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Entropy and specific heat capacity
I have seen the equation $S(T_2)=S(T_1)+C_p\ln(T_2/T_1)$ where $C_p$ is the molar heat capacity at a constant pressure. I understand that this assumes that the temperature range is sufficiently small that the constant pressure heat capacity does not vary significantly over it.
My question pertains to the derivation of this expression. One of the first steps is substituting $q=C_pdT$ into $dS=\frac{q_{rev}}{T}$. I am not 100% sure as to why these enthalpies are equated. Is it because the heat exchanged by a system at a constant pressure is always reversible heat, or is this not true?
If the above were true, it would explain to me why I never see an expression involving constant volume heat capacity. If the above is not true, could someone also explain why suh a form does not exist?
Finally, is the equation $S(T_2)=S(T_1)+C_p\ln(T_2/T_1)$ only applicable where the pressures are the same at the initial and final temperatures? On the one hand, the derivation assumes a constant pressure in the infintesimal changes from the initial to final temperature (and of course uses a constant pressure heat capacity!); however on the other hand entropy is a state function. So I would think that if one could find how entropy varies wit pressure at a constant temperature, and the change from the final to the initial state involved both a temperature and pressure change, then I would think that the above expression can be used to find the change in entropy due to the temperature change only, and then the additional factor from the pressure could be used?
Thank you in advance.
EDIT: For the second part, from having a quick think about the reversible heat enxchange in and isothermal process, I think $S(T_2,p_2)=S(T_2,p_1)+R\
 \ln(p_1/p_2)$. So then one could construct a Hess-like cycle to go from one temperature and pressure/volume, to another temperature and pressure/volume. The use of the cycle and step-wise calulcation works because entropy is a state function I think?
Hopefully I answered the second part of my query (I would appreciate if someone could verify that it is correct), although I am still not sure about why the heats used are the same in the derivation, and whether the expression really does refer to the same pressure at the initial and final temperatures, even though the process in getting from these two temperatures can involve pressure changes (again, thinking about $S$ being a state function, so the process itself when changing states does not matter).

It is not true that heat exchanged at constant pressure is always reversible.
But, if you want to determine the change in entropy from thermodynamic equilibrium state 1 at $(T_1,P)$ to state 2 at $(T_2,P)$, you need to forget entirely about the actual irreversible process path that took you from state 1 to state 2.  It is of no further use.  You instead need to focus exclusively on the two end states.  And, in order to apply the entropy equation, you need to devise a reversible path between these exact same two ends states.  Any reversible path will do, because they will all give the same value for the entropy change.
In the case of your example, you can choose a constant pressure reversible path for convenience. So, for such a path, from the first law we know that $\mathrm{d}H = \mathrm{d}q$. But, now how do we figure out a reversible path between the two states in which the system temperature is changing reversibly along the path?  If we contact the system with at constant temperature reservoir held at the final thermodynamic equilibrium temperature $T_2$, that path would not be reversible because there would be a finite temperature difference between the system and the reservoir over most of the path. But, now, suppose we contact the system, not with a single constant temperature reservoir, but with a sequence of constant temperature reservoirs, each at a slightly different temperature over the range $T_1$ to $T_2$. This would guarantee that, over the entire path, the system would differ only slightly from the reservoir it is currently in contact with. So, in this case, $\mathrm{d}q_\mathrm{rev} = \mathrm{d}H = C_p\,\mathrm{d}T$, where $\mathrm{d}T$ is the change in the system temperature over the present increment of heat exchanged. We use the equation
$$\mathrm{d}H = C_p\,\mathrm{d}T$$
because of the precise definition of constant pressure heat capacity in thermodynamics:
$$C_p\equiv\left(\frac{\partial H}{\partial T}\right)_P$$
So now, for the change in entropy, we have:
$$\Delta S = \int{\frac{\mathrm{d}q_\mathrm{rev}}{T}} = \int_{T_1}^{T_2}\frac{C_p \,\mathrm{d}T}{T}$$
So the key thing to remember here is that, if you want to determine the change in entropy for an irreversible process, you first need to determine the initial and final end states (say, using the first law of thermodynamics) and then devise a reversible path between these same two end states to calculate the integral of $\dfrac{\mathrm{d}q_\mathrm{rev}}{T}$ for that path.

Hi, I was wondering why you used an exact differential for change in heat instead of an inexact differential? Otherwise it looks great. Thanks for the answer!

@cngzz1 How would you represent the difference between an exact differential and an inexact differential?

I would use the lower case Greek letter delta q instead. Here's a link on inexact differentials (to see the symbol involved): https://en.m.wikipedia.org/wiki/Inexact_differential

@cngzz That is just an artifact of how you learned it.  I choose not to use that kind of notation.

Ok I understand. Thanks for your reply.


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What is the difference between balancing a redox reaction in acidic solution vs basic solution?
I'm asked to balance the potassium permanganate and hydrogen peroxide reaction. I misread the problem and thought that the reaction was in acidic solution. This is the result that I've got:
$$\ce{2MnO4 + 5H2O2 + 6H+ -> 2Mn^2+ +5O2 +8H2O}$$
What do I have to change to balance the equation in a basic solution?.

The best approach is not to think in terms of changing this equation, but to just start over from scratch.

You basically just have to add hydroxide ions to either sides of your equilibrium.
$$\ce{2MnO4 + 5H2O2 + 6H+ + \color{blue}{6OH-} -> 2Mn^2+ +5O2 +8H2O + \color{blue}{6OH-}}$$
$\ce{H+}$ and $\ce{OH-}$ ions can be collectively re-written as $\ce{H2O}$, so we have:
$$\ce{2MnO4 + 5H2O2 + \color{blue}{6H2O} -> 2Mn^2+ +5O2 +8H2O + 6OH-}$$
Finally, we can cancel out the excess water molecules which leaves us with:
$$\ce{2MnO4 + 5H2O2  -> 2Mn^2+ +5O2 +2H2O + 6OH-}$$

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Why is ice less dense than water?
I know the traditional explanation, which says that ice has large spaces between $\ce{H2O}$ molecules because hydrogen-bonding gives it an open structure. But what does the open structure have to do with hydrogen-bonding? Why isn't a similar phenomenon observed in other species which exhibit hydrogen-bonding, like $\ce{HF}$ or $\ce{NH3}$?

See this post.

This certainly doesn't answer your question, but FYI it's a myth that expansion upon freezing is unique to water. There are any number of organic compounds for which this is the case, also some pure elements which clearly to not hydrogen bond like gallium, antimony, germanium, silicon and more.

What I don't get is why the hydrogen bonding somehow causes the ice to have an open structure.

The following is an image of the hexagonal crystaline form of ordinary ice (Ice I$_h$) taken from S.S. Zumdahl, Chemistry, 3rd ed., copyright © 1993 by D.C. Heath and Company:

Note that the dashed lines represent hydrogen bonds. Liquid water actually has a similar "open" structure also due to hydrogen bonding. But in the case of liquid water, the hydrogen bonds are not rigid and semi-permanent as in ice. So imagine that in the image above, the hydrogen bonding network collapses. This is what happens when enough thermal energy is present to break the rigid hydrogen bonds resulting in melting. Clearly, once this crystaline structure is no longer forced into place by the rigid hydrogen bonding in ice, it can collapse into itself, resulting a greater density of water molecules.
Thus the liquid form of water, although engaged in transient hydrogen bonding, is not as open and expanded as when held into it's solid form by the rigid, semi-permanent hydrogen bonding.

As an addition to the other answers, note that the "honeycomb" structure that is responsible for the lowered density upon freezing is not sacrosanct.  It can be collapsed, without melting, by high pressure starting at about 200 MPa.  These are the high pressure ice phases, about a dozen of which are known. All those in equilibrium with the liquid, apart from the low pressure Ice $I_h$ phase, are more dense than the liquid with which they are in equibrium; so the melting point of water starts rising once we get to the threshold of Ice $III$ at about -22°C and 210 MPa.
We can also go the other way, generating ice structures even more open and lower in density than Ice $I_h$.  Such phases are not realized in pure water but appear in clathrates such as the well known methane clathrate.

Hydrogen bonds hold water molecules in place in solid phase
Structure of ice is a regular open framework of water molecules arranged like honeycomb
When it melts framework collapses and the water molecules pack closer together, making liquid water more dense

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Dependence of the angle of rotation on the wavelength of plane polarized light
Wavelength difference is a big deal, I know. It can solely change the whole interaction between the chiral molecule & the light. But I am not sure what's the mechanism by which light of different wavelengths produces different angles of rotation.
So, why and how wavelength matters?

The rotation of plane polarised light by a solution of, say, sucrose depend on the ability of the oscillating magnetic filed of the light to induce an electric dipole moment in the molecule and the ability of the oscillating electric field of the light to induce a magnetic dipole moment.
For these interaction to have any magnitude it is supposed that the electrons in a molecule move in a helical path, or, alternatively that there are two linear dipoles generated by electron motion which are in planes that are at some angle to one another. Although this model is clearly artificial it emphasises the fact that movement of charge must follow somewhat crooked pathways under the influence of the radiation. The theory of this is very complicated but the result from a quantum calculation is relatively straightforwards and is that the molecular rotation M at wavelength $\lambda$ is given by
$$M_\lambda = a\sum_i \frac{\lambda_{0i}^2X}{\lambda^2-\lambda_{0i}^2}$$
where i represents all the electronic states of the molecule and the wavelength $\lambda_{0i} = c/\omega_{0i}$ where $\omega_{0i} = (E_i-E_0)/\hbar$ is the frequency of the $i^{th}$ excited state at energy $E_i$. The parameter $X=\mathrm{Im}(\mu_{m_{0i}}\cdot\mu_{e_{i0}})$ is the value imaginary part of the complex dot product of the induced magnetic and electric dipole moments and a  are a set of constants independent of wavelength.
From this formula it is seen that the polarisation rotation depends on the wavelength. (The dependence of rotation angle on wavelength has been called optical rotatory dispersion.) We can also conclude

that experiments should be performed where the molecule has little absorbance but that the rotation will be larger as an absorption band is approached. (in an absorption band elliptically polarised light can be formed)
The strength of any optical transition is not important as the dot product X depends on induced moments and the angle between them. This also means that weak optical transitions can be as important as strong one in rotation the polarisation. If the induced dipoles are perpendicular then the rotation vanishes as the dot product is zero.
That many excited states i can be involved and their effects can cancel to some extent so the value of the rotation is hard to predict.
The polarisation rotation of two mirror image molecules are equal and opposite in size and if a molecule is identical with its mirror image polarisation rotation must be zero and the term X is zero.
When the wavelength $\lambda < \lambda_{0i} $ i.e. the wavelength crosses a transition then the signal changes sign and this is called the Cotton effect.

