text:
What is the difference between molar conductance and molar conductivity?
I'm learning (or at least trying to learn) about electrochemistry, but a major obstacle to that, is that different books I refer use different terms for the same symbols. So in a last ditch attempt to clear stuff up, I've resorted to Chem.SE.
So here's what I intend to do; I'll list out everything I think I've understood, as well as pose a couple of questions regarding some of them. I'd really appreciate it if someone would take the time to go through what I've listed out, checking them for errors and then clearing those queries which I've got. So here I go

Symbols used:
Resistance ($R$) , Resistivity or specific resistance ($\rho$), Conductance ($C$), Conductivity or specific conductance ($\kappa$), Area of cross-section of the electrode ($A$), distance between the electrodes ($L$),

and the REALLY confusing bit, Molar conductance according to some books and Molar conductivity according to others and one book uses both terms, both represented by $\mathrm{Λ_{m}}$

Now,
$$
R = \rho \frac{L}{A}\\
R = \frac1C\\
$$
Therefore
$$\frac1\rho = \frac1R\cdot\frac{L}A = C\cdot\frac{L}A = \kappa$$
Now if I've got this right, then,

Conductance is the degree to which the solution conducts electricity.
Conductivity is the conductance per unit volume of the solution; it may also be considered as the concentration of ions per unit volume of solution.
Molar Conductivity is the conductance of the entire solution having 1 mole of electrolyte dissolved in it.

Q1. So what's Molar Conductance?
Q2. Is there a difference between Molar Conductivity and Molar Conductance?
Also, according to Ostwald's Dilution Law, greater the dilution, greater the dissociation of the electrolyte in solution.
Regarding dilution of an electrolyte solution, this is what I've understood

As dilution increases, Conductivity (ion concentration per unit volume) DECREASES.
As dilution increases, Molar conductivity (Conductance of 1 mole of electrolyte in the total solution) should INCREASE in accordance with Ostwald's Law

Q3. How does dilution affect Molar Conductance?
Q4. How is Conductance affected upon dilution?
I suppose if the above statements are proof-read and the queries answered, I might get fairly good idea about this....
Also if you feel there is are any additional points worth mentioning, by all means go ahead and put it in the answer.
And finally, if anyone could recommend a decent site that deals with the above-mentioned terms and concepts in a fairly lucid manner, it'd be appreciated.

Yes, terminology is often used inconsistently. Start with the Wikipedia article https://en.wikipedia.org/wiki/Molar_conductivity which will answer most of your questions.

Let me add kudos for both diligence to spot such a difference and curious enough to determine the difference. Using terms inconsistently or unnecessarily using a different form of the term drive me crazy. Another huge annoyance is variables. Very frequently a variable x will be written one way in the text, but another in the equation $y=2x$. So is x=$x$ ?!?

To me, it looks like there is nothing useful called "Molar Conductance". The only useful quantity in concentration terms would be Molar conductivity.

I can understand your frustration. The use of terminology is often inconsistent and confused (much to my chagrin). I think you've got the general idea, the conductance ($G$) can be defined as follows:
$$G = \frac{1}{R}$$
i.e. the ease with which a current can flow. As you said,  $$R = \rho \frac{l}{A}$$
one can now identify, $$G = \kappa\frac{A}{l}$$ where the conductivity $$\kappa = \frac{1}{\rho}$$
Molar "any quantity" always has the dimensions (it is helpful to think in terms of dimensions) $\text{"quantity" } \mathrm{mol^{-1}}$
so, it follows molar conductivity $$ \Lambda_m = \frac{\kappa}{c}$$ where $c$ is the molar concentration. It is useful to define molar conductivity because, as you already know, conductivity changes with concentration.
Now, to address the effect of change of concentration on molar conductivity, we need to consider the case of weak and strong electrolytes separately.
For a strong electrolyte, we can assume ~ 100% disassociation into constituent ions. A typical example is an $\ce{MX}$ salt like $\ce{KCl}$
$$\ce{MX} \rightleftharpoons \ce{M^+} + \ce{X^-}$$ The equilibrium constant for this reaction is
$$ K = \frac{[\ce{M^+}][\ce{X^-}]}{[\ce{MX}]}$$ and thus, with decreasing molar concentration of the electrolyte, equilibrium shifts towards the disassociated ions.
At sufficiently low concentrations, the following relations are obeyed:
$$\Lambda _{m}=\Lambda _{m}^{0}-K{\sqrt  {c}}$$
where $ \Lambda _{m}^{0}$  is known as the limiting molar conductivity, $K$ is an empirical constant and $c$ is the electrolyte concentration (Limiting here means "at the limit of the infinite dilution").
In effect, the observed conductivity of a strong electrolyte becomes directly proportional to concentration, at sufficiently low concentrations
As the concentration is increased however, the conductivity no longer rises in proportion.
Moreover, the conductivity of a solution of a salt is equal to the sum of conductivity contributions from the cation and anion.
$$ \Lambda_{m}^{0}= \nu_{+}\lambda_{+}^{0}+\nu _{-}\lambda _{-}^{0}$$
where:
$ \nu _{+}$ and  $\nu _{-}$ are the number of moles of cations and anions, respectively, which are created from the dissociation of $\pu{1 mol}$ of the dissolved electrolyte, and $\lambda _{+}^{0}$, $\lambda _{-}^{0}$ are the limiting molar conductivity of each individual ion.
The situation becomes slightly more complex for weak electrolytes, which never fully disassociate into their constituent ions. We no longer have a  limit of dilution below which the relationship between conductivity and concentration becomes linear.
We always have a mixture of ions and complete molecules in equilibrium. Hence, the solution becomes ever more fully dissociated at weaker concentrations.
For low concentrations of "well behaved" weak electrolytes, the degree of dissociation of the weak electrolyte becomes proportional to the inverse square root of the concentration.
A typical example would be a monoprotic weak acid like acetic acid (again, from your graph):
$$\ce{AB} \rightleftharpoons \ce{A^+} + \ce{B^-} $$
Let $\alpha$ be the fraction of dissociated electrolyte, then $ \alpha c_0$ is the concentration of each ionic species. And $(1 - \alpha)$, and $(1 - \alpha)c_0 $ gives the fraction, and concentration of undissociated electrolyte. The dissociation constant is:
$$K = \frac{\alpha^2 c_0}{1-\alpha}$$
for weak electrolytes, $\alpha$ is tiny, so the denominator is nearly equal to one so, $$ K \approxeq \alpha^2 c_0$$ and $$ \alpha = \sqrt{\frac{K}{c_0}}$$ (like I said earlier)
for conductivities, one can now write the following relation
$$\frac{1}{\Lambda_m} = \frac{1}{\Lambda_m^0} + \frac{\Lambda_mc}{K (\Lambda_m^0)^2}$$ This fits the curve seen in your graph.
Caveat, all these arguments hold for dilute solutions. Things get out of hand at high concentrations, and one has to account for some additional phenomenon (for example, acetic acid will form hydrogen bonded dimers).
Anyway, long story short, conductivity increases with increasing dilution (though differently for strong and weak electrolytes). From the definitions I outlined at the very start, guessing how conductance changes is trivial.

Let's get the terms right, firstly, because that seems to generate some confusion.
Conductance=ability of a component to conduct electric current (it might also refer to heat and others, but we'll stick to the ones in your example for now).
It depends on: the material of your component (that is actually its conductivity), the length of the component (l) and the cross section of the component (A). It's unit measurement is S(from conductivity)*m(from length)/m2 (from area) which yields S/m.
Conductivity is the ability of a material to conduct electric current, regardless of its dimensions. It is also called specific conductance. And it's the reverse of resistivity, measured in S (siemens, mentioned also above).
If I have a cell/component of 1 sqm cross section and 1 m length, conductance and conductivity would be equal in value. If these are different values for the component Im computing, I adjust the conductivity of the material in general with the dimensions, and get the conductance of my particular component (such as a conductor, piece of metal, electrolyte environment etc)
So I would say about a metal that it has high conductivity and about a rod (or other specific piece) of the same metal that is has a (assumably) high conductance.
Now, moving to the questions:
Q1: Molar Conductance = conductance of all the ions produced by ionization of 1 g mole of an electrolyte when present in V mL of solution. Mathematical relation is miu=k * V ; k is the conductivity=specific conductance
Q2: Molar Conductivity=lambda= k/ molecular concentration which makes it equal to molar conductance per mole. So this is pretty much the difference between them, if we could name it as such.
Note that these refer to ionic conductivities, for electrolytes.
Q3: The variation of molar conductance with concentration can be explained on the basis of conducting ability of ions for weak and strong electrolytes. A strong electrolyte is a compound that is totally dissociated in water in its component ions while a weak one is an electrolyte where we have present the compound in itself, as well as component ions separately. I remind you that ionic conductivity/conductance both rely on the presence of ions in solution.
For weak electrolytes: as we increase the volume (thus the dilution), we encourage the equilibrium to shift towards dissociation. Therefore, we will have more ions in solution and a implicit increase in molar conductance. It basically means that 1 mole dissolved into 10 mL of water gives out less ions than 1 mole of the same, in the same conditions, in 100 mL of water.
For strong electrolytes, the observed effect is the same (molar cond. increases with V increasing), but the reasons are different. In concentrated solutions (less solvent) the strong opposingly charged ions tend to attract each other. Give them enough space  (i.e increase V) to move freely of these attractions and you get an increased molar conductance.
Q4: Conductance, we said is conductivity* cross area/ length of component, and the reverse of resistance. So conductance=current intensity/tension; The intensity of the current in turn, is directly proportional with ion concentration (no of ions/volume) and ionic mobility (nevermind this for a bit). Therefore, if we have lower ions concentration (same ions/bigger volume), we will have a smaller current and hence, a decrease in conductance.
Hope these help with sth. If you have some follow-up questions, please feel free. In the meanwhile, I will think of some resources to send you as trustworthy source.
Cheers.

There is no separate mathematical relation like μ = κ* V. And it looks like there is no useful term called Molar conductance (μ). The actual relation is : Molar conductivity(Λ) = κ* V = κ/c , The V is not arbitrary, it is volume of solution containing 1 mol of ions in a cell of indefinite area  BUT Having electrodes 1 cm apart, then only  κ* V will make some fixed sense.


text:
How and why can more volatile acids be made with sulfuric acid?
I have read that sulfuric acid has low volatility. Then it states that more volatile acids can be made using sufuric acid and their corresponding salts, because of the low volatility.
Can you help me understand the chemistry behind the second statement?

The gist is that you could take concentrated sulfuric acid, dilute it 50:50 by volume with water and add sodium chloride. You could then boil off HCl. Not an efficient way to get HCl but it would work.

What does volatility have to do here?

