Why is oxygen oxidized with lead instead of being reduced? Oxygen is more electro negative yet in the redox reaction

$$\ce{2PbO(s) -> 2Pb(s) + O2(g)}$$

my text book says lead is reduced and oxygen is oxidized. How can this be? It doesn't make sense to me. Oxygen is usually $2-$ while lead is usually $2+$.

Your confusion stems from that fact that you are trying to look at what reactions are favorable instead of the reaction as written.

Oxygen is readily reduced because it is a strong oxidant. You can check any reduction half reaction with oxygen on the left side. Chances are that the reduction potential will be fairly positive.

In the reaction you have written, oxide is being oxidized to oxygen. Nowhere (including your textbook) do we claim that this reaction is favorable or spontaneous in the forward direction. In fact, if you consult a table of free energies of formation (example), you would find an entry like:

$$\ce{PbO (s, red)}, \Delta G^{\varnothing}_{f}=-189.3\ \mathrm{kJ}\ \mathrm{mol}^{-1}$$
or
$$\ce{PbO (s, yellow)}, \Delta G^{\varnothing}_{f}=-188.5\ \mathrm{kJ}\ \mathrm{mol}^{-1}$$

These values indicate that indeed, the reverse reaction is the one that's favorable, the one where oxygen serves as an oxidizing agent and is reduced.

However, that is not the reaction as written. The notation of chemistry allows us to describe reactions that are hypothetical or unfavorable.


How can rocks absorb CO2? In the beginning of the earth, there was at a certain point so much $\ce{CO2}$ in the air that most of it was absorbed by rock. The consequence of that was the snowball earth, because the atmosphere was empty; all the heat of the earth couldn't be more absorbed by $\ce{CO2}$, so it was getting cold. But how can rock ($\ce{SiO2}$?) absorb $\ce{CO2}$?

rock ($\ce{SiO2}$)?

Rock is not strictly $\ce{SiO2}$. Rocks are composed mostly of silicates: minerals such as feldspars $\ce{CaAl2Si2O8}$ or $\ce{(Na,K)AlSi3O8}$, olivine $\ce{(Mg,Fe)2SiO4}$, pyroxene $\ce{CaMgSi2O6}$, etc.

There are two ways in which rocks can absorb carbonates, one being a more direct form of the other.

Weathering of rocks by carbonic acid. $\ce{CO2}$ reacts with $\ce{H2O}$ to form acid and bicarbonate: $\ce{CO2 + H2O -> H+ + HCO3^-}$. This acid would then trigger: $\ce{4KAlSi3O8 + 4H+ + 2H2O -> 4K+ Al4Si4O10(OH)8 + 8SiO2}$. This is the reaction of orthoclase (a type of feldspar, common in granites) to form kaolinite (a clay, aka dust or mud). All of the resulting cations ($\ce{Ca2+, Mg2+, K+, Na+}$, etc. and the bicarbonate end up washing to the oceans, where they precipitate carbonates: solid rocks composed of mostly calcite ($\ce{CaCO3}$) and dolomite ($\ce{CaMgCO3}$), thus overall removing  $\ce{CO2}$ from the atmosphere.
Directly reacting of $\ce{CO2}$ with solid rock. All of the above basically happens in-situ, directly transforming solid silicate rocks to solid carbonate rock. When this happens, minerals such as magnesite ($\ce{MgCO3}$) or siderite ($\ce{FeCO3}$) also occur. This is actually one of the way to do industrial carbon sequestration: instead of putting $\ce{CO2}$ in the atmosphere, for example in power stations, you inject it down to the rock to lock it up in solid minerals.


Ground state term symbol for Cerium (Ce) I need to calculate the term symbol of the ground state configuration of Cerium.
The electron configuration is: $[Xe] 4f^1 5d^1 6s^2$
Since the d and f electrons are in different subshelles, they are non-equivalent electrons and Pauli exclusion principle doesn't apply, so:
$l_1=2$;$l_2=3$
using addition of angular momentum:
$L=5,4,3,2,1$ and $S=1,0$
Using Hund's rules, the ground state term symbol should be the one with $S=1$ and $L=5$, this is $^3H$.
However, NIST lists $^1G$ as the ground state.
Any help would be really apreciated.

Iván, you are right on your determination using Hund's rules. However, for heavy elements, with strong spin-orbit coupling, the LS coupling assumed by Hund's rules may not apply. In this case, the j-j coupling may give better agreement with experiment.

There are extensions of Hund's rules for j-j coupling [1].

If you check NIST Atomic Spectra Database Levels Data, you will see that the wavefunction for the ground state of the Ce atom is 55% $^1\text{G}^\circ$ and 29% $^3\text{H}^\circ$. This mixture between singlet and triplet is due to the spin-orbit coupling.

[1] T. E. H. Walker, J. T. Waber, Modified Hund’s Rule for jj Coupling. Physical Review A. 7, 1218–1224 (1973). https://dx.doi.org/10.1103/PhysRevA.7.1218


What is the connection between vapor pressure and intermolecular force explanations of boiling? I've had boiling explained to me in two ways, and I'm having trouble understanding how they connect.

The first way is that a liquid is held together by the intermolecular forces, and boiling involves breaking those bonds/interactions. This explains why, for instance, the boiling point of water is higher than that of methane, but not why boiling point is pressure dependent.

The second way is in terms of vapor pressure. Boiling occurs when bubbles form, and in order for a bubble to form and not be crushed, its internal pressure must be equal to the atmospheric pressure. From this view, the boiling point is the temperature at which the vapor pressure equals the atmospheric pressure. This explains why boiling point drops as pressure drops.

My issue is understanding how these coexist. The way I think of it, the strength of the intermolecular interaction should be the same no matter the pressure, so the temperature to break those interactions should be constant. Also, you can boil a liquid just by lowering the pressure, but I don't see how this affects the intermolecular interactions in any way.

Consider boiling as the rapid escape of molecules.

In slow evaporation and in rapid boiling, only those molecules with sufficient kinetic energy are able to leave the liquid. As each "fast" molecule leaves, it subtracts its high KE from the average KE pool of the liquid, i.e. cooling it.

For evaporation to take place, there has to be a chance for molecules to escape and not to be replaced by those in the gas above the liquid. In fact, if the pressure is high (below the critical temperature), then the gas can condense to the liquid, faster than the liquid can evaporate to the gas.

For boiling to take place, there is the further restriction that localized temperature rise takes place faster than in the main body of the liquid, so that specific locations overheat, e.g. where heat flux is highest. Where pressure is a factor, then boiling may be prevented until the pressure is relieved, as in "black smokers". The water exiting the vent may be at 200 C or more, but does not boil unless the critical temperature is exceeded.


CuSO4.5H2O >> CuSO4 + 5H2O : chemical or physical reaction? In a textbook it says it's a chemical reaction which involves breaking bonds.
But I don't think so since I think water molecules are just trapped in the crystal. What's the right answer and why ?

Copper II forms many types of complexes with water. The aquo ion is $\ce{[Cu(H2O)6]^{2+}}$ where the oxygen ligates to the metal atom. The structure is a distorted octahedral one, two bonds are longer than the other four. (The Jahn-Teller effect$^*$ is behind the fact that two ligands are not bound as strongly as the other four). It is possible with ammonia to form various hydro complexes; $\ce{[Cu(NH3)(H2O)5]^{2+}}$... $\ce{[Cu(NH3)5(H2O)]^{2+}}$. Similar complexes form with chloride ions. Copper sulphate is coloured in solution (due to $\ce{[Cu(H2O)6]^{2+}}$ ion) and in the crystal; the anhydrous crystal is colourless and this seems to indicate that the waters are bound, due to metal to ligand transitions, as in other transition metals. (The reason that sulphate is not significantly bound to the copper in aqueous solution, as it is in the crystal, is due to the vast excess of water (water is 55.5 M) which will displace the sulphate.)

In the crystal of $\ce{CuSO4.5H2O}$ the metal is, as usual, 6 coordinated, with four water oxygens in the plane and one oxygen from a sulphate occupying each axial position. The extra ($5^{\text{th}}$)  water molecule is hydrogen bonded between the second oxygen of the sulphate and a bound water molecule in the plane. This same water's oxygen is also $\ce{H}$-bonded to a sulphate oxygen bound to the next $\ce{Cu}$ atom. The $\ce{H}$ bonds are not quite of equal length  0.287 and 0.294 nm respectively.

(Fifth water oxygen thick black circle, sulphate hashed, $\ce{Cu}$ small circle, other four oxygens, large circles. Source: Wells 'Structural Inorganic Chemistry')

So bonds are broken, $\ce{Cu}$ to oxygen, also oxygen to oxygen and oxygen to sulphur hydrogen bonds as water is removed.

$^*$ The Jahn Teller theorem shows that any non-linear molecule in a degenerate state will undergo a distortion which will lower its symmetry and split the degenerate state. The theorem does not indicate what the distortion will be, or how big it will be. Nevertheless it is extremely important in describing the geometry of metal complexes.


Explanation for increase in the number of oxidation states for transition elements from Sc to Mn of the 4th period From Sc, to Mn, the number of oxidation states increases from one (Sc) to seven (Mn). The explanation for this is because the unpaired 3d electrons can be lost along with the 4s electrons during bonding.

Should this be the case? Wouldn't it be easier to lose an electron when it is paired because of interelectronic repulsion in the orbital? Could someone explain this? Thanks!

The oxidation state of a transition metal is an accounting device.  It's not the number of "lost electrons".  In potassium permanganate, which has Mn (VII), the manganese atom shares its seven 3d electrons with the ligand (oxygen) in what are essentially covalent bonds, and the "positive charge" associated with the metal is not very different than what it is for ionic calcium compounds.

Regarding "paired electrons are less likely to be involved in bonding than unpaired electrons" -- both paired and unpaired electrons in a transition metal can participate in bonding, provided they are in the outermost shell.  The paired electrons in inner shells are too deep in energy to take part in bonding.


USNCO Question - Determining solute using spectrophotometer Below is a question from the 2017 USNCO:

The concentration of which approximately 0.01M solution could be most accurately determined by a visible spectrophotometer (or colorimeter)?

(A) $\ce{Mn(NO3)2}$

(B) $\ce{Co(NO3)2}$

(C) $\ce{Zn(NO3)2}$

(D) $\ce{Pb(NO3)2}$

The answer sheet states B is the correct answer.

I researched the colours of the ions in solution, and found:

Mn is pink
Co is pink
Zn is clear
Pb is clear

I was unsure how to progress from there.

How should I approach this question in an exam if there was no reference book?

An aqueous solution of $\ce{Mn^2+}$ is only faintly pink, whereas that of $\ce{Co^2+}$ is much more colored:


Computing of strengthening of π-interaction I need some practical advise. What would be the easiest approach to compute strengthening or weakening of π-interaction in a bond with multipole-bond character. I'm not really fond of going through coefficients of Kohn–Sham orbitals. I did it on occasion, but it's a lot of work and I just do no have time. Unfortunately the molecules I need to compare have low symmetry ($\ce{C1}$) so I can't make use of symmetry to divide $\sigma$ and $\pi$ contributions. I'm not quite sure but I remember to read that the energy decomposition analysis in ADF could be used to divide in $\sigma$ and $\pi$ components. Let me make it clear, I'm not interested in comparing bond lengths, I need quantitative measure for the strength of $\pi$ interaction: either energy component or relative contribution of $\sigma$ and $\pi$ components orbitals.