'strength of any optical transition' what do you mean by it?
"it is supposed that the electrons in a molecule move
in a helical path" is this assumption rational and legitimate?

'strength of transition'  just means how intense the absorption spectrum is, i.e. how big the extinction coefficient is. The 'helical path' thing is just a model to understand the effect. Linear dipoles at an angle work just as well and is more intuitive.  It simply means that the radiation induces some complicated electron response that can be broken down into a helical path among other motions.

Would you refer to me any paper or link where I can learn more about this helical path. Still didn't get the hold of it.

The text book by Kuhn & Forsterling ' Principles of Physical Chemistry' section 8.11 has a brief explanation but an old book W. Kauzmann 'Quantum Chemistry' has the all the gory details of how a chiral molecule can have a spiral path of induced  electron motion.

Any internet link?

Light travels at different speeds in different materials.  The ratio of the speed at which light travels in a vacuum compared to the speed light travels in a given material is given by the index of refraction ($n$) of the material.
$${{n}=\mathrm{\frac{speed~ of~ light ~in~ vacuum}{speed ~of~ light~ in~ material}}}$$
The index of refraction is a basic property of a material.  Further, the index of refraction of a material is wavelength dependent.
The beam of plane polarized light that we use in our polarimeter experiments is created by combining a beam of left circularly polarized light with a beam of right circularly polarized light.  Those two beams interfere with each other to produce the plane polarized light we use in the polarimeter experiment. It is important to note that both the right and left circularly polarized beams are chiral as they trace out a right and left handed helix.
If the material in the polarimeter is not chiral, then the angle of the plane polarized light will not be rotated as it passes through the sample. However, if a chiral material is placed in the path of a plane polarized beam, then the left circularly polarized component of the beam will be rotated a different amount than the right circularly polarized component (the 2 beams are chiral and interact differently with the chiral sample) and we observe a rotation of our light beam as it passes through the polarimeter. In other words, the index of refraction of the left circularly polarized light ($n_L$) is different from the index of refraction of the right circularly polarized light ($n_R$) in chiral media.
Since $n$ itself is wavelength dependent, both $n_L$ and $n_R$ are also wavelength dependent and the observed rotation from our sample will change as we change the wavelength of our light beam.
This property of $n_L$ and $n_R$ is the basis of the Cotton effect.  If you would like to read more about the Cotton effect and its use in chemistry see this earlier answer.

"combining a beam of left circularly polarized light with a beam of right circularly polarized light" would you elaborate it?

@Mockingbird Circularly polarized light can be used to create plane polarized light and vice-versa, see here for example. Therefore, a plane polarized beam of light may be treated as the combination of two circularly polarized in-phase beams of the same frequency and each with half the amplitude of the resultant plane polarized beam.  Viewing a plane polarized beam this way makes it easier to understand its interaction with chiral molecules.

So it means the plane polarized light becomes circularly polarised light as the 2 circularly constituent polarized lights get out of phase. But, the resultant electric wave vector would be rotating, so the measured rotation angle will depend on the length of the polarimeter, right?

@Mockingbird Very close. The plane polarized light does not "become" circularly polarized.  Rather the plane polarized light can be viewed as the superposition of 2 circularly polarized beams (one right circularly polarized and the other left).  Since the 2 circularly polarized beams interact diasterotopically with  the chiral media (they have different indices of refraction in chiral media), the plane of polarization rotates as the beam travels through the chiral sample.  So yes, the observed rotation is dependent upon polarimeter length.

So, chemists use a specific length of the polarimeter to compare optical activity?

They convert the experimentally observed rotation into the specific rotation or the molecular rotation. Both of these are independent of path length and concentration.

Ah! another question,  why left and right circularly polarized light have different index of refraction?

They differ only when they pass through chiral media.


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If concentration of K+ ions inside the cell is almost equal to concentration of Na+ ions outside of the cell why these ions cross the membrane?
I'm a bit confused about electrochemical gradients of sodium and potassium ions. They have similar concentrations and are kind of similar ions but their distribution is different across the membrane. If that is true why they cross membrane. Why don't these gradients somehow cancel out? Why is there even need for sodium to enter the cytosol when potassium ions which are also positive are already inside?

While there is a good chance that you'll get a good answer here, I suspect that you might actually have better luck on Biology.SE.

Simple osmosis.

In order to understand why this happens, you have to recall the permeability characteristics of the plasma membrane. It seems that the movement of ions to and fro the cystol is counterproductive however this is not so. Actually there is unequal distribution of ions in the extracellular fluid and cytosol

Extracellular fluid is rich in Na+ and chloride ions (Cl–). In
  cytosol, however, the main cation is K+, and the two dominant anions
  are phosphates attached to molecules, such as the three phosphates in
  ATP, and amino acids in proteins.

Differences in intracellular and extracellular concentrations of ions result primarily from:

the Na +/K + pump and
the permeability characteristics of the plasma membrane.

Ions pass through the plasma membrane through ion channels. The two major types of ion channels are leak channels and gated ion channels.
i. Leak Channels
Leak channels, or non-gated ion channels, are always open and are responsible for the permeability of the plasma membrane to ions when the plasma membrane is unstimulated, or at rest. The plasma membrane is more permeable to K +and Cl- and much less permeable to Na+ because there are many more K + and Cl- leak channels than Na+ leak channels in the plasma membrane.
Because the plasma membrane typically has more K+ leak channels than Na+ leak channels, the number of potassium ions that diffuse down their concentration gradient out of the cell into the ECF extracellular fluid ECF is greater than the number of sodium ions that diffuse down their concentration gradient from the ECF into the cell. See below

The plasma membrane is relatively permeable to K + and much less permeable to Na + and to negatively charged molecules found inside of the cell. Consequently, positively charged K + tend to diffuse out of the cell through K + leak channels, leaving the negatively charged molecules behind, thus polarizing the membrane. The membrane is at equilibrium when the tendency for K + to diffuse out of the cell is resisted by the negative charge of the molecules inside the cell.
ii. Gated Ion Channels
These channels open and close in response to small voltage changes across the plasma membrane. The movement of ions into or out of the cell changes the charge difference across the plasma. membrane, which causes voltage-gated ion channels to open or close. Voltage-gated channels specific for Na+ and K + are most numerous in electrically excitable tissues
The Na+/K+ Pump
The differences in K + and Na + concentrations across the plasma membrane are maintained primarily by the action of the Na + –K + pump. Through active transport, the Na+/K + pump moves K+ and Na+ through the plasma membrane against their concentration gradients. Potassium ions are transported into the cell, increasing the concentration of K + inside the cell, and Na + are transported out of the cell, increasing the concentration of Na + outside the cell. Approximately three Na+ are transported out of the cell and two K + are transported into the cell for each ATP molecule used. These pumps help maintain the resting membrane potential by pumping out Na+ as fast as it leaks in. At the same time, the Na+/K+ ATPases bring in K+.
In summary it is important to note, there is uneven distribution of ions in and outside the cytosol.
Because the plasma membrane has more K+ leak channels than Na+ leak channels, the number of K+ ions that leave the cell is greater than the number of Na+ ions that enter the cell. As more and more K+ ions leave the cell, the inside of the membrane becomes increasingly negative and the outside of the membrane becomes increasingly positive. The inability of most anions to leave the cell also contributes to the negativity of the resting membrane potential. These trapped anions cannot follow K+ out of the cell because they are attached to non-diffusible molecules such as ATP and large proteins.
Nevertheless, sodium ions do slowly diffuse inward, down their concentration gradient. If not controlled, such inward leakage of Na+ would eventually destroy the resting membrane potential. The small inward Na+ leak and outward K+ leak are offset by the Na+/K+ ATPases (sodium pump)
This diffusion accross the plasma membrane is necessary because:

There is uneven distribution of ions which needs to be corrected.
The characteristics of the plasma membrane has lot to do with how transport of substances occur ( It is important to note that not only ions use this path, other substances vital for cell growth have to use this path as well, therefore strict control is necessary)

Hope this helps
References

Anatomy and Physiology (Seely)
Principles of Anatomy and Physiology (G. Tortora, B. Derrickson)


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Why increase in enthalpy leads to increase in entropy?
Given statement: Enthalpy of adsorption is the strength at which molecule (say $\ce{H2}$) is bound to the adsorbent (say some metal center). For a free $\ce{H2}$ molecule, the entropy contribution is from translational, rotational and vibrational motion. After the adsorption, the translational and rotational motions will not contribute to the entropy change $\mathrm{\Delta S}$. Thus the only contribution to $\mathrm{\Delta S}$ is from the vibrational motion of $\ce{H2}$ which will decrease for increase in adsorption strength.
As per the given statement, increase in enthalpy should decrease the entropy. However, Enthalpy-Entropy correlation states that increase in enthalpy leads to increase in entropy. How this is true?

Could you explain why you think that the enthalpy increases? Isn't the absorption exothermic as, for instance, stated here (--> decrease of enthalpy) ?

Why, recalling  the change of Free / Gibbs enthalpy expressed as $\Delta{}_R G = \Delta_RH - T\Delta_RS$, both changes of enthalpy and of  entropy (multiplied by  temperature) are taken into consideration?  Currently your question reads like one were enough, which I doubt.