What is boiling off in the reactions I outlined? Sulfuric acid or hydrochloric acid? Isn't the first one to boil off "more volatile"?

An acid according to the Brønsted-Lowry definition is any substance that can release $\ce{H+}$ ions. By having done so, it becomes a Brønsted-Lowry (conjugate) base which can accept $\ce{H+}$ ions. This is shown in equation $(1)$:
$$\begin{array}{ccccc}\ce{HA} & \ce{<=>} & \ce{H+} & + & \ce{A-}\\
\text{acid} & & \text{proton} & & \text{conjugate base}\end{array}\tag{1}$$
As already hinted, a Brønsted base is any substance that can accept such an incoming $\ce{H+}$ ion (reaction $(2)$)— as such, an acid will always react with a base and vice-versa; it is not possible for only an acidic reaction to happen.
$$\begin{array}{ccccc}\ce{B} & + & \ce{H+} & \ce{<=>} & \ce{HB+}\\
\text{base} & & \text{proton} & & \text{conjugate acid}\end{array}\tag{2}$$
Let me quickly quote part of the question:

[M]ore volatile acids can be made using sulphuric acid and their corresponding salts

The corresponding salt of a more volatile acid could also be termed its conjugate base. So the reaction we are talking about includes the conjugate base of one acid and sulphuric acid as a second acid. To exemplify things, I will use hydrochloric acid as the first acid whose conjugate base is chloride. If we thus mix chloride ions (e.g. in sodium chloride) with sulphuric acid, the following equilibrium will result:
$$\ce{Cl- + H2SO4 <=> HCl + HSO4-}\tag{3}\\[1.2em]
K = \frac{[\ce{HCl}][\ce{HSO4-}]}{[\ce{Cl-}][\ce{H2SO4}]} = \frac{K_\mathrm{a} \left ( \ce{H2SO4} \right )}{K_\mathrm{a} \left ( \ce{HCl} \right )}$$
The equilibrium constant $K$ is the ratio of the acid constants of hydrochloric and sulphuric acid. Since hydrochloric acid ($\mathrm{p}K_\mathrm{a} \approx -8$) is more acidic than sulphuric acid ($\mathrm{p}K_\mathrm{a} \approx -3$), we expect the equilibrium to be shifted towards the reactant side — which would be true for a closed, pressurised system. Once we open the system, and allow for gases to be exchanged, the picture changes. Compare equation $(\text{3'})$, wherein the aggregate states of the reaction partners are mentioned.
$$\ce{Cl- (s) + H2SO4 (l) <=> HCl (g) ^ + HSO4- (s)}\tag{3'}$$
Hydrochloric acid is a gas and can diffuse away while sulphuric acid is a liquid and thus cannot diffuse away. (And cannot drip away since we are using a proper beaker of some kind.) If we constantly remove $\ce{HCl}$ from the reaction mixture by letting it diffuse away, we are exercising pressure onto the system, and per Le Châtelier’s principle the product side (one of whose species we are removing) will now be favoured.
This general principle applies to any acid, as long as it is more volatile than sulphuric acid, i.e. has a lower boiling point under the pressure used. If the second acid happens to be liquid rather than gaseous, a simple distillation will suffice to remove it from the reaction mixture.

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text:
how can electrons exchange from 3d orbital to 4s if their energy levels are not equal?
I was taught that electrons only exchange in orbitals of equal energy or degenerate orbitals. The energy of 3d orbitals is 5 and that of 4s is 4(by n+l rule), but then how is it that in the case of copper and chromium electrons exchange from 3d orbital to 4s?

Where do your energy numbers come from? Which unit are they in?

What does "exchange" even mean... There is no promotion of electrons in Cr and Cu, that is just how their ground state is... It is not like God created all Cr atoms as 4s2 3d4 and then the Cr atoms realised "hey it makes more sense for me to exchange one 4s electron to 3d" so they moved one and are happier now.

A sort of simplified answer but here goes...
Without some other excitation, electrons only "fall in" towards the nucleus. This is sort of like the fact that rocks roll down hill not up hill. So once an electron "falls in" it has excess energy that must be shed. It can either give the energy off as a photon or transfer the energy to one or more other electron(s) which then go into some excited state (Auger effect).
The transitions occur  according to selection rules. If the selection rules don't allow a particular transition then it is forbidden. If the atom is in free space with no other perturbations, then forbidden transitions would not occur. However for molecules, in liquids and gases, because of vibrational and rotational microstates such a transition may actually occur but it will be relatively weak in intensity to the transitions allowed by the selection rules. Likewise forbidden transitions may occur in solids because of the vibrations of the atoms within the crystal which break the symmetry of an atom, or its excited state.
Now if an atom or molecule is excited by an external energy source, then an electron can move from a lower energy orbital to a empty higher energy orbital. Again these transitions obey the selection rules.

so as not to confuse/clarify, perhaps you should mention that rotational and vibrational levels  only occur in molecules and don't involve electrons anyway. Breaking selection rules is usually a result of interaction by induced or external magnetic & electric fields, such as by spin-orbit coupling, Zeeman & Stark effects. Collisions can also induce transitions.

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text:
Why is the formation of NaCl an exothermic process?
My book says,

When NaCl forms, what really happens is that $\ce{Na+}$ and $\ce{Cl-}$ come together and form an ionic bond and thus NaCl is formed and it is a quite exothermic reaction.

Why and from where is this energy coming from? Ok Ok, I read that every system wants to reduce energy, but where literally is this energy coming from, is it because of some electrons or from the nucleus? Or from where?

To be honest I think this is bordering on a physics.SE question. You are essentially asking "why is it more favourable for positive and negative charges to be next to each other". Answer: electrostatic forces.

@orthocresol Who says that? I never asked for that! I only asked about the exact reason that why, when an electron goes from Na to Cl, there is a HUGE release of energy?

Because sodium becomes positively charged, chlorine becomes negatively charged, and positive and negative like being next to each other. No need to shout at me.

Haha, It's who whose shouting now. What do you mean by they "like" to be next to each other?  This ain't philosophy my friend!

@orthocresol True, but OP’s doing his best to ask it in a chemical way …

Why and from where is this energy coming from?..

If we go much deeper it can be explained physically by the famous equation (energy mass equivalence) $ E = M C^2 $. So the energy comes from the mass defect which gets converted to energy.( I hope you know what mass defect is). As we know that energy released in a chemical reaction is very small in comparison to a nuclear reactions, so the mass defect in chemical reactions is also very small in comparison to nuclear reaction. So in general we do not take into consideration the mass defect in a chemical reaction(since it can be neglected) So we generally explain evolution of energy on the basis of other energies but all of these point again to energy mass equivalence.

The negative standard enthalpy of formation of sodium chloride (i.e. ‘the formation is exothermic’) is primarily due to the formation of a crystal lattice. This crystal lattice is a macroscopic arrangement of ions in a cubic fashion with each sodium ion surrounded by six chlorides and each chloride ion surrounded by six sodium ions.
If it were not for the generation of this lattice and its inherent stabilising properties (a very good charge distribution, charges neatly alternate, contact between species of identical charge is strongly inhibited) the formation of sodium chloride would be much less exothermic.
Note most importantly, that it is not merely the generation of sodium cations (that process is endothermic) or the generation of chloride anions (that process is only weakly exothermic) that causes the overall exothermicity of the entire reaction.

"Note most importantly, that it is not merely the generation of sodium cations (that process is endothermic) or the generation of chloride anions (that process is only weakly exothermic) that causes the overall exothermicity of the entire reaction" . This is exactly , what my question is. I want to know that what causes the sodium conversion to sodium anion to loose energy? I know it's getting stable, but what really is causing it to loose energy?

@AaryanDewan That’s not exactly the question you asked, but here is the answer: Electron and nucleus attract each other. To strip away the electron, you need to overcome the attraction. That costs energy. Note that calling metal cations ‘stable states’ is typically wrong: If something lost an electron by ionisation (as metals do if they become cations) that will always cost energy and thus metal cations are never more stable than the pure metal.

The answer is lattice energy.
The enthalpies involved in the formation of ionic compounds can be understood by Born-Haber cycle.
The following steps are involved in the formation of $\ce{NaCl}$ for example.
$$\ce {Na(s) -> Na(g)}\tag{1}$$
$$\ce{Na(g) -> Na+ + e-}\tag{2}$$
$$\ce{1/2 Cl2(g) -> Cl(g)}\tag{3}$$
$$\ce{Cl(g) + e- -> Cl^{-}(g)}\tag{4}$$
$$\ce{Na+(g) +Cl^{-}(g) -> NaCl(s)}\tag{5}$$
You can see that energy is required in the steps 1-3 whereas energy is evolved in 4th and 5th steps.
The energy evolved in the 5th step is called the lattice energy of the ionic compound and this only is responsible for the overall process being exothermic.
Now why lattice formation is exothermic? This is because when a lattice is formed, new attractive forces are set up that are dominant over the new repulsive forces set up. As it is known that attractive forces decrease the potential energy of a system(or increase the stability), so energy is released as heat.

Thanks for the answer! But, Please tell me the exact reason for the reduce of energy. Born-Haber cycle and Lattice energy are words, not explanations!

At the first step in the formation of NaCl, the solid Na should be converted to gaseous Na by the process sublimation so there comes sublimation entalpy, secondly the gaseous Na should lose one electron to become a Na+ ,so there comes the role of ionisation entalpy and if we see the case of chlorine molecule at first they need to dissociate to 2 cl and of electron affinity they should get an electron becoming stable. At last we need to bring both of them to an equilibrium position such that the released energy is sufficient for them to make a bond (lattice energy) then in the same Na+and Cl-ions the electron and the protons in the nucleus attract so some energy is again spend against that. Thus while looking overall we can under that the process of formation of NaCl is an exothermic process


text:
How to store gas
For an experiment with an Arduino I want to put some helium in a balloon, and I want to be able to put more gas in it, electronically. My idea was to do this with the help of a small gastube, ( < 1 L) and connect this gastube, with pressurized Helium in it to the balloon, with a valve in between, which I then again can control with the arduino. However, I can't find such a gas tube/ gas cylinder. Any help or comments on how or where to find such a cylinder?
Thanks!

Your local party supply store will probably rent you a compressed helium cylinder. I have seen helium for balloons sold in the same size tank that is used for propane grills. You would need some kind of pressure regulator between the cylinder and your apparatus.


text:
Saltwater Desalinization
If a bucket of seawater is put into a centrifuge, would the spinning force be enough to move the salt in the solution toward the outer wall of that centrifuge?

This will not work. Centrifugation will only separate heterogeneous mixtures by density. For example, when centrifuging blood, the cells and platelets will separate to the bottom of the tube because they are not in solution. Depending on the lipid concentration in the plasma, a lipid layer may phase separate (not in solution) and move to the top.
The salt in sea water is dissolved in solution. It will not separate. Even if the centrifugation would produce a concentration gradient across the solution, diffusion would counteract that gradient.