I welcome any suggestions concerning methodology and/or software that can help.

First off, definitely check out ETS-NOCV, which I believe is a fairly recent addition to ADF which seems to do exactly what you're interested in. Other NBO-esque software may also suit your needs - I seem to recall NBO5 in Gaussian can generate sigma and pi natural bonds.

Regarding doing things the energy decomposition way:

If you have not already seen it, this paper goes about doing (in ADF) what you want to do:

http://pubs.rsc.org/en/Content/ArticleLanding/2012/DT/c2dt31845h

(disclosure - I am a member of the Stranger group)

I did this for a related system a while back but have forgotten the exact details, however fragment bonding analysis is a nifty technique that allows estimation of the energy of single bonds in a molecule as well as decomposition by bond type.

Broadly, you cut your molecule into chunks on either side of the bond you're interested in and generate fragment orbitals from them. (that is, you use the TAPE21 files for these molecular fragments as a basis for the calculation of the full molecule). These fragment orbitals subsume all of the interactions within the fragments so that the total bonding energy only consists of interactions between the fragments.

From here, you can actually remove atomic orbitals from the fragment orbitals. If you prune the orbitals relevant to pi bonding, you can estimate the pi and sigma components by relative energy.

Here's where you should start: ADF manual - Analysis


What is a covalent semiconductor? Aren't all semiconductors' atoms bond in a covalent bond? How is a "covalent semiconductor" different from other semiconductors?

No, most semiconductors are ionic.

I'm not sure I've seen the term "covalent semiconductor" before. Wikipedia has an article on covalent superconductors.

I think the main point would be that many if not most semiconductors are inorganic solid-state compounds. Consequently, most of these (e.g. III-V semiconductors are ionic solids, not really covalent ones. Consider GaAs or InP, which are typically classified as ionic.

That is the solid state electronic structure can be considered mainly composed of cations and anions (unlike organic chemistry where the electronic structure is considered mainly of neutral atoms).

(Incidentally, defining whether some of these compounds have purely ionic or purely covalent interactions is somewhat futile, since most bonding interactions are both somewhat ionic and covalent.)


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 $\mathrm{Z}$.


Vapour pressure in an enclosed container with liquid at normal conditions Can vapour pressure happen in an enclosed container with liquid at normal conditions?

I guess the answer is yes, but I don't understand why some liquid on the liquid surface may process sufficient kinetic energy to escape into gas phase.

The kinetic theory of gases is based on the idea that particles are moving randomly. Even if all particles started motion with exactly the same speed, as they interact with each other and the walls of the container their velocities would become randomized. The kinetic energy of particles at a given temperature is an average; some particles are moving much faster, and can escape from a liquid. These faster-moving particles, removed from the liquid, leave the liquid a bit cooler. Thus, evaporation puts a liquid into a gas and chills the liquid.

For a more complete explanation, see this video.


How do you handle manganese nitrate hexahydrate? So I wanna do an experiment using manganese nitrate and I bought manganese (II) nitrate hexahydrate $\ce{Mn(NO3)2.6H2O}$. But when I received it, it was kinda watery already (most part is still crystallized but there was also liquid in the bottle). After inspecting the bottle, it says that I need to refrigerate at 5 degree Celsius. But after I refrigerate it becomes icy hard, I cannot take anything out. So I am now wondering how to handle this compound. I am afraid if I use the liquid part it will have various concentration, and it is barely impossible to use the solid part because it's so hard. Thank you.

Manufacturers of chemicals can make life interesting for first-time users. What a boon the CRC Handbook is! Without even listing $\ce{Mn(NO3)2.6H2O}$, the CRC Handbook can solve your problem. It gives the melting point of $\ce{Mn(NO3)2.4H2O}$ as $\pu{25.8^\circ C}$. Undoubtedly, the melting point of the hexahydrate is even lower, but maybe not by much. In hot water, the tetrahydrate is "infinitely" soluble.

Heat your jar of hexahydrate to 30C in warm water until all the crystals are melted and withdraw what you need with a pipette. You might plan ahead and fill several small bottles with the hexahydrate for later use. I doubt that the material needs to be stored cold.


Why is ammonium a weak acid if ammonia is a weak base? $\ce{NH3}$ is a weak base so I would have expected $\ce{NH4+}$ to be a strong acid. I can't find a good explanation anywhere and am very confused. Since only a small proportion of $\ce{NH3}$ molecules turn into $\ce{NH4+}$ molecules, I would have expected a large amount of $\ce{NH4+}$ molecules to become $\ce{NH3}$ molecules.

First, let’s get the definition of weak and strong acids or bases out of the way. The way I learnt it (and the way everybody seems to be using it) is:

$\displaystyle \mathrm{p}K_\mathrm{a} < 0$ for a strong acid
$\displaystyle \mathrm{p}K_\mathrm{b} < 0$ for a strong base
$\displaystyle \mathrm{p}K_\mathrm{a} > 0$ for a weak acid
$\displaystyle \mathrm{p}K_\mathrm{b} > 0$ for a weak base

Thus strong acid and weak base are not arbitrary labels but clear definitions based on an arbitrary measurable physical value — which becomes a lot less arbitrary if you remember that this conincides with acids stronger than $\ce{H3O+}$ or acids weaker than $\ce{H3O+}$.

Your point of confusion seems to be a statement that is commonly taught and unquestionably physically correct, which, however, students have a knack of misusing:

  The conjugate base of a strong acid is a weak base.

Maybe we should write that in a more mathematical way:

  If an acid is strong, its conjugate base is a weak base.

Or in mathematical symbolism:

  $$\mathrm{p}K_\mathrm{a} (\ce{HA}) < 0 \Longrightarrow \mathrm{p}K_\mathrm{b} (\ce{A-}) > 0\tag{1}$$

Note that I used a one-sided arrow. These two expressions are not equivalent. One is the consequence of another. This is in line with another statement that we can write pseudomathematically:

  If it is raining heavily the street will be wet.

  $$n(\text{raindrops}) \gg 0 \Longrightarrow \text{state}(\text{street}) = \text{wet}\tag{2}$$

I think we immediately all agree that this is true. And we should also all agree that the reverse is not necessarily true: if I empty a bucket of water on the street, then te street will be wet but it is not raining. Thus:

  $$\text{state}(\text{street}) = \text{wet} \rlap{\hspace{0.7em}/}\Longrightarrow n(\text{raindrops}) \gg 0\tag{2'}$$

This should serve to show that sometimes, consequences are only true in one direction. Spoiler: this is also the case for the strength of conjugate acids and bases.

Why is the clause above on strength and weakness only true in one direction? Well, remember the way how $\mathrm{p}K_\mathrm{a}$ values are defined:

$$\begin{align}\ce{HA + H2O &<=> H3O+ + A-} && K_\mathrm{a} (\ce{HA}) = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}\tag{3}\\[0.6em]
\ce{A- + H2O &<=> HA + OH-} && K_\mathrm{b} (\ce{A-}) = \frac{[\ce{HA}][\ce{OH-}]}{[\ce{A-}]}\tag{4}\end{align}$$

Mathematically and physically, we can add equations $(3)$ and $(4)$ together giving us $(5)$:

$$\begin{align}\ce{HA + H2O + A- + H2O &<=> A- + H3O+ + HA + OH-}&& K = K_\mathrm{a}\times K_\mathrm{b}\tag{5.1}\\[0.6em]
\ce{2 H2O &<=> H3O+ + OH-}&&K = K_\mathrm{w}\tag{5.2}\end{align}$$

We see that everything connected to the acid $\ce{HA}$ cancels out in equation $(5)$ (see $(\text{5.2})$) and thus that the equilibrium constant of that reaction is the autodissociation constant of water $K_\mathrm{w}$. From that, equations $(6)$ and $(7)$ show us how to arrive at a well-known and important formula:

$$\begin{align}K_\mathrm{w} &= K_\mathrm{a} \times K_\mathrm{b}\tag{6}\\[0.6em]
10^{-14} &= K_\mathrm{a} \times K_\mathrm{b}\\[0.6em]
14 &= \mathrm{p}K_\mathrm{a} (\ce{HA}) + \mathrm{p}K_\mathrm{b} (\ce{A-})\tag{7}\end{align}$$

Now let us assume the acid in question is strong, e.g. $\mathrm{p}K_\mathrm{a} (\ce{HA}) = -1$. Then, by definition the conjugate base must be (very) weak: $$\mathrm{p}K_\mathrm{b}(\ce{A-}) = 14- \mathrm{p}K_\mathrm{a}(\ce{HA}) = 14-(-1) = 15\tag{8}$$

Hence, our forward direction of statement $(1)$ holds true. However, the same is not true if we add an arbitrary weak acid to the equation; say $\mathrm{p}K_\mathrm{a} (\ce{HB}) = 5$. Then we get:

$$\mathrm{p}K_\mathrm{b} (\ce{B-}) = 14-\mathrm{p}K_\mathrm{a}(\ce{HB}) = 14-5 = 9\tag{9}$$

A base with a $\mathrm{p}K_\mathrm{b} = 9$ is a weak base. Thus, the conjugate base of the weak acid $\ce{HB}$ is a weak base.

We realise that we can generate a weak base in two ways: by plugging a strong acid into equation $(7)$ or by plugging a certain weak base. Since the sum of $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b}$ must equal $14$, it is easy to see that both cannot be strong. However, it is very possible that both the base and the acid are weak.

Thus, the reverse statement of $(1)$ is not true.

$$\mathrm{p}K_\mathrm{a}(\ce{HA}) < 0 \rlap{\hspace{1em}/}\Longleftarrow \mathrm{p}K_\mathrm{b} (\ce{A-}) > 0\tag{1'}$$


Possible E2 products of S-2-bromo-2-methyl-1-phenylbutane I have been asked to predict wether the following reaction will proceed by elimination or substitution.

$\hspace{2.4cm}$

I believe that the reaction would proceed by the $\mathrm{E2}$ mechanism. I am confused, however, as to what products what would be formed from this reaction. Currently I believe that there would be a total of five products: 2 sets of E/Z isomers, and a product with a double bond with no carbon substituents on one end (i.e along the dashed bond). Is this the case?

The major products will be (E)-2-methyl-1-phenyl-1-butene and (Z)-2-methyl-1-phenyl-1-butene, formed in a 1:1 ratio. There is are reasons both for why elimination occurs specifically across the first and second positions, and as for why the product is formed in a 1:1 ratio.

The hydrogen on position 1 are called benzylic hydrogen, and are much more acidic than aliphatic hydrogen due to the resonance stabilization of the benzyl anion. This will cause these hydrogen to be removed much more quickly than the others, resulting in a majority of 1,2-unsaturated products. This is of course not to say that the other products are not formed, but they are to a lesser extent.

The products are formed in a 1:1 ratio because there are two ways for the $\ce{H}$ and $\ce{Br}$ to be anti-periplanar. This is easily seen when constructing the Newman projections of the starting material:

$\hspace{3.6cm}$

Upon extraction of the hydrogen, the electron pair forms a $\pi$-bond with the adjacent carbon, ejecting the $\ce{Br-}$, and locking in the stereochemistry.


