text:
Reduction of α,β-unsaturated nitro compounds
Would it be possible to reduce a α,β-unsaturated nitro compound to a saturated amine using for example $\ce{NaBH4}$? I know these kind of compounds will be reduced by hydrogenation using a $\ce{Pt/C}$ catalyst, $\ce{LiAlH4}$ or freshly prepared Raney Nickel, but I'm looking for a reducing agent that is easier to handle than those.
The book from which I'm learning, Clayden et al., does not really go into much detail on reduction of nitro compounds or unsaturated alkanes, other than that hydrogenation is a good approach to both and that the nitro group is one of the most easily reduceable functional groups. But how about a α,β unsaturated compounds then? How would I asses whether some reducing agent is able to reduce both simultaneously?

Perhaps you could have a look at :
http://www.sciencedirect.com/science/article/pii/S004040390060148X

@jori I feel that the electron-withdrawing nature of the nitro group would make reduction with electrophilic reducing agents less likely but reduction with nucleophilic reducing agents more likely. Perhaps something strong such as LAH would do the trick.

I'm not aware of a $\ce{NaBH4}$-based reduction of nitroalkenes to saturated amines, such as in:

It is however possible to reduce nitroalkenes to saturated hydroxylamines using $\ce{BH3*THF}$ in the presence of catalytic amounts of $\ce{NaBH4}$. (DOI)
If there's a chance to perform the desired transformation in one step via catalytic hydrogenation at normal pressure: go for it!
The handling of these reactions is nothing to be afraid of.
Typically, I performed hydrogenations in a standard 2- or 3-neck flasks and used ordinary balloons (think childrens' birthday) as gas storage. Advice: Stuff a piece of rubber tube into the inlet of the balloon and secure it with teflon tape.

Either attach this to the glas olive of a valve attached to your flask
or take a plastic syringe, cut off the base, stuff the the plastic cylinder into the rubber tube. Fill the balloon and attach a needle. Hydrogen may now be fed to your reaction mixture through a rubber septum cap.

UPDATE
I still doubt that the intended reduction by $\ce{NaBH4}$ is possible.
There is however another established route from $\alpha,\beta$-unsaturated nitroalkenes to amines which is worth to be mentioned:
Condensation products of aromatic aldehydes and nitroalkenes are reduced by iron in hydrochloric (or acetic) acid to the corresponding arylalkanones.

These ketones can be converted to primary amines by reductive amination in methanol (or ethanol) in the presence of (excess) ammonia and, of course, hydrogen over Raney Nickel or Urushibara Nickel.

In the course of this reaction, the imine intermediate is in situ reduced to the desired primary amine.
The resulting amines are valuable precursors in the synthesis of natural products: Their conversion to imines or amides and subsequent Pictet-Spengler or Bischler-Napieralski cyclisations furnish tetrahydro- and 3,4-dihydroisoquinolines.

You mean like this: https://www.youtube.com/watch?v=PC4DT4ier2Q ? $\ce{BH3}$ is impossible to handle safely in amateur setting, way worse I think than Raney Nickel or even LAH. Perhaps Urushibara nickel will do the job, which is said to be more easy to handle (read non-pyrophoric). For small scale reactions your idea with the balloon might work well indeed. Do you have a link to a somewhat more extensive description of the technique?

I agree with this, you should not avoid catalytic hydrogenation... Nitro groups are easily reduced, but NaBH4 alone won't make it. Another nice feature of sodium borohydride!

I also saw relatively cheap Parr hydrogenators on eBay (some in the range of 400-600$). These things seem pretty heavy and loud and I'm a bit weary of the explosion hazards, but they can be used for larger amounts and require less work. What level of expertise is required to handle such things anyway? Do you need special training? (see e.g. http://ebay.to/1BMFHHP)

@Jori Yes, like in the video, although I usually evacuated the flask through a larger diameter using a piece like this. In the video, the tube attached to the balloon looks like a commercially available piece, I always chopped something together from material from the lab drawers ;) The balloon technique works nice, but I never did this on a kilogram scale. Btw, Raney Nickel isn't that bad as long as you handle it with care and keep it wet. It really is pyrophoric :D

@Jori The ebay offer doesn't seem to include the reaction flask. You might want to decide for yourself whether you want to run a hydrogenation at 60 psi (4 bar).

@KlausWarzecha Thank you for the edit, that is very helpful. Can you comment on why $\ce{Fe/HCl}$ reduces the $\alpha,\beta$-unsaturated nitroalkene to a keton instead of a saturated nitroalkane or a (un)saturated amine? What exactly do you mean by: "You might want to decide for yourself whether you want to run a hydrogenation at 60 psi (4 bar)."? I'll wait until the bounty period is over (to maximize the attention) and then reward you the bounty :-).

@Jori I can't comment on the exact mechanism of the reduction - I don't know which intermediates are involved. Concerning the higher pressure I just wanted to point out that higher pressure involves higher risks in the case the reactor bursts! Which doesn't mean that the risks can't be handled. Back in the days I worked in a group that performed electrolysis of metals under 150 atm CO. But that was in stainless steel autoclaves and the reaction was carried out in a concrete fume hood without a solid roof but 5 cm steel doors ;)

@Jori >Can you comment on why Fe/HCl reduces the α,β-unsaturated nitroalkene to a keton instead of a saturated nitroalkane or a (un)saturated amine?  || because enamines tend to isomerize into imines that react with water producing ammonia and ketone. You should reduce the double bond first.

@permeakra Ah of course. $\ce{Fe/HCl}$ is selective for the nitro group?

There are numerous reports on transfer hydrogenation, where hydrogen is transferred from strongly reducing agent onto double bond or other groups.
In particular, Raney nickel was utilized in reduction of both nitrogroups to amines and alkenes to alkanes, using formic acid for the former and isopropanol for the latter. I guess, that combined process is also likely to work.

$\ce{NaBH4}$ will only reduce the $\ce{RCH=CHNO2}$ to $\ce{RCH2CHNO2}$. An additional reducing agent, such as $\ce{Zn/HCl}$, can reduce the nitro group to the amine. A catalyst + $\ce{NaBH4}$ could do the whole thing, but you'll have to look at the literature.
Catalytic hydrogenation, using $\ce{Pd/C}$ under moderate conditions (around 5 bar $\ce{H2}$, 30-90 °C), can reduce $\ce{RCH=CHNO2}$ directly to  $\ce{RCH2CHNH2}$.

I have done the reduction of RCH=CHNO2 to RCH2CHNO2 using NaBH4 in the presence of column chromatography grade silica gel. It works well. Hydrogenation over Pd/C is poor giving a mixture unless you do it in the presence of ammonia.


text:
How to determine which of two metals will have the higher temperature after applying the same amount of thermal energy?

Two metal samples, $\mathrm{X}$ and $\mathrm{Z}$, of the same mass and initially at $\pu{25 ^\circ C}$, are heated so that each metal receives the same amount of thermal energy, which metal will have the highest final temperature?
  Specific heat capacities:
  $c(\mathrm{X}) = \pu{0.35 J//g*K}$,
  $c(\mathrm{Z}) = \pu{0.895 J//g*K}$

Does the specific heat capacity matter in the final temperature or not?

By definition, the heat capacity is the energy required for a body to be warmed by $\pu{1 K}$. For the specific heat capacity it's the same but normalized by the mass. The higher the heat capacity is, the more energy will be required to warm it.
The metal $\mathrm{X}$ having the lowest heat capacity will have the higher temperature if it receives the same amount of energy than the metal $\mathrm{Z}$.

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Why does the aqueous solution of sodium peroxide turns red litmus into white?

A white solid is either $\ce{Na2O}$ or $\ce{Na2O2}$. A piece of red
  litmus paper turns white when it is dipped into a freshly made aqueous
  solution of white solid.

Identify the substance and explain with the help of balanced
  equation.
Explain what would happen to the red litmus if the white solid
  were the other compound.

From the solution and theoretical knowledge, I know the following:
$$\begin{align}
\ce{Na2O2 + 2 H2O &-> 2NaOH + H2O2}\\
\ce{Na2O + H2O &-> 2NaOH}
\end{align}$$
The aqueous solution of $\ce{Na2O}$ will convert red litmus into blue (basic property), so, the substance is $\bf{not}$ $\ce{Na2O}$ for sure.
But, how will it be $\ce{Na2O2}$? Why does the aqueous solution of sodium peroxide turns red litmus into white?

Your analysis seems right and you remember correctly that colour change of litmus of litmus paper will indicate the pH value of the solutions into which it is dipped.
That is, unless the dye (litmus) is irreversible transformed to another compound which no longer shows a pH dependent colour.
As the experiment states:

A piece of red litmus paper turns white [...]

With other words: Litmus was bleached.
This bleaching typically is the result of an oxidative transformation of a (former) dye.
Here, the only conceivable oxidant in solution is hydrogen peroxide. Consequently, the initial compound in question must have been $\ce{Na2O2}$.

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text:
Why do only ketones exhibit keto-enol tautomerism?
Why can't esters or acids or their derivatives exhibit tautomerism? I mean if they did, then there would be a conjugation between lone pair electrons of oxygen and pi electrons adding to their stability. Then why don't they exhibit similar tautomerism?

Though I do not have a reference, I believe that some esters would exhibit similar or the same behaviour as ketones. At least $\alpha{}$-protons of esters are far more acidic than their esterless counterparts, which proves a similar effect is applicable on esters.

Esters, amides and acids do exhibit tautomerization to the enol form, but the enol content in these compounds is just much less than that found with aldehydes and ketones. Why?
The keto-enol process is an equilibrium; to understand where an equilibrium lies, we must compare the relative stabilities of the products (enol) and reactants (carbonyl compound).
Let's compare some resonance structures for an ester and a ketone and their corresponding enols.

The ester is stabilized by resonance structures such as A, whereas in the case of the ketone, the analogous resonsnce structure A' contributes, but not nearly as much; so ester carbonyls are stabilized compared to ketone carbonyls.
In the enols, the ester resonance structure B is not particularly favorable due to the negative charge on carbon and positive charge on oxygen, just the reverse of what we would want based on electronegativities.  At least in the ketone case resonance structure B' has a positive charge on hydrogen, so resonance structure B' contributes more than B (this is why carbon-carbon double bond stability increases with increasing alkylation).  As a result the enol resulting from the ketone is stabilized compared to the enol resulting from the ester.
Just where the carbonyl-enol equilibrium lies will depend on the relative stabilities of the carbonyl compound and the enol.  As we've just discussed, the ester carbonyl is stabilized relative to the ketone carbonyl, this will push the equilibrium further to the carbonyl side for the ester.  Also, the enol from the ketone is somewhat more stabilized than the enol from the ester, this will push the equilibrium to the enol side for the ketone.
So both factors, starting material (the carbonyl compound) stability and product (enol) stability will work to increase the enol content for ketones and reduce the enol content for esters.

In esters why can't the positive charge be on adjacent carbon instead of oxygen? That way the newly formed pi bond and the lone pair electrons on oxygen would be in conjugation adding to the stability of ester. Why isn't that possible?

To draw a resonance structure for the ester with the positive charge on the carbon adjacent to the carbonyl carbon you would need to have hydride ion ($\ce{H^{-}}$) as part of the resonance structure.  Hydride ions are very high energy, consequently resonance structures involving them are very high energy and do not contribute significantly to the overall description of the molecule.  BTW, if the above answer was helpful, please mark it as accepted. Thanks.

