text:
Stabilization by resonance
My textbook (NCERT Chemistry. Part I. Textbook for Class XI [1, p. 106]) states that

Resonance stabilizes the molecule as the energy of the resonance hybrid is less than the energy of any single canonical structure

by giving two examples: $\ce{O3}$ and $\ce{CO2}.$
While I can see that clearly in case of $\ce{O3}$ molecule, I can't see that how resonance lead to lowering of energy as in case of $\ce{CO2}$ as two of its canonical structures have formal charge separation, while the one without formal charge separation seems to be more stable than the other two and hence the overall resonance hybrid.
It looks to me that the structure of $\ce{CO2}$ without formal charge separation is the correct one but spectroscopic data suggests that it has to be a resonance hybrid.
What am I missing?
References

Ashok Srivastava; Kalyan Banerjee; Gautam Ganguly; Naresh Yadav; Mathew John; Rajender Chauhan. Chemistry. Part I. Textbook for Class XI; NCERT: New Delhi, 2006. ISBN 81-7450-494-X.

What the book means is probably that the actual structure of CO2 has a bond length shorter than that of a "typical" C=O bond (in carbonyls, around 123 pm; in CO2, 116 pm; in CO, in which a pure triple bond is present, 113 pm), so it has a partial triple bond character, as the canonical forms with charge separation suggest

@The_Vinz I like your about me column :)


text:
How many transition elements are there?
I searched the internet and most of sites say there are 38 transition metals. But our textbook says that group 12 or the group of zinc can not be counted as transition metals, which already puts the number to 37.
Moreover, Sc, La and Ac also can't be counted as transition metals. That puts the number to 34. Y being in the same group as Sc should also be cancelled as transition metals making the number 33.
These are all my thoughts and haven't been varied by any reliable sources. And also it's based on our textbook. I would like to know what is wrong with these thoughts.
Note: The book considers thorium a d block element. Thus, according to the book the number of d block elements is 41.

In my point of view, the transition elements are all those belonging to the "small columns" of the table, whatever their chemical properties. So, they contain atoms from Sc to Zn, and the same in the next lines.

AFAIK, whole group 12 doesn't fit criteria for belonging to transition elements. Why isn't Sc or Y transition element?

According to definition ... To become transition element an elements STABLE ION must have unpaired electron in d orbital. The stable ion of Sc is 3+. In this ion there is no electron in d orbital. For this reason, Sc and Y is not transition element. This is the theory given in our book.

Related:  https://chemistry.stackexchange.com/questions/119066/is-scandium-considered-a-transition-metal?r=SearchResults.  Note that the emergence of organometallic chemistry enters into the reckoning, as demonstrated with scandium.

"The book considers thorium a d block element." That depends on your point of view. See this answer for a discussion of what happens in the metal versus gaseous atoms.

Shakespeare apparently said "What’s in a name? That which we call a rose, by any other word would smell as sweet." Rule no. 1, don't trust the internet until and unless you have some idea about the credentials of the website (and this comes with experience).
The usage of the term transition metal is nothing but historical. You can stick to IUPACs definition (https://goldbook.iupac.org/terms/view/T06456)

An element whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete d sub-shell.

By that zinc should not be a transition element. The definition has nothing to do with paired electrons as you wrote in the comments.
Look at the history of word, this is from 18th century from the unabridged OED.

transition element  n. Chemistry  [after Russian perexodnyj èlement (see transitional element n. at transitional adj. and n. Compounds)] (originally, now historical) any of the nine metallic elements forming group VIII of Mendeleev's periodic table; (now) any of a large class of metallic elements occupying the central block of the periodic table (groups IV b–VIII, I b, and II b, or 4–12, the d-block), which are characterized by partly filled d orbitals, commonly show variable valency and an ability to form coordination compounds, and form many coloured compounds; often extended to include also those elements having partly filled f orbitals (the lanthanides and actinides).