Thank you Ben. Excellent and simple explanation. And as I expected. :-(

Sorry. Desalinization is done commercially by distillation or reverse osmosis.


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How to determine the oxidation states of the metal atoms when both cation and anion are coordination complexes?
How to determine the oxidation states of the metal atoms when both cation and anion are coordination complexes?
For example, how to determine the oxidation states of $\ce{Pt}$ in the following compound?:
$[\ce{Pt(NH3)4}][\ce{PtCl6}]$

You just have two different oxidation states. Pt in the cation is +2, Pt in the anion is +4. Saying that the average oxidation state is +3 is technically correct, but isn't really useful (imo) and can potentially be misleading.

Related: IUPAC name of a salt made of both cationic and anionic complexes

There is something called an average oxidation state, which is typically not very meaningful chemically, but can (usually) easily be calculated by just knowing the structural formula. In your case, since chlorine is the most electronegative element (save nitrogen), you can assume it to be in $\mathrm{-I}$ state while nitrogen of course would have $\mathrm{-III}$ and hydrogen $\mathrm{+I}$. Thus to get platinum’s average oxidation state just sum up to the compound’s charge:
$$\begin{align}2 x + 4 \cdot (-3) + 3\cdot 4 \cdot (+1) + 6 \cdot (-1) = 0\\
2x - 12 + 12 - 6 = 0\\
2x -6 = 0\\
2x = 6\\
x = 3\end{align}$$
Which gives you an average oxidation state of $\mathrm{+III}$.
But as I said before, average oxidation states do not carry any physical or chemical meaning and can only serve as exercises to students and pupils. Take for example tetrathionate’s ($\ce{S4O6^2-}$) average sulphur oxidation state of $+\frac{5}{2}$.
Instead, one should always attempt to allocate per-atom oxidation states you cannot do that without knowing the charge of each of those complexes, and you need to do it per complex; ideally by drawing it, heterolytically cleaving all bonds (or homolytically for homoatom bonds) and then counting electrons. In the case of your compound — tetraamminplatinum(II) hexachloridoplatinate(IV) — you should arrive at the numbers in brackets in the IUPAC name I gave you, knowing that the cation has $2+$ charge and the anion $2-$.

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text:
ppm as hexane/propane/methane etc
This question is primarily related to stack testing data.
Whenever stack testing is being conducted and the parameter being measured is VOC concentration, it is often reported as ppm (as hexane) or ppm (as propane) and so on.  My question is: what exactly does it mean physically to say ppm as hexane vs. ppm as propane or heptane and so on?
My apologies if it's a stupid question.  I think my question probably gets into a little bit about how a flame ionization detector actually works.

ppm is parts per million.  So 1 mg in 1 kg is 1 ppm.  Also, 0.1 percent is 1000 ppm.

@DarrenRhodes not necessarily. If it were ppm by volume (which for gases is the same as per mole) it would be different, hence the need to specify whether the metric is mass/mass or vol/vol. And, in this case, the result might be normalised to a specific hydrocarbon, in which case it might be total voc mass/kg of gas.


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Atomic radius definition disproved - Bohr's model to Schroedinger's wave equation
According to the Schrödinger wave equation, Bohr's model of atomic orbitals is disproved and a new formalism that describes atomic orbitals is introduced.
It follows that the definition of atomic radius - "the distance between the center of the nucleus to the outermost orbit" - doesn't make sense, given that the new theory does not describe atomic orbitals as discrete entities in space.
If an orbital has no exact boundary, then what exactly is the definition of atomic radius and how can I calculate it?

The most probable distance of the distribution is used.

I'm not sure of the answer that's why I'm answering in the comments. The atomic radius is measured in following forms:1) Covalent radius 2)Metallic radius 3)Vander waals radius. Measurement of any of these radius does not involve the concept of orbitals.  It involves the distance between the atoms in bonded state. The half of the bond length gives the approx radius. That's why the radius of halogen is much larger than expected(as Vander Waals radius is measured in their case).

How do you measure radius of elements in their elemental stste???

I'm going to keep this simple and not give a book length explanation. A Cl atom would have many kinds of radii. First what would be the radius of a Cl atom out in free space. Second what would the nuclear distance be between Cl atoms in the $\ce{Cl2}$ molecule? Half that distance would be another radius. Third what would be the distance be between a $\ce{Cl^-}$ ion and various cations such as $\ce{Na^+}$, $\ce{K^+}$, $\ce{Rb^+}$, $\ce{Ca^{2+}}$ and so forth. // For diatomic gases you could measure the distance by the rotational moments of the molecules. For solids you'd use x-ray diffraction.

As you may know, atomic orbitals are wave functions, solutions of the Schrödinger equation for an atomic system.
In a perfectly spherical system you may express an orbital as a function depending on the distance from the nucleus ($r$) and two angles ($\phi$ and $\theta$).
[If you pick a particular $r$, the angles $\phi$ and $\theta$ work in a way similar to how we locate points on the surface of the earth (latitude and longitude).]
This being said, the image below may clarify things and answer your question (source):

The graph deals only with s atomic orbitals of hydrogen, the 1s highlighted in red.
See the x-axis? It is measured in multiples of $a_0$ a.k.a. Bohr radius,
$$a_0 = \frac{4 \pi \epsilon_0 \hbar^2}{m_e e^2} = \frac{\hbar}{m_e c \alpha}$$
This is the radius obtained from Bohr's theory for the 1s orbital of hydrogen.
How can it the useful as a unit in this graph?
Well, the square of a wave function is a probability distribution, $\rho(r, \phi, \theta) = \Psi^2 (r, \phi, \theta)$.
If you had a dice with probability distribution $\bar{\rho}(n) = 1/6$ (a fair dice) for each $n = 1,..., 6$, you would get, throwing that dice, on average,
$$\sum_{n=1}^6 \bar{\rho}(n) \times n = \sum_{n=1}^6 \frac{n}{6} \\= \frac{1}{6} + \frac{2}{6} + \cdots + \frac{6}{6} = 3.5$$
So $3.5$ is the expected value of throwing a dice.
See how the last equality is just the usual average we're familiar with?
But the first one is most useful in our context.
Since our variables $r$, $\phi$ and $\theta$ are all continuous, in order to get an average radius, we integrate instead of sum:
$$\int_0^{2 \pi} \int_0^{\pi} \int_0^\infty [\rho(r, \phi, \theta) \times r] r^2 \sin(\phi) d r d \phi d \theta $$
$r^2 \sin(\phi) d r d \phi d \theta$ is just the volume element in spherical coordinates (in our dice example we had an implicit $1$ there as "volume"), what matters is just what is in squared brackets.
Now 1s ($n = 1$, $\ell = 0$) orbitals do not depend on angles, they are perfectly spherical.
That means that $\Psi_{n \ell m_\ell}(r, \phi, \theta) = R_{n \ell} (r)$ with $n = 1$, $\ell = 0$ and $R_{n \ell} (r)$ is some function describing the radial electron wave function.
Our integration simplifies to:
$$\int_0^{2 \pi} \int_0^{\pi} \int_0^\infty [R^2_{n \ell} (r) \times r] r^2 \sin(\phi) d r d \phi d \theta \\
= \int_0^{2 \pi} \int_0^{\pi} \sin(\phi) d \phi d \theta \int_0^\infty [R^2_{n \ell} (r) \times r] r^2 d r \\
= 4 \pi \int_0^\infty [R^2_{n \ell} (r) \times r] r^2 d r \\
= \int_0^\infty [ (4 \pi r^2 R^2_{n \ell} (r)) \times r ] d r $$
This surely looks like the expected value of throwing a dice!
The variable being averaged is $r$ and the new probability distribution is $\rho_\text{radial} (r) = 4 \pi r^2 R^2_{n \ell} (r)$, called radial probability distribution.
That's what's being plotted in the figure above.
If you go further and carry out the last integral ($R_{1 0} (r)$ can be found here), you'll see the answer is just $a_0$, the Bohr radius.
That's expected, after all, since Bohr's theory was quite successful for the hydrogen atom.
Nothing has been disproved.
Thus, the atomic radius of an atom can be obtained from quantum mechanics by calculating the expected value of the distance between the outermost electron and the nucleus, as MaxW has pointed out.

I am unaware of ANY definition of atomic radius that states it is the Bohr radius, please provide the citation for your source for that claim.
The Bohr model works fairly well with the H atom in/near the ground state, why do you claim it is "disproved"? Again, please provide the citation. It is true that the Bohr model fails (badly) for more complex atoms (is this what you mean by "disproved"??)
As already noted, "atomic radius" has a number of different definitions. Interatomic distance depends on temperature and pressure, as you probably know. You ask what the definition is, and we can't answer without you also describing the context - what use would you make of it? (There is no one definition that fits all uses.) The crudest measure would be to determine the volume of a single atom (or molecule) based on its density.
g/cc→moles/cc→cc/mole→cc/particle
You could then assume each unit volume is a cube and its radius is ½ of the length of one edge, or you could assume that the units are spheres and have maximum packing ( = π/(3√2)) of that volume.
X-Ray or neutron (etc.) diffraction determines the structure and spacing of crystals, which can easily give interatomic nearest neighbor distance in solids. In some liquids and amorphous (and semi-crystalline) solids too (using synchrotrons). The concept of radius breaks down in considering ions in solution, since ions are generally enveloped in a solvation sphere so that anion to cation distance is a statistical value determined by both of the ion-solvent interactions. There is also a definition of cross-section (πr²) for atoms in the vapor phase (gasses and plasmas). And once covalent interactions become significant, hard sphere models of structure aren't going to work very well at all.
Comparing elements crystal structures at STP (see Wikipedia Periodic_Table_ (crystal_structure)) indicates that most of the elements you will encounter have near optimal packing, unfortunately the elements of the "most" chemical interest to us (Groups 13-17, aka "the p-block" elements) rarely have optimal packing, so the above rough calculation isn't very good for them.


text:
Can you get drunk by just smelling an empty vodka bottle?
Is it possible to get drunk but just smelling an empty vodka bottle?
I got this empty Skull Vodka bottle from my mother which I thought looked really cool but it still smells a lot like vodka in it. I just want to make sure this will not affect me in any way.

No. You will be fine.

Hmm... was the bottle empty before or after you started "smelling"?

Wash it with water, and let it dry. It will not smell of vodka anymore.

Technically, you could get drunk from smelling vodka. Like everything else that you can sense smelling, you uptake a little bit of the vodka (mainly the alcohol, i.e. ethanol component) by smelling it.
But those quantities are really minute. You would have to take a hot vodka vapour bath and stay in there for quite some time until you realise something. Remember that the amount you need to drink to get drunk varies, but is certainly somewhere above two shots. Now imagine how long it takes for ‘two shots’ worth of vodka had to condense in your nose and mouth from breathing — almost forever.

