

The book says the correct answer is (B) with such an explanation:

First ionization energy is a periodic trend that increases up and to
  the right on the periodic table. After looking at the position of
  these elements on the periodic table, carbon is clearly the best
  answer-choice (B).

However, a chart I found on Google Images (Ionization Energy Chart) says that Cl has the greatest first ionization energy. What is right then?

my special chart and web sites - please [edit] your post to cite your sources for this data

Find better textbook...

Your book must be wrong. NIST gives these values (in eV) as 11.260300 for C and 12.96763 for Cl. Even not reaching that website, I would say that a halogen would have a higher ionization potential than a group 14 element of a previous period just because there are no reason for the latter to not have a higher value, as for instance N has (half filled energy sublevel is relatively more stable and electron would leave this atom harder).
Source
Kramida, A., Ralchenko, Yu., Reader, J., and NIST ASD Team (2014). NIST Atomic Spectra Database (ver. 5.2), [Online]. Available: https://physics.nist.gov/asd [2017, August 23]. National Institute of Standards and Technology, Gaithersburg, MD.

Take a look at the first chart in https://en.wikipedia.org/wiki/Ionization_energy.
Note that the Ionization Energy is not a smooth function of atomic number, and note specifically that IE on each row (of the Periodic Table) is not monotonic (that is, it zigs and zags). The Periodic Table organizes the elements by atomic number and to some extent by orbital occupancy but I know of no property which is always correctly predicted using it. Its advocates say it shows trends. I don't dispute that but I don't find it useful because you can't trust it to give the right prediction anywhere close to "most" of the cases. Using the typical confidence level of 5%, for it to be a reasonably accurate tool, then it should predict rankings for Periodic trends (see:https://en.wikipedia.org/wiki/Periodic_trends) 95% of the time, or 19 times out of 20. You can see from the IE vs At.No. chart that in the first 20 elements it prediction is wrong 4 times When comparing adjacent elements. In other words, it's really not very good at these predictions, but better'n nuttin'.

A simple way to solve this question is to know the pattern of increasing and decreasing ionization energy on the periodic table. Ionization energy is the energy required to remove an electron from a gaseous atom or ion. Basically how it works is that ionization energy increases from bottom to top on a family (vertical) and increases from left to right on a period (horizontal).
Looking at the periodic table, we usually compare fluorine with other elements around it. Carbon is two boxes away while chlorine is one. Usually this is how you would solve ionization energy questions. However, there are some exceptions, and the vertical increase is different that horizontal increase.
Hope this somewhat helps!
Source: http://www.lenntech.com/periodic-chart-elements/ionization-energy.htm


text:
Why isn't cyclobutane planar?
In short, cyclobutane adopts a certain, non-planar, configuration in order to reduce ring strain caused by the fact that carbons are $\mathrm{sp^3}$ hybridised. That much is understandable.
In an exercise it was suggested to propose another kind of hybrisation for the carbons in order to justify a (theoretical) planar configuration. I thought that carbons could be $\mathrm{sp}$ hybridised. This way, $\mathrm{p}$ orbitals could form the $\ce{C-C}$ bonds and $\mathrm{sp}$ orbitals could form the $\ce{C-H}$ bonds.
I checked the answer and I was right. However, I couldn't think of something that could justify why this particular structure is not possible. That exactly is my question.

It's definitely possible. In my opinion, the more important question is: why is it less stable than what is actually seen?

The only problem I could think is the weaker C-C bonds. When they are formed form sp3 orbitals they are much stronger than when they're formed from p orbitals. Maybe that's the issue? Or maybe the position of hydrogens? I don't know I'm only making assumptions.

Why do you think cyclobutane has $sp^{3}$ hybrids?

@Zhe My testbook did not suggest otherwise. In order to explain ring strains, it compared the angles of tetrahedral strucutre with the ones presented in real cyclopropane, cyclobutane etc, and it seemed rational to me that that was the case. Is it not so?

So, I've seen cases where people have argued that cyclo_propane_ should be $sp^{2}$ hybridized. The same arguments might apply in this case.

I understand what you mean. You might be right. The textbook is for undergraduates so it most likely simpifies theories.

