Can I use sodium hydroxide (NaOH) to remove oxides from metals? Sodium hydroxide dissolved in water is a very strong base. I think this could mean it could de-oxidate the metal oxides during re-ox reaction. Basically, I ask whether the following happens:
$$\ce{NaOH + Fe2O3 -> Fe + something}$$

I have read that it does reduce carbon dioxide producing sodium carbonate and water:

$$\ce{2 NaOH + CO2 -> Na2CO3 + H2O}$$

Would that work with metals? If yes, which ones?

Your second reaction isn't a redox reaction. When carbon dioxide is dissolved in water, two equilibriums are forms:

\begin{eqnarray}
    \ce{H_2CO_3 + H_2O &<=> HCO_3^- + H_3O^+} \\
    \ce{HCO_3^- + H_2O &<=> CO_3^{2-} + H_3O^+}
  \end{eqnarray}

Remember that water also is in equilibrium:

$$\ce{2H_2O <=> OH^- + H_3O^+}$$

Adding $\ce{NaOH}$ will — according to Le Chatelier's principle, the equilibriums are disturbed, and more carbonate ($\ce{CO_3^{2-}}$) is formed.

If you're interested in a nice redox reaction involving iron(III) oxide, aluminium is much better choice.

The short answer is no. Sodium Hydroxide is a soluble base (alkali) and Iron (III) oxide is a base too. An acid will react with metal oxides forming a salt and water. For example, if a sample of Iron(III) oxide was reacted with HCl, Iron chloride and water would form. It's possible to extract the pure Iron by electrolysis. Although, I am not sure that Iron Chloride is soluble so another acid may be more suitable or electrolysis could be performed using molten iron chloride.


How does H's ionization energy relate to its transition energy (Bohr's Model)? I am currently reviewing some material about orbital energy levels. In my review book there is a short snipet that reads:

  The IE of $\ce{H}$ from its ground state ($n=1$) is $1312\ \mathrm{kJ/mol}$. Because of the squaring of the principle energy level, the ionization of an $\ce{e-}$ in $\ce{H}$ from the $n=2$ level is $1/4$ of that value ($328\ \mathrm{kJ/mol}$). The transition energy from the $n=1$ level to $n=2$ level is the difference between the two values, $984\ \mathrm{kJ/mol}$.

I am confused about how exactly an atom's ionization energy relate to its transition energy. If IE is the energy required to pluck off the first e- from an atom and TE is the energy required to change between different orbitals, does this mean that an atom with greater IE will need to absorb a greater amount of energy to jump orbitals?

Lastly, is there a equation that directly ties in both ionization energy with transition energy?

Ionization energy of some arbitrary atom is not related whatsoever to its transition energy. Also, there are many different transitions (even in a hydrogen atom), each with its own specific energy. It is just that this particular transition of this particular atom happens to have the energy of 3/4 its ionization energy, one of the few things the Bohr model managed to get right. You can't generalize that to other atoms and other transitions.

How exactly an atom's ionization energy relate to its transition energy?

I think it is best to clear up what the terminology of the question is so that we can address the correct answer.
The ionization energy is the direct measure of the electric interactions between the positively charged  nucleus center of an atom, and the negatively charged electrons around it. Ionization energy is like that invisible force you can feel between two magnets close together. If you were to measure how much effort is required to split the magnet apart till they are effectively not interacting, that would be like measuring the ionization energy and an electron from an atom. One way to supply the energy to pull off (knock-out) an electron,  can be by shining light (in the case of the first ionization energy for a hydrogen atom, $1312\ \mathrm{kJ\ mol^{-1}}$, where $\mathrm{kJ\ mol^{-1}}$ is amount of energy required = $15.60\ \mathrm{eV}$ (another unit of energy that express the amount of electric attraction the negative electron feels toward positive nucleus), and in terms of focused energy carried by light (electromagnetic waves, generated by moving charges or current) that is $= 91.18\ \mathrm{nm}$ electromagnetic waves, which we call  X-ray light.)

What is a transition energy?

The intuitive way to think about a transition state is to try and recall a time when you were in danger or had to make snap decision that brief instant before hell breaks lose is like a transition state. Something you have to build up eneough energy to get over. In the Bohr model, the transition state is when an electron moves from one orbital to another. The HUGE thing about this model that made it work is that Bohr assumed the energy between transition states in atoms (i.e the differences between the energies measured for each electron in an atom and the next) is quantized, meaning that it take it can be certain values only. Like having a light switch with 4 positions. It can have different levels but switches can just hover in between, either one or off .The transition energy is the energy required to move electrons between transition state. It is like when water is being heated. You are delivering energy into it by heating and nothing seems to happen until it reaches $100\mathrm{ºC}$ and then it boils.

Bohr related the model for the hydrogen atom and thus approximation for inonization energies for electron in hydrogen energy levels is give by.

Bohr model is given by $[ \Delta  E = \frac{m_ee^4}{8\varepsilon_0^2h^2} \lgroup\frac{1}{n_2} - \frac{1}{n_1} \rgroup$, where $n_i = 1,2,3,..]$

Where $n_1$ and $n_2$ are the quantum principle numbers between each quantized state. ie. $n_1 = 1$, means level 1, $n_2 = 2$, level 2 etc..

Even thought we know Bohrs model is incorrect, the quantization of energy as boundaries for electrons  is the basis of all modern models. It is not atoms that absorb energy in the case of ionization energy,

Misconception is that atoms absorb energy when its actually the electrons of nucleus particles that do. In the case of ionization energy, it its the electrons themselves that are being energized to the point of boiling off. Thus the valence electrons, which the ones furthest from the nucleus with lowest ionization energies, have lower ionization energies than core electrons (those closer to nucleus)

McQuarrie DA & Simon JD (2015) Physical Chemistry: A molecular approach . Viva Books


How did the first amino acids appear on Earth? Is there proof or a theory that shows that the first amino acid syntheses on Earth were influenced by a change in heat, gravity, radiation, or something similar?

The universe cooled after the Big Bang. There was certainly an epoch when the universe’s overall temperature decreased from 200 °C to −40 °C throughout a very large volume. There is your test tube (more like a Florence flask) for abiogenesis and then the panspermia reservoir.

Before you suspect the preceding is clever, how and when (what ambient vacuum temperature) did elements heavier than boron arise?


Pressure Cooker at High Altitudes and Low Altitudes I just have a doubt if I'm right or wrong about this. Please explain more if possible.
Also, I have some questions regarding boiling point and vapour pressure.

 High Altitude Cooking
At high altitudes, the atmospheric pressure is lower than that at sea level, so the boiling point at high altitudes is quite low, which means water boils very fast and at low temperatures. The food inside it does not get enough heat to get cooked and thus food is difficult to cook at high altitudes.
Using a pressure cooker at such conditions helps increase the boiling time as the pressure inside the pressure cooker increases due to the vapour produced inside it. The boiling point increases and thus the time taken to reach the temperature increase and the cooking is better because the food is getting heat for a long period of time.

At Low Altitudes
The boiling point of liquid at low altitudes is more than that at high altitudes, so we do not have to worry much about less time being available for cooking the food. In fact, we try to decrease the cooking time in order to get food cooked faster. Even less amount of water can produce a good vapour pressure and get equal to the pressure produced inside the cooker. But less amount of water takes a lesser time to reach a particular temperature than more amount of water ( as $Q=mc{\delta}t$ ) and thus lesser the mass, lesser the time  to get the required heat, hence faster the cooking.

Questions
1. Have I understood this perfectly or is there something more to it?
2. Does the vapour pressure of the water inside the cooker have to become       equal to the vapour pressure produced due to itself in order to boil?
3. How does pressure cooker increase time for cooking at high altitudes and decrease time for cooking at low altitudes being basically the same thing?
4. The Critical temperature is the temperature at which gas just becomes liquid. The boiling point is the temperature at which all the bulk of the liquid starts turning into gas. How are Critical and Boiling Points different? How are they similar?Is there any mathematical relation between them?

Edit - As 1e9dB's answer points out I made a huge mistake in this answer. The inside pressure of the pressure cooker is ambient pressure plus the extra pressure of the valve. The pressure cooker does not cook at an absolute pressure independent of altitude.

You have it all backwards. The pressure cooker allows the pressure inside the cooker to be above the sea-level atmospheric pressure which also increases temperature inside the cooker and thus decreases cooking time. So the critical factor is the increase of the boiling temperature of water as a function of the pressure inside the cooker.

The pressure cooker raises the pressure inside the cooker to the same absolute pressure, regardless of the altitude at which the cooking is being done. A typical value for the internal pressure would be about 2 atmospheres absolute which makes the internal cooking temperature about 120 °C. (The steam is pushing against the pressure regulator (weight) to escape, not the outside atmosphere. It is like the safety valve on an old steam locomotive that keeps the pressure at a certain level so that the pot doesn't blow up.)

The inside temperature of a pressure cooker when steaming (120 °C) is way below the critical temperature of water (about 374 °C) so that factor doesn't enter into the cooking time at all.

Some qualifications and clarifications:

Altitude affects average air pressure, it's quite possible for pressure to be higher than 1 bar (101 kPa) at 1 km above sea level.
You should be careful to clearly state what the heat flow is.

In general "heat" is a vague term which is often used to describe energy or the flow of energy. The energy required to increase the temperature of a substance depends on the mass of the substance. The rate of temperature increase depends on the heat flow (as well as the mass and other things (such as heat loss)). Assuming the same heat flow, on average a given mass of water will boil sooner (and at lower temperature) at higher altitude.

Most food cooks faster at higher temperature. So, low altitude cooking can be faster if the limiting variable is the boiling of water. Likewise, high altitude cooking will be slower if the boiling of the water is the limiting variable. This will have little to do with frying an egg or roasting meat.

Your question mixes (confuses?) time and temperature. In an open container, a given amount of water will boil quicker at high altitude (on average, with a fixed rate of energy input). The temperature rise will be the same as at low altitude, until the boiling point is reached. At the boiling point (which will be lower at higher altitude (on average)) the temperature of the liquid becomes almost constant until (almost) all of the water has boiled away. IF this temperature is the temperature the food is cooked at, then higher altitude means slower cooking.

=========

The critical point of a substance Tc, is the temperature and pressure (Pc) at the "end" of the vapor-liquid curve. It is the temperature above which no liquid phase can exist, regardless of pressure. (It is the highest pressure at which both liquid and gas can co-exist.) I think of it as the temperature at which (and above which) the liquid and gas are indistinguishable.

The Wikipedia article has a diagram showing a 'typical' phase diagram. [see Critical point (thermodynamics)]. That diagram shows many different regions. If you count carefully, and include the dashed lines, you can count at least 11 different areas. The major problem with this is that the dashed lines don't represent real phases, and also the names given in that diagram are arbitrary. For instance, there's no technical difference between vapor and gas. Similarly, liquid phase and compressible liquid aren't inherently different.

