text:
Why is the ionic radius of Al(3+) smaller than that of Li+?
I was examining the ionic radii of some ions from this site for a school assignment. I noticed a weird anomaly in the ionic radius of $\ce{Li+}$  as compared to that of $\ce{Al}^{3+}$.
The ionic radius of $\ce{Li+}$ is about 2/3 as large of that of $\ce{Al}^{3+}$. How is this possible, since the electronic configuration of $\ce{Al}^{3+}$ is $[1s^2, 2s^2 2p^6]$ whereas that of $\ce{Li+}$ is $[1s^2]$? It appears that despite the presence of an extra shell in $\ce{Al}^{3+}$, it has a smaller ionic radius than $\ce{Li+}$.
The answers in this post suggests an exactly opposite outcome. All the answers says that $\ce{Mg^{2+}}$ is above $\ce{Ca^{2+}}$, thus it has a smaller radius. By applying  the exactly same logic i can say $\ce{Li+}$ is above $\ce{Al^{3+}}$ so it must have less ionic radius but that is not the case.
As the below picture suggests,(taken from the same post).

I think it is because of diagonal relationship. As $\ce{Mg^{2+}} = \ce{Li+}$ because of diagonal relationship and $\ce{Mg^{2+}} \gt \ce{Al^{3+}}$ because ionic radius decreases for cations along the period by going left to right.
Am I correct?

The ionic radius is not the radius of the particle; it is the radius of its influence.

Al^3+ is shorter because it can attract anions closer.

http://chemistry.stackexchange.com/questions/49863/comparing-radii-in-li-be2-mg2-al3-and-na

@user34388 Why does Li+ has more radius of influence ? Is it because of diagonal relationship ?

Please don’t ‘nag’ us to answer your question. We are all volunteers volunteering our free time. If you want to increase the attention your question gets, you should add a bounty. However, please pay attention to the caveats presented therein.

The size of the 1s orbital in $\ce{Li+}$ and $\ce{Al^3+}$ is not necessarily the same. In fact, they are quite different because of the much larger effective nuclear charge in $\ce{Al^3+}$. One can easily look up the wavefunction for the 1s orbital and see the radial dependence on $Z_\mathrm{eff}$.
Therefore, merely looking at the electronic configuration cannot tell you anything about the size. If that was the case then one would expect $\ce{Cl-}$ and $\ce{K+}$ to have exactly the same size, which is clearly not true.

All the answers says that $\ce{Mg^{2+}}$ is above $\ce{Ca^{2+}}$, thus it has a smaller radius. By applying  the exactly same logic i can say $\ce{Li+}$ is above $\ce{Al^{3+}}$ so it must have less ionic radius but that is not the case

You are missing the point here entirely. Mg and Ca are in the same group which is what makes them directly comparable. Li is not "above" Al any more than F is above Na.

I think it is because of diagonal relationship. As $\ce{Mg^{2+}} = \ce{Li+}$ because of diagonal relationship and $\ce{Mg^{2+}} \gt \ce{Al^{3+}}$ because ionic radius decreases for cations along the period by going left to right.

The diagonal relationship is something that is observed, not a fundamental principle of chemistry. It still has to be rationalised. Saying that it is "because of the diagonal relationship" is akin to saying "electrons move very fast because they have a high velocity".
The diagonal relationship holds because 1) you add an extra shell of electrons going vertically from $\ce{Li+}$ to $\ce{Na+}$, which decreases $Z_\mathrm{eff}$ and increases ionic radius; and 2) you go horizontally from $\ce{Na+}$ to $\ce{Mg^2+}$, which increases $Z_\mathrm{eff}$ (same number of electrons but one more proton) and decreases ionic radius.
Going one more step horizontally to $\ce{Al^3+}$, one can see that $Z_\mathrm{eff}$ should increase again. So your thoughts on the matter are correct, but they just do not get to the crux of the matter.
The origin of the much smaller ionic radius of $\ce{Al^3+}$ is therefore directly attributable to the much larger $Z_\mathrm{eff}$ in $\ce{Al^3+}$ on the 2p valence electrons, compared to that on the 1s valence electrons in $\ce{Li+}$.
Be careful that the ionic radius is not always particularly well-defined and the values can vary from one source to another depending on how they are experimentally obtained.

So, $Z_{\text{eff}}$ is the cause of the problem. There are  many factors deciding physical properties, How would i know where does which apply ? Sorry for so much trouble and thanks for the well detailed answer.

By reading and learning more. There's no way around it - practice makes perfect.

No doubt about that. Can we obtain ionic radii  from just quantum theory equations ?

You would have to have a strict definition of the ionic radius, and the definitions that I am aware of do not involve quantum calculations on free ions, but I assume that if you make up a definition, it would be possible with the computational power we have now to determine ionic radii. Don't quote me on it, though.

Isn't half the distance between a cation and a anion is the ionic radius of cation and anion ? is anything wrong with this definition.

That has nothing to do with quantum theory though. That is the definition one might use when carrying out X-ray crystallography on a solid. Which is why I said "the definitions that I am aware of do not involve quantum calculations on free ions".

The main difference between $\ce{Li+}$ and $\ce{Al^3+}$ is their charge. Aluminium has three times lithium’s charge.
When calculating orbitals, the first step is to set up a Schrödinger equation and then to see what the Hamiltonian operator is made up of. If you didn’t understand that sentence, just rephrase it in your mind to ‘what is the energy of an electron and which forces does it encounter?’ The most important force here is the nuclear charge, which creates a spherical potential energy gradient — the closer to the nucleus an electron is, the lower its potential energy. (You can think of this like a mountain — the lower you are the less potential energy you have.)
The next step is to solve (or approximate — we cannot solve it for systems with more than one electron) the Schrödinger Equation. Its solution will give you orbitals, spherical wave functions. To draw the typical picture of an orbital, we need to define a cutoff value first; this could be ‘the electron has a probability of $95~\%$ of being found within this boundary.’ And only after that can you talk about orbitals and their sizes.
Remember again that the effective nuclear charge is pretty much the most important thing in the life of a (core) electron. Doubling or tripling it increases the forces the electron is subjected to proportionally. Thus, the larger the charge the closer the electrons will come to the nucleus, or the quicker the wave function decays. This is also the reason why atomic radii decrease across a period: a higher effective nuclear charge exercises greater force on the surrounding electrons and draws them closer towards itself.

This part was the theory. What about the practice? Well, ionic radii are even less well defined than covalent radii; it basically comes down to obtaining a crystal structure of a specific salt and measuring distances. So all that the smaller ionic radius tells us is that anions can approach aluminium even closer than they can lithium. Again, this should come as no surprise, as the anions (being charged bodies by themselves) are also more strongly attracted by a triple positive charge.


text:
Can a drink with such composition have 0 caloric value (whether manufacturer lied)?
I bought a drink which have next statements:

The nutritional value, g/100 ml:
      no proteins, fats, carbohydrates.
Energy value (calorific value):
      kJ / 100 kcal to mg: absent.

Can a drink with such composition have 0 nutritional/calorific value ?
Composition:

prepared drinking water,
flavoring base "Cola"

dye: caramel,
acidity regulator: phosphoric acid,
caffeine - no more than 150 ml/g,
natural flavor enhancer,

sweetener

aspartame,
acesulfame potassium,
sodium saccharin,
sodium cyclamate,

preservative sodium benzoate.

It contains a source of phenylalanine.

Ice water has a negative caloric value.

@MaxW That piqued my curiosity. Do you have a source/explanation?

You have to keep in mind the bane of all introductory chemistry students: excess precision and rounding. Every measurement has error bars, even if those error bars are implicit. It's senseless to report a value that has (much) more precision than your error bars.
A value that's measured as zero isn't necessarily exactly zero. ... it's only "mostly zero".
The particular rounding rules that are used in nutritional labels depend on the specific regulatory agency, but in the US, the FDA provides rounding rules for nutritional labels. It counts anything less than 5 (dietary) Calories per serving as being zero. Likewise, anything with less than 0.5 g of protein, fat or carbohydrate can be listed as having zero grams.
This is how Tic-Tac breath mints can be advertised as being "zero calorie" despite being made almost entirely of sucrose (table sugar): a serving contains less than 5 Calories, and so can be rounded to zero on the label.
In your case, there's a few things which could contain calories. Carmel color is made from sugar, and may contain residual saccarides. Aspartame is a methyl ester of a dipeptide, which is converted in the body to its constituent amino acids, providing a small amount of calories. (Gram for gram, aspartame would contain almost the same amount of calorific energy as sugar does. It's about 200 times sweeter than sugar, though, so on an equivalent sweetening basis it has much fewer calories.) Depending on what the "natural flavor enhancer" is, that might also contain carbohydrates, proteins or fats. (Natural flavors are often fats or fat-soluble, so flavorings often contain trace amounts of dietary fats.)
For a single serving (and for 100 ml) there's only a small amount of carbohydrate, only a scant amount of amino acids, a negligible amount of fat, and an insignificant number of calories. They're a rounding error. But it is true that if you had enough of the drink, you could indeed have an appreciable number of calories, carbohydrates, protein, etc. It's just that it would come along with enough water to make actually consuming that amount physically unfeasible.

The only foodlike ingredient I see there is caramel, and there might be too little of it to matter.
So I would say this drink can have 0 nutritional value.
Not that that's a bad thing. Water has 0 nutritional value too, and all the other ingredients are probably common in drinks.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Van der Waals pressure
I was solving some problems and I had a doubt a problem since the resulting pressure gives me a negative value. The problem was the following:

Could $100~\mathrm{g}$ of $\ce{N2}$ in a $200~\mathrm{mL}$ container exert a pressure of $25~\mathrm{atm}$ at $28~^\circ\mathrm{C}$ if it behave like an ideal gas? If your answer is no, what pressure would exert the gas, knowing that the van der Waals constants for $\ce{N2}$ are $3.11~\mathrm{\frac{L\,atm}{mol^2}}$ and $0.076~\mathrm{\frac{L}{mol}}$.