Pardon me but I didn't understand why an hydride ion comes into the picture. Just like in keto-enol tautomerism a H+ ion will come out and then later form a bond with oxygen, right?

You commented, "Just like in keto-enol tautomerism a H+ ion will come out and then later form a bond with oxygen, right?".  Yes, that is right.  But in your first comment you asked, "In esters why can't the positive charge be on adjacent carbon instead of oxygen?"  Try writing a mechanism to create such a structure, I think you will find that a hydride ion is required somewhere in the process.

Yeah you are right about the +ve charge on the adjacent carbon issue. But can there be a double bond between ester carbon and the adjacent carbon due to tautomerism between those 2 carbons and oxygen?

Yes, that is the enol form of the ester.

Good answer, I would have written something similar myself if you had not done it first.


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Why is osmium the densest known element?
Why is osmium so dense despite there being heavier elements after it in the periodic table?

Is there any agreement as to whether Iridium is denser or Osmium? They're very close.

The density of an element is related to how many atoms can be placed in a given volume and the weight of the nuclei.  Therefore, the smaller the atomic radius of an atom and the higher the atomic number of the nucleus, the greater the density of the element. The very small atomic radius of osmium results in a small metal-metal separation.  This small atomic separation along with osmium's relatively high atomic number gives rise to osmium's high density.
The small atomic radius can be attributed to the following factors:

f orbitals are very diffuse and therefore result in poor screening of the electrons further out.  In the case of osmium ($\ce{[Xe] 4f^{14} 5d^6 6s^2}$) the poor screening by its 4f orbitals leads to a contraction in the n=5 and n=6 orbitals.
Due to osmium's high atomic number, relativistic effects come into play.  These relativistic effects have been discussed previously on SE Chem for gold and mercury.  A general discussion as well as the mathematics describing the relativistic effect can be found in these earlier posts.  Basically, for heavy nuclei the s electrons must circulate at relativistic speeds in order to maintain a stable orbit (not fall into the nucleus).  Therefore the electron mass increases and the s orbital radius decreases (the p orbital radius also decreases, but to a lesser degree).

The orbital contraction caused by these two effects leads to a small atomic radius for osmium.  Therefore, short metal-metal bonds result.  This is reflected in the small unit cell volume for osmium (27.96 cubic angstroms).  For comparison the unit cell volume for lead is 121.3 cubic angstroms! As a result, more osmium atoms can pack into a given volume compared to other elements
Osmium's relatively high atomic number along with its small atomic radius, as explained above, gives rise to osmium's high density.

That, and the fact that it has a close-packed structure.

Do the relativistic effects reduce the OVERALL size if the atom or just the inner s orbitals (and slightly p)? so basically, is this effect more significant it increasing the mass of the atom or decreasing it's size?

@Michael  Lead has a face centered cubic crystalline structure while osmium crystallizes in a hexagonal close packed structure.  In both structures the lead and osmium atoms occupy around 74% of the volume, yet the densities of lead and osmium are very different (11.34 and 22.61 g/cm^3 respectively).  So I don't see how the crystal structure, per se, plays into the answer.  I fall back to unit cell size, which reflects osmium's small radius, and atomic weight. If I'm missing something, please elaborate.

@user3764899  The relativistic effects reduce the size of all orbitals that have significant density at the nucleus.  Therefore, all s and, to a lesser degree, p orbitals are contracted.  Osmium's 6s orbital (its outermost orbital) is consequently contracted, so overall osmium's size (atomic radius) is reduced.  Look here and notice how osmium's radius sticks out like a sore thumb.  Its radius is by far the lowest for its atomic number.

@ron true. I wasn't saying that it's the only reason - but it definitely helps. A small size does not help if the structure is not close-packed.

Adding a discussion of packing fraction for various crystal structures would make this answer an A+. https://en.wikipedia.org/wiki/Atomic_packing_factor

The argument about cell volumes is fallacious. Looking at molar volumes gives a more reasonable perspective. Pb = 207.2/11.34 = 18.27 cm^3/mole. Os = 190.23/22.59 = 8.421 cm^3/mole. So Os has a molar volume which is 46.09% of that for Pb.


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What type of compound/element might be in these glowing tiles
I saw these tiles and these as well while surfing the net, they obiviously glow when their electrons are exited to a higher state and then de-exite over a long period to give off visible light.
My question is which chemical might get exited by visible light and them emit back visible light for a long time.

Glow in the dark materials are materials that exhibit phosphorescence. The most common materials used for this effect are zinc sulfide and strontium aluminate, doped with various elements that affect the colour of luminescence.
Phosphorescence is similar to fluorescence, where an electron is excited to a higher energy state, but instead of dropping back immediately by emitting a photon, an intersystem crossing event occurs where the electron's spin flips resulting in a triplet state (S=0 to S=1 in the figure). Because the electron has the same spin as the other electron it left behind in the ground state, the transition to the ground state is forbidden and thus occurs much more slowly. In practice, a given material usually exhibits both fluorescence and phosphorescence. Materials that are very phosphorescent have a high probability of intersystem crossing so that when light is shined on them, a large population builds up in triplet states that will release light slowly. What makes a given material have a high intersystem crossing probability is pretty complicated, but it tends to be higher in molecules with large atoms—even things like fused silica glasses are phosphorescent, but very weakly.

Source


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Good resource for chemical equivalence
I need a good resource that explains the concept of chemical equivalence and its various application in physical chemistry. I have not found anything worthwhile from a Google search.

i have two sources, but it might not be that they're available on your continent?

@ADG: Mention them nonetheless. They might be available online.

Anyways they are Indian books: RC Mukherjee and P Bahadur(Better get an old version of subjective one, which has many solved and unsolved examples)


text:
Why does dimethyl ether have a bond angle bigger than 109.5 degrees?
Why does dimethyl ether have a bond angle bigger than 109.5 degrees, in contradiction with VSEPR theory?

Due to replusion between the two methyl groups, according to Competition Science Vision, October 2008, page 1033.

Related: http://chemistry.stackexchange.com/q/18697

In case of $\ce{Me2O}$, the positive deviation of the bond angle can be explained on the basis of steric crowding of two methyl groups.
There are other contradiction too like this:
In case of $\ce{(SiH3)2O}$, the huge bond angle increase is mainly due to back bonding and partly due to steric repulsion.

Let us compare the $\ce{C-O-C}$ bond angle in dimethyl ether (110.3°) with H-O-H bond angle in $\ce{H2O}$ (104.5°).
The bond angle in dimethyl ether is greater due to the following reason. The $\ce{CH3}$ functional group is many times heavier than the H atom (both act as side atoms). So the effect of lone pair - lone pair repulsion in $\ce{H2O}$ is pronounced in $\ce{H2O}$ whereas due to heavy molecular weight the repulsive forces aren't able to fully force $\ce{CH3}$ functional groups to compress them.


text:
Why don't electrons pair up in an oxygen radical?
In an oxygen radical (produced by the break down of O3 for instance), there are 2 unpaired electrons in the outermost subshell which make the molecule unstable. Why don't these electrons pair up which would make the atom more stable (I think)?

Pairing up of electrons in this case is unfavourable due to increased inter-electron repulsion; the average spatial separation of the electrons is higher in the triplet state where the two electrons are unpaired. Regular oxygen molecules also exist in a triplet state and have unpaired valence electrons; they are said to be biradicals.


text:
A 4.00 L sample of gas is cooled from 67 °C to a temperature at which its volume is 2.60 L
A $4~\mathrm{L}$ sample of gas is cooled from $67~\mathrm{°C}$ to a temperature at which its volume is $2.6~\mathrm{L}$. What is this new temperature? Assume no change in pressure of the gas.

Hint: Use Ideal gas equation:

$$\large PV=nRT$$

Given pressure is constant, can you tell the final temperature from this relation?

Then $\frac{V_\text{1}}{V_\text{2}} = \frac{T_\text{1}}{T_\text{2}}$ can be derived using the relation given by @ADG


text:
Why does N₂ react with O₂ to Form NO at high temperatures?
This also raises questions that I have about the Haber Process which produces ammonia ($\ce{NH3}$) from molecular nitrogen ($\ce{N2}$) and hydrogen ($\ce{H2}$).
I have heard multiple times that bond between diatomic nitrogen is one of the strongest bonds in nature due to the fact that it is a triple covalent bond that fills the valence shells of both atoms.
I understand that at high temperatures it is possible to break this bond, but I don't understand why the resulting Nitrogen atoms wouldn't simply return to their previous bonds as the temperature cooled.
For example, I read that lightning can result in this reaction:  $\ce{N2 + O2 -> 2NO}$
Why would the atoms not return to their original bonds since they would be more stable in that manner? Is bonding indiscriminate at high energy levels? Completely random and dependent on luck?

$\Delta G = \Delta H - T \Delta S$
In the case of the $\ce{N2 + O2 -> 2NO}$ , $\Delta H$ and $\Delta S$ are both positive, so the reaction is thermodynamically favorable at high temperature (such as in lightning) but not at low temperature.
If the temperature drops to room temperature after NO is formed, it is thermodynamically favorable for NO to decompose to nitrogen and oxygen.
However, that NO is unstable at room temperature tells us nothing about the rate of the decomposition reaction.  In fact there was an interesting 40 year study showing very little decomposition of NO sealed in glass tubes over that time period.  The authors' calculations show that without a catalyst the timescale of decompsition could be $10^{29}$ years!

So the NO does return to N2 and O2 after the temperature returns?

In Earth's atmosphere, the NO formed by lightning, reacts with O2 to form NO2, which then forms HNO3.  So in our environment, no NO does not return to N2 and O2, because it reacts with more O2 before that can happen.

The Haber process usually uses metal catalysts to help the bonds break. It runs at extremely high temperatures and pressures, enough to cause worry that it will burst steel vessels. At these conditions breaking $\ce{N2}$ and $\ce{H2}$ can and does happen. It's true that the nitrogen-nitrogen bond is broken, but once that happens ammonia forms quickly.


text:
Does iron(III) sulfate react with copper?
As my understanding, there should be an oxidation-reduction reaction:
$$
    \ce{2 Fe^3+ + Cu → 2Fe^2+ + Cu^2+}
$$
However, I always see the process using $\ce{FeCl3}$ to etch copper, but I never see people using $\ce{Fe2(SO4)3}$ to do this. Is it because such reaction does not occur, or is it because $\ce{FeCl3}$ is acidic in aqueous solution?

In $\ce{FeCl3}$ etching of copper, the chloride ion is extremely important.
Chloride ions coordinate to $\ce{Fe^3+}$, $\ce{Cu+}$ and $\ce{Cu^2+}$.
See Copper Etching in Ferric Chloride Ind. Eng. Chem., 1959, 51 (3), pp 288–290 for the relative concentrations of the particular chloride complexes.
In contrast, if $\ce{Fe2(SO4)3}$ were used, there would only be aqueous ions.
At the copper surface, there is only a one-electron oxidation of $\ce{Cu}$ to $\ce{Cu+}$.  As the above article explains, without coordination of chloride, the $\ce{Cu+}$ would be essentially insoluble.
The rate limiting step of the etching reaction is diffusion of ions from the surface of the copper.  Coordination of chloride provides solubility and enables diffusion.  This is the main reason that $\ce{FeCl3}$ is needed.