The vodka need not condense anywhere. It can be taken directly into the blood by the lungs. This is the basis of devices which create breathable mists of liquor, and it does not take long to get drunk from those.

@Nij I know, condense is the wrong word but I didn’t know a better one. Absorb maybe? Also, what kind of a throughput do those devices have?

No. You won't get drunk from the scent of an empty bottle. If you could, it wouldn't be necessary to drink the vodka!


text:
What's causing my energy calculations to be wrong?
I'm calculating the energy density of certain flammable compounds in MJ/kg. To do this I compare the sum of the bond strengths of the reactants to the products and then extrapolate the energy density from the difference in the molar binding energy of each compound.
For methane my result was almost exactly 50 MJ/kg while the Wikipedia result says it's 55 MJ/kg.
For ethanol my result was 28.6 MJ/kg while Wikipedia says it's 26.4 MJ/kg
I don't expect my calculations to be spot on but I didn't expect them to be this off.
What are the possible reasons for my calculated energy density to be quite different than the supposed actual values?

Do you assume the product water being in liquid form or vapour form? The value of 55.5 MJ/kg for methane includes water in liquid form. The corresponding value for water in vapour form is 50.0 MJ/kg.

So the hydrogen bonds in the liquid water contribute to the number found on Wikipedia? If so then do you have any idea what happened with my ethanol calculation? Thanks for the feedback.

As far as I know, the formula involving bond enthalpies is approximate and valid only when all reactants and products are in gaseous form. This is because the formula neglects the intermolecular forces of attraction. SO the results you got I would say are approximately correct considering you obtained them from bond enthalpy calculations.


text:
Is there such a thing as a Boron-Carbon-Lithium alloy?
I am trying to create an alloy with heat resistance for a project. Is there such a thing as a boron-lithium-carbon alloy, and are there any benefits to it?

In the study Compound Formation in the Systems Lithium-Carbon and Lithium-Boron, Journal of the American Ceramic Society, Volume 50, Issue 10, October 1967, Pages 520–523
molten lithium metal was added to $\ce{B_{4.0}C}$ in proportions ranging from 7 to 71% lithium.

a  substitution  reaction  evidently  occurs  between
  lithium  and  boron  carbide  which  leads  to  the  formation  of
  lithium  boride  and  lithium  carbide

Chemistry does not work quite like that. Going around the periodic table and poking elements at random (or just for the sake of their atomic number, disregarding other properties) is hardly leading to any good.
As for the $\ce{Li-B-C}$ ternary system, I strongly suspect that it has not been studied in full. (Good to know it has been studied at least partially, but there might be many more unknowns to it.) Also, I suspect that there are some compounds containing all three elements (you may call them alloys if you like, though most chemists wouldn't), but those would be unsuitable for your project because of other requirements which you didn't specify. Do you want your alloy to be stable in the air, for starters?

Ok. I should have gotten a little more information beforehand. I was looking for an inert element, that was very light, preferably with some heat resistance. I was using Lewis dot structures to try to form the compound. Thanks for the answer

Yeah, I thought so. Well, these things are going to be light for sure, but hardly inert.


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what's the difference between sand and quartz?
What's the difference between sand and quartz?
Is quartz more crystalline form of sand? Do they have the same composition or my be not? or my be the two words is similar? Do they contain more than silica?
Thanks.

I'm voting to close this question as off-topic because the answer is very easily Googled.

atleast you gave a reason for the thumbs down, I h8 it when they dont give a reason.

Sand is a poorly-defined mixture, not a necessarily a pure substance. Some sands are fairly pure $\ce{SiO2}$, the same chemical formula as pure quartz, but in some areas of the world, sand can be made of entirely different chemicals, such as calcite.
Quartz is a particular crystal form of $\ce{SiO2}$, but $\ce{SiO2}$ also forms the minerals tridymite, cristobalite, coesite, and stishovite, amorphous silica, etc. It's like having a collection of Legos or TinkerToys that have the same pieces, but can be assembled in different formats.

Thanks a lot for the informative explanation, so quartz is only one compound (silica) and sand contain more than one compound not only silica

Possibly... some sands are reasonably pure SiO2, others are fairly random mixtures.

quote from wikipedia

The composition of sand varies, depending on the local rock sources and conditions, but the most common constituent of sand in inland continental settings and non-tropical coastal settings is silica (silicon dioxide, or SiO2), usually in the form of quartz.

Sand is a heterogeneous mixture
Quartz is compound

You might want to expand on this - it's a borderline HW question and your answer should at least include the relevant text from the Wikipedia entry for quartz (and not just the one for sand).

I think you can do a better job, shall you?

I can certainly spell better - editing your answer now.

you converted my ego into a heterogeneous mixture


text:
What is the perfect definition for chirality?
Why is chirality defined differently for organic and inorganic compounds?
Why are inorganic compounds deemed to be optically active if they have more than one of the same ligands attached to the central metal atom, but glycine is optically inactive for having two $\ce{H}$ atoms attached to the carbon atom?
Finally, what should be the criterion for chirality (optical activity) and the perfect definition for it?
Could anyone explain how $\ce{$cis$-[PtCl2(en)2]^2+}$ is optically active, but the $trans-$isomer is not?

Definition of chirality is rooted in symmetry. Chiral molecules are dissymetric (i.e lacking $S_n$). I can provide a more detailed answer with some examples later

Chirality is not defined differently for organic and inorganic compounds.

@Jan a carbon with four different groups is chiral but a metal with four different ligands may not be chital. In complex compounds the metal can have same ligands and be chiral but a carbon with two same groups is achiral.

@JM97 That is a simplification at best. Chirality is never defined by the number of groups attached to a specific atom. It is always defined either by the mirror image or the improper rotation definition. (The two definitions are mathematically identical, though.)

The correct definition of chirality is given in the IUPAC gold book as follows:

chirality
  The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane,  σ = S1, a centre of inversion,  i = S2, a rotation-reflection axis, S2n). If the object is superposable on its mirror image the object is described as being achiral.

From this we can deduce, that all organic compounds, which have tetrahedral carbons with four different ligands, are chiral. This is a rule of thumb and an easy way to recognise chirality.
Glycine is optically inactive as it has an internal mirror plane:

Often we teach this as a rule or a definition, because it is so easy to comprehend and to see.
In inorganic chemistry, we usually deal with more complex structures, where it is not necessary, that all ligands have to be different. One example for this is the cation in tris(ethylenediamine)cobalt(III) chloride:

Another interesting example is (Λ/Δ)-cis-dichlorobis(ethylenediamine)cobalt(III) chloride.

I think Martin has provided an excellent answer, and I would like to supplement it with a few additional details and examples that might prove insightful.
So as I already mentioned in the comments, the definition of chirality is rooted in symmetry. A chiral compound can contain no improper axis of rotation ($S_n$), which includes planes of symmetry ($S_1 = \sigma$) and inversion center ($S_2 = i$).
Chiral molecules are always dissymmetric (lacking $S_n$) but not always asymmetric (lacking all symmetry elements except the trivial identity). However, asymmetric molecules are necessarily always chiral.
And thus, chiral molecules have isomers that are non-superimposable mirror images.
This is the definition of chirality and holds true for all compounds, either organic or inorganic.
Chirality can manifest in several ways, and most commonly we talk about point chirality.
This leads us to the notion of a stereocenter--any point in a molecule, though not necessarily an atom, bearing groups, such that interchanging of any two groups leads to a stereoisomer.
A chiral centre is a stereocenter consisting of an atom holding a set of ligands (atoms or groups of atoms) in a spatial arrangement which is not superimposable on its mirror image.
The concept of a chiral centre generalises the concept of an asymmetric carbon atom, i.e., a carbon atom bonded to four different entities, such that an interchanging of any two groups gives rise to an enantiomer. This is probably what you've been taught thus far, and is an easy shorthand.
The above defintions apply not only to tetrahedral carbon atoms, but to any other type of central atom with an appropriate set of bound groups or ligands.
Numerous sulfur derivatives exhibit pyramidal bonding where the non-bonded electron pair located at the sulfur atom acts as a fourth ligand. The same would apply to nitrogen derivatives (tertiary amines), but usually these compounds rapidly interconvert through an trigonal planar transition state (pyramidal inversion) and thus prevent separation into enantiomers.
As already mentioned by Martin, glycine is optically inactive because it has an internal mirror plane. In other words, it would not possess a non-superimposable mirror image. I would encourage you to draw/model the molecule and see for yourself.
The  $cis-\ce{[PtCl2(en)2]}^{2+}$ looks something like this:

This has a non-superimposable mirror image:

The $trans-$ isomer looks something like this (replace $\ce{Co}$ with $\ce{Pt}$):

This molecule clearly possesses a mirror plane, (just like glycine) and would this not have a non-superimposable mirror image, and is thus achiral.
In these propeller like complexes, the stereocenter is clearly the metal atom at the center.
All these examples exhibit point chirality, but this is not the only way in which chirality manifests itself in molecules.
Going back to organic chemistry. $\ce{1,1'-Bi-2-naphthol}$ (BINOL) looks something like this:

This molecule doesn't possess an asymmetric tetrahedral carbon, however it still has two non-superimposable mirror images (given below) and is therefore chiral. This is an example of axial chirality.

Similar to this is a notion of planar chirality that can be observed in compunds like $\ce{(E)-Cyclooctene}$ (pictured below with its mirror images):

Next, you talk about optical activty. I get the feeling that you use the terms interchangeably. However, optical actvity is not the same as chirality. Chirality derives from the symmetry and geometric arrangement of the molecule, and often results in optical activity.
Optical activity is derivative in nature; chiral compunds tend to be optically active. It is is derived from the interaction of chiral materials with polarized light.
I would like to note that it is possible for a compound to be chiral, yet have an optical activity that is hard to measure (cryptochirality)
Update: I thought I'd include a few words on helical chirality, something I wished to talk about initially, but decided against it fearing my post was already way too long. However, as Martin brought it up, I thought I should, like he suggested, through it in here for completeness sake.
Helices are chiral as they can exist in enantiomeric left- or right-handed forms. Typical examples for helical strutures are provided by the helicenes (benzologues of phenanthrene). With four or more rings, steric hinderance at both ends of these molecule prevents the formation of planar conformations, and helicenes rather adopt non-planar, but helical and enantiomeric structures with $C_2$ symmetry. Pictured below are the two enantiomeric forms of 6 helicene.

Additionally, all the examples given here depict manifestation of elements of chirality in molecules, however, chirality can also manifest itself in supramolecular assemblies. For example, the naturally occurring DNA double helix is twisted in a right handed fashion.