I have converted the 100 grams of $\ce{N2}$ to moles of $\ce{N2}$, and with that I've replaced the data in the van der Waals formula, using $a = 3.11~\mathrm{\frac{L\,atm)}{mol^2}}$ and $b = 0.076~\mathrm{\frac{L}{mol}}$. The result was a negative pressure, and I am pretty sure that's impossible (or at least that's what my chemistry teacher said).
This is what I did:
$$\begin{align}
P &= \frac{nRT}{V-nb} -\frac{an^2}{V^2}\\
  &= \frac{(3.57~\mathrm{mol})(0.082\mathrm{\frac{L\,atm}{K\,mol}})
    (301.1~\mathrm{K})}{(0.2~\mathrm{L})
   -(3.57 ~\mathrm{mol})(0.076\mathrm{\frac{L}{mol}})}
   -\frac{(3.11~\mathrm{\frac{L~atm}{mol^2}})
    (3.57~\mathrm{mol})^2}{(0.2~\mathrm{L})^2}
\end{align}$$
The result was:
$-1626.19~\mathrm{atm} - 9.9075~\mathrm{atm} = -1636.0975~\mathrm{atm}$

I think we need more details about what you did. With what you've given I cannot tell where your mistake is.

I just didn't want it to be taken as "homework", but I alredy edited the post.

Ok. Firstly, the constant $a$ has units of $\mathrm{L^2~atm~mol^{-2}}$ (not $\mathrm{L~atm~mol^{-2}}$), but I am guessing it is just a typo on your part. The issue seems to arise from the values given for $a$ and $b$. Wikipedia gives the constants as $1.352$ and $0.0387$ respectively (converted to the same units as yours). Using those values I get a pressure of 997.3 atm.

As soon as you got $V<nb$, you should have stopped and asked yourself "What on earth am I doing?" The van der Waals formula is not supposed to work in these conditions, not even approximately. As soon as you are there, you could be dead sure something's seriously amiss (like, a typo in the problem statement, or one of the values is in the wrong units, or whatever).

Let's get the obvious out of the way: The answer to that exercise is yes. Hence it is completely unnecessary to do the second part. With the parameters given in the exercise I calculated a pressure of about 440 atmospheres using the ideal gas law.
But then since we're all here to learn, let's also have a look at the second part. The good news is that I can't see anything wrong with how you used the equation. The problem is already pointed out by orthocresol in the comments, which I will quote here for safekeeping (emphasis by me):

Ok. Firstly, the constant $a$ has units of $\mathrm{L^2~atm~mol^{-2}}$ (not $\mathrm{L~atm~mol^{-2}}$), but I am guessing it is just a typo on your part. The issue seems to arise from the values given for $a$ and $b$. Wikipedia gives the constants as $1.352$ and $0.0387$ respectively (converted to the same units as yours). Using those values I get a pressure of 997.3 atm.

I would assume that nobody caught this mistake earlier since you are not actually supposed to do that part of the question.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why is the energy of an electron always negative?
I just wrote my chemistry exam a few hours ago. However, I am still confused about the following question:

Why is the energy of an electron in an orbital always negative?

I wrote that from the following equation
$$E_n = -R_h\left(1-\frac{1}{n^2}\right),$$
we can see that the energy will always be negative.
Is my answer correct?

It is right, but does not actually explain anything.

Energy of electron in n th orbit.

Since Rydbergs  constant is negative

Energy too negatuve

Rydberg constant is positive. It is that minus before the whole thing that makes it negative. So you basically say that energy is negative because it has a minus before it. Which, like I said, is true, but not much of an explanation.

Negative with respect to what?

Technically it is a physics.se question.
Commonly energy of electron in a molecule/ion is measured relatively to free electron in vacuum, which is assumed to be zero. Since electron in orbit needs to have energy lower than that in vacuum, or it shall leave the particle, it is lower than zero. Typically it happens thanks to interaction with positive charge of atomic nucleus.
Thus, the answer for your question is: because otherwise it shall leave the particle.

First of all, regarding your comment, Rydberg's constant is not negative. If it were, then the equation you gave would have a positive energy all the time.
To be precise, Rydberg's constant is $$R_{\infty}=\frac{m_ee^4}{8\epsilon_0^2h^3c}$$ where all those constants have their usual meanings. Importantly, all of those constants are manifestly positive.
Whether the energy is positive or negative is entirely arbitrary. The energy of an electron can also be formulated as, $$E_n=R_h(1-\frac{1}{n^2}).$$
The only difference between the equation you wrote down and the one I just wrote down is that the above equation sets the zero of energy at the ground state of the hydrogen atom. You'll see that when $n=1$ the energy is zero and as $n\to\infty$ the energy approaches the Rydberg constant which is the ionization energy of the hydrogen atom.
When you say you wrote your exam I assume you mean you took your exam, not that you were writing the exam? Haven't heard that expressions before, but I suppose it makes sense. Maybe you were supposed to write that the negative energy represents the fact that the electron in a hydrogen atom is more stable (lower in energy) when associated with a proton than when isolated by itself.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What does a number in brackets after another number mean? I.e. 21(1)cm-1
What does the number in brackets mean in these two examples?
$$21(1)\ \mathrm{cm^{-1}}$$
and
$$1.0(3)\times10^{-7}$$

The results of measurements and other numerical values of quantities are often given with an associated standard uncertainty. A numerical value and the associated uncertainty may be expressed as shown in the question:
$$\begin{align}
y&=21(1)\ \mathrm{cm^{-1}}\\[6pt]
&=a(b)\ \mathrm{cm^{-1}}
\end{align}$$
where
$y$ is the estimate of the measurand (e.g. the result of a measurement) expressed in the unit $\mathrm{cm^{-1}}$,
$a$ is the numerical value, and
$b$ denotes a standard uncertainty expressed in terms of the least significant digit(s) in $a$.
It is important to note that the given uncertainty refers to the least significant digits of the given numerical value. For example, in the expression
$$l=23.4782(32)\ \mathrm m$$
the $(32)$ represents a standard uncertainty equal to
$$u(l)=0.0032\ \mathrm m$$
Many physical constants are also given in this form. For example, the previously recommended value for the molar gas constant $R$ as given by NIST from 25 June 2015 until the new value became available on 20 May 2019:
$$R=8.3144598(48)\ \mathrm{J\ mol^{-1}\ K^{-1}}$$
where the $(48)$ represents a standard uncertainty of
$$u(R)=0.0000048\ \mathrm{J\ mol^{-1}\ K^{-1}}$$
This form is in accordance with various current standards, in particular

JCGM 100:2008 Evaluation of measurement data – Guide to the expression of uncertainty in measurement
ISO/IEC Guide 98-3:2008 Uncertainty of measurement   Part 3: Guide to the expression of uncertainty in measurement (GUM:1995)
ISO 80000-1:2009 Quantities and units – Part 1: General

The Guide to the Expression of Uncertainty in Measurement (GUM) also shows other permissible forms:

7.2.2 When the measure of uncertainty is $u_\mathrm c(y)$, it is preferable to state the numerical result of the measurement in one of the following four ways in order to prevent misunderstanding. (The quantity whose value is being reported is assumed to be a nominally 100 g standard of mass $m_\mathrm S$; the words in parentheses may be omitted for brevity if $u_\mathrm c$ is defined elsewhere in the document reporting the result.)
1) “$m_\mathrm S=100{,}021\,47\ \mathrm g$ with (a combined standard uncertainty) $u_\mathrm c = 0{,}35\ \mathrm{mg}$.”
2) “$m_\mathrm S=100{,}021\,47(35)\ \mathrm g$, where the number in parentheses is the numerical value of (the combined standard uncertainty) $u_\mathrm c$ referred to the corresponding last digits of the quoted result.”
3) “$m_\mathrm S=100{,}021\,47(0{,}000\,35)\ \mathrm g$, where the number in parentheses is the numerical value of (the combined standard uncertainty) $u_\mathrm c$ expressed in the unit of the quoted result.”
4) “$m_\mathrm S=(100{,}021\,47\pm0{,}000\,35)\ \mathrm g$, where the number following the symbol $\pm$ is the numerical value of (the combined standard uncertainty) $u_\mathrm c$ and not a confidence interval.”

Note that item 2) corresponds to the form given in the question.
Concerning item 4), however, the GUM notes

The ± format should be avoided whenever possible because it has traditionally been used to indicate an interval corresponding to a high level of confidence and thus may be confused with expanded uncertainty (…).

Furthermore, ISO 80000 notes

Uncertainties are often expressed in the following manner: $(23{,}478\,2\pm0{,}003\,2)\ \mathrm m$. This is, however, wrong from a mathematical point of view. $23{,}478\,2\pm0{,}003\,2$ means $23{,}481\,4$ or $23{,}475\,0$, but not all values between these two values. (…)

From Wikipedia:

In metrology, physics, and engineering, the uncertainty or margin of error of a measurement, when explicitly stated, is given by a range of values likely to enclose the true value. This may be denoted by error bars on a graph, or by the following notations:

measured value ± uncertainty
measured value $^{+uncertainty}_{−uncertainty}$
measured value (uncertainty)

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Order of reactivity towards nucleophilic reagents considering alkyl halogenides, acyl halides and aromatic halogenides
Considering alkyl halogenides (1°,2°,3°,4°), acyl halides, and aromatic halogenides and their combiations what is the order of reactivity towards nucleophilic reagents and why (considering inductive effect, mesomeric effect, etc.)? I know its a basic question that I could have searched for but I didn't find a source that compares all of the above.


text:
How is the existence of carbon monoxide justified, in spite of the absence of mono-atomic oxygen in nature?
I have been taught that oxygen is found as $\ce{O2}$ in nature.
Then, how does carbon monoxide form if oxygen cannot be found as $\ce{O1}$ in nature?

Well, you just break the $\ce{O2}$ apart. Ordinarily that's not very favourable, but if carbon is next door to it and says "I want to bind to oxygen", then it can happen.

High temperature may also help to break it apart.

Which allotrope of carbon would you like to discuss?

Very similar question - http://chemistry.stackexchange.com/questions/34022/how-can-we-balance-one-half-of-an-oxygen-molecule-or-any-other-diatomic-in-a-c

Limiting reagengt -- here oxygen -- may lead to the formation of CO instead.  And $\ce{CO2}$ once formed, may react with further carbon to CO, as well.