A word of warning: hybridisation is a theory to explain bonding, not too predict it. In this case, one should be extremely careful with hybridisation at all. The theory definitely breaks down here, and probably makes things a lot more complicated than necessary.

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

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

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

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

The gas-phase equilibrium conformation, as derived from electron diffraction and spectroscopic experiments, is nonplanar with a dihedral angle of about $35^\circ$ and an inversion barrier of only $\pu{6 kJ mol^-1}$. The puckering of the molecule is observed in Raman-spectra, and the corresponding absorption band
  is still present in the solid at $\pu{113 K}$. Theoretical calculations using ab initio methods also result in a bent conformation as the energetic minimum.
As one would expect from the shallow potential of the ring inversion, there are various planar and bent derivatives, and it does not seem possible to predict the ring conformation from the substitution pattern.

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

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

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

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

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

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

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

Bibliography

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

Interesting, your hybridization numbers would suggest that 1) the $\ce{C-H}$ bond in cyclobutane is somewhat more acidic than the $\ce{C-H}$ bond in cyclopropane and that 2) cyclobutane should give a positive bromine-water test as does cyclopropane.  Do you know if either of those two characteristics are actually observed with cyclobutane?

authors: Αντώνιος Κελεσίδης


text:
CO : CO₂ ratio when given amounts of carbon and oxygen react with each other

By the reaction of carbon and oxygen a mixture of $\ce{CO}$ and $\ce{CO2}$ is obtained. What is the composition of the mixture by mass obtained when 20 grams of $\ce{O2}$ reacts with 12 grams of carbon?

I wrote the equation as
$$\ce{4C + 3O2 -> 2CO + 2CO2}$$
and identified $\ce{O2}$ as the limiting reagent then the masses of $\ce{CO}$ formed will be 11.66 grams and mass of $\ce{CO2}$ formed will be 18.33 grams, which gives the ratio as 7:11, but the answer is given as 21:11.
Can somebody please tell me where am I going wrong?

What was your motivation for writing the equation that way?   That automatically assumes an (incorrect) answer.

First and foremost refer to the tip given in the answer of @ChesterMiller: use separate reactions for each given process, and don't rely on limiting reagent concept when more than one reaction is possible:
\begin{align}
\ce{\underset{$x$}{C} + 0.5 \underset{$0.5x$}{O2} &-> \underset{$x$}{CO}} \tag{1} \\
\ce{\underset{$y$}{C} + \underset{$y$}{O2} &-> \underset{$y$}{CO2}} \tag{2}
\end{align}
let $x$ mol be the amount of carbon participating in forming carbon monoxide $\ce{CO}$, and $y$ mol -- the amount of carbon participating in forming carbon dioxide $\ce{CO2}$.
We have the following amounts of reactants in the system:
$$n(\ce{C}) = \frac{m(\ce{C})}{M(\ce{C})} = \frac{\pu{12 g}}{\pu{12 g mol^-1}} = \pu{1 mol}$$
$$n(\ce{O2}) = \frac{m(\ce{O2})}{M(\ce{O2})} = \frac{\pu{20 g}}{\pu{32 g mol^-1}} = \pu{0.625 mol}$$
where $m$ -- mass; $M$ -- molecular weight.
Taking into account reactants ratios, the following system of equations is relevant:
\begin{align}
\begin{cases}
x + y &= 1 \qquad &\text{total amount of carbon}\\
0.5x + y &= 0.625 \qquad &\text{total amount of oxygen}
\end{cases}
\end{align}
Solving this system of equations, one gets $x = 0.75$ (mol) and $y = 0.25$ (mol). Hence the masses of the products are:
\begin{align}
m(\ce{CO}) &= x \cdot M(\ce{CO}) &= \pu{0.75 mol} \cdot \pu{28 g mol^-1} &= \pu{21 g} \\
m(\ce{CO2}) &= y \cdot M(\ce{CO2}) &= \pu{0.25 mol} \cdot \pu{44 g mol^-1} &= \pu{11 g}
\end{align}

Is this actually a plausible reaction?  Could we also get more $CO_2$ but some unburnt $C$?

@badjohn The way the problem is stated (conditions were such that only $\ce{CO}$ + $\ce{CO2}$ are presented in the system) suggests only this solution.

@andselisk Yes but it raised this related question in my mind.  Maybe, I should start a new question.

You should have written:
$$\ce{C + O2 -> CO2}$$
and
$$\ce{C + \frac{1}{2}O2 -> CO}$$
Then you should have determined what fraction of the carbon reacts by by the first reaction and what fraction of the carbon reacts by the second reaction.

How can I find that?

Let f be fraction of C that forms CO2 and let (1-f) be the fraction of C that forms CO.  Solve for f such that 20 grams of O2 are consumed.


text:
Why would red fuming nitric acid become white fuming nitric acid if left out at low temperature?
I'm reading this answer which refers to both red fuming nitric acid (RFNA) and white fuming nitric acid (WFNA).
Those two Wikipedia articles say:

Red fuming nitric acid (RFNA) is a storable oxidizer used as a rocket propellant. It consists of 84% nitric acid (HNO3), 13% dinitrogen tetroxide and 1–2% water. The color of red fuming nitric acid is due to the dinitrogen tetroxide, which breaks down partially to form nitrogen dioxide. The nitrogen dioxide dissolves until the liquid is saturated, and evaporates off into fumes with a suffocating odor. RFNA increases the flammability of combustible materials and is highly exothermic when reacting with water.
White fuming nitric acid (WFNA) is a storable liquid oxidizer used with kerosene and hydrazine rocket fuel. It consists of nearly pure nitric acid (HNO3). WFNA is commonly specified as containing no more than 2% water and less than 0.5% dissolved nitrogen dioxide or dinitrogen tetroxide.
WFNA as an oxidizer has somewhat less performance than red fuming nitric acid (RFNA) but is considerably safer (though extremely corrosive), as it has little to no dissolved nitrogen tetroxide, which is an extremely toxic and volatile chemical. If not inhibited, it will form nitrogen tetroxide on contact with most metals and some organic materials. WFNA can be converted from RFNA by simply leaving the RFNA out in low temperature for a couple of hours. (emphasis added)

I'm asking about that last sentence; why would red fuming nitric acid convert to white fuming nitric acid by leaving it out at low temperature?

The first two paragraphs you are quoting answer your question from the third one. By cooling down the solution with nitrogen dioxide, one shifts the equilibrium towards the formation of nitrogen tetroxide: $$\ce{\color{gray}{N2O4} <=>[^ $T$][v $T$] 2\color{brown}{NO2}}$$

@andselisk I'm having trouble understanding how RFNA is a "storable oxidizer" if the $\ce N_2 \ce O_4$ breaks down. Is it the partial pressure of $\ce N \ce O_2$ that keeps it stable, and letting the $\ce N \ce O_2$ escape leads to loss of $\ce N_2 \ce O_4$ by driving your reaction to the right, or is it the evaporation of the $\ce N_2 \ce O_4$ itself?

"Breaks down" makes it sound like a permanent change, but it's reversible. When stored, they reach an equilibrium based on temperature and stay there.

Ehhm I'm not sure we are on the same page. Both RFNA and WFNA are oxidizers not because of the nature of nitrogen oxides dissolved, but because both are mainly $\ce{H\overset{+5}{N}O3}$. If you purge the system, say, when distilling nitric acid, then yes, the equilibrium between $\ce{NO2}$ and $\ce{HNO3}$ will shift, and at some point you get rid of dissolved gas (no more brown fumes).

@andselisk the distinguishing feature between RFNA and WFNA seems to be the amount of dissolved $\ce N_2 \ce O_4$; ~13% and <0.5% respectively. When one "leav(es) the RFNA out in low temperature for a couple of hours" to make WFNA, what exactly is happening? This is the page that I and my question are on. If you can clearly explain what is happening, then please consider posting it as an answer. Thanks!

If my first comment (which I thought says its all) doesn't deliver, I'm afraid I fail to understand the question and therefore I cannot give a satisfactory answer.

@andselisk it may be sufficient to a trained chemist, I don't know. But I'm still learning. Posting an answer would give you more room to explain to me and future readers, and an opportunity for me to see how well your answer is received by others.

@andselisk: I see the same problem as uhoh. So the equilibrium shifts and... where do the substrates go, so that only nearly pure $HNO_3$ remains? You assume we know what is vented, what is dissolved, what condenses as liquid or evaporates, and/or which reactions take place and what conditions inhibit given transition. We don't. Do you need to store RFNA in heated rooms, etc?

You can convert RFNA to WFNA by removing dissolved nitrogen oxides ($\ce{N2O4}$ and $\ce{NO2}$) from the RFNA.
Given the volatility of $\ce{N2O4}$ and $\ce{NO2}$ it might be possible to just let the gases evaporate from the solution at low temperature. However, since the boiling point of $\ce{N2O4}$ if about $21.69~\pu{°C}$, I do not think that this is a very practical method.
Another path that is theoretically possible is the reaction of residual water with excess $\ce{N2O4}$ and $\ce{O2}$ from air to $\ce{HNO3}$.
$$\ce{2 H2O + 2 N2O4 + O2 -> 4 HNO3}$$
Again, to me this seems not to be a very plausible way. Since the Wikipedia article does not give a source for its claim, we cannot decide which of the two paths, if at all, is the correct one.

Why does a room-temperature boiling point (21.69ºC for N2O4 and 21.2º C for NO2) make it impractical to "just let the gases evaporate"? It would seem the opposite to me. Nitric acid's boiling point is some 60 degrees higher.

I would assume that significant amounts of $\ce{N2O4}$ and $\ce{NO2}$ would stay in solution. Maybe bubbling nitrogen through it would help to evaporate, but this was not part of the question.

Thank you for taking time to post an answer! I understand there will be a balance between $\ce N \ce O_2$ and $\ce N_2 \ce O_4$, but I'm wondering; does Henry's law tell us that if the partial pressure of both gases drops to zero above the liquid's surface, then the gases will continue to evaporate until their concentrations in solution also drops to zero?

I'm still not sure of the answer to "Why would red fuming nitric acid become white fuming nitric acid if left out at low temperature?" Is it possible to add a "Because..." sentence? Thanks!

Nitric acid produces red fumes when its concentration is above 80%, and white fumes when it is above 90%. In the tutorial below, which shows how to make 95% nitric acid, a test at the beginning of the video shows the acid fuming red. This acid initially had a concentration of 95%, but as it is unstable, it decomposes with time, producing water which dilutes it. At the end of the video, we can see that the freshly produced acid make white fumes, which guaranties that the concentration is at least 90%.
How to make 95% nitric acid from household items

RFNA contains up to 20% of dissolved NO2 and less than 1% of water. Used mostly as rocket propellant oxidizer.

$\ce{NO2}$ and $\ce{N2O4}$ mixture is an equilibrium mixture which depends on temperature. At low temp $\ce{NO2}$ dimerises more, so equilibrium will be more shifted to $\ce{N2O4}$ (colourless).


text:
Why does dichloridobisethylenediaminecobalt(III) have only three optical isomers?

Sorry for the bad drawing and having $\ce{NH}$ instead of $\ce{NH2}$
In the first structure, the chlorine atoms are in trans position on the plane.
In the second structure, the chlorine atoms are in cis position on the plane.
In the third structure, one chlorine is in the plane and the other chlorine is occupying one axial position.
Now all three of these will have one enantiomer making the number of isomers in total $6$.
So $\ce{[Co(en)2Cl2]+}$ should have $6$ optical isomers, but the answer in the book says that $\ce{[Co(en)2Cl2]+}$ has $3$ optical isomers. How?

Compound 1 has a plane of symmetry and does not have enantiomers. Compounds 2 and 3 are enantiomers of each other.

@Zhe In 2 both chlorine is on the plane and in 3, one chlorine is at axial position. So how 2 and 3 are enantiomers of each other ? I thought bonds don't break in optical isomerism.

(1) in octahedral geometry there is no inherent difference between "equatorial" and "axial" sites.  Simply by 90° rotation of the molecule you can interconvert "equatorial" and "axial" sites. So your argument for why the second and third are not enantiomers doesn't work. (2) why don't you try mirroring the second structure and seeing whether it is the same as the third? You may have been taught to do that with organic compounds.

I suggest building a model...

(3) the first structure is not an optical isomer of the other two. So it is incorrect to say that there are three optical isomers of [Co(en)2Cl2]+. (4) the first structure is not chiral and hence it does not have an enantiomer. (5) All in all, it seems you need to take it slowly and go back to basics (check every mirror image to see whether it is superimposable) instead of coming up with shortcuts of your own for determining chirality, since these shortcuts aren't giving you the correct answer.

@orthocresol (1) : I thought that but I was not sure. (2) : I tried doing that but as I told I did not (not sure) know that there are no axial or equitorial bonds in octahedral. (3),(4) : I think I am missing something but is chiral centre even possible with two same ligands. At least in Organic molecules all the functional groups attached to the chiral carbon should be different, right ?

That only applies to tetrahedral centers. Since this is an octahedral complex, that does not apply.

In simplistic terms, we can see $\ce{[Co(en)2Cl2]^+}$ as an $\ce{[MA4B2]}$ pseudooctahedral complex. We can initially consider the $\ce{A\bond{->}M}$ and $\ce{B\bond{->}M}$ bonds identical. Trying this out on a model octahedron or molecular modelling kit will show you that only two possible different configurations exist: cis and trans.
Here, however, we have one of the ligand types actually being a briding ligand. Because the bridge is too short to allow for a trans configuration, the two coordination sites of these bridging ligands will always be cis to each other.
For the complex in which the two chlorido ligands are trans to each other, only one possible arrangement of the $\ce{en}$ ligands is possible. At first approximation, the entire complex should have $C_\mathrm{2v}$ symmetry if not more: a $C_2$ axis with two perpendicular $\sigma_\mathrm{v}$ planes; one bisecting the $\ce{en}$ ligands through the $\ce{C-C}$ bond, the other being orthogonal and mapping each $\ce{en}$ onto the other. Due to the presence of planes of symmetry, this complex cannot be chiral.
In the complex in which the two chlorido ligands are cis to each other, one could think of up to three potential arrangements for the bridges; however, as I pointed out, it is not possible to have the two coordinating atoms of one $\ce{en}$ ligand be trans to each other, reducing the total down to two. These two possibilities do not differ in whether one of the chlorides is in an ‘axial’ position (remember, we are assuming equal bond lengths initially so there is no defined ‘axial’ position). Rather, in one case you have a leftwards-turning screw and in the other case a rightwards-turning screw if you assume the $\ce{en}$ ligand to be a plane and wanted to screw the entire thing into your paper.
Neither of these two structures have a plane of symmetry (or higher symmetry) and both are therefore optically active. Careful analysis will reveal, that they are in fact mirror images and thus enantiomers of one another. They are called the Δ and Λ isomers, respectively. All three isomers are attached in the image below.

Figure 1: isomers of $\ce{[Co(en)2Cl2]^+}$. From left to right: trans, cis-Λ, cis-Δ.
Thus, there are three isomers of the compound, two of which are optical isomers of each other (the third is a geometrical isomer).
Finally on the topic of whether axial bonds are actually longer, i.e. whether there is a Jahn-Teller type distortion or not. The answer is no. $\ce{Co^{III}}$ is a $\mathrm{d^6}$ complex. It could be assumed to be high or low spin. In case it accidentally ends up to be low spin, the $\mathrm{t_{2g}}$ orbitals are fully populated while the $\mathrm{e_g}$ orbitals are entirely unpopulated; there is no need for Jahn-Teller distortion.
In the potentially more likely high-spin case, one of the $\mathrm{t_{2g}}$ orbitals is fully populated while all other orbitals are half-populated. We might expect a Jahn-Teller distortion to remove the degeneracy of the $\mathrm{t_{2g}}$ orbitals. However, this distortion would be minor at best: the $\mathrm{t_{2g}}$ orbitals do not interact well with the ligands since they are oriented to point in-between the ligand orbitals. (On a group theory basis, the $\mathrm{t_{2g}}$ orbitals are nonbonding.) Therefore, any symmetry reducing effect will be very minor at best and not lead to a noticeable distortion. Thus, it is reasonable to assume near-identical bond lengths and no principal (‘axial’) direction.