While DNA certainly has inherent chirality from it being a monodirectional helix, it is also chiral due to the inherent asymmetric carbons of deoxyribose.


text:
What databases or repositories are there that have generalized SMILES for functional groups?
Does a resource exist that has the SMILES formula of the generalized pattern that represents the group?
For instance: I have the aldehydes group, which has the pattern: -CHO.  Is there a resource that lists each functional group with the generic pattern that defines the group and its SMILES representation?
I'm hoping for, at the very least, getting the SMILES formula for the functional groups represented in this pdf: EPA Chemical Compatibility Chart
Follow up question: Is SMILES still relevant or are there better ways to represent chemicals?

Aldehyde is just C=O

@user34388 Except that C=O matches ketones and carboxylates in addition to aldehydes. The trick with functional group specification is in excluding things that shouldn't be matched, rather than forming things which do. This is why SMARTS, rather than SMILES, tends to be better for substructure searching. SMARTS is basically just SMILES with additions which help in querying.

Thanks @R.M., would you like to post this as an answer so I can give you credit (and build upon it in the comments?)

@lilott8 It's not an answer to your question, though, only some comments about the sort of answers which are likely to be useful.

@R.M. no, ketones are C(=O)

@user34388 I think I understand your confusion. Text searching is a lousy way to look for SMILES-based substructures. While many ketone SMILES do contain C(=O), they don't need to: C1CCCCC1=O is a perfectly valid SMILES for cyclohexanone. Likewise, while aldehydes often contain C=O, they don't need to: O=CC is a valid SMILES for acetaldehyde, as is CC(=O)[H]. Conversely, CCOC=O (ethyl formate) is not an aldehyde. As such, pretty much any decent substructure search program that takes SMILES does not do text searches, and will give back ketones, esters, etc. for a query of C=O.

If it helps, you have a list of common chemical groups and their SMARTS at http://www.daylight.com/dayhtml_tutorials/languages/smarts/smarts_examples.html

It's now a mini graph network object but works in a simple way. Has no dependencies, and if you want you can copy and paste the smiles however.
https://github.com/Sulstice/global-chem/blob/master/global_chem/miscellaneous/open_smiles.py
Code wise, you can follow the tutorial - something like this:

gc = GlobalChem()
node = gc.get_node('open_smiles')

node.get_smiles()
node.get_smarts()

Here's the repository: https://github.com/Sulstice/global-chem

I personally find it educational (even cathartic) to type out my own SMILES and make my own resources to fit my own purpose. For example, I have written out all 20 amino acids in such a way that I could just append them to generate full peptides, proteins from the AA letter sequences, simply by  appending the SMILES blocks ("macros") that I made for myself.
If you want patterns to detect (match) them in bigger molecules, you have a problem because you can't really use regex like pattern matching on SMILES strings. SMARTS was invented for that (see other answer above, which was right on and, to the comment, which I can't reply to, the provided link explains the relationship between SMARTS and SMILES).
But anyway all structural chemical pattern matching requires recreating the graph model of the molecule.
SMILES is great for writing out structures and it's really educational to just write out your own "macros" that you can then put together to larger structures.

I actually think this should be part of the curriculum taught in organic chemistry


text:
Cleaning butyl rubber
In almost every butyl rubber product (stopper, gloves, masks) a thin white, powdery spots appear over time. I have tried scraping it, cleaning it with detergent, etc. But it comes back a day or maybe three after cleaning it. What is causing this? Is debris sticking to the rubber? Is the butyl rubber reacting in some way?

These are made one time use only, Don't use these products again and again. Once used must be discarded, that's what we used to do in our lab.

@Khan I think he is talking about door stoppers, half face protection masks that with replaceable  filter cartridges, and thick gloves.

According to this MSDS, Astlett Rubber, Inc.'s BUTYL RUBBER BK-1675N product contains multiple stabilizers and antiaglomerates, each of which is described as a white powder. I suspect it is the leaching or other means of migration of these stabilizer's to the rubber surface that causes the white spots. The following is an excerpt from the above MSDS:

Butyl rubber contains moisture, stabilizer (Agidol-2 or Irganox-1010)
  and antiaglomerate (calcium stearate or zinc stearate). Stabilizer and
  antiaglomerate are not discharged into the atmosphere during
  processing butyl rubber into finished goods.
Stabilizer Agidol-2 is a white crystalline combustible powder. Flash
  point in the open crucible is 172°C, ignition point is 208°C, the
  standard temperature of spontaneous ignition is 352°C. Agidol-2 is
  permitted for use as a stabilizer of rubber that is used in food
  industry, in water  pipelines, and children toys production.
Stabilizer Irganox-1010 is white crystalline combustible powder.
  Ignition point is 288°C. Meets all requirements for use as a
  stabilizer in rubber th at is used in medicine.
Zinc Stearate is white crystalline powder or paste. The standard
  temperature of spontaneous ignition is 900°C. Maximum permissible
  concentration in work area is 4 mg/m3.
Calcium Stearate is a white powder. The standard temperature of
  spontaneous ignition is 560°C. Maximum permissible concentration in
  work area  is 10mg/m3.


text:
How to get rid of bubbles?
Certain solutions, such as aqueous SDS or soapy waters, after being shaken evolve a lot of bubbles. These bubbles disappear over time, but it can take minutes or sometimes hours (eg. for 20% SDS).
Is there anything that can be done to accelerate the popping of these bubbles? For instance, would heating the solution affect the rate of bubble collapse?
Note: I am interested in methods that do not alter the chemical properties of the solution. For instance, I know that adding certain substances can affect bubbles, but if I added those then the solution would be something different from what I wanted in the first place. It is assumed that procedures such as heating are not sufficiently energetic to cause a noteworthy reaction.

Centrifuge the sample?

Without addition of chemicals? Try out ultrasound (the same physical principle opticians use to clean your pair of glasses).  If you whish, this may be handled even contact less (some background and here), to mention examples.


text:
What are angular nodes?
I know that a radial node is a value of $r$ where the probability of finding an electron is zero, and nodal planes are planes where the probability of finding an electron is zero, but I haven't been able to find a good definition of an angular node. What are they exactly?

Nodal everything are places where probability of finding an electron is zero. Some of them are planes, some are not.

http://chemistry.stackexchange.com/questions/8655/angular-and-radial-nodes I typed a new answer to that question. I hope it answers yours sufficiently as well.


text:
Water-gas shift reaction without catalysts
I'm designing an experimental gas generator. The details are not really important here, but I want to convert a mixture of carbon monoxide, hydrogen and $\ce{H2O}$ gas to a safer form. I know the water gas shift reaction is capable of doing so, but I want to know if it works without a catalyst.
I do understand that the WGSR is an equilibrium reaction, and therefore will not get rid of the carbon monoxide completely, but I hope to at least reduce the amount. This setup will, of course, be done outside and I will be careful not to allow unburned $\ce{CO}$ gases to escape anywhere.

Edit: Someone pointed out to me in the comments that I can remove the carbon dioxide, a useless product of the WGSR, within the reaction vessel to shift the equilibrium towards the products. So the second paragraph is incorrect.

You can use sodium arsenite sscrubber to reduce the CO2 conc. thereby causing equilibrium to shift almost towards products side.

@JM97 That's toxic and carcinogenic -- I'd be trading one poison for another there. Perhaps I can use sodium carbonate.


text:
Is it possible for halogens to show oxidation states less than -1?
Halogens like $\ce{Cl,Br,I}$ generally show oxidation states between $-1$ and $+7$. However, is it possible that they can show an oxidation state beyond $-1$ (e.g. $-3$)?
I was thinking of some compound in which chlorine is bonded with a less electronegative element and forms more than one bond with such element(s).
Is it theoretically possible?

Cl, Br & I typically have oxidation state of -1, expect with compounds containing more electronegative atom than these (N, O & F).

'Typically' they show -1, but can they show beyond that? That is my question

would you like to tell, what is meant by beyond? be precise, just write it instead of letting people to guess

I'm sorry if you didn't understand what I meant by beyond. I meant can they show something like -3,-5 or -7

I don't think so,,,

Yes, they can show oxidation states lower than -1. For example in hyperlithiated compounds, like $\ce{Li3Cl}$, the chlorine would formally have an oxidation state of -3.
However, a word of caution, oxidation states are a tricky thing. They are determined by adopting the hypothetical view that the compound you are looking at consists only of single-atomic ions, e.g. in the case of $\ce{Li3Cl}$ of 3 $\ce{Li+}$ and $\ce{Cl^{3-}}$. The hypothetical charges of these "ions" are then your oxidation states. Clearly, this view has nothing to do with the real bonding situation in the compound and oxidation numbers correlate very poorly with the real electron distribution. It is simply a useful little device for keeping track of redox reactions. So, the -3 oxidation state of chlorine in $\ce{Li3Cl}$ does not mean that chlorine has a 10-electron valence shell. The real bonding situation will be somewhat more complicated.

I wager it's more like Li has 1/3 there.

I agree. We may have structures in which the lithium does not transfer its electrons fully to the halogen. Moreover, the proposed compounds seem to be only computational results, not experimentally realized.


text:
electron affinities of Be and Ca
I have found in the textbook that Be has negative electron affinity while Ca has positive why? how can I compare the electron affinity values of these two.?

You can compare by comparing electronegativities of each of them. In general electron affinity is directly proportional to electro negativity.

explain a bit..


text:
Hydrogen gas' reducing property
I just went though a reaction wherein hydrogen gas is liberated and hence restricts the product formed to ferrous and not ferric salt. I just want to know how H2 gas gets its reducing property? It has a moderate bond strength also if I am not wrong.

"Liberation of hydrogen prevents the formation of ferric chloride."

When HCl reacts with Fe it is H+ which gets reduces and not H2 gas.

H+ reduces or gets reduced to H2?

H+ gets reduced to H2. Also the first comment is my textbook statement please see my views in my answer.

H+ gets reduced to H2. OK. But how can H2 in its gaseous state, reduce Fe+3?

Iron on reaction with HCl forms Fe(II) salts and not Fe(III) salts because even if Fe(+3) is formed, as we can see in electrochemical series Fe(+2) will be favoured ie. H2 which gets can reduce Fe(+3) to Fe(+2). So thermodynamically Fe(+2) salt is favoured. If you want to make Fe(+3) salt then you can react Fe and Cl2 directly.

How H2 gets its reducing property?

H2 gas will react with FeCl3 to form FeCl2 and HCl. I don't know the exact mechanism but it can be explained well on the basis thermodynamic data available.

I agree with the answer. Just thought to say this- 'not in every condition though. At some conditions,may be the reverse reaction would be favoured(by Nernst equation)'

@AmritanshSinghal but the reaction is carried in a closed inert atmosphere of N2 so H2 evolved will not get liberated and react with Fe(+3).