In formation of carbon monoxide, a coordination covalent bond is involved. Read https://en.m.wikipedia.org/wiki/Coordinate_covalent_bond

(Almost) all reactions are not simple additions of compounds to form bigger compounds, but involve breaking of bonds and forming of bonds.
As is with (almost) other reactions, the reaction $\ce{2C + O2 -> 2CO}$ proceeds by breaking the bonds between $\ce{C}$ (it is heavily bonded in nature) and the bonds within $\ce{O2}$, and then re-combining as $\ce{CO}$ among other byproducts. (Note: this is extremely simplified but should  be enough for your level.)
In conclusion, here is a very simplified mechanism:

$\ce{C -> C*}$
$\ce{O2 -> 2O*}$
$\ce{C* + O* -> CO}$

Learn to distinguish between the elements — which always form some sort of elemental structure — and atoms of those elements that can also form compounds with other atoms of other elements.
When your teacher tells you that ‘oxygen typically occurs as $\ce{O2}$’, then they are making a statement about elemental oxygen i.e. a compound formed solely with oxygen atoms. Carbon monoxide is a compound that is formed of one oxygen and one carbon atoms — a statement that is (generally) true for elemental oxygen no longer holds true for compounds that include atoms of oxygen.
Likewise, elemental sulfur is typically $\ce{S8}$ but a number of sulfur compounds contain 1, 2, 3, 4, … sulfur atoms. And a similar case can be made for phosphorus, one modification of which is $\ce{P4}$.

In order to form $\ce{CO}$, first you must break the $\ce{O2}$ bond.
Even if $\ce{O2}$ bond energy is as large as nearly 500 kJ/mol , this bond can be broken via lightening in the atmosphere, elevated temperatures or catalytic effect.
When the proper catalysts interact with the $\ce{O2}$ molecule, bond can be broken with less activation barrier than in gas phase via interaction between molecule and potentially energetic surface/nanoparticle sites. This interaction makes easier decomposition via heterolytic or homolytic bond cleavage depending upon a magnitude of the force applied by the catalyst.  If the carbon atom exist in the environment and attraction between oxygen atom is very strong together with the thermal excitations or pressure difference or any other effect which enables oxygen to desorb from/diffuse on the catalyst, $\ce{CO}$ molecule can form.

From the article "Combustion" on Wikipedia (after googling "combustion"):
"Thermodynamically, the chemical equilibrium of combustion in air is overwhelmingly on the side of the products. However, complete combustion is almost impossible to achieve, since the chemical equilibrium is not necessarily reached, or may contain unburnt products such as carbon monoxide, hydrogen and even carbon (soot or ash). Thus, the produced smoke is usually toxic and contains unburned or partially oxidized products."
So, if you have educts like pure carbon C and pure oxygen O_2, they will always react, if you provide the energy to overcome the reaction barrier, because the product is far more stable than pure C or pure O_2.
EDIT:
In the reaction process the O2 binds to pure carbon and gets split up process, because it is thermodynamically better to form two CO molecules.
Also, although you may not find single oxygen atoms in nature, O_2 is an aggressive gas, because it reacts spontaneously with most elements and "oxidzes" many materials over time. It may not be spectacular as fluorine, but it is still a very reactive gas.

I don't think you should answer when you are drunk... jokes aside, you didn't actually answer the question. Your first two paragraphs are not related and your last paragraph is a regurgitation of the question. How does $\ce{CO}$ which have only one oxygen atom form from $\ce{O2}$ which has two oxygen atoms?

@user34888 well, I guess I thought of smth, but forgot to write it down


text:
What are the differences and similarities between an aliphatic and an alicyclic compound?
I read that alicyclic compounds are compounds that are both aliphatic and cyclic and i know that aliphatic compounds are acyclic while alicyclic compounds are cyclic but are there any other differences and similarities between them and what is exactly an aliphatic compound?

Aromatic compounds (wiki) (about.com) (study.com) (chemguide) (khan)
Aromatic compounds are compounds that contain one or more aromatic rings.
Each aromatic ring, in its unsubstituted form, has the chemical formula $\ce{C_nH_n}$ (e.g. benzene), or with one or more of the $\ce{CH}$ groups replaced with an atom from Group V or Group VI, usually $\ce{N}$ (e.g. pyridine) or $\ce{S}$ (thiophene) or $\ce{O}$ (furan). They are usually stabilized by resonance.
Aliphatic compounds (wiki) (about.com) (study.com)
Aliphatic compounds are hydrocarbons (wiki, compounds containing only $\ce{C}$ and $\ce{H}$) which are not aromatic.
An example would be hexane (wiki).
Alicyclic compounds (wiki)
Alicyclic compounds are aliphatic compounds that contain one or more rings.
An example would be cyclohexene (wiki).
Gallery
Benzene, aromatic (courtesy Introduction to Organic Chemistry):

Furan, aromatic (courtesy Wikipedia):

Cyclohexane, alicyclic (courtesy Jarad Chemicals):

Cyclohexene, alicyclic (courtesy Exporters India):


text:
Difference Between Betaine & Zwitterion?
what is difference btw Betaine & Zwitterion, Both having positive and negative charge in a molecule.

How to categorize Zwitterion and Betaine? or both are same.

From Wikipedia:

A betaine (BEET-ah-een, /ˈbiːtɑːˌiːn/) in chemistry is any neutral chemical compound with a positively charged cationic functional group [...] which bears no hydrogen atom [...]. A betaine thus may be a specific type of zwitterion.

it mean Betaine is a class of Zwitterion, But why it called  Betaine? Also, there is hydrogen atom in Betaine in a given formula,

@N.ALodhi The cation in the given betaine is bonded to three $\ce{Ph}$ (phenyl) groups, which means no hydrogen atom.


text:
Why do square pyramidal molecules have only two vertical mirror planes?
I would have expected molecules like IF5 to have 4, with two planes along the bond axes, and two planes bisecting the bonds. This is similar to how square planar molecules have four vertical mirror planes. However, apparently there are only two planes. Why is this? What am I misunderstanding?

who says there are only 2?

@DavePhD My inorganic book

There are two types of mirror planes. Two of each type for a total of four, as Dave has described.

Either your book is wrong or you are misunderstanding what it is trying to say.
There are 4 vertical planes of symmetry.
See for example GROUP THEORY AND ITS APPLICATIONS IN CHEMISTRY, SECOND EDITION , section 1.5.2
See especially:

Vertical planes are further classified into two, namely $\sigma _v$ and $\sigma _d$ The $\sigma _v$ planes fall along the bonds and the $\sigma _d$ planes bisect the bond angles in molecules. The subscripts v and d stand for vertical and diagonal. However, both $\sigma _v$ and $\sigma _d$ are vertical planes


text:
What chemicals could be added to human urine to cause harmful chemicals to precipite out?
What chemicals could be added to human urine to render it drinkable with no harmful side effects?
Urine is salty and contains other dissolved solids. Thus, drinking untreated urine would result in osmotic pressure actually removing water from the body.
So, what chemicals would be needed to cause the salts and other dissolved solids to precipite out or otherwise be rendered harmless?
I am thinking about a tablet, capsule or drops that would render human urine drinkable within 5 minutes or so.

Possible duplicate of https://chemistry.stackexchange.com/questions/58528/what-chemicals-would-need-to-be-added-to-human-urea-to-make-it-a-nutritional-sub

Yes, that is my original question. Should I consolidate the two questions or address them separately?

If they are separate - one has to do with urea, this one is a little more broad - then they can remain as-is. It boils down to: if this question and the other ones are "the same," then one of them needs to go. If they are demonstrably different, they remain separate. You can simply clarify if they are dupes or not.

I'm still interested in any ideas for transforming the chemicals in urine into a nutritional substance, or at least nutritionally neutral.

There isn't any magic chemical that would make urine drinkable in the way that you're describing. You can reformulate the question 100 ways and there still won't be any such magic chemical.


text:
Synthesis of diclofenac
I found a synthesis of diclofenac, and the last step with the addition of hydrazine and $\ce{KOH}$ seems interesting.

However, I couldn't find the mechanism behind this reaction, therefore I proposed one forming $\ce{N2}$.
Could you help me find the right mechanism regarding this type of reaction or correct mine?

This reaction is a Wolff-Kishner reduction [Organic Chemistry Portal, Wikipedia] which accomplishes the exhaustive reduction from a ketone to an alkyl group. Usually Wolff-Kishner requires forcing conditions, but I think this substrate could react under more mild conditions. The mechanism explains why.

The mechanism begins with the condensation of hydrazine onto a ketone (details not shown) to give a hydrazone. Under basic conditions, this hydrazone is deprotonated at nitrogen to give an anionic intermediate. In this case, the negative charge can be delocalized onto oxygen, resulting in an enolate structure. Typically, the negative charge is only shared between a nitrogen and carbon, so this substrate gives a particularly stable intermediate. Protonation of the enolate at carbon gives the first C-H bond necessary to form the product. A second deprotonation at nitrogen gives a similar flow of electrons to form another enolate structure, this time with cleavage of the C-N bond and release of nitrogen gas. Another C-protonation gives the lactam precursor to diclofenac. Cleavage of the amide with hydroxide (details not shown) gives the target.


text:
What are the physical manifestations of radial nodes?
I realize that nodes (both angular and radial) are areas with zero probablility of finding an electron.  I realize that plotting the square of the radial eigenfunction for an orbital will give the probability density of finding an electron at a given distance from the nucleus.
So for the 1s orbital (zero nodes), the graph of the square of the eigenfunction will be one hump.  For the 2s orbital (one node) there will be a small hump followed by a large hump with a zero point in between them (talking about the sqaure of the eigenfunction not the eigenfunction itself).
For the 1s orbital, the peak of the hump will represent the most probable radius the electron will be found from the nucleus.  The probability will lower as the radius increases or decreases from the most probable radius (similar to gaussian distribution), and the electron will have zero probability of being found within the nucleus of the atom (0 radius).
And from what I understand the radial node(for the 2s orbital) is physically manifested at the point where the 1s orbital and 2s orbital meet (my instructor used the analogy of russian nesting dolls), and on the NON squared eigenfunction plot of the 2s orbital, the node is represented where the wave function crosses the x axis going from +y to -y.
So here is what Im really wondering (please correct me on anything stated previously if I am mistaken).  Is the first (smaller) hump on the squared radial eigenfunction (probability density function) for the 2s orbital the probability that the electron is in the 1s orbital (I believe this would be called an excited state if im not mistaken)?  This would make the most sense to me, as it would be saying that it is possible for a 2s electron to be in either the 1s orbital or 2s orbital, but not in between (where the wave function goes from +y to -y)
Edit: Perhaps it is incorrect to say an electron from the 2s orbital has the possibility of being in the 1s orbital.  What I really mean is the electron may occupy the same space as the 1s orbital, but since orbitals are just 3d areas where an electron is most likely to be found perhaps I am not incorrect in saying that a 2s electron occupies the 1s orbital since the 1s orbital is technically just a region in space.  My guess is that the probability of a 2s electron occupying the same space as the 1s orbital is lower than it occupying the space of the 2s orbital due to repulsion by the electrons that already occupy the 1s orbital.
Here is a picture representing everything ive tried to explain so far:

You have it all backwards. The point where the 1s orbital and 2s orbital meet is not important at all, and in particular has nothing to do with the radial node of 2s. (These are different points, in fact.) When an electron sits on 2s, well, this means it sits totally, 100% on 2s and not on 1s. It does not care at all about 1s. It does not even "know" whether 1s exists. True, that electron spends part of its own existence in the region of the first hump where most of 1s also resides, but then again, every orbital overlaps with every other. It's not like orbitals own the space they occupy.

related: http://chemistry.stackexchange.com/questions/53926/can-the-regions-of-1s-and-2s-subshells-overlap?rq=1

I realize that plotting the square of the radial eigenfunction for an orbital will give the probability density of finding an electron at a given distance from the nucleus.