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text:
Common ion effect and acids
If I have $0.01~\mathrm{M}$ $\ce{HBr}$ in $0.09~\mathrm{M}$ $\ce{KBr}$, why would the presence of $\ce{Br-}$ ions from the $\ce{KBr}$ not reduce the dissociation of $\ce{HBr}$ based on Le Chatelier's principle? Or would it?

The $\ce{Br-}$ would make less $\ce{HBr}$ dissociate, But $\ce{HBr}$ is such a strong acid ($\mathrm{pKa}$ is $\mathrm{-9}$) that still only a trivial amount would be molecular.

As pointed out be DavePhD, in reality there will be no real detectable change in the dissociation of $\ce{HBr}$ since it is such a strong acid. I will attempt to give a more theoretical answer to this question using equilibrium constants.
We have the following chemical equations (note that we assume $\ce{KBr}$ completely disassociates) :
$$\ce{HBr \rightleftharpoons H+ + Br-}$$
$$\ce{KBr -> K+ + Br-}$$
We have two equations:
$$\ce{K_{a}\ = \ \frac{[H^+][Br^{-}]}{[HBr]}}$$
$$\ce{[Br^{-}] = [Br^{-}_{HBr}] + [Br^{-}_{KBr}]}$$
The extent of disassocation of $\ce{HBr}$ is given by value of $\ce{[Br^{-}_{HBr}]}$.
By substituting the second equation into the first equation and rearranging:
$$\ce{K_{a}\ = \ \frac{[H^+]([Br^{-}_{HBr}] + [Br^{-}_{KBr}])}{[HBr]}}$$
$$\ce{[H^+][Br^{-}_{HBr}]\ =\ K_{a}[HBr] - [H^+][Br^{-}_{KBr}]}$$
$$\ce{[Br^{-}_{HBr}]\ =\ \frac{K_{a}[HBr] - [H^+][Br^{-}_{KBr}]}{[H^+]}}$$
From this equation, it can clearly be seen that the disassociation of $\ce{HBr}$ decreases by adding $\ce{KBr}$. However since the Ka value is ${10^9}$, the second term in the numerator is negligible compared to the first term and the equation can be reduced to:
$$\ce{[Br^{-}_{HBr}]\ \approx \ \frac{K_{a}[HBr] }{[H^+]}}$$
As you can see, this expression is independent of the concentration of $\ce{KBr}$ which illustrates DavePhD's point that it really won't effect the disassociation of $\ce{HBr}$.


text:
Strong Base combined with Sp2 carbons in Nucleophilic Substitution and Beta-Elimination Reactions

When we are given this problem, there is a certain problem. We have a strong base, but hydrogen on the carbon of the phenyl. E2 is impossible. Also, SN2 is not possible since we have a hindered base.
The only product we could get is SN1. My teacher put up the answer in handwritten black in the above picture.
However can SN1 reactions happen when there is a strong base in the solution? Would that strong base be the nucleophile or the solvent tert-butyl alcohol be the nucleophile?

The strong base is capable of being a nucleophile in this sort of situation; deprotonating tert-butyl alcohol will increase its nucleophilicity. In an SN1 reaction the overall rate is only dependent on the loss of a leaving group, so a strong base/poor nucleophile will react via SN1 as long as there are not plausible E1 pathways (e.g. sp3 bonded hydrogen alpha to carbocation).


text:
Where does the Fe(III) ion go?
Iron(II) sulphate solution is added to a test-tube. Following this, potassium manganate solution is added gradually to the test-tube as well. The observation seen is that the green solution turns pink at first, then potassium manganate turns colourless as manganese(VII) is oxidised to manganese(II).
But Fe(II) is oxidised to Fe(III) ions as well, which are brown in colour, so shouldn’t the solution turn brown instead of colourless?
I have seen these observations experimentally in the lab but still did not understand. I know I’m probably thinking about it the wrong way so any help would be much appreciated.

In solution, the iron(III) will be hexaaqua iron(III), 6 water molecules coordinating the iron ion, with the oxygen atoms being at the vertices of an octahedron.
Hexaaqua iron(III) has a high spin d5 electron configuration.
All the d-d transitions are spin forbidden.
Hexaaqua iron(III) is nearly colorless.

Your expectations are not completely unfounded. Even without the addition of special ligands, aqueous solutions of Fe(III) often appear yellow since they contain the ions $\ce{[Fe(OH)(H2O)5]^2+}$ and $\ce{[Fe(OH)2(H2O)4]+}$.
However, your experiment takes place under acidic conditions. (I guess, sulfuric acid was added.) This is apparent because the purple permanganate is reduced to the colourless $\ce{Mn^2+}$ ion:
$$\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}$$
Under neutral conditions, permanganate would be reduced to dark brown manganese(IV) oxide:
$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O}$$
Under the given acidic conditions, the colourless complex $\ce{[Fe(H2O)6]^3+}$ is stable:
$$\ce{[Fe(OH)2(H2O)4]+ + 2H+ <=> [Fe(OH)(H2O)5]^2+ + H+ <=> [Fe(H2O)6]^3+}$$
Therefore, the solution appears colourless.

This answer mentions about colours of aqueous iron solutions and how to test them (only)
$$\ce{10 FeSO4 + 2 KMnO4 + 8 H2SO4 -> 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O}$$
Iron(II) salts are usually very pale green in aqueous solution and iron(III) salts are yellow in aqueous solution. You can test what form of iron is present by adding thiocyanate, which gives red precipitate exclusively with iron(III): Place a drop of test solution on a spot plate place and add one drop of 1 per cent ammonium thiocyanate solution. A deep-red colouration may appear. Nitrites should be avoided in acid solution as they form nitrosyl thiocyanate $\ce{NOSCN}$, which yields a red colour, disappearing on heating similar to that with iron(III)
$$\ce{Fe^3+ + 3 SCN- ->Fe(SCN)3}$$
And iron(II) can be detected most reliably with 2,2'-bipyridine reagent where deep-red complex bivalent cation $\ce{[Fe(C10H8N2)3]^2+}$ is formed in mineral acid solution. Test a drop of the faintly acidified test solution with one drop of the reagent on a spot plate; a red colouration may be obtained.

Do you mean to say say that the solution will not turn colourless when the reaction is complete? Because according to my chemistry book it does (though it is not explained why). And I conclude from your answer that Fe(III) ppt are brown and Fe(III) in solution are yellow,am I right?

@Sara I meant to say that there must be colour, but as it is not there, i just gave you the method to check for which state of iron is there

@ADG Why do you say "there must be colour", considering that for high spin Fe(III) all d-d transitions are spin forbidden and Laporte forbidden?

@DavePhD you're last comment was above my knowledge, sorry


text:
Identify structure of compounds from their reactions and molecular formula

Compound P ($\ce{C7H6O}$) on refluxing with aqueous ethanolic $\ce{KCN}$ solution for 1 hour, forms Q ($\ce{C14H12O2}$). Q on oxidation gives R ($\ce{C14H10O2}$). R on refluxing with aqueous $\ce{NaOH}$ solution followed with acidification forms an $\alpha-$hydroxyacid S ($\ce{C14H12O3}$). What are the structures of the compounds?

My effort:
P has DBE (double bond equivalent) equal to 8, so 2 benzene rings (not possible with 7 carbons) or some cyclic or many double bonds might be necessary. Now some enolate reaction might be happening with $\ce{KCN}$, like  aldol or Perkin or Knoevnagel or Stobbe or Claisen or benzoin etc. Oxidation doesn't affect the formula so it might be becoming an isomer (that too a functional one). After that a Cannizaro might be the thing.

You weren't far from resolution. The substrate is benzaldehyde, first reaction benzoin condensation, so Q is benzoin. It is later oxidated to benzil - R ($\ce{C14H10O2}$) (not ($\ce{H12}$) - in oxidation formula must change). Finally benzil undergoes benzilic acid rearrangement, so S is benzilic acid.
As for how one could get to the solution... Compounds having 6-8 carbons and highly unsaturated are usually benzene derivatives. In this case benzaldehyde is only common thing with this formula (it was unlikely to be, for example, tropone). First reaction must be dimerisation, next oxidation, and hydration.


text:
Why doesn't nitric acid have a positive charge?
From the diagrams I've seen $\ce{HNO3}$ has a dative covalent bond so why doesn't it have a positive charge?

Do you mean an overall positive change or a positive charge on the nitrogen atom?

Wouldn't a positive charge on the nitrogen cause an overall charge on the molecule?

Not necessarily, see the nitro group.

One side of the $\ce{HNO3}$ molecule does have a positive charge, but this is neutralized by the negative charge resulting from the Oxygen atom with only one bond to the nitrogen. These charges cancel leaving an overall neutral molecule.


text:
How to find the valence of transition elements?
Let's take Iron for example

As we can see there are 4 desolate electrons in normal state. Can you tell me which electron moves to make the valence 2 and then 3. We have learnt to find the valence only using the configuration formula.

For the 2nd and 3rd rows for 4b-7b groups the highest oxidation state equal to the group number is preferable over all else and for the 1st row for 4b-7b groups the highest oxidation state is equal to the group number.
Unfortunately, for all other information you'll need case-by-case approach (i.e. plainly remember it).

Indeed, in some graduate inorganic classes, a common question is to give the oxidation states, common oxides, and a few other items for every transition metal.

The two 4s electrons are lost to form the $\ce{Fe^2+}$ ion and then the spin paired 3d electron is lost to form the $\ce{Fe^3+}$ ion. As a general rule the 4s electrons are lost before the 3d (at least for period 4 elements) and spin paired electrons are lost before unpaired ones.

where are these electrons gone?

The electrons are transferred to another atom or ion to form either another (usually negative) ion, or a neutral atom.
E.g. The reaction of iron with an acid such as hydrochloric acid:
$\ce{2Fe + 6H+ -> 2Fe^3+ + 3H2}$


text:
When will the mass of the reactant increase while being heated up?
Thermo-gravimetrc analysis, often refered to as TGA, measures the mass of an unknown solid substance as it heats up, of course for the elemental analysis of the very substance.  When the substance is heating up, due to special chemical or physical reactions, the mass is decreasing. e.g.: Dehydration. Anyhow, we know that in special circumstances the mass of the substance will increase. When? That is the question.
( This is not quantum physics! I am talking about rest mass, if you had wondered. As a result, the increase will have nothing to do with Einstein's relativity formulas.) The diagram below is an example of some normal polymer tested in normal conditions, no relations with the exceptions!

update: There also is a special type of gas that is blown above the surface of the substance.  The system is designed in a way that doesn't allow the gas to "interfere" with the measurements. And I just got lucky to find another substance experimented. I estimate it is going to be a liquid alkanol, maybe ethanol. Diagram two displays the experiment's results.
I know it seems a little bit vague, but all the info that I provided at above is all we have got to answer the question.

By "mass" you mean density aka mass per cm3 or do you mean literally total mass of some plastic object? Like Klaus written bellow,there is nothing from nowhere,if that gas reacts with that polymer and puts some of its atoms into it,that will be the source of the extra mass.If you mean density,it could perhaps be some kind of polymer that gets denser whem it transitions from solid to liquid phase.I have never heard about any polymer like that,I am not even sure if it is possible,thats my best guess given your description.