I guess you are correct.

@AmritanshSinghal inert atmosphere is required to prevent O2.


text:
Effect of pKa value of acid component in Mitsunobu Reaction?
Mitsunobu Reaction is used to established C-S, C-C, C-N, C-O bond formation by condensation of acid component (Nuclephile) with primary and secondary alcohol in the presence of triphenyl phosphine (Ph3P) and diisopropyl azadicarboxylate (DIAD).
Literature suggest that Mitsunobu reaction is only possible if The nucleophile (or pronucleophile) is normally a relatively acidic compound containing an O-H,
S-H, or an N-H group with pKa<15, preferably below 11.
My question is that, why Mitsunobu reaction can't occur with acidic compound having Pka value more than 15 (pKa>15)?

I figured out the answe
The desired Mitsunobu reaction proceeds through path (a). On the contrary, in case of the reaction with less acidic HA (pKa>15) , the hydrazo anion (4) attacks the alkoxyphosphonium (5) directly to afford alkylated the hydrazine derivative (6) as a by-product path(b).
Since the anion (4) is not efficient in deprotonating the weakly acidic HA. In order to overcome these drawbacks, the acidic component should have pKa<15 to get desire product.


text:
Gaussian parameters for IP and EA calculation
I have already calculated ionization potentials (IP) and electron affinities (EA) for a whole bunch of organic molecules (from around 20 to 25 heavy atoms, mostly carbon atoms, just a few oxygen or nitrogen atoms here and there). A few molecules are actually a little smaller (6 heavy atoms). I have scrupulously checked the geometry of each and every molecule using the following DFT parameters in gaussian 09:
# freq=noraman cphf=noread b3lyp/6-31g(d) geom=connectivity
 integral=grid=ultrafine scf=maxcycle=1000

and made sure that I got the stationary point in every case.
Then I follwed a published paper in which they applied Koopman's thoerem approximation using HOMO and LUMO energies as IP and EA values, respectively. My results look pretty reasonable in terms of comparing the molecular structures between themselves, and also in terms of the obtained experimental results from a biological system. However, I understand that one can always question the accuracy of a chosen computational method...
I would like to ask you if it would make sense for me to double check the computational results I got with a potentially more accurate method (basis set, functionals), and which one would you recommend in this case? It just so happens that I will have access to a pretty decent cluster (12 cores, RAM 12 GB, HDD 500 GB) with like 1000 hrs/core computation time.
Thanks in advance for any help you can offer.

Koopman's theorem heavily relies on error compensation in the first place, but it is a very good approximation. In almost all cases you should double check your calculations with different methods and basis sets for possible inconsistencies. This discussion of DFT may be very helpful: http://chemistry.stackexchange.com/q/27302/4945

Could you post a DOI to the paper you followed?

If you can afford to do so, it is always a good idea to compare calculations between at least two methods. Good results with B3LYP/6-31G(d) in particular are based upon favorable error cancellation, and B3LYP was not even parameterized against any data in particular (B3PW91 was parameterized for thermochemistry and the PW91 correlation functional was replaced with LYP without changing any other parameters). So, double-checking is wise!
Modern perturbation theory-based approaches have become much more affordable due to the density fitting (DF) and resolution of the identity (RI) approximations (same thing, different name), combined with spin-component scaling which may or may not improve things. The scaling is now roughly $O(N^4)$ and a small prefactor with basis set size; compare this with B3LYP at $O(N^3)$, exact MP2 at $O(N^5)$, CCSD at $O(N^6)$, and CCSD(T) at $O(N^7)$.
Energies are one of the most obvious improvements of range-separated or long-range-corrected (LRC) density functionals. Here the authors find that LC-$\omega$PBEh deviates little from $G_{0}W_{0}$, which is a workhorse for IE and EA calculations. Perhaps the most common new functional that is seeing quite a bit of use in the computational organic chemistry literature is M06-2X, which is shown here to perform even better than most approximate coupled cluster methods for this set of molecules (which are all closed shell!):

With regards to the basis set size, there is only one correct direction to go and that is up. You may be able to afford CCSD(T)/6-31G(d) calculations, but the results will be meaningless due to a severely deficient basis set for anything but the simplest calculations. See here for a recent discussion about the use of basis sets in DFT versus wavefunction methods. Wavefunction methods have always required larger basis sets for quantitative results, especially when not performing a basis set extrapolation. This would mean something like Dunning's correlation-consistent basis sets, augmented with diffuse and polarization functions, such as aug-cc-pVDZ or aug-cc-pVTZ if one can afford it. An interesting and effective truncation of these basis sets (they become very large) is given by Truhlar's "calendar" basis sets.
Less well-known are the requirements for LRC-DFT, which due to modern parameterization schemes are larger basis sets and larger integration grids for the XC contribution, often larger than what is default. Even Pople's triple-zeta sets based on 6-311G will offer some improvement, but the Ahlrichs def2 and Jensen's pcseg-n are probably better.
Finally, a word about the cost of LRC-DFT. Although its formal scaling cost is identical to traditional global hybrids such as B3LYP, in practice range-separated functionals may be 2-3x as expensive due to how the different contributions are calculated and then combined, depending upon the package being used.


text:
Analysis of salts using color change when heated
In order to detect basic radicals one of the preliminary test done is dry test tube heating. My book says that on doing so salts that would appear black when hot and cold are supposed to be Mn, Co, Ni or Cu salt. But does it refer to white salts turning black during heating or salts of any colour including black when it says that they appear black after heating?

This may help-http://www.ncert.nic.in/ncerts/l/kelm207.pdf

@JM97 The problem is not the language of the book neither the colour of the sample that it could be after heating but the condition which may apply(I suppose) to the salts before we heat and observe the colour change.
Do these inferences apply to salts of any colour transforming into the mentioned colours or only white salts?

The cations you mentioned in my knowledge do not form any black color salts. But technically, any color salt which may turn black might be from these cations.

@SanjuktaGhosh could you upload the picture of that paragraph?

AmritanshSinghal Thank you.
@JM97 I'm sorry. I don't have the book at hand, I borrowed it from the library and had returned it. Its from my records that I'm actually quoting.

What is "basic radical" even supposed to mean?

@Mithoron The obsolete (19th century) terms “acidic radicals” and “basic radicals” describe anions and cations, respespectively, in salts. They can still be found in some textbooks concerning qualitative inorganic analysis (in particular in India).

No when cold they can be of different colours as can be seen from the following table:


text:
Examples of molecules with D4 and C9 point groups
Trying to come up with examples of (hypothetical) molecules/shapes that have $D_4$ and $C_{9h}$ point groups.
For $D_4$, I thought a cyclobutane structure, but with 4 substituents coming off of the ring all pointing in the same direction. However, I'm not sure if this is $D_4$ because in a book I am reading it says that all the $D_n$ point groups show chirality, which this molecule does not.
For $C_{9h}$, I really can't think of any shape/structure that doesn't just end up being $D_{9h}$. $C_9$ would be a 9-bladed propellor shape, although I'm not really sure how to draw this, and I don't see how I could alter that to get $C_{9h}$.
Thanks

For nearly any point group which is actually feasible (including $D_4$), look up the Symmetry gallery. Your all-cis tetra-substituted cyclobutane would be $C_{4v}$, BTW.
As for $C_{9h}$... well, I don't quite believe this is possible, but probably we may come up with something that would look almost realistic.
A substituted nanotube fragment will do, I think.

Surely it is hypothetical, so what? Cyclo-$\ce{C18}$ is hypothetical as well, and in my opinion, has lower chances to be stable. Also, its symmetry is kinda dubious.

why is cyclo-C18 hypothetical? wasn't it first observed in 1989? http://science.sciencemag.org/content/245/4922/1088

No, it was not. These guys made some calculations, then prepared some precursors, then saw in the mass spectra some $\ce{C18}$ which is probably cyclic, though nobody knows for sure. That's a circumstantial evidence at best.

@http://iopscience.iop.org/article/10.1086/500110/pdf maybe you will be more convinced by this 2006 gas phase spectrum, although these author say it is a cumulene ring with alternating bond angles, D9h.

a 2011 book says "With this correction, the minimum is the cyclo-polyacetylene C9h. However, there is a tendency for HF as well as BLYP to artificially yielding lower symmetry structures. Thus, the most stable structure of the cyclic C18 is still debatable" https://books.google.com/books?hl=en&lr=&id=hu3FCgAAQBAJ&oi=fnd&pg=PA343&dq=cyclic+c18+c9h&ots=1TUtcWwKlu&sig=5CmIe2xxsGMV6zG24Oq3HJempos#v=onepage&q&f=false

OK, well, so this thing does exist, after all. It is its exact geometry (and consequently symmetry) that is hypothetical.

I'm wondering if anyone tried to make your structure, closest I found is http://pubs.acs.org/doi/abs/10.1021/ja807126u "For [9]cycloparaphenylene, the situation is
slightly more complex. Reduced symmetry due to an odd number of
phenyl rings results in a lowest energy conformation in which the
dihedral angle varies between 18°, 30°, 31°, and 33° around the carbon
nanohoop (Figure 3b)"

That's another story. Paraphenylene structure is all wobbly; mine is rigid.

so will it twist less, or not at all? The unsubstituted version of your structure is structure number 180 here http://pubs.rsc.org/en/content/articlehtml/2015/cs/c4cs00366g , but I don't think anyone succeeded in making it yet

Yes, I believe chances are pretty high that it will not twist. (Not that the molecule would be easy to make, though.)

The cyclo-C18 polyyne is $C_{9h}$ according Carbon Rich Compounds II: Oligoacetylenes at page 46
The molecule is a planar 18-carbon ring with alternating short and long bonds (triple and single bonds) and bond angles alternating between 157 and 163 degrees.

Isn't it $\ce{D}_{9\ce{h}}$? Forgive me if I am wrong.

You might want to attach this image.

Now that's what I call a stretch. We'll have to tell the difference between $C_{9h}$, $D_{9h}$, and probably other options, based on some minuscule, nearly invisible deviations. In the absence of other options, though, this might do.

@user34388 it that accurate, or is like scheme 8.4 here: https://books.google.com/books?id=kP99Ee7Qq4oC&pg=PA354&dq=c18+polyyne+c9h&hl=en&sa=X&ved=0ahUKEwi94dWjlpTPAhWDNSYKHXtbAS8Q6AEIJjAC#v=onepage&q=c18%20polyyne%20c9h&f=false

@DavePhD I have no idea, I just used ChemSketch to draw it. You could draw it yourself.

ChemSketch is no good at all for drawing things like this.