A gas above Tc is indeed "supercritical", by definition, however. As temperature is increased towards Tc, the properties of the liquid phase converge towards the equivalent properties of the gas (vapor). Keep in mind that when speaking of these phase diagrams, we are speaking of systems at equilibrium. Each of the two solid lines represent the points of P and T at which two phases can co-exist at equilibrium. The triple point, when it exists for a substance, is the point at which all three phases can co-exist (and shouldn't be confused with Tc).

Of course, in the real world, I'm sitting here typing with a glass of liquid water, ice, which no doubt has water vapor over it. However, if I were to wait long enough, the ice would melt, and I'd be left with liquid and vapor (gas), and if I were to wait a really really long time, the liquid would all evaporate (unless I sealed the glass). So, the real world only works "near" equilibrium, rarely exactly at it.

The phase diagram works by describing which phases exist at a given T and P. It also enables you to predict how much increase (or decrease) in P or T is necessary to move the system to a different region in the phase space. I believe it is most useful to think of Tc as the point at which there is 'no difference' between gas and liquid, rather than the point at which liquid "just becomes" gas. It is incorrect, imho, to think of a supercritical gas as a liquid, but certainly it has some properties (depending on pressure) which are similar to ones we associate with liquids.


How does solubility of corrosion product affects rate of corrosion? Why is it that if the corrosion product are soluble, then the rate of corrosion increases while insoluble corrosion products cause rate of corrosion to decrease?(How exactly would insoluble corrosion product protect the material?)

The process in general is called passivation. It means, that the surface of a metal is corroded and the corrosion product forms a protective layer. If the corrosion products are hardly soluble, it therefore blocks the metal beneath it from further corrosion. If the corrosion products are soluble, this protective layer can be washed away and the material is exposed to further oxidation.

In iron, for example, the initially formed iron hydroxides are somewhat soluble in water. The formed rust is flaky and friable, which basically means it is removable by little or no stress, i.e. rain can break the surface and expose more metal to the oxidising agent. It therefore usually always decomposes completely.

In another example, copper does form such a protective layer of copper oxides. This is also called patina. These layers are hard(er) to remove and protect the underlying metal from further oxidation. This is also the reason, why for example many church roofs and the Statue of Liberty are green.

This method is often used to protect metals from corrosion, i.e. the formation of the protective layer is induced chemically.

Corrosion is nothing but actually the hydration of the metallic compound.
For example if you take rusting of pure iron you have to stick water molecules around it to make it rust or corrode. And we know that solubility is greatly affected by hydration hence the substance with more solubility corresponds to more corrosion as compared to one less solubility.


Bonds between metals and non-metals My teacher in college says that bonds between metals and nonmetals are ionic.

$ \ce{Metal - Metal} $ $\Rightarrow$ Metalic bond

$\ce{Non metal - Non metal}$ $\Rightarrow$ Covalent bond

I have to write about $\ce{CuCl2}$, and found in Wikipedia that if you subtract electronegativity you get what bond it is:

$$\chi (\ce{Cl}) = 3.16, \quad \chi (\ce{Cu}) = 1.90$$

$$\chi = 3.16 - 1.90 = 1.26$$

$\chi < 0.4 \Rightarrow$ covalent non-polar

$0.4 < \chi < 1.7 \Rightarrow$ covalent polar

$\chi > 1.7 \Rightarrow$ ionic

So it should be really a covalent bond. Which one is correct?

You should be careful with simple associations such as "metal + non-metal = ionic bond". These tend to throw out the idea of understanding the chemistry involved in favour of rote memorization. Note for example that mixing caesium metal with gold will produce a salt instead of an alloy, caesium auride ($\ce{Cs^+ Au^{-}}$). Mixing barium metal and platinum can also produce salts, though their structures are somewhat more complex. One can also argue that there is significant ionic character in solid xenon difluoride, even though both atoms are non-metals.

The idea of using electronegativity to determine covalent/ionic character is also meant as a helpful guide, not as a strict rule with black-and-white limits. Firstly, all bonds have both ionic and covalent character; both concepts are an oversimplification, and in reality it is more correct to say that a bond has a certain contribution from each type of bonding. This means there is a smooth transition from compounds with mostly ionic character and those with mostly covalent character. Also, the inequalities you mention rely on Pauling electronegativities. Electronegativity is surprisingly still a hotly debated topic, as we continue to search more general, more fundamental and more precise ways of defining it. Pauling electronegativities are based on empirical thermodynamic data regarding bond energies after applying a certain equation that was "picked", not derived from scratch. The values are particularly poorly defined for transition elements, such as the $\ce{Cu}$ in your problem. You get some not-so-easy to explain situations, like $\ce{HF}$ as a gas that is a borderline ionic compound.

Finally, in light of these comments, the answer to your question is that bonding in $\ce{CuCl_2}$ (I'm pretty sure that's what you actually meant to write) has intermediate characteristics between a purely ionic and a polar covalent bond, with similar contributions (though pinpointing which is highest sounds like an exercise in futility). A good way to study it more in depth is to analyze Fajans' rules. After a little self-calibration, you can get a good feel for the degree of ionicity and covalency of a compound. Some further but less certain evidence (lots of caveats!) for the intermediate character of $\ce{CuCl_2}$ can be found by looking at the substances' melting and boiling points ($\pu{498°C}$ and $\pu{993°C}$ [decomposition], respectively, according to Wikipedia). They are both quite high compared to substances with polar covalent bonds (dimethylformamide boils at around $\pu{150°C}$), but rather low compared to substances with very ionic bonds ($\ce{NaCl}$ boils over $\pu{1400°C}$).

we aspect that the bond formed between Fe and Cl is ionic because Fe is metal and Cl is  non metal but charged on Fe is +3 and on Cl is-1 so  polarisation of chlorine ion by Fe+3 ion takes place but also with presence of  d orbital in valence  shell of Fe+3 ion polarisation also takes place  to maximum extent  and covalent charachter  arises in  ionic compound FeCl3.


Why aren't neutralisation reactions reversible? Let's take
$$\ce{3HCl + Fe(OH)3 -> FeCl3 + 3H2O}$$
This would be a normal one-way neutralisation reaction.

How about hydrolysis of ferric chloride (technically iron(III) chloride)?

$$\ce{FeCl3 + 3H2O <=> 3HCl + Fe(OH)3}$$

As this is reverseable, we should be able to right it with the reactants and products flipped.

$$\ce{3HCl + Fe(OH)3 <=> FeCl3 + 3H2O}$$

Now that is nothing but our original reaction …
Why aren't neutralisation reactions reversible?

Also, does anyone have any good links to chemical equilibrium as in comments?

Neutralization reactions are reversible. In theory, at least, even if not so much in practice, all reactions are reversible.

If you look at things on a molecular scale, there's a principle called microscopic reversibility. In effect, there's no physical reaction mechanism you can create that's a one-way door. Maxwell's demon doesn't exist, and you can't get around that no matter how many steps you chain in whatever order, or which particular reagents you use.

What you can do, though, is set things up so it's exceedingly unlikely that the reaction proceeds in the other direction. That's what's happening with reactions we normally deem "irreversible". There's something about them which makes it difficult, in practice, for products to go back to reactants.

One common feature is the size of the equilibrium constant. If a particular reaction has an equilibrium constant that's really big, the equilibrium for that reaction is going to be vastly tipped to the product side. So much so, that the reactant side of the reaction will not be present in appreciable amounts. (For example, this is why combustion is viewed as irreversible. The reaction is so favorable toward the products that only parts per million or parts per billion of the reactants will be present at equilibrium.)

Another method for promoting "irreversibility" is for the products to partition to another phase. For example, if the reaction forms a gas or a precipitate, that product is removed from the reaction volume, and the concentration of the products drops, pulling the reaction toward the products. Note that even in this multi-stage setup, each step is reversible. That evolved carbon dioxide molecule could re-dissolve in solution, and then react backwards, but it's terribly unlikely for it to do so. You've basically fixed one of the products at a very low concentration, which draws the equilibrium toward the product side.

Precipitates work much the same way, as the activity of the solid is fixed by surface behavior, and does not increase because you increase the amount of solid you add. (The rate of reaction may increase by adding more solid, but the equilibrium of the reaction does not.)

So in your examples, the reason you get hydrolysis or neutralization is due to how you set up the reaction conditions. Have a large excess of HCl? You drive things toward neutralization. Have a way of getting rid of HCl (for example by driving it off as a gas, or reacting it with a base)? You drive things toward hydrolysis.

In this neutralization reaction, Fe(OH)3 forms a dirty green ppt. which is insoluble in water.This demonstrates one point as to why neutralization reactions are irreversible.

There maybe other points as well, such as specific temperature and pressure conditions,etc.

Please feel free to  my answer, thnx!


What is the differences between partial pressure and vapour pressure? Was looking at Henry's law and Raoult's law
constants and there seemto be lots of
equations involved.
Henry's law involves partial pressure and the latter involves the vapor pressure.
Wondering what the difference is?

In a mixture of gases, each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the volume of the mixture at the same temperature.

What does this mean? For example, if we have a mixture of gases $A$, $B$ and $C$ in an isolated room, then, according to Dalton's law, the pressure exerted by the gases will be the sum of their partial pressures :
$$P = p_A + p_B + p_C$$
where $p_A$, $p_B$ and $p_C$ are the partial pressures of each gas. Also, if we have a moles of $A$, b moles of $B$ and c moles of $C$, we can express the partial pressure of each gas as below:
$$p_A = \frac{a}{a+b+c} P$$
$$p_B = \frac{b}{a+b+c} P$$
$$p_B = \frac{c}{a+b+c} P$$

  Vapor pressure or equilibrium vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.

As the definition states, if we took an equilibrium as $\ce{A(l) <=> A(g)}$, where $A(l)$ and $A(g)$ are the liquid and the vapor state of the compound $A$ . The vapor pressure of $A$ is the pressure exerted by the gaseous state ($A(g)$) over the liquid state ($A(l)$).
Now, if we have a mixture of $A$, $B$ and $C$, we will have the equilibrium between the liquid and gaseous state : $\ce{(A, B, C)(l) <=> (A, B, C)(g)}$ and the vapor pressure $P_v$ (I noted $P_v$ so there won't be any confusion between it and the total pressure $P$ I mentioned before).
According to Raoult's law, the vapor pressure $P_v$ is the sum of the products between the vapor pressures of each compound taken separately and the molar fraction of the compound in the mixture (i.e. in the solution) :
$$P_v = x_A p_{v^A} + x_B p_{v^B} + x_C p_{v^C}$$

This may look just like Dalton's law, but there are different pressures involved.
The partial pressure of a gas is the pressure exerted by a gas in the volume occupied by a mixture of gases, while the vapor pressure of a gas is the pressure exerted by a gas over its condensed phase.
Although :

  The vapor pressure that a single component in a mixture contributes to the total pressure in the system is called partial pressure.