This isn't quite right.  The square of the radial function will give the probability density for s-orbital, but this is not the probability of finding the electron at a given distance from the nucleus, because the number of points a given distance from the nucleus increase with $r^2$.  To get the plots in Fig. c, the radial function squared in multiplied by $r^2$.
For example, for 1s, the probability density is greatest at the nucleus, but by multiplying by $r^2$ when r=0, a radial probability density of zero is obtained.
Furthermore, as you can see from the graphs on pages 187 and 188 here, for 2s and 3s the most probable location (highest probability density) is at the nucleus (r=0). Again, only by multiplying by $r^2$ when r=0, is a radial probability density of zero is obtained.
So "Fig. a" in the OP is misleading, the central region of 2s and 3s should be shaded much more intensely than the outer region(s).

So here is what Im really wondering (please correct me on anything stated previously if I am mistaken). Is the first (smaller) hump on the squared radial eigenfunction (probability density function) for the 2s orbital the probability that the electron is in the 1s orbital (I believe this would be called an excited state if im not mistaken)? This would make the most sense to me, as it would be saying that it is possible for a 2s electron to be in either the 1s orbital or 2s orbital, but not in between (where the wave function goes from +y to -y)

No, the 2s electron is only in the 2s orbital.

So why does the probability distribution curve have a smaller 1st hump (smaller radius) than the 2nd hump (larger radius) for the 2s orbital if the probability that the electron is closer to the nucleus is higher?

@Ki11akd0g the probability that the electron is in a given infinitesimal volume element (probability density) is highest at the nucleus, but the probability that it is a given distance from the nucleus (radial probability density) is at the outermost hump, primarily because a spherical shell have infinitely more points than a single point, and because a bigger radius shell has a greater area than a small radius shell. .

I actually understand the concept of nodes now, what was misleading is the color patterning depicted on thw images.  I was considering the 2s orbital to contain a 1s orbital inside.  The 2s orbital and 1s orbital are completely separate, although they may overlap and occupy the some of the same space, they are separate entities.

AND I understand what you meant by the probability distribution vs the radial probability distribution.  The probability of finding an electron decreases as you move further from the nucleus (on a straight line path away from the nucleus).  But there are alot more points at which the radius equals a specific value the further you move away from the nucleus (there is a greater area of points at r=2 than r=1).  So its the product of the probability of actually finding an electron at a given distance from the nucleus and the greater area at which that probability applies to.  THANKS SO MUCH!


text:
What does the symbol ϕ mean in skeletal formulae?
I was given this formula by my A-level teacher and was curious as to what it means. It seems to be the Greek letter phi. I have searched online and have not been successful. Any help would be greatly appreciated.

Didn't you already post this then delete it when I told you the answer...?

Sorry, I meant to ask what the whole of the formulas name is? I gather that the symbol is an old version of Pi but does the equation it is in have a proper name. I apologize, I worded the previous question incorrectly and to avoid any confusion I deleted it. I appreciate your previous reply and did find it helpful but unfortunately it was not the answer I was looking for, for this question but thank you anyway.

It just means any aromatic group.. it doesn't mean anything specific in the same way that R can be used for anything. I also re-iterate that it isn't used nowadays and doesn't form part of the a-level syllabus

@l1zzy23 Please note that, according to IUPAC recommendations, it is generally not acceptable to use circles to indicate electron delocalization in adjacent fused rings, since such diagrams are at best ambiguous in terms of the character of the shared fusion bond between the rings.

According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), the Greek lowercase letter phi $(\phi)$ may be used to represent a phenyl group.

It is acceptable to use the Greek lowercase letter phi $(\phi)$ to represent a phenyl group. That abbreviation has a long history, and ambiguity is unlikely since that letter is rarely used for other purposes in chemical structure diagrams.

Note that this traditional use is only considered “acceptable”, i.e. not “preferred”.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred name for the hydrocarbon parent ring component is the retained name chrysene. The indicated numbering of this polycyclic hydrocarbon is is the result of the application of the specific criteria used to number fused ring systems:

Therefore, the complete name for the entire compound given in the question is 1,3,7,9-tetraphenylchrysene.


text:
Why is the melting point of KBr higher than that of CsCl?
I thought that when compounds have similar forces and charges, the one with the higher molecular weight has the higher melting point. If that is true, then why is the melting point of $\ce{KBr}$ higher than that of $\ce{CsCl}$?

Why, the charges are indeed similar, but the distances between them are not. Hence the forces are different, too.

see this:-http://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound

It may be helpful considering molecular weight for say $\ce{KBr}$ vs $\ce{KCl}$ or $\ce{CsCl}$ vs $\ce{CsBr}$, and actually melting point would go down with increasing molecular weight, but it actually has nothing to do with molecular weight despite the trend. It would be more helpful to look at lattice energy, electronegativity, electron affinity, and atomic radii, but its a little more complicated to compare a $\ce{Cs}$ salt of a halide with a $\ce{K}$ salt of a different halide as you've changed more factors.  Although chlorine has a higher electronegativity and smaller atomic radius than bromine, caesium has an even larger atomic radius than potassium (relative to the size difference between chlorine and bromine) as well as a lower electronegativity than potassium.  All of these factors will affect the lattice energy and therefore the melting points.  But, like I said, when comparing two similar salts, make sure one of the elements stays constant.  Its much more tricky to compare $\ce{KBr}$ with $\ce{CsCl}$ than it is to compare ($\ce{KBr}$ with $\ce{KCl}$) or ($\ce{CsBr}$ with $\ce{CsCl}$), or even ($\ce{KBr}$ with $\ce{CsBr}$) or ($\ce{KCl}$ with $\ce{CsCl}$).  When comparing $\ce{KBr}$ with $\ce{CsCl}$, you must determine what will affect melting point more: the ($\ce{Cs}$ vs $\ce{K}$) difference or the ($\ce{Cl}$ vs $\ce{Br}$) difference.


text:
Is a compound composed out of gold, manganese, xenon, krypton, and oxalate possible?
Please keep in mind while reading this that I do not intend to create this, it is part of the backstory for a video game I am producing.
I created a compound based on knowledge from my chemistry class and Wikipedia.
The formula is: $\ce{AuKr4Xe2C2O4^2-C2O4^2-MnO4-}$
The 2D structure looks like:

And the 3D Structure is:

I believe the IUPAC name would be Aurodixenichexakryptic permanganate dioxalate (please correct me if I am wrong).
I know the molar mass is $\pu{1089.7413 amu}$.
It is a thick, black liquid similar to Vantablack, as in it absorbs $99.99\%$ of the light hitting it, making it seem two-dimensional. It is odorless and tasteless.
I determined the melting point to be $0.01~\mathrm{K}$ and boiling to be $\pu{7538384392 K}$. The density under normal conditions is $\pu{89.45 g mL-1}$.
I was hoping someone could tell me is all the information thus far was valid and if someone could provide me other key facts about the compound (i.e. alternate names, $\mathrm{pH}$, $\mathrm{pOH}$, $\mathrm{p}K_\mathrm{a}$/$\mathrm{p}K_\mathrm{b}$, etc.)

As I mentioned at the beginning, it is for a fictional video game. I just wanted some basis in fact.

I don't understand the downvotes. Yes, this is kind of ridiculous, but still a legitimate question.

If you want some basis in fact, change the element symbols to something different, which is not in our periodic table at all. That would be better.

@Michael "I determined the melting point to be 0.01K and boiling to be 7 538 384 392K. The density under normal conditions is 89.45 g/mL." is far from legitimate.

That stuff yearns to go back to its standard states. This would be an explosive. It would probably delete the person synthesizing it, along with his apparatus, his laboratory and the general viscinity. This wouldn't "boil" - it would blow.

No. Not possible. The oxalate ion can be -2 NOT -4.

No, this is not possible.
Actually, if I would have to think of the most unlikely chemical conceivable, that would be it. Let's see why:

Krypton is a noble gas that doesn't bond to anything. All of the known krypton compounds can be counted on one hand, and most of them contain fluorine. Putting krypton in a large molecule like this just can't be.
This is the same for xenon. Even though it's slightly more reactive than krypton, it still has problems forming bonds with other stuff and even when it does it's extremely unstable.
Gold is another noble element that doesn't like bonding to stuff.
Permanaganate is a very strong oxidiser. This is not something you want to have around organic stuff (such as your oxalate). They are simply not stable together.

I don't know how you derived the data for it, but it seems completely unreasonable. A boiling point of 7538384392 K? How about no way? Even if you somehow could create this compound (god knows how), I would expect it to be so unstable it would decompose completely at ambient temperature. And by "ambient" I mean ambient on a Kuiper belt object.

If you attached a few atoms of einsteinium to the compound it would be even more unlikely. :D

Interestingly, tetraxenonogold(II) cation is known. It is not totally impossible that it could form a salt with permanganate. Then for a fictional game's purpose, you can assume a clathrate compound of krypton and a double salt of tetraxononogold permanganate - oxalate. This is as close to a real chemistry as it can get.

@vapid then again - permanganate and oxalate together in one compound? Good luck keeping that stable.

@Michael Why not? There are many examples of compounds which have both oxidizing and reducing components. Sure, most of them are explosive, but can be stable at ambient conditions. Besides, oxalate is not THAT powerful reducing agent. Permanganate on the other hand will oxidize most of organic compounds, therefore any organic permanganate will be unstable and potentially explosive. But they do exist, and many are stable at room temperature.

@vapid read OP's question again. Usually mixed reductant-oxidiser are stable when they're in solid form. He wants the compound to melt at 0.01 K. Ambient conditions, on earth, are 273 degrees above that! So yea, some compounds are stable but not this one.