There's nothing from nowhere: Heating of a polymer in vacuum will never lead to an increase in mass.
In the presence of an acidic catalyst and water (or alkanols, etc.), however, polyesters might partly undergo cleavage and/or transesterification with a net mass uptake.

The only "special " circumstance I can think of is when the polymer can actually absorb or adsorb moisture or just any gas when it is heated and thus "increasing" the mass somehow.


text:
Why does coordinate covalent bond form?
Coordinate covalent bonds are bonds on which both electrons from one atom.
But why does this happen? Some may think it is because one of the bonding atoms have strong electronegativity. But experimental evidence suggests otherwise.
For example, carbon monoxide. The EN of carbon is 2.6, and oxygen 3.4. This may lead you to think that carbon will donate a pair of electron, but on Princeton.edu:

Carbon monoxide (CO) can be viewed as containing one coordinate bond and two "normal" covalent bonds between the carbon atom and the oxygen atom. This highly unusual description illustrates the flexibility of this bonding description. Thus in CO, carbon is the electron acceptor and oxygen is the electron donor.

Nevertheless, in some cases like NO2, it does follows that the less electronegative atom domates electron. But why??

That simply follows from the fact that oxygen normally cannot have three bonds. With three bonds it has to donate electrons, else it would end up with more than 8 electrons in its outer shell.

We can share the women, we can share the wine
We can share what we got of yours, cause we done shared all of mine

"Jack Straw" Grateful Dead
But seriously now, the most important thing in the Princeton.edu article (which explains at the bottom that its content comes from Wikipedia) is
"The distinction from ordinary covalent bonding is artificial"
Does $\ce{H2}$ come from a hydride and a proton or from two hydrogen atoms?  The reason a molecule is stable is independent of a particular path by which it is made.
For $\ce{CO}$, it is only in your mind that oxygen brought more wine (electrons) to the party, but what really matters is that there is enough wine (electrons) at the party for everyone to be satisfied.
$\ce{CO}$ is isoelectronic with $\ce{N2}$ and $\ce{CN-}$; the reason for bonding is the same for each.

Electronegativity affects coordinate covalent bonds; indeed atoms which accept electrons should be "electron hungry" enough to take them. However it stems mainly from that the acceptor doesn't have complete electron shell, thus acts as Lewis acid and donor acts as base.

If electronegativity is a factor in the reason why coordinate covalent bonds occur together with the incomplete electron shell, then why doesn't a coordinate covalent bond occur in CO2 rather than the well-known double bond?

Electronegativity is important factor in hydration of first group cations.  You seems to have some weird misconception; this one ? Or another?


text:
Abundant substances which I may use like coupling agents
What is the most abundant substances which I may use like coupling agent for Wood plastic composite with HDPE polymer? I don't want maleated polyolefin because they are not so abundant in my country.

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

If you don'y want to use maleated polyolefin (the best coupling agent for this composite), you can use oxidized polyethylene, ethylene propylene rubber, ethylene-co-glycidyl methacrylate or polyolefin elastomer.


text:
How does higher carbonyl compounds disolve in water?
We know that higher the carbon, the more is hydrophobic the compound nature is, Still when we take a compound like sugar , a hand of sugar, it readily dissolves in water? can someone explain me reg this

Sugar contains a lot of hydroxyl groups. So due to hydrogen bonds, sugar can readily dissolves in water.


text:
If the hydroxide ion is such a poor leaving group then why does water dissociate?
Water dissociates into the hydroxide and hydronium ions but, why does this happen if the hydroxide ion is such a poor leaving group. Alcohols don't undergo $\ce{S_{N}1}$ reactions because the hydroxide ion doesn't dissociate so why does it happen in water?

Only 1 in 10,000,000 water molecules dissociate.
HI, HBr and HCl fully dissociate, and I-, Br-, and Cl- and are much better leaving groups.

what is it exactly that makes the hydroxide ion such a bad leaving group?

Weak bases (conjugate bases of strong acids) are good leaving groups.  Strong bases are bad leaving groups.  H-, CH3-, etc. would be even worse than OH-.

Is SH- a good leaving group?

I would expect SH- to be a poor leaving group, but not quite as bad as OH-, based upon the pKa of H2S being 6.9.  I don't have any personal experience with it though.

@DavePhD You are avoiding the issue there I think, because the next logical question would be: "Then why is hydroxide such a strong base?". This question is more or less equivalent to what OP asked.

@Jori I don't follow what you are saying.  Strong bases are bad leaving groups.  H-, CH3-, and NH2- are even stronger bases than OH-, so they are even worse leaving groups.  I-, Cl-, and Br- are very weak bases, so they are much better leaving groups than OH-.

@DavePhD I meant that when OP asked why SH- is a bad leaving group answering with: "because it is not very acidic" is just moving the issue around a bit. It is like saying well SH- is not really acidic because it is a bad leaving group. It is a bit nitpicking from me and I do not question your abilities in anyway, but you see my point? If you truly want to understand why e.g. OH- is a strong base you should go into detail about the structure of OH-. That is, a better answer would be "well OH- is strongly basic because the oxyanion is a hard nucleophile and has a large electronegativity.

@yJori, yes I understand now, your explanation is more complete

Water dissociating into H+ and OH- isn't the same concept as free OH- leaving. A more accurate comparison would be forming NaOH by dropping Na metal in water, because in that case, OH- leaves water to become part of a new molecule as opposed to just ionizing in solution for a second before it becomes full water again.
Alcohols are very capable of SN1. Water is the leaving group as it is obviously protonated first. Benzyl alcohol to benzyl chloride, t-butanol to t-butyl bromide, etc.


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Why is it necessary to consider infinitesimal changes in p,V,T for H,U and G given that they're state functions?
State functions such as $G$ only depend on the state of the system and are not dependent on the "path" that took the system to that state (which would be the case for work, for example, which is not a state function.
We know that:
$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T$$
So…
$$\mathrm{d}G=\left(\frac{\partial G}{\partial p}\right)_T\mathrm{d}p+\left(\frac{\partial G}{\partial T}\right)_p\mathrm{d}T$$
Consequently, by comparing coefficients:
$$V=\left(\frac{\partial G}{\partial p}\right)_T$$ and $$-S=\left(\frac{\partial G}{\partial T}\right)$$
Just taking the equation involving $V$ now to save time and space:
$$\int_{p_1}^{p_2}V\mathrm{d}p=\int_{p_1}^{p_2}\left(\frac{\partial G}{\partial p}\right)_T\mathrm{d}p$$
Using the perfect gas equation and integrating leaves the result:
$$G(p_2)-G(p_1)=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$
But if $G$ is independent of the path taken to get to the final state, why shouldn't I use the equation
$$\Delta G=V(p_2-p_1)?$$

It is important to make clear any requirement that a variable be constant for each equation you write.
To go from
$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T$$
to
$$\mathrm{d}G = V\,\mathrm{d}p,$$
you would need to assume temperature is constant.
And to go from
$$\mathrm{d}G = V\,\mathrm{d}p$$
to
$$\Delta G = V(p_2 - p_1)$$
you would need to assume volume is constant.
If temperature and volume are both constant, the pressure is constant also.
So your final equation is only saying: when temperature, pressure and volume are constant, there is no change in Gibbs free energy.

When you write $\mathrm{d}G=V\,\mathrm{d}P-S\,\mathrm{d}T,$ you are talking about the differential change in $G$ between two closely neighboring thermodynamic equilibrium states, one at $(T,P)$ and the other at $(T+\mathrm{d}T,P+\mathrm{d}P).$
So, when you integrate this equation, you must integrate over a path comprised to a continuous sequence of thermodynamic equilibrium states.  Over such a path, at constant temperature, the molar volume $V$ changes with pressure $P,$ and this needs to be taken into account in the integration.


text:
Confused about parity of atom and chirality
OK. I'm working in computational chemistry and the problem that bothers me is:
Given a molecule, I know that it has a central asymmetric atom with some parity which is either "Anticlockwise" or "AntiAntiClockwise" (or a really stupid way to say clockwise). I have the molecule drawn on a 2D canvas (they are in one plane).
For this example you could imagine 4 atoms. I have to specify which atoms lie in a different plane (3D perspective) and I can't seem to understand why is the parity important in that case? I'm using the SMILES O[C@@](F)(Cl)Br notation.

I'm really confused about this concept and it seems like the internet doesn't offer me a lot of information about it. So please tell me how is parity related to the plane the atom lies in? How can I map the edges and dashes to the SMILES  of this structure?

Maybe you should be more specific.  What molecule is it?  4 total atoms or the central atom plus 4 more?

@DavePhD a total of 4 atoms.

I really would like to be more specific but I have no idea how to explain this to myself.

so a central atom with 3 other atoms at vertices of equilateral triangle approximately?

@DavePhD I hope this edit will help you udnerstand the problem

BTW, who's calling this "anti-anti-clockwise?" I can't find any use of that online.

My collegues >.>.It's killing me to see this in my code but oh well..life is life :D

This is really a cheminformatics problem. It's a reflection of the chirality or atomic stereochemistry about the atom you mention.
There are multiple ways of specifying chiral information in 2D and 3D.
As mentioned by Dave SMILES uses a clockwise and counter-clockwise from left to right in the string.
In 2D, we usually use wedge and hash notation. The problem is that some users use wedges and hashes for graphical appearance and not for stereochemical information (e.g., pseudo-3D look). So file formats also add atom parity to indicate the actual stereochemistry of atoms.
The "parity" idea has to do with the relationship between the atoms as they are ordered in the file, vs. the stereochemistry around the central atom. As indicated in the other answer, these are enantiomers, even though they use the same "@@" SMILES stereochemical indicator:

$\ce{O[C$@@$](F)(Cl)Br}$
$\ce{O[C$@@$](F)(Br)Cl}$

Note that the atom ordering is different, so the "parity" will be different. For your example, $\ce{O[C$@@$](F)(Cl)Br}$ I get "1" for the MDL parity.
From the MDL specification (Appendix A: Stereo Notes):

Number the atoms surrounding the stereo center with 1, 2, 3, and 4 in order of increasing atom number (position in the atom block) (a hydrogen atom should be considered the highest numbered atom, in this case atom 4). View the center from a position such that the bond connecting the highest-numbered atom (4) projects behind the plane formed by atoms 1, 2, and 3.
Sighting towards atom number 4 through the plane (123), you see that the three remaining atoms can be arranged in either a clockwise or counterclockwise direction in ascending numerical order.

For more details, you might want to look at the Open Babel documentation, OpenSMILES or the MDL specification.

This anwser seems a lot better than mine.  I only know about chirality, not parity. I could only think of gerade and ungerade.

Definitely keep yours - it's short, succint, and correct. Anyone who wants the full-winded explanation can read mine. The main problem is that there are umpteen ways to specify chirality in 0D or 2D chemistry formats and people keep inventing new ones. :-)

In SMILES notation, the order of the remaining three atoms corresponds to clockwise order when viewed in the direction looking from the first listed atom toward the carbon center.
So you have written $\ce{O[C@@](F)(Cl)Br}$.  Looking in the direction from the O atom toward the C atom, the clockwise order of the remaining atoms is F, Cl, Br.
You could write the other enantiomer as $\ce{O[C@@](F)(Br)Cl}$.