Examples of electronically conductive foods? There are many foodstuffs that are ionically conductive (for example, anything moist and salty). What about foods that have a high electronic conductivity?

The only thing I can think of would be something which has been charred – presumably the char, being mostly carbon, would have a fairly high conductivity.

I wonder why nobody mentioned the famous vodka with gold flakes. Admittedly, this hardly qualifies as food, and the amount of gold is minuscule, but in the absence of any solid examples with true metallic character I think it may deserve a curious look.


Why is Promethium the only radioactive Lanthanide? Why is it that only Promethium (Pm) is the only radioactive lanthanide?

I was trying to figure out what percentage of all radioactive elements in the Periodic Table the f-block accounts for, when it struck me that Promethium was the only radioactive lanthanide shown. I found this a bit odd, since I expected to see a couple of other radioactive elements there as well, especially since Promethium is (somewhat) near the middle of the series.

Lanthanoids are not generally supposed to be radioactive, with exception of mentioned very long lived radioisotopes.

The longest half-life radioactive nuclides of lanthanoids ( in years )

  $\ce{^{150}Nd}:\ \pu{6.7e18}$
  $\ce{^{151}Eu}:\  \pu{5e18}$
  $\ce{^{148}Sm}:\  \pu{7e15}$
  $\ce{^{144}Nd}:\   \pu{2.29e15}$
  $\ce{^{152}Gd}:\  \pu{ 108e12}$
  $\ce{^{147}Sm}:\ \pu{106e9}$
  $\ce{^{138}La}:\ \pu{ 102e9}$
  $\ce{^{176}Lu}:\ \pu{ 38.5e9}$

Promethium is rather a victim of circumstances there was no free position to have at least one stable isotope. It is related also to the Liquid drop model.

There are few semi-empirical rules, related to proton and neutron fermionic pairing within the nucleus model:

Elements with the odd proton number have maximally 2 stable isotopes.

2 Isotopes of adjacent elements with the same nucleon number, like $\ce{^N_{A}X, ^N_{A+1}Y}$, are not both stable. ( But some of these unstable ones  have very long half-life as seen above.)

Isotopes not close enough to the Valley of stability are not stable, as they have too little or too many neutrons.

The stability of isotopes of similar nucleon number generally decreases in order

both proton/neutron numbers are even
one number is odd, the other is even.
both numbers are odd.

For Pm and Tc, both have odd proton numbers what gives them disadvantage.

All their possible stable isotope candidates are already taken by their  neighbors as energetically preferred variants. Therefore all isotopes of given 2 elements beta decay to nuclei of adjacent elements with less energy.


Nomenclature Precedence Order for Unsaturation and Hydrocarbon Chain When naming branched alkenes/alkynes, the numbers of the carbons where the branches and multiple bonds are located are supposed to be the lowest possible.  Do the branches take precedence over the multiple bonds, as for example, 2-methyl-3-pentyne?  If the counting of the carbons for this molecule is flipped so that the triple bond is lower in number, it would be 3-methyl-2-pentyne.  Is the first way correct?

Unsaturated bonds take precedence over "branches". Consider these bonds as functional groups and name the compound, it will make you number carbons correctly.

Also the compound you mentioned above can't possibly exist as it has 5 bonds on $\ce{C_3}$.


Stability of dialkyl substituted alkenes The stability order of alkyl substituted alkenes with respect to heat of hydrogenation was given as
$$\ce{R_2C=CH2} > \ce{RHC=CHR}~(\textit{trans}) > \ce{RHC=CHR}~(\textit{cis}).$$
While I understand that trans alkene is generally more stable than cis, I don't understand why $\ce{R_2C=CH2}$ is the most stable. Shouldn't there be repulsion between the bulky alkyl groups which leads to decrease in stability?

Following is an image of hyperconjugative effect. We can see a $\ce{\sigma}$-orbital delocalising electron pair with an empty $\ce{\pi^\ast}$-orbital.

The explanation to your problem can be given by deciding whether the effect is "centralized" or "decentralized". Let me explain it by an example.

A quick scene:
Suppose, you and your friend is sitting across a boat ($\ce{C_{you}=C_{friend}}$) in the middle of a river, and both of you have to reach on either bank soon (a process similar to delocalization, probably). Assume the methyl groups to be oars you need to move across the river.

In $\ce{R2C_{you}=C_{friend}H2}$, both the oars are with you. So, only you can row the boat and hence both of you can reach to the bank soon. Technically, the $\ce{\beta-H}$'s are at one carbon, so the effect can happen only on it, hence "centralized" and super-fast.
But, in $\ce{RHC_{you}=C_{friend}HR}$, both of you have the oars. So, both of you can row the boat and therefore both of you will reach to either bank a bit later. Technically, the $\ce{\beta-H}$'s are distributed equally among the carbons, so the effect can happen on either side, hence "decentralized" and not as fast as in the previous case.
As per your concern on steric factors, you may know that "alkyl" is referred to a straight hydrocarbon chain, and it isn't as bulky as to create "steric hindrance".
I guess that makes some sense :)

EDIT: The above stuffs explains how is first isomer the most stable. I'd previously said that "steric effects" doesn't matter, but in case of cis and trans isomers, it matters and it's the only thing that differentiate them (along with polarity). Due to this, cis isomer is least stable among the three.
Reference table of $\ce{\Delta H^°_f}$: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/heats-formation.html#Alkenes (in #C = 4)


Will heating alumina (Al2O3) to its melting point yield pure Al? My inorganic chemistry is at a basic level, so my question is, if I heat alumina ($\ce{Al2O3}$), to its melting point around 2000 C, will I wind up with pure Al?  I think the oxygen will release from the $\ce{Al2O3}$ when the temperature approaches 2000 C.

I bought 5 kilograms from a friend who works at a refinery near my town (he works in logistics and does not have a chemical background) with 1.75$ of almost pure alumina ~99.95%, and I want to know if I melt that white powder will I obtain 2.5kg of 99.95% aluminium?

You've figured out one half of the Hall-Heroult process. Just melting alumina on its own won't liberate the oxygen. The reason is difficult to explain without getting into chemical thermodynamics - one way to think about it is that the lowest energy state is one where oxygen is bound to alumina - where the electrons from alumina have been "taken" by the oxygen. In diatomic oxygen, the electrons are "shared" (the bonds are covalent) and so the total energy is higher. This is a very brief and not-very-detailed explanation, but the end result is that oxygen and aluminum, if given the opportunity, would rather form a bond than not.

When you add thermal energy, unless it changes that balance (it doesn't, at least not until you get way past the melting point), all you are doing is making it more possible for the reaction to happen. If you kept going, you could eventually add enough thermal energy to completely dissociate the alumina, at which point (depending on how you did it) you might be able to remove the oxygen and recover the aluminum.

However - the amount of energy it would take to do that is extremely large. Instead, it turns out to be much easier to simply force oxygen to give its electrons back to aluminum using an electric potential. When you do that, you electrolyze the compound and create diatomic oxygen at the anode and aluminum at the cathode.

The Hall-Heroult process takes this another step to make this work at an even lower temperature (1000 C) by dissolving the alumina in molten cryolite. As a result, it is one of the most (maybe the most) efficient industrial process for producing aluminum, and is responsible for the widespread availability and low cost of aluminum in modern times (and also for the existence of Alcoa).

Your idea about using solar energy isn't bad - although you might be surprised at the amount of energy you need to produce aluminum - both to melt it, and to electrolyze it. One thing you might look into is using solar thermal heating (with a collector, for example) to melt the alumina/cryolite mixture, and then use photovoltaics just for the electrolysis step. That would be quite a bit cheaper than using photovoltaic electricity alone to run an electric heater.

Industrially, the cryolite mixture is kept molten by passing a large electric current through it. The electricity is usually produced by power plants, since the amount required is so large (15 kWh/kg according to Alcoa, and that's just for the electrolysis step!). Although coal and gas power isn't renewable, the advantages are a relatively low cost per unit of power (which might not stay true in the next few decades) and on-demand power generation (likely to always be a big advantage over solar).

There is also nuclear, hydroelectric, and geothermal power - all of which provide large amounts of on-demand power very inexpensively, once the plants are built.


Energy level of hybrid orbitals in the molecular orbital energy diagram I would like to know what is the level of energy of a hybrid orbital?

For instance, lets consider the $\ce{N2}$ molecule. According to its geometry, we know that there is an orbital 2p and 2s that are going to form 2 sp orbitals.
 In the molecular orbital energy diagram, where should we place the these 2 sp orbitals? Would they be in the middle of a 2p and a 2s? And which p orbital do we take to take the middle ($p_x$,$p_y$,$p_z$, $\pi$ or $\sigma$?) Will the sp* be higher or lower that the $2p_x$ and $2p_y$?

When we have an sp hybrid orbital, it is usually made of the s orbital and the p orbital that points in the bounding axis ($p_z$). It's energy will be the mean of the energy of the initial orbitals.

In the example of $\ce{N2}$, it is essential to bear in mind that each sp won't form an $\sigma$ and an $\sigma^*$ orbital. Only the one pair of the sp orbitals, the one that overlap the most, will do. The other pair will contribute to the $\ce{N2}$ lone pairs because there is nearly no overlapping. The 2 others bonds that form the triple bonds are made with the 2 left orbitals $p_x$ and $p_y$.

By the way, it is not necessary to use hybrid orbitals for the molecular orbital theory.

Credits to Philipp and Ben Norris.


Rearrangement of a gas law equation For example, let us that say we have a question where we have to use the ideal gas law and that the moles and temperature are constant. Then the gas law can also be written as $$nRT=PV$$ However, according to my solutions manual, this can also be written as $$nRT=P_1V_1=P_2V_2$$

My question is, how are we justified in rearranging the equation like this?

If one rearranges the ideal gas law equation, you can obtain the following (assuming $n$ and $T$ are non-zero):

$$\frac{PV}{nT} = R$$

$R$ is a constant, and there are in fact infinitely many possible sets of values $(P, V, n, T)$ that satisfy the equation. Let $(P_1, V_1, n_1, T_1)$ denote one such set, and let $(P_2, V_2, n_2, T_2)$ denote a second one. Then, we have:

$$\frac{P_1V_1}{n_1T_1} = R$$
$$\frac{P_2V_2}{n_2T_2} = R$$

So, by the transitive property:

$$\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$$

You can let $n$ and $T$ be constant as well (as in your specific example), but the principal is the same and leads to an analogous result, since any product and/or ratio of constants will itself be a constant.


How can I make a fairly saturated copper (II) sulfate solution by heating? I have an investigation to do for my chemistry class. I'm looking at copper complexes and using ammonia to create a precipitate in the copper (II) sulfate solution, then redissolving the precipitate to complete the ligand exchange. When I tried the experiment the first time, I was unable to make a good solution. I dissolved about $3.5$ grams of $\ce{CuSO4·5H2O}$ into $200$ mL of water, and couldn't get any more to dissolve. I titrated in $1$ mL of ammonia and the solution turned cloudy (I assumed this was the precipitate). After $4$ mL I stopped, as the solution hadn't changed at all. This was all carried out at room temperature.