@Michael there are many liquid explosives known - it's not that if you mix any oxidizer with any reductor they go 'KABOOM' instantly - it is all about kinetics. As for the OP's question - I wouldn't worry about permanganate + oxalate, but rather that menacing ring of noble gas atoms...

A compound needs to have an overall neutral charge. The formula has a -5 charge so it is not a compound.
The supposed 2-d structure has a -20 charge and therefore doesn't correspond to the formula.

One more No from a topological standpoint. There is no known molecular structure of this topology (assuming any bond any any atom, preserving connectivity only). The closest one would be some derivative of existing  benzo[1,2:4,5]dicyclobutene [1]:

Also, tetraoxo-manganese unit bound in a shown fashion suggests a coordination polymer (e.g. neither a molecular structure, nor a monomer, see e.g. [2]) and therefore cannot be rationalized. If you would ask me, I'd go with something looking a bit more plausible and a bit crazy, like

Good game design suggests catchy and artistic, yet believable world. That's why  first Thief and Deus Ex were such successful games. It's hard to break the fourth wall when player feels the universal physical and chemical laws were violated, and your molecule and especially its physical properties make player scream "no waaay!", followed by Alt+F4.
References

Lawrence, J. L.; MacDonald, S. G. G. Acta Cryst B, 1969, 25 (5), 978–981 DOI: 10.1107/S0567740869003293.
Lv, D.-Y.; Gao, Z.-Q.; Gu, J.-Z.; Liu, J.-Z.; Dou, W. Transition Met Chem 2011, 36 (3), 275–281 DOI: 10.1007/s11243-011-9466-2.


text:
Ammonium in NH4Cl
I am learning the basics of bonding and reading about valence and lone electron pairs. I am confused on why nitrogen would bond 4 atoms of hydrogen such as in ammonium chloride. To me nitrogen has a valence of 3 so it can only form 3 bonds yet in NH4Cl it forms 4 bonds. Why would this be? Is this a concept of electronegativity? Furthermore in my book it states that the ions of NH4Cl are NH4+ and Cl-. Obviously the NH4+ has a positive charge but is this because it gives one electron to the Cl atom in the bond NH4Cl? Very basic but trying to grasp concepts.
When NH4Cl is in water, what pulls these two apart to create the cation NH4+ and Cl-?  Are the bonds that these form with water stronger than the bonds between the NH4 and Cl so much so to pull them apart?

$\ce{NH3}$ has a lone pair, so it forms a dative bond with $\ce{H+}$ by donating that pair to the bond between them.

As you have suspected, water pulls them apart. Water is polar, meaning that it has an unbalanced slightly positive charge near the $\ce{H}$ and an unbalanced slightly negative charge near the $\ce{O}$. These charges are responsible for dissolving.
Actually, the process of dissolution for is endothermic specifically to $\ce{NH4Cl}$, meaning that it takes more energy to dissolve it more than it gives. Other water-soluble chemicals like $\ce{NaCl}$ usually gives more energy in dissolving than it takes, making it exothermic.
Then what pulls them apart? Entropy. Search for "Gibb's free energy" for more information.
$\ce{NH4+}$ is not particularly stable as this is reversible:
$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$

NH4Cl is example of a compound having dative, covalent and ionic bond.
As far as your question is concerned N make 4 bonds by sharing 3 electrons with three H atoms and giving a pair of electrons to H+(dative bond) and whole molecule acquire a charge of +1.As far as dissociation of NH4Cl in water is concerned you will need knowledge of entropy, hydration energy and bond enthalpy etc.

NH3 is a base and when it is reacted with hydrogen chloride gas, hyrdochloric acid it form NH4Cl (NH3 + HCl → NH4Cl).
NH3 form NH4+ upon accepting one proton from HCl, as HCl is strong acid, it exist in the form of H+ and Cl- ion. Nh4Cl is formed due to electrostatic force btw NH+ and Cl-.
There is no loss and gain of electron, you are confusing it with ionic in which one atom loss electron and other atom gain electron to form bond( There is no loss of electron from NH4+ and again of Electron by Cl-).

Two ways of understanding. 1: H gives electron to Cl and then N to H, so indirect electron transfer. 2: proton transfer instead of electron transfer as you stated.

If you react with hydrochloric acid instead of hydrogen chloride gas (who would use hydrogen chloride gas instead of the acid), the electron transfer is more succinct.

@user34388, NH3 it self is a gas, unless it's in the form of solution.

So if you combine NH3 + HCL the bonding is done by a proton transfer or indirect electron transfer and if so, what "H" gives the Cl the electron?

@user34388 I think i found the answer, do you agree with this here: https://imgur.com/j61r86o

I agree. deleted that paragraph.

@Atticus283blink, very useful comment,


text:
How to calculate the number of atoms of each element in a urea?
I have a stoichiometry question which I answered incorrectly and I am not sure where I went wrong. The question is as follows:

Urea, $\ce{(NH2)2CO}$, is used for fertilizer and many other things. Calculate the number of $\ce{N}$, $\ce{C}$, $\ce{O}$, and $\ce{H}$ atoms in $1.68 \times 10^4~\mathrm{g}$ of urea.

First, I found the total atomic weights of each element in this compound to find the percent composition:

N: 28.02 amu
H: 4.032 amu
C: 12.01 amu
O: 16 amu
Total atomic weight of urea: 60.062

Then, I found the percent composition of urea:

N: 46.65%
H: 6.71%
C: 20.00%
O: 26.64%

Then, I converted $1.68 \times 10^4~\mathrm{g}$ into moles by dividing this amount in grams by 60.062, and found that it is approximately 279.71 moles.
After that, I took percentages of this to calculate the number of moles of each element in the sample of urea (for example, 46.65% of 279.71 to find the number of moles of nitrogen).
Finally, I multiplied each of these numbers in moles by Avogadro's number ($6.022\times10^{23}$) to obtain the number of atoms. However, when I checked my answers in my textbook, I was wrong by a significant amount.
\begin{array}{lrr}
       & \text{My answer}    & \text{Correct answer}\\\hline
\ce{N}:& 7.83 \times 10^{25} & 3.37 \times 10^{37}\\
\ce{H}:& 1.13 \times 10^{25} & 6.74 \times 10^{26}\\
\ce{C}:& 3.37 \times 10^{25} & 1.69 \times 10^{26}\\
\ce{O}:& 4.49 \times 10^{25} & 6.74 \times 10^{26}\\\hline
\end{array}
Do you know where I am going wrong?

Something is off with the answer from the text: the number of moles of nitrogen given above is on the order of 100 trillion.

the answers given are just wrong. From the molecular formula there should be as many oxygen atoms as carbon atoms. There should be 4 times as many hydrogen atoms as carbon atoms. There should be twice as many nitrogen atoms as carbon atoms.

I think that whoever added all the answers into the chart may have accidentally messed up the books answer for atoms of oxygen, because I am positive that when I answered the question the number of atoms of oxygen was equal to the number of atoms of carbon that is listed (1.69×10^26) which is correct.  The only incorrect answer was for Nitrogen and only the exponent was wrong.

There are $\approx 280 \times 6.0\cdot10^{23}\times 8 $ atoms. N atoms are present as 2/8 of these H as 4/8, and C=O=1/8,

You don't need complicated calculations once you know the number of moles of urea. everything else is just that number time the number of atoms in a single molecule of each element in a single urea.

You were correct in converting the grams of urea to mol of urea, but the next step would be to convert the mol of urea to molecules of urea using avogadro's number (which converts mol to molecules or atoms).  Then you want to multiply the molecules of urea by the number of atoms of the element in question that are present in Urea.  So for example there are 2 atoms of N per molecule of Urea, therefore multiplying the molecules of urea you calculated earlier by 2 will yield the total number on nitrogen atoms present.  As was mentioned in the comments, the value for N atoms present given by your book was incorrect and should have been 3.37 × 10^26 atoms of Nitrogen.  See if you can calculate the correct number of each of the remaining elements, as the answers given were correct.

do not stuck into percentage,this method is incorrect.here by formula you can say that one mole of urea contain 2 mole of N, 4 mole of H,1 mole of C and! mole of O.
Also the no. of moles you calculated(for urea) are 279.71,thus no. OF moles of N will be 2x279.71 that of H will be 4x279.71 and so on, since now you know the number of moles of each atom you can now calculate no. of atom of each element.

why downvote???

I see a couple subtle errors in your calculations, both of them involving units.
First, when you get the number of moles (279.71) you have the number of moles of molecules.  To count individual atoms you have to multiply the percentage composition (which I will say more about, below) by the total number of moles of atoms.  So, convert your moles of molecules to moles of atoms by multiplying by the eight (coubt them) atoms per molecule.  So you have, if I follow significant figures, 2237.7 moles of atoms in your urea sample.
Next is the matter of percentages you use.  You used mass percentages.  You need molar percentages.  So instead of multiplying 46.25 mass% nitrogen by those 2237.7 moles, you should multiply 25 mole% nitrogen, this being two nitrogen atoms divided by eight total atoms in the molecule.  You could also have multiplied the two nitrogen atoms per molecules times 279.71 moles of molecules directly, saving a conversion step (a good idea is to try both ways and see that they are equivalent).  Do the same for the other elements.
Finally, you or the book misrecorded one of the answers.  The exponent in the nitrogen atom count should be 26, not 37.


text:
Do molecular orbital diagrams exist for ionically bonded compounds and molecules?
Do molecular orbital diagrams exist for ionically bonded compounds and molecules?

The boundary between ionic ans covalent may not be as succinct as you think...

Why is this being close voted? It is not primarily opinion-based nor is it unclear.

@orthocresol user34388's answer was from http://www.lisgar.net/magwood/molecular%20orbitals.htm which gives an explanation of the differences between HF and LiF molecular orbital diagrams

Molecular Orbital theory is a simplified approach to describe chemical bonding based on symmetries and orbital overlap. In principle, it makes no assumptions on the type of bonding, i.e., covalent, ionic, Van der Waals. This distinction is more a construct to understand the strength of the bond.
In case of a covalent bond, the electrons that make up the bond are shared by both nuclei, whereas in an ionic bond the bonding originates mostly from the Coulomb interaction, that is, the electrons are mostly located on a single nucleus. In MO theory, this can be understood in terms of the interaction strength. When the atomic orbitals that make up the bond have similar energy (and same symmetry), there will be a strong interaction and the resulting MO will have character of both AOs. In other words, the electrons will be shared more or less equally by both atoms. When there is a large difference between the AOs, the resulting bonding MO will have mostly the character of the AO which is lowest in energy and as such the electrons in this MO stay mostly around this atom. This resembles an ionic bond.


text:
Do electrons hybridize only during bonding?
Do electron orbitals hybridize on their own in an solitary atom or do they only do this only when bonding is a posibility? If hybridization does occur in atoms without bonds, then is it random? Do the electrons mix and match as they tumble around in space?

To set things right first - hybridisation is only a mathematical process, a human construct. It does not involve any physical movement of electrons. Therefore hybridisation cannot "happen" in any sense. Trying to make hybridisation happen is like trying to make a half-empty glass "turn" into a half-full glass.

To expand on orthocresol's excellent comment, you confusion likely comes from a common misconception.

Electrons and orbitals are not the same things. Electrons are fundamental particles with particle-wave duality. The behavior of electrons can be modeled with wavefunctions. Orbitals are mathematical constructs based on the probability density of the wavefunctions.

Technically, orbitals do not hybridize because orbitals do not exist. Since orbitals are mathematical constructs, they do not correspond to any physically observable entity. Electrons are the physically observable entity. Electrons do not hybridize. Electrons interact quantum mechanically to form bonds when their wavefunctions have appropriately matching energy and symmetry. Electrons then adopt new wavefunctions for their new situation. These multielecton wavefunctions can be intractable.
Hybridized orbitals represent a model that simplifies some of the mathematics by replacing the canonical hydrogenic atomic orbitals with hybridized atomic orbitals that are oriented in directions that match common molecular geometries. Orbital hybridization is a model that is thus only useful when describing bonding. If there is no bonding, hybrid orbitals are a nonsensical model.


text:
PIB wave functions and the effects of e- delocalization on UV absorption in conjugated systems
I'll try to keep this short and concise. What made me start thinking about this question was the material we have been covering regarding PIB wave functions in my quantum chemistry class, and material we have been covering regarding ozone absorption in my climate class. I would like to try to understand the relationship between the increased wave function domain that delocalization yields for conjugated pi electrons, and the effect that this has on the wavelengths of light that they can absorb. Like ozone, I understand that most other conjugated pi systems have strong UV absorption, so I would assume there is some correlation there.
Specifically though, I would like to know if this can be explained (at least to some extent) by a simple QM model like the 1D PIB, and if how electrons' wave functions are altered when their spacial domain is extended across the whole molecule by delocalization can explain UV absorption in conjugated pi systems.

You can use the PIB model as a very approximate way of thinking about energy levels in conjugated ($\pi$ electron) molecules. Absorption in triatomics is much more tricky, however, and usually a Walsh diagram is necessary.
The general idea is that as the conjugation gets larger the absorption wavelength for electronic transitions gets smaller and so shifted to the red part of the visible spectrum. (A similar argument applies to particles on a ring for aromatics)
The energy levels of a PIB vary as $E_n \approx (n/L)^2$ where n is the quantum number and L the length of the box, or conjugation. Take this to be sum of all conjugated bond lengths. Thus as L gets larger the energy levels become closer together. Secondly as the number of conjugated bonds increases the number of quantum levels filled also increases, as there are more electrons. The optical transition is from the highest filled level to the next one up (HOMO-LUMO), the energy difference is $E_{n+1}-E_n$ at each value of L. If you take the length of the box to be the number of bonds m of length a, then $L=ma$ it is easy to show that the absorption spectrum shifts to the red as the conjugation increases, although you should definitely not expect accurate values in terms of energies.


text:
Accuracy of FP-LAPW vs Plane Wave Pseudopotential: examples
I am trying to learn about FP-LAPW software. I would like to know how well FP-LAPW outperforms Pseudopotential PW calculations. What are some properties/calculations/quantities in which FP-LAPW gives consistently better results than Pseudopotential PW theory? I know that FP-LAPW should give better results on almost everything (after all, it is an all-electron calculation). I am specifically looking for calculations where FP-LAPW gives much better results than Pseudopotential PW.

In all cases FP-LAPW will be better than Pseudopotential + Plane Waves calculations. However, FP-LAPW calculations are better at calculating properties which involve core electrons (e.g. core level shifts, hyperfine structures). Such calculations are often not possible with Pseudopotentials + Plane Waves, or are more complicated.This is because PP+ PW calculations do not treat core electrons explicitly.  Furthermore,  FP-LAPW calculations do not depend on pseupotentials. Although one can test pseudopotentials to check for transferability, there is no guarantee that the pseudopotential works for the system that you are simulating. As a result, "artifacts" due to pseudopotentials can be avoided with FP LAPW.


text:
test to differentiate between NaCl and zinc hydroxide
I had conducted a Salt spray corrosion test on Zinc coated bolts. The test result showed white deposits over the bolts. I want to confirm whether its zinc hydroxide or simply the sodium chloride salt that got deposited. So any  test to differentiate between NaCl and zinc hydroxide?

NaCl is soluble in water, zinc hydroxide is not.

You can use the large difference in water solubility to determine whether the residue is sodium chloride or zinc hydroxide. Sodium chloride is very soluble in water and zinc hydroxide is insoluble in water. You can do the test as follows:
1) Brush or scrape a portion of the residue onto a piece of paper or card stock.
2) Transfer the residue onto a dark colored plate or bowl.
3) Add enough water to cover up the residue and swirl gently.
Any white solid that remains should be zinc hydroxide as the sodium chloride will dissolve away.


text:
How is NH4Cl a salt?
So what I know is, a salt is an ionic compound between a metal and a non-metal which exchange electrons. Like in $\ce{NaCl}$, $\ce{Na}$ is the metal and $\ce{Cl}$ is the non-metal, $\ce{Cl}$ hogs up the valence electron of $\ce{Na}$ and so the atom becomes charged.
But in the salt $\ce{NH4Cl}$, neither nitrogen, hydrogen nor chlorine are metals. All of them are non-metals. How can it be a salt?

This is not the correct definition of salt. The IUPAC definition of salt is "a chemical compound consisting of an assembly of cations and anions". In fact there are many salts which do not fit the definition you give.

oh, so NH4 is a cation and Cl is an anion, thats why?

No. $\ce{NH4+}$ is the cation and $\ce{Cl-}$ is the anion.

that is what I said :/

No that's not what you said. These little "+" and "-" are important, they make a great deal of sense.

@orthocresol But according to this IUPAC definition, all acids and bases, e.g., NaOH in solid phase will fall under the definition of salt!

If we take the Lewis view of acids and bases then it is clear that ammonia has a lone pair which can be donated to a Lewis acid. The proton is the Lewis acid in this case.
Thus ammonia and a proton are able to react to form an ammonium cation. If the proton was obtained by breaking the H-Cl bond in hydrogen chloride (this can either be done by dissolving hydrogen chloride in water to form a dilute solution of hydrochloric acid or by the gas phase reaction of an amine and hydrogen chloride) then the anion required for electroneutrality of the system will be the chloride anion.
If the water concentration is very low then solid ammonium chloride can form while if plenty of water is present then the ammonium and chloride ions are well solvated by the water and exist as a solution of ions in water.
It is interesting to point out that ammonium chloride can decompose on heating to form ammonia and hydrogen chloride. When the vapours cool down then the ammonium chloride will reform. This causes ammonium chloride to be able to sublime with ease when heated.


text:
Chirality of substituted adamantanes
Is this compound chiral?

I can't tell where the chiral centres are because this polycyclic backbone is too confusing. I tried to draw a mirror image, but I got really confused and couldn't tell whether it was superimposable. And I don't have a molecular modelling set so I can't build a model.

Yes, this compound is chiral.
The polycyclic backbone is called adamantane. It has $T_\mathrm{d}$ symmetry, meaning that as far as chirality goes, it behaves like a perfect tetrahedron, somewhat like methane does.
It also has the interesting property that if you extend the C-F, C-Cl, ... bonds inwards, they will all meet at the same point. Those red dotted lines form a perfect tetrahedron.

Now, since the adamantane backbone behaves exactly like a methane molecule - why don't we just replace it with one? [n.b. this is not 100% rigorous but works here.]

Hopefully you'll agree that this is chiral. It basically behaves like it has a chiral centre right in the middle of the adamantane backbone.
If you really wanted to be sure, you could build a mirror image and verify that they are non-superimposable. (Hint: they are.)
Each of the four substituted carbons is connected to four different groups, and theoretically one would expect $2^4 = 16$ stereoisomers. However, because three of the four groups (the polycyclic backbone) are held in a rigid, fixed geometry, this number is reduced to 2.
There is more discussion of this class of compounds here.

I find it fairly obvious that every quaternary carbon (which is tetrahedrally coordinated) has all different substituents and thus breaks all symmetry. However, I find this answer very nicely illustrated, and it makes a good point for more complicated cases.


text:
Migration of ethyl group in pinacol rearrangement

For this substrate, will the methyl or ethyl group migrate preferentially? I know that the migratory aptitude of Et is greater than that of Me, but I saw a lecture where the prof said migrating Et will break the bonds and that only shifting of Me is possible. Is it right? What will be the product formed?

There is no clear cut answer as a migratory group is concern. You are right, Ethyl group has greater tendency to migrate than Methyl group. But the group which will stabilized the carbocation will move, this is the rule of thumb.

Would make nice exercise in computational chemistry. Even semiempirical methods should give correct answer

Is it true Et shift is not allowed here?

The two possible compounds resulting from the pinacol rearrangement of your substrate, 3,4-dimethylhexane-3,4-diol, are as follows.

A quick Reaxys search throws up a paper in which this pinacol rearrangement was investigated.1 I quote directly:

Four different investigators have studied the pinacol-pinacolone rearrangement of 3,4-dimethyl-3,4-hexanediol. It was finally demonstrated by Nybergh that, contrary to the conclusions reached by the three previous investigators, the rearrangement involved the preferential migration of an ethyl group rather than a methyl group.

The paper describes how the authors carried out the reaction and analysed the mixture to determine the ratio of the two products shown above. In 1956, there was much less technology than we enjoy now, so no NMR and no HPLC. It turns out that they had to hydrogenate the products to make a mixture of the corresponding alcohols and figure out the ratio by IR spectra. The result: $70\mathrm{-}73\%$ of 3-ethyl-3-methylpentan-2-ol (corresponding to the ethyl shift), depending on the form of the starting pinacol used.
Unfortunately, I can't access the (very old) journals that claim that the Me shift happens preferentially.
References
(1) Reeve, W.; Karickhoff, M. The Pinacol Rearrangement of meso- and dl-3,4-Dimethyl-3,4-hexanediol. J. Am. Chem. Soc. 1956, 78 (23), 6053–6054. DOI: 10.1021/ja01604a028.


text:
Can orbitals penetrate the same space other orbitals occupy?
Do electrons in outer orbitals penetrate orbitals occupied by lower energy electrons? If there is no electron in the orbital, do the orbitals still penetrate one another? For example, in the $\ce{Na}$ atom, only one electron is in outermost 3s orbital does the 3p orbital also penetrate the 3s?

Can you explain what you mean?

It's orbital not orbit.

What do you mean by penetrate? "Penetration" describes the proximity to which an electron can approach the nucleus. In a multi-electron system, electron penetration is defined by an electron's relative electron density (probability density) near the nucleus of an atom. Electrons in different orbitals have different wavefunctions and therefore different radial distributions and probabilities of existing around the nucleus. Roughly speaking.

Sodium does not have a 3p electron in the ground state.
Only in an excited state would there be a 3p electron.
Every orbital of an atom penetrates every other orbital of an atom, as at most 2 dimensional regions (planes, spherical shells, conical shells, points) have zero probability density.

Usually it is considered that the orbitals are there and have the same properties (shape, energy, penetration etc.) whether or not electrons are present. Electrons are added to levels according to set of rules, such as the Pauli Principle, Hunds Rules and these then determine bonding / anti-bonding, magnetic properties (via total electron spin), dipoles in molecules, etc.

And, further complicating things, the nice single electron orbitals we all know and love from hydrogen are not the same as the orbitals in mulit-electron atoms because of the presence of the other electrons. One of the more amazing points is how close the real orbitals are to the single electron orbitals. (Hans Bethe and Ed Salpeter managed to write an entire book on just the two-electron atom).


text:
Fixed cell parameters, TINKER
I am using TINKER package to calculate the external reorganization energy for pentacene crystal. But I have problems with that - I've generated .xyz file of 5 molecules with VESTA and I need to perform the calculation on those five molecules (image below) with periodic boundary conditions (PBC).

How can I find the proper size of cell and keep it fixed for calculations? When I use keywords A-axis, B-axis, C-axis and alpha, beta, gamma and assign lattice vectors and angles to them after optimization or minimization for the charged system like that apparently cell gets broken and molecules are distorted out of box:

I'm not sure I see in your image how the molecules (or cell) are "broken."

If you have the unit cell above, then you can specify the size of the box based on A, B, C, $\alpha$, $\beta$, $\gamma$ as you indicate. I don't believe that Tinker will change the cell parameters during a run.

@Hutchison - sorry, maube the word broken is not the best, bus the point is thay molecules are no longer in the box, because the size of the box was set so they fit accuratelly in it. And in the image you can see that they for example are two of them in z axis (the one along the pentacene).

btw, I suggest this site for finding reference structures of simple compounds (follow the link). https://www.google.ru/search?q=pentacene+site%3Acrystallography.net&oq=pentacene+site%3Acrystallography.net

It's a little hard to tell from your second visualization, but to me, it simply seems as if the molecules have shifted somewhat with respect to the periodic box.
A unit cell or periodic boundary "box" is simply a frame of reference. There's no walls. The molecules don't "see" it.
It's merely an indication that there's translational symmetry. In short, in the left-hand view of the second image, the molecule "breaking the box" is also wrapped around into the box by translational symmetry.
I don't know what visualization tool you're using, but if you display more than one unit cell, you'll easily see that the molecules "wrap" around.


text:
Why are antibonding orbitals higher in energy?
I found an explanation on Quora.com
https://www.quora.com/Why-are-antibonding-orbitals-more-destabilized-or-higher-energy-when-theyre-between-two-atoms-of-very-different-electronegativities
I wasn't satisfied with their reason about electrons being inside the two nuclei region.
Why exactly does electrons being shared between atoms reduce the energy? What's the quantum mechanical and mathematical argument? What's the quantum mechanical and mathematical explanation of the octet rule?
Electrons being shared sounds like repulsion forces would increase. Anti bonding sounds like a way to reduce electron repulsion by the coulomb force.

Why are you mixing octet rule with it? Also answer there isn't bad and while asking about QM you talk about Coulomb... What do you really want?

QM and Coulomb.

@Nim If you want to see the quantum mechanical details of how bonding (via orbital interaction) works on a basic level have a look at around the first 20 pages of this book. You will need some knowledge of QM though.

You shouldn't stop asking "why," but really explaining why is really, really complicated. It sort is like learning to run. First you learn to waddle around, then you can walk, and then you can run. Starting off such an explanation with quantum mechanics isn't throwing you in the pool it is more like throwing you in the middle of the Pacific Ocean.
Start with this. I assume that you're happy with the shell structure of atoms. Now we want to make molecules.
Consider the reaction: $$\ce{H + H -> H2}$$
Two hydrogen atoms each with a $\ce{1s^1}$ electron bond to form a "molecule" with 2 electrons in some sort of orbital. There is no shell model for a molecule, so how do we conceptualize the two hydrogen orbitals combining? A straightforward approach is the linear combination of atomic orbitals (LCAO). So we have a $\ce{1s}$ orbital from the first hydrogen and a $\ce{1s}$ orbital for the second hydrogen. Remember that each of these orbitals can hold two electrons. We now, with some nativity, just assume that the energy of atomic orbitals must be conserved. So how much in energy the molecular orbital is lower than the hydrogen orbital then the other orbital must exist with that much "antibonding" energy. But for $\ce{H2}$ there are only two electrons, thus the antibonding orbital is empty and the net result is that a $\ce{H2}$ molecule has less energy than two lone H atoms and the molecule will stay together.
Now consider the reaction:
$$\ce{He + He -> He2}$$
Both He atoms have two $\ce{1s}$ electrons. Thus when we try to form a $\ce{He2}$ molecule both the bonding and the antibonding orbitals are occupied. The net result is that the atoms in the $\ce{He2}$ molecule have no net affinity to stay bonded together. Notice also that this would not be true for a $\ce{He2^+}$ ion. With two electrons in the bonding orbital and only one in the antibonding orbital the ion would be "stable" in free space - at least until the ion bumped into some other atom or molecule in interstellar space.
Now using LCAO is naive. However it works well for a lot of chemical reactions. The really nice thing is frankly that the mathematics is fairly simple. Think of Newton's gravity equation. It worked well enough to put men on the moon, but it isn't exact. Now with GPS satellites the GPS clocks must be corrected for relativity. Scientists knew about relativity before we went to the moon of course, but the expanded mathematical treatment just wasn't necessary.
So does that answer your question in enough detail so that you can accept the explanation?
Edit - adding information about quantum mechanics
The easiest treatment of atomic orbitals is to assume hydrogen like orbitals and tweak the value of the orbitals based on the interactions of electrons. The gist here is that the simplest mathematical treatment assumes that the electrons move around the nucleus in defined orbits like the planets move around the sun. It was quickly realized that this sort of modeling was woefully inadequate.
Even with this simple treatment there is a massive fundamental problem that bedevils all such calculations. The three body problem in physics. If our solar system just had the sun and the earth, then it would be possible to use Newton's gravity formula to solve the equation which predicts the movement of earth exactly. However when the rest of the solar system is added, then an exact solution is no longer possible. We can solve the movement numerically by essentially calculating where the planets will be after some time $t$. So based on where all the bodies are in the solar system we calculate a vector for earth and all of the other planets and advance the time by an hour. We repeat the calculation 525,960 times and we can predict where all the planets will be exactly one year from now.
Now skipping several other mathematical models which offered better solutions than the planetary model, we arrive finally at the Schrödinger equation which gives quantum mechanical calculations for orbitals. The high level results are threefold. First the Schrödinger equation is subject to the same three body problem as before so an exact solution is only possible for a system with one electron and the nucleus. Second the electron orbitals are not discrete paths through 3D space but rather smeared out probability functions. Third different types of suborbitals (s,p,d and f) have different shapes.
Now with the more realistic model for atomic structure we can again combine atomic orbitals to get molecular orbitals, with the new constraints introduced by quantum mechanics. But as we did before, when we create a molecular bonding orbital from two hydrogen atoms, then we get an antibonding orbital as well. Thus the same explanations used for hydrogen and helium bonding as before still apply, however the actual values of the bonding energies do change.

I'd be remiss if I didn't point out something else. Going back to the the motions of the planets in the solar system we also stumble into something known as chaos theory in mathematics. Because of all the interactions small errors in the mass or velocities of the planets would result in much different results for the calculations. This is important in chemistry because the ethane molecules in a gas aren't a static entities, but rather dynamic ones. The molecules are rotating and the atoms are vibrating. Molecules are bumping into one another causing rotations and vibrations to change.
Now trying to model all of this in one model is too complex for us to handle yet computationally. So chemists build models for different aspects of the problem and then try to piece together an overall picture.
In the olden days, AAA would print a route map for someone wanting to drive from Augusta, Maine to Tampa, Florida. They didn't draw the whole route on a massive roll, but rather pieced the route together. Each page showed the pertinent decisions relative to that stretch of road. So a page in a large city might be a few miles, but out in the country a page might be a 100 miles.
So rather like AAA, professional chemists don't have one massive bonding model that we keep hidden from new students. Rather we describe atomic orbitals and molecular bonding in a number of different ways depending on the situation. By piecing the overall model of a molecule together we can understand predict overall behavior.

are you unwilling to provide the quantum mechanical explanation?

LOL- see https://en.wikipedia.org/wiki/Quantum_mechanics

which specific section deals with chemical bonding? I looked at the infinite potential well and the particle in the box part. I don't see the connection between infinite potential wells and chemical bonding.  Is there a connection between infinite potential wells/particle in boxes and chemical bonding?  I know that particle in the box problems are part of physical chemistry studies.

@Nim as far as chemical bonding is concerned, p in box is a far too primitive model even for the simplest molecule, H2. It is only really applicable to long conjugated systems and even then it is not very accurate. It is more relevant in other parts of physchem e.g. translational partition function in statmech but that has nothing to do with what you are asking. Our current QM understanding of bonding is derived from obtaining approximate solutions to the Schrodinger eqn for the molecule in question. It is far more complex than the introductory QM models (p in box, harmonic oscillator etc.)


text:
Can thionyl chloride convert two equivalents of carboxylic acids to acyl chlorides?
Specifically, I've outlined the arrow pushing heuristic for getting from a carboxylic acid to an acyl chloride in red. Now the mechanism outlined in green is the standard route for what I'm told happens with the release of SO2 gas. However, how stable is this intermediate and is it possible that it undergoes conversion twice as I've drawn in orange?