The vapor pressure still refers to the pressure of a gas over its liquid (or solid) phase in a phase transition (condensed phase$\ce{<=>}$non-condensed phase), while the notion of partial pressure can be extended to a mixture of gases only.

All the definitions are linked to Wikipedia, and they can really cause confusion throughout chemistry learners..

Partial pressure is the pressure exerted by the individual gas in the mixture of different gases at the same temperature.

Vapour pressure is the pressure exerted by the condensed(cooling) gas by lowering the temperature.


Hybridisation of hexaamminecobalt(III) complex Does $\ce{[Cr(NH3)6]^3+}$ have $sp^3d^2$ or $d^2sp^3$ hybridisation?

Oxidation state of $\ce{Cr}$ is $+3$. The electronic configuration is $3d^3$ and the $d$ electrons occupy $t_{2g}$ orbitals. If the electrons from the ligands were to occupy the $e_g$ orbitals the stability of the complex decreases. So shouldn't the hybridisation be $sp^3d^2$?

It has d2sp3 configuration since NH3 is a strong field ligand which causes the electron to pair up and use the inner d orbitals(3d) .


Effect of dopant size on XRD peak shift If we add an impurity atom (doping) to the crystalline struture of another material and take the XRD of the doped material what kind of peak shift can we expect. For the better description of my question i have added an image below

In the image black bold line ($2\theta $) is the standard XRD peak of a material; blue ($2\theta ^{L}$) and green ($2\theta ^{R}$) dotted lines are the peak shift caused due to the doping of impurity atom. Can you tell me if there is any relationship between dopent size and peak shift (right/left dotted lines).

Doping, not to confuse with alloying, introduces a low amount of impurities or more generally point defects in a crystal. The overall periodicity of the system is thus not changed and no shift should be expected in XRD. However doping introduces some disorder that makes XRD peaks broader.

Sometimes, dopants can go to the interstitial site of the target material. We can say this makes some distortions happen.That actually change the crystal structure. Then we can see the certain peak position changes.

Also, misalignment can make peak position changes.So we have to operate the machine properly. calibration is very important.

Strain and stress of the materials can make XRD peak shift as well.

Good luck.


Which is the most stable out of the following linked cyclohexadienes? I am unable to find the answer to this question. I was guessing it should be (2) due to symmetry but my book says it is (3) without any explanation.

I know that stability is determined by how low the energy of formation is. Does this have something to do with the different sorts of dienes that are present? Like, one type of diene would have less energy than another?

If nothing else, it would be helpful if someone could indicate the names of these compounds, because I don't even have the faintest idea where to start searching for more information.

You must be aware about the concept of resonance. Similar ly over here as you can see in option number 3 there is longer conjugation between double bonds then compared with any other option because Three double bonds are in conjugation and this conjugation supplies stability to the structure

Structure $3$ is stabilized through resonance.

Double bond - Single bond - Double bond, can perform resonance.

More the number of resonating (or canonical) structures, more the stability.

Structure (2) also exhibits resonance, but it has less number of resonating structures than structure $(3)$.

Therefore, structure $(3)$ is the most stable one.

If you are not familiar with the concept of resonance, read it. You will encounter many situations where resonance will have a direct link with the stability of certain structures.


Chemical to remove nail polish but retain ink I have a 1954 team signed New York Yankees baseball with 5 Hall of Famers on it. When I was little and showed it to my friends, the signatures started to smudge a little so my Mother put clear nail polish on it to protect it. Is there a chemical I can use to eliminate the polish without hurting the signatures?

It would depend mainly on the chemical composition of the ink (permanent marker? standard ballpoint?)

If it was signed with standard ballpoint, you have a problem, because the main film forming / thickening agent used for ballpoint pen ink is the same used in the majority of nail polishes - nitrocellulose. This means that a solvent that removes nail polish (ethyl acetate, for example) will also remove the ink.

Nail polishes use solvents such as butyl acetate. If the solvent dissolves the ink, it would have destroyed the signatures when the polish was applied. Therefore you should try to use a similar solvent (e.g., ethyl acetate) to get rid of the polish. Use a cotton swab soaked in the solvent and try it first on the least valuable signature. If it works you can apply it to the rest of the baseball, but be warned that the signatures may appear more faded afterwards.


Which is more stable? CF2 (singlet) or CF2 (triplet) Generally, we know that triplet carbene is more stable than singlet. But in my book, it is written that singlet difluorocarbene is more stable than triplet difluorocarbene.
What's the reason for it?

So generally singlet is less stable as triplet has half-filled orbitals which leads to stability.

However whenever a group that can give  electrons by back donation (examples being lone pair carriers like $\ce{-OH},$ $\ce{-NH2},$ etc. are attached, they stabilise the singlet by back donation of their lone pair into the vacant orbital of the singlet carbene.

This is similar to back bonding in $\ce{BF3}.$


First point of difference rule 1,6-dimethylcyclohex-1-ene According to the first point of difference rule, second one should be correct right?

First of all, the ‘lowest sum rule’ mentioned in the picture in the question does not exist in IUPAC nomenclature.

As already explained in a related answer, the most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

The compound that is given in the question doesn’t have any principal characteristic group that could be expressed as a suffix. Therefore, Rule (c) is not relevant in this case.

Thus, a low locant is assgined first to the double bond according to Rule (e). Therefore, the name of the parent structure without any further substituents is cyclohex-1-ene. (Without further substituents, however, the locant ‘1’ would be omitted.)

Finally, low locants are assigned to the remaining substituents as a set according to Rule (f).

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.

Thus, the correct name is 1,6-dimethylcyclohex-1-ene rather than 2,3-dimethylcyclohex-1-ene since the locant set ‘1,6’ is lower than ‘2,3’.

No, the right one is incorrect as the numbering starts at the double bond, it is suppose to start at the first substituent.


The equivalent weight of HNO3 in the following reaction The equivalent weight of HNO3 in the following reaction is ‘x’, then the value of x/21 is

3Cu+8HNO3⟶3Cu(NO3)2+2NO+4H2O

in this I atempted by using redox formulas but my answer is comming 1 but correct answer is 4

The term equivalent weight is used in conjuction with normality (which is quite obsolete, except in the industry). Anyway, whenever you determine equivalent weight, you have to know beforehand whether you are considering a given "molecule" as an acid or as an reducing/oxidizing agent. This is where the confusion sets in.

Nitric acid as an acid is a monoprotic acid: It's equivalent weight is the same as its formula weight

Calculating the equivalent weight of nitric acid as an oxidizing agent is another story, it depends on how the redox equation is written.

From your equation nitric acid is acting as an oxidizing agent because there is no nitrate ion on the right hand side and the oxidation state of N is different on both sides:

$\ce{HNO3}$ is converting to NO. In the answer above, you already have the equation

$\ce{ NO3−}$ +3e => NO

In this case, the equivalent weight of nitric acid is 1/3 (formula weight).

As you can see equivalent weight is very much dependent on the way you write the equation. This will come with practice.

"Equivalent weight" has been depreciated by IUPAC because it is ambiguous in some cases. To me this is one of them. The relevant half cell reaction is:

$$\ce{NO3−(aq) + 4H+ + 3e−  <=> NO(g) + 2H2O(l)}\quad\quad E = +0.958$$

Thus there are $\ce{4H+}$ but only $\ce{3e-}$. So is the equivalent weight 1/4 or 1/3 as much as the molarity?


Artificial nose sensors I'm am doing research in creating an artificial nose. I need to know how this process works by someone with domain expertise; essentially, what technologies are available currently to detect a broader range of particles in a closed system. My understanding so far goes like this on a high-level overview: particle comes into contact with the olfactory cells and particles from the environment have a custom signature which we associate as smell of a banana or orange, now how to do this artificially or how is this done artificially?

"What technologies are available currently to detect a broader range of particles in a closed system" a LOT of technologies allow this: spectrophotometry, NMR, GC-MS, etc. These technologies give a lot of information about the molecular structure (but not about the molecular  function).

The hard part in your idea is trying to replicate the sensivity of a nose, which is, at the moment, impossible as far as I know.

Trying to deduce via a robotic nose the smell of a novel compound is a task that could, theorically, be made if that "nose" had exactly the same receptors and "circuits" of our nose, since receptors interact in a unique way with molecules, and that is the source of all those unique smells that we sense.

A more approximate approach, doable with current technologies, could be a database-based approach: a lot of "smelly" molecules could be categorized and described in terms of smell.
The analyzer (one of the machines mentioned earlier, for instance) could identify the molecule(s) involved, and the database could provide information about the smell in that environment.

The problem? A sistem like this doesn't work like a nose at all! Any unknown molecule couldn't be "smelled".

Maybe in the future, with the improvement of technology, a molecular docking study will be performed in seconds, and that, when all the proteins associated with smell will be known, could allow us to know the exact way a molecule affects our receptors, our brain and our perception, hence creating a working "digital nose".

And finally, let me add a note: you said "particles from the environment have a custom signature which we associate as smell of a banana or orange". One might argue wether that "signature" is in the molecule or in the nose itself. While many compounds are sensed in a similar way within a population, there are genetic mutations between species and even individuals. Puppies are known to be appealed by cat feces, for instance: is that just a behavioural thing, or their perception of the smell of cat's feces is different from ours?
Some people like the smell of truffles, others hate it: it's (at least partially) because of the individual ability of detecting molecules.

The point is not that of detecting a "signature", but in detecting molecules and interpreting them via our specific "signature".
If olfactory proteins didn't exist, smell wouldn't exist: it's not a physical property of the molecule, but the result of an interaction with who "smells".

Don't be fooled: that's not a philosophical sentence: light wouldn't cease to exist if we had no visual receptors. But for sure colours wouldn't exist, since also colours exist only because we are evolved to detect light as a mixture of three different chemical responses (each associated with one of the RGB colours).

I think the major gas sensing technologies are NDIR(cheap), semiconducter type(cheapest), fuel cell type(cheap) and conductive polymers(expensive)


When I dissolve sugar in my cup of tea/coffee, does it become a liquid? When I dissolve sugar in my cup of tea/coffee, does the sugar go from being a solid to being a liquid?

Related: https://physics.stackexchange.com/questions/5562/what-keeps-the-sugar-suspended-in-the-tea

When the sugar dissolves, it is broken up into individual particles; each molecule is surrounded by water molecules. Essentially, you are trying to decide if a single molecule is a solid or liquid when dissolve.

The best fit is a liquid since the molecules are free to move independently and will take on the shape of the container.