The molecule is not 2 dimensional, it is 3 dimensional, in this case tetrahedral. Since the vertexes have different CI priorities, the molecule will spin one way in space, this is visualized by placing the lowest priority atom in the back and assigning priorities to the remaining. This spin causes rotation of polarized light and other effects.
If you are familiar with chirality, which is an extension of Group Theory, it would seem to answer the questions you are having, unless I totally misunderstood what it is you are asking.


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Why is hydrogen bonding more significant than any other interaction between dipoles?
Solvents can be categorized as polar and protic. I understand that to be protic there had to be a significantly polar bond involving a hydrogen atom. However, how is the a hydrogen bond different in any way to another dipole-dipole interaction and therefore, what's the need for the distinction between polar and protic solvents?
When considering the hydration enthalpy $\triangle_{hyd} G^{\theta}$ it is not sufficient to simply use the Born equation:
$$\triangle_{solv} G^{\theta}=-\frac{N_Ae^2z^2}{8 \pi \epsilon_0}(1-\frac{1}{\epsilon_r})$$
BECAUSE this only considers Coulombic interactions (i.e hydrogen bonds are not solely coulombic interactions). My textbook seems to suggest that hydrogen bonds are somehow different and more significant. My question is why? Surely there should be no distinction between different polar bonds; what makes $\alpha^{\delta-}-H^{\delta +}$ special?

Hydrogen bonds often have a significant covalent component. This can be seen in many aspects of the hydrogen bond: it is directional, unlike Coulombic interactions, and the bond length is generally significantly shorter than predicted by Coulombic models. There is great variation in the covalent character of hydrogen bonds depending on their environment, with e.g. bond lengths between 2.2 and 4.0 Angstroms known.
Other intermolecular interactions have significant covalent components; there is nothing special about hydrogen per se but it is the most important example.

Indeed, for hydrogen bonds, you can consider some of the covalent character like a 3-center two-electron bond.


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Does more branching lead to lower boiling point in alkanes?

List the following alkanes in order of increasing boiling point.
A. $\ce{CH3(CH2)4CH3}$
  B. $\ce{(CH3)2CHCH(CH3)2}$
  C. $\ce{CH3CH2CH(CH3)CH2CH3}$
Answer: $C<B<A$, low to high.

Why is it that B has not a lower boiling point than C? Is it not more branched? I believe branching decreases surface area, leading to less intermolecular interactions, and to a higher boiling point.

Possible duplicate of http://chemistry.stackexchange.com/questions/6944/why-do-branched-chain-compounds-have-lower-boiling-points-than-the-corresponding?rq=1

Is there a mistake here?

B has more branching and this means that molecules of B pack less efficiently and so there are fewer Van der Waals forces between them and so B has the lowest boiling point. On the other hand A is a linear molecule and so packs efficiently, has stronger VdWs forces and so has the highest boiling point. As @DavePhD's data shows $\ce{B < C < A}$

$A$ n-hexane, boiling point 69 degrees C.
$B$ 2,3 dimethyl butane, boiling point 58 degrees C.
$C$ 3-methyl pentane, boiling point 63 degrees C.
$B<C<A$


text:
Is there a substance that would react with seawater to produce a dense, insoluble, nonporous precipitate?
Is there a substance that would react with seawater to produce a dense, insoluble, nonporous precipitate? Alternatively, is there a hygroscopic substance that will absorb water and expand into a dense/tough non-permeable mass? This is a material to seal off a small inner diameter pipe once water intrudes into it

A superabsorbent polymer sounds appropriate for this application, and the wikipedia article suggests they are used for "blocking water penetration in underground power or communications cable".
Will the water be removed from the pipe again after it has intruded, and if so do you need the seal to remain impermeable? I would expect this kind of polymer to dry out and shrink with time in the absence of water and it would not necessarily then be able to swell again with the same performance.


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In which beta-elimination reaction do the structures of the carbocations formed influence the reactivity?

I believe there is a mistake. If we go back to the mechanisms, E2 does not go through a carbocation. So, there should not be any impact.
E1 does however. So, E1 should be circled. Is there a missing point or argument in my logic.


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Why does microwave heat up things so much more quickly than visible light?
I wonder why microwave has more "energy" to heat things up than visible light does. Microwave has a much longer wavelength, therefore should supposedly has less energy than visible light.
But does wavelength and frequency really have to do with how fast a certain type of electromagnetic radiations give an object heat, or it is something to do with how a microwave oven is built?

Photon for photon the effect should be greater with visible photons provided the molecule does not have a very high fluorescence/phosphorescence yield. This is because all the energy absorbed has to end up as heat. Also, for example, a 500 w light bulb mainly generates heat and only a tiny, tiny fraction of this energy as light but this is not true of microwave ovens. So the comparison is more difficult that just looking at the power consumption. 500 W of laser light is the sort of thing that burns holes through walls.

Microwaves don't have more energy, they just resonate at the frequency that causes molecular bonds to rotate. This specifically applies to dielectric molecules, molecules like water that have electric dipoles.
From Wikipedia:

Water, fat, and other substances in the food absorb energy from the microwaves in a process called dielectric heating. Many molecules (such as those of water) are electric dipoles, meaning that they have a partial positive charge at one end and a partial negative charge at the other, and therefore rotate as they try to align themselves with the alternating electric field of the microwaves. Rotating molecules hit other molecules and put them into motion, thus dispersing energy. This energy, when dispersed as molecular vibration in solids and liquids (i.e., as both potential energy and kinetic energy of atoms), is heat. Sometimes, microwave heating is explained as a resonance of water molecules, but this is incorrect such resonances occur only at above 1 terahertz

For reference, energy is related to wavelength by the formula:
$$E=\frac{hc}{\lambda}$$
and
$$\lambda=\frac{c}{f}$$
where

$\lambda$ is wavelength
$c$ is the speed of light
$f$ is frequency
$h$ is Plank's constant
$E$ is energy

You can see by the first equation that as $\lambda$ gets smaller $E$ gets larger.

Molecular bonds do not rotate, rather the whole molecule rotates in microwave spectroscopy.

The resonance frequency of water bonds is at the frequency of microwaves, which causes them to move and the resulting friction causes the increase in heat, according to my Physics professor.

The key difference with microwaves is the amount of energy emitted
Microwaves don't have more "energy" than visible light per photon. But this is irrelevant in the case of a microwave oven.
There are two reasons for this but the first one one (as described in the existing answer) is actually a distraction: microwaves are relatively efficient at exciting molecular rotations and vibrations (especially of ions) so dumping their energy as heat in the object they are interacting with (usually your food).
But this is a distraction from a more significant point: microwave ovens push out a lot of microwave energy, usually a little under a kilowatt. This is about 10 times the energy radiated by a traditional incandescent (visible) lightbulb. If you could dump that much visible light into a small volume and your food absorbed it, it would cook. But you are more likely to end up with a very hot box and food that wasn't cooked efficiently. Many electric heaters (radiating red and near infra-red photons) push the same sort of power as a microwave, but they radiate it over a much larger area and don't cook you because the energy is dumped over a very much larger volume. Unless you stand too close.
Electric grills are basically just very powerful incandescent bulbs tuned to emit most of their radiation in the near infra red. They emit similar power to a microwave (~kW) but don't cook food as efficiently as infra red is mostly absorbed by the surface of the food they heat (microwave energy penetrates more deeply allowing food to warm from the inside. Interestingly very salty foods may not cook as well in a microwave as surface effects (where the radiation couples strongly with the dielectric fields caused by salty ions) absorb more of the microwave energy near the surface.

While it's 10x the total energy radiated by an incandescent bulb, only 2% of a lightbulb's energy is radiated in the visible spectrum. 1 kW of visible light is about as much as is produced by 500 light bulbs.

@romkyns I should have said "visible and infrared light" (since the wattage rating includes the full range of emitted energy). But the point is that the overall energy flow in a microwave is much larger.

You should revise "microwaves are relatively efficient at exciting molecular vibrations". This is basically wrong.

microwaves excite molecular rotations not vibrations.

@M.Farooq Yes, I was oversimplifying and have revised that sentence. But the idea that the only heating mechanism is coupling to molecular rotations is also wrong. Direct interaction with ions is also a big component in microwave heating and sometimes the dominant one.

@porphyrin Yes, the primary coupling is with molecular rotation not vibration, but microwaves also strongly excite the movement of free ions and this is sometimes the dominant mechanism of heating. But neither is a key issue in understanding the difference between microwaves and visible light. The key issue is tot total amount of power dumped in the enclosed space.

Microwave heating is basically dielectric heating. There is no need to invoke ionic movement, otherwise pure water would never heat up in a microwave oven. https://en.wikipedia.org/wiki/Dielectric_heating

@M.Farooq This explains one major ways microwaves heat food (and the major way they heat pure water). But it is far from complete in real mixtures where ionic components also exist. This is why salty foods often cook much more strongly on their surface and not evenly. See this article.

Right, salts decrease the temperature, pure water would heat more quickly as compared to salty water. Please note that "exciting molecular rotations and vibrations (especially of ions) so dumping their energy" is still problematic because microwaves do not excite any normal mode vibrations like IR. It would be better to say translation for ionic conduction.

@M.Farooq I've no idea what you mean by "salts decrease the temperature" in this context. Also the detailed mechanism of microwave heating is mostly irrelevant here: it is the power levels that matter which is why I have not tried to be very precise about the mechanism.

I was referring the effect mentioned here, that if we have salty water, the temperature rise is not as great as it would be if it were pure water. https://www.sciencedirect.com/science/article/pii/S1878535211000189


text:
How to calculate the average size of cubic crystals of KCl powder?
The total area of $\pu{0.5 g}$ sample of $\ce{KCl}$ powder is measured to be $\pu{1620 cm2}$. If the powder consists of tiny cubic crystals, what is the average size of these crystals?
I'm studying for a physical chemistry final and am not sure how to do this practice question.
I know that the density is
$$\delta = \frac{m}{A},\label{one}\tag1$$
where $m$ is the mass and $A$ is the surface area, so do I just find
$$A=\frac{m}{\delta}$$
to find the average "size"?
Or do I use \eqref{one} and then find out what the unit cell structure of $\ce{KCl}$ powder is and use that to estimate what the surface area of the crystal is?

For a cube, the surface area is given by the formula $6a^2$ where $a$ is the length of one side. The volume is given by $a^3$. Surface area per volume is
$$\frac{6a^2}{a^3}= \frac{6}{a}.$$
We know the total surface area to be $\pu{1620 cm^2}$. We can calculate the total volume from the mass of the $\ce{KCl}$ and its density.
\begin{align}
  V &= \frac{m}{\rho}\\
  \rho(\ce{KCl}) &= \pu{1.98 g//cm^3}\\
  V &= \frac{\pu{0.5 g}}{\pu{1.95 g//cm^3}}\\
    &= \pu{0.3 cm^3}
\end{align}
An average crystal should have the same ratio as the total. So we set the two equal and solve for the length.
\begin{align}
  \frac{6}{a} &= \frac{\pu{1650 cm^2}}{\pu{0.3 cm^3}}\\
  a &= \frac{6 \times \pu{0.3 cm^3}}{\pu{1650 cm^2}}\\
  a &= \pu{9E-4 cm}
\end{align}


text:
Which Lewis Structure for BeCl2 is more commonly seen?