I know the two best ways to make a more concentrated solution are time and heat. I don't have that much time, as I'm on winter break and I'll need to complete the experiment as soon as possible when I go back next week. It would probably take a week to get the solution concentrated enough, so I've decided to use heat to help the dissolution along. After looking around the internet, I found very little that was helpful. One page outlined how to make a supersaturated solution, which was not what I wanted. It advised heating to $100^{\circ}$ Celsius, so I suspect that will be too hot for my experiment. A video done by students showed them making a solution at $50^{\circ}$ Celsius, but they used copper (II) oxide. I'm not sure if that makes the procedure significantly different.

I'm not sure what temperature is necessary for this to work. If all else fails, perhaps I'll change my topic to an investigation of the relationship between heat and solubility rates. I'll finally know how to make a saturated solution in a reasonable amount of time.

Other notes: I'll be using a hot plate (Fisher Scientific, 11-500-4H) for the heating, not a Bunsen burner. The ammonia is stock/commercial, and the concentration isn't on the label.

Thanks in advance.

I suggest to use a combination of heating, dissolving, shock cooling and filtrating off the precipitate.

As was already stated, copper(II) sulphate dissolves slowly and is better soluble in the heat. The exact temperature doesn’t matter, but why not just go up to $80$ or $100~\mathrm{^\circ C}$? This will also make the dissolution faster. You can look up the solubility of copper(II) sulphate to calculate how much you would need and use some more to be on the safe side.

Then, cool the solution down rather quickly (but not too quickly or the glass will crack). This will initially create a supersaturated solution but the supersaturated bit will crystallise out rather quickly. You can filter off this crystalline precipitate to get your saturated copper(II) sulphate solution. Do not filtrate too quickly, or the solution will still be too warm and slightly supersaturated.

Then finally remember that if you add too much ammonia, rather than a precipitate of $\ce{Cu(OH)2}$ you might get the redissolved $\ce{[Cu(NH3)4(H2O)2]^2+}$ tetraammindiaquacopper(II) complex. You can tell this by the intensive dark blue colour.


Have I identified the IUPAC name of this ketone correctly? $\ce{(CH3)2-CH-CO-CH2-CH2-OH}$

The IUPAC name I came up with was 1-hydroxy-4-methyl-3-pentanone because:

Ketone has higher priority over alcohol
The lowest locant set is achieved by numbering the C bonded to -OH group as 1.

But the answer given states it is 4-methyl-3-oxo-1-pentanol which violates the first rule mentioned above.

Please clarify the discrepancy between my reasoning and the textbook answer.

Your assessment is correct. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  16. Ketones
  17. Hydroxy compounds (includes alcohols)
  (…)

However, locants are placed immediately in front of the part of the name to which they refer. Therefore, the correct name for the compound that is given in the question is 1-hydroxy-4-methylpentan-3-one.


Identifying orbital given electron probability density plot Being a high-school student, I only have a qualitative understanding of the wave mechanical model of atom.

Below are my inferences regarding the plot shown in the above picture:

The concerned orbital has $3$ radial nodes (I'm counting the places where the value of the y-coordinate is zero excluding the ones at the extreme left and right).

Hence, the orbital should satisfy $n-l-1=3$. The orbitals satisfying the condition are $4s, 5p, 6d, 7f,\ldots$.

Are my inferences correct?

I think your inference is correct, though the plot looks very awkward. There should be an exponential decay at large values of $r$, but in the plot the curve plunges sharply onto the $x$-axis. Also the shapes of each peak should not be so symmetric, but instead skewed towards the left.


Which of the following elements forms a water-soluble, alkaline oxide? An element burns in oxygen to form a solid oxide which dissolves in water. The pH of the resulting solution is greater than seven. Which one of the following could the element be?

A. sulphur

B. phosphorus

C. magnesium

D. sodium

E. carbon

The right answer is sodium. But I don't really understand why it can't be magnesium as well.

solid oxide which dissolves in water

Magnesium oxide $\ce{MgO}$ is (practically) insoluble in water.

You might be interested to know that burning of magnesium was the classical flash light in the early days of photography:  Wikipedia.


VSEPR theory, chemical bond and quantum mechanics VSEPR theory correctly predicts the shapes of many symmetry-broken molecules such as $\ce{H2O}$ and $\ce{NH3}$. Take $\ce{NH3}$ for example. In VSEPR theory, the nitrogen atom is (approximately) at the center of a tetrahedron, the three $\ce{N-H}$ bonds point to three of the four vertices of the tetrahedron, and the lone pair of nitrogen points to the $4$th vertex. But quantum mechanically speaking, the electrons should all be delocalized in the entire $\ce{NH3}$ molecule. How do I unify the two pictures to understand the concept of chemical bonds and VSEPR theory in quantum mechanics? Does VSEPR correspond to some kind of trial wave function (e.g. antisymmetrized geminal power (AGP))?

Note: when I say how to understand chemical bonds in quantum mechanics, I mean in the chemical bond description of molecules, electron pairs are localized at the bonds while quantum mechanics again says everything can be delocalized. So it's the same discussion as VSEPR v.s. QM. If there are only two atoms and one bond, the quantum mechanical meaning of the chemical bond is clear.

VSEPR is a simple and generalised approach based (mainly) on empirical observation. It is a great theory in predicting, in a first order approximation, the geometrical shape of molecules. That is a lot harder with other methods. It is of course based on a physical foundation, not only empiricism. You might want to explore electron-domains and the Pauli principle in this context.

There have been a lot of post-rationalisations that cause more harm than good, because they clearly go beyond the limitations of the theory. One of the most miss-taught ones is the involvement of d-orbitals in hybridisation schemes, but that comes from the lack of understanding of some instructors, and its confusion with valence bond theory.
One of the most prominent examples where it (basically) fails is outlined in my answer here: Are the lone pairs in water equivalent?  At this point it should be noted, that the general topology of the electron density is quite well reproduced.

With that in mind, a reconciliation of this empirical approach with quantum theory to understand bonding is dangerous, if not futile. It should be used for providing a reasonable guess for a molecular structure and its principle explanation. Anything beyond that might lead to wrong conclusions. It cannot, in any way, be used to generate a guess for a wave function, because it is not based on it.

You will have to use another method for that. Often confused with VSEPR is the valence bond theory (VBT). It cannot be stressed enough, that the two are completely independent (even at its crudest level). There has been a lot of development in that field, and what is often taught as VBT can only be classified as a very crude approximation. It is in principle an exact theory, but carried out at the theoretical limit is not as easily understood as that what is taught commonly. (Just have a look at resonance, and its misconceptions as are outlined here: What is resonance, and are resonance structures real?)
Another way to describe bonding is molecular orbital theory (MOT), which already comes with the necessary properties of electron delocalisation. Unfortunately, this theory is more difficult to get started in, and does not provide an easy picture to follow. The good thing about this is, that it doesn't get much more complicated at its theoretical limit.
The two approaches are at their respective theoretical limit (VBT decribes electron correlation via resonance structures, MOT need multiple determinants for this) equivalent.

Understanding bonding is not easy and it is by far not without controversy. Even at approximate VBT or MOT there can be many misconceptions, and incorrect deductions, conclusions, rationalisations. For everything in the "grey" areas, there are opinions, interpretations, and opinions about interpretations.
In any and every case one should always be aware of the limitations of the model used. One should also be critical about the found results. One should always expect everything to be a lot more complicated than expected (point in case: CO2).
If you keep all of that in mind: VSEPR is an awesome model system.


Is rotation of a mirror image allowed in order to determine chirality? Doesn't rotation alter the mirror image of the molecule? So, chiral molecules are those which have non-superposable mirror images.

trans-1,2-Dimethylcyclopentane is chiral; 2-butanol is chiral; both don't superpose on their mirror image.

Then my book cited 2-propanol as achiral. How?

The mirror image doesn't seem to superpose with the molecule. How can it be achiral then?

Then the book provided a pic where the author showed how 2-propanol is achiral; he rotated the mirror image & then superposed! Done! 2-propanol is achiral!

But really how can rotation be allowed? After all the mirror image is a stereo-isomer; if it is rotated, wouldn't it be changed?

I'm not getting, why rotation is allowed before superposition. Not every-time, the mirror image is rotated before superposition.

Can anyone please explain how rotation is allowed before superposition?

When checking out chirality or achirality you are always allowed and supposed to rotate the molecule freely in all dimensions. Essentially, what you are trying to prove, is that the mirror image of a molecule can be achieved by rotation alone.

The thought behind this is simple. Chirality is macroscopically proven by optical activity. If you reduce the experiment to its bare basics, you have a polarised light beam that passes through a solution and hits a single molecule to be deflected either to the right (dextrorotatory) or to the left (levorotatory). A second molecule it hits which has the same chirality will move it in the same direction by the same amount. If, however, it hits a molecule which is exactly the mirror image of the first, the effect will be cancelled and the plane of polarisation is moved back into the vertical. In a solution containing only one enantiomer of a chiral substance, the beam will never hit a mirror image and thus always be deflected in a certain direction.

If we continue this thought experiment, we realise that molecular motion in solution is random. Thus, a molecule can — if we give it conciousness — move around freely and let it try to assume exactly the mirror image of the previous one. Therefore, we are allowed and encouraged to rotate (but not to use another mirror image or inversion!) our molecule until it ‘fits’.

By the way, most if not all achiral compounds can also be rotated before you apply the plane of symmetry. Oftentimes if not always, you can rotate them in a way that you see the plane of symmetry present in the molecule itself. If you managed to do that, you won, because just use a plane parallel to that one for mirror imaging and the image will automatically be identical to its template without further rotation.

The plane of symmetry in isopropanol cuts through the $\ce{CHOH}$ bit of isopropanol, dissecting each of those four atoms into two half-spheres. One methyl group will be transformed exactly onto the other by this plane.


Why doesn't a C4H8 alkane exist? This structure:

     H   H
     |   |
 H - C - C - H
     |   |
 H - C - C - H
     |   |
     H   H

Which would be $\ce{C4H8}$, like butene, but an alkane, because it's functional group is C - C.

It does exist. The formula $\ce{C_{n}H_{2n+2}}$ is only applicable to straight chain alkanes. They can't have any degrees of unsaturation, which is defined as a multiple bond or a ring.


Coordinate bonds between molecules For the following question

Which molecules react to form a dative/coordinate bond?
A. $\ce{CH4}$ and $\ce{NH3}$
B. $\ce{C2H2}$ and $\ce{Cl2}$
C. $\ce{NH3}$ and $\ce{HF}$
D. $\ce{Cl2}$ and $\ce{HF}$

the answer scheme says C is correct, but I do not see how two already complete molecules can form coordinate bonds. $\ce{NH3}$ has a lone pair of electrons, $\ce{HF}$ having none. How can a coordinate bond between this lone pair and $\ce{HF}$ happen? I am confused.
The others are also equally confusing.

Option C is correct.

You could start by looking up the definition of a coordinate bond.

Here's a tiny excerpt from the above link:

  A coordinate bond is a kind of 2-center, 2-electron covalent bond in which the two electrons derive from the same atom.

In your case, as you have correctly noted yourself, the $\ce{NH3}$ molecule will have an "lone pair" of electrons. However, your judgement that $\ce{HF}$ has none is quite incorrect– the fluorine atom has $3$ lone pairs. (not that it's very relevant to the question, but you can look up the structure of $\ce{HF}$ here)

Since the $\ce{NH3}$ is a lewis base, its lone pair of electrons would abstract the $\ce{H}$ from $\ce{HF}$.