When sugar dissolves into tea or coffee, the liquid transforms the sugar into a liquid so it can fit in with the liquid and slide in with the molecules. If you try to evaporate the water for long enough, you will turn the sugar back into a solid.


Brown ring from the anion detection of NO3- Today in our practicals I observed that when I did the wet test for nitrate (Making a 1 ml solution of salt solution+$\ce{FeSO4}$ solution+hot concentrated $\ce{H2SO4}$);
The brown ring appeared for a very short time and then disappeared.

But later on I came to know that it was nitrate only. What may be the possible reason for this temporary appearance of a tinge of a brown colour and then its sudden disappearance?

The brown ring formed in the confirmatory test for the nitrate ion is due to the complex called nitroso ferrous sulphate, or pentaaqua nitrosyl iron (II) sulfate with the formula $\ce{[Fe(H2O)5NO]SO4}$.
The concentrated sulfuric acid decomposes this complex and brown ring disappers in a few seconds when $\ce{H2SO4}$ moves to the upper layer. This is why slow addition of the acid (through the sides of the test tube) is advised.

brown ring formed due to a complex formed by reaction between Fe+2 and NO3- .complex is of formula [Fe(H2O)5NO]SO4 .The ring is unstable because Fe is present in +1 oxidation state thats why H2SO4 is added slowly because it can oxides Fe and can break the ring


Complexes formed by lanthanides and actinides A very simple test for the presence of alcohols in the lab involves adding ceric(IV) ammonium nitrate solution which gives a pinkish red colour in the presence of alcohol.

When I looked up the structure of this compound it turns out to be quite interesting. So a couple of questions about lanthanides came up in my mind (a Google search only comes up with very complicated research papers):

Wouldn't this compound be highly unstable since there 12 donor atoms (forming an icosahedral structure) surrounding it (but it is quite stable)?
Is the formation of complexes in lanthanides and actinides similar to that of transition metals? Do $\ce{f}$ orbitals undergo crystal field splitting just like $\ce{d}$ orbitals?
Are higher coordination numbers like 12 more common (than octahedral and tetrahedral structures) among lanthanides and actinides because they have more orbitals for bonding, or is this compound an exception?

After a lot of digging around, I finally found something about this in J.D. Lee's Concise Inorganic Chemistry. I thought I'd post whatever I found over here anyway since it's very interesting:

Answer to question 2

Complex formation by lanthanides is different from that of actinides. In lanthanides, the $\mathrm{4f}$ orbitals are well shielded by the larger $\mathrm{5d}$ and $\mathrm{6s}$ orbitals and are deep inside the atom. So $\mathrm{f}$ orbitals do not participate in any bonding and complex formation is similar to that of transition metals. However in the actinides, the $\mathrm{5f}$ orbitals extend outwards and participate in bonding much more easily making the interactions much more complex than in the transition metals.

Answer to question 3

Higher coordination numbers are apparently extremely common among f-block elements (strange right?).
Octahedral (6) and tetrahedral (4) structures are very rare except when bulky ligands are present. Most common coordination numbers among lanthanides are

7 (Capped trigonal prismatic) yttrium acetylacetonate hydrate (Yttrium seems to be grouped with the f-block elements due to similar properties)
8 (Square antiprismatic and Dodecahedral) cerium acetylacetonate and holmium tropolonate
9 (Tri-capped trigonal prismatic) nonaaquaneodymium(III) complex $\ce{[Nd(H2O)9]^3+}$

10 (very complex) and 12 (icosahedral) are seen only in the larger elements like cerium and thorium with small ligands like $\ce{NO3-}$ and $\ce{SO4^2-}$

The actinides also commonly form exotic structures like chained tricapped trigonal prismatic in $\ce{[ThF8]^4-}$ and $\ce{[PaF7]^3-}$ distorted cubic in $\ce{[PaF8]^3-}$

The text mentions that the nature of bonding in $\ce{[Ce(NO3)6]^2-}$ is still not understood because it would imply bond orders of less than 1 or participation of f orbitals.

Could anyone explain anything about the first question since there is nothing mentioned on the stability of these compounds.

They form complexes compound but their tendency to form complexes is less their transition metal ions .the lanthanide ions spite a high charge have low charge density due to their large size.they form stable complexes with chelating ligands such as EDAT ,oxime beta-ketones.


Why doesn't H₄O²⁺ exist? I know this question has been asked previously but I cannot find a satisfactory explanation as to why is it so difficult for $\ce{H4O^2+}$ to exist. There are explanations that it is so because of $+2$ charge, but if only that was the reason then existence of species like $\ce{SO4^2-}$ should not have been possible.

So, what is exactly the reason that makes $\ce{H4O^2+}$ so unstable?

I myself was always confused why $\ce{H3O^+}$ is so well-known and yet almost nobody talks of $\ce{H4O^2+}$. I mean, $\ce{H3O^+}$ still has a lone pair, right? Why can't another proton just latch onto that? Adding to the confusion, $\ce{H4O^2+}$ is very similar to $\ce{NH4+}$, which again is extremely well-known. Even further, the methanium cation $\ce{CH5+}$ exists (admittedly not something you'll find on a shelf), and that doesn't even have an available lone pair!

It is very useful to rephrase the question "why is $\ce{H4O^2+}$ so rare?" into "why won't $\ce{H3O^+}$ accept another proton?". Now we can think of this in terms of an acid-base reaction:

$$\ce{H3O^+ + H+ -> H4O^2+}$$

Yes, that's right. In this reaction $\ce{H3O^+}$ is the base, and $\ce{H^+}$ is the acid. Because solvents can strongly influence the acidity of basicity of dissolved compounds, and because inclusion of solvent makes calculations tremendously more complicated, we will restrict ourselves to the gas phase (hence $\ce{(g)}$ next to all the formulas). This means we will be talking about proton affinities.

Before we get to business, though, let's start with something more familiar:

$$\ce{H2O(g) + H+(g) -> H3O^+(g)}$$

Because this is in the gas phase, we can visualise the process very simply. We start with a lone water molecule in a perfect vacuum. Then, from a very large distance away, a lone proton begins its approach. We can calculate the potential energy of the whole system as a function of the distance between the oxygen atom and the distant proton. We get a graph that looks something like this:

For convenience, we can set the potential energy of the system at 0 when the distance is infinite. At very large distances, the lone proton only very slightly tugs the electrons of the $\ce{H2O}$ molecule, but they attract and the system is slightly stabilised. The attraction gets stronger as the lone proton approaches. However, there is also a repulsive interaction, between the lone proton and the nuclei of the other atoms in the $\ce{H2O}$ molecule. At large distances, the attraction is stronger than the repulsion, but this flips around if the distance is too short. The happy medium is where the extra proton is close enough to dive into the molecule's electron cloud, but not close enough to experience severe repulsions with the other nuclei.

In short, a lone proton from infinity is attracted to a water molecule, and the potential energy decreases up to a critical value, the bond length. The amount of energy lost is the proton affinity: in this scenario, a mole of water molecules reacting with a mole of protons would release approximately $\mathrm{697\ kJ\ mol^{-1}}$ (values from this table). This reaction is highly exothermic

Alright, now for the next step:

$$\ce{H3O^+(g) + H+(g) -> H4O^2+(g)}$$

This should be similar, right? Actually, no. There is a very important difference between this reaction and the previous one; the reagents now both have a net positive charge. This means there is now a strong additional repulsive force between the two. In fact, the graph above changes completely. Starting from zero potential at infinity, instead of a slow decrease in potential energy, the lone proton has to climb uphill, fighting a net electrostatic repulsion. However, even more interestingly, if the proton does manage to get close enough, the electron cloud can abruptly envelop the additional proton and create a net attraction. The resulting graph now looks more like this:

Very interestingly, the bottom of the "pocket" on the left of the graph (the potential well) can have a higher potential energy than if the lone proton was infinitely far away. This means the reaction is endothermic, but with enough effort, an extra proton can be pushed into the molecule, and it gets trapped in the pocket. Indeed, according to Olah et al., J. Am. Chem. Soc. 1986,  108 (5), pp 1032-1035, the formation of $\ce{H4O^2+}$ in the gas phase was calculated to be endothermic by $\mathrm{248\ kJ\ mol^{-1}}$ (that is, the proton affinity of $\ce{H3O^+}$ is $\mathrm{-248\ kJ\ mol^{-1}}$), but once formed, it has a barrier towards decomposition (the activation energy towards release of a proton) of $\mathrm{184\ kJ\ mol^{-1}}$ (the potential well has a maximum depth of $\mathrm{184\ kJ\ mol^{-1}}$).

Due to the fact that $\ce{H4O^2+}$ was calculated to form a potential well, it can in principle exist. However, since it is the product of a highly endothermic reaction, unsurprisingly it is very hard to find. The reality in solution phase is more complicated, but its existence has been physically verified (if indirectly).

But why stop here? What about $\ce{H5O^3+}$?

$$\ce{H4O^2+(g) + H+(g) -> H5O^3+(g)}$$

I've run a rough calculation myself using computational chemistry software, and here it seems we really do reach a wall. It appears that $\ce{H5O^3+}$ is an unbound system, which is to say that its potential energy curve has no pocket like the ones above. $\ce{H5O^3+}$ could only ever be made transiently, and it would immediately spit out at least one proton. The reason here really is the massive amount of electrical repulsion, combined with the fact that the electron cloud can't reach out to the distance necessary to accommodate another atom.

You can make your own potential energy graphs here. Note how depending on the combination of parameters, the potential well can lie at negative potential energies (an exothermic reaction) or positive potential energies (an endothermic reaction). Alternatively, the pocket may not exist at all - these are the unbound systems.