@IvanNeretin C9h is 0.3 kcal/mol lower than D9h, has alternating bond angles of 157 and 163 degrees http://pubs.acs.org/doi/abs/10.1021/ja00120a026

Yeah, I got that. Like a cog wheel with nine asymmetric cogs, only the cogs are so small, and their asymmetry is smaller yet, so when you look at it, you pretty much see just a round wheel.


text:
How to illustrate London forces?
This may be too subjective for this forum, but I'll take my chances.
For the purposes of a "quick and dirty" illustration of London forces, I'm looking for best practices and creative ideas.  For something like dipole-diple attractions, there are some largely-agreed-upon conventions:

But for London forces, there is a shortage of good ways to illustrate them in notes or on the board.  Personally, I have used these in the past:

But there must be other, clearer, more clever means of illustrating London forces. I've searched the questions here on London forces but I don't see that this issue has been addressed.
EDIT: I'm interested in ways of showing modest organic molecules interacting via London forces.  The vast majority of illustrations found via Googling show only rather small molecules or atoms.  It seems to me that the illustration ought to convey the notion of the molecules at close range (or snuggling, as I generally tell students), since London forces are impossible otherwise.

You can literally google it: https://www.google.com/search?q=london+forces&tbm=isch

@user34388 If you are thinking about the diagram from Purdue, don't you think it is wrong to distort the electron cloud?  Electrons moving w/i their orbital don't control the probability distribution of the orbital do they?  That style also doesn't scale to even slightly larger molecules in a practical way.

Then use a diagram which does not distort the cloud? What you drew is quite good.

Related https://chemistry.stackexchange.com/questions/74780/why-do-alkenes-with-more-surface-contact-have-greater-london-forces/74792#74792


text:
What is meant by the "probability density" of finding an electron?
Could someone please explain to me in layman's terms what probability density of finding an electron means, just as probability means chances of finding an electron. With due respect, please don't answer the formula or the distribution curves as I understand and can plot those, I just need the definition.

Probability density means probability per unit volume, ie. If you multiply probability density with volume then you will get the probability of finding ē in that volume.

The formula IS the definition.

In quantum mechanics we often need to find the probability at some position $x$. However, $x$ is a continuous variable with an infinite number of values so it does not make sense to ask what is the probability of being at some exact position , say $x=1/4$, as this will be vanishingly small. Instead we imagine that the probability is that of being at position x to $x + \mathrm{d}x$ where $\mathrm{d}x$ has some very small value. If $f(x)\,\mathrm{d}x$ is that probability then $f(x)$ is the probability density which must be a real non-negative number. It turns out that quantum mechanics postulates that the probability density is given in terms of the wavefunction is $|\psi|^2$.

In quantum mechanics, we clearly know that electron isboth a wave and a particle.Due to this characteristic feature of the electron, being a wave it can exist at numerous places at an instant. So the probability distribution function in relevance with the Schrödinger wave equation are the chances of finding an electron in the region which needs to be studied.

So imagine you have a friend who's an alcoholic, and your phone is broken. You know he's not home, so he's probably at his favorite bar. If you had an aerial map of the area and had to guess where he was, you'd draw a dot on his favorite bar stool [most likely place he is], a circle around the bar building [still could be here], and a dotted line encompassing all of the shrubs, taco truck, and parking lot around the bar [he MIGHT be getting a taco, beat up in the parking lot, or passed out in the bushes].
So in this little parable, your friend is an electron and you have drawn a probability density.
Extra credit: the reason we have to use a probability density is that, for complex mathematical reasons [Heisenberg uncertainty principle] you can't know your friend/electron's velocity AND position, so you're making a mathematical guess about where your friend/electron is, and then describing that mathematical guess in terms that are confusing to us plebes. But really it's just a heat map. Also, depending on what element/number of electrons/orbital you're dealing with, you can get different shapes. Don't be confused - it's just like knowing more friends each with their own different bar, and they don't like each other so they don't overlap.

Forgot to mention - the name for the mathematical description of the probability density is the schrodinger equation, in case anyone throws that at ya.

This is certainly a colorful answer, but might not be the most appropriate for SE ;)


text:
IUPAC nomenclature of coordination complexes
Where can I get the latest rules on IUPAC nomenclature of coordination complexes?
I have J D Lee book which is also assumed to  have outdated rules.

You can refer Solomon and fryhle textbooks if you want

I think IUPAC's Red Book in the 5th edition by 2005, is the most recent reference of this topic, IR-9 (pages 142 onward) about coordination compounds.  Corrections of errors in the print are published here.

In addition the correct reference to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) given in the answer by Buttonwood, you can find a short summary in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.


text:
What are the products of the reaction of NH4Br (aq) + Hg2(NO3)2 (aq)?
Predict the products of the following reaction based upon solubility rules. (Separate substances in a list with a comma. Include states-of-matter under the given conditions in your answer). The reaction is:
$$\ce{NH4Br (aq) + Hg2(NO3)2 (aq) -> ?}$$
I can not figure out what the products of this reaction are. I would think $\ce{NH4(NO3)(aq)}$, $\ce{Hg2Br2(aq)}$, but this is not correct.

According to this table, Mercury (I) salts are somewhat insoluble, so $\ce{Hg2Br2 (s)}$ would be precipitated as a solid.

According to this table, mercury (I) bromide ($\ce{Hg_2Br_2}$) is somewhat insoluble (among other mercury (I) salts), so it gets precipitated as $\ce{Hg_2Br_2(s)}$ instead of $\ce{Hg_2Br_2(aq)}$, making the overall equation this:
$$\ce{2NH4Br(aq) + Hg2(NO3)2(aq) -> 2NH4NO3(aq) + Hg2Br2(s)}$$

Mercury (I) ions are insoluble because every orbital is full, making it unable to form ligands with water.


text:
Amount of Cl- needed to precipitate AgCl given Ksp and [Ag+]

Why is $0.25 \ \mathrm{mL}$ the correct answer in this problem?
I used the Ksp to find the molarity of Chlorine to be $3.47 \times 10^{-2} \ \mathrm{M}$.
Then I used the formula
$$\text{Molarity}_{\text{initial}} \times \text{Volume}_{\text{initial}} = \text{Molarity}_{\text{final}} \times \text{Volume}_{\text{final}}$$
to find the volume of Chlorine, which turned out to be $0.21 \ \mathrm{mL}$  for me. However, my answer does not match the correct answer.
What am I doing wrong?

Review the calculations

1.8/7.2 = 0.25 exactly

The needed molarity for chlorine  that you found is wrong.

First, solve for [$\ce{Cl-}$] using the expression for K$_{\mathrm sp}$ and the given molarity for $\ce{Ag+}$:
$$K_{\mathrm sp} = 1.8\times 10^{-10} = [\ce{Ag+}][\ce{Cl-}]$$
which rearranges to
$$[\ce{Cl-}] = {1.8\times 10^{-10}\over 7.2\times 10^{-5}} = 2.5\times 10^{-6}\, \mathrm{M}$$
Note that the units for K$_{\mathrm sp}$ are actually $\mathrm{M}^{2}$: I have omitted that from the math above but dimensional analysis tells us that the answer for $[\ce{Cl-}]$ is in units of molarity ($\mathrm{M}$).
Because we have 100 $\mathrm{mL}$ of $\ce{Ag+}$ solution, divide the $[\ce{Cl-}]$ molarity by 10 to get the actual number of moles of $\ce{Cl-}$ needed (here, I multiply by 0.1 to achieve the same result):
$$\mathrm{mol}\,\ce{Cl-} = (2.5\times 10^{-6}\,\mathrm{mol}\cdot\mathrm{L}^{-1}) \times 0.1\,\mathrm{L} = 2.5\times 10^{-7}$$
We are given a solution that is 0.001 $\mathrm{M}$ $\ce{Cl-}$, so we can get the volume needed in liters:
$${2.5\times 10^{-7}\,\mathrm{mol}\,\ce{Cl-}\over 10^{-3}\,\mathrm{M}\,\ce{Cl-}} = 2.5\times 10^{-4}\,\mathrm{L}$$
Finally, convert the above quantity to $\mathrm{mL}$:
$$\left({1000\,\mathrm{mL}\over\mathrm{L}}\right)\cdot\,2.5\times 10^{-4}\,\mathrm{L} = 0.25\,\mathrm{mL}$$

Where did you get the value 0.1 Liters from when trying to convert the molarity of Cl to moles?

@ChrisB - That's from the 100 $\mathrm{mL}$ (or 0.1 $\mathrm{L}$) of solution containing the $\ce{Ag+}$, and we need a 1:1 ratio of moles of each species. Since the units of molarity are $\mathrm{mol}\cdot\mathrm{L}^{-1}$, we need to divide by 10 or, as I did, multiply by 0.1.


text:
How do ionic solution conduct?
Do ionic compounds use ions to conduct? Or do they use electrons?
Also why don't they conduct in solid form and only in liquid or gaseous forms? I mean electrons can still travel in solid forms.
Do their ionic bonds break in the liquid and gaseous forms?
Thanks for answering

I am assuming you mean electricity, not heat.

yes electricity

Ionic solids do not conduct because the atoms are typically bound too tightly to a crystal lattice for anything to function as mobile charge carriers.  Mobile charge carriers are available in metal because metals can easily change oxidation states allowing electrons to move; whereas, most ions cannot easily change state.
Ionic liquids on the other hand have both mobile cations and anions which can serve as charge carriers possibly resulting in moderate conductivity.
As far as bonding the bonds in ionic liquids are constantly breaking and reforming just as in all liquids.
There are no bonds between gas particles.  Gasses are poor conductors; however, if the gas ionizes turning into plasma then the electrons are free to be charge carriers making the gas a conductor.

but why are gases poor conductors?? And do ionic compounds conduct through electrons or the ions?

Ionic compounds are either an ionic solid or an ionic liquid.  Ionic solids do not conduct.  Ionic liquids can through movement of ions.   Ionized plasma can conduct through movement of electrons.

Why are gasses poor conductors of electricity is a different question entirely. Since it is already on the physics exchange.  http://physics.stackexchange.com/questions/38631/can-gases-conduct-electricity

Also. to be an electrical conductor means something with charge is free to move.  If no charge can move then electricity cannot conduct itself through a medium.

but if gases are ionized...can't they just conduct through ions?? What is the need of conduction through electrons?

In the classical states (solid, liquid, and gas) there is little net charge.    Ionized gas has a net charge so it is a mixture of electrons,  ions, and neutral particles.   I'm guessing, but it seems some ions would drift according to the flow of the electrons, so it may be possible to have both.  I'm do not claim to be an expert on plasma physics.

First of all ionic compounds are not bounded by bonds they are bound by electrostatic forces of attraction. Secondly if the ionic compound is an electrolyte which generally is, the ions separate. The ions require a medium to polarise them in short to separate them liquid and gaseous mediums thus help as they cant self ionise in their solid state. Hope it helps.