$\ce{NH4+}$ has four equivalent bonds. You could call one of them the "dative" bond, but there's really no way to distinguish them from each others in general cases.

The concept of dative bond comes here just because the bond between nitrogen and the hydrogen from $\ce{HF}$ involves 2 electrons (like all single bonds), both of which are contributed by the nitrogen atom (unlike most single bonds).

(Image source)


Is there a way to contain fluorine gas for long term so that it can be visually observed? Bromine, chlorine and iodine can all be sealed in a glass container for display without the elements reacting with the glass.

But if you try to seal fluorine in glass I believe it will react and fog the surface making it harder to see the gas, right?

What's the best way, if at all possible for long term containment of fluorine for display?

As @JonCuster mentions, some materials are pretty resistant to fluorine gas at room temperature.

But out of curiosity, I checked Theodore Gray's website. He's made an effort to have a "periodic table" table with actual elements. He also has a beautiful book, called (naturally) The Elements.

  Almost anything placed in the path of a stream of fluorine gas will spontaneously burst into flame. This includes things like, oh, say glass, steel, and other things not normally thought of as flammable. It is, therefore, fairly difficult to have a sample of it in an element collection. I used to have a statement here that there was no transparent container that could hold fluorine for any length of time, but I was corrected by a man who has figured out how to do it.

There's more here but I'll summarize. Basically, the problem with glass is the $\ce{Si-O-H}$ bonds, which will auto-catalytically react with $\ce{F2}$ to give $\ce{HF}$, and that etches the glass. So with a high quality quartz tube, a lot of annealing, and a fluorocarbon grease to coat the glass and minimize reactivity, it works.

Interestingly, I always had the idea that fluorine gas was a light blue-green. Instead, it's brownish.


What material is synthesized in the rubber egg demo? One of the popular do-at-home chemistry demonstrations is placing an egg in a container of vinegar for two days. After that period most if not all of the eggshell has dissolved and the outer part of the egg has congealed giving the egg a 'rubber' like appearance and feel.

So chemically the vinegar (acetic acid) reacts with the egg protein (albumen).

Chemically speaking, what is this product?

The shell dissolution reaction is the calcium carbonate in the egg being dissolved by the acetic acid, resulting in release of $\ce{CO2}$ and formation of calcium acetate.

$$\ce{CaCO3 + 2 CH3COOH -> Ca(CH3COO)2 + CO2 + H2O}$$

The rubberization of the egg white is primarily a protein denaturation reaction.  Albumin in egg whites is a globular protein, meaning each molecule is a single chain that is tightly bundled into a ball-like ("globular") form.  In this form the molecules can stay nicely solvated and the egg white is liquid.  However, the low pH from the acid (and also possibly the high concentration of calcium acetate) causes each chain to unfurl and then to bind with other neighboring chains, forming a complicated cross-linked tangle of molecules.  This tangle is no longer a liquid but is a rubbery solid-like material.

So the products are primarily calcium acetate, $\ce{CO2}$, and denatured albumin.


Common Ion Effect Calculations Given a solution with a $\ce{S^{2-}}$ (sulfide ion) concentration of $2.76 \times 10^{-15}$ mol per litre, and $K_s(\ce{FeS})=4.90 \times 10^{-18}$, calculate the solubility of $\ce{FeS}$ in this solution.

I calculated this by letting the concentration of $\ce{Fe^{2+}} = x$, therefore the concentration of $\ce{S^{2-}} = 2.76 \times 10^{-15} + x$.

Therefore $4.90 \times 10^{-18} = x(2.76 \times 10^{-15} + x)$.

Upon solving this as a quadratic equation, I arrive at the result of $2.14 \times 10^{-9}$ mol per litre.
[EDIT: $2.21 \times 10^{-9}$ mol per litre]

The answers to this problem state to ignore the contribution of $\ce{S^{2-}}$ from the $\ce{FeS}$ as it is negligible, and state that the answer is actually $1.78\times10^{-3}$ mol per litre.

Which method is correct?

The value of the answer you were given is incorrect, both in its value and its explanation of what can be ignored.

As you note:

$K_{sp} = 4.90\times10^{-18}=[\ce{Fe^{2+}}][\ce{S^{2-}}]$

Just eyeballing the equation, we would expect a solution of $\ce{FeS}$ in pure water to contain equal quantities of iron(II) and sulfide ions, at a concentration of $2.21\times10^{-9} M$ (the square root of the solubility product.)  This is on the order of a million times more concentrated than the solution described in the problem, so we can ignore the sulfide concentration in the original solution.  For the purposes of this problem, the iron sulfide is dissolving in pure water.

Additionally, solving the quadratic equation you provided gives much the same result.  The solution rounds to $x = 2.21\times10^{-9} M$; it only differs in the sixth place from the square root of the solubility product.  It's easy enough to make a small numerical error with scientific notation and solving a quadratic; I have to assume that's what happened to you here.

The only way that your instructor's answer would be correct would be if the question intended us to understand that there was some chemistry going on to keep the concentration of sulfide ions at the given concentration.  Without that, it just doesn't make sense.


Schrödinger's equation and spin quantum number As spin quantum number cannot be derived from Schrödinger's  equation, it cannot predict opposite electron spin.
I mean to ask that how do we obtain the information conveyed by the spin quantum number when we don't speak of it in this theory.

Good question.

This was one indication that there had to be something more than just the Schrodinger equation. Another problem which was perceived quite quickly is that the Schrodinger equation treated time differently than the spatial dimensions which is not in the spirit of special relativity. Thus, Paul Dirac constructed what we now call the Dirac equation. When you solve the Dirac equation in a Coulomb potential, spin, and hence the fine structure of the hydrogen atom energy levels in an electric field, falls out naturally.

This proves to be fairly technical and I won't try to do any of the mathematics for this (I don't completely understand it all), but you can find solutions to the Dirac equation for some common cases here.

As orthocresol points out, we first observed the intrinsic angular momentum of the electron in the Stern-Gerlach experiment, so people guessed what this would look like at first (this is how it was used in the Schrodinger equation) and Dirac came along and explained that people had guessed correctly. This can also be seen from the fact that the Dirac equation reduces to the Schrodinger equation in the non-relativistic, low energy case.


Have there been no advances in the determination of effective nuclear charges since Clementi and Raimondi in the 60s? Effective nuclear charge is a very important concept in chemistry, and is the basis for the qualitative explanation of many observed chemical and physical properties, including several periodic trends. Hence it seems to me to be of significant importance that these values are accurately determined (not only for neutral atoms, but for monoatomic ions too). Slater's rules for the calculation of effective nuclear charges, from the 1930s, are in practically any university-level general or inorganic chemistry book, though they are heavily outdated and inaccurate, being mostly of interest due to their simplicity and their historical value. Many books go on to mention updated values for atomic screening constants, calculated by Clementi and Raimondi in 1963 and 1967. This updated table of effective nuclear charges is significantly more interesting, but not used as often probably because all the values for each orbital in each atom must be looked up individually.

However, it's been 50 years since Clementi and Raimondi's first publication on the topic, and I have seen no clear reference to a more recent improved calculation. Surely it is not because the values they found are not without flaws. For example:

Screening constants have only been calculated up to $Z = 86$, leaving out a portion of the table, in particular the actinides which have interesting electronic properties
No relativistic terms whatsoever were considered in the calculation of the screening constants, which is deleterious especially for atoms beyond about $Z \simeq 70$
The calculations are not completely reliable as they are, having not converged well especially some lanthanides, creating odd kinks in screening capacities for electrons in the $4f$ subshell
The calculations converged to inaccurate electron distributions in several atoms, such as $[\ce{Kr}]4d^85s^2$ for $\ce{Pd}$ instead of the actual $[\ce{Kr}]4d^{10}$

By now we should be able to do their calculations again with much more accuracy and much more speed, both due to the development of physical and mathematical theory behind the calculations, and due to the stupendous increase in affordable computational capacity. So why has there been no improvement to effective nuclear charge calculations? Has no one bothered? Or have there been several attempts, none of which have gained wide acceptance since then (sort of like what happens with electronegativity)?

I would say the most important reason why no development for effective nuclear charges is that the concept of effective nuclear charges has fallen to disuse in the recent years.  I can think of two important reasons:

Effective nuclear charges is intended to be used to explain electronic structure of isolated atoms.  To this day some atoms can still pose a challenge, but those are the atoms that experience heavy relativistic effects and would need other special treatments than shielding effect along.  However, for the most part, chemists are not interested in atoms anymore.
Its theoretical framework is too simple.  Chemist need their results to be extremely accurate.  A 0.1 eV change can easily be a game changer in an organic reaction.  That, unfortunately, cannot be achieved by a model that has only one adjustable constant.

However, the idea still lives on.  For example, pseudo-potential or Effective Core Potentials can be considered as more general and modern version of effective nuclear charge concept.  Instead of simply changing the charge but keeping the Coulomb potential unchanged, ECP provides the entire effective potential of core electrons. Valence electrons and sometimes even sub-valence electrons has to be excluded from such treatment to be able to produce results with sufficient accuracy. These pseudo-potentials are actively developed and widely used today, can provide chemically relevant accuracy in complex molecular systems and may qualify as "improvement to effective nuclear charge", although they are evolved into different species already.


Replacing gas in a chamber Is there a way to replace a gas (air) for example with another gas (say, helium) in a gas chamber? It does not need to be a complete replacement, but there would need to be a significant amount of the new gas. Thanks!

Chemists who need to control which gases are in their vessels (e.g. those who work with air sensitive compounds) do this routinely.

There are two ways to achieve gas replacement. The more reliable way is to work with vacuum-lines that allow switching the input to a vessel between a vacuum and an inert gas. First the air is removed and then the inert gas introduced.

Often, though, this is overkill. In many cases where liquids are involved the liquid needs to be degassed (to remove air dissolved in the liquid). In many cases it is good enough to bubble the inert gas through the liquid and the vessel for long enough to sweep out the unwanted gases. Bubbling the inert gas through the liquid and allowing it to escape the vessel is good enough if done for long enough.

If it isn't important to remove all traces of the original gas, this is usually an effective technique and gets more effective the longer the inert gas is allowed to flow.


What happens if I make a hole in a container with superheated water? Superheated water is achieved by heating water above 100 degree Celsius under high pressure.

Now, suppose the temperature at which the water is getting heated is around 250 Celsius and sufficient pressure is provided. If, I now make a small-hole in the container, the pressure will decrease obviously, but what will happen to the superheated water?

Will it come out in the form of vapor with lot of energy or some explosion might take place or just about something else?

If you have water at a temperature of $T=250\ \mathrm{^\circ C}$, you need a pressure of approximately $p=39.2\ \mathrm{atm}=4.0\ \mathrm{MPa}$ to keep it liquid. Liquid water at this temperature and pressure has a specific enthalpy of $h_0=1086\ \mathrm{kJ/kg}$.