EDIT: I've done some calculations of proton affinities/stabilities on several other simple molecules, for comparison. I do not claim the results to be quantitatively correct.

$$
\begin{array}{lllll}
\text{Species} & \ce{CH4}  & \ce{CH5+}    & \ce{CH6^2+}     & \ce{CH7^3+}     & \ce{CH8^4+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{Yes}  & \text{No} \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 556 & -246 & -1020   & N/A   & N/A  \\
\end{array}
$$

Notes:

Even without a lone pair, methane ($\ce{CH4}$) protonates very exothermically in the gas phase. This is a testament to the enormous reactivity of a bare proton, and the huge difference it makes to not have push a proton into an already positively-charged ion.
For most of the seemingly hypercoordinate species in these tables (more than four bonds), the excess hydrogen atoms "pair up" such that it can be viewed as a $\ce{H2}$ molecule binding sideways to the central atom. See the methanium link at the start.

$$
\begin{array}{lllll}
\text{Species} & \ce{NH3}  & \ce{NH4+}    & \ce{NH5^2+}     & \ce{NH6^3+}      \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 896 & -410 & N/A   & N/A    \\
\end{array}
$$

Notes:

Even though the first protonation is easier relative to $\ce{CH4}$, the second one is harder. This is likely because increasing the electronegativity of the central atom makes the electron cloud "stiffer", and less accommodating to all those extra protons.
The $\ce{NH5^{2+}}$ ion, unlike other ions listed here with more than four hydrogens, appears to be a true hypercoordinate species.  Del Bene et al. indicate a five-coordinate square pyramidal structure with delocalized nitrogen-hydrogen bonds.

$$
\begin{array}{lllll}
\text{Species} & \ce{H2O}  & \ce{H3O+}    & \ce{H4O^2+}     & \ce{H5O^3+}      \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 722 & -236 & N/A   & N/A    \\
\end{array}
$$

Notes:

The first series which does not accommodate proton hypercoordination.
$\ce{H3O+}$ is easier to protonate than $\ce{NH4+}$, even though oxygen is more electronegative. This is because the $\ce{H4O^2+}$ nicely accommodates all protons, while one of the protons in  $\ce{NH5^2+}$ has to fight for its space.

$$
\begin{array}{lllll}
\text{Species} & \ce{HF}  & \ce{H2F+}    & \ce{H3F^2+}     & \ce{H4F^3+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 501 & -459 & N/A   & N/A    \\
\end{array}
$$

Notes:

Even though $\ce{H3F^2+}$ still formally has a lone pair, its electron cloud is now so stiff that it cannot reach out to another proton even at normal bonding distance.

$$
\begin{array}{lllll}
\text{Species} & \ce{Ne}  & \ce{NeH+}    & \ce{NeH2^2+}    \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{No}   \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 204 & N/A   & N/A    \\
\end{array}
$$

Notes:

$\ce{Ne}$ is a notoriously unreactive noble gas, but it too will react exothermically with a bare proton in the gas phase.
Depending on the definition of electronegativity used, it is possible to determine an electronegativity for $\ce{Ne}$, which turns out to be even higher than $\ce{F}$. Accordingly, its electron cloud is even stiffer.

$$
\begin{array}{lllll}
\text{Species} & \ce{H2S}  & \ce{H3S+}    & \ce{H4S^2+}     & \ce{H5S^3+}     & \ce{H6S^4+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{Yes}  & \text{No} \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 752 & -121 & -1080   & N/A   & N/A  \\
\end{array}
$$

Notes:

The lower electronegativity and larger size of $\ce{S}$ means its electrons can reach out further and accommodate protons at a larger distance, while reducing repulsions between the nuclei. Thus, in the gas phase, $\ce{H2S}$ is a stronger base than $\ce{H2O}$. The situation is inverted in aqueous solution due to uniquely strong intermolecular interactions (hydrogen bonding) which are much more important for $\ce{H2O}$.
$\ce{H3S+}$ also has an endothermic proton affinity, but it is lower than for $\ce{H3O+}$, and therefore $\ce{H4S^2+}$ is easier to make. Accordingly, $\ce{H4S^2+}$ has been detected in milder (though still superacidic!) conditions than $\ce{H4O^2+}$.
The larger size and lower electronegativity of $\ce{S}$ once again are shown to be important; the hypercoodinate $\ce{H5S^3+}$ appears to exist, while the oxygen analogue doesn't.

$\ce{H4O^{2+}}$ would be much smaller than $\ce{SO4^2-}$. The charge density is too high much higher.

The oxygen in water already becomes electron-deficient when it forms $\ce{H3O+}$. In your (not completely, see below) hypothetical $\ce{H4O^{2+}}$, the additional proton barely finds any electron density to make a covalent-ish bond with.

Whereas for the sulfate ion, you just have to distribute some extra charge over a relatively large molecule.

So $\ce{H4O^{2+}}$ exists, at least in theory, but it is extremely unstable. It's properties have been detailed (with links to scientific papers) here: What is the Ka of OH- and Kb of H3O+?


Methods to prepare lactic acid from acetaldehyde I got one method [here](using acetaldehyde and carbon monoxide)1.

But is there any alternative method (using specific catalyst)?

A fairly simple two step synthesis involves nucleophilic addition of $\ce{HCN}$ to the carbonyl group and then aqueous hydrolysis of the nitrile to yield the carboxylic acid product.

Acetylene + $\ce{HCN-Ba(CN)2}$ → vinyl cyanide

Vinyl cyanide - hydrolysis → acetaldehyde cyanohydrin

Acetaldehyde cyanohydrin - dil HCl → lactic acid


Wavenumbers of cyclic ketones Rank the compounds below in order of increasing C=O absorption frequency in IR spectrum (top - lowest wavenumber, bottom - highest wavenumber):

Question: Would I count compound #1's cyclopentane or all 11 carbons because wavenumber increases with 5C or smaller rings.

If I count #1 as a cyclopentane for this purpose then #3 has the lowest wavenumber and #2 has the highest.

Here is nice comparison of all 3 ring sizes. Octahydro-1H-inden-1-one (either cis- or trans-) carbonyl strechings fall between cyclohexanone and cyclobutanone, thus, fused ring is not a factor (it is to some extend because of sunstitution).

#1  Octahydro-1H-inden-1-one C=O stretch from reference

cis 1730 $\text{cm}^{-1}$

trans 1754 $\text{cm}^{-1}$

#2 - Cyclobutanone NIST Listing C=O stretch at 1790 $\text{cm}^{-1}$ from spectra

#3 - Cyclohexanone NIST Listing C=O stretch at 1740 $\text{cm}^{-1}$ from spectra


How can I undo the tarnish on my aluminium headphones? I thought of cleaning my headphones, which has an outer aluminium casing, with Dettol liquid which I believe has strong oxidising agents. This was a bad idea. It left a tarnish on the aluminium and I'd like to know if I can undo this.

I would try any sort of weak acid that wouldn't cause damage to other parts of the headphones.

Citric Acid (found in lemon juice), Acetic Acid (found in vinegar), and Phosphoric Acid (found in cola) are probably all safe bets for sticking on a cloth and wiping off the casing. The primary difficulty you have here is the issue that comes from the fact submerging your headphones in anything would likely damage the electronics inside. You can rig up a shallow flat pan full of acid and resting the headphones on it's side in there, though I'd take caution.

My friend once asked me to clean their silver chain. I used H2O2 as a school chemistry book suggested and it all turned black. They brought it to a jewelry store and the chain was cleaned ... with H2O2/NH3. A friend of mine brought a mixture of H2O2 and NH3 from lab to home to clean some silverware. Premixed, in one bottle. These two chemicals started to react when they were going home by bus. Having a bag that makes a whistling sound and stinks like ammonia while going by bus is not fun.

DIY stain cleaning using lab chemicals often go wrong and rarely worth the trouble.


Why stepwise dissociation for acids but not hydroxide bases? My son asked this over the weekend and I didn't have a good answer: how come with polyprotic acids we assume that the release of each $\ce{H+}$ is progressively harder, but with bases like $\ce{Ca(OH)2}$ we write $\ce{Ca^2+}$ and $\ce{2OH-}$ as the products?

What I know about, say, sulfuric acid, seems to suggest that when we put calcium hydroxide in water we should get $\ce{CaOH+}$ and a single hydroxide, and it should be much harder to release that second hydroxide.  What am I missing?

This is not a general rule. Much like with acids for some bases we write progressive capture of one proton at a time. It all depends on the nature of the acid/base.

In the case of $\ce{Ca(OH)2}$ you don’t have a molecule like $\ce{HO-Ca-OH}$ or something like you can assume for $\ce{HO-SO2-OH}$ (sulfuric acid). Rather, you have an ionic compound composed of two $\ce{OH-}$ ions per $\ce{Ca^2+}$ ion. Once we dissolve this, all the ions are released at the same time resulting in $2~\mathrm{mol}$ of $\ce{OH-}$ per mole of $\ce{Ca(OH)2}$ dissolved.

As a different example take sodium oxide $\ce{Na2O}$. If you slowly add water to that compound the oxide ion will grab one proton at a time resulting in $\ce{2 NaOH}$ first before a second set of protons will generate water. This is much the same mechanism as for diprotic acids. (Note, however, that the oxide ion is such a strong base that it is immediately protonated with a single equivalent of water — much like sulfuric acid is such a strong acid that it will immediately release one proton on contact with water.)

If I would conclude from Jan's statement

Rather, you have an ionic compound composed of two $\ce{OH-}$ ions per $\ce{Ca^2+}$ ion. Once we dissolve this, all the ions are released at the same time resulting in 2 mol of $\ce{OH-}$ ion per mole of $\ce{Ca(OH)2}$ dissolved. [$\ldots$]

then it would be not that difficult to give the following two reactions to answer the original question.
1.
$$\ce{Ca(OH)2 (aq) <=> Ca(OH)+ (aq) + OH- (aq)}$$
2.
$$\ce{Ca(OH)+ (aq) <=> Ca^2+ (aq) + OH- (aq)}$$

Water is already in an ionised-equilibrium state, so there may be availability of protons and hydroxides in the (universal) solvent

Why not try to rearrange the above equation by taking the help of protons contributed from water, let's do it.

Rearranging 1st equation with the help of, protons and hydroxides, on either sides of equation; we get:
$$\ce{Ca(OH)2 (aq) + H+ (aq) <=> Ca(OH)+ + H2O}$$

Again, rearranging 2nd equation with the help of, protons and hydroxides, on either sides of equation; we get:
$$\ce{Ca(OH)+ (aq) + H+ (aq) <=> Ca^2+ + H2O}$$

It would now be easy for us to conclude why the reaction must proceed harder progressively; because as soon as we see the modified the form of reaction 1&2: we notice the repulsive interaction between $\ce{Ca(OH)+}$ ions and $\ce{H+}$ ions. So one may conclude with this pointed statement:

In the case of polyprotic acids we assume that the release of $\ce{H+}$
ions is progressively harder.

said by Asker, has justified why must not "strong bases like Ca(OH)2", "directly form  $\ce{Ca^2+}$ and $\ce{2OH-}$ as the products?", further added by Asker.
Jan says

This is not a general rule.