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13C NMR of bromobenzene ipso carbon shielding
I've got a $\ce{^{13}C}$ NMR spectrum for bromobenzene. The ipso carbon is significantly more shielded (appearing upfield) than the two ortho carbons. Why is this? Wouldn't the electron-withdrawing effects of the halogen be enough to deshield the ipso carbon?
Does it have to do with the fact the $\ce{Br}$ is a deactivating group? I see that on an NMR database site, the ipso carbon for the $\ce{^{13}C}$ spectrum of fluorobenzene is deshielded. Does it have to do with the size difference between fluorine and bromine?

Hello and welcome to Chemistry.SE. It's a good idea to take the short [tour] to learn more about how this site works. You can also visit the [help] if you have any questions. Best of luck with your question!

The principle influence on chemical shift for 13C is the paramagnetic term, which essentially describes the electron density in the p orbitals, and this largely mirrors the influence of electronegativity that is seen in 1H chemical shifts.
However, there is something called the 'heavy atom effect' where the effect of substitution runs counter to electronegativity. This is the case for Bromine, Iodine, Tellurium etc. This comes from an increased level of diamagnetic shielding introduced by the electrons from the halogen atom.
To understand why this is the case, we can consider that the overall shielding on the ipso carbon relative to a methyl susbituent. Even though the bromine is more electronegative than the methyl group, it is so large that it still contributes significant electron density around the ipso carbon nucleus, and hence provides greater overall shielding than a methyl group.
While trends in electronegativity are useful tools to justify observations on chemical shift, it is always good to relate the effects of electronegativity, and other properties, on what is happening to the electron density around the nucleus on question. Ultimately, chemical shift is a measurement of electron shielding around the nucleus.


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Comparing acidity of anilinium ion and para-fluoro anilium ion

Which of these compounds - anilinium ion and para-fluoro anilium ion - is more acidic?

I thought the answer would the be first one since $\ce{-F}$ is an ortho-para activating group, so the electron density at the carbon at which $\ce{-NH3+}$ is attached would be more. I know that $\ce{-F}$ has a strong inductive effect, but since this is para with respect to $\ce{-NH3+}$, I thought I could say +M dominates. However the answer given was the second one. Why is this so and when can I neglect the inductive effect of halogen substituents in benzene to favour the mesomeric effect caused by them?

The best way to compare acidities of organic compounds is by drawing out the conjugate bases and figuring out which one is more stable. Molecules with more stable conjugate bases are more acidic, so doing the same here:

The conjugate bases are aniline and para-fluoro aniline respectively. The $\ce{-NH2}$ group shows +M (mesomeric effect), so on drawing out the resonance structures we find that the second resonance structure is more stable as the resulting negative charge is inductively withdrawn by fluorine group.
For halogens the inductive effect dominates over their positive mesomeric effect, especially in the case of fluorine (because it is small and the lone pairs experience a high effective nuclear charge,in case of large atoms like chlorine, bromine and iodine, the outer electrons are held weakly)

So the only place I can consider the +M effect is if I have electrophilic substitution happening?

The halogens are only weak activating groups,so +M effect can be considered in electrophilic substitutions. In general their  inductive nature is more dominant than their ability to show +M effect

But the inductive effect is virtually non-existent for the 3rd carbon atom and beyond, whereas the +M effect on o/p would still be present.

@xasthor if you draw the second resonance structure,the negative charge will appear directly on the carbon bonded to fluorine. Here inductive effect is dominant.

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Enzyme kinetics calculation
How to calculate or determine enzymatic activity?
The only information that I have are the substrates concentrations and the absorbance.

Welcome to Chemistry StackExchange. To get an answer you will need to add some more details. The only thing that is definite at present is that concentration is directly proportional to absorbance.

If it is a spectrophotometric assay, you will need to know the extinction coefficient of the product and/or substrate you are monitoring at the wavelength of the assay.  Almost certainly you will need to calculate the intial rate of reaction.     Principles of enzyme assay and kinetic studies


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How to state something as strong or weak oxidising agents?
It is said that conc. $\ce{H2SO4}$ is a strong oxidising agent. Why is it regarded as strong? In case of acids, we say something that fully ionizes in an aqueous solution is a strong acid. But what differentiates a strong oxidiser from a weak one?

If you want a quantitative measure of oxidizing ability, that's the redox potential. BTW, H2SO4 is a weak oxidant (though a strong acid).

Above which value of redox potential it is regarded as strong

I should have added "IMHO". There is no objective and universally accepted definition of what is weak and what is strong oxidant. As to me, HNO3 is strong.

@Ivan what about $\ce{O2F2}$ lol? Strong would be insufficient to describe its oxidizing power. :)

Stating that a chemical is a strong oxidizing agent by itself is quite vague, and is always with respect to some other chemical. You might hear that $\ce{KMnO4}$ is a strong oxidizing agent quite commonly in organic chemistry, although thats only true while you're comparing most of the organic compounds and its reagents, like Fehlings Reagent for example. However, if you compare $\ce{O2F2}$ and $\ce{KMnO4}$, then $\ce{KMnO4}$ would seem to be a weak oxidizing agent compared to the excessive oxidative ability of $\ce{O2F2}$. You can now clearly see how oxidative strength is a relative term.
Its possible to quantify the oxidative power of a substance. Its measured as the voltage developed between the substance and reduced products when performed in aqueous medium and is called Electrode Potential. You can learn more about it over here:
https://en.wikipedia.org/wiki/Electrode_potential


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Why is polymer B more dense than polymer A?
The official solution is that A has branched / side chains, weak Van der Waals forces between the polymer chains, less compact packing than B, and large inter-chain distances.
B has no branched chains, hydrogen bonds between the polymer chains, compact packing, and small inter-chain distances.

What does a branched chain mean? How does this influence the density? Doesn't B also have side chains?

What does 'compact packing' mean in this case? Does it mean that separate chains are packed closely together, or that in a single chain, the molecules are closer together?

First polymerisation proceeds through free radical mechanism, which may cause intermediate radicals to abstract H from other parts of growing chain, thus creating new branches in the process.
Refer Wikipedia for detailed mechanism.
On the other hand, second polymerisation proceeds through condensation reaction, where no such thing is possible (condensation can occur only at terminals).
Furthermore, the chains of B interact through H-bonding, which is much stronger than van der Waal interaction of polymer A. This causes chains to come closer and increases the overall density of polymer.

So branching means that if I have 'branched' polyethene, I'll have a Hydrogen replaced by a chain? But wouldn't that make it denser than before? If it were unbranched, there'd be large distances between two chains because of the weak Van der Waals forces, but now, since the two chains are actually linked, they will be closer, and will thus take lesser space?

PP is very rarely made via FRP (free radical polymerisation), the result is a some gunk that looks rubbery, is actually liquid however. iPP via Z-N and derived catalysts can be made basically defect-free. It still doesn't crystallise as perfectly as nylon, because it needs to make a helical structure.

“Branched chain” simply means that the main carbon backbone has other, smaller carbon backbones coming off of it. You’re correct, a hydrogen atom would have to be removed from a carbon atom in order to attach another polymer chain. Generally, branched chains decrease the density of a polymer because they fill space less efficiently than a perfectly straight chain. Imagine a pile of firewood - what would have more empty space? A pile of perfectly straight logs, or a pile of brush?
Polymer B does not have side chains. “Side chains” refers to units off the main chain with a carbon backbone, while the non-carbon atoms in polymer B are more accurately referred to as “functional groups”. Here, the N-C=O is known as a “ketone”.
Compact packing refers to the packing between chains, which is what determines the density of the polymer. Effects of interactions between chains are MUCH more significant than the small differences in interatomic bond length that can occur when you start adding functional groups.
As mentioned by jonsno, the hydrogen bonding of polymer B pulls the chains closer together than the VDW forces of polymer A.


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A numeric definition which determines volatile and non-volatile compounds?
Important conditions: stripping by air from water.
Hypothetical example: Let’s say if Henry’s constant is e.g. 0.005, everything below is non-volatile, everything above volatile.
The problem is I cannot find any study which determines which any constant that can tell me this compound is considered volatile, or not.
I am trying to determine which compounds during aeration of industrial wastewater will strip fast/as first.
I have tables of Henry's constant but I cannot determine which is the border of "normally volatile compounds".

https://en.wikipedia.org/wiki/Volatile_organic_compound While this is only discussing organics, it does show what constitutes "volatile" depends on what regulatory body you are following.


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Are esters and amides soluble in cold acids/bases?
The main problem I have is that we say amines are soluble in acids because they are bases, and they will thus react with acids. But if A reacts with B, how does that imply A is soluble in B? Yes, esters and amides do not react with cold acids/bases, but why does that imply that they are insoluble? Ethanol does not react with water, but it's soluble in it.

This could be six to twelve separate questions, combinatorially mixing (esters/amides/alcohols) and (cold/hot) (acids/bases).

Edited. Please see the new question.


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Does electroplating fall off if I disconnect electrolytic cell?
So I understand the basic mechanism of electroplating. I want to ask is there any chance that the electroplating may fall off if I disconnect the electricity supply? As a consequence, should I recommend the plated object to be taken out of the electrolytic solution before disconnecting the electricity supply?
Specifically I want to know if disconnecting electrolytic cell can be the sole reason why the plating on an electrode may fall off.

If the plating falls off when the power is cut in solution, then it would fall off after being taken out of solution as well.

@JonCuster Does it actually ever fall off?That's my question...

Electroplating can - bad surface prep, too high of a stress built up, etc. Just like any other coating.

@JonCuster Assuming I have done everything properly... can the electroplating fall off only due to disconnection of electrolytic cell?

No.$\mathstrut$


text:
Why are a calico strainer and chloroform used in a British Pharmacopoeia assay of milk of magnesia?
The following is an assay for milk of magnesia in the British Pharmacopoeia (emphasis added):

Dissolve the sodium hydroxide in $\pu{150 cm^3}$ of purified water, add the light magnesium oxide, mix to form a smooth cream and then add sufficient purified water to produce $\pu{2500 cm^3}$. Pour this suspension in a thin stream into a solution of the magnesium sulfate in $\pu{2500 cm^3}$ of purified water, stirring continuously during the mixing. Allow the precipitate to subside, remove the clear liquid, transfer the residue to a calico strainer, allow to drain and wash the precipitate with purified water until the washings give only a slight reaction for sulfate. Mix the washed precipitate with purified water, dissolve the chloroform in the mixture.