If you open the container, the pressure drops to the ambient pressure of $p=1\ \mathrm{atm}=101\,325\ \mathrm{Pa}$. A part of the hot liquid water flashes to steam. The temperature drops to the boiling point ($T=99.974\ \mathrm{^\circ C}$) at the new pressure. A new equilibrium of liquid water and steam is established at the new temperature and pressure. The corresponding specific enthalpy of the steam is $h_\text{steam} = 2676\ \mathrm{kJ/kg}$; the specific enthalpy of the liquid water is $h_\text{liquid} = 419\ \mathrm{kJ/kg}$.

The flashing process is very quick, hence we may assume that there is no significant heat exchange with the environment; i.e. we may assume the following enthalpy balance in order to estimate the vaporized fraction $x$:

$$\begin{align}
h_0 &= x \cdot h_\text{steam} + (1-x) \cdot h_\text{liquid}\\
x &= \frac{h_0 - h_\text{liquid}}{h_\text{steam} - h_\text{liquid}}\\
x &= \frac{1086\ \mathrm{kJ/kg} - 419\ \mathrm{kJ/kg}}{2676\ \mathrm{kJ/kg} - 419\ \mathrm{kJ/kg}}\\
x &= 0.296
\end{align}$$

This means that approximately $29.6\ \%$ of the superheated water flashes to steam when the container is opened.


In which solvent is AgBr least soluble? The solute $\ce{AgBr}$ is least soluble in

  A $\ce{H2O}$
  B $1.0~\mathrm{M}\ \ce{FeBr3}$
  C $1.0~\mathrm{M}\ \ce{CaBr2}$
  D $1.0~\mathrm{M}\ \ce{AgNO3}$

I'm confused as to why the answer is B. Is it because $\ce{Ag}$ has low solubility with $\ce{Br}$, and the higher the concentration of $\ce{Br-}$ ions the less $\ce{Ag}$ will dissolve?

While solubilities are typically quoted in $\mathrm{g/l}$ or related units, the actual physicochemical idea behind it is an equilibrium constant known as the solubility product. In general, dissolution and precipitation can be considered two directions of the following equilibrium:

$$\ce{m A^n+ (aq) + n B^m- (aq) <<=> A_mB_n (aq) <=>> A_mB_n (s) v}\tag{1}$$

The outer sides are both preferred with respect to the central one; thus, the central compound is usually ignored, giving the following equilibrium constant:

$$K_\mathrm{sp}^{-1} = \frac{a(\ce{A_mB_n})}{a(\ce{A^n+})^m a(\ce{B^m-})^n}\tag{2}$$

Notes concerning equation $(2)$:

technically, this would be a precipitation equation; the solubility product is derived from the opposite equation hence the superscript $-1$.
activities should be used here, and not concentrations. This makes a difference as it allows for an activity of a solid (which has no concentration in solution — being solid outside of the solution).

Going the dissolution pathway, the activities of $\ce{A^n+}$ and $\ce{B^m-}$ are in the numerator and the solid’s activity is in the denominator. Thankfully, the activity of any solid compound is always $1$ per definition. That gives us $(2')$:

$$K_\mathrm{sp} = a(\ce{A^n+})^ma(\ce{B^m-})^n\tag{$2'$}$$

Which is the way the solubility product is typically represented. Now since this is an equilibrium process, Le Chatelier’s principle influcences it. Most importantly, you can add any one reactant to generate more of the product and vice-versa. If you take a look back at $(1)$, you’ll notice that the dissoluted ions enter the equation separately. Thus, we can use higher concentrations of one ion to reduce the overall solubility since then less of the other ion can remain dissolved.

This already tells us that all of options B through D result in a lower solubility since bromide ions (or silver ions) are dissolved. But how do we compare the three? The key here is that a $\pu{1M}$ solution of a salt does not necessarily mean that all its ions have a concentration of $\pu{1M}$. Instead, for $\ce{CaBr2}$ two moles of bromide are in solution for each mole of salt so if $c(\ce{CaBr2}) = \pu{1M}$, the concentration of bromide will be $c(\ce{Br-}) = \pu{2M}$.

Therefore, the solubility of $\ce{AgBr}$ will be lower in $\ce{CaBr2}$ (twice the concentration of bromide) and lower still in $\ce{FeBr3}$ (thrice the concentration of bromide). Hence B is your answer.


Hydrogen bonding in staggered vs eclipsed conformations In my first semester of organic chemistry, we learned that staggered conformations have lower energy/higher stability.

However, my thought is that if you have interacting OH groups, the hydrogen bonding would make the eclipsed conformations more stable than the staggered conformation. Is this be the case? Why or why not?

For example, a small molecule containing two carbons, each with an alcohol group (or more attached).

--
Related question:
In a 1,3-diaxial interaction, would multiple OH groups (or anything that exhibits hydrogen bonding) lend to stability?

Disclaimer: I am not a computational chemist, I am completely new to this stuff. Just a humble student of @pentavalentcarbon

  However, my thought is that if you have interacting OH groups, the hydrogen bonding would make the eclipsed conformations more stable. Is this be the case? Why or why not? For example, a small molecule containing two carbons, each with an alcohol group (or more attached).

I looked at two model compounds, ethane, and ethane-1,2-diol. Also, I did preliminary molecular mechanics based geometry optimisations using the MMFF94s force field in Avogadro.

I performed a CCSD(T) calculation in psi4 using the aug-cc-pVDZ basis. For ethane, obviously two conformers were identified:

$$\begin{array}{cc}
\hline
\text{conformer} & \text{Energy (in hartrees)} \\ \hline
\text{eclipsed} & -79.5971\\
\text{staggered} & -79.6027\\
 \hline
\end{array}$$

The difference in single point energies between the staggered and eclipsed forms $\Delta E_{\text{staggered-eclipsed}} = -0.0056$ which means the staggered form is lower in energy by about $14.7\ \mathrm{kJ\ mol^{-1}}$ (or $3.5\ \mathrm{kcal\ mol^{-1}}) $

Now, for ethane-1,2-diol. A first preliminary calculation, and random rotor search in Avogadro gave me three distinct conformers (although, two were very close in energy), however, every trial after that just produced two. They are given below:

$$\begin{array}{cc}
\hline
\text{conformer} & \text{Energy}\ (\mathrm{kJ\ mol^{-1} }) & \ce{OCCO}\ \text{bond angle} \\ \hline
\text{conformer 1} & 72.0914 & 59.0266\\
\text{conformer 2} & 85.534 & -179.967\\
 \hline
\end{array}$$

I have attached some screen shots below:

Now, hydrogen bonding is not explicitly treated in this calculation, but based on my discussion with @pentavalentcarbon the Leonard-Jones term in the forcefield (and to a lesser extent the electrostatic term) does capture these interactions.

So clearly, some stabilisng influence is evident in the structure, because if I were to argue on sterics alone, conformer 2 should have had lower energy than conformer 1. As an additional test, I measured the distance between $\ce{H}5$ and $\ce{O}2$ and found it to be 2.62 Angstroms. This is indeed suggestive

Now, using the geometries of conformer 2 and conformer 1, and that of an additional conformer with completely elcipsed oxygens that I obtained from Avogadro, I performed another CCSD(T) calculation is psi4 (again in the aug-cc-pVDZ basis). The table given below summarises my results:

$$\begin{array}{cc}
\hline
\text{conformer} & \text{Energy (in hartrees)} \\ \hline
\text{conformer 1} & -229.744410904091524
\\
\text{conformer 2} & -229.740781143788695\\
\text{completely eclipsed} & -229.725483329882337
 \\
 \hline
\end{array}$$

Again, the difference in single point energies between conformer 1 and conformer 2, $\Delta E_{\text{1-2}} = -0.0036297603$, means that conformer 1 is more stable by about $9.5399\ \mathrm{kJ\ mol^{-1}}$ (or about $2.277\ \mathrm{kcal\ mol^{-1}})$

This is probably due to an intramolecular hydrogen bond, as suggested, which has an overall stabilising effect on the gauche form of ethylene glycol (conformer 1). This is thus one of the instances in which the anti conformation (conformer 2) is higher in energy relative to the gauche conformation.

Finally, the difference in single point energies between conformer 1 and the eclipsed conformer, $\Delta E_{\text{1-eclipsed}} = -0.0189275742$, this implies that conformer one is more stable by about $49.694\ \mathrm{kJ\ mol^{-1}}$ (or $11.877\ \mathrm{kcal\ mol^{-1}}$)

Thus, I would say the energy penalty of eclipsing is still much higher than any stabilisation that can be gained from hydrogen bonding in this conformation. Shifting the angle between the two $\ce{OH}$ groups by about 60 degrees relieves this eclipsing strain while still allowing for the formation of a hydrogen bond, and is by far the best compromise.

To be honest, I wanted to scan over several dihedrals and make a potential energy plot, and calculate an exact barrier to rotation but faced two hurdles: a) couldn't write down the z-matrix properly, and couldn't write the appropriate loop in the program. b) the calculations that I did perform and have quoted here, took a while to run and based on that, I am guessing what I had in mind would have been a little too ambitious for my poor,  little laptop.

Addendum @Jan pointed out in the comments that the gauche effect is responsible for the stability of the gauche conformer, which to be fair, I did not think about at all because I was so focused on the "hydrogen bond", however, it is worth looking into. Here's a post that discusses it (answered by @Jan himself ;) )


How are atoms manipulated? I was wondering how scientists are able to handle atoms. They are very small, but surely people are able to interact with them somehow? The Large Hadron Collider is one example. Also, they try and make new elements by mixing atoms together. How are they able to handle something so small?

In general, one usually doesn't work directly with single atoms, though there are ways to do that as well, usually involving scanning probe microscopy (basically poking at atoms with very sharp probes to either manipulate them or just image them).

Atoms become much easier to work with when they are are ionized and can be manipulated by electric or magnetic fields (atoms have neutral charge so the Lorentz force can't be used to control them). Even then, most techniques deal with small groups of ions rather than individual ones. Ions are relatively easy to manipulate and we can easily trap them, accelerate them into beams, and focus them. The LHC is, of course, a very impressive (and expensive) example of this. The way it works is that packets of ions are accelerated by electric and magnetic fields to form two ion beams, which are then focused into a very small area and made to intersect, allowing the ions to collide and all the cool particle physics stuff to happen.

Manipulation of ions is also used in countless other scientific applications: semiconductor manufacturing, surface analysis, medical analysis, radioisotope dating, to name a few.


How do I make PDB 1HIG's data usable for producing ball-and-stick models? PDB: 1HIG, when opened using Accelrys DS Visualizer, Avogadro and Jmol produces a series of dots in space and cannot be displayed as 3D ball and stick models. Is there any way to overcome this problem?

This is the output from Accelrys DS Visualizer 4.0:

This is the output of Avogadro:

This is the output of Jmol:

This specific PDB file only contains the coordinates of all the $\ce{C_{\alpha}}$ atoms of the protein. It simply does not contain enough information to display an actual 3D structure of the protein.

The file does contain secondary structure assignments, so it should probably be possible to create a cartoon display of the protein, but that would require more than a simple PDB viewing program. But there is no way to create a ball-and-stick model with the data contained in that file.


How to write electronic configuration of p-orbitals? For example, the electronic configuration for the element boron can be writen as:
1s2 2s2 2p1
However, it is can be written more accurately by taking the different p-orbitals into account:

The electronic configuration with the different p-orbitals starting from boron ascending in atomic number:

B - 1s2 2s2 2px1
C - 1s2 2s2 2px1 2py1
N - 1s2 2s2 2px1 2py1 2pz2
O - 1s2 2s2 2px2 2py1 2pz2
F - 1s2 2s2 2px2 2py1 2pz2

Each type of 'p-orbital' can hold a maximum of 2 electrons. But the way they are assigned to which orbital is not always systematic e.g. the electrons do no wait till, pX orbital is full to be able to fill the next one (I know this has something to do with energy levels and trends in periodic table).
Can someone tell me how this works and how I can use it to be able to write electronic configurations for p-orbitals. It is much easier just using the summation of  px + py + pz = pX; but my teacher sometimes writes it out fully...

Electrons don't pair up in orbitals before each is singly occupied by Hund's rule of maximum multiplicity.
Writing out px, py, pz fully or using p^n notation (n=no of electrons in a p orbital) is a matter of context and circumstance rather than the standard convention; you might use the 'full version' when explaining bonding through hybridization but omit it while stating the electronic configuration of an atom generally. Both notations are quite interchangeable.


Why do increasing concentrations of zinc chloride not linearly decrease the freezing point? I did an experiment on how different concentrations of $\ce{ZnCl2}$ affect the freezing point of demi water. My results were that till a concentration of about $10\ \mathrm{mol\,dm^{-3}}$ the freezing point decreased (more or less in a linear manner). After that at concentrations of $15~\mathrm{mol\,dm^{-3}}$ the freezing point started to increase again (it again approached $0~\mathrm{^\circ C}$).

The graph below shows the freezing point depression of different $\ce{ZnCl2}$ solutions. The $x$-axis is the concentration of $\ce{ZnCl2}$ in $\mathrm{mol\,dm^{-3}}$, the $y$-axis is the freezing point depression in kelvin.

Why does the freezing point depression initially decrease, and then increase again after a certain concentration has been reached?

You've reached the other side of the eutectic.

See, that graph of lowering freezing temperature is always a part of a bigger picture.

Once you are past that minimum, it is no longer freezing point that you are measuring. Instead, it is the point where the solution becomes saturated and $\ce{ZnCl2*6H2O}$ (or whatever hydrate is stable at those conditions) starts crystallizing.

You'd see a similar pattern with pretty much any soluble salt. Zinc chloride, if anything, is good in that it dissolves immensely well, so the graph is wide, and its features can be seen more clearly.


Why does pouring carbonated beverages onto fruit cause them to fizz? I have noticed that making mimosas, the champagne fizzes more than just pouring it into a flute. Likewise, pouring beer onto citrus fruit or strawberries fizzes a lot. My theory is that the pH drop associated with the acidic fruit causes the solubility of CO2 to decrease and hence, the fizz. Can anyone verify this?

All carbonated drinks are oversaturated solutions of $\ce{CO2(g)}$. Pouring over fruits lead to contact with many residual air bubbles, which serve as  fizzing places for $\ce{CO2(aq) -> CO2(g)}$.

It is more or less the same as the ( previously already opened ) drink shaking, where violent fizzing occurs as well.

It is a similar process activation, like if you bend the activation chaff in a pocket heater with molten $\ce{CH3COONa \cdot 3 H2O}$, or if you pour overcooled water from a freezer ( see YouTube ) and watch fast ice growing.


Why is the units of kcat 1/s? I understand that $k_\text{cat}$ measures the turnover number of an enzyme. This measure is therefore a quantity of molecule conversions per unit of time. I suspect that my problem is more that of a lack of maths but why is the unit expressed in $\mathrm s^{-1}$? Why not just seconds?

The turnover frequency is a rate. (It's actually the kinetic rate of the reaction in saturating substrate concentration, normalized to the amount of enzyme.)

It might be helpful to think about something like a factory that makes a certain number of widgets in a certain number of time. How would you naturally describe how fast the factory could make widgets? Most people would express how fast the factory works in terms of "widgets per hour". It's the same with enzymes - it's most natural to describe how fast the enzyme works in terms of "turnovers per second". It's just that the "turnover" is implied and is viewed as "dimensionless", so it's left off, leaving you simply with "s-1" as a unit.

(Note that even though the "turnover" is left off, it's still very important to know it's there, and to know what, exactly, one "turnover" is. In complicated reactions there is sometimes different reasonable ways to count what "one turnover" is, and different definitions could result in a 2, 3, 4, 5, etc. fold difference in value. Those vexing numeric differences become easy to explain once you realize this researcher is using the formation of one molecule of this product as "one turnover", but that researcher is using the disappearance of one molecule of that substrate as "one turnover".)

Naively, one could invert the ratio to come up with a value in time, rather than in 1/time. For example, if a factory makes 5 widgets per hour, it takes them 0.2 hours to make a widget ... sort of. One issue you run into is that that's the time on average. In a stochastic process like enzyme turnover, it's highly unlikely that any single turnover will take exactly the time specified. Some will take longer, some will be quicker. Only for a large number of turnovers in aggregate does that time value have meaning. Saying that "a turnover takes 5 s" gives a false impression about individual turnovers in a way that saying "this enzyme works at 0.2 turnovers per second" doesn't.

Additionally, in the case of multi-step reactions, saying that "a turnover takes 5 s" may be incorrect. The entire pathway may take much more that that. It's just that once the pathway has reached steady state it's only the rate limiting step which takes 5 s.  For example, take a pathway of 20 steps, 19 of which take 4 s each, but one which takes 5 s. It's going to take (19*4 + 5) = 81 seconds for any given substrate molecule to start at the beginning of the pathway and make it to the end. However, once the pathway is full and each step is handing off products to the next step, the pathway will be producing product at a rate of 1 every 5s. (This, of course, doesn't apply to enzymes where all reactions take place at the same active site, but it's a consideration in the general case, which explains why 1/s is the preferred unit for reaction rates.)


Is it possible to have anions of metals? I've been wondering if it is possible to have negatively charged metal ions. Most ions tend to have electron configuratons of noble gases. So then is it possible to obtain metal ions from metal atoms by changing their electron configuration to that of the next noble gas in the periodic table? (e.g. something like Be(6-))

Yes.  And, ironically, the most clear-cut examples involve alkali metals.  In these compounds you have the cation, often an alkali metal itself, stabilized in a complex so that electrons can be transferred to the recipient alkali metal (Na or heavier, based on today's known compounds), and the latter is a full-fledged alkali metal anion.


Hydrogen reactions with catalysts I have a question about reactions involving hydrogen using catalysts. I'm a layman, so forgive me if my questions seem elementary.

Reacting hydrogen plus oxygen and a flame reacts, unleashing energy (in the form of an explosion) and creating water.

My questions:

Why does it need a "flame or spark "? Is there something unique
about the fire that causes the reaction? Or is it merely the heat of the flame? In other words, if you put hydrogen and oxygen in, say, a
pressurized vessel and added energy in the form of heat (but not a
flame) would it still react and create water? Would it do it without an explosion?
I understand catalysts "speed up" the reaction, or put another way,
reduces the amount of energy required to cause it. I've read that in
this particular reaction ($\ce{H2 + O2}$) copper works as a
catalyst. Is there a way to determine by looking at the chemical
formulas if other catalysts would work as well or better or is it
just trial and error? For example, could you tell if silver would
also act as a catalyst just by looking at the formula? Or would you
need to actually add silver and see for yourself?
When a catalyst is used, is it inexhaustible or does it get "used
up"? If you were using copper to react these two, could you use it forever? If not, is there a way to "recharge" the catalyst?

Thanks in advance, I'm trying to learn about this particular reaction for a project I'm working on ($\ce{H2 + O2}$) but without a formal science background, some of the scientific papers I'm reading are a little hard to grasp.

Adding a bit of information for your question a). The reactin we are looking at is:

$$\ce{2 H2 + O2 -> H2O}$$

As you can clearly see, the final product water has one oxygen atom for 2 hydrogen atoms, while in the beginning, we only had diatomic molecules. This means that during the reaction, at least one fo the diatomics had to break to single atoms. In this case, this happens with the reaction:

$$\ce{H2 -> 2H^.}$$

From this step, we have a nice reaction scheme called a radical chain reaction. However, we first need to break the $\ce{H-H}$ bond somehow, and this is what flame is doing. If you break a significant amount, then you don't need to provide flame anymore, the reaction is self-supporting — the flame is ignited.

Now, a catalyst is doing exactly this. Many metals, including copper, can adsorb hydrogen molecules to the surface, and this also results in breaking the $\ce{H-H}$ bond — so the explosion can start this way too.


What exactly is a solubility product, and what is its use? What exactly is a solubility product, and what is its use?

I've come across this term while reading, and don't have access to anyone who can help me understand this concept.

If we take small particles of a sparingly soluble salt, say AgCl, then one or two particles may dissolve in water and a saturated solution is obtained pretty soon. Upon adding extra AgCl to the solution precipitation of AgCl starts. At this stage, you'll find an equilibrium is established between the dissolved and the undissolved AgCl.

$$\ce{AgCl(s) <=> Ag+ (aq.) + Cl- (aq.)}$$

And for this equilibrium if you're to find the equilibrium constant, let's call it  $K_{sp}$, you get:

$$ \ce{K_{sp} = [Ag+] \cdot [Cl-]} $$

(Both the concentrations are the ones taken at equilibrium.)

This $K_{sp}$ is called solubility product. It's useful to predict whether or not more solute can be dissolved in your solution or not. We do this by comparing it to something which is called the ionic product.


Solvent for recrystallization of benzoic acid? Which is best solvent for recrystallization of benzoic acid?

Water
$\ce{CH3Cl}$
$\ce{CCl3COOH}$
$\pu{0.1 M}\ \ce{NaOH}$

I understand the solute/solvent should be soluble at high temperature, and insoluble at room temperature. But how to make the specific arguments for eliminating the alternatives?

It can be difficult to guess at what a given compound will crystallise out from, and often there is a lot of trial and error involved.

Generally, the most 'common' type of recrystallisation involves finding a solvent in which the compound of interest is sparingly soluble at room temperature, but highly soluble at elevated temperature such that slow cooling of a saturated solution affords recrystallised material.

If you look up the entry for benzoic acid in Purification of Laboratory Chemicals, Armagero and Perrin, several solvents are suggested:

  Benzoic acid has been crystallised from boiling water (charcoal), aqueous acetic acid, glacial acetic acid, benzene, aqueous EtOH, petroleum ether (b 60-80C), and from EtOH solution by adding water.


Which has greater N-O bond length: NO or NOCl? I came across this question. According to me, $\ce{NO}$ should have a shorter bond length as it has a triple bond, while $\ce{NOCl}$ has a double bond.
But the answer is given that $\ce{NO}$ has greater bond length than $\ce{NOCl}$. Why?
Is the answer wrong

Also relevant:  the nitrogen-chlorine bond lengths:

$\ce{NOCl}=196\text{ pm}$ Wikipedia

$\ce{NCl3}=176\text{ pm}$ Wikipedia

The oxygen in nitrosyl chloride is donating its lone pair to the antibonding orbital in the nitrogen-chlorine linkage, weakening that bond but adding more pi bonding between nitrogen and oxygen to become like a triple bond.

If we were supposed to compare bond lengths using "simple" principles of bonding, this is a poor choice.  Nitrosyl chloride presents more than meets the eye.


Equilibrium constant I know that equilibrium constant depends on temperature and pressure. But according to Le Chatelier's principle when we change pressure of reagents the equilibrium is the same. Why?

The position of equilibrium depends on temperature and pressure, as Nick Burns describes in his answer. The equilibrium constant depends on temperature only. The equilibrium constant is not a measure of the position of equilibrium, but rather which positions (representing different compositions) along the the reaction coordinate surface are equilibrium positions. Changing the temperature changes the shape of the reaction coordinate surface. Changing the pressure just moves you to a different point on the surface.

Example: consider a hypothetical reaction in which two molecules of gas react to form a third.

$$\ce{A(g) + B(g)<=>C(g)}$$

The reaction quotient $Q$ is simply the ratio of the product of the partial pressure(s) of the product(s) to the product of the partial pressure(s) of the reactant(s). $Q$ can have any value representing different points on the reaction coordinate surface.

$$Q=\frac{P_C}{P_A P_B}$$

The equilibrium constant represents special values of Q in which the reaction coordinate surface has a free energy minimum. For the sake of our example, let's make the equilibrium constant an easy value to work with, say $K=2$.

$$K=Q_{eq}=\frac{P_{C,eq}}{P_{A,eq} P_{B,eq}}=2$$

Since we are talking about gases, we also have the constraint the sum of the partial pressures of the reagents must equal the total pressure of the system. Again, for ease, let us set this value to an easy value to work with, say $P_T = 7 \text{ kPa}$.

$$\frac{P_{C}}{P_{A} P_{B}}=2$$
$$P_A + P_B + P_C = 7$$

Now we have two equations of three variables, which means there are an infinite number of possible values of the partial pressures of the reactions components that fit these two conditions. One is $P_A = 1$, $P_B =2$, and $P_C =4$. Another equally valid combo is $P_A = 2$, $P_B=1$, and $P_C=4$. The position of the equilibrium in this case is better given by a ratio of the sum of the partial pressure(s) of the product(s) to the sum of the partial pressure(s) of the reactant(s):

$$\frac{P_C}{P_A + P_B}=\frac{4}{3}$$

If we change the total pressure to, say $P_T = 4$, we change the allowable combinations of $P_A$, $P_B$, and $P_C$ that can get us to the equilibrium condition of $Q=2$. One allowable combination (of an infinite number) is $P_A=1$, $P_B=1$, and $P_C=2$. The position of the equilibrium is now:

$$\frac{P_C}{P_A + P_B}=\frac{2}{2}=1$$

Le Chatelier's principle suggests that When pressure decreases, the equilibrium shifts toward the side of the reaction with more equivalents of gas.

If we consider changing the situation from $P_A = 1$, $P_B =2$, $P_C =4$, and $P_T=7$ to a situation with $P_T=2$, we can do some math to figure out new values of $P_A$, $P_B$, and $P_C$ that fit this new situation. First, we set the partial pressures of the compoents to $\frac{2}{7}$ of their original value, and calculate Q.
$$P_A =1\times \frac{2}{7}=\frac{2}{7}$$
$$P_B =2\times \frac{2}{7}=\frac{4}{7}$$
$$P_A =4\times \frac{2}{7}=\frac{8}{7}$$
$$Q=\frac{P_C}{P_A P_B}=\frac{\frac{8}{7}}{\frac{2}{7} \times \frac{4}{7}}=\frac{\frac{8}{7}}{\frac{8}{49}}=49$$

Since $C>K$ the equilibrium will shift to decrease $Q$. This can happen by decreasing $P_C$ and increasing $P_A$ and $P_B$. Since there is an reaction equation relating these three components in a 1:1:1 stoichiometric ratio, the change must have the same value vut in different directions:

$$K=2= \frac{\frac{8}{7}-x}{\left(\frac{2}{7}+x \right)\left( \frac{4}{7}+x\right)}$$

Rather than goof up the algebra, I let WolframAlpha do it. The solution for $x$ is:

$$x=\frac{1}{128}\left(-19 \pm \sqrt{681} \right)\approx \{+0.2534..., -1.6106...\}$$

The first answer for $x$ is nonsense, since it would put one partial pressure less than $0$. If we change the partial pressures appropriately, we get new values of

$$P_A = \frac{2}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.5391...$$
$$P_B = \frac{4}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8248...$$
$$P_C = \frac{8}{7}-\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8894...$$

The position of equilibrium is

$$\frac{0.8894...}{0.5391...+0.8248...}\approx 0.65$$

which is less than the value of $1.33$ from when $P_T=7$ and different from the value of $1$, when the partial pressures were different (but still met the criteria than $P_T=2$ and $Q=2$).

Changing the temperature changes the values of $Q$ which define the equilibrium state:

$$K=e^{-\Delta G^\circ/RT}$$


How can I eliminate the harmful gas produced from the heating/manipulation of galvanized steel? When welding on steel with a galvanized coating, hexavalent chromium is yielded from the chemical reaction. This gas is something that I would like to eliminate before I begin welding, possibly with some liquid or mixture. Any ideas?

Galvanizing by definition is zinc coating. The zinc boils off ( as 1700 F)  is approached and immediately oxidizes to ZnO , a white powder. ZnO has the unusual characteristic of causing zinc chills = brass founders ague = oxide shakes,  IF inhaled while less than one hour old . After that is is pretty harmless and used in various lotions and creams and rubber products.  So avoid the fumes when welding. However , one recovers from "chills" in about 24 hours so not toxic ( I can verify no lasting affects as I had "chills" many decades ago). There is another potential problem ; zinc can cause liquid metal embrittlement of steel . This occurs where the zinc is not exposed to air as if two surfaces are placed together and the edges sealed by welding. I don't know where you think chromium is coming from other than steel made with scrap ( the majority) typically contains about 0.04 % tramp chrome ( and Ni ,and Cu , and Mo). So essentially all steel  welders in the world are exposed to this tramp chrome.


How to assign priority groups of tartaric acid? How do we assign priority groups of tartaric acid? Counting outwards from any of the chiral carbons we get the same atom.

First you have to rank the molecules around the two chiral centres of tartaric acid according to Cahn Ingold Prelog principle. (I have added the principle in below after this 3 points)
After ranking the molecules according to Cahn Ingold Prelog principle, then orient the molecule such that the lowest priority group is directed away from you. The three remaining groups then project toward you.

If you intend to find the configuration of the molecule (i.e. R or S), then let me add this point:

If the three groups projecting towards you are ordered from highest priority to lowest priority (i.e from rank 1 to rank 3) in clockwise manner, then the configuration is "R". If the three groups projecting towards you are ordered from highest priority to lowest priority (i.e from rank 1 to rank 3) in counterclockwise manner, then the configuration is "S". I mentioned "from rank 1 to rank 3" - because the 4th one is oriented away from us and is not considered while checking the order.

Cahn Ingold Prelog principle:

Rule to prioritize the groups attached to a chiral carbon:

Prioritize the four atoms, or groups of atoms, attached to the chiral center based on the atomic number of the atom that is bonded directly to the chiral center. The higher the atomic number, the higher the priority.

Like say you have carbon and lets the 4 atoms attached to it are: $\ce{H},\ce{OH},\ce{Cl},\ce{Br}$.
So here the carbon is chiral, as all the 4 atoms attached to it are different and now for ranking, we rank it on the basis of atomic number. Here as Br has highest atomic number, It would be ranked 1, then Cl would be ranked 2, then OH will be ranked 3 (oxygen will be considered as it is directly attached to the chiral carbon) and then hydrogen will be ranked last.

Atoms participating in double/triple bonds are considered to be bonded to an equivalent number of similar "phantom" atoms by single bonds. This is done just for the purpose of the rule. See this pic for better understanding:
If two or more of the atoms that are bonded directly to the chiral center are the same, then prioritize these groups based on the next set of atoms (i.e. atoms adjacent to the directly bonded atoms)

Lets take an example here. Say to the carbon I have $\ce{CH3},\ce{COOH},\ce{Cl},\ce{H}$ attached.
Here first we will compare the atomic number of the atoms which are directly attached to the chiral carbon i.e. C, Cl and H. From these as Cl has highest atomic number, hence it is ranked 1. Then, for the carbon, as for both the groups i.e $\ce{CH3}$'s and $\ce{COOH}$'s, it is the carbon atom which is directly connected to the chiral carbon. Hence, we need to look for the next set of atoms. $\ce{CH3}$ next set of atoms are H, H & H; while $\ce{COOH}$'s next set of atoms are O, H and O-C (see pic provided in rule 2). So from here easily we can say $\ce{COOH}$ is ranked 2, then $\ce{CH3}$ (i.e. rank 3) and then H.


R/S configuration of bridging carbon in bicyclic system I am trying to figure out the R and S configuration of bridging carbons in bicyclic systems.
In the image, there are two bicyclic systems and depending on the position of the Br substituent, the R or S configuration of the bridging carbons change. How is the configuration actually determined? I don't understand how the prioritization works in this case, as two substituents are similar. Thus, the reason must be the adjacent stereocenter... But I am not able explain it properly.

Several days ago I offered a solution to the R/S assignment of the bridgehead carbons of (1R,2S,4R)-2-bromo-2-methylbicyclo[2.2.2]octane 1. During the middle of the night I realized that the method was faulty because the solution would not accommodate the (1S,2R,4S) enantiomer  3 but would rather predict (1R,2R,4R) which clearly is a non-existent diastereomer. The shortcoming lay in not applying, in a heirarchical sense, CIP rule 4b (RR/SS>RS/SR) over rule 5 (R>S).
Focusing on bromide 1, the configuration of C2 is "S" with Br>C1>C3>C9. Structure 2 is structure 1 from a different perspective to which we will refer during the ensuing discussion. To determine the configurations of the bridgehead carbons, C1 and C4, a digraph is the method of choice. In the "Digraph of C1" one starts at the non-duplicate atom C1 and traces the six possible paths back to duplicate atoms C(1) that are each attached to three phantom atoms of atomic number zero. The vertical chain has the top priority while the hydrogen (not shown) at C1 has the lowest priority. The two horizontal chains are mirror images and they must be assigned to the second and third priorities. In these chains C2 has been predetermined as having the S-configuration. Locate C4 in the left chain. It has the R-configuration because C3>C8>C5. Locate these atoms in structure 2 and convince yourself. [I use my hands. Point your right thumb from C4 to H and your fingers will point from C3 to C8 to C5]. In the right hand chain C4 has the S-configuration because C3>C5>C8. Check these priorities in structure 2.
Now the left hand chain is designated as RS and the right hand chain as SS. Since RR/SS>RS/SR, the order of carbon atoms attached to non-duplicate C1 is C2>C6>C7. Locate these atoms in structure 2 and convince yourself that C1 does indeed have the R-configuration!
Look at "Digraph of C4". Now that you are versed in the method, the priorities around non-duplicate C4 are C3>C8>C5 because C3-chain>SS>RS>H. C4 also is of the R-configuration! <continued>

The digraphs below confirm the assignments to the enantiomer 3 (4), (1S,2R,4S)-2-bromo-2-methylbicyclo[2.2.2]octane. I'll leave the analysis to the reader.