Much like with acids for some bases we write progressive capture of one proton at a time. It all depends on the nature of the acid/base. In the case of Ca(OH)2 you don’t have a molecule like HO−Ca−OH or something like you can assume for HO−SO2−OH (sulfuric acid).
Rather, you have an ionic compound composed of two hydroxide ions per a calcium ion. Once we dissolve this, all the ions are released at the same time resulting in 2 mol of hydroxide per mole of $\ce{Ca(OH)2}$ dissolved. As a different example take sodium oxide $\ce{Na2O}$.
If you slowly add water to that compound the oxide ion will grab one proton at a time resulting in $\ce{2NaOH}$ first before a second set of protons will generate water. This is pretty much the same thing that happens to diprotic acids. (Note, however, that the oxide ion is such a strong base that it is immediately protonated with a single equivalent of water — much like sulfuric acid is such a strong acid that it will immediately release one proton on contact with water.)*
Jan statement is much correct and easily explains my asserted warning at last sentence under the conclusion.
Conclusion:
So let me again repeat the whole equation but with a slight change, this time for $\ce{H2SO4}$, a strong acid:

Rearranging 1st equation with the help of, protons and hydroxides, on either sides of equation; we get:
$$\ce{HO−SO2−OH (aq) + OH- (aq) <=> O−SO2−OH- + H2O}$$
Again, Rearranging 2nd equation with the help of, protons and hydroxides, on either sides of equation; we get:
$$\ce{O−SO2−OH- (aq) + H+ (aq) <=> O−SO2−O^2- + H2O}$$

Note that ${\rm p}K_{\rm a}$ and ${\rm p}K_{\rm a}$ are like constants whose value does not change until either temperature or some other thermodynamic variable changes (do not confuse with Kinetic Variables, acid dissociation constants have nothing to do with it) are changed (https://en.wikipedia.org/wiki/Acid_dissociation_constant)
TL;DR;

Firstly, determine the Acid Dissociation Constant constant of the given base, acid or mixed salt.

Secondly, determine the Thermodynamic conditions and values of its variables to help you go into detail of the reaction (not necessary, just knowing that it is constant or not, is enough)

Thirdly, write the whole dissociation reaction (only, because those that does not dissociate or whose dissociation constant is too small is generally ignored, i.e., Weak acids and Weak Bases and their mixed salts, can be ignored).

Finally, interpret the interactions between the filtered out dissociation-reactions, then cross out those that are against any of the 1st three rules. You may also filter out on the basis of what are you actually trying to obtain from these reactions. Just keep in mind, that the fundamental rules of Ionic Equilibrium in the context of whole Chemistry. must not be violated.
You will be having only the most important and relevant reactions. Just a few steps.


What solvents work on aloe vera? I'm looking for a solvent which can fully dissolve dried aloe vera gel. I assume the main problem is the constituent polysaccharides.

Hot water, the "universal solvent", should work nicely.


Why no thermal expansion for camping tents? I know that solid materials expand when heated, which is called thermal expansion.

But what is happening to tent square or tent canvas of (plastic?) camping tents? I have noticed, that during the day, when it’s hot, there is a lot of tension in the skin of a tent. But when it gets colder (when raining or during the night), the tension is less and the fabric gets sloppy.

What is the reason?

The surface of a tent consists mostly of water-resistant polymers such as latex or plastic. These polymers are disordered molecules such as spaghetti, so they have unlike crystals a high entropy.

Gough (1805) and Joule (1859) studied the thermal behavior of rubber. They made the following observations, which are known as Gough-Joule effects:
• a taut with constant weight rubber shrinks when it is heated
• rubber emits heat when it is stretched

The cause of the rubber elasticity is fundamentally different than the elasticity of crystalline materials. In contrast to a crystal, the atoms are not regularly arranged in a rubber, but they form long chains (polymers) which are interconnected and form a network. Clearly, one can think of it as a plate of cooked spaghetti. This difference in microstructure are reflected in the mechanical properties of rubber. When the rubber network is loose, very many different configurations (configurations) of the chain between two connection points are possible - more disorder (larger entropy S).

The air inside the tent heats up and expands, the tent, while not completely airtight still acts like a balloon expanding with the heated air inside it and as the air cools down it shrinks, deflating the tent/balloon.


Dissolving sand (or something similar) in a bottle of water? I would like to create an illustration to demonstrate something during a talk. The idea goes that there is sediment in a bottle of water that sits at the bottom until shaken up then adding another substance to the water makes the sediment dissolve or appear to vanish.

My question is; What would make this possible? What materials or chemicals would work?

To clarify: it sounds like you want a layer of insoluble sediment, so that if you shake the bottle, the water looks cloudy. Maybe it settles back out quickly, or maybe the water stays cloudy.

Then, you add something and shake the bottle again. This time, the sediment disappears.

There are lots of possibilities here. One that comes immediately to mind is lime (calcium hydroxide), which is only slightly soluble in water. Add acid (vinegar if you're timid and don't mind adding a large quantity, hydrochloric acid if you want fast results), and it will be converted to a highly soluble salt.

Chalk (calcium carbonate) would work, too, but it would fizz (produce bubbles of gaseous carbon dioxide) when you added the acid.

If you can say a little more about the context of your demonstration, we might be able to provide better ideas. Is this actually about solubility, or is it a metaphor for something else?

You need (1) Any solid in powdered form (2) Recommended: Any other product that would change the nature of the solid to make it soluble in the solvent (let's inaccurately call it "solubility enhancer"), (3) Any solvent in which this solid would dissolve without producing gases or unsoluble co-products.

You would pick the solid and solvent based on:

A) How much solid you would like to dissolve in the liquid. The Ksp (Solubility Product Constant) would be a good indicator of this, telling you how many molecules you can dissolve in a given volume of solvent at the given temperature you'd like to do your demo. If you want to have a lot of sediment that would disappear completely, you would need a compound that is highly soluble in that solvent and at that temperature (in the form resulting from the reaction with the solubility enhancer if you chose to use one).

B) How quickly you want it to dissolve (i.e. "dissapear"). You can influence on this independently of the Ksp in at least to ways: The first is how finely ground your solid is (as it influences its contact surface), you could go from pellets down to "nano-powders", the second is how the reaction itself will influence the conditions determining the speed of the reaction: the solubility of many compounds increases with temperature for instance, and some reactions give out energy (called "exothermic" reactions), which means that you could chose a reaction that increases it's own rate of reaction as it proceeds (carefully within the limits necessary not to increase the container's temperature or internal pressure beyond what would be safe for your demo).

I would recommend one setup whereby:

(1) The solid to be dissolved is first set at the bottom of your container
(2) The solid to be dissolved is covered by a very thin layer of another solid densely packed in such a way as to insulate the first layer as much as possible from the solvent. This solid won't dissolve but the relative quantity is so small that no one will notice.
(3) The solvent is then very carefully added as not to mess up the two layers of powder
(4) You keep the recipient untouched, preferably right were you need it for the demo
(5) Right before your demo you carefully add to the solvent the chosen "solubility enhancer"

As you start shaking the bottle, the insulating solid layer will disperse, allowing for the underlying solid powder/pellet? mass to contact (1) the "solubility enhancer" it will react with to create a more soluble product (2) the solvent it will dissolve in. Now based on the solvent, the solid, and the product you make it react with, you could be shaking for a dozen seconds, or for a couple hours. :)


What are the parameters of protonation of a molecule? Adding an acid replaces the $\ce{-MgBr+}$ with an $\ce{H+}$. Why does this happen? Is $\ce{H+}$ a better electrophile than $\ce{MgBr+}$?

Or, to put it another way, why doesn't $\ce{MgBr}$ stays attached to $\ce{O}$?

We are taking about electrophilicity of ions.
It basically depends upon:
1. charge on the ion(+)
2. density of plus charge on it

So comparing $\ce{H+}$ and $\ce{CH3+}$, the former has a higher plus charge density so that's why it will be more attracted towards 'O'.


What defines the specific enthalpy of atomization? 1 mol of $\ce{F2}$ is turned into $\ce{2F}$ (atoms).

Is this the specific enthalpy of atomization, or two times that?

ChemGuide says that 'The standard atomisation enthalpy is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state.'

For example,
\begin{array}{rc}
\ce{1/2 Cl2 (g) -> Cl (g)} & \Delta H_{a}^{0} = +122~\mathrm{kJ/mol} \\
\ce{1/2 Br2 (l) -> Br (g)} & \Delta H_{a}^{0} = +112~\mathrm{kJ/mol} \\
\ce{Na (s) -> Na (g)} & \Delta H_{a}^{0} = +107~\mathrm{kJ/mol}
\end{array}

Enthalpy of atomisation is energy required to break 1 mole of E-E bonds.

Eg: $\ce{1 F2 -> 2 F}$ .............................$\Delta H_{atomisation}$


Basicity of nitrogen heterocyclic compounds Question :
The order of basicity for the following compounds is :

The logic I've applied is that since 4 has a conjugated system, the lone pair of N will be delocalised and hence would be the least basic. Conjugate systems/Resonance is not seen in the other three compounds
I'm not entirely sure about how to differentiate between the other three. For morpholine 2, If we see the stability of the conjugate acid formed, the O would exert (very less) -I effect, which should destabilize the positive charge on N (on formation of conjugate acid). Now between 1 and 3, 1 has a five other C atoms attached to the N, while 3 has four. Now CH3 is a +I group, hence more CH3 in 1 should stabilize the positive charge on N more than 3.
I don't think Inductive Effect should matter much, considering Inductive Effect becomes very ineffective with increase in chain length. But this is the only reasoning that strikes me. So ultimately, order of basicity becomes 4 < 2 < 3 < 1. Is my reasoning and the final answer correct?

Nitrogen inversion is not relevant to this discussion. A faster rate of inversion does not correspond to a greater delocalisation in the electronic structure, as lone pair inversion is not equivalent to resonance. The lone pair does not exist as a delocalised smear; it exists in two different localised forms.

The PDF on nitrogen inversion in the comments does not even bring up the topic of basicity, and quite rightly so.

Inductive effects explain why morpholine 2 is less basic, and resonance effects explain why pyrrolidone 4 is even less basic. However, since the pKaH's of piperidine 1 and pyrrolidine 3 are so similar (11.22 and 11.27 respectively), I think it is pointless to try and qualitatively rationalise a difference in their basicities. It is like trying to explain why the temperature in Spain one day is 11.22 °C and the temperature the next day is 11.27 °C. The relative basicity could well be different if you switch the solvent, which would invalidate any reasoning based on - for example - s-orbital character of the nitrogen lone pair, which I think is the only plausible qualitative argument here.

Simply looking at the data, though, the order you want is 4 < 2 < 1 < 3.

I would like to answer it,

KEY POINT: More electron density on N, more availability of the lone pair, more good base.

$(1)$ The nitrogen is directly connected inside the ring, surrounded by lots of carbon atoms. N being more electronegative than C, snatches the electrons towards itself as the nearest carbon atoms shows +I effect, it drops down as we move away from N. So, electron density on N increases, which means the lone pair is more available for donation. Hence, it's a good base!

$(2)$ The ring is the same but this time, a highly electronegative O came now. As it is more electronegative than N, it will snatches the electrons towards itself from the downside ring. Thus, decreasing the electron density on N. Hence, it's now no more willing to donate its lone pair. It is more tightly bound to the N as the repulsions from the electrons around it are decreased. Hence, not so good base.

$(3)$ Everything still the same, but the number of carbon atoms reduces to $4$ around it. Hence, +I effect reduces compared to $(1)$. So, electron denisty reduces comapred to $(1)$. Basic character reduces compared to $(1)$.