What is a calico strainer, and why is it used here? Is it better than using, say, Büchner filtration?
As well, why is chloroform used (see emphasized text, above)?

Are you certain that chloroform is used in the assay, in my BP , milk of magnesia is assayed with hydrochloric acid, and a complometrix titration with sodiun edetate

Pour this suspension in a thin stream into a solution of the magnesium sulfate in 2500 cm3 of purified water, stirring continuously during the mixing. Allow the precipitate to subside, remove the clear liquid, transfer the residue to a calico strainer, allow to drain and wash the precipitate with purified water until the washings give only a slight reaction for sulfate. Mix the washed precipitate with purified water, dissolve the chloroform in the mixture <-- this last line.

Ok I understand where you got chloroform from.
This is not the assay of milk of magnesia, but the extemporaneous preparation.
In extemporaneous preparations chloroform is used as:

a general anaesthetic
a carminative
flavouraing agent
preservative

So in the case of mik of magnesia it has two main roles that is a carminative and a preservative.
Hope this helps

Thank you so much! But, why is the calico strainer used? In the experiment I had done, we had to use the buchner funnel instead even though the monograph had called for the calico.. would this affect the results in any way?

They essentially perform more or less similar functions, though a bunchner funnel is more efficient as it additionally dries up quickly the precipitate (if vacuum setup is used)

Chloroform is used as an emulsion stabilizer.
The calico strainer is used as it is less prone to clogging than a buchner funnel.
The mix will clog frits and normal filter paper.

Milk of magnesia has a pH ~10.5, perhaps not high enough to react with chloroform - or would it? Or what other process results in dispersion stabilization?


text:
What is the molecular shape of Sulfur Pentafluoride Anion?
$\ce{SF6}$ is an octahedral shape which makes perfect sense. $\ce{SF5^+}$ arranges 5 pairs of electrons in a trigonal bipyramidal structure. But what if you add a lone pair to this? Will the electron pairs arrange into an octahedral shape, but due to extra repulsion of lone pair, or will it have some distortion?
Will it be similar to $\ce{BrF5}$ structure?

Yes, $\ce{SF5^-}$ does infact have an octahedral shape. It's structure would be

Yes, it is similar to the structure of $\ce{BrF5}$

That's not octahedral, it's derived from octahedral but the geometry is square pyramidal. I realise the question asked about the disposition of electron pairs, but it's not good to reinforce this slightly sloppy phrasing.

More imformation fore above question
It is not Octahedral because octahedral formed in 0 lone elctrones and 6 bond electron pairs
It should be squre pyramidal according to this reason,
Valence electrones by sulfur atom-6
Electrons by 5 florin atoms-5 (1*5)
(- )charge electron-1
Therefore total electrons-6+5+1=12
Therefore repultion items-12÷2=6
Bond pairs between 5 Florine atoms and sulfur atom-5 (because the bond is SF5-)
Therefore lone electron pairs-6-5=1
So molecular shape: squre pyramidal(5,1)
therefore it is same to the molecular shape of BrF5.

I have deleted the other answer since they were duplicates by different user accounts. You may want to contact the team to merge your accounts.


text:
How to rationalise the major product formed in a nucleophile promoted epoxide cleavage reaction?
The Problem is to find major product of reaction as given below (the star is a labelled carbon):

The two possible reactions that come to my mind are:

Nucleophile promoted epoxide cleavage reaction, ① in image below
SN2 substitution of chloride as a strong nucleophile $\ce{C2H5O^-}$ is being used, ② in image below

I thought that both these products should be formed, but to my surprise the answer is given as product ③ (in image), in which it appears that after ring opening of epoxide to ① they they have substituted the chlorine by using $\ce{OH-}$.
So my questions are:

Why is the product ③ major product, why not ①?
Is the product ② also formed and if no, then why not? If yes, then on what basis is product ③ the major product? Shouldn't product ② be also formed proportionately as in the final structure both product ② and product ③ appear very similar to me.
Can anyone please give me hints on how to solve such types of question when to form epoxides as in product ③ (even though epoxide is a ③ membered ring with ring strain also if $\ce{C2H5O^-}$ is left in solution it may again try to open the epoxide ring) and when not to form epoxide as in product ① which doesn't have any ring strain.
It would be great if anyone could also suggest the approximate percentage of products ①, ②, ③. However it is not necessary a simple explanation is sufficient.

Considering question (B) it seems as none of the products should be stable enough. How about the product which has OEt groups at the terminal C's and an OH in the middle? That seems to be not vulnerable to any more substitutions.

With respect to my earlier comment, I can't say whether this product will be obtained by an epoxide ring opening from pdt. 3 or pdt. 2.

Let's do this step by step and figure your questions out.
First, what a nucleophile needs is positive terminal in the compound.

The other carbon attached to oxygen is not as positive as the one where the attack has been done due to inductive effect of $\ce{-CH2}$ adjacent to it. The product (1) is not formed as there is no source providing $\ce{H+}$ to form alcohol. If possible a very small amount  may be formed by $\ce{H+}$ extraction from $\ce{EtO-}$. Next the intramolecular nucleophilic substitution takes place.

I do agree that product (2) is formed, but as a minor product. Here what we observe is that there is neighboring group participation in the mechanism. Generally these neighboring groups increase the rate of the reaction by switching on abilities like dipole in the bond. Whenever we observe compounds with such epoxides or sometimes even alcohols, we better think of such consequences which can help us predict the product.

But the epoxide formed should be further substituted by an OEt- to give a product with OEt groups at the terminal c's and an OH group in the middle.

That's true but there is no specific source that provides H+. A minor product tends to form indeed.


text:
Why doesn't Calcium Carbonate Burn?
I was doing my Science Fair Project and I tried to burn Calcium Carbonate. But the match went out as soon as it touched the Carbonate. I am wondering why the Carbonate isn't burning. I later tried to use a torch and a lighter but nothing worked. My science fair project is due tomorrow and I've looked everywhere on the internet but nothing really explained why the calcium carbonate didn't ignite.
Do I need HEAT or something like that to make it burn?
(I'm a 5th grader so please take it easy on me)

What reaction? Why would it burn?

The only reaction you will get by heating calcium carbonate is decomposition to calcium oxide and carbon dioxide.

HEET? You mean the anti-freeze/water-removal thingy?

@paracetamol nahh, its most probably HEAT. (probably misspelt in the hurry to submit question)

@Pritt The OP does mention previous attempts to set the thing on fire (such as, use of a torch, lighter,etc). Ultimately ending the question by asking if "HEAT" is necessary to do the trick seems... highly unusual.

@paracetamol Adding anti-freeze to help combustion? Quite unlikely.

@Pritt There're quite a few anti-freeze agents out there that are highly flammable ._.

@paracetamol like ethylene glycol right? I think that one is quite flammable.

Do you mean combustion of calcium carbonate?

Calcium Carbonate cannot burn. Its already in its highest oxidation state (+2 for Calcium and +4 for Carbon), and cannot be oxidized (combusted) to a higher oxidation state. However, calcium carbonate does decompose on heating to quicklime and carbon dioxide.
$$\ce{CaCO3 -> CaO + CO2}$$
This doesnt give out any flames in the first place.
Now let's talk about how to salvage your project. Why don't you try to heat the calcium carbonate, and show that carbon dioxide is evolved by using a burning match stick, or by using lime water? $\ce{CO2}$ extinguishes the burning flames and forms a milky precipitate with $\ce{Ca(OH)2}$ (limewater). This can be used to show that perhaps egg-shells and marble have carbonate content in them.

Why don't you try to heat the calcium carbonate... The kid's a fifth-grader, not sure if he can get his hands on the 800+ degrees Celsius crucible for his project ;)

@paracetamol Since the kid doesn't have access to the "800°C crucible", another alternative could be to boil at lower temperatures with strong acids. $$\ce{CaCO3 + CHA -> CaA2 + H2CO3}$$ and the carbonic acid that forms will anyway equilibrise to carbon dioxide and water. $$\ce{H2CO3 <=> CO2 + H2O}$$.


text:
What is melting? Which bonds do we break to melt something?
To melt diamond, we have to break the covalent bonds, which we can consider 'intermolecular' because it is one giant molecule.
To melt Methane, we have to break the van der Waals (intermolecular) forces.
For $\ce{NaCl}$, ionic bonds, which are intermolecular as well in a sense.
But I've read that melting graphite also requires breaking the covalent $\ce{C-C}$ bonds rather than just breaking the van der Waals forces between the layers.
And for cross-linked polymers, I've read that melting involves breaking only the bonds linking the two polymer chains.
So I guess what I want is a precise definition of melting. Does it involve breaking the weakest bonds? The intermolecular forces? The intramolecular bonds?
When we melt a molecule of, say, $\ce{H2O}$, we imagine molecules of $\ce{H2O}$ roaming around with little intermolecular forces, as most hydrogen bonds have been broken.
What do we 'imagine' when we melt diamond, or a cross-linked polymer, or graphite, or ionic compounds like $\ce{NaCl}$?

Have a look first for Lindemann theory (or Lindemann Criterion) for melting. To find out if diamond /graphite melts before decomposing you will have to find a phase diagram. Some types of polymers melt e.g. polythene, many types do not, e.g. bakelite It all depends on their molecular structure.

I'm only a highschool student. I looked at it, but all I can make out is that melting occurs when the average thermal vibrations of atoms is greater than the inter-atomic distances. I cannot relate this to the question.

Ignoring your (not necessarily helpful) distinction between bond types, the main point is that a liquid is a distinct thermodynamic phase. Above the melt temperature, it is preferred, having a lower Gibbs free energy. There is still 'bonding', with inter-atomic forces holding the liquid together.

So melting involves breaking the intermolecular forces? What if we have a polymer with partial cross-linking?