I wasnt the downvote, but I do believe that it was because you said the medium polarises the molecule... which is not true it just dissolves it

Would solvates be the correct word then?


text:
Primitive cells and their variations
Is there any intuitive way(without involving higher maths) to explain possible variations in seven primitive types of (unit)cells? I would like explanation in context of why end centred is possible only for orthorhombic and monoclinic but not in other unit cells.

Related: http://chemistry.stackexchange.com/questions/47834/why-can-a-face-centered-cubic-lattice-not-be-redrawn-as-a-body-centered-tetragon/47868#47868

I will not consider hexagonal cell in this answer.
Unit cell is generally parallelepiped, which has three sides $a$, $b$ and $c$ and three angles between them $\alpha$, $\beta$ and $\gamma$. The sides can be f.e. of equal length and angles can be same and equal to $90^\circ$, this way you will have cubic unit cell. You can try to vary sides and angles and you will discover, there is only those seven types of unit cells.
When I way learning this I found useful third chapter in "Foundations of material science and engineering". At least there is nice table and images to show those cells. You can find it by typing "Crystal structures and crystal geometry" into Google.
Well, now answer last question. Let's say we have cubic base-centered (more correct than end-centered, I think) cell. From top it would look like this:

I draw only two cubic base-centered cells (I can add more if it will be more clear) next to each other. However, as you can see, there is one smaller cell (red) inside of our base-centered lattice (green) and that is simple tetragonal cell (If you draw it in three dimension, you will see, that the height is $\sqrt{2}$ times greater). As it has lower volume, this is the primitive cell, not that base-centered.

I have a vague feeling I've said this before, but anyway: every centered lattice can be decomposed into some other primitive lattice, just like you did.


text:
Difference between continuous and line spectrum of hydrogen
The difference between a continuous and a line spectrum is well understood in that a clear cut boundary is present in the case of a line spectrum. When will continuous spectrum be obtained and when a line spectrum for example  in  hydrogen? What method can be used to observe either or both these spectra?

See Wikipedia articles on Emission Spectrum https://en.wikipedia.org/wiki/Emission_spectrum and Black Body (Radiation) https://en.wikipedia.org/wiki/Black_body

I will quote part of this answer:

Quantum mechanics dictates that the energy levels of atomic hydrogen are discrete--just from the way the math works out. The reason why hydrogen's peaks are so spaced out is that it has so few degrees of freedom (just the one electron). Xenon, in contrast, has many closely-spaced peaks because it has 54 electrons (so it has many more accessible energy levels than hydrogen).

The implication is that to a certain resolution, line spectra appear to become continuous once the energy level spacing and the excitation energies become close together. If you look closely enough, they are still discrete.
However, there comes a point where the electron that's being excited is no longer bound to the atom; that would be the first ionization potential (IP) or ionization energy. In the case of the hydrogen atom, that is the Rydberg constant. See this answer for how this arises mathematically.
For more details on the hydrogen atom, see here:

The infinity level represents the highest possible energy an electron can have as a part of a hydrogen atom. If the electron exceeds that energy, it is no longer a part of the atom. The infinity level represents the point at which ionization of the atom occurs to form a positively charged ion.

At this point it is important to distinguish between hydrogen's absorption/excitation spectrum and its emission/deexcitation spectrum (taken from here):

Since energy is inversely proportional to wavelength, the higher-energy excitations are toward the left of the spectra. This is visual confirmation that as energy level spacing decreases, lines in spectra get closer together. Past that point, what do the spectra look like? Certainly there will be peaks at the ionization energy, which is at a wavelength of about 91 nm. Applying more energy will also strip the electron from the proton, but the excitation from $n = 1$ to $n = \infty$ will still occur at the same energy, and likewise for the absorption from $n = \infty$ to $n = 1$. So there is only one peak in the hydrogen emission and exitation spectra directly resulting from the interaction between the electron and the proton (related). Any potential continuous band structure at higher energies due almost entirely to the free electron in the continuum, or you could think of any point in that spectral region being composed of two contributions, one being the fixed electron-proton interaction energy and the other being related to the kinetic energy of the free electron.
Note that this is for a single atom; for a collection of atoms approaching a bulk phase, you may have continuous spectra under other situations. Any time there are free electrons, energy levels may be continuous. There is also above threshold ionization, but without doing any calculations I am not sure how intense these effects are in terms of directly contributing to UV/vis spectra; if the $x$-axis is electron energy in eV and the $y$-axis is related to excitation probability, the effect is most likely negligible in practice.

I apologize if I misunderstand, but it seems like you are asking two questions.

Why does the hydrogen spectrum have so few, widely spaced peaks?
Why do spectral peaks have width?

Here are answers:

Quantum mechanics dictates that the energy levels of atomic hydrogen are discrete--just from the way the math works out. The reason why hydrogen's peaks are so spaced out is that it has so few degrees of freedom (just the one electron).
Xenon, in contrast, has many closely-spaced peaks because it has 54 electrons (so it has many more accessible energy levels than hydrogen).
Spectral lines are broadened for a number of reasons (both internal to the system--perhaps states are mixing, and external--the pressure on the system has some effect on line widths).
However one type of broadening that is inherent to all systems (including simple, little hydrogen) is something called "lifetime broadening," and it is a result of the uncertainty principle. The principle puts a lower bound on the product of the uncertainty of the energy and the uncertainty of the lifetime of any quantum state, like the first excited state of hydrogen (2p), for example. Hydrogen's 2p has a lifetime of ~1.6 ns which gives it a peak width (an uncertainty in the energy of the transition) of ~0.000004 eV.

(see: http://farside.ph.utexas.edu/teaching/qmech/Quantum/node122.html )


text:
Cubic crystals anisotropy
My textbook states

"Cubic crystals contain four three-fold axes, converting the ellipsoid to a sphere."

What indeed is the ellipsoid and sphere in the lattice? I know the meaning of three fold axes but unable to figure out what and how ellipsoid is converted to sphere.

From the limited excerpt of your book (title? author?) it is likely it just deals with the interaction of linearly polarized light with condensed matter.  If there is near- and far range order (as regular consecution of building elements), the solid is crystalline.  Among the crystaline states, there are (Bravais) lattice systems, that differ in symmetry, i.e. because they possess different symmetry elements, some are of higher symmetry (like cubic) than others (like monoclinic).  And vectorial physical properties of the material are coherent with the crystalline symmetry met.

(image source).
By means of an index ellipsoid, you may map the orientational dependance of physical properties, in other words, anisotropy.  In the case of mineral- and more general in polarized light microscopy, such an ellipsoid may be used as representation to map refractive indices along three directions orthogonal to each other. Now

either the three axes of the indicatrix differ from each other: optical biaxial system, characteristic Bravais lattice of lower symmetry: orthorhombic, monoclinic and triclinic.
two of the three axes of the indicatrix are identic, the third is different: optical mono- or uniaxial system, characteristic for hexagonal, trigonal and tetragonal Bravais lattice.
all three axes of the optical indicatrix are identic, in other words, the material is optical isotropic: this is characteristic for the cubic Bravais lattice.  (And: for the state of randomness, the glasses).

So in short: it is the number, the degree, and the relative orientation of the symmetry elements of the cubic Bravais latice that renders the more general case of an ellipse into the special case of a sphere.  And this means, the question should be "cubic crystal's isotropy".
Further information may be found for example here.

The ellipsoid refers to the size of the anisotropic parameters in different directions at right angles to one another and along some axis, say refractive index, being different in two directions compared to the third. This 'ellipsoid' of refractive index may be elongated like a rugby ball (prolate ellipsoid) or squashed like a doughnut (oblate ellipsoid). In a cubic crystal the value is the same in all three directions hence a sphere.  A cube has four 3-fold axes, look down in a diagonal direction along opposite corners.


text:
Relative atmospheric lifetimes of carbon monoxide and methane
In atmospheric chemistry, the lifetime of most substances is dictated by their reactions with the hydroxyl ($\ce{OH}$) radical. The rates of these reactions are listed by NIST as about $\mathrm{1.5e^{-13} cm^{3}\  mol^{-1}\ s^{1}}$ for carbon monoxide ($\ce{CO}$) and $\mathrm{6.7e^{-15}\ cm^{3}\ mole^{-1}\ s^{1}}$ for methane ($\ce{CH4}$). With atmospheric $\ce{OH}$ concentrations of about $\mathrm{1.2e^6\ mole\ cm^{-3}}$, this leads to an atmospheric lifetime of about 2 months for $\ce{CO}$ and 4 years for $\ce{CH4}$. This is indeed what's observed, so the relative reaction rates are definitely experimentally true.
My question is: what's the theoretical explanation for the $\ce{CO}$ reaction being faster? I would expect the $\ce{C-O}$ triple bond to be much more stable than any $\ce{C-H}$ bond. In atmospheric chemistry texts I've seen, the reaction goes like
$$\ce{CO + OH -> CO2 + H*}$$
$$\ce{H* + O2 -> HO2}$$
I've also come across notes from a pure chemistry class that mentions a different reaction with an intermediate:
$$\ce{CO + OH -> OCOH*}$$
$$\ce{OCOH* + O2 -> CO2 + HO2}$$
I can't get my head around how the intermediate would/should increase the reaction rate, though, or how something like electronic configuration could lead to $\ce{CO}$ being more susceptible to $\ce{OH}$ than $\ce{CH4}$.

The dominant oxidation (via hydroxyl radical) pathways for CO don't have any intermediate species as unstable as the methyl radical, which is step one for methane oxidation via the hydroxyl radical.

Can you explain a bit more? How would the fact that the methyl radical is so unstable mean that CH4 oxidation is slower instead of faster? Is it that it requires more activation energy to get from CH4 -> CH3 than from CO to OCOH*?

Sorry so late getting back, but the last sentence in your comment basically gets it.  The methyl radical is extremely unstable so there is a large energy barrier associated with the initial step in methane oxidation, the abstraction of a hydrogen to form the high-energy state methyl radical.  That's just a wordy form of you last sentence.


text:
Is it possible for a carbon atom with a negative formal charge to have ONE bond and ONE lone pair?

I was completing some problems from my textbook and found that the answer of this problem seemed incorrect to me.
From my limited understanding, carbon atoms with a negative formal charge must have 5 electrons (3 bonds and 1 lone pair), but this answer didn't match up with this rule.
I would really appreciate it if someone could tell me what I'm missing here.

The hydrogen atoms are not shown, so this is really $\ce{CH3CH2CH2CH2-}$ where the rightmost carbon atom has a lone pair and a formal negative charge.
The species is the n-butyl anion:

Oh my god, I completely forgot about that! I feel like such an idiot right now haha. Thank you so much!

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@Martin-マーチン Alright, removed and noted.