$(4)$ Less basic than $(3)$ as N got a $\ce{sp^2}$ hybridized carbon around it which is more electronegative than the previous. Hence, that +I effect reduces. So, a weak base than $(3)$ . But, the O here isn't connected directly to the ring. So, it will not effect any other C atom to perform their +I effect. Hence, the density on N reduces a bit, whereas in $(2)$, it reduces a lot as the O here will effect every C to not to give their best at +I effect, it will also snatches electron density around N a bit. Hence, it's the least Basic. So, correct order:

$\ce{(1)>(3)>(4)>(2)}$

Actually I do not at all like chemistry, but i don't know why i posted this as my ever first unexpected answer on this site. This is probably because this topic is my favourite by mistake from my class $12$ chapter: Amines.


Harmonic frequencies (IR) and basis set size Usually vibrations (harmonic), derived from harmonic approximations are bigger than for experimental values. The question is, how does this depend on the size of the basis set. Many papers give that as a statement in their introduction without any explanation.

My idea would be that, if I use variational methods like DFT, then my energy will always be bigger than the real energy, as long as I use a finite basis set. And also increasing the set gives better results in DFT. The harmonic frequencies I determine computationally are derived from the Hessian matrix (or better the force constants for the harmonic oscillator) and this matrix is the second derivative of the energy, which is much too big for small basis sets.

So can I say that, because a smaller basis set will give me a bigger energy, the derived frequency will have to be bigger as well?

No, you cannot say that larger energies will necessarily give larger frequencies.

Consider, for instance, a hypothetical (and purely fictional) method of determining the electronic energy of a system for a given geometry. Call this set of nuclear coordinates (geometry) $\mathbf{G}$, and our hypothetical method is a function which acts on these coordinates to produce an energy, i.e. $$f(\mathbf{G})=E$$

This method, however, is a very special kind of method. It is one which gives an absolutely unreliable energy. It gives you a number that has absolutely no basis in reality and can't be trusted at all. But, one thing we know about this method is that the error, which is truly horrendous and unpredictable, is constant! From this fact, it follows that this method predicts the correct frequencies, the experimental ones. That is, at the minimum of the potential predicted by our hypothetical method and the actual minimum, the difference between our energy, $E_0$, and the true energy, $\mathcal{E_0}$, is,
$$E_0-\mathcal{E_0}=\epsilon$$
or,
$$f(\mathbf{G})-\mathcal{E}_{\mathbf{G}}=\epsilon$$
For all real methods, however, we would need to write the above equation as,
$$h(\mathbf{G})-\mathcal{E}_{\mathbf{G}}=\epsilon({\mathbf{G}})$$

The point of all this being that the hypothetical method I described was not variational or anything else you might desire out of a quantum chemical method. It only predicted energies for which the error term was constant. Whereas the real method, $h(\mathbf{G})$, predicts energies for which the error term is a function of the geometry $\mathbf{G}$.

The concise way of saying this is that frequencies depend only on the curvature of the potential well and not on the depth of the potential well.

With that established, we have answered the question that smaller basis sets, with their larger energies, do not necessarily give higher frequencies.

For instance, if you look at Sinha and Wilson's paper,[1] you will learn of something very empirical which computational chemists do sometimes that bothers me a lot but is illustrative for our purposes. In that paper, a big set of 42 organic molecules is taken, and the harmonic frequencies are calculated at a bunch of different methods and basis sets, then these frequencies are compared with the experimental values. A least squares fit is then performed to come up with scaling values to map the harmonic frequency onto the experimental frequency. As you say, these constants are numbers less than one, say 0.95. The reason these constants are less than one is what bothers me about doing this. Experimental values are anharmonic frequencies. We don't live in a harmonic world. Anharmonicity shifts the frequency down, by lowering one of the potential walls to accommodate bond-breaking, hence we need to scale our harmonic frequencies downwards. So we're basically comparing apples with oranges and there's no way to be sure that this will work in all cases.

More to your question, in this paper, the correction factors do not increase monotonically with basis set size as you suggested (this is the same as frequencies decreasing with basis set size). I won't reproduce their numbers here, because I don't know if that's frowned upon, but for both MP2 and B3LYP, the cc-pVnZ and aug-cc-pVnZ (n=D,T,Q) basis sets have T < D < Q for the scaling factor. So, the triple-zeta basis set invariably predicts a lower energy, but predicts a higher frequency.

Even more interestingly, if one only considers low-frequency modes, which are particularly poorly described by normal modes (these are torsions usually), the ordering becomes Q < T < D for the scaling factors. That's really a failure of the harmonic model more than anything, but it illustrates the point well that lowering the energy really doesn't guarantee you anything in terms of frequencies.

To address your comment, the reason there is any dependence on basis set size is because the error away from the equilibrium structure will change with basis set size, and hence the curvature of the potential changes which affects the frequency.

[1] Pankaj Sinha, Angela K. Wilson, J. Phys. Chem. A, 2004, 108 (42), 9213-9217. DOI: 10.1021/jp048233q

smaller basis set would give a bigger energy that is sum of BO electron energy,but the frequency is derived from this energy


Extracting silver from Tollens test When we add tollens reagent (diamminesilver(I) cation, $\ce{[Ag(NH3)2]+}$) to glucose it forms a silvery mirror on the walls of the container. Since the sides are coated with actual Silver,

is there any possible reagent/method to take the silver out from the sides of the test tube.?
will it be in its pure form or will there be organic/inorganic  impurities?

Silver's pretty soft—it's possible to just scrape it off with a metal spatula or something. Also, not all of the silver deposits on the glass. You might be able to adjust the conditions of the reaction to make more silver precipitate into the liquid where it's easier to get at. You can, of course, redissolve the silver in nitric acid or the like.
Since the byproducts of the reaction are soluble, it should be pretty pure, depending on how pure the starting materials are. If other easily reducible metal cations are present, you might get some impurities from that.

As a side-note, this isn't a particularly efficient way of getting elemental silver. The typical films produced in a test tube probably contain a couple hundred milligrams at most. Even a whole undergrad lab's worth of tubes is probably not worth recovering.

1) Silver is pretty easily dissolved in concentrated nitric acid (attention, nitrogen oxides fumes). The resulting solution may be gently evaporated to get rid of excess of nitric acid and silver nitrate may be then reduced with zinc and then washed from excess zinc with diluted sulfuric acid. Alternatively, silver nitrate may be used as is for similar tests in future.

2)As Michael noted, it is pretty pure.


Is iron in the brown ring compound in a +1 oxidation state? In the standard brown ring test for the nitrate ion, the brown ring complex is:
$$\ce{[Fe(H2O)5(NO)]^{2+}}$$

In this compound, the nitrosyl ligand is positively charged, and iron is in a $+1$ oxidation state.

Now, iron has stable oxidation states +2 and +3. Nitrosyl, as a ligand, comes in many flavours, of which a negatively charged nitrosyl is one.

I see no reason why the iron doesn't spontaneously oxidise to +3 and reduce the $\ce{NO}$ to −1 to gain stability. But I don't know how to analyse this situation anyway. I think that there may be some nifty backbonding increasing the stability, but I'm not sure.

So, why is iron in +1 here when we can have a seemingly stable situation with iron in +3?

According Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of
Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction
Revisited

The correct structure is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$

For many years it was thought that iron was reduced to $\ce{Fe^{I}}$ and $\ce{NO}$ oxidized to $\ce{NO+}$, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin $\ce{Fe^{III}}$ ($S=5/2$) antiferromagnetically couples with $\ce{NO-}$ ($S=1$) for an observed spin of $S=3/2$.

The oxidation state of Fe is +1 and not +3 as most answers have stated
NO+ is a strong ligand from crystal field theory(not so sure but please let me know if it is wrong ) hence Fe is in +1 Oxidation state

And as far as@Davephd answer is considered even in +1 state the spin of the compound is 3/2 as NO+ is a strong ligand

Also NO is a paramagnetic substance .so the unpaired electron it can give up to Fe to reduce it from Fe +2 to +1 and No becomes NO+

Fe+ in presence of NO+ the electonic configuration changes to 3d(2,2,1,1,1)(i dont know how to edit it so please forgive me) and so in the complex sp3d2 hybridisation takes place
And so the spin still remains 3/2 here


Determining whether a compound is ionic or molecular in nature $\ce{HCl}$ is a molecular compound because there is a covalent bond between $\ce{H+}$(a proton) and $\ce{Cl-}$ (a chloride ion)
$\ce{NH4NO3}$ is an ionic compound because there is an ionic bond between $\ce{NH4+}$ (an ammonium ion) and $\ce{NO3-}$ (a nitrate ion) even though all the atoms comprising the molecule are non-metal.
Then, which of the following is true?
$\ce{H3C-COOH}$ is an ionic compound because there is an ionic bond between $\ce{C2H3O-}$ (an acetate ion) and $\ce{H+}$ (a proton)? OR
$\ce{H3C-COOH}$ is a molecular compound because there is a covalent bond between $\ce{C2H3O2-}$ (an acetate ion) and $\ce{H+}$ (a proton)?

How do you tell if a bond between the two ions are molecular or ionic? (regardless of whether they are polyatomic or monoatomic ions)

In principle, when studying a structure like HCl, CH3COOH, or any other compound, the first thing to do is drawing the symbol of the atoms in a reasonable geometry on a sheet of paper. Then consider the outer electron layers : put enough points around each symbol to get a reasonable Lewis structure. For example, one point near $\ce{H}$ and six points around $\ce{O}$. Then try to join atoms with their neighbors to make covalencies (or doublets). When all atoms are joined to neighbors, count the number of electrons around each atom, taking into account that the two electrons included in a covalence between A and B  are at the periphery of both atoms A and B. Check all atoms. If an atom respects the octet law, don't modify it : its structure is covalent. If an atom A does not respect octet law, try to remove one electron from its outer layer, and send this electron to the B atom, which becomes negatively charged. The covalency disappears from the region between A and B, and the bond AB becomes ionic with a positive charge on the A atom, which becomes $\ce{A+}$, and a negative charge on $\ce{B}$ which becomes $\ce{B-}$.
Using this method, you will see that $\ce{HCl}$ and $\ce{CH3COOH}$ are both made of covalencies, because all their atoms are respecting the octet rule. You will also see that $\ce{NaCl}$ may first be drawn in a covalent structure. But then the sodium atom does not fit the octet rule. It will if an electron is removed from the sodium atom, to be given to the chlorine atom, producing then $\ce{Na+}$ and $\ce{Cl-}$ ions.
Hydroxides contains two sorts of bonds. Hydroxides like $\ce{NaOH}$ are made of covalent bonds between $\ce{O}$ and $\ce{H}$ atoms, and of ionic bonds between $\ce{Na^+}$ and $\ce{O-}$.
Oxygenated acids are a special case. They may get two sorts of bonds between $\ce{A}$ and $\ce{B}$. For example sulfuric acid $\ce{H2SO4}$ can be drawn with this method to give first a covalent structure with two bonds $\ce{H-O}$, two bonds $\ce{O-S}$, and two double bonds $\ce{S=O}$. This is OK for all atoms except the central Sulfur atom, which is surrounded by six doublets or six covalences. At this point, some chemists are thinking that it is a case of hyper valence. Some think that the octet rule may be obtained by removing the second bond from each double bond from the sulfur atom, and sending it to the outer Oxygen atom. This makes a central sulfur atom charged $2+$ (respecting the octet rule) and two oxygen atoms charged $1-$. Both theories can be discussed.
The only problem with this reasoning is the fact that it does not work properly  for atoms of the third column. For example, $\ce{BF3}$ should be ionic making $\ce{B^{3+}}$ and $\ce{F-}$ ions. Indeed $\ce{BF3}$ remains covalent, and it is hard to justify. Maybe, on may admit that Boron atom prefers to stay covalent rather than to loose three electrons. Three is probably too much.