Liquids and solids both have intermolecular forces, a liquid must as it has a meniscus, i.e it is not a gas. The  difference liquid to solid is one of the amount of energy available. Liquids have enough energy that molecules can become mobile and rapidly move past one another whereas in a solid this does not normally happen. Chemical bonds do not need to break to cause melting or to vaporise a substance.

But then what would be happening in a sample of melted diamond if the covalent bonds are unchanged?

This is a good question. Unfortunately there are several criteria by which chemists indentify whether a process is melting or not. One of them is called the Lindemann Criteria which says:
"Crystals are considered to melt, when the vibrational amplitude becomes half of the interatomic spacing in the crystal lattice."
What does it mean? Normally, at temperatures greater than 0K, atoms possess kinetic energy. Infact, temperature itself is a measure of the kinetic energy of the constituent atoms. The kinetic energy of an atom is related to the temperature as:
$$\text{Kinetic Energy}=\frac{3}{2}k\text{T}$$
where $k$ is a constant called Boltzmann Constant, with a value of $1.38×10^{-23}\text{J/mol}$. Atoms in a solid are characterized by their fixed mean positions. They only vibrate within a certain boundary around them. Since they have kinetic energy (and velocity as well), they tend to leave their current location, but repulsive forces from the other atoms push it back to its original position. In this way, you could consider atomic bonds like "little springs".

The extent to which the atom displaces itself from the mean location is called it's vibrational amplitude. As we increase temperature, the atoms will have more velocity, and can consequently displace itself further from the mean position. Lindemann defines the melting temperature as the temperature at which the amplitude becomes half of the space between two adjacent atoms of the crystal.
Yet another criteria for defining the melting point is the Born criteria that says:
"Crystals are considered to melt, when the shear modulus approaches zero"
You might wonder, what is this shear modulus? It's infact, a measure of how much tangential stress a solid object can handle, and the tangential strain caused by it. Tangential forces on an object are forces that act parallel to the surface of the object, hence the name. See the picture below that shows a tangential force applied on an originally cuboidal object:

Tangential (or shearing) stress is defined as:
$$\sigma _{shear}=\frac{F_{shear}}{A}$$
where $F_{shear}$ is the tangential (shearing) force, and $A$ denotes area of surface in question.
Considering the picture above, shearing strain would be defined as:
$$\epsilon _{shear}=\frac{\Delta x}{h}$$
There exists a unique proportionality between the stress applied and strain produced, and is given by:
$$\frac{\sigma _{shear}}{\epsilon _{shear}}=S$$
where $S$ is called the shear modulus. Its a constant for a given material and a given temperature.
In a physical sense, liquids are considered as substances that cannot withstand a tangential stress. When a tangential stress is applied, liquids simply keep on increasing the strain, even under small stress. This itself is a useful way to consider when a solid melts, when its shear modulus becomes zero!
Now, let's consider how we consider melting for the different examples that you have asked for:
Diamond: Diamond has a structure of covalent bonds arranged tetrahedrally to each carbon. Each bond has it's characteristic bond length. Applying Lindemann condition over here would be considered advisable , so that melting point is where vibrations of carbon atoms is half of the $\ce{C-C}$ bond distance. Ultimately some of the $\ce{C-C}$ would break and the molten mixture would largely consist of radicals of varying sizes, that permit free movement between themselves.
Graphite: While each layer of graphite is only weakly bonded to the other layer by van der Waals' forces, each layer (called graphene) itself is a quite a large molecule. You can consider a liquid as a large number of small molecules, so clearly graphene doesn't look like this. To melt it, you would have to break some of the $\ce{C-C}$ bonds so that we can produce small molecules that can move freely. We apply Lindemann condition in this case. The molten mixture would be similar to that of diamond.
Branched polymers: Remember how I told to consider liquids as small molecules that can move freely sliding among themselves? We consider the same thing over here. The branch bonds are comparatively weaker than the rest of the bonds in the polymer, and these are the ones that will break when heated. As far as bond-breakings are concerned, we use the Lindemann condition. While the unbranched polymers are big molecules, they are still small enough to exhibit liquid character at the high temperature required to break the branched bonds. The molten mixture would consist of the straight-chain polymers in radical forms.

Your answer tells me what melting is from a very broad perspective, but I can't apply it to actually find out what I want to know.

Oh, that's a rather unique way to give an answer. I shall be waiting!

Thank you for the answer! The biggest follow-up question I have is what does it mean for a molecule to be 'small'? We say graphene isn't small, but polymer chains are. How do we draw the distinction? If we have a sample of a branched polymer consisting of 100 chains and every 2 chains are joined to each other (so there are 50 pieces), why do we need to break the branch bonds rather than the Van Der Waals forces between the pieces for melting? That would allow the pieces to move freely. Lastly, aren't Van Der Waals forces weaker than the branch bonds? If so, why aren't these broken first?

@Saad there is a lot that goes on. Youre right, my explanation using 'big' is quite vague, while what you must consider are the van der Waals' attractions. If the pieces are large enough for the attractions to make it solid, it will be solid. This is not for network solids. In case of graphite, van der Waals' bonds are broken first, but the graphene sheets still remain in solid state. You must break a few $\ce{C-C}$ bonds to get it to liquid state.

For network solids, we have small molecules bonded to each other, so breaking the Van Der Waals forces would cause no difference as the molecules would stay the same distance apart, and these forces would soon reform. Is this correct? Also, could you answer my 'pieces' question?

No, network solids have only one giant molecule. Molecular solids are the ones consisting of the small molecules bonded by van der Waals' attractions.

I cannot understand your "pieces" question? Can you explain where your doubts are?

If we have a sample which consists of 100 polymer chains, and every two chains are cross-linked to each other, for example if we have chains A,B,C,D, then we have that chains A and B are linked to each other, and chains C and D are linked to each other. Now, if we heat up this sample, what will happen? If we treat this as just samples of a molecule which consists of two polymer chains linked to to each other, then only the Van Der Waals forces between two 'pieces' ( 2 chains + branch ) will break for melting. But if this is the case, then the melting point should remain unchanged by crosslinks

@Saad Thats a good point. However its not really possible to polymerise in such a way. It's always all linear, or all branched.

My exam board had a question in which it asked 'Why does increasing the cross-linking increase the melting point? '. Isn't that the same thing?

@Saad Crosslinking links all the chains together. Definitely it increases the melting point. You can't just connect only two strands. If you want to branch, you connect all, or none at all.

But then why does it use the term 'increasing' the cross-linking? Cross-linking would suffice if it can only be fully connected.

Let's be precises, branching is not the same as crosslinking. Related, but not the same.

It is possible to only partially crosslink, the crosslinking process is a chemical reaction and thus it can be manipulated by changing temperature and reactant concentrations to reach different endpoints. If you have a 5000 US gallon reactor of polymer and you don't add enough crosslinking agent, you should expect only partial crosslinking of the bulk polymer. As you increase the number of chains that are crosslinked, you create more bonds that must be broken to form a new phase (in this case a liquid phase) and thus the melting temperature goes up.

But why do we have to break more covalent bonds if we want to form a liquid? Could you explain your answer in light of my 'pieces' question which refers to a sample of a partially cross-linked polymer?

@J.Ari I'm not sure if you're correct or Pritt, but if partial cross-linking is possible according to you, could you explain my 'pieces' question? ( 10 comments up )

@Saad - Why do you think melting point is unaffected by van der Waals forces?
\@ Pritt - http://pubs.acs.org/doi/abs/10.1021/ma048335eAs (an example piece of literature on partially linked polymers) I am also quoting my experience in industrial scale polymer manufacturing when my operators fail to follow the recipe and the quality chemists quarantine a batch for "low degree of cross linking", i.e. partial cross linking.

@J.Ari I'm assuming the molecule used in branching is small, so the Van Der Waals forces aren't really changed. Would the melting point change regardless? Why?

If someone could explain this to me, I would be extremely grateful.

Polymers tend to not conform to the small molecule approximation. Still, two branched polymers, with their increased surface area and definite structure, can form stronger interactions with each other. This is because the atoms are in a defined order and when the van der Waals forces act on them, the effect is cumulative.


text:
To add acid or base to water
There is a rule stating that we shall add a strong acid to water, and not the other way because of safety; if we would add water to the acid, the reaction could be dangerous (boil).
What about adding strong bases and water? Which is the safer order to mix them?
I would like to know the safer addition orders, and the justification, for water+acid or water+base(?)/base+water(?). Thank you!

Same guidelines. Another big issue is that if something splashes, it'll be whatever you're adding into, so much safer to add to water.

The reason for adding strong acids or bases to water rather than the other way around is that the dissolution/reaction of these compounds with water tends to be very exothermic and can result in splattering or even boiling of the strong acid or base, particularly if water is added to the concentrated acid or base.
On the other hand, if you slowly add the acid or base to water, you will never have a concentrated acid or base present to splatter as it will be rapidly diluted by the much larger amount of water. Even if you accidentally add the acid or base too quickly, at least the splattering will primarily be of water or dilute acid/base rather than splattering concentrated strong acid or base. It is still important to do the addition slowly so that you only slowly increase the acidity/basicity of the solution. This way, the rate of reaction (and thus also the heat evolution) will happen at a safe level and greatly reduce the risk of splattering.
As a side note, this concept does not only apply to the addition of strong acids and bases to water. While mixing any chemical reagents (unless otherwise specified of course), you should add the more reactive or concentrated solution to the less reactive or more dilute solution, for the same reasons as described for the addition of strong acids and bases to water.

I was thinking about that... and I predict that there are some exceptions... so you should rewrite your last paragraph, to advert that there are exceptions! For instance, image chain reactions between a common and a "more dangerous" material...

@sissi_luaty , I'm sorry but I don't understand what you are asking of me. Could you give an example? Should I take the last paragraph out so that the answer sticks to your question of just acids and bases. I'm glad to help, I just need some clarification.

The fact is that the words "in general" can be tricky in this context... it's not clear from your text if the last rule is for all the cases or for most of the cases. - after pondering about that I suppose that there are some exeptions...

Do not remove the extra part, but please clarify that (because one exception could mean a bang!) with a minor edit... thank you!

Yea, I've gotten into the habit of (over) using terms like "in general" because there are so few absolutes in chemistry! I can definitely see how that backfires here and makes people wonder what the exceptions are. But I honestly can't imagine a case in which you would do things the other way around, that is adding a less concentrated reagent to a more concentrated one. That's the kind of thing that if there is an exception out there, it would need to be made perfectly clear on the bottle, in training, the SOP, etc. I think the thing for me to do here is word it in more definitive terms.

@sissi_luaty , I edited the last sentence per my discussion in the previous comment. Let me know if you have any other concerns...


text:
Can a base like OH- attack aldehyde and ketone to create hydrate instead of enol?
I don't know if this question is kind of vague. But i was wondering like in the reaction of saponification of ester, where you have a base like OH- directly attacking the carbonyl carbon, can you also have a base like OH- attack the carbonyl carbon atom and then protonate the O- atom to create like a hydrate? Or does the base prefer attack hydrogen of alpha carbon of ketone and aldehyde? I am kind of confused about when a enol is create and when a hydrate is created since both of them can be formed by a base. So, hence I ask this question.

Ester hydrolysis and the soap-making process
In aqueous solution, esters are subject to hydrolysis to the corresponding carboxylic acid and alcohol components:

Hydrolysis is,  the reverse of the esterification reaction. The addition of
base accelerates ester hydrolysis because the nucleophile is a hydroxide ion, which is much more reactive than a water molecule.

Enol formation
The a-protons of carbonyls, are somewhat acidic, because the negative
charge on the conjugate base is stabilized by resonance with the carbonyl oxygen. In the formation of an enol, a base abstracts an a-proton from a carbonyl compound, and that same proton (or a proton on a nearby acid group) is
transferred to the carbonyl oxygen. The deprotonated form of an enol is an enolate.

However in this scenario  I don't think an enolate will form in this case.

enol formation is favored in acidic environments rather than basic

The preffered route would be the first one above.
Edit:
To avoid a bit confusion:

To avoid generalisation, the R1 group given on the ester can be anything from -C(CH2)3,  in this case 2 alpha carbons and no alph hydrogen. (No enol possible)
an alpha hydrogen may be there but reaction conditions won't favour enol formation.

Hope this clears doubts

Thanks for the response. But which compound do you mean by "no alpha proton on the carbon adjacent to the carbonyl group" Ester, aldehyde and ketone all have alpha carbon.

@TLo, esters do form enols but, this hydrolysis reaction there wouldn't be an enol intermediate, due to reasons mentioned. These are typically seen in aldol addition reaction

Why enolate formation requires acidic medium although the base is attacking the hydrogen?

@SurajS A base is a good nucleophile and tends to attact the carbonyl carbon  directly

"Please not that I am making reference to the above reaction only (hydrolysis of esters). Some compounds can form enols in both mediums however, I am trying not to overgeneralise*


text:
How come electrons in the d-subshell are considered valence electrons in transition metals?
Similarly, why are f-subshell electrons considered valence electrons in the lanthanides and actinides?
I was under the impression that the outermost electrons, and therefore the electrons with the highest principal quantum number, made up the valence electrons.
Also, if the d-subshell electrons are considered valence electrons in transition metals, how come they're not considered valence electrons for elements in Groups 13-18?
I understand that the highest d-subshell is of higher energy than the highest s-subshell but I fail to see the logic in how it would be considered a valence electron.

The valence electrons are the ones responsible for engaging in chemistry and corresponds to the orbitals which are highest in energy. These electrons are the most easily accessed because they are not buried in the energy well of an atom. For main group elements, the highest principal quantum number p and s orbitals are the highest in energy, as you said, but for the transition metals, the partially filled d orbitals are the highest in energy. Generally, whichever orbitals are being filled last as you go across the periodic table are the ones that should be considered the valence electrons.
Another thing to note, as elements are ionized, the orbitals with higher orbital angular momentum drop in energy faster than orbitals with lower angular momentum. So, in the transition metals, the n d orbitals drop much more than the (n+1)s and p orbitals. Since, transition metals have <10 electrons (the number it would take to fill up all the d orbitals), all of them reside in d orbitals (and none in the s and p) and so these are the highest occupied orbitals and therefore the ones that engage in chemistry.

When you say valence shell, you have to pay careful attention to the word valence.
What does valence mean? Wikipedia says a bit about it's etymology:
"The etymology of the word "valence" traces back to 1425, meaning "extract, preparation", from Latin valentia "strength, capacity", from the earlier valor "worth, value", and the chemical meaning referring to the "combining power of an element" is recorded from 1884, from German "Valenz".
Clearly, valence refers to the combining power of the element.
Which shell decides the combining power of an element? The $(n+1)s$ or $nd$ subshell? To further prove my point, consider these examples:

Iron has its most prominent oxidation states (valencies) are +2 and +3. The respective electronic configurations are $\ce{[Ar]}3d^6 4s^0$ and $\ce{[Ar]}3d^5 4s^0$. The $\ce{Fe^{3+}}$ is extra stable since it has a half filled $d$ subshell. Does the $s$ orbital play any role here? Absolutely not.
In acidic medium, the permanganate ion ($\ce{MnO4^-}$) is a powerful oxidizing agent, reducing itself to $\ce{Mn^{2+}}$. Why $\ce{Mn^{2+}}$? Recall that it's electronic configuration is also $\ce{[Ar]}3d^5 4s^0$. Is it because of the $s$ shell? Think carefully.

While you may have heard that valence shell means the outermost shell, it's not always the case, and never the case in transition and inner transition metals.


text:
Can aldol reaction occur during α-halogenation of ketone?
I know both aldol and $\alpha$-halogenation reaction well.
I know that in and aldol reaction, the carbonyl containing molecules like ketones and aldehydes attacks themselves when base like $\ce{OH-}$ attacks $\alpha$-proton.
The same thing also happens in $\alpha$-halogenation of these carbonyl containing molecules.
So my question is that in presence of halogen, before the double bond of the enol can attack halogen atom of $\ce{Br2}$, $\ce{Cl2}$, $\ce{I2}$ in $\alpha$-halogenation, can aldol reaction reaction occur by double bond of the enol attacking another carbonyl containing molecules?

Hello, I have edited your question using MathJax. Learn more on how to use it here: https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here

@PrittBalagopal You can (and should) also edit out unnecessary "statements of weakness", like thanks and similar salutations.

Carbonyl compounds, in general are quite bulky and poorly electrophilic They only react when no other better electrophile is around. The major products would definitely be the $\alpha$-halogenated ketone, while some minor amounts of aldol may form.


text:
1H NMR of p-methoxy phenol, 32 possible lines?
This is a follow-up question, emerging from answering this one.
The Silverstein book 1 aims to rise awareness that not every 1H NMR signal that on first inspection may look like a doublet actually is a doublet.  Triggered by counting possible number of lines of signals for p-methoxy phenol

-- which would yield just 16 (two ddd) I came across this part of the book about p-chloro nitrobenzene:

(loc. cit., p. 175)
Question: For the sake of understanding this part I leave out a potential simplification of the 1H NMR spectrum if some of the centres of the signals of the aryl H's accidentally may coincide, or some parts of the multiplets overlap.  For a moment, I assume each aryl H's multiplet may be recorded without disturbance of the signals by any other aryl H.
Is my understanding of this section correct, in theory there were up to 32 (thirty-two) lines observable, as each of the aryl H's couples with the other three H's relatively in ortho, meta, para position, hence four ddd?
On the next page, the book actually displays an experimental 1H NMR spectrum of p-chloro nitrobenzene

and assuming the more intense peaks could be a partial overlap of two lines (not all of them are smooth), I "count" only 16 lines. As if -- a speculation follows -- obviously the p-coupling, as the one with the lowest coupling constant, was not-at-all resolved in this acquisition.
1 Spectroscopic Identification of Organic Compounds, Wiley, 6th edition, 1998, ISBN 0-471-13457-0

I think your theory is correct, but my tests were inconclusive. You can try your own hand at simulating the spectrum, for instance with Hans Reich's program DNMR. It resulted in the following:

This is the downfield half of the spectrum. I did not attempt to replicate the experimental spectrum exactly, it's only to illustrate. I used the following parameters:

$\sigma_A = \pu{7.5 ppm}$
$\sigma_X = \pu{8.19 ppm}$
$J_{ortho} = \pu{8 Hz}$
$J_{meta} = \pu{3 Hz}$
$J_{para} = \pu{0.5 Hz}$

On to your question. If you increase the J sufficiently, while retaining symmetry among the J, you get...

... 20 peaks! (only one set of multiplets is shown, so the total number of peajs is twice the number visible above) . I continued tinkering with the values, and breaking the symmetry among the couplings led to ...

... 26 peaks!
Presumably breaking the symmetry in the shifts will increase the number of peaks further, but surprisingly this had just the opposite effect.
But this is not the most illuminating approach. Better would be better to dig into the QM.


text:
Why is my Cu2O Cuprous Oxide Not Red
A thin film sample of polycrystalline Copper(I) Oxide (Cuprous Oxide, $\ce{Cu2O}$) on quartz looks reddish in the center but bluish/grayish at the sides where the film is thicker.
The same thin film on a Silicon Oxide wafer looks greyish, with a slight tint of blue.
XPS analysis shows that the bulk of the film contains no traces of Copper(II) that can be detected within the instrument's detection limit ($\pu{0.1-0.5 at.\%}$?).
The $\ce{Cu2O}$ film is grown by CVD using a metal-organic precursor and water.
What can be causing the gray color of the $\ce{Cu2O}$? Can it be the morphology/porousity of the $\ce{Cu2O}$? If it is trace amounts of contaminants, what are some good guesses?

The Merck Index says of Cu2O: Color may be yellow, red or brown depending on the method of preparation and particle size. I don't know if your greyish color on the silicon oxide wafer could really be a pale yellow; it seems like a long-shot from your description. The slight tint of blue seems like it must be from oxidation to some form of Cu(II). Some Cu(II) complexes can be pretty intensely blue (although more commonly are green), so that you might see it at levels below the XPS DL. Anyway, there's a couple long-shot ideas on your interesting/difficult problem. And welcome to Chemistry.SE!