One is more stable according to its charge. The other one does not satisfy as much as the first does the formal charge rule but has a complete octet.

It forms a polymer with bridging Cl atoms.

Beryllium chloride has a significant ionic contribution to its electronic structure. A quick calculation, DF-BP86/def2-SVP, reveals that the charge of beryllium is $\mathbf{q}(\ce{Be})=0.8$ and of chlorine $\mathbf{q}(\ce{Cl})=-0.4$, based on natural bond orbital analysis.
The Lewis structures best describing this are depicted below. The percentage indicates how much Lewis character this model actually describes based on the before mentioned calculation.

Of course this is just an approximation and the truth lies somewhere in between, like Eljee already mentioned. For these bonding situation, resonance is unavoidable. In quantum chemistry these phenomena are usually very well described by valence bond theory.
Another possibility, which is quite complementary, is looking at the structure with molecular orbital theory. Since the molecule has $D_\mathrm{\infty{h}}$ symmetry, you can observe $\pi$ bonds, that are delocalised over the whole molecule. And these are only depicted in the right hand side structure above. Below are the valence orbitals of $\ce{BeCl2}$, DF-BP86/def2-SVP.

So you can see, that one Lewis structure is not enough to actually describe the bonding in that molecule well, but as a first approximation, the left hand structure above should be sufficient.

The two structures drawn are both contributing to a total distribution of the charge. These are called Resonance forms. In some cases it means the electrons are equally distributed over a molecule, due to symmetry. In the case of your molecule the contribution of the two forms will most likely not be equal, but the truth lies somewhere in between the two structures you have drawn.


text:
Determining the direction of chemical reactions based on the activity series

Based on the activity series (table of standard reduction potentials), indicate whether or not the following reactions would occur in the direction written.

Don't worry about what the activity series is, I don't even know what it is so it isn't important. I'm more confused on the procedure and how to find the answer. Would you just find the oxidation states, then find the half reactions according to what the oxidation states are and find the voltage according to the reduction potentials sheet, then be able to tell whether it was spontaneous therefore whether it would run in the direction it was written?
Here are a few of the first questions, the easy ones.
$$\ce{Ag + KNO3 → AgNO3 + K}$$
$$\ce{Zn + AgNO3 → Ag + Zn(NO3)2}$$
$$\ce{Li + H2O → LiOH + H2}$$
$$\ce{Cu + FeSO4 → Fe + CuSO4}$$

It would be useful to include the reduction potentials with the question. Write out the half reactions, find the potentials and see if their reaction goes in the same direction. If its opposite, flip the sign. When potential has a positive value, its considered favored in that direction.

Without the value of standard reduction potentials, you cannot answer this question.


text:
Determine the molecular formula of the compound using percentage composition
I’ve attempted to solve this problem with percent composition, but it says the answer is wrong. Is it a $pV=nRT$ problem, and if so, can I use the units as they are given?

A $\pu{0.02867 g}$ sample of gas occupies $\pu{10.0 ml}$ at $\pu{290.0 K}$ and $\pu{1.10 atm}$. Upon further analysis, the compound is found to be $38.734 \%~ \ce{C}$ and $61.266\%~\ce{F}$. What is the molecular formula of the compound?

I keep on getting $\ce{CF}$ for the answer via percent composition.

You're actually on the right track. Looking at the percent composition, you've correctly identified that the ratio of $\ce{C}$ to $\ce{F}$ atoms is 1:1, however, you cannot assume that the formula is just $\ce{CF}$ (which isn't a known compound), it could be any compound with that ratio, $\ce{C2F2}$, $\ce{C3F3}$, $\ce{C4F4}$, etc.
The way to narrow it down is to use the gas data. You are right that you need to use the ideal gas law. If you look at the equation, you have everything except the number of moles, so you're interested in finding $n$:
$$n=\frac{PV}{RT}$$
As for the units, that depends on what units you have in $R$. If you have $R$ in $\mathrm{J/(mol\cdot K)}$, then $P$ needs to be in $\mathrm{Pa}$ and $V$ in $\mathrm{m^3}$, if it's in $\mathrm{L\cdot atm/(mol\cdot K)}$, then $P$ should be in $\mathrm{atm}$ and $V$ in $\mathrm{L}$. You can either convert what you have to match $R$ or convert $R$ to match your variables; it doesn't really matter.
Once you have the number of moles, you can calculate the molar mass of the compound from the sample mass:
$$m_\mathrm M=\frac{m_\mathrm s}{n}$$
Then, since we know the ratio of $\ce{C}$ to $\ce{F}$ is 1:1, the molar mass will be some multiple of the mass of $\ce{CF}$ which is $ 31.0091\ \mathrm{g/mol}$, so just divide the molar mass by that and you'll know $n$ in $\ce{C_nF_n}$.
Also, while not a sure way if fake compounds are possible answers, not all compounds with formula $\ce{C_nF_n}$ are known ($\ce{CF}$, for instance), so you can narrow down your choice of answer.


text:
Is hydrogen bonding generally defined to include only three period two elements?
Traditionally, hydrogen bonding has been defined to only include interactions between a positively polarized hydrogen and three period-two elements: nitrogen, oxygen, and fluorine.
Why was this definition proposed? A common question is why can't chlorine, by the aforementioned definition, hydrogen bond?
Does this have anything to do with hard and soft acid-base theory? Hydrogen is a hard acid, and chlorine is a comparatively soft base ... and the period two elements are all relatively hard?

I don't think there is any such traditional definition requiring $\ce{N}$, $\ce{O}$ or $\ce{F}$.
For example, in table 7 and the discussion thereof in Hydrogen Bonding Annual Review of Physical Chemistry Vol. 22: 347-385 the hydrogen bonding in the following species are discussed:

$\ce{ClHCl^-}$
$\ce{BrHBr^-}$
$\ce{IHI^-}$
$\ce{BrHCl^-}$

as well as related neutral radicals.
Hydrogen bonds in $\ce{S-H}$ containing compounds are discussed.  These are just examples, others that do not involve $\ce{F}$, $\ce{O}$ or $\ce{N}$ are also discussed.

The following is copied (footnotes omitted) from Definition of the Hydrogen Bond (IUPAC Recommendations 2011) Pure and Applied Chemistry Volume 83, Issue 8 (Aug 2011)

2. DEFINITION
The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a
  molecular fragment $\ce{X-H}$ in which $\ce{X}$ is more electronegative than $\ce{H}$, and an atom or a group of
  atoms in the same or a different molecule, in which there is evidence of bond formation.
A typical hydrogen bond may be depicted as $\ce{X-H\bond{...}Y-Z}$, where the three dots denote the bond.  $\ce{X–H}$ represents the hydrogen bond donor. The acceptor may be an atom or an anion $\ce{Y}$, or a fragment or a molecule $\ce{Y-Z}$, where $\ce{Y}$ is bonded to $\ce{Z}$. In some cases, $\ce{X}$ and $\ce{Y}$ are the same. In more specific cases, $\ce{X}$ and $\ce{Y}$ are the same and $\ce{X-H}$ and $\ce{Y-H}$ distances are the same as well leading to symmetric hydrogen bonds. In any event, the acceptor is an electron rich region such as, but not limited to, a lone pair of $\ce{Y}$ or π-bonded pair of $\ce{Y-Z}$.
The evidence for hydrogen bond formation may be experimental or theoretical, or ideally, a combination of both. Some criteria useful as evidence and some typical characteristics for hydrogen bonding, not necessarily exclusive, are listed below, numbered E# and C#, respectively. The greater the number of criteria satisfied, the more reliable is the characterization as a hydrogen bond.
2.1 List of criteria
For a hydrogen bond $\ce{X-H\bond{...}Y-Z}$:

(E1) The forces involved in the formation of a hydrogen bond include those of an electrostatic origin, those arising from charge transfer between the donor and acceptor leading to partial covalent bond formation between $\ce{H}$ and $\ce{Y}$, and those originating from dispersion.
(E2) The atoms $\ce{X}$ and $\ce{H}$ are covalently bonded to one another and the $\ce{X-H}$ bond is polarized, the $\ce{H\bond{...}Y}$ bond strength increasing with the increase in electronegativity of $\ce{X}$.
(E3) The $\ce{X-H\bond{...}Y}$ angle is usually linear (180º) and the closer the angle is to 180º, the stronger is the hydrogen bond and the shorter is the $\ce{H\bond{...}Y}$ distance.
(E4) The length of the $\ce{X-H}$ bond usually increases on hydrogen bond formation leading to a red shift in the infrared $\ce{X-H}$ stretching frequency and an increase in the infrared absorption cross-section for the $\ce{X-H}$ stretching vibration. The greater the lengthening of the $\ce{X-H}$ bond in $\ce{X-H\bond{...}Y}$, the stronger is the $\ce{H\bond{...}Y}$ bond. Simultaneously, new vibrational modes associated with the formation of the $\ce{H\bond{...}Y}$ bond are generated.
(E5) The $\ce{X-H\bond{...}Y-Z}$ hydrogen bond leads to characteristic NMR signatures that typically include pronounced proton deshielding for $\ce{H}$ in $\ce{X-H}$, through hydrogen bond spin–spin couplings between $\ce{X}$ and $\ce{Y}$, and nuclear Overhauser enhancements.
(E6) The Gibbs energy of formation for the hydrogen bond should be greater than the thermal energy of the system for the hydrogen bond to be detected experimentally.

2.2 Some characteristics of hydrogen bonds

(C1) The $\mathrm{p}K_\mathrm{a}$ of $\ce{X-H}$ and $\mathrm{p}K_\mathrm{b}$ of $\ce{Y-Z}$ in a given solvent correlate strongly with the energy of the hydrogen bond formed between them.
(C2) Hydrogen bonds are involved in proton-transfer reactions ($\ce{X-H\bond{...}Y -> X\bond{...}H-Y}$) and may be considered the partially activated precursors to such reactions.
(C3) Networks of hydrogen bonds can show the phenomenon of co-operativity, leading to deviations from pair-wise additivity in hydrogen bond properties.
(C4) Hydrogen bonds show directional preferences and influence packing modes in crystal structures.
(C5) Estimates of charge transfer in hydrogen bonds show that the interaction energy correlates well with the extent of charge transfer between the donor and the acceptor.
(C6) Analysis of the electron density topology of hydrogen-bonded systems usually shows a bond path connecting $\ce{H}$ and $\ce{Y}$ and a (3,–1) bond critical point between $\ce{H}$ and $\ce{Y}$.

I have been taught that H bonding involves only these three elements from two professors and it's also the introductory textbook definition. That being said, it is the textbook definition, and it's more than likely oversimplified. But why might textbooks limit their discussions to these three elements?

The professors/authors probably want to convey that if you see O-H, N-H and F-H you should be thinking about hydrogen bonding, while otherwise you usually don't need to consider it.  But certainly a review article in a good journal takes priority.  The review article also says "A large body of infrared spectroscopic evidence now
shows that acetylenic protons form hydrogen bonds" and cites 7 references.

how exactly is hydrogen bonding defined?

@Dissenter There was a 1939 seven volume series on S-H hydrogen bonding http://pubs.acs.org/doi/pdf/10.1021/ja01266a052, that explains that it has been "shown definitely that a hydrogen attached to a highly negatively substituted carbon atom" can participate in a hydrogen bond, so it was already know by 1939 that hydrogen bonds are not limited to F, O, N.