Doesn't all group 12 elements (not just Zn) fail to satisfy the definition of transition elements?

Right, as per IUPAC, Group 12 is not a transition metal group.

First and foremost, d-block elements and transition elements are two different entities.
These terms become not exchangeable as you start using current IUPAC definition of transition element (see M. Farooq's answer).
Older textbooks included entire d-block, including Group 12 elements, into transition elements cohort because historically the name "transition" referred to the generic chemical properties ranging in-between those of elements from s- and p-blocks, at the same time ignoring lanthanoids and actinoids.
I think I found quite a strict yet simple definition that is also aligned with the IUPAC's one, and allowing to actually count the transition elements [1, p. 1357]:

transition element.
(transition metal).
Any of a number of elements in which the filling of the outermost shell to eight electrons within a period is interrupted to bring the penultimate shell from 8 to 18 or 32 electrons.
Only these elements can use penultimate shell orbitals as well as outermost shell orbitals in bonding.
All other elements, called “major group” elements, can use only outermost shell orbitals in bonding.
Transition elements include elements 21 through 29 (scandium through copper), 39 through 47 (ytterbium through silver), 57 through 79 (lanthanum through gold), and all known elements from 89 (actinium) on.
All are metals.
Many are noted for exhibiting a variety of oxidation states and forming numerous complex ions, as well as possessing extremely valuable properties
in the metallic state.

Following this definition, there are nine transition elements in 3d-series, nine in 4d-series, 23 in 5d-series (including lanthanoids, the inner transition elements) and [currently] 30 in 6d-series (provided there are 118 elements and oganesson is the heaviest known element), totaling 71 transition elements (2021):

Adapted from Periodic table, not highlighted.svg by Crossover1370
Strictly speaking, there is little known about electron configuration for elements with $Z > 108,$ so calling out an exact number of transition elements is likely going to be considered a bold move for the foreseeable future.
Reference

Hawley’s Condensed Chemical Dictionary, 16th ed.; Larrañaga, M. D., Lewis, R. J., Lewis, R. A., Hawley, G. G., Eds.; John Wiley & Sons, Inc: Hoboken, New Jersey, 2016. ISBN 978-1-118-13515-0.

"Strictly speaking, there is little known about electron configuration for elements with  Z>108..." - What can we say about Cn (since it belongs to the group 12)? Its atomic number (Z = 112) is greater than 89 (actinium) which indicates that it is a transition metal, but also greater than 108. So, is there an uncertainty regarding it (probably the reason why textbooks don't mention if Cn is a transition metal or not)?

@Pal Predicted electron configuration suggests that Cn is a transition element that follows the aforementioned trend, and to prove it otherwise one would need to use experimental methods, which is pretty much impossible at the moment considering how unstable the element is.

@andselisk If we restrict ourselves to periods 4 to 6, is it okay to refer to elements in column 12 as post-transition elements instead? Or to be even more specific, d-block post-transition metals?

The term "transition" has a geometric origin. The periodic table contains big and small columns, The big columns are number $1$, $2, 13, 14$, etc. The small columns are number $3$ to $12$, and they are all situated in the "transition" zone between the big columns $2$ and $13$. So $\ce{Zn}$ and the other elements of the column $12$ are called transition elements.

source: Stack Exchange

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text:
NMR Coupling in an aromatic system via π system
I was presented today with the following molecule:

and asked the splitting pattern of $\ce{H_a}$ given that up to $\mathrm{^3J}$ and $\mathrm{^4J}$ coupling is allowed.
I thought that $\ce{H_a}$ couples to $\ce{H_b}$, $\ce{H_c}$ and $\ce{H_d}$ given that

there is a $\mathrm{^3J}$  coupling to $\ce{H_b}$ (Clockwise)
there is a $\mathrm{^4J}$  coupling to $\ce{H_c}$ (Clockwise)
there is a $\mathrm{^4J}$  coupling to $\ce{H_d}$ (Anticlockwise)

Hence the splitting would be ddd.
However, the answer is apparently dd for the splitting of the $\ce{H_a}$ signal.
My thought was that anticlockwise, the $\mathrm{^3J}$ coupling is not possible due to the absence of a π bond in that direction down to $\ce{H_d}$ hence the coupling through the π system would not be possible.
Am I correct or is there another reason for no $\mathrm{^3J}$ coupling to $\ce{H_d}$?