Maybe you are not seeing the bulk color of the oxide but light interference effects.  See https://en.wikipedia.org/wiki/Thin-film_interference.

@OscarLanzi The film is about 500-1000 nm thick, probably not light interference effects?

Not sure if this is correct, but while the $\ce{Cu2O}$ is being made, a certain amount of black-colored $\ce{CuO}$ may also have formed.

@Athena of the 500-1000nm thickness is true, yes that's a little thick for interference.  Was an idea.


text:
Are stronger bonds always shorter?
This question relates to a few earlier questions I have seen on the site (particularly this one about the bond strength of heavier isotopes). Does a bond necessarily have to be shorter to be stronger?
To my understanding, length and strength are definitely correlated, but I don't think it is necessarily a one to one correspondence. The examples I brought up in the comments of ron's answer in the linked question related to data tables that seemed to show no difference in bond length for deuterium substituted bonds despite significant changes in energy. (HCl NIST, DCl, data table)
To expand on my question, if length and strength are always directly related:

How do we explain experimental and theoretical cases where the length doesn't differ with bond energy?
If there is some direct relationship between length and strength, what is that relationship?

For the sake of not being too general, I'm looking for an answer focused on comparing bonds of the form $\ce{A-B}$ and $\ce{A-C}$ (similar to the examples I asked about in the linked question). So for a more specific case related to the linked question, does substituting a protium for a deuterium always make the bond shorter in addition to being stronger?

Most of the time, shorter bonds mean more effective overlapping between the orbitals, which makes stronger bonds, although alot depends on the element and size of orbitals.

Short answer is a definite no. I would have to dig around a bit to find where I read that, though...

Is your question only meant to be for a given pair of elements? i.e. do you mean, "if the A-B bond in compound 1 is shorter than the A-B bond in compound 2, does that mean the first bond is stronger"? Or do you mean to include comparisons between different bonds A-B and C-D?

@orthocresol I would say my question mainly relates to comparing A-B to A-C, where C could be the same as B or could be different. Comparing entirely different bonds (A-B to C-D) presumably adds in other factors that affect the overall length.

Anytime I see "always" in a question about chemistry I get skeptical. It seems such propositions always lead to exceptions and/or "clarifications" of the question.

@MaxW that was the reason I asked the question. I was confused in the case of HCl vs DCl (and really deuterium substituted bonds in general) because all the evidence I could find seemed to suggest that while DCl had a stronger bond than HCl(greater bond energy), there was no evidence from calculated or experimental values that it was any shorter. But ron's answer to a previous question and a lab report I had to grade seemed to suggest that a stronger bond should always be shorter.

This is an interesting question and depending on how you define bond strength the answer is different. Let us for simplicity consider only diatomic molecules and let us assume that the electronic potential between the two atoms is well described by a Morse potential
\begin{align}
V(r) &=D_e \left( 1 - \mathrm{e}^{-a(r-r_e)} \right)^2,& \text{with } a &=\sqrt{\frac{k_e}{2D_e}}.
\end{align}
Here $D_e$ is the depth of the potential (at the minimum position $r=r_e$) and $k_e$ is the (harmonic) force constant. The potential depth is related to the dissociation energy $D_0$ by
\begin{align}
D_0 &= D_e - \frac{1}{2}\omega_ehc,& \text{with } \omega_e &=\sqrt{\frac{k_e}{\mu}},
\end{align}
where $\omega_e$ is the harmonic wavenumber, $\mu$ is the reduced mass and $h$ and $c$ have their usual meaning.
We can define the strength of the bond by the magnitude of the dissociation energy $D_0$ or by the "spring constant" $k_e$ of the bond. In your example with deuterium you implicitly assumed the former definition. In the Born-Oppenheimer Approximation the potential (i.e. the Morse potential) does not depend on the mass of the atoms (they are assumed to be infinitely heavy) and the potential is the same for $\ce{H2}$ and $\ce{D2}$. However, because deuterium is heavier than hydrogen, the harmonic frequency is lower and therefore the dissociation energy is larger ($\ce{D2}$ has a smaller zero point energy). Using this definition, the bond length is not easily related to the bond strength (as the reduced mass of the system plays a role as well).
As stated earlier, we can also use $k_e$ as a measure for the bond strength. The larger the value of $k_e$, the steeper the harmonic part of the potential well and the more localized the nuclear wavefunctions will be. In other words, if we take $k_e$ as a measure of bond strength, then stronger bonds are indeed shorter.
As stated by @porphyrin, the separation between $k_e$ and $D_e$ is not very strict (see also the formula's above) and the explanation above implicitly assumed a constant $D_e$, just as we assumed constant $k_e$ for the different isotopologues.

It is appealing to try to separate force constant and dissociation energy but in the Morse potential they are linked as $\alpha = \omega_e\sqrt{2\pi^2c\mu/(Dh)}$

You are right of course. I've made an edit to make the assumption of a constant $D_e$ more explicit.

@Paul I like this answer. The one thing I'm a little confused on is the bond length's relation with $k_e$. While the well could be more localized, if it is centered around the same point (i.e. reaches a minimum at the same $r_e$) wouldn't the bond length be the same?

@Tyberius The minimum of the potential does not necessarily coincides with the expectation value for $r$ of the different vibrational states. In general the wavefunctions have more intensity toward the shallower part of the potential at larger values of $r$. The shift from $r_e$ for each vibrational state $|v\rangle$ is $\langle v|r-r_e|v\rangle$ and depends on $k_e$. However, this is not an easy expression. See this paper: http://inaesp.org/PublicJG/gallas_morse1980.pdf

I would like to add a few cases featuring A-A type bonds, where bond strength and bond length orders follow quite an anomaly.

$\ce{N-N}$ and $\ce{P-P}$ bonds. The lone pairs present on nitrogen atoms repel each other and the large size of phosphorus atoms facilitates for charge distribution. This makes the $\ce{N-N}$ bond weaker than the $\ce{P-P}$ bond. This also explains why nitrogen portrays a lesser tendency for catenation. To sum up, both bond-length and bond-strength follow the same order, i.e. $\ce{N-N \lt P-P}$.
Both strength and length follow the same order in these two cases as well:

$\ce{O-O \lt S-S}$
$\ce{F-F \lt Cl-Cl}$

Lastly, I would like to highlight a third case, $\ce{C-C}$ and $\ce{Si-Si}$ bonds. On comparing the strengths, $\ce{C-C}$ bond takes the lead, and that's why we're all (atleast I am) a fan of organic chemistry! As for the length, $\ce{Si-Si}$ bond is longer.

I know this is kind of an old question, but I thought this would make for a valuable contribution as there is a canonical relationship between bond length and bond force constant, which is a measure of bond strength.
This relationship is something which very early chemists studied quite a bit because as more experimental data was gathered, they obviously wanted a way to make sense of it. The simplest rule which was found empirically, and is quite a useful qualitative tool, is known as Badger's rule [1], after Richard M. Badger who discovered the relationship in 1933.
The rule states:
$$k_e(r_e-d_{ij})^3=1.86\cdot10^5$$
with $k_e$ given in dynes/cm (old unit for force) and $r_e$ in angstrom.
There are two important points here. First, such an empirical relationship actually does exist and it is quite accurate. This can be plotted linearly with a simple rearrangement, and this is done in Badger's paper and in many other papers if you dig around.
The real important point here, however, is the constant $d_{ij}$. The $i$ and $j$ represent the row of the periodic table. Hydrogen is counted as it's own row (I guess helium would be included too). This is very powerful because it means that within each row, and within each combination of rows, the force constant and equilibrium bond length are characterized by the use of a single parameter. Badger even says that the whole reason for finding this relationship is so that the strength of bonds can be easily characterized using $k_e$. Note that this term $r_e-d_{ij}$ is described as an effective distance of approach for the nuclei.
You can also imagine how useful this would be when one couldn't simply calculate the bond lengths to within a tenth of an angstrom without thinking about it. If one had the vibrational frequency, you could estimate the force constant and hence determine the approximate equilibrium bond length.
Strictly speaking this equation was derived from diatomic data, but it can also be applied successfully to polyatomic molecules[2], though it is kind of inconsistent for transition metals from what I understand.
Some researchers revisited Badger's rule as recently as 2000 with the hopes of improving the rule as well as testing how it performs with charged species and other unusual diatomics[3]. Basically what they found is that without significantly complicating the form of the law, the rule cannot be significantly improved. The average error in determining $k_e$ from their set of over 100 diatomics was about $17\%$. Which isn't that bad considering they had $\ce{He_2^{2+}}$ and things like this in the set.
Basically, this rule relating bond length to bond strength, as measured by force constant, is very robust for diatomics (it probably never goes the other way), and is a very good rule of thumb in polyatomic systems. The whole question of isotopes actually could be described by a law like this if corrections were made either for anharmonicity or to use the vibrationally-averaged coordinate rather than $r_e$.

[1] Badger, R. M. (1934). A relation between internuclear distances and bond force constants. The Journal of Chemical Physics, 2(3), 128-131.
[2] Badger, R. M. (1935). The relation between the internuclear distances and force constants of molecules and its application to polyatomic molecules. The Journal of Chemical Physics, 3(11), 710-714.
[3] Cioslowski, J., Liu, G., & Castro, R. A. M. (2000). Badger's rule revisited. Chemical Physics Letters, 331(5), 497-501.


text:
Maximum density of heavy water at 11.6 degree Celsius?
I came across a question asking me the temperature at which $\ce{D2O}$ will have maximum density. I didn't really know, so I checked the answer, and I found that the answer is $\pu{11.6^oC}$. But why ? Why is the density of $\ce{D2O}$ maximum at $\pu{11.6^oC}$? I can't find the reason anywhere...

It's possible to explain why there is a maximum at all, but justifying why the maximum occurs specifically at the temperature it does is almost hopeless. The best you can do is probably "because that's how our Universe is set up", or more precisely, "because the Standard Model has the dimensionless constants that it does".

The maximum density of $\ce{H2O}$ occurs when cooled to about $\pu{4^oC}$. At this point, the thermal motion of the molecules has slowed down enough such that water molecules can start to orient themselves in a manner more resembling that found in the expanded crystalline ice structure, and density thus starts to decrease with decreasing temperature. This pre-freezing expansion happens at a somewhat higher temperature for $\ce{D2O}$ due to the greater hydrogen bonding strength of $\ce{D2O}$ at any given temperature.
Also note that $\ce{D2O}$ freezes at about $\pu{4^oC}$, so clearly the maximum density of $\ce{D2O}$ must occur at a greater temperature than that of $\ce{H2O}$. In the case of $\ce{H2O}$, the temperature of maximum density occurs at about $\pu{4^oC}$ above it's freezing point, where for $\ce{D2O}$ the temperature of maximum density occurs at about $\pu{7^oC}$ above it's freezing point.

I understand, what you are trying to explain when you say 'molecular vibrations have slowed down'  but this is not correct; the frequency is the same irrespective of temperature. What would be more accurate is to explain that as the freezing temperature is  approached regions of structured H bonds have a chance to form without thermal motions disrupting them by too much.  On balance (H bonds expansion vs. thermal motion causing contraction)  the density is increased. On freezing the H bonds are now 'fixed' into a larger lattice and then occupy a larger volume than in the liquid.

@porphyrin , yea, I didn't have that all right. I made some changes, thx for the heads up.


text:
Does the laser side of laser fluorescence spectroscopy have to be so complicated?
I understand the working principle behind zapping some gas and looking at spectral emissions, and I can see why photodetection is so difficult. But I thought that pointing a laser into a vacuum chamber would be the easy part. So I ask: what the heck is all this equipment that the laser has to go through before it even gets to the sample?

Actually it is not that complicated. It looks like the experiment is of the pump-probe type as there is a delay line and a retroreflector. Divergence of beam has to be controlled, then has to be split into two, possibly polarisation changed and beams refocussed, height adjusted etc. to get it into the sample.

@porphyrin - so, it is really that simple, huh? Said tongue in cheek, because that is what I first thought too. Somehow in my memory it seemed easy to set that kind of thing up too! To the OP, what looks complicated at first glance becomes quite easy with a little experience...

@Jon - not simple by any means just not that complex!  I have always found  that postgrad students soon become fully adept at using such instruments and unfazed at making their own improvements :)

I think the point is that somebody new sees a bunch of individual components, while an experienced person sees functional units.

"Divergence of beam has to be controlled, then has to be split into two, possibly polarisation changed and beams refocussed, height adjusted etc. to get it into the sample."....OK, but why does the laser have to be manhandled in so many different ways? I'm aiming a laser into a chamber; do I really need to control divergence and polarization over just a few feet? And where would I find a primer on the functions of all of these components? (links to equipment suppliers are pretty useless as explanations are usually flimsy)

The mucking around with the beam is all about improving the signal to noise ratio in the detector.


text:
Why is bromine produced at the anode when aqueous sodium bromide undergoes electrolysis?
The electrode potential for Bromine is $\pu{+1.07V}$ whereas for Oxygen to be produced it is $\pu{+0.40V}$.
$$\ce{O2 +2H2O + 4e- -> 4OH-}$$
As the electrode potential for bromine is higher, it means bromine should be reduced rather than $\ce{OH-}$ ions.
I'm aware that the concentration affects the electrode potentials, but the question simply states 'an aqueous solution of NaBr' with no reference to concentration whatsoever, so I can only assume standard solutions.
Also, why do we not look at the electrode potentials of other possible reactions at the anode, like;
$$\ce{O2 + 4H+ + 4e– -> 2H2O}~~~~~~~(E=\pu{+1.23V})$$ or
$$\ce{HO2- + H2O + 2e– ->3OH–}~~~~~~(E=+\pu{0.88V})$$ or
$$\ce{H2O2 + 2H+ + 2e– -> 2H2O}~~~~~~~ (E=\pu{+1.77V})$$.
Why are these not viable?

You don't have a standard solution of $\ce{OH-}$ in there, otherwise things might have turned differently.

But all I'm told is that it is aqueous. How do I judge if it is a standard or non-standard solution?

What is a standard solution, come to think of it?

A solution where the concentrations of any ions is 1M

Right. Now, you've been told that it is aqueous solution of what?

Sodium Bromide. So the concentrations of the Hydrogen and Hydroxide ions will be much lesser than 1?
But then why will Hydrogen ions still be reduced at the cathode?

That's right! As for the reduction of hydrogen, there is really no other option. Na's potential if way too far away. Besides, even should Na be reduced, it would react instantly with water. Bromine is not like that.

The answer below suggests that water will be reduced at the cathode. And if the concentration of H+ ions in the solution was low enough (not sure how to do this without adding a base), wouldn't Sodium be reduced? Lastly, why do none of the other reactions occur? Suppose we have a very dilute solution of NaBr. Would we get oxygen by the first reaction, or would the rest also occur to some extent?

Sodium won't be reduced; you can't make the concentration of H+ low enough for that. As for the other reactions, some of them might occur to some minor extent, depending on the conditions.

I see.. thanks! Could you clarify what you mean by conditions? I'm unclear on this in general.
If we have two possible reactions, and one has a higher cell potential, will the other one not occur at all, or will it occur to a lesser extent? Also, is there some sort of approximate value which determines if one reaction will occur? For example, something like ' the difference in electrode potentials between the two reactions in question must be more than 1V for the reaction to occur even under non-standard conditions'

The reference book values of potentials are only good for the standard solutions (1M of everything). There is so-called Nernst equation which tells you how the potential will change in non-standard concentrations.

Oh, I know about that, but I'm asking what do I do if I'm not given the concentrations of the ions, and still wish to make a prediction? Like, if the difference between two electrode potentials is greater than some emperical value, like 0.30V, then we can say that unless the concentrations are absurdly high or low, the higher one will undergo reduction.

This is  how I see it:
In the electrolysis of NaBr, water is reduced at the cathode. This occurs because water is more easily reduced than are sodium ions. This is reflected in their standard reduction potential.
At cathode reduction of water occurs:
$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH-(aq)}$$
And hydrogen gas is produced
At the anode, where oxidation occurs, the standard oxidation potential of water is –1.23 volts, while that for bromide ions is -1.07 volts.
The production of bromine itself has a negative electrode potential.
One of the half-reactions must be reversed to yield an oxidation.
$$\ce{Br^2- -> Br2 + 2e-}~~~~~~~ Eº -1.07$$
Remember that when one reverses a reaction, the sign of Eº (+ or –) for that reaction is also reversed.
This means that bromide ions are more easily oxidized than water.
It is important to note: When current begins to flow, the distribution of ions around the electrodes changes, and the equilibrium electrode potentials no longer accurately apply.
I think other factors also apply;

Concentration
Temperature
Nature of ions

Why is water reduced at the cathode? Shouldn't it be Hydrogen ions since they have an electrode potential of 0 V? Also, could you write the reaction in which water is reduced? The 1.23V reaction I know has Oxygen being reduced, not water.

I added the equation showing reduction of water at cathode

The electrode potential for the reduction of water is -0.83V whereas for H+ ions, it is 0.00V. As such, shouldn't Hydrogen ions be reduced? Also, why are you considering the +1.23V reaction of water rather than the +0.40V one?

And where are you  generating the hydrogen ions from in the first place?  Please show the reaction that you are referring to. Also read Ivan Neretin's comments they essentially answer your additional questions

The hydrogen ions come from the water present
H2O <---> H+ +OH-


text:
Why is a CO single bond stronger than a NO single bond?
Why is a CO sigma bond stronger than a NO sigma bond? I would think that since nitrogen and oxygen are of similar size and have similar atomic orbitals (because they are adjacent in the period), their MO "overlap" would be superior.
Does it have something to do with their similar electronegativities "fighting" over the electrons?

No, it doesn't. It is just the extra electrons. Think of F... or better yet, think of Ne; it has the same size and same orbitals, so why is there no $\ce{Ne2}$?

@IvanNeretin Octet rule. I'm not sure what point you are making here. C-O sigma bond has a greater bond energy than a N-O single bond. What "extra electrons" are you referring to?

How many valence electrons are there in a CO molecule? How many are in NO?

What "single bond"? You mean sigma bonds in CO and NO molecules?

@Mithoron exactly

@Mithoron Could it be that the larger electronegativity difference between C and O means that there is more negative charge sitting on the oxygen, thus making a stronger electrostatic bond between the two atoms?


text:
What is the oxidation state of the element Oxygen in the molecule FONS, assuming that it does indeed exist
What is the process to determining charges regardless of the molecule? Is there some sort of rule book?

Look at the bonds, then at the electronegativities. There is nothing as complicated as to justify writing a whole book on the subject.

After ranking them based on electronegativity values from lowest to highest;  S, N, O, F, what do I do? I don't understand how any one of these will have a positive charge. @IvanNeretin

Basic concepts are better learned on the examples of existing compounds. Think of sulfuric or nitric acids. What are the formal charges of S and N in them?

S has +6 and N has +5? @IvanNeretin

That's right. Now what's strange about them having positive oxidation states? Nothing.

How are the atoms bonded together?

@gsurfer04 Its a hypothetical molecule.

Are you aware of constitutional isomers?

Note: Since this is a homework question, I will not give a complete answer to this question (Site Policy).
I will assume that the molecule is:

Although the molecule is highly strained, lets assume it exists.
Use the simplified formula to find Oxidation number:
$$\text{Oxidation Number}=(\text{Num. of bonds formed with electronegative atoms})-(\text{Num. of bonds formed with electropositive atoms})$$
I'm sure you can continue from here.

How did you come up with this equation?

Well, its the only one that satisfies all their valencies :) @XHyper

How did you create the lewis dot structure in the first place? I'm still struggling with that.

@XHyper Nitrogen forms 3 bonds, Oxygen and Sulfur each 2, and Fluorine forms 1. Try any other structure, this is the only one.

Thus, Oxygen would have (1) - (1) = 0 oxidization charge? @Pritt Balagopal