Understanding energy level splitting of an octahedral complex from spin-orbit coupling, 1st and 2nd order zeeman effects

To answer part b it is necessary to consider the symmetry properties in an octahedral field. The spin orbit Hamiltonian is $H_i=\zeta\sum_i\bar l_i\cdot \bar s_i$ where $\zeta$ is a constant characteristic of the radial part of the d orbitals. The orbital angular momentum, which may be written in the form (for the $x$ component) as $\displaystyle l_x =-i\hbar \left(y\frac{\partial }{\partial z}-z\frac{\partial}{\partial y}\right)$ (and similarly for $l_y$ and $l_x$ by rotating subscripts) transforms as does the $T_{1g}$ irreducible representation in an $O_h$ field.

If a term has an orbital symmetry $X$ it will only be susceptible to spin orbit coupling if the symmetry product $X\cdot T_{1g}\cdot X$ contains the totally symmetric representation $A_{1g}$. In other notation $\langle i|H |j\rangle \ne 0$ or $\int \psi_i H \psi_f d\tau \ne 0$). Thus the $E_g$ levels are unaffected in point group $O_h$ because $E_g \cdot T_{1g}\cdot E_g$ does not contain $A_{1g}$ but the $T_{2g}$ interact ($T_{2g} \cdot T_{1g}\cdot T_{2g}$ does  contain $A_{1g}$).

Notes: It is assumed that the potential energy effect of the ligands is only small compared to that of the central field of the ion, thus the spin-orbit coupling is a small perturbation on the energies.

The full derivation is complex and involves using double symmetry groups.

The free ions have SO couplings $^2\mathrm D_{3/2} =-3\zeta_d/2$ and $^2\mathrm D_{5/2} =\zeta_d$. (The formula you use if for the free ion).  In the octahedral field the calculation is rather complicated but the result is that $T_{2g}$ levels are split by $3\zeta/2$ in the ratio you show.

In the remaining parts of your question similar symmetry arguments should indicate what levels are split, but only a detailed calculation will produce the energies that they are split by.


Van der Waals nonspecific interaction definition

The definition of the van der Waals force that I like (and apparently so too do the folks who have contributed to the Wikipedia article on the topic) is much more *specific about what is and is not a van der Waals force. This definition originates in the International Union of Pure & Applied Chemistry Compendium of Chemical Terminology

The van der Waals force is the sum of the attractive or repulsive forces between molecules (or between parts of the same molecule) other than those due to covalent bonds, the hydrogen bonds, or the electrostatic interaction of ions with one another or with neutral molecules or charged molecules.

Thus, a van der Waals force is a catch-all term to include all other very weak interactions between atoms and molecules:

Dipole-dipole interactions between permanent dipoles
Dipole-induced dipole interactions
Induced dipole - induced dipole interactions (London dispersion)
$\pi-\pi$ stacking
Halogen bonding
Repulsive forces derived from atoms trying to be inside another atom's van der Waals radius (Steric strain and torsional strain)


The exact nomenclature of alkyllithiums is an inorganic problem (cf. 2005 Red Book) and in there one finds as an example (IR-10.3.2, p 229; repeated in P-69.3, p 1043 of the 2013 Blue Book)

$\ce{LiMe}$ lithium methanide (compositional name)
$\ce{[LiMe]}$ methyllithium
$\ce{[(LiMe)4]}$ tetra-µ3-methyl-tetralithium
$\ce{(LiMe)_n}$ poly(methyllithium)

so, the name you choose will depend on what structural features you want to emphasise. Note that the Red Book does not prescribe "preferred" IUPAC names, and the Blue Book explicitly refuses to assign PINs (p 1036):

Neither preferred IUPAC names or preselected names (see P-12.2) for organometallic compounds involving the transition elements (including the Group 3 elements) and Groups 1 and 2 elements, excpet for 'ocene' compounds, are noted. This determination will await consideration by a task group on inorganic and/or organometallic nomenclature.

Obviously, from a practical point of view, the name "methyllithium" is more commonly used and more easily recognised. In line with this, we will therefore look for a name of the form "_yllithium". The use of "methyl" instead of "methanide" is permissible because (p 1037, 2013 Blue Book)

Because of a long established tradition, names for substituent groups as defined in Section P-29 can also be used as ligand names.

and hence the question becomes: what is the preferred prefix for the following group?

This is a purely organic question, so can be answered with the 2013 Blue Book. The relevant rules are as follows:

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'.
P-57.1.5.1 Preferred prefixes derived from cycloalkanes are of the 'cycloalkyl type' (see P-29.2); preferred prefixes derived from cyclic compounds other than cycloalkanes are all of the 'alkanyl type'.

The preferred IUPAC name of the parent hydride, according to Hantzsch–Widman nomenclature (P-22.2.2.1), is 1,3-dithiane. Correspondingly, the preferred prefix is 1,3-dithian-2-yl:

Finally, returning to the Red Book:

IR-2.2.3.2 Parentheses are used in names in the following ways:
(a) To enclose substituent group or ligand names in order to avoid ambiguity, for example if the substituent group or ligand names contain multiplicative prefixes, such as (dioxido) or (triphosphato), or if substitution patterns would otherwise not be unambiguously specified, or if the substituent group or ligand name contains numerical or letter descriptors [...]

we can conclude that a suitable name would be (1,3-dithian-2-yl)lithium.


Why do most explosives have symmetrical chemical structures?

The main reason to synthesize symmetric molecules are related to their synthesis:

Some symmetry is due to the nature of the molecular structure itself, like RDX and HMX. This is related to tricyclic structure of hexamethylenetetramine. Similarly, for Tri/tetramers of peroxyacetone. For these structures, they are created rather spontaneously (HMX with some help) than by directed multistep synthesis.
Some symmetry is due to reaction thermodynamics and kinetics, that does not prefer asymmetrical structures. E.g. TNT is produced easily by direct toluene nitration, as methyl and nitro groups direct coming groups to 2,4,6 positions, forming the symmetric molecule. An asymmetric 2,3,4-TNT is not created this way, its synthesis would be much more complicated, expensive and product may not be safe to produce.

Many asymmetric molecules are not stable enough to be synthesized, or are not chemically stable enough to be kept.

For the keeping stability, Picric acid and its salts are famous for their disintegration products making the explosive oversensitive.

Some symmetry is due low control over the reaction quotient, combined with the advantage of reaction undergoing all steps. See nitroglycerine being trinitrate ester, PETN being tetranitrate ester.

If they are symmetrical, their instability is evenly distributed over the molecule, so they are relatively stable, compared to eventual asymmetric variant, considering their explosive potential.
That allows their synthesis, safe and cheap production, safe manipulation, safe aging.

If they were asymmetrical, their instability would be localized and concentrated to some molecule locations and the molecule would be much less stable.

Additionally, at big asymmetry of bigger molecules, there is question, why to keep the "non-explosive molecule part"?

Consider 2,3,4,5,6-pentanitrodiphenyl versus 2,3,4,5,6-pentanitrotoluene.

Another example of an asymmetrical unstable explosive is the renowned $\ce{NI3 . NH3}$

Summary

I would summarize the answer as kinetic and thermodynamic instability, together with synthesis complexity, discriminates existence and production of asymmetric explosives, comparing to much broader group of non-explosive compounds.

Additionally, the symmetry origin is often coined in symmetry of molecules chosen to be converted to explosive, or to be the structural basis for explosive.

As their symmetry makes promise to form an efficient explosive. E.g. pentaerythritol is perfectly symmetrical alkohol with 4 OH groups to be nitrated.


Calculating stoichiometry of a reaction between xenon and fluorine

Xenon and fluorine were mixed in (1) a 1:1 ratio by volume and sealed in a vessel attached to a barometer. After some time, crystals were found to have formed on the walls of the vessel and the barometer showed that (2) the pressure had dropped to 70% of the original pressure. Analysis of the gases remaining indicated (3) 4 parts xenon to 3 parts fluorine. Construct a balanced equation for the reaction. Assume constant temperature throughout.

You have properly analyzed (1) and (2). It remains to analyze (3). We started with ten parts of gas, 5 parts xenon and 5 parts fluorine. From (2) and (3), we see that 4 parts xenon and 3 parts fluorine are left in the gas phase. Hence 1 part xenon ($\ce {Xe}$) and 2 parts fluorine ($\ce {F2}$) must have reacted to form crystalline $\ce{XeF4}$.

The balanced reaction is left as a simple exercise.

The choice of starting with ten parts of gas (as opposed to fifty parts, or a hundred parts) is arbitrary. Since we're given only ratios, we can choose an arbitrary number of parts of gas and work with that throughout. Choosing ten makes the math simple.


Number of diastereomer pairs of 1,3-dichloro-1,2,3-triphenylpropane

I basically agree with Ron's answer, but had to draw all of the possible structures to confirm it. The complication is that carbon-2 is non-stereogenic but it may be chirotopic depending on the configuration of the neighboring atoms. The IUPAC Gold Book calls this a pseudo-asymmetric carbon atom. Carbon-2 is not a stereocenter, because it does not have four different "things" attached to it. However, its configuration is important relative to carbons-1 and -3. Below are all of the possible combinations.

Out of the eight possibilities, we see there are only four unique compounds (names generated in ChemDraw):

A: (1R,2s,3S)-1,3-dichloro-1,2,3-triphenylpropane

B: (1S,3S)-1,3-dichloro-1,2,3-triphenylpropane

C: (1R,2r,3S)-1,3-dichloro-1,2,3-triphenylpropane

F: (1R,3R)-1,3-dichloro-1,2,3-triphenylpropane

The small r and s are essentially used to indicate the stereochemistry at carbon-2 relative to carbons-1 and -3. Note that in compounds B and F, the stereochemistry at carbon-2 is inconsequential as in both cases the phenyl group is syn to one chloride and anti to another chloride. I haven't been able to find an adequate explanation of assigning the r/s designations, but I believe that the R configured substituent receives higher priority than the S configured substituent, and Cahn–Ingold–Prelog system is followed from there.


Pale blue species forming during electrolysis of NaHCO3

The blue color was caused by copper being stripped from the wire. btw, when your hair turns orange, it is because it is being stripped of color, among other other things (chemical bonds), which will cause it to be brittle and also absorb any hair dye to the extreme that it feels waxy...I'd suggest you cut off a piece of hair for a test before dunking your whole head in bleach (chlorine will turn it greenish).

P.S. Use nitric acid to dissolve Cu.


Determining solubility of a gas in a molten salt

If you know the phase diagram of the gas, and can calculate the exact pressure the gas should have given the volume outside of the molten salt, you can use the measured pressure/temperature to determine any difference in pressure. Then you can calculate the number of moles of gas that must be absorbed to generate the pressure difference.
If the gas is relatively inert (which it probably needs to be if it wont react with molten salt), you can probably use the ideal gas equation:
$$P_0V = NRT$$
If you compare to the actual pressure, you can determine the number of moles absorbed by the salt:
$$\delta N = (P_0-P_\mathrm{exp})V/(RT)$$


How to characterize gaseous ethylene that forms as byproduct?

If you want something simple and don't mind less than 100% certainty then you could bubble the gas through Bromine Water - in the presence of an alkene the bromine water would be decolourised from brown when reacting with the ethene.  Of course this does just indicate the presence of any alkene so isn't exactly specific! There is more about this on this website.


Is using melting/boiling point good to measure volatility?

According to this Wikipedia page:

  In chemistry and physics, volatility is the tendency of a substance to
  vaporize. Volatility is directly related to a substance's vapor
  pressure. At a given temperature, a substance with higher vapor
  pressure vaporizes more readily than a substance with a lower vapor
  pressure.

So, I'm not sure how much information on volatility you will get from melting point measurements.  Ideally, the enthalpy of vaporization would be the best measurement, but, if it fits your needs, the boiling point at ambient pressure is a simple way to determine whether there is a significant difference in the volatility of various compounds.


How is the specific heat of an alloy obtained?

The enthalpies and calorific capacities of certain mixtures are tabulated in standard references. If you do not have such data, you can use the following approximation:

  Rule 1: For a mixture of gases or liquids, calculate the total change in enthalpy as the sum of its changes in the pure components of the mixture. In fact, the enthalpy changes associated with component mixing are neglected, which is an excellent approximation for gas mixtures and similar liquid mixtures such as pentane and hexane, but is bad for different liquids, such as nitric acid and Water.

  Rule 2: For highly diluted solutions of solids or gases in liquids, neglect the change in enthalpy of the solute. This approach is better, as the solution is more diluted.

  The calculation of the enthalpy changes for heating or cooling a known composition mixture can often be simplified by estimating the calorific capacity of the mixture as follows:
  \begin{equation}
    C_{p,m}(T) = \sum_i y_i\cdot C_{p,i}(T)
\end{equation}

  Where:

  $C_{p,m}(T)$ Calorific capacity of the mixture

  $y_i$ Mass or mole fraction of the i-th component

  $C_{p,i}(T)$ Calorific capacity of the i-th component

Reference:

Richard Mark Felder and Ronald W. Rousseau. Elementary Principles of Chemical Processes. John Wiley & Sons Inc.; 3rd Revised edition. 702 pages.


How to explain the increase in current with the increase in the concentration of responsible ions?

The electrical current is carried by all ions in the solution, including the $\ce{SO_4^{2-}}$/ $\ce{HSO_4^-}$, $\ce{NO_3^-}$, and $\ce{PO_4^{3-}}$/ $\ce{HPO_4^{2-}}$/ $\ce{H_2PO_4^-}$ ions, and not just by $\ce{H+}$ and $\ce{OH-}$.  So, even though the $\ce{OH-}$ concentration is decreasing with increasing acid concentration, like you say, it's much more than made up for by the increased concentration of the anions from the acids.

EDIT (in response to a comment): The contribution of ions that don't participate in the electrochemistry differs depending on the system.  For example, in $\ce{NaOH}$ the $\ce{Na+}$ effectively just distributes to balance the charge from the concentration gradient in $\ce{OH-}$:

$\ce{H2O}$ is reduced to $\ce{H2}$ and $\ce{OH-}$ at the cathode, increasing the $\ce{OH-}$ concentration there
$\ce{OH-}$ is oxidized to $\ce{O2}$ and $\ce{H+}$ at the anode, decreasing the $\ce{OH-}$ concentration there
Thus, $[\ce{OH-}]$ changes from high at the cathode to low at the anode
In order for the solution to maintain electroneutrality, something has to balance the higher concentration of negative charge at the cathode, and in this system $\ce{Na+}$ is the major contributor.  (Water self-ionization can't provide enough $\ce{H+}$ to maintain the charge balance.)

In other systems such as $\ce{H3PO4}$, the acid-base equilibration behavior of an electrochemically inert species can contribute to the charge carrying function:

Adjusted to pH 7 with, say, NaOH, phosphoric acid exists mainly as $\ce{H2PO4-}$ and $\ce{HPO4^{2-}}$
Water is oxidized at the anode, producing $\ce{H+}$, which reacts with $\ce{HPO4^{2-}}$ to form $\ce{H2PO4-}$
Water is reduced at the cathode, producing $\ce{OH-}$, which reacts with $\ce{H2PO4^{-}}$ to form $\ce{HPO4^{2-}}$ and $\ce{H2O}$
So, $\ce{HPO4^{2-}}$ has a higher concentration around the cathode than the anode, and $\ce{H2PO4^{-}}$ has a higher concentration around the anode than the cathode.  These concentration differences result in a net transfer of charge between the two electrodes.
All of the negatively charged phosphate ions are balanced out by a corresponding distribution of positively-charged $\ce{Na+}$, similar to the $\ce{NaOH}$ case.


Required temperature for the reduction of iron oxide using carbon

The direct reduction with carbon is only possible if the iron oxide ($\ce{Fe2O3}$, $\ce{Fe3O4}$, or $\ce{FeO}$) is present as a liquid.

$$\ce{FeO$_n$ + $n$C -> Fe + $n$CO}$$

Accordingly, the required temperature is very high. Note the melting points of $\ce{Fe2O3}$ (1539 °C), $\ce{Fe3O4}$ (1597 °C), $\ce{FeO}$ (1377 °C), pure $\ce{Fe}$ (1538 °C), and the eutectic of iron containing 4.3 % carbon (about 1150 °C).

In the blast furnace process, solid iron oxides at lower temperatures are mainly reduced with carbon monoxide.

$$\ce{FeO$_n$ + $n$CO -> Fe + $n$CO2}$$

The required carbon monoxide is produced from the exothermic reaction of carbon with oxygen

$$\ce{2C + O2 -> 2CO}$$

and from the endothermic Boudouard rection of carbon with carbon dioxide.

$$\ce{C + CO2 <=> 2CO}$$

A temperature well above 700 °C is required to favour the formation of carbon monoxide. Hence, a part of the carbon has to be burned in order to generate the required heat and to produce carbon monoxide.

Hot fine-dispersed iron would be quickly oxidized in air (for example, steel wool can burn in air) whereas fused iron is only oxidized at the surface of the liquid. Note that the pig iron produced in the blast furnace process has a high carbon content, which significantly lowers the melting point.


How does the cyclisation of the 3,6-dimethylocta-2,6-dien-4-yne occur?

First add water to the triple bond a ketone would be formed as the molecule is symmetrical hydroxy group can be attached yo any of the two carbons after that it would undergo tautomerism to form a ketone. Now the second step cyclisation occurs as follows:


I think the key concepts that are explainable at this age level are:

Atoms are made of protons, neutrons, and electrons
Protons have positive charge, electrons have negative charge, neutrons have no charge
Opposite charges attract
Atoms can gain or lose electrons, but not protons or neutrons
Atoms hold onto their electrons more or less tightly depending on which atom it is (you won't be able to really explain why, but you could just show them a periodic table and explain that non-metals (the right side) like electrons more than metals (the left side))
Atoms can "steal" electrons from each other
Electrons can be "knocked off" of atoms by high-energy photons (light or EM radiation)

The difficulty with explaining these things by analogy to regular-life experiences is that there is no regular-life experience that captures the whole picture. This means you are limited to activity-type demonstrations, videos/pictures, or analogies for specific pieces of the puzzle.

I would recommend a mixture of all three approaches. Kids like hands-on stuff, so maybe you could explain the basic concepts with very simple analogies, then do an activity where the kids are given "electrons" (tennis balls, maybe) and are assigned to be metals or non-metals. Then you could have them give each other electrons and count charges as they go. If you had containers that could hold 8 electrons, you could even represent valence shells and the octet rule - metals could start with two or three electrons, non-metals with 5, 6, or 7, and the metals would need to empty their shells while the non-metals would need to fill them.

This would demonstrate a very important concept that is hard to teach in the abstract: Electrons don't disappear.

Since you are also interested in teaching about ionization by radiation, you could assign each student a number representing their ability to "hold on" to an electron. Here is where you might be able to use a shopping-type analogy - only instead of students giving up electrons as money, they would give them up in exchange for money. In this case, the money represents the energy, and you would be giving them the energy (money) needed to remove (sell) an electron. Atoms that charged a higher "price" (require more energy) would wind up with more money at the end of the exercise.

There are many variations on these ideas that you could use, but I think anything that covered the seven points I mentioned above while simultaneously developing an intuitive understanding that energy and electrons are conserved could be called successful.


There is a considerable amount of debate as to whether d-orbitals of main group elements (I assume we’re talking about them here; the discussion is rather pointless for d-block elements) are used in hybridisation or not.

Originally, the octet expansion, a.k.a. using them, was postulated by Linus Pauling because he was unhappy to use charge separation to describe molecules such as sulphur trioxide. However, it is entirely possible to draw Lewis structures that do not violate the octet rule and hence do not require d-orbital hybridisation for practically every case. There may be more mesomery required in some cases and less in others.

I’m not too good at repeating the arguments of supporters of main group elements’ d-orbital hybridisation theory, because I belong to the other group. The main argument against d-orbital participation is, as you noticed, that they are far removed on an energy scale; even behind the next s-orbital (3s, 3p, 4s, 3d) — hence the order of elements in the periodic table. I also have troubles understanding why phosphorous would only allow one d-orbital to hybridise, sulphur two and chlorine and the heavier halogens three. This makes no sense to me. (It seems to make sense for others, though.)

In my opinion you are absolutely correct in asking that if a 3p-element can use 3d orbitals, why should it not use 4s for hybridisation, and consequently why 2p cannot use 3s. Assuming 3d orbitals to be used, I could not answer your question satisfactorily.


Searching for a solvate that changes its solvation property

http://nvlpubs.nist.gov/nistpubs/jres/086/jresv86n2p181_A1b.pdf
http://www.phasediagram.dk/binary/magnesium_sulfate.htm

Sodium sulfate hydrates,

http://iopscience.iop.org/0022-3727/41/21/212002/fulltext/
http://www.springerimages.com/Images/Chemistry/1-10.1007_s00269-008-0256-0-0

DOI:10.1209/0295-5075/102/28003
http://ej.iop.org/images/0295-5075/102/2/28003/Full/epl15377f5_online.jpg
http://www.phys.tue.nl/nfcmr/PhD-Saidov-2012.pdf

====

http://patentimages.storage.googleapis.com/US7638109B2/US07638109-20091229-D00001.png
http://www.genchem.com/properties.asp
sodium carbonate


Is it possible for Cytosine and Hypoxantine to make three hydrogen bonds?

I strongly believe that it won't be possible for these compounds to form three hydrogen bonds.

Revisiting the basic definition of a hydrogen bond:

  A hydrogen bond is an electrostatic attraction between two polar
  groups that occurs when a hydrogen (H) atom covalently bound to a
  highly electronegative atom such as nitrogen (N), oxygen (O), or
  fluorine (F) experiences the electrostatic field of another highly
  electronegative atom nearby.

Such electronegative atoms as oxygen  and nitrogen with free lone pairs are potential hydrogen bond acceptors.

Hydrogen atoms attached to very electronegative atoms like O and N have strong partial positive charge and these are potential hydrogen bond donors

These two crucial requirements for hydrogen bond formation, have thus been summarised:

  In order for a hydrogen bond to occur there must be both a hydrogen
  donor and an acceptor present. The donor in a hydrogen bond is the
  atom to which the hydrogen atom participating in the hydrogen bond is
  covalently bonded, and is usually a strongly electronegative atom such
  as N,O, or F. The hydrogen acceptor is the neighboring electronegative
  ion or molecule, and must posses a lone electron pair in order to form
  a hydrogen bond

Here is an impression of how these two would bond:

Note the "X" illustrating the missing requirement(s) to form a 3rd H-bond

missing potential donor group (electronegative atom) just C present.

Acknowledgements

Hydrogen Bonding
Voet & Voet Biochemistry: Genetic code

.


Is (2R,3S)-3,4-dimethylpentan-2-ol feasible to be dehydrated with POCl3 and pyridine?

I am uncertain whether or not you are assuming that the E2 product of a "cis elimination" is the (E)-alkene. Alternatively, you were given the geometry of the alkene and were asked to provide a mechanism for its formation. In either case, let's go through the steps.
You are correct that the alcohol 1 reacts with phosphorus oxychloride to form derivative 2a in addition to pyridine hydrochloride. Recall that there is free rotation about the C2-C3 bond in 2a, which is shown as its Newman projection in 2b in the anti-periplanar conformation ready for E2-elimination. Clearly, this elimination leads to (Z)-3,4-dimethylpent-2-ene. (Z)-4. If indeed it is the stereoisomer (E)-4 that is the target, then one can invoke SN2 displacement on 2a by chloride to invert the configuation at C2 leading to chloride 3a. Now E2-elimination of 3b leads to (E)-4.


Can luciferin + ATP → luciferyl adenylate + PPi be reversed?

In theory every chemical reaction is reversible.

The reaction that you mention does not emit any light yet, that comes later with the following reaction:
$$ \text{Luciferyl adenylate} + \ce{O2 ->} \text{Oxyluciferin} + \text{AMP} +h\nu $$

Since the reaction you mentioned occurs under loss of pyrophosphate ($\ce{PP_{i}}$) which is where the energy was stored beforehand, I guess it's somewhat hard to get it back.

It's almost like hoping that
$$ \ce{2CO2 + 3H2O -> C2H5OH + 3O2} $$
occurs under room temperature. Which it doesn't, because $\ce{CO2}$ is such a thermodynamical black hole. I think the same applies to your question.


There are two half reactions:

elemental copper to cupric ion
nitrate ion to nitric oxide

The copper half reaction should be easy to balance, but the reduction of nitrate to nitric oxide is more complicated. start with what you know:

Nitrate goes to nitric oxide, one to one.
Then balance oxygen with water since it is aqueous, presumably.
Then balance hydrogen with protons since it is acidic from nitric acid.
Finally, balance the half reaction with free electrons, which should be three electrons for a single nitrogen atom in the reactants to be consistent with nitrogen going from +5 to +2.

There are different numbers of of electrons in the copper oxidation and nitrogen reduction half reactions at this point, so you have to multiply each half reaction to get the numbers of electrons in each equal to cancel out. The free protons should be converted into nitric acid by adding the appropriate amount of nitrate ion to each side. your final balanced equation should have a ratio of elemental copper to nitric acid of 3 to 8, respectively.


Is supercriticality dependent on partial presssure, or total pressure?

There is no such thing as supercritical properties. By looking at a gas under given pressure and temperature, you'll never be able to tell is it supercritical or not.

If you try to lower temperature or pressure and see what happens, then yes, the interaction is more complex, and a mixture of two compounds does indeed have a joint critical point different from those of both pure compounds; moreover, the very meaning of a critical point becomes different and more complicated.


What is the source of discrepancy in the calculation of Standard State Gibbs Free Energy change of the reaction of HCl (l) dissolving in water?

Using the $\ce{\Delta{G^{\circ}}}$ value of liquid $\ce{HCl}$ ($-131.17$ kJ/mol) is wrong. Simply put, we are indeed interested in gaseous $\ce{HCl}$. The normal boiling point of $\ce{HCl}$ is $-85^\circ$ C. On the other hand, the standard state of Gibbs Free Energy is $25^\circ$ C for what you are calculating. (Having liquid $\ce{HCl}$ entails water being below that temperature as well, in which case it would freeze. Additionally, the equilibrium constant for the dissociation of $\ce{HCl}$ is also measured at $25^\circ$ C). Finally, though this is unrelated to your question, if $\ce{HCl}$ was liquid, it would not be included in the $K_a$ expression.

Obtaining the $\ce{\Delta{G^{\circ}}}$ value of $-95.30$ kJ/mol for $\ce{HCl}$ $(g)$, we find that $\ce{\Delta{G^{\circ}}}$ for the reaction is $-35.87$ kJ/mol. Applying the equation $K = e^{-\ce{\frac{\Delta{G^{\circ}}}{RT}}}$, we find that $K_a$ for $\ce{HCl}$ is $1.9 * 10^6$ (within magnitude of your given value).

What this looks like physically is that $\ce{HCl}$ gas is bubbled up into water, and it dissolves (dissociates) into the water as it mixes.


Why are some elements more abundant than others in the universe?

WHAT MAKES HYDROGEN ABUNDANT IN UNIVERSE:

After few minutes of creation of the universe, protons and neutrons began to react with each other to form deuterium, an isotope of hydrogen. Deuterium, soon collected another neutron to form tritium. Rapidly following this reaction was the addition of another proton which produced a helium nucleus.

Sources say that there was one helium nucleus for every ten protons within the first three minutes of the universe. After further cooling, these excess protons would be able to capture an electron to create common hydrogen. Consequently, the universe today is observed to contain one helium atom for every ten or eleven atoms of hydrogen.

WHAT MAKES NEON STAND 4TH ON THE ABUNDANCE LIST:

The star in its first stage is powered by a series of nuclear fusion reactions that convert hydrogen to helium. The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two γ rays, and a great amount of energy. It then takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium.
Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. The after reaching high temperature the helium-4 nuclei begin to fuse, producing beryllium-8.

The neon production:

  Beryllium-8 has low neutron-to-proton ratio makes it very unstable. So
  it reacts with a third helium-4 nucleus to form neon, magnesium and
  carbon. Neon production doesnt stop here. Again from the reaction of
  carbon with helium-4 to initiate a series of reactions, neon, oxygen
  and some more elements are produced. So in the stages of stars, neon
  is produced multiple times in huge amounts making it one of the
  abundant element in the universe.


Which is the major product formed on nitration of benzyl methyl ether?

1) $\ce{NO+2}$ electrophile.

Yes

  2) Benzene ring is attached to an activating group, so meta positions
  are enriched and electrophile has chances of attacking ortho/para
  positions.

The effect of the substituent in your example is probably similar to the effect of a methyl group (e.g. replacing a methyl hydrogen with a methoxy group won't have a big effect on the electron donating ability of the substituent).  Look at the relative rates of nitration in the following figure.

(image source)

We see that in toluene the ortho and para positions react 43 and 55 times (these two numbers are rather close) faster than in benzene, while reaction rate at the meta position changes little from benzene.  This is consistent with the pi-donating ability of a methyl group adding electron density to the ortho and para positions making them more attractive to attack by an electrophile.

  3) ortho position is more sterically hindered, hence para- major
  product and ortho minor product.

Look at the relative rate for the ortho position in t-butyl-benzene, here steric factors definitely play a role in slowing down attack at the ortho position. By comparison, the methyl group (and your substituent) is not very large and may slightly reduce the reaction rate at the ortho position in toluene (43 vs. 55 rather than both numbers being 55) or your example.

Another factor to consider is statistical in nature, there are two ortho positions, but only one para position. So, if everything else were equal, we would expect twice as much ortho product as para product.

In summary,

the electron donating effect of the substituent increases the rate at the ortho and para positions to a similar degree
steric factors might reduce the rate at the ortho position a small amount
statistical factors favor twice as much attack at the ortho position compared to the para position

This statistical effect is the largest of the three and is consistent with the ortho isomer being the major product.

One final cautionary note. When we discuss reaction rates we are really discussing kinetics - which product will form the fastest. However, electrophilic aromatic substitution is an equilibrium process. If you allow the reaction to run at high temperature, or for a long time, then thermodynamics can take over and we may wind up with a product composition based on relative product stability rather than kinetic factors.


Reaction of carbonyl cyclic diene with Grignard reagent

The given answer makes no sense with the substrate actually presented (but see "alternative possibility" below). You will get phenol.

If you start with phenol then, of course, the Grignard reagent just yanks off the hydroxyl hydrogen, only to have that hydrogen restored by the acid workup.  Now suppose you were to have the keto form cyclohexa-2,4-dien-1-one.  This tautomerizes to phenol precisely because the methylene hydrogen is more strongly acidic than it would be as a hydroxyl hydrogen in phenol.  So the Grignard reagent would if anything, go even faster for the methylene hydrogen in cyclohexa-2,4-dien-1-one than it would go the hydroxyl hydrogen in phenol - unless the ketone enolizes first.

That latter issue is academic.  Either way you get the same phenolate ion product, which upon acid workup gives phenol.

An alternative possibility

If the substrate structure in the question was misdrawn, then the given answer could be correct. Cyclohexa-2-en-1-one, with only one double bond in the ring conjugated to the carbonyl group, does not have any (relatively) strongly acidic hydrogens and so the Grignard reagent will add to the carbonyl group.  After acid workup the ketone is thereby converted to the tertiary alcohol in the given answer.

Grignard reagents do not react with carbon-carbon double bonds, so after adding to the carbonyl group or extracting an acidic proton that's all there is.


Note: I was employed at DuPont in the 1990's, and my understanding of Teflon(R) cookware manufacturing is based on proximity to product research and development and not from hands-on experience with PTFE formulations.

Teflon(R) polytetrafluoroethene coated materials are manufactured in several steps.  The PTFE coating is specially formulated as a waterborne dispersion containing surfactants and pigments.
Versions of Teflon(R) coatings sold in the early 1990's used a primer/topcoat system to improve adhesion of the polymer film to the substrate.  These formulated coatings are applied to the cookware and the pan is fired at high temperatures to form the PTFE film.
Coatings can be applied to both grit-blasted and smooth substrates.
The color(s) observed on PTFE coated cookware results from interactions of the pigments, polymers and surfactants during the high-temperature curing process.


My textbook states that the energy of the p orbitals will split in a magnetic field. (Physical Chemistry, McQuarrie, p.210)


Can not find structural formula with compounds' IUPAC name

The name given is not standard.

Fluoran itself is a spiro compound with two components:

(it's strictly an isobenzofuranone, but I'm lazy to type that multiple times.)

Because of the nomenclature rules of spiro compounds (which I'm sure someone will come along and tell you all about), the atoms in the benzofuran system are labelled 1-7 and those in the xanthene system are labelled 1'-10', with an apostrophe.

Atoms 1 and 2 of the benzofuran system are both within the furan ring. If I am not wrong, atom 1 is the carbonyl carbon, and atom 2 is the ring oxygen. The given name, that starts with "1,2-benzo", therefore makes no sense as you cannot fuse a benzene ring to atoms 1 and 2 of the benzofuran system.

After a little bit of Googling, I came across this compound: 6'-(Diethylamino)-1',2'-benzofluoran. I think with the apostrophes, it makes much more sense.

The numbering within the xanthene system parallels that of anthracene:

Having "1',2'-benzo" means that you have an additional fused benzene ring to carbons 1' and 2'. The 6'-substituent also makes sense. Chances are if the paper you were reading labelled the xanthene atoms with 1-10 without apostrophes, the 6-amino group should be a 6'-amino instead. Your desired structural formula is probably

ChemDraw's name to structure generator gives this molecule when you input 6'-(N-ethyl-N-isobutylamino)-1',2'-benzofluoran. The amino group, which is just an ordinary substituent, should probably be a prefix to "benzofluoran" which is the parent compound.

I am sure in due time someone will come along and give you all the official IUPAC rules but I figured that if you just wanted the structural formula, my not-very-rigorous explanation would suffice.


A function for getting the nuclide's nuclear binding energy?

Nuclear physics is very complicated and full of unsolved problems. For example, we seem to understand rather poorly the effects of three-body forces between nucleons, and describing them quantitatively is probably out of the question. Therefore, I expect that nuclei binding energies for most isotopes are solely determined empirically and have no good theoretical curve to originate from. Maybe there are statistically fitted curves to the data we have, though they may well not have a simple form and likely are only good approximations for relatively small intervals.

Edit: As it turns out, equations indeed exist, though as expected they are quite unwieldy and are semi-empirical.


H-NMR Equivalent Protons and Coupling of Glycerol

This might be confusing since both ends of the molecule are the same. The difference is that the alcohol points to only one side of the molecule.

So, while the hydrogen atoms on carbons 1 and 3 on paper have the same environment, they are not chemically equivalent when looking at the whole molecule.

Concerning your second question: The chemical shift of protons on heteroatoms such as Oxygen, Sulfur, or Nitrogen is highly dependent on the recording conditions (concentration, temperature, solvent) of the NMR. Frequently, the signals are broadened and show no coupling (broad singlett). The reason for this is the rapid intermolecular proton exchange. If there is a rapid exchange between two chemically non-equivalent groups within a molecule (like in this one), only a single, averaged signal is observed.


Mechanism of enolether formation with ethyl orthoformate in estr-4-ene-3,17-dione (Djerassi's norethisterone synthesis)

Usually a catalytic amount of acid is used in these reactions and they are run in the presence of alcohol (ethanol if you want to prepare the ethyl enol ether). Pyridine is typically added at the end of the reaction to recapture the acid catalyst.

The first step involves protonation of the carbonyl oxygen.  Protonation of the C-3 oxygen produces a resonance stabilized carbocation where the charge is delocalized over the carbonyl and the adjacent double bond (protonation at the C-17 oxygen does not have an adjacent double bond,  hence protonation at the C-3 oxygen is favored).  The nucleophilic alcohol oxygen reacts at the carbonyl carbon (highest positive charge density) to produce a hemiketal.  Loss of water gives rise to the enol ether product.  The eliminated water is captured by the orthoformate and generates more alcohol.

$$\ce{HC(OEt)_3 + H2O -> HCO2Et + 2EtOH}$$

Some argue that the reaction actually proceeds through the ketal (rather than the hemiketal) since ketals can be isolated in these reactions.  Others argue that just because the ketal is formed, doesn't prove it is actually involved in the enol ether formation reaction.  IMO, lacking definitive evidence, I opt for the simpler hemiketal mechanism.


Standard electrode potential of a galvanic half cell is zero at equilibrium at standard conditions?

There is a difference in the meaning of $\Delta E^\circ$ and $E^\circ$. $\Delta E^\circ$ actually denotes the change in standard electrode potential while $E^\circ$ denotes standard electrode potential.

$E^\circ$ is always constant for a given redox couple (eg. $\ce{Zn^2+|Zn}$). So no matter what  concentration of $\ce{Zn^2+}$ you take in a certain electrode, the standard electrode potential of $\ce{Zn^2+|Zn}$ is always $-0.76\ \mathrm{V}$.

Since $E^\circ$ is always constant, it never changes. So the value of $\Delta E^\circ$ (change in $E^\circ$) would always be zero and so it is quite a useless quantity to consider.

For $\Delta E$ however, consider the Nernst equation,
$$E = E^\circ - (RT/nF)\ln Q$$
Change in $E$
$$\Delta E = \Delta E^\circ - \Delta(RT/nF)\ln Q$$
therefore
$$\Delta E = -(RT/nF)\Delta\ln Q$$

At equilibrium $\Delta E$ is zero because there is no change in the reaction quotient $Q$ of the reaction (therefore $\Delta \ln Q$ is zero).


Burning different isotope combinations of hydrogen and oxygen, how much does the energy vary?

Let me add to Paul's excellent answer with a simple thermodynamic calculation.

Hydrogen oxidation.
$$\ce{H2 + 0.5 O2 -> H2O(g)}$$
This reaction has a standard free energy change of about -229 kJ/mol.  Using $\Delta G^\circ = -RT \ln K$, I get the equilibrium constant for this reaction of $K=1.2\times10^{40}$.
Deuterium isotope exchange between water and hydrogen.

A 1976 paper gives the value for this reaction.

$$\ce{H2O(g) + HD -> HDO(g) + H2}$$
The value given is $K = 3.5$, which by the standards of isotope exchange reactions is very large. Using $\Delta G^\circ = - RT \ln K$, we can calculate that the standard free energy change of this reaction is $\Delta G ^\circ =$ -3.1 kJ/mol.
Add the above reactions.
$$\ce{HD + 0.5 O2 -> HDO(g)}$$
We can add the free energies of the reactions to get that $\Delta G = $-231 kJ / mol.

Thus, burning $\ce{HD}$ gas provides about 1.4% more free energy than burning $\ce{H2}$ gas. And this is by far the largest difference that we would observe for any common stable isotope.  For nearly any isotopic exchange reaction that involves other isotopes, such as $\ce{^{13}C}$ or $\ce{^{18}O}$, the relevant equilibrium constants will be $0.9 \lt K \lt 1.1$ or so.  Only deuterium vs. protium (i.e. $\ce{D}$ or $\ce{^2H}$ vs $\ce{^1H}$) would often have values outside that range.  Thus for isotopes like $\ce{^{13}C}$ or $\ce{^{18}O}$, the difference will be closer to 0.1%, or, as isotopic geochemists like to say, 1 ‰ (i.e. one per mille).


Why do frying pans stick less with heat?

Heat creates steam bubbles, which hold the egg away from the pan until it is somewhat cooked - see Leidenfrost effect.


Will nitrate be able to oxidise mercury?

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

\begin{align}
\ce{NO3− (aq) + 2 H+ + e− &<=>  NO2 (g) + H2O} &\quad &E_0 = \pu{+0.80 V} \\
\ce{Hg2^2+ + 2 e− &<=> 2Hg (l)} &\quad &E_0 = \pu{+0.80 V} \\
\ce{Hg^2+ + 2 e− &<=> Hg (l)} &\quad &E_0 = \pu{+0.85 V} \\
\ce{2 Hg^2+ + 2 e− &<=> Hg2^2+} &\quad &E_0 = \pu{+0.91 V} \\
\ce{NO3−(aq) + 4 H+ + 3 e− &<=> NO (g) + 2 H2O (l)} &\quad &E_0 = \pu{+0.958 V} \\
\end{align}

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.

Now if you also consider the Nernst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.


It should work but it is not guaranteed and will depend on application. Drug efficacy is sensitive to the polymorph of the drug and for that reason, a slight change in manufacturing requires a complete redo of any certification process. I assume you are not in the manufacturing business, but it is something to consider and be aware of that your results may vary just by this simple change.


Graphite: Thermal and electrical conductivity

You are wrong to say that graphite has free electrons.  The electrons are bound to sheets of aromatic rings--delocalized but not free. In electrical conduction, the electron pairs are able to jump from bond to bond on the sheet and between sheets. It is worth noting that graphite has anisotropic conduction where between sheets conduction is lower than conduction along a sheet.

For thermal conduction, energizing an atom does not transmit the energy via free electrons like a metal would.  instead, you are limited to a predominately phonon driven heat transfer again, but now the structure is less dense and the carbons are less rigidly bonded.

Diamond as your correctly identified has a rigid structure that can easily conduct phonons but not electrons which gives good thermal, but not electrical conduction.


Yes, carbodiimides exhibit axial chirality similar to allenes. While there are significant differences in the allene - carbodiimide systems (see below) there are also some similarities. For example, just like in amines and imines the barrier to racemization in generally carbodiimidesis low, typically in the 6-10 kcal/m range.

Anet did a study on 1,3-diisopropylcarbodimide ("Energy barrier of racemization in diisopropylcarbodiimide", Frank A. L. Anet, J. C. Jochims, C. H. Bradley
J. Am. Chem. Soc., 1970, 92 (8), pp 2557–2558, DOI: 10.1021/ja00711a064).

In the chiral compound each isopropyl group would contain two diastereotopic methyl groups split by the isopropyl proton so we would expect to see 2 doublets. On the other hand, if racemization is rapid all methyl groups become equivalent and we would then expect to see only one doublet. Experimentally it turns out that around -150 C the doublet broadens into 2 doublets giving rise to a barrier height for racemization around 6.7 kcal/m.

Important Notes:

1) Just like with amines and imines, placing very electronegative groups on the carbodiimide nitrogens (fluorine, for example) raises the barrier to racemization.  Incorporating the carbodiimide structure into a small ring (see here for an example) can also raise the barrier to enatiomerization.  Both techniques have been used to permit isolation of chiral carbodiimides.

2) Carbodiimides are not linear, but rather the $\ce{N-C-N}$ angle is around 170°.  This is attributed to the terminal nitrogens being more electronegative than the central carbon and consequently affecting the hybridization of the central carbon (e.g. changing it from $\ce{sp^1}$ to $\ce{sp^{x}}$ where 1            < x < 2)

3) Racemization in carbodiimides can be achieved by two different mechanisms

by traditional nitrogen inversion involving a planar nitrogen atom, and/or
by traditional cis-trans isomerization about the $\ce{C=N}$ double bond

both mechanisms are thought to play a role in carbodiimide racemization.


Both accuracy and precision are desired for proper data analysis. They are not mutually exclusive. This is heavily dependent on the person and the measurement tool as well. You must have heard of the idiom "A bad workman quarrels with his tools"... this is true in science as well. Even if you have million dollar instrument, and the operator is unskilled, the data will be garbage. The premise that it is easier to make mistakes is not valid and generally RSD of 1-5 % is routine in scientific measurements. An RSD of 20% would be considered too high.


Meaning of Cu1.5Al in Cu-Al phase diagram (I understand all other indexes incl. fractions like Cu0.75Al0.25)

As the page you referenced mentioned, things can get a bit tricky.

I will start with your example of $\ce{Cu_{0.75}Al_{0.25}}$ - that is not just $75%$ Cu and $25%$ Al - that phase has a particular crystal structure, with the prototype (basically the first identified alloy with that crystal structure) of W (tungsten). So, it has the arrangement of atoms in the unit cell in the same places as pure tungsten, but $1/4$ of them are Al, and $3/4$ of them are Cu. There is no ordering of these atoms on the W lattice sites, or else the symmetry would be different, and the prototype might just be $\ce{Cu3Al}$.

For $\ce{Cu_{1.5}Al}$ it is similar, but perhaps even more confusing.  The 'Published phase label' is $\ce{Cu_{1.5}Al}$, so that is what it is usually called in the literature.  The 'formula' is $\ce{Cu_{1.7}Al}$, which just means when people started referring to the phase they didn't have an accurate composition measurement.  Then, the 'Prototype' is $\ce{Co_{1.75}Ge}$.  That means that the first similar crystal structure was measured in the Co-Ge system.  The non-integer subscript(s) indicate that the unit cell has vacancies in the ideal structure (without looking it up, I would speculate that the Co sites are $7/8$'s occupied).

To further complicate matters, if you look at the Co-Ge phase diagram you will find that the phase label is now $\ce{Co_{1.67}Ge}$, Formula is $\ce{Co_{1.5}Ge}$, and the prototype is still called $\ce{Co_{1.75}Ge}$.

So: crystal structures can be complicated, identification of composition has been getting better over time (folks have been looking at the Al-Cu system for a long time!), and if you really want to know what is going on you probably have to dive into the x-ray diffraction literature and look at the published atom positions in a given phase.


Separating specific components of a PUREX raffinate

You have chosen a difficult challenge, a word of advice unless you are working in a nuclear site I would advise you in the strongest terms possible not to attempt a separation of radioactive things in your kitchen, basement, garden shed or some other home lab.

OK warning over, now the idea of adding flouride to create a percipitate of lanthanide (and post plutonium actinides) is a nice idea but it will make any further processing of these elements harder. I also worry that any traces of plutonium or other elements such as zirconium might become incorperated into the lanthanide flouride perceipitate.

I would suggest that you consider solvent extraction as the separation method, I would suggest that you consider the TRPO process chemistry from China. If you were to dilute the PUREX raffinate to reduce the nitric acid content to about 0.5 M, then shake the PUREX raffinate with about 30 % (v/v) of cyanex 923 in an aliphatic kerosene then you will have a organic solution which is a mixture of lanthanides, zirconium, iron and residual plutonium / uranium.

I would suggest stripping (back extracting) the organic phase with strong nitric acid, the plutonium, uranium and zirconium will stay in the organic phase while the lanthanides will go back into the aqueous phase.

In some ways a better alternative would be to use TODGA or a malonamide, a solution of one of these reagents in aliphaltic kerosene will extract the lanthanides from PUREX raffiniate without adjustment of the nitric acid content of the raffinate. You can then strip the lanthanides (plus the americium / curium) from the raffinate. These types of processes have been tested at ITU in their hot cells using PUREX raffinate formed from real fuel.

In some ways I would want to use dilute nitric acid as a stripping agent for the lanthanides from a diamide based process. You can then reextract the lanthanides from the dilute nitric acid with 30 % DEHPA (bis-2-ethylhexyl hydrogen phosphate) in kerosene. In turn back extract this new organic phase with nitric acid, if you are careful it is possible to get some separation of the lanthanides into the heavy and light lanthanides by selective stripping with different concentrations of nitric acid.

Another method which would work would be to use the TRUEX process chemistry from the USA, this is based on CMPO in tributyl phoshate modified kerosene.


Here is the Walsh diagram depicting all the valence molecular orbitals (a diagram showing how individual molecular orbitals change in energy due to bending around the central atom). Oxygen has 6 valence electrons, so ozone has 18 electrons in total. If we start on the right where ozone would be linear, we can see that all the orbitals up to the $2\pi_\mathrm u$ orbitals (don't worry about why they are named this) are doubly occupied and the two $2\pi_\mathrm u$ orbitals are both singly occupied.

Note: 6a$_1$ and 2b$_1$ have the incorrect sign on the outer p-orbitals.

If we bend the molecule slightly (moving to the left on the diagram), we can see that there is a favorable interaction between the p-type orbitals on the end as well as between the black of the central p-type orbital and the black of the outer p-type orbitals (Someone drew the $\mathrm{6a_1}$ and $\mathrm{2b_1}$ orbitals wrong, flip the central orbital). This lowers the energy of the molecule.

So why doesn't it keep bending? As a good first-order approximation, we can estimate the relative energy of configurations by the highest energy orbital (provided the other orbitals don't change too much). We can see that if we keep bending, the $\mathrm{1a_2}$ orbital and the $\mathrm{4b_2}$ orbital start rising in energy, eventually rising above the $\mathrm{6a_1}$ orbital. Thus too much bending will be unfavorable, and thus ozone prefers a bond angle of around $117^\circ$.

This same diagram can be used for other molecules, such as $\ce{CO2}$. Try using it to figure out why $\ce{CO2}$ is linear.


Why are there two Hydrogen atoms on some periodic tables?

The two hydrogens are the same, but some periodic tables show hydrogen in both places to emphasize that hydrogen isn't really a member of the first group or the seventh group.

Hydrogen is a diatomic gas in it's elemental state, which is different from the other group one metals (and similar to the group seven elements).  At the same time, hydrogen usually loses its electron, similar to sodium and the other group one metals.


The simplest way to look at this trend is through VSEPR theory, which produces a very good qualitative understanding of molecular geometry even if it is not compatible with modern molecular orbital theory. However, VSEPR is not the whole story for the heavier Group V and Group VI hydrides.

According to VSEPR, electron domains (i.e. bonds and lone pairs) are arranged around the central atom in geometries that minimize electrostatic repulsion between the electron domains. These geometries also happen to be solutions to the Schrödinger equation for the set of molecular orbitals as linear combinations of atomic orbitals, and in the molecular orbital case, they arise not from electrostatic repulsion, but strictly from the mathematics of the linear combinations producing a minimum energy self-consistent set of orbitals.

VSEPR is qualitatively simpler for small molecules. All of the binary hydrides in Group IV, V, and VI have steric number four around the central atom. This means that the number of lone pairs and the number of bonding domains add up to four. These species can also be described using the $\ce{AXE}$ method, where $\ce{A}$ denotes the central atom, $\ce{X}$ denotes an atom bonded to $\ce{A}$ and $\ce{E}$ denotes a lone pair of electrons.

The Group IV hydrides all have four bonds and no lone pairs. We describe this arrangement as $\ce{AX4}$.
The Group V hydrides all have three bonds and one lone pairs. We describe this arrangement as $\ce{AX3E}$.
The Group VI hydrides all have two bonds and two lone pairs. We describe this arrangement as $\ce{AX2E2}$.

There are several factors that affect the degree of repulsion:

Lone pairs - Lone pair domains are more diffuse and have higher repulsion than bonding domains.
The nature of $\ce{X}$ - Larger and/or more electropositive atoms have higher repulsion.
Bond order - Bonds with higher bond orders have higher repulsion than bonds with lower bond orders.

Since $\ce{X=H}$ in all cases, we only need to worry about the first factor: lone pairs.

The arrangements $\ce{AX4}$, $\ce{AX3E}$, and $\ce{AX2E2}$ all have four electron domains, and thus the base electron domain geometry group is tetrahedral.

The differences in bond angle between $\ce{CH4}$, $\ce{NH3}$, and $\ce{H2O}$ originate from the increasing number of lone pairs:

$\ce{CH4}$ has no lone pairs: the bond angles are the ideal tetrahedral bond angles of $109.5^\circ$.
$\ce{NH3}$ has one lone pair: the bond angles are contracted a little because of the grater repulsion of the lone pair to $107.8^\circ$.
$\ce{H2O}$ has two lone pairs: the bond angle is contracted further due to the repulsion of two lone pairs to $104.5^\circ$.

  All of the Group IV hydrides will have perfect tetrahedral geometry due to having four bonds to the same atom and no lone pairs. Thus $\ce{CH4}$, $\ce{SiH4}$, $\ce{GeH4}$, $\ce{SnH4}$, and $\ce{PbH4}$ all have bond angles of $109.5^\circ$.

For Group V, the bond angles are:

$$\begin{array}{c|c}
\ce{AH3}&\text{angle}\\ \hline
\ce{NH3}&107.8^\circ \\
\ce{PH3}&93.5^\circ \\
\ce{AsH3}&91.8^\circ \\
\ce{SbH3}&91.7^\circ \\
\ce{BiH3}&90.5^\circ \\ \end{array}$$

Notice that there is a large decrease from $\ce{NH3}$ to $\ce{PH3}$ and then very small changes for $\ce{AsH3}$, $\ce{SbH3}$, and $\ce{BiH3}$.

You could make the argument that the larger orbitals of the heavier elements will have high repulsion than the smaller orbitals of nitrogen. This is where VSEPR breaks down. VSEPR works well for some compounds because there is significant $s-p$ mixing when constructing the molecular orbitals. This mixing need not happen for the heavier elements. One molecular orbital solution for $\ce{PH3}$, for example, indicates very little contribution from the $\ce{P}\ 3s$ orbital. We can describe the $\ce{P-H}$ bonds as predominantly $\ce{P}_{3p}\ce{-H}_{1s}$ bonds. Thus the geometry around $\ce{P}$ is determined mostly by the orientation of the $p$ orbitals.

If we look at the Group VI hydrides, we see the same trend: a large decrease from $\ce{H2O}$ to $\ce{H2S}$ and then very little change. The same argument can be made for primarily $\ce{A}_{p}\ce{-H}_{1s}$ bonding in the heavier Group VI hydrides as for the heavier Group V hydrides.

$$\begin{array}{c|c}
\ce{AH2}&\text{angle}\\ \hline
\ce{H2O}&104.5^\circ \\
\ce{H2S}&92.1^\circ \\
\ce{H2Se}&91.0^\circ \\
\ce{H2Te}&90.0^\circ \\
 \end{array}$$


Oh, I was just reading about this. This excerpt points out the factor contributing to the chemical shift observed in alcohols

  The extent of hydrogen bonding varies with concentration, temperature, and solvent, and changes in the degree of hydrogen bonding can cause substantial shift changes. This is very evident in the nmr spectrum of ethanol taken at different concentrations in CC14 (Figure 9-29). The hydroxyl resonance will be seen to move upfield at the lower concentrations. This is the result of decreasing association by hydrogen bonding through equilibria...

Source

So basically the reason (or a reason) for alcohols moving upfield due to hydrogen bonding, which results in decreasing association.

I looked at your link and there are a few notable things. They mention *variable, highlighting that it is not necessarily constant based on different conditions, which connects to hydrogen bonding. For carbon-2, we have the normal ~1.7 for methine protons + 2.5 from oxygen of alcohol, coming to be a pretty close value. So no, it's not "why on earth would the electrons on the carbon-2" but what on earth is going on with that alcohol?


The rate of evaporation decreases because the activity of the solvent in the liquid phase decreases, while the activity in the vapor phase (or solid, for that matter) stays the same. This is also the reason for the increase in boiling point and the decrease in melting point of the solvent.

If we add a volatile solute, it will evaporate too, thus changing the solvent activity in the vapor phase as well. The resulting behavior can't be characterized universally; you'll have to consult the binary phase diagram of boiling for the particular system, and these can be pretty diverse.


Why does the tautomerism of purine favor 9H-purine?

The given purine has two structures as shown below.

In $\ce{9H–Purine}$ (figure 2), lone pairs are far apart from each other (3 and 7). Comparatively in $\ce{7H–Purine}$ (figure 1) lone pairs are very near each other (3 and 9) . Nearness of lone pairs in $\ce{7H–Purine}$ could leed to greater repulsions between them. This could make it energetically unfavorable in comparison to $\ce{9H–Purine}$.

This could be the reason why $\ce{9H–Purine}$ is more favored tautomer compared to $\ce{7H–Purine}$.


The net result is $\ce{C6H5OH + Zn -> ZnO + C6H5H}$, but under the reaction conditions (zinc DUST and strong heating (b.p. of phenol $= \pu{182 ^\circ C}$)), I suspect the reaction mechanism involves the zinc surface rather than individual $\ce{Zn}$ atoms or $\ce{Zn^{2+}}$ ions. And the zinc surface will be a complex oxide. I have read that the reaction is not clean and produces several byproducts, typical of high temperature reactions.

The acidity of phenol is not great enough to dissolve zinc (at least, not easily), but it could adsorb on the zinc/zinc oxide surface. First step: probably a phenolation of an oxygen to phenolate and hydroxide:

$\ce{Zn - Zn - ZnO - Zn - Zn- (surface) -> Zn - Zn - Zn (OH, OC6H5) - Zn - Zn}$.

At high temperature, the $\ce{H}$ on the hydroxyl could reach the phenyl group in a five-membered ring and go off as benzene ($\ce{C6H5H}$), leaving $\ce{Zn - Zn - ZnO - ZnO - Zn}$.

Magnesium and zinc have some similarities (ionic radius, oxide coat), but magnesium is significantly more reactive. I suspect that magnesium in phenol would react at lower temperatures and give magnesium phenolate and hydrogen: $\ce{Mg + 2 C6H5OH -> Mg(OC6H5)2 + H2}$, no benzene. In this case, the acidity of phenol would be enough to dissolve the magnesium; fine powder (pyrophoric) would not be needed.


What is the lower bound to the temperature at which a fire can burn?

Fires which burn below about 400°C are known as cool flames, and this is a well known and widely researched phenomenon. Most hydrocarbons and many alcohols can produce cool flames, and the conditions under which they burn depends on the oxygen content available. The most common encounter that most people have with the cool flame effect is through incomplete combustion of a petrol engine, leading to engine knocking.

The lowest recorded cool flame temperatures are between 200 and 300°C; the Wikipedia page references n-butyl acetate as 225°C. You can read a lot more about cool flames on that page. One thing to note is that cool flames are very hard to visibly detect at lower temperatures - both heat and light being two of the by products of the combustion process.

NASA have conducted a number of experiments in space on low temperature combustion of nano droplets of heptane, and this is an interesting area of science still being investigated.

An interesting point to make relates to your question of whether you could hold your hand in the flame and not burn yourself. Clearly the answer is no, but for some cool flames it may take noticeably longer to feel the pain of heat. Paper starts to discolour at about 150°C, but may not burn as ignition temperature is between 220-250°C. For reference, candle flames are about 600-1400°C.


Why do you need a strong base for a titration

There is no reason why you cannot use a weak base to titrate a strong acid.  The calculation is probably considered a bit more complex but I think the real reason is a practical one.  For the sake of simplicity, I swapped your question around just a bit, and the figure below shows the titration of a strong base (blue) weak base (reddish) and very weak base (dashed line) with a strong acid (say, for example 1 M $\ce{HCl}$).

In each case the concentration of the acid is the same (1.0 M) and the concentration of the base is 0.100 M.  The volume of the base titrated is 100 mL.  For the standard strong acid/strong base scenario, one finds a well defined equivalence point.  When you transition to a weak base, however, there is additional information presented in the titration curve, namely the pKb of the base, which is a measure of its strength.  The pKb of the base can be obtained by finding the pH at the half-way mark between no acid added and the equivalence point (and if you know a bit of calculus, this is the point of minimum slope prior to the equivalence point). I've marked the pKb of the reddish titration curve with a dot, and the pKb is 9.  As the base gets weaker (less willing to accept a proton) the region before the equivalence point becomes less well defined.  In an extreme example, I have shown a simulated titration curve of acetic acid (a very poor base indeed!) with a strong acid.  Notice how the equivalence point in this curve would be very hard to pick out if we didn't know where we were looking.

So the bottom line is this: the calculations aren't harder (one uses the Henderson–Hasselbalch equation and this is a standard exercise in a first year analytical chemistry class) but the data one obtains from the experiment are harder to analyze.  And in practice, we do perform weak/strong acid/base titrations when we are interested in determining the strength of the acid or base under consideration.


Are IR signals affected by nearby atoms?

As was already said,

  IR works on bond lengths. So if the difference is negligible, instrument noise is going to dwarf anything useful you'll pull out.

There will be a difference, and you can definitely detect it computationally, but by normal experimental means (benchtop FT-IR or ATR, not a gas-phase experiment) background noise might the same order of magnitude as any frequency shift you'd see.

For example, the carbonyl stretching frequency and carbonyl bond lengths calculated using B97-D/6-31G(d):

$$
\newcommand{\AA}{\unicode{x212B}}
\begin{array}{llll}
           & \text{bond length} \, (\AA) & \text{frequency, calc.} \, (\text{cm}^{-1}) & \text{frequency, expt.} \, (\text{cm}^{-1}) \\ \hline
\text{pentanal}   & 1.218 & 1772 & 1727 \\
\text{2-hexanone} & 1.223 & 1753 & 1718
\end{array}
$$

Technically, you can distinguish between these in an experiment, but one would be more interested in the presence/absence of a large aldehyde band. This works due to the sharpness of the carbonyl stretch.

Closer to what I think you're asking about is the influence on vibrational frequencies of groups far from the area of substitution. These particular examples are tough because the normal modes of hydrocarbon-like species tend to couple the motion of many $\ce{C-H}$ stretches together. With that in mind, here are the highest-frequency $\ce{C-H}$ stretching modes, which happen to belong to the terminal $\ce{CH3}$ groups. The bond length is for the hydrogen eclipsed with the carbonyl.

$$
\newcommand{\AA}{\unicode{x212B}}
\begin{array}{lll}
           & \text{bond length} \, (\AA) & \text{frequency, calc.} \, (\text{cm}^{-1}) \\ \hline
\text{pentanal}   & 1.102 & 3044 \\
\text{2-hexanone} & 1.102 & 3038
\end{array}
$$

"Wow, that's amazing, you can resolve it!" Not this time. Compare the full spectra for pentanal/valeraldehyde (top) and 2-hexanone (bottom) and try to perform the peak assignment. Is it it the 2960/2962 pair? 2935/2936?

Experimental values/images taken from the Spectral Database for Organic Compounds.


IUPAC naming for compounds containing carboxylic acid and anhydride together?

Acids are senior to anhydrides; therefore, a compound that contains both functional groups is named as acid.

The acid is substituted using the usual principles of substitutive nomenclature. The prefix ‘oxo’, denoting $\ce{=O}$, is used to indicate a carbonyl group when the group cannot be cited as suffix. The group $\ce{R-CO-O-{}}$ is indicated by an acyloxy prefix (e.g. ‘acetyloxy’, ‘propanoyloxy’, etc.).

For example: 3-(acetyloxy)-3-oxopropanoic acid

(Cyclic anhydrides, however, are named as heterocyclic pseudo ketones.)


Does concentration of salt increase or decrease rate of rusting?

In general, salt (particularly NaCl) will increase the rate of corrosion (rusting).

To understand why, consider metallic iron $\ce{Fe}$ which rusts (oxidises) to iron(II) oxide $\ce{Fe2O3}$ in the presence of oxygen $\ce{O2}$ and water $\ce{H2O}$.

$$\ce{4Fe +3O2 +6H2O->4Fe(OH)3}$$

Corrosion (rust) is a 'redox' reaction, which means it involves reduction of oxygen into hydroxide ions and oxidation of metallic iron to iron cations.

For any redox reaction to take place, electrons are transferred. In this case, from the metallic iron which loses electrons to oxygen. The presence of salt (or any electrolyte) in the water accelerates the reaction because it increases the conductivity of water, effectively increasing the concentration of ions in the water and so increasing the rate of oxidation (corrosion) of the metal.

The situation is complicated by the fact that salt dissolved in water actually reduced the amount of dissolved oxygen present in the water. This is because the water molecules are attracted to the dissolved ions from the salt (solvation), which has the tendency to decrease the weak affinity of non-polar oxygen molecules to water, thereby deriving dissolved oxygen out. So if the metal is totally submerged in water, higher salt concentrations can actually reduce the rate of corrosion, if the water is not aerated.

So it depends upon the amount of oxygen getting into the water and onto the metal. For example, underwater plants can release oxygen in salty water which will accelerate the rate of corrosion of the submerged metal. Of course, if the metal is not completely submerged in salt water but only partially submerged, or is affected by salt water 'spray', it will be still be exposed to oxygen and will rust faster.


Why are calcination and roasting processes carried out at a temperature just below the ore's fusion temperature?

The purpose of roasting a/o calcination is oxidation a/o decomposition of original ore to form metal oxides.
It is desirable to keep the intermediate product solid to obtain porous product. This allows easy access for further reduction by carbon monoxide by coke or charcoal reduction process. If oxides were in form of a solidified melt, the reduction arrangement would be problematic.
Additionally, some ores or oxides have too high melting point to be practical or easily possible to melt them.


Will a Galileo's Thermometer containing liquid paraffin explode in my car if the temperature is above 35 degrees Celsius?

Do not confuse the flash point with the autoignition temperature.

I don't know what atmosphere you have inside the thermometer, but either way I think it's hardly the case that you reach the autoignition temperature of paraffin (gasoline undergoes autoignition at 246-280 °C).


Why don't electrons pair in [Fe(NH3)6]2+?

(If you know the basics, go to last paragraph and skip the introduction part)

There are two types of ligands. Strong field ligands and weak field ligands. Those which only cause a small extent of crystal field splitting is called weak field ligand and those which cause large splitting is called strong field ligand.

Why does crystal field splitting take place?

Let's take an octahedral complex. The $d_{x^2-y^2}$ and $d_{z^2}$ point along x,y,z axis and the $t_{2g}$ orbitals point between the axis. When the 6 ligands approach along the three axes, the energy of the $e_g$ orbitals increase more than the increase in the energy of $t_{2g}$ orbitals. Hence the d-orbital splits into two. Filling each electron in $t_{2g}$ orbitals decrease energy by $0.4\Delta_0$ and filling electrons in $e_g$ increase energy by $0.6\Delta_0$

Coming back to your question, you know how electrons are filled in the orbitals (according to hund's rule of course). But how does electron pairing take place in your case?

Let the energy required to pair an electron be P.
Let's say there are 4 electrons. The two arrangements are one electron in each sub-orbital (no pairing) and 4 electrons all in $t_{2g}$ itself (instead of 4th electron going to $e_g$, its paired up in $t_{2g}$).
In the first case, the total energy is $-0.6\Delta_0$ and in second case, its $-1.6\Delta_0+P$. So, only strong field ligands can form the second arrangement. $P$ value is constant for a given metal ion.

Finally, ammonia is weak field ligand but in certain cases, it acts as a strong field ligand (example: for Cobalt). For Fe, ammonia is a weak field ligand. Hence no pairing takes place.


Electronic configuration of period 3 element from its successive ionization energies

Yes, your reasoning is logical and makes sense, but the question must have a typo because those IE values are the exact match for boron B, an element from period 2 right above aluminium Al (data from [1, p. 10-204], converted from $\pu{eV}$ to $\pu{kJ mol-1}$):

\begin{array}{lccccc}
\hline
\text{Element} & \mathrm{IE_1} & \mathrm{IE_2} & \mathrm{IE_3} & \mathrm{IE_4} & \mathrm{IE_5} \\
\hline
\ce{B} & 801 & 2427 & 3659 & 25022 & 32822 \\
\ce{Al} & 578 & 1816 & 2745 & 11580 & 14842 \\
\ce{Ga} & 580 & 1980 & 2965 & 6102 & 8299 \\
\ce{In} & 559 & 1821 & 2704 & 5210 & \\
\ce{Tl} & 589 & 1971 & 2878 & & \\
\hline
\end{array}

Another hint that it's not aluminium is the absolute value of the jump in energy between $\mathrm{IE_3}$ and $\mathrm{IE_4}$, which is too high for latter group 13 elements.
Probably the author meant group 13, not period 3, but that's just a speculation at this point.

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.

Appendix

Source code for the plot, pgfplots:

\documentclass{standalone}
\usepackage{pgfplots}
\usepackage{siunitx}

\begin{document}

\begin{tikzpicture}
    \begin{axis}[
        title={Successive ionization energies of group 13 elements},
        xmin=1, xmax=5,
        ymin=0, ymax=35000,
        xlabel = $i$,
        ylabel = {$\mathrm{IE}_i$/\si{\kilo\joule\per\mole}},
        minor tick num = 1,
        mark = *,
        grid,
        legend cell align = {left},
        legend pos = north west,
    ]

    \addplot coordinates {
        (1,801)
        (2,2427)
        (3,3659)
        (4,25022)
        (5,32822)
    };
    \addlegendentry{Boron}

    \addplot coordinates {
        (1,578)
        (2,1816)
        (3,2745)
        (4,11580)
        (5,14842)
    };
    \addlegendentry{Aluminium}

    \addplot coordinates {
        (1,580)
        (2,1980)
        (3,2965)
        (4,6102)
        (5,8299)
    };
    \addlegendentry{Gallium}

    \addplot coordinates {
        (1,559)
        (2,1821)
        (3,2704)
        (4,5210)
    };
    \addlegendentry{Indium}

    \addplot coordinates {
        (1,589)
        (2,1971)
        (3,2878)
    };
    \addlegendentry{Thallium}

    \end{axis}
\end{tikzpicture}

\end{document}

Converted to PNG with the following command:

magick convert -density 120 G13_ionization_energies.pdf -quality 100 G13_ionization_energies.png


Why fossil fuel is too dangerous?

Combustion of fossil fuel does not only release $\ce{CO2}$ but many other harmful substances(pollutants) like oxides of sulphur ($\ce{SO_{x}}$), oxides of nitrogen($\ce{NO_{x}}$), Carbon Monoxide($\ce{CO}$), etc...

oxides of sulphur ($\ce{SO_{x}}$):

When sulphur containing fossil fuel is burnt, then oxides of sulphur is produced. Sulphur Dioxide($\ce{SO2}$) is generally produced.

Low concentration of $\ce{SO2}$ causes diseases like asthma, swelling and irritation of respiratory tract.

High concentration $\ce{SO2}$ leads to stiffness of flower buds, which eventually fall off!

Oxides of Nitrogen($\ce{NO_{x}}$):

When in automobile engine, fossil fuel is burnt at high temperature significant amount of nitric oxide($\ce{NO}$) and nitrogen dioxide($\ce{NO2}$) is released.

$\ce{NO2}$ is lung irritant. it is highly toxic to living tissue.

$\ce{NO2}$ decrease the rate of photosynthesis, cause respiratory diseases in children, etc.

Carbon monoxide($\ce{CO}$):

$\ce{CO}$ is produced due to incomplete combustion of carbon. It is major constituent in smoke of vehicles.

$\ce{CO}$ binds with haemoglobin and form carboxyhemoglobin, which is 300 times more stable than usual oxygen-haemoglobin! If concentration of carboxyhemoglobin reaches 3 to 4% then oxygen carrying capacity of blood reduces.

Few more!

$\ce{SO2}$ with smoke and fog is responsible for classical smog(London smog)

$\ce{NO2}$ and $\ce{NO}$ is responsible foe photochemical smog(Los-Angeles smog)

Update:

I was reading more about environmental chemistry and i discovered that $\ce{NO_{x}}$ is responsible for 40% depletion  of concentration of Ozone layer!

  why are scientist more concerned about fossil fuel instead of CFC?

As commented by LDC3 The production of the more hazardous CFC's has been banned.

There are 95 derivatives of chloro and bromo fluorocarbon which are listed as ODS(Ozone Depleting Substances).

On 16th September, 1987 93 countries sign Montreal Protocol and accept not to use ODS.

Reference: GSEB textbook pdf


Why is Nitric Acid more reactive and caustic than Sulfuric or Hydrochloric?

Nitric Acid is indeed what most often is referred to as a "strong acid" (at ambient temperatures), and its $\mathrm{pK_a} \approx -1$, i.e nitric acid in diluted solution is almost always fully dissociated except in extremely acidic solutions.

First of all, how does it compare to other "strong acids", for example sulfuric acid

$\ce{HNO3 + 2 H2SO4 ⇌ NO2+ + H3O+ + 2HSO4−}$

Clearly, in this reaction sulfuric acid acts as the "acid", while nitric acid acts as the "base", hence sulfuric acid is indeed a stronger acid.

Reactions of dilute nitric acid with some metals are typical of an acid, namely:

$\ce{Mg + 2 HNO3 → Mg(NO3)2 + H2\uparrow}$

Other metals that react in this fashion are manganese and zinc.

You do correctly note that a special feature of nitric acid's reactivity is its strength as an oxidising agent, and this is also evident in its reactions with metals, for example. However, do note, that the exact details depend on the concentration of the acid and the nature of the metal

For example, nitric acid can oxidise non-active metals such as copper and silver, however, the products depend on temperature and the acid concentration.

For instance, copper reacts with dilute nitric acid at ambient temperatures with a 3:8 stoichiometry:

$\ce{3 Cu + 8 HNO3 → 3 Cu^2+ + 2 NO + 4 H2O + 6 NO3−}$
The nitric oxide produced may react with atmospheric oxygen to give nitrogen dioxide.

However, with more concentrated nitric acid, nitrogen dioxide is produced directly in a reaction with 1:4 stoichiometry:

$\ce{Cu + 4 H+ + 2 NO3− → Cu^2+ + 2 NO2 + 2 H2O}$

Metals like chromium, iron and aluminium readily dissolve in dilute nitric acid, but the concentrated acid forms a metal-oxide layer that protects the bulk of the metal from further oxidation; thus, concentrated nitric acid serves to passivate these metals towards further reactions. Common metals that are passivated by concentrated nitric acid are iron, cobalt, chromium, nickel, and aluminium.


Short answer: It does, but unstable and hard to detect reliably..

Long answer.

First of all, we should look at structure of carbon dioxide and silicon dioxide. It is pretty easy to see that silicon forms polymeric dioxide with no double bonds, while carbon forms molecules with two double bonds. It is not an anomaly, obtaining silicon compounds with double bonds $\ce{Si-E}$ is tricky at best. This is attributed to larger atom size and more diffuse $p$-orbitals for silicon, making formation of double bonds unfavoured in most circumstances.

Now, let's look at carbon monooxide. It is a molecule with bonding very similar to that in dinitrogen molecule. The carbon atom has formal negative charge and oxygen has formal positive charge. However, because of high electronegativity of oxygen, electron density is pulled back towards oxygen, so the resulting dipole moment is diminutive. The molecules contains not double, but triple bond, making formation of similar molecule for silicon highly unfavorable. Still, silicon monooxide can be detected in gas phase.

Now, why solid phase with composition $\ce{SiO}$ is not formed is a more interesting question. Unfortunately, such questions not always have an easy answer, as rules guarding stoichiometry of solid phases is often base on arcane concepts of atom packings and electron count. It is entirely possible, that such a phase would be obtained one day in some exotic conditions. For clarity, I would use sterical considerations. $\ce{Si-O-Si}$ fragment is roughly linear, while $\ce{Si}$ atom favors tetrahedral coordination. Going from here, only two simple phases with no dangling bonds can be formed, based on diamond structure. It is structures with either $\ce{Si}$ or $\ce{SiO4}$ units in the structure nodes. It IS an oversimplification, as silicon dioxide forms many rather esoteric crystal structures, because $\ce{Si-O-Si}$ fragment isn't perfectly linear, but is is close enough in my opinion.

Do not rely on oxidation states in prediction of existence of covalent compounds,   +2 oxidation state for silicon is a thing. It exists in $\ce{Si6Cl12}$, adopting structure similar to cyclohexane.


Why does it have two CAS numbers? As I know, the CAS number should be unique to each substance.

Yes, that’s exactly the principle and also why two CAS reference numbers are associated with $\ce{FeO(OH)}$. SciFinder gives 20344-49-4 to be the ‘boring’ laboratory iron oxyhydroxide, while 1310-14-1 is mentioned as the mineral goethite. Goethite is the α form of $\ce{FeO(OH)}$; β (akaganéite), γ (lepidocrocite) and δ (feroxyhyte) forms exist, too.

Since each of these four minerals has a well-defined structure, much better defined than simple $\ce{FeO(OH)}$, they should (and will) each get a single CAS number for unambiguous referencing. At the same time, the generic CAS number 20344-49-4 also exists for lab-grade non-mineral $\ce{FeO(OH)}$.

So what you are experiencing is not a violation of the one-substance-one-CAS-number rule but rather a confirmation of its strictness.


What causes the blackening of light bulbs

A 14 page article (https://www.jstage.jst.go.jp/article/jieij1917/6/4/6_4_398/_article) describes the blackening of incandescent light bulbs as being totally due to evaporation of the filament. Platinum had been used as far back as 1840, and Edison's first patent was for a carbon filament (1879), but by this time (1922), tungsten was the metal used in almost all filaments.

Blackening of carbon-filament bulbs has not been a significant problem, due to the short commercial lifetime of carbon filaments. Any carbon deposition would likely be from sputtering of the filament, not evaporation. Blackening is a problem with tungsten filaments, and not so much from evaporation of metal, but from evaporation of tungsten oxides formed from adventitious water or oxygen, followed by deposition on the cool bulb surface. Inert gas is used instead of vaccuum to impede the mobility of tungsten atoms. But then, tiny amounts of oxygen or water vapor in the bulb can cause severe blackening; getters such as zirconium are used to combine with oxygen. Large bulbs used for theater lighting often contained some tungsten powder; when the bulb darkened, the bulb was shaken to abrade off the tungsten and increase the brightness. https://en.wikipedia.org/wiki/Incandescent_light_bulb#Bulb_blackening

Halogen bulbs use iodine to react with the deposited tungsten, which vaporizes as the iodide and is redeposited on the hot filament, where it decomposes. Halogen bulbs are smaller than the usual incandescent bulb and they run hotter filaments to keep the glass hotter, so that iodine can react there with the deposited tungsten and carry it back to the filament.

Your description of Hammer adding an anode to catch electrons emitted by the hot filament is correct, but I suspect that the original thinking went something like this: "We know that electrons are emitted by the hot filament. As a result, the filament must become positive; thus some metal atom at the surface might be repelled, carrying a positive charge (which would be neutralized as it passes thru the electron cloud). This metal atom might deposit on the cool bulb, producing a darkening. Perhaps we can prevent this repulsion/evaporation by making the filament negative so as to retain the positive metal ions." Apparently, this did not work, but only increased the electron emission, which could then be described as a bug or a feature.  OK, let's call it a feature and name it after Edison.


What are the major environmental drawbacks with using methane as a combustable fuel?

Methane has many lovely properties. It's a very dense source of energy. It is plentiful, it seemed to be cheap to extract (until we worked out the real cost), and similarly seemed to be cheap to transport long distances. It can be stored for many months in vast quantities.

But.

It's a greenhouse gas - a powerful one at that. On a 100-year horizon, it has about 20 times the global warming potential of $\ce{CO2}$. That needn't be a problem, if natural-gas extraction had no leakages, and distribution systems had no leakages.  But they do have leakages, which means the supply chain worsens global warming.

Furthermore, when methane is burnt, $\ce{CO2}$ is released into the atmosphere, worsening global warming even further.

We've learnt to handle methane's other main problems: that it's colourless and odourless, so can just accumulate unnoticed waiting to explode, if you let it; and that it's also dangerous if handled carelessly in any other way.

It also generates $NO_x$ when burnt in air, and they are local as well as global pollutants. It is possible to have flue-scrubbers to remove the $NO_x$.

That doesn't mean that that methane is permanently bad.  It can be created in a carbon-neutral fashion (from biological sources, or synthesised from atmospheric  $\ce{CO2}$), and if leakages are minimised, offers a potential useful form of energy storage. (note that in these forms, it would be an energy & storage vector, not a net energy source, so its economic value would be very different - but that's economics, not chemistry)


How to find the combining valence of an anion when it exhibits variable valency?

First of all, sorry for the (very!) long post!

Determining the Number of Valence Electrons

The number of valence electrons of atom can be determined by the group in which the atom is in. For example, Na is in group 1A, and thus has one valence electron (1+ charge). Chlorine is in group 7A and thus has 7 valence electrons (1- charge). A quick (but very simplified) way of determining if an element has a positive or negative charge is to look again at its group. Groups 1A, 2A, and 3A are positive; while groups 5A, 6A, and 7A are negative. An element in group 4A can be both.

Ionic Compounds:

If the charges are the same, no "balancing" is necessary:

if Na (1+) were to bond with Cl (1-) it would form NaCl.

If the Charges are different, a quick way to determine the formula unit is to assign each element the charge of the other:

if B (3+) were to bind with Cl (1-) it would form BCl3: Boron get the 1 from Chlorine; Chlorine gets the 3 from Boron.

Covalent Bonding

Covalent Bonds are a little trickier and has more contributing factors. It seems like you're having trouble with Lewis-Dot diagrams so that's what I'll address. In order to correctly draw the Lewis-Dot diagram of a compound, you have to determine the formal charge of each atom. For demonstration purposes I'm going to use CH4.

The formal charge of an atom can be determined by:

Number of Valence Electrons - non bonding electrons - number of bonds

So in our example, the formal charge of Carbon is :

4 Valence electrons - 0 non bonding electrons - 4 bonds = formal charge of 0.

Then Hydrogen:

1 Valence electron - 0 non bonding electrons - 1 bond = formal charge of 0.

So if we add the formal charged together we get 0, which makes sense because the charge of the compound is 0.

In general, the more compounds that can have a formal charge of 0, the more likely it is that that particular structure will occur, as it will be more stable. If the compound is a poly atomic ion, the more electronegative atom should get the negative formal charge (electronegativity increases as you move up the group and to the right in a period).

One more significant detail is the ability of some atoms to expand the octet, which can occur if the element is in the third period or beyond. In this case, the charge of the atom can be made 0 (or whatever appropriate) if more bonds are made than typically seen.

i.e. in ICl4-, Iodine, though typically only making one bond, bonds to four Chlorine atoms. In this case formal charge = 7 valence electrons - 4 bonds - 4 non bonding electrons = 1-, the charge of the polyatomic ion. While Cl is more electronegative, only I can expand its octet, thus I has the negative formal charge. The 4 non bonding electrons on Iodine were determined by counting the total number of valence electrons (7 from I + 4*7 for Cl + 1 from the charge = 36) and adjusting accordingly.

There are other exceptions such as free radicals and atoms that can have less than a full octet, but the "rules" above should be sufficent guides for basic Lewis Dot structures.

Hope that covered everything :)