@Dissenter Definition of the hydrogen bond (IUPAC Recommendations 2011) Pure and Applied Chemistry Volume 83, Issue 8 (Aug 2011) http://www.degruyter.com/view/j/pac.2011.83.issue-8/pac-rec-10-01-02/pac-rec-10-01-02.xml : The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a
molecular fragment X–H in which X is more electronegative than H, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation.

@Dissenter "But why might textbooks limit their discussions to these three elements [N, O, F]?" I'm sure that hydrogen bonding through N-H and O-H hydrogen bond donors are prominently discussed because of their biological significance and their role in explaining physical properties (e.g., boiling point and polarity).

@DavePhD - the point about bond strength getting stronger (in the context of H-bonding) as the angle approaches 180 degrees - I've heard this before, but why is this the case?

My book* has  this  graph of boiling points of various hydrogen-element compounds. $\ce{NH3}$, $\ce{H2O}$, and $\ce{HF}$ are the only ones that violate the trend, so it would make a bit of sense to call the interaction responsible only for those three and nothing else.

(*) Chemistry The Central Science by Theodore E. Brown, H. Eugene H LeMay, Bruce E. Bursten, Catherine Murphy.

With only a single binary hydrogen compound, hydrogen bonding is seen only with N, O, or F as you show.  But in more complex compounds or at least two compounds in combination, hydrogen bonding has become known with most nonmetals.  See the other answers.

In a rather different vein, hydrogen in water can donate weak hydrogen bonds to ... carbon!  The standard electrostatic model cannot work, but hydrogen bonding can also be rendered as a molecular orbital interaction and that does work with the methane-water interaction (Russ. J. Phys. Chem. 2016, 90 (10), 1978–1985).  Not just a chemical curiosity, this interaction contributes to the stability of the methane-water clathrate found on sea bottoms and in natural gas processing (see methane clathrate on Wikipedia).


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Why is covalent bonding stronger than electrostatic attraction?
I guess if we look at this problem from a "conjugate" perspective then the conjugate of a covalent bond is two elements with electrons lying around. On the other hand, with two electrostatically attracted things, the lack of a bond leaves two separate, charged particles.
But that still doesn't answer the question; why are electrons less stable than ions? I guess in the case of an ion though there are still electrons but also a nucleus the stabilize the electrons.

What makes you think ionic bonds are weaker than covalent bonds? It seems to me that ionic bonds are stronger than covalent bonds.

think about this question... does it take more energy to separate ionic compounds into their non ionic elemental state or does it take more energy to separate two covalent bonded atoms into their non ionic elemental state... I know apples and oranges ... NaCl vs H2O?

But can two ions come together to form a covalent bond? In that case, the "conjugate" is no longer two neutral atoms.

Ionic bonds can easily be as strong as covalent ones.
First off, let's be clear that almost everything has some ionic and some covalent character. Moreover, it's not an either-or situation. Some bonds have both strong ionic and strong covalent character.
For now, I'll stick to simple cases.
Let's consider $\ce{LiF}$, which seems a prototypical ionic bond. The interatomic distance is $\pu{156 pm}$ according to NIST. That gives a potential energy of $\pu{903 kJ/mol}$. Now, that assumes a full electron is transferred from $\ce{Li+}$ to $\ce{F-}$, which isn't the case (more like $0.8 - 0.9$). This simple calculation also ignores Born repulsion which is similar to the short-range Van der Waals repulsion between atoms.
A more complete analysis J. Chem. Educ. 2000, 77 (8), p 1076 gives a much better agreement. The real bond dissociation energy is $\pu{577 kJ/mol}$ (Source, PDF via the Internet Archive). A simple spherical ion model from the paper gives the bond enthalpy as $\pu{608 kJ/mol}$.

calculations based on a simple ionic model reproduce the bond energies of $\ce{LiF(g)}$ and $\pu{NaF(g)}$ to within $5\%$.

If we look up covalent bond dissociation energies, the triple bond in dinitrogen is $\pu{945 kJ/mol}$. Ok, that's perhaps stronger, but $\ce{LiF}$ only has a "single" charge and clearly some ionic compounds have greater charge transfer.
More "prototypical" covalent bonds include $\ce{C-H}$ ($\pu{410 kJ/mol}$) and $\ce{O=O}$ ($\pu{497 kJ/mol}$), both less than $\ce{LiF}$.
Now, as I said earlier, it's not the case of a bond being only covalent or only ionic.
To quote from R. Gillespie J. Chem. Educ. 2001, 78 (12), 1688:

The large atomic charges in $\ce{BF3}$ and $\ce{SiF4}$ are important to understanding the great strength of these bonds—which are the two strongest known single bonds, with bond dissociation enthalpies of $613$ and $\pu{565 kJ/mol}$, respectively. In comparison, the $\ce{C–C}$ bond in ethane has a bond dissociation enthalpy of only $\pu{345 kJ/mol}$. The great strength of the $\ce{BF}$ and $\ce{SiF}$ bonds arises from both the relatively large amount of shared density and the large atomic charges. So it is not strictly correct to say that these bonds are “predominately ionic”; if we are forced to use these terms we have to say that they are both strongly covalent and strongly ionic.

Finally, you can of course treat both covalent and ionic bonds using MO theory. For example, an approximate picture of $\ce{LiF}$ under MO theory might be something like this:

We readily see $\ce{LiF}$ has very high ionic character. For all intents and purposes, the "bonding" $\sigma$ orbital is localized on the $\ce{F}$ atom, there's very little energy difference between the $\pi$ and $\sigma$ here.

Your numbers are wrong for LiF, It should be 577KJ/mol. see: http://www.nist.gov/data/nsrds/NSRDS-NBS31.pdf

I like the care you have taken in choosing the examples in your answer, but I am also a little concerned that you are simply calculating the potential (I assume you just calculated the PE for separation of two charges from the distance) for LiF, while you used actual bond dissociation energies for $\ce{N2}$. That does not appear to be a fair comparison. However, I will change my answer so that it uses LiF as well as I think it is a better example to use.

How do you reconcile these two statements: "they are both strongly covalent and strongly ionic" "We readily see LiF has very high ionic character." You don't seem to stick to simple cases.

How does BF3 have strong ionic and covalent character? Aren't shared e- density and highly localized charges mutually exclusive? Or are you saying that there is a lot of shared e- density between B and each of the Fs but also B and each F takes on a significant degree of partial charge?

The article by Gillespie argues you should avoid the phrase "ionic character" because "ionic" and "covalent" aren't mutually exclusive. It's pretty clear from dipole moment in diatomic LiF and quantum calculations that ~0.9 electrons are transferred. OTOH, B-F and Si-F clearly have both shared electron density AND electrostatic contributions, so both are important in the bonding in these compounds.

Some bonds can be purely covalent, e.g., H-H, O=O, etc. LiF is almost completely ionic. But there are definitely species where both contributions are relevant. Moreover, I think it's silly to say "covalent bonds are inherently stronger" since there are clearly pretty strong ionic interactions on par with C-C, C-H, C-F, etc.

The real problem with "atomic charges" is that you cannot rigorously define them, so chemists will be eternally debating these concepts.

Maybe the article by Gillespie was right and we shouldn't use these terms.

Are u sure your MO diagram is correct? Because due to the large difference in electronegativity, wouldn't there be an inability of the atoms to have such interaction?

@TanYongBoon - $\ce{LiF}$ certainly exists and has an interaction - as indicated above, it has a measurable dissociation energy. I would say it's mostly ionic in character, which is probably what you're indicating.

I like the answer but I disagree. You might want to discuss it. Orbital coefficients are related to the share of electrons. On ionic compounds there is no electron share (not totally of course). So there is no bonding on this sense. However, there is a strong coulombic attraction.

Using your "conjugate" way of seeing two covalently bonded atoms as electrons, let's try $\ce{O_2}$ which gives you "$\ce{2O^{\bullet}}$". If you think about how the electrons are arranged in terms of electron configuration, you will see that $\ce{O^{\bullet}}$ has $\ce{[He]}2s^2 2p^4$ which is not a particularly stable electron configuration (if you only know the "secondary school electron configuration" you should recognise this as 2,6). Not having a full electron shell (2,8) means this is unstable.
If you do the same with $\ce{NaCl}$, you get $\ce{Na+}$ and $\ce{Cl-}$. Both of these have full shell electron configurations ($\ce{Na+}$ is $\ce{[Ne]}$ or 2,8; $\ce{Cl-}$ is $\ce{[Ar]}$ or 2,8,8). So the ions are indeed more stable in this sense.
However, relative stability of ionic and covalent bonding cannot be simply compared using your "conjugate" method, because you need to make sure that you are making the comparison under similar conditions. For example, you cannot really say $\ce{NaCl}$ is less stable than $\ce{O2}$ because NaCl dissociates into ions in water while $\ce{O2}$ does not, because the surrounding water molecules have a very strong dipole which will interfere much more strongly with ionic bonding (which is electrostatic) than covalent bonding in a homonuclear diatomic (which has no reason to dissociate into neutral radicals in water). You can possibly compare hardness of $\ce{NaCl}$ versus diamond, i.e. carbon that is purely covalently bonded, but graphite would not make such a good comparison because between the layers of graphite there are only weak Van der Waals interactions (and it is therefore softer because you are overcoming VdW interactions not covalent bonds). There are also special cases such as ionic salts with a particularly low melting point, and there are many examples of weak covalent bonds. You also want to pick examples with few confounding factors, such as large differences in electronegativity which can give ionic contributions to covalent bonds, or large differences in size and effective charge which can give covalent contributions to ionic bonding.
I will answer your question using two examples of giant lattices that show near-ideal ionic and covalent bonding: $\ce{LiF}$ versus diamond (with acknowledgements to Geoff Hutchison for the choice of LiF). Lithium, fluorine and carbon are all in period 2, so avoiding further complications due to differences in size such as comparing to $\ce{NaCl}$. I am "taking a shortcut" here by comparing boiling points, which is a bulk property that you can use for a giant lattice, because it is difficult to find a pure ionic diatomic.
$\ce{LiF}$ has a boiling point of 1909 K, while diamond sublimes at 3900 K. By this measure, it would appear that covalent bonding is stronger than ionic bonding in this case.
In $\ce{LiF}$, the distance between the lithium and fluoride ions is 202 pm*, while in diamond the distance between carbon atoms is 154 pm. In $\ce{LiF}$, the two oppositely charged ions are further away than the (positively charged) nuclei of any two carbon atoms in diamond that is held together by (negatively charged) electrons between them. What I am trying to point out is that, both forms of bonding involve interactions of positive and negative charges, but in covalent bonding the charges are much closer together, so the attractive electrostatic forces are much greater. This is why in this particular comparison, diamond comes out with stronger covalent bonding than the ionic bonding found in $\ce{LiF}$.
As I indicated in the third paragraph, there are many cases where the bonding within an ionic substance can be stronger or weaker compared to a covalent substance. This is why I have emphasised that everything that is discussed is only true for this particular comparison. It will be unwise to claim that one type of bonding is overwhelmingly stronger than another, because it always depends on what exactly is being compared. Given the near infinite combinations of atoms that can give rise to bonding that is ionic, covalent, and a mixture of the two, it would be difficult to make a full catalogue of covalent and ionic substances and compare these against each other.
*computed by taking half the distance of the cubic lattice.

"It will be unwise to claim that one type of bonding is overwhelmingly stronger than another". Definitely agree!

Short answer:
To quote Mehrdad: You're basically comparing an intermolecular force with an intramolecular force.
Ionic bonds are two charged atoms sitting next to each other with minimal overlap of their orbitals.
Covalents bonds are two atoms becoming one molecule with significant changes in orbital structure.
It is not always clear cut: there are atoms with part-ionic and part-covalent character or two atoms sitting next to each other with moderate overlap of their orbitals.
Here is a good explanation (for solids): Are metallic/ionic bonds weaker than covalent bonds?
Long answer:
Electrostatic bonds by definition are incapable of any significant bond formation. Bond formation involves the formation of a lower energy "bonding orbital(s)" out of two (or more) higher energy ordinary orbitals on adjacent atoms within the molecule. The formation of bonding orbitals gives covalent bonding orbitals much lower energy (not smaller, see diagram) than non-bonding orbitals which exist in the unbound state. This energy difference is larger than is possible for merely electrostatic attractions. Why is this the case? Because the electrons in bonding orbital are better able to counteract the positive charges of the adjacent nuclei. This makes sense because the negative charge on one molecule in an electrostatic arrangement is only able to neutralize the positive charge from its own nuclei. Bonding orbitals that are only present in covalent bonds share those electrons between two nuclei and thereby better neutralize the charges of both nuclei, which results in a lower energy for the bonded system than is possible in an ionic system.

Image Source

What's inherently better about bonding orbitals vs electrostatic attraction?

So...stabilizing electrons across multiple nuclei vs positive attracts negative ... What's the difference?

-1. The electrostatic attraction is small? Calculate the Coulomb interaction in $\ce{LiF}$ (ionic, + small nuclei) or $\ce{Al^{+2}O^{-3}}$ as part of $\ce{Al2O3}$.

@GeoffHutchison Al2O3 has a mixture of ionic and covalent character. Without the covalent character the bonds would be much weaker to the point of being non-competitive with covalent bonds. The same goes for LiF.

Everything has some covalent character, but I think it's silly to insist the covalent contribution is so high. I'll find numbers later.

I added comments to my answer above, but simple ionic analysis gets the bond enthalpy of LiF to ~5%. I have found no discussion of LiF as a covalent interaction, even in the diatomic.

It really depends on which covalent bonds and ionic bonds we are comparing.
Generally, the strength of a bond, ionic or covalent, depends on two things:

The atomic radius of the elements which take part into the bond
the difference in electronegativity between those 2 elements

I think this answer suffers from the fallacies of the question. If not you should explain each of the factors you mentioned.

There is nothing to be explained.Those 2 things come from theory.What else I forgot to mention is how many chemical bonds the elements form and if there are any bonds where the bonded pair comes from the same orbital which is of course an anti bond.

I mean where there are 2 bonded pairs and their electrons come from the same orbital for example an atom uses 2 electrons from its 1s orbital and the bonded atom uses 2 electrons from its 1s orbital.


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Periodic trends: why is effect of protons greater than electrons?
Why is it that adding protons has a greater effect than electron-electron repulsion on periodic trends like atomic radius and ionization energy (assuming # of shells constant)? It seems that if protons and electrons have the same charge and electrons are actually closer to the electrons in question (e.g. for ionization energy), that their effect would be the same?

This is because the attraction is not between each proton and it's corresponding electron, but between each electron and the nucleus. Since the number of protons increases the charge of the nucleus, but the charge of the electron stays constant, the attraction is greater. The attraction of the electron to the nucleus depends on the charge of the nucleus, and it is independent of the number of electrons in "orbit"
A great analogy for this is the solar system. (Gravity works very similarly to electrostatic force.) It won't matter if suddenly there is a mother earth orbiting in the outskirts of the solar system, Earth would be unaffected by that change, and we would still orbit the sun at the sae distance. However, if the sun suddenly becomes noticeably heavier, the planets will each be pulled in with a stronger force, which reduces the orbit radius. This will happen equally to each individual planet. It doesn't matter how many planets we have, The pull of the earth is only affected by the mass of the sun and the mass of earth.
This applies to the atom if you think of the nucleus as the sun and each electron as a planet. The force attracting the electron to the nucleus is only affected by the charge of the nucleus as each electron carries the same charge, and the charge is not the collective  charge of all electrons, but the individual charge. Hence all each electron perceives is the increase in the charge of the nucleus that is increasing its attraction to the nucleus.

But if electrostatic force is analogous to gravitational force, then adding an electron would be equivalent to adding a planet the mass of the sun that repels (rather than attracts) the other planets. Why would adding one proton to the nucleus exert a greater force than adding an electron if their mass is not what matters -- the charge stays the same?

@Jess You would be adding a proton which attracts the other electrons because the protons are positive whereas the orbiting electrons are all negative. If you are saying that the electron will repel the other electrons in orbit, then you are correct and this also plays a role in determining the size of the orbital. This is why the Ionic radius is different from the atomic radius.

So electron repulsion plays a role-- but why is the effect of adding protons greater? i.e when you move across a period, ionization energy increases, justified by the increasing # of protons added to the nucleus. But electrons are also being added simultaneously, with the same charge -- why doesn't the added repulsion cancel out the extra attraction?

Because electron density is diffuse, so the "shielding" of the nuclear charge by other electrons is imperfect. See, for example discussion on the effective nuclear charge.

Jonathanjaya has done convincing explanation; i would just say the success on the part of nucleus is "Unity brings strength" ; protons together build up a heavy positive centre instead of acting as a collection of individual protons, where as electrons remain"divided" and victory is that of the nucleus, which can retain all electrons overcoming  the comparatevely weak electron-electron repulsions; now if you wonder how protons can stay together in spite of their repulsion, you have to think about short range forces which are charge independent.

I have been looking for a clear, thoughtout explanation on this exact question for a few years now. @Jonathanjaya, I've incorporated this into my lessons. I hope that is agreeable to you:).


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Finding pH of tri-protic acid
I am studying for my final and I am given this problem:

A solution of sodium phosphate is made from 10.5 g sodium phosphate in
  150 mL of water. What is the pH of this solution

Given: Ka1, Ka2.
The solution says that we can simply calculate this pH with the following eqution:
$$pH=\frac{pK_{a1}+pK_{a2}}{2}$$
However this makes absolutely no sense to me, as this is the pH at the equivalence point.  What am I missing as towhy this equation can be used?

amphoteric
A chemical species that behaves both as an acid and as a base is called amphoteric. This property depends upon the medium in which the species is investigated: $\ce{H2SO4}$ is an acid when studied in water, but becomes amphoteric in superacids.

from the IUPAC goldbook.

For ampholytes like $\ce{NaH2PO4}$ the $\ce{pH}$ is independent on the concentration (in first approximation). Consider the reactions of that are possible:
\begin{align}\ce{
H2A-    + H3+O &~<=> H3A     + H2O  \tag1\\
H2A-    + H2O  &~<=> HA^{2-} + H3+O \tag2\\
HA^{2-} + H2O  &~<=> A^{3-}  + H3+O \tag3\\
}\end{align}
As the first approximation, we will assume that everything happening in equation $(3)$ is negligible, as is will only very little influence the concentration of $\ce{HA^{2-}}$. Therefore we can focus on the acidity constants $K_{a_1}$ of $(1)$ and $K_{a_2}$ of $(2)$:
\begin{align}
K_{a_1}  &= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-})   }{c(\ce{H3A})}  \tag4\\
K_{a_2}  &= \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \tag5\\
\end{align}
We will just go ahead and multiply these equations, since they are coupled, simultaneously happening processes. Then we will cancel whatever cancels.
\begin{align}
K_{a_1} \cdot K_{a_2} &= \frac{c(\ce{H3+O})\cdot c(\ce{H2A-})   }{c(\ce{H3A})} \cdot
  \frac{c(\ce{H3+O})\cdot c(\ce{HA^{2-}})}{c(\ce{H2A-})} \\
K_{a_1} \cdot K_{a_2} &= c^2(\ce{H3+O})\cdot  \frac{c(\ce{HA^{2-}})}{c(\ce{H3A})} \tag7\\
\end{align}
Now another major assumption is that the reactions $(1)$ and $(2)$ are happening to the same extent and therefore we can approximate
$$c(\ce{HA^{2-}})=c(\ce{H3A})\tag8$$
and rewrite $(7)$ as
\begin{align}
c^2(\ce{H3+O}) &= K_{a_1} \cdot K_{a_2}\\
c(\ce{H3+O})   &= \sqrt{K_{a_1} \cdot K_{a_2}}\\
\ce{pH}        &= \frac12\left(\mathrm{p}K_{a_1}+\mathrm{p}K_{a_2}\right).
\end{align}
However, this equation only works as a first approximation. Usually the coupled equilibria are very complex and in most instances a computer needs to be involved to calculate it accurately. More information on polyprotic acids.


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How does the oxidation of NaI by sodium hypochlorite to give I+ work?
I'm in trouble with the oxidation of NaI by $\ce{NaClO}$, the reaction should give $\ce{I+}$ which then reacts by electrophilic substitution with an aromatic complex. As of now I've come to:
$$\ce{ ClO- (aq) + 2I- (aq) + 2H+(aq) -> I2 (s) + Cl- (aq) + H2O (l)}$$
But I can't then find a way to obtain then the electrophilic $\ce{I+}$ that should react.

I think the reaction should be done in basic solution, not acidic solution.  "One equivalent each of sodium iodide (2.77 g) and sodium hydroxide (0.74 g) was added, and the solution was cooled to 0 C. Aqueous sodium hypochlorite (34.50 g, 4.0% NaOCl) was
added dropwise over 75 min at 0-3 C" J. Org. Chem. 1990,55, 5287-5291.
So perhaps:
$$\ce{ ClO- (aq) + I- (aq) + H2O (l)  -> I+ (aq) + Cl- (aq) + 2OH- (aq)}$$

Edit: after looking at the cited article more closely, the hydroxide was only being added because of the acidity of the iodophenol product, so you could write
$$\ce{ ClO- (aq) + I- (aq) + 2H+(aq)  -> I+ (aq) + Cl- (aq) + H2O (l)}$$
also.

You'd think the active species was iodine monochloride.  I+ on its own looks improbable, it's too Lewis-acidic to survive on its own, especially in water.

@AbelFriedman In section 16.2.10 of Inorganic Chemistry by Gopalan, it says $\ce{I+}$ can stablized by solvent molecules and there are species such as $\ce{I_2+}$, $\ce{I_3+}$, etc.


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Maximum coefficient of friction between nylon and polyethylene
For a research project I found out last minute that I need the coefficient of friction between nylon and Ultra-high-molecular-weight polyethylene. After hours of scouring google nothing useful turned up.
Since I actually need the maximum and not the precise value, I came here to ask what that could possibly be and if there is any literature on that. Basically I need to find out what the maximum friction could be, caused by a Dyneema (polyethylene) rope on a plastic nylon pulley.
I was hoping there is some sort of "rule" what the maximum coefficient of friction could be for 2 smooth plastic surfaces.