No. The byproduct of the reaction of thionyl chloride with an acid is unstable and it decomposes very quickly. And even if the compound would be more stable, it would be probably less reactive than thionyl chloride.

Cool, thanks. Why did you put your response as a comment instead of an answer?


text:
Structure of chromium trioxide
According to most sources, $\ce{CrO3}$ has the following Lewis structure (courtesy ChemSpider):

However, it is also not planar (courtesy Wikipedia):

Implying that there is a lone pair somewhere.

What is the electron configuration of the $\ce{Cr}$ atom?
Which orbital is responsible for the three double bonds?
Where does the lone pair come from?
How would its MO diagram look like?

Possible duplicate of What is the color and structure of CrO3?

@ringo Not even close. None of my questions here is answered there.


text:
Why is iodine the heaviest bonded element in circumstellar/interstellar space?
As I understand, the heaviest bonded element found in space to date is iodine, in the form of iodide in sodium iodide. This is the only element heavier than iron on this list, and that makes sense to me because the nuclear fusion inside stars only continues until iron. The only way that heavier elements are formed is when a star is near death and the energy is so high in the center of the star that more nuclear fusion occurs, and then the elements are flung into space.
If this is true, then wherever they found the sodium iodide should be traces of more heavier-than-iron elements. Why then hasn't an element such as, say, bromine been found? Especially when it behaves so similarly to iodine.
This had stumped me for a while and information on the topic is sparse, so I'd just like some input on if it may be a limitation on observation of space or an astrochemical scenario where somehow iodine is more likely to form bonds in space than its heavier-than-iron counterparts

Why "bonded"? Is any atom not bonded except the noble gases (and some hydrogen)?

I disagree with the assertion that Sodium Iodide has been found in circumstellar or interstellar space.
Sodium Iodide is not on the list Molecules in Space kept updated by University of Köln.
Instead it is likely that someone misunderstood an article such as Interstellar NaI absorption towards the stellar association ARA OB1, where NaI does not mean sodium iodide, but instead means neutral atomic sodium.
Also, LaO (lanthanum oxide) has been found in stellar atmospheres, and lanthanum is heavier than iodine. See table 5 here: http://www.ucl.ac.uk/phys/amopp/people/jonathan_tennyson/papers/311.pdf
as well as the 1948 article On the Occurrence of Lanthanum Oxide in S-Type Stars.

By "bonded", you mean in the form of a molecule or molecule ion? Only gaseous molecules can be found in space, because they have sharp spectroscopic lines. Heavy compounds condense on-/into stellar dust, which makes them mostly indistinguishable.
Iodine is heavy, but also forms only weak bonds, and such has high vapour pressure, simply speaking. Also true for any iodine compound, check the boiling points of the sodium halides.


text:
What are some problems that arise from using LCAO (Linear Combination of Atomic Orbitals)?
What are some problems that arise from using LCAO (Linear Combination of Atomic Orbitals)?

Well, the biggest problem is not getting the right answer... I assume you have something more specific in mind?

Do you mean in comparison with plane waves, Wannier functions, or GTOs vs. STOs, ...?

@pentavalentcarbon I thought that basis sets didn't matter whether we are using LCAO or not...Isn't this the case? Also, I am getting sort of confused. What do people generally have in mind when they say LCAO in computational chemistry? Is it initialization of the wavefunction? Or how basis sets are arranged to get the wavefunction (even after initialization)?

@JonCuster Or is it how we solve the one-particle problem?


text:
What is responsible for making the bonds identical?
Here is an MO diagram of methane found on the internet:

There are 4 bonds here, one involving the 2s of carbon, three involving the 2p of carbon, and they have different energies.
However, I am told that the four bonds are actually equal.
Then what made them equal? Also, I am told that hybridisation does not exist.

Re: "I am told that hybridisation does not exist." Think of the Kekulé structure for benzene. Benzene doesn't "vibrate" between the two double bond models. Bonding using Molecular Orbit Theory shows that benzene is really aromatic. So MOT is a both a more sophisticated and more realistic model for bonding, but it is more complicated.

The hydrogens are all equivalent. When thinking about MOT, you have to throw almost everything you thought you knew about bonding out the window. The electrons are not localized between any two pair of atoms—they are truly molecular orbitals. This graphic nicely shows the phasing of the molecular orbitals of methane:
$\hspace{4cm}$
[source]
The lowest energy orbital is a linear addition of the carbon $\mathrm{2s}$ orbital and all four of the hydrogen $\mathrm{1s}$ orbitals. The pair of electrons occupying this orbital is not specific to any one hydrogen, and bonds all the atoms in methane together (even the hydrogens to the other hydrogens). The other bonding orbitals are of degenerate energy and result from linear combinations of the carbon $\mathrm{2p}$ orbitals and the hydrogen $\mathrm{1s}$ orbitals.
Naturally a molecular orbital that results from a combination of a $\mathrm{s}$ orbital will be of lower energy than one than one that results from a $\mathrm{p}$ orbital.

May I ask which software you used to draw the diagram?

Also, how did you obtain the energies?

I like your explanation. A simple rule taught which may be too simple is that a lower energy MO is a stronger bond and these are (geometrically) shorter...why doesn't the MO theory match the geometry of the shape? or am I overlooking something in your explanation?

The general answer is that the bond lengths  between any two types of atoms in a molecule will be the same if exactly the same forces forming the bond act on each atom, i.e. those resulting from electrostatic interaction between electrons and nuclei. Thus, in a molecule such a methane we expect all C-H bonds to be the same length, as there are only one type of  C-H bonds, in other molecules we expect each C-C, C=C, or S=O bond etc.  to be the same as others of the same type, unless other nearby atoms perturb them causing a change in electrostatic interactions and so bond length. We know by experiment that the typical, or average, values for CC single, double or triple bonds are approx 154, 134 & 120 pm respectively (& approx 140 pm in aromatics), but actual values differ very slightly between different types of molecules as can be observed by x-ray crystallography, or for small molecules, microwave spectroscopy in the vapour phase.
A conventional molecular formula can be misleading, for example, in $\ce{HNO3}$  one NO bond has a different length to the two others because the structure is $\ce{HONO2}$ and so one oxygen is bound as H-O-N, the other two as N=O. There are clearly many other examples such as this.
In the case of methane, and numerous other molecules such as $\ce{H2Be, H2O, NH3 SO2 NO3- }$, benzene  etc., bonds are the  same length because there are only two types of atoms present and the molecules have some symmetry. However, there are very important exceptions, e.g. $\ce{[CoF6]^{3-}}$  where there are also only two types of atoms. Unequal bond lengths occurs in many other complexes with one type of ligand, such as $\ce{[Cu(H_2O_6)]^{2+}}$, where the degeneracy of an electronic state plays a role in stabilising the energy by what is called the Jahn-Teller Effect. Organic molecules such as cyclobutadiene ($\ce{C4H4}$) also show JT distortions.
The Jahn-Teller theorem states that any non-linear molecule in a degenerate electronic state (orbit or spin) will be unstable and by distortion will lower its symmetry and split the degeneracy. It does not state what the distortion will be, i.e. how the molecule’s geometry will change or how big this change will be, just that it will occur and have the effect of lowering symmetry and energy. (It also states that if there is a centre of inversion this will be maintained). The figure shows, in a very schematic way (as energy gaps are not to scale), what happens to d orbital energy levels for an octahedral $\ce{ d^9 Cu^{2+} }$ complex.

Note that the average orbital energy remains the same after applying the JT interaction (right), the doubly degenerate $e_g$ levels rise and fall equally, and with the triply degenerate $t_{2g}$ levels, two levels fall by 1/3 of the splitting, the other rises by 2/3.
As can be seen in the $\ce{ d^9 }$ complex there is no stabilisation of the $t_{2g}$ levels as they are all filled, but there is a stabilising energy of the $e_g$ by $\delta _e/2$ and this is achieved by lengthening two opposite bonds in this octahedral complex. The two axial ligands are approx 238 pm long and the four equatorial ones (those in a square plane) are approx 195 pm. The effect of Jahn-Teller splitting can often be observed in the visible absorption spectrum of complexes. In the $\ce{V^{IV} d^1 complex VCl4}$, the JT effect occurs and at low temperature this molecule forms a elongated tetrahedral structure.

So which interesting question are you actually answering? Not the one presented here, unfortunately.

yes actually it it is an interesting question, but so as to not to appear patronising I've removed this comment.  I answer the question generally in the first paragraph, then point out that in molecules such as the OP mentions in his question, there are cases when bond lengths can be different even though at first sight one would expect them to be the same.

Oh, I too believe this question here is interesting. However, it is specifically asking for methane, and why the bond lengths are equal while the MO diagram seems to suggest inequivalent bonds. Unfortunately, even with your edit you do not address that adequately; especially since 70 or so % of your answer deal with other compounds that don’t need to be mentioned.

The title to the question is quite general about bonds;  then asks about methane. I answer the general question as it also applies to methane; then as I say in my previous comment, I point out that its not as simple as using MO blindly as it will give the wrong answers in some cases. It seems very restrictive and possibly misleading  just to answer wrt methane when there are other molecules to which following the same treatment will produce wrong answers. I can see we probably wont agree, however.

@Jan lol calm down, i think that the idea of answering in general instead of specific is quite interesting, although s/he really didn't talk about MO diagrams.


text:
Iodine/ Magnesium Exchange reaction?
Traditional Grignard reagents are made by reacting alkyl halide with Magnesium metal. However, the preparation and scope of their application have been limited because of low functional group tolerance.
Until recently, Knochel and coworkers have demonstrated the functional-group tolerance of magnesium-halogen exchange, which is now the method of choice for the preparation of highly functionalized organomagnesium reagents at low temperature as shown in the example

Turbo Grignard reagent ( isopropyl magnesium chloride) is used to change iodine or bromine with Magnesium.
what is reaction mechanism for exchange between iodine and magnesium?

Note that iPrMgCl itself isn't TurboGrignard. TurboGrignard is iPrMgCl•LiCl

http://chemistry.stackexchange.com/questions/18846/grignards-do-transmetallation Thermodynamics basicity & kinetic basicity