TL;DR The correct answer is:
$\ce{H3C−COOH}$ is a molecular compound because there is a covalent bond between $\ce{C2H3O2−}$ (an acetate ion) and $\ce{H+}$ (a proton).
Compounds consisting of non-metals are usually considered molecular (or covalent compound). The given compound $\ce{H3C−COOH}$ is acetic acid (or ethanoic acid) and consists of only non-metals, i.e, Carbon, Hydrogen and Oxygen. There is a difference of less than 2.0 in electronegativity on Pauling scale between the Carbon and Hydrogen. This difference results in an electrons being shared between the non-metals. The shared pairs or bonding pairs is covalent bonding.

The Carbon–hydrogen bond and the Carbon–carbon bond are covalent bond as the carbon shares its outer valence electrons with up to four hydrogens/carbons, as shown in the Lewis structure. Infact, all the carboxylic acids having $\ce{R−COOH}$ are covalent.
In general, covalent compounds do not conduct electricity when dissolved in water. Though acetic acid is covalent it dissociates into acetate ions and hydrogen ions. Acetate ions are responsible for conducting electricity Thus, acetic acid is molecular (or covalent) as a pure liquid (and doesn't conduct) BUT, when it is dissolved in water, they become Ionic Solutions (some of the molecules break apart and form ions).
Please note that bonding is a scale, with ionic on one end and covalent on the other. Bonding has both ionic and covalent character. It really just depends where the bonding electrons are located - between the atoms (covalent), or on the negatively charged ion (ionic).
I also found these videos which may be helpful: Is CH3COOH (Acetic acid (Ethanoic acid)) Ionic or Covalent/Molecular? and CH3COOH Lewis Structure (Acetic acid) by Wayne Breslyn.
Learn more about covalent bonds at Khan Academy and Wikipedia.


How can you predict a double displacement reaction with no precipitate? Is it possible to predict a double displacement reaction between two ions where the products remain in aqueous solution? I see how you can predict precipitation reactions if you know whether the product(s) is/are soluble, but that doesn't apply to all possible double displacement reactions.

If there is no precipitate, then there is nothing to drive the reaction. Both possible products would be soluble, and so there is no net reaction (it's just the reactant ions in solution).

Prediction, you say, is possible if you know the reactivity series like this for an elemtary level.


Why does the trimethylammonium cation have a larger -I effect than the ammonium cation? Why does $\ce{N(CH_3)_3^+}$ have a larger $-I$ effect than $\ce{NH_3^+}$?

Since the methyl group is good at donating electrons it would stabilize the charge on nitrogen atom, decreasing its potential to withdraw electrons hence its $-I$ effect.

I can't figure out why its the other way around. Can someone help me understand this?

$\ce{N(CH3)3+}$ has more -I than $\ce{NH3+}$
, this irony happens because if you consider
$\ce{N-CH3}$ bond and $\ce{N-H}$ bond,  which is more polar?
Of course $\ce{N-H}$ bond will be more polar  due to more difference in electronegativity  value,  so that implies electron density will be more on $\ce{N}$ in $\ce{NH3+}$ than $\ce{N}$ on $\ce{N(CH3)3+}$.
And therefore since electron density is less on $\ce{N}$ of $\ce{N(CH3)3+}$ , it will have greater -I effect.

The reason is that the methyl  though gives electron via +I effect in case of H it is easy to cleave the N-H bond for nitrogen and release proton and take the electron pair from hydrogen but with C it being more electronegative than H the same is not possible .


Do ligands in ionisation isomerism interchange? The text says:

  This form of isomerism arises when the counter ion in a complex
  salt is itself a potential ligand and can displace ligand which can
  then become the counter ion.

For other structural isomers that I studied in coordination compounds I simply thought that those  compounds (structural isomers) separately exist. In fact I think of structural isomers in coordination chemistry this way only that these are independently existing with different properties but same molecular formula so we study them.

But the text above is confusing it seems that they convert like resonating structures or conformation isomers.

What is true? If they do interchange then isn't this a contradiction to Werner's theory that postulates that secondary valencies are non ionisable.

note:the below problems is related to one of the comments:

    Please Consider this problem-"Give evidences that [Co(NH3)5Cl]So4 and [Co(NH3)5 So4]Cl are ionisation isomers".

  Here we test by dissloving them in water and reacting them separately
  with silver and barium ions  to test for counter ions.One gives
  AgCl and other BaSo4 as ppt. .Now if there is equlibrium like keto-
  enol tautomerism then only one of the two isomers should be stable in water
  then why such kinds of evidences are given?

Actually, in the ionisation isomerism the ligands do not interchange. They get displaced. Thus one ion becomes counter ion for the other. When one has look at the molecular formula it seems the ligands in the coordination sphere are interchanged with the ligands outside the sphere. So, this does not cross out Werner’s theory as there is displacement of ion and not just interchange.


Do endothermic reaction have higher energy products than reactants? If a reaction absorbs energy does  that mean it has a net gain of energy?

The simple answer is yes - as energy is required to be absorbed for the reaction, the products have more energy than the reactants.

Once you give enough energy to reactants to cross the activation energy barrier it would react to Form the compounds, if the net energy is released, then the products will have lower potential energy or else (as in your case) the potential energy of products will be more than that of reactants


Colour intensity of transition metal complexes Even though they are both $\mathrm{d^7}$ complexes, the colour of $\ce{[Co(H2O)6]^2+}$ is pale whereas that of $\ce{[CoCl4]^2-}$ is more intense. Does the octahedral or tetrahedral geometry matter in determining the molar absorption coefficient of the two species?

At first, it is necessary to ask why some transition metal complexes are colourful at all — remember that alkaline metals, alkaline earth metals and earth metals usually form colourless solutions, e.g. sodium, magnesium, aluminium. The transition metals’ colour stems from the d-electrons they have and the shape of the complexes they create. A typical orbital scheme of a 3d-metal in an octahedral complex will look something like this (image originally taken from Prof. Klüfers web scritptum to his coordination chemistry course at the LMU Munich):

On the left are the metal orbitals, from top to bottom it is 4p, 4s and 3d. On the right are the ligand group orbitals, assuming six ligands each only donating electrons in a $\unicode[Times]{x3C3}$ fashion only. Any remaining metal electrons will be filled into the $\mathrm{t_{2g}}$ orbitals which correspond to $\ce{d}_{xz}$, $\ce{d}_{xy}$ and $\ce{d}_{yz}$ and into the $\mathrm{e_g}$ orbitals corresponding to $\ce{d}_{z^2}$ and $\ce{d}_{x^2-y^2}$

The energy difference marked as $10\,\mathrm{Dq}$ in the image would then be the energy of a photon required to excite an electron from $\mathrm{t_{2g}}$ to $\mathrm{e_g}$; this photon would thereby be absorbed, which is where we get our absorption coefficient from. It is usually in the visible range of light.

There are a few quantum mechanical principles that allow or disallow excitation from one energy level to another. One of these is the spin-allowance rule: Excitations are only allowed if the spin of the entire system remains unchanged. (So excitation cannot involve a spin exchange.) This is why $\ce{Mn^{II}}$ or high-spin $\ce{Fe^{III}}$ complexes are generally colourless: all d-electrons are of identical spin so there is no freedom to be excited.

A second rule is the Laporte-rule whose formal definition I never properly understood, but which states that d to d excitation is forbidden in centrosymmetric species such as octahedra. This rule would mean that all d-element complexes should be colourless. However, complexes are not static but the ligands vibrate back and forth, often in a non-centrosymmetric way. Whenever a photon of the correct energy hits a complex that is not ideally centrosymmetric, the excitation is allowed and the photon absorbed. Because this is a rather rare phenomenon, the colour of such a complex is weaker than it could be.

The $\ce{[Co(H2O)6]^2+}$ case is more complicated again, because in a perfectly centrosymmetric world it would be degenerate: Since it is $\ce{d^7}$, two spin-down electrons share three $\mathrm{t_{2g}}$ orbitals. It would be favourable to somehow make these orbitals dissimilar. This can happen by distorting the octahedral shape in a Jahn-Teller manner: The two water ligands on the $z$ axis are moved slightly further away, creating a slight stabilisation of all orbitals which contain a $z$ component, most notably $\ce{d}_{xz}$ and $\ce{d}_{yz}$. But since the final product is still centrosymmetric, this paragraph only explains the wavelength of absorption slightly better and can be disregarded.

Next question is why the chlorido complex of cobalt(II) is not octahedric but tetrahedric. This is due to a hypothetical $\ce{[CoCl6]^4-}$ being unbalanced in terms of large chloride ions around a small cobalt ion that only has a low charge to combat the $-6$ the chlorides have to offer. The tetrahedral complex $\ce{[CoCl4]-}$ has everything that the octahedral $\ce{[Co(H2O)6]^2+}$ has except the centre of symmetry. Because centrosymmetry is missing, the Laporte rule is no longer valid and every photon of the correct energy can excite electrons. More excitation equals a stronger extinction.

Also, the colour of both complexes is notably different because the orbital scheme is, too. $\mathrm{t_2}$ and $\ce{e}$ are now much closer together and their order is flipped. Combining that with chloride ions that usually generate low values of $10\,\mathrm{Dq}$, the energy difference is even smaller therefore the excitation wavelength longer therefore the colour we see blueshifted (complementary colour).

In octahedral there is inversion centre so centre of symmetry is there so there is no chance of p-d mixing.. So they are laporte forbidden..... But in tetrahedral complexes there is no centre of symmetry.. So there is a chance of p-d  orbital mixing.. Therefore these are laporte partly allowed.. So intensity of color is more. In case of CoCl4  2- than ur octahedral complex