Please note that using MathJax in the title creates issues; \ce{…} should only be used for chemical compounds and reactions. Also, I added the compound name as image metadata (it's better to fill in something meaningful in place of "Enter image description here"), and I suggest you do the same in your future post as it makes finding your post much easier.

@andselisk why does it create issues in the title? is mathJax not compatible with other browsers or other mobile devices?

See here and there. Besides, in this case there is literally no justification for typing π with MathJax as it can be easily inserted as a Unicode character.

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text:
Are Phenol and Alchohol both hydrocarbons?
In many books there are said that Alchohols are hydrocarbons, and some says they are not. On the other hand Phenols are said aromatic hydrocarbons. It is well known that Phenols are not Alchohol. Which is correct and why?

Hydrocarbons do not contain oxygen atoms.

So, how Phenol can be aromatic hydrocarbon?

Can you write a brief about that as an answer please? Please. :), I'm a school student learning organic chemistry at a primary stage.

Can you name the books that say such nonsense, so that others might be warned about them?

Phenols are hydroxyderivates of aromatic hydrocarbons, with hydroxy group attached to an aromatic benzene ring, or more generally aren carbon ring.
Alcohols are hydroxyderivates of alifatic hydrocarbons, with hydroxy group attached to a saturated carbon atom.
Hydrocarbons contain hydrogen and carbon, nothing more. Phenols and alcohols contain also oxygen, so they are not hydrocarbons.


text:
Can this sp2 lone pair resonate?
In methyl isocyanate, $\ce{CH3CNO}$, shown below,

 it seems that the nitrogen lone pair can resonate with the carbon oxygen double bond, generating a resonance structure something like acetonitrile oxide:

In that case, why is the geometry around the nitrogen in the methyl isocyanate compound bent and not linear? Can the lone pair even resonate like that?
I remember in Klein's organic chemistry, it is stated that

Whenever an atom possesses both a π bond and a lone pair, they will not both participate in resonance

In methyl isocyanate, $\ce{CH3CNO}$, shown above, it seems that the nitrogen atom has both a π bond and a lone pair. So perhaps the sp2 lone pair on nitrogen atom cannot resonate with the carbon oxygen double bond? I am not sure.

I don't understand the title. Vinyl is CH2=CH-. It has no lone pair !

@Maurice I mean vinyl as in lone pair on a double bond (such as in N)

Whenever an atom possesses both a π bond and a lone pair, they will not both participate in resonance

The sentence means that only one, either the  bond or the lone pair will participate in resonance if the atom has both.

For example, in a pyridine, the nitrogen has a lone pair and it is also attached to a  bond.
Here, however, only the  bond participates in resonance because it is in the plane perpendicular to the molecule, while the lone pair is not.

Methyl isocyanate is best represented by a group of three resonance structures:

Here is the 3D structure of the most significant resonance structure of methyl isocyanate:

So C and O are both sp2 hybridized?

The first Carbon is sp3, the second sp. The oxygen atom must have either sp2 or sp hybridization, because it
needs a p orbital to participate in the C–O π bond. This oxygen atom has three
attachments (the carbon and two lone pairs), so we use sp2 hybridization.

Did this answer ur question? @DrPepper

kind of? I know in amides the N looks sp3 but it is actually sp2 because it is conjugated to the carbonyl, so I was wondering why that doesn't happen here. The N looks like it should be sp2, but if this resonance works, it would be sp because of conjugation to the C=O in H3CNCO

@DrPepper Well, look at the most significant structure, N is hybridized sp2.  When an atom that appears to be sp3 hybridized can participate in resonance, that atom is in fact sp2 hybridized. This is because resonance requires an open p-orbital to occur. Sp3 atoms have 4 sp3-hybridized orbitals but no open p-orbitals, and therefore cannot participate in resonance, while sp2-hybridized atoms can.

yes but why doesn't resonance drive the N to be sp hybridized?

If the lone pair on a nitrogen (or on any atom) participates in resonance, then that nitrogen (or whatever) atom must be 2-hybridized so that the the lone pair is in a p-like orbital to ensure appropriate symmetry for -overlap.

Yes, but in the case of H3CNCO, if the lone pair on N were to be in resonance with the carbon oxygen double bond, then you would need an additional p orbital on N, making it sp hybridized (one p orbital bonds with C and one p orbital for the lone pair in resonance). However, N is clearly not sp hybridized. Why is that?


text:
3-methyl-4-oxobutanenitrile or 3-formylbutanenitrile?

Which name is the Preferred IUPAC Name of the above molecule? 3-methyl-4-oxobutanenitrile or 3-formylbutanenitrile?
I feel the second one is the PIN, since it's shorter than the first one. But, I haven't been able to navigate the IUPAC Blue Book, regarding such cases.

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

According to Rule P-44.1.1, the priority list of substituent groups has to be considered first in order to decide the parent chain for this compound. You have already done that in both of your suggested possibilities: The principal characteristic group that is expressed as a suffix is the nitrile group ($\ce{-(C)N}$).
The next rule (P-44.3.2) is looking at the longest chain. Therefore, the parent structure is a butanenitrile.
The following rules (P-44.4.1.1 and P-44.4.1.2) would be looking at multiple bonds, which is not relevant in this case.
Next, Rule P-44.4.1.8 is looking for a lower locant for the attached group that is expressed as a suffix (here: nitrile); however, the nitrile group has the lowest locant ‘1’ in both suggested cases; therefore, this rule doesn't help to make a choice.
The next rule (P-44.4.1.10) again is not relevant in this case.
Finally, Rule P-45.2.1 is looking for the structure that has the maximum number of substituents cited as prefixes to the parent structure. Therefore, the preferred IUPAC name is 3-methyl-4-oxobutanenitrile (two substituents) rather than ‘3-formylbutanenitrile’ (only one substituent).

By way of comparison, if you add more substituents to the methyl group, the principal chain can switch, for example 4,4-dichloro-3-formylbutanenitrile (three substituents) rather than ‘3-(dichloromethyl)-4-oxobutanenitrile’ (two substituents).

The Preferred IUPAC name is
3-methyl-4-oxobutanenitrile.
Reference: Pubchem
Explanation:
This compound has two functional groups. The seniority table says that the Nitrile group has a higher precedence over the Aldehyde/Formyl group. While picking the longest chain, we have the following set of rules (From Wikipedia):

It should have the maximum length.
It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.
...

Here, the longest chain in both cases has 4 carbon atoms. The chain should also contain the maximum number of substituents of the suffix functional group. We then choose the chain containing the oxygen atom attached to it, thereby obtaining methyl as a substituent on C3, giving us the name.
Note that oxo can be used even for an aldehyde (From Wikipedia)

If the compound is a natural product or a carboxylic acid, the prefix oxo- may be used to indicate which carbon atom is part of the aldehyde group; for example, $\ce{CHOCH2COOH}$ is named 3-oxopropanoic acid.

This logic is extendable to include any functional group that is above the aldehyde group in the seniority table.
Answering the doubt you posed regarding 3-formylpentanenitrile in the comments, here, the pentane chain has the most number of carbon atoms and also contains the suffix functional group attached to it, i.e. formyl, which is what makes it the PIN.

"The chain should also contain the maximum number of substituents of the suffix functional group". Isn't the suffix functional group nitrile? If yes, then both the longest chains have maximum number of nitriles, i.e. 1. So, how can we decide which longest chain to choose?

By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. Here, the possibilities are that it has nitrile + formyl or nitrile + oxo groups, in which case we consider the oxo group one.

The preferred IUPAC name 3-methyl-4-oxobutanenitrile is correct; however, the explanations are wrong. Oxo and formyl are not suffixes and have nothing to do with the principal characteristic group that is expressed as a suffix.


text:
Why is this reaction spontaneous in standard conditions?
To study if this reaction is spontaneous I proceed in this way:
$$\ce{2H_{3}O^{+} +Fe ->H_{2} +Fe^{2+} + 2H_{2}O}$$
First I write the two semireaction:
$$reduction: \ \ \ce{2H_{3}O^{+} +2e^{-} -> H_{2}} $$
$$oxidation: \ \ \ce{Fe -> 2e^{-} +Fe^{+}} $$
Then I write the scheme of the cell:
$$\ce{Fe|Fe^{2+}||H_{3}O^{+}|H_{2}}$$
I know that:
$$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}$$
$$E^{0}_{cell}=E^{0}_{H_{3}O^{+}|H_{2}}-E^{0}_{Fe|Fe^{2+}}=-E^{0}_{Fe|Fe^{2+}}=-0.44$$
But also:
$$\Delta G^{0}=-nFE^{0}_{cell}$$
So $$\Delta G^{0}>0$$
And the reaction is not spontaneous, while my textbook says it is, so I must be wrong. Someone could find my error?

Note that it is preferred not to use MarhJax/Latex formatting of titles on Chemistry SE site to avoid search problems.

It has been a while since I took electrochemistry so I'm not sure about the formal procedure that you're using.
You wrote the correct overall reaction.
$$\ce{2H3O+ + Fe -> H2 + Fe^{2+} +2H2O}\tag{1}$$
and you almost wrote the correct half cells reactions (should be $\ce{Fe^{2+}}$ in equation 3):
$$reduction:  \ \ \ce{2H3O+ + 2e− ->H2}\tag{2}$$
$$oxidation: \ \ \ce{Fe -> 2e^{-} +Fe^{2+}}\tag{3} $$
Then you wrote the scheme of the cell: $$\ce{Fe|Fe^{2+}||H_{3}O^{+}|H_{2}}\tag{4}$$ And an equation for the overall cell potential.
$$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}\tag{5}$$
Your problem is between equations 4 and 5. To subtract $E^{0}_{left}$ the potential must be written as the reduction potential for the half cell. So either you're supposed to be:

Adding the values for the right and left reactions in equation 5

-- or--

Writing both half cells as reduction potentials in equation 4.

The gist here is that the tabular values are always written as reduction potentials. So for iron:
$$\ce{2e^{-} +Fe^{2+} -> Fe }\quad\quad E^{0}_{red} = -0.44\tag{6} $$
I was taught to reverse the sign for an oxidation reaction.
$$\ce{Fe -> 2e^{-} +Fe^{2+}}\quad\quad E^{0}_{ox} = 0.44\tag{7} $$
Then add the reduction and oxidation potentials to get the overall cell potential.
$$E^{0}_{cell}=E^{0}_{red} + E^{0}_{ox}\tag{8}$$

The main mistake is rather algebraic and forgetting the electrochemical conventions. When we write this equation, we follow the convention that all electrode potential values will be inserted right from the table with their signs intact, whether it is an oxidation or reduction. Don't change the sign of the electrode potential value. The sign of the electrode potential is invariant.
$$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}$$
So from the table, the Fe(II)/Fe(0) couple is -0.44 V
So, $$E^{0}_{cell}=0.00-(-0.44)= +0.44V $$
when $E^{0}_{cell}$ is positive, delta G value is negative.

I do believe that this is the way that is used in current teaching. The "improvement" is to get the sign change, $-(-0.44)$ on paper rather than doing it mentally.

In old school teaching, the equation was E0cell=E0right+E0left, and the student was asked to change the sign. This was the Gibb's or American convention. Rest of the Europe followed Ostwald's approach, which is taught today.

LOL - My first chemistry class was in 1968 when I was in the 10th grade. "Old school" enough?

That is exactly the time I was talking about. Glad that you saw the golden era of science in the US.

Your mistake is that $E^\circ_\ce{Fe/Fe^{2+}} = \pu{-0.44 V}.$ So that $-E^\circ_\ce{Fe/Fe^{2+}} = \pu{+0.44 V}.$

no... $ E°_{Fe^{2+}/Fe}$ = - $0.44$ V. The table values are always written as reductions.


text:
Why are Newman Projections drawn with 120 degrees when actually sp3 carbons are 109.5 degrees?
Title says the question, pretty much.
For example, in the Newman projection of ethane, the carbon hydrogen bond angles are $120$ degrees, not $109.5$ degrees, as they actually are in an $\text{sp}^3$ hybridized center. Why is this so, and does it have any effect on the arguments traditionally derived from Newman projections?

why the downvote? this is a legitimate question

I also thought to downvote as for is at least bizarre

It is a projection. Depending on the orientation of a water molecule with respect to the projection plane, the projected angle can be anywhere from zero to 180 degrees. Similarly, the angles in ethane projected to a 2D plane will be different from 109.5 degrees unless the three atoms making up the angle are in the plane:

Other projections (Fisher projection) make the angles appear as right angles. In either case, the 3D angles remain at 109.5 degrees.


text:
Epoxide with solvent vs with water

The answer given is an epoxide.
But is an epoxide stable at room temperature?
And also epoxides tend to open in presence of water. Even if methyl alcohol is treated as a solvent, then too methyl alcohol is more acidic than water. Then if in water we expect the alpha-bromide alcohol, then why not in methyl alcohol?

Epoxides are a lot more stable than you seem to think, they do not routinely open in water. The product in this case is styrene oxide which is widely commercially available, stable in MeOH solution, and may be distilled at atmosphere pressure.

Would we get the same product at room temperature with water?

The starting material is poorly soluble in water, I doubt you would get any reaction unless you addded a phase transfer catalyst


text:
Calculating concentration of a powder in a liquid
I'm trying to determine how to calculate proper measurements for a 5% concentration of a vitamin C ester powder (ascorbyl palmitate) mixed in a 50/50 carrier oil mixture (jojoba oil/squalane oil) with 2% vitamin E oil (α-Tocopherol).
The powder is measured in mg and the oils in mL.
Densities:

Jojoba oil = 868 mg/mL
Squalane oil = 810 mg/mL
α-Tocopherol = 950 mg/mL
Ascorbyl palmitate = 1.2 g/cm3

I want to create a concentration of 5% powder in 118 mL of oil. However, the total volume of the entire mixture should not exceed 118 mL, so I would need to also know how much oil to subtract before adding the powder to get the exact final volume desired.
How would I correctly calculate both the quantity of powder to mix in and the proper starting volume of the combined carrier oils?


text:
Symmetry of an FO
I want to ask a question about the symmetry of an FO for $\ce{[CuL2]^-}$ where L is a $1e^-$ donor sigma bonding ligand (e.g. Me).
For the ligand orbitals, I produced the following orbitals shown below:

and also used the character table to assist me in assigning symmetry.

I assigned the axes according to the principal axes, which was in the z-direction:

This is the working I did for the bonding FO:

Under E, no change so give $1$
Under $\ce{2C_{\infty}}$ no change so give $1$
Under $\ce{\infty \sigma_{v}}$ no change so give $1$
Under $i$ no change so give $1$
Under $2S_{\infty}$ no change so give $1$
Under $\ce{\infty C_2}$ no change so give $1$

Hence the overall symmetry is $\ce{\sigma _g^+}$
However, applying the same working for the antibonding combination above:

Under E, no change so give $1$
Under $\ce{2C_{\infty}}$ no change so give $1$
Under $\ce{\infty \sigma_{v}}$ no change so give $1$
Under $i$ change in phase so give $-1$
Under $2S_{\infty}$ change in phase so give $-1$
Under $\ce{\infty C_2}$ no change so give $1$

The overall symmetry I got was $\ce{\sigma _u^-}$ but according to the diagram, it is $\ce{\sigma _u^+}$.
Where was I mistaken?
The entire MO diagram is shown below for reference:

Under inversion 'i' the antibonding orbital is -1, so it has to be  'u' , also the $C_2$ is -1 so this points to $\sigma_u^+$ this being the only possible combination.

@porphyrin please explain briefly in an answer if possible. I thought that because the principal axes is the $\ce{z}$ axes going to the right, rotating in a $\ce{C2}$ fashion i.e. 180 degrees on it, there isn't a change in phase. I agree with the inversion as passing through the centre of the molecule and out the other side with the same distance, I discover a change in phase but I cannot see the $\ce{C2}$ being $-1$ Is it that the $\ce{C2}$ lies along the $x$ and $y$ axes and this is the $\ce{C2}$ you are referring to?

yes; the $C_2$ is in the xy plane as shown in the  answer by vik1245 below.

@porphyrin yes yes I answered my own question! Thanks anyway for the help to make me find it! Upvote my answer if you feel it's sufficient to help other users as well.

What are you using FO as an abbreviation for? I've never come across that abbreviation before and I can't find it online.

This diagram from Molecular Orbitals by David Willock will be sufficient to explain the matter:


text:
clarification on background limits in electrochemistry
I’m teaching myself electrochemistry, and this question is from the general introduction of the book I’m using. According to Bard’s Electrochemical Methods book, in a given cell, the background limits ‘are the potentials where the cathodic and anodic currents start to flow at a working electrode when it is immersed in a solution containing only an electrolyte added to decrease the solution resistance (a supporting electrolyte).’ I don’t fully understand the significance of the italicized, but my main question comes from the following figure, which shows that the onset of H+ reduction and Br- oxidation (Where the background limits are) occur at different potentials. This is counter-intuitive for me, since, according to introductory electrochemistry from gen chem courses, one half of a reaction cannot occur without the other for a given applied potential.

How can this occur?

I like M. Farooq's answer below, but a quick recommendation, Bard&Faulkner is not the best to self teach electrochemistry. I would recommend:
https://www.amazon.com/Electrode-Dynamics-Oxford-Chemistry-Primers/dp/019855690X/ref=sr_1_4?keywords=electrode+dynamics&qid=1577775622&sr=8-4
https://www.amazon.com/Electrode-Potentials-Oxford-Chemistry-Primers/dp/0198556845/ref=sr_1_1?keywords=electrode+potentials&qid=1577775607&sr=8-1

the background limits ‘are the potentials where the cathodic and anodic currents start to flow at a working electrode when it is immersed in a solution containing only an electrolyte added to decrease the solution resistance (a supporting electrolyte).

Bard's book is notoriously complex. Some bright scientists are not good at communication. Richard Feynman was an exception.
All Bard is saying that the background limits define the potential window [see flat line in your figure] where electrolysis does not occur on the working electrode. The background limits are dependent on the electrolyte used, it is also dependent on the solvent and the electrode material as well.
Now keep in mind that modern potential sweep experiments there are three electrodes in a cell. There is a reference electrode, in your case, an Ag/AgBr electrode. Then you have a working electrode, Pt/H+ and then the writer is quiet about the third one, which is called the counter electrode. Most of the time it is understood that it is their. It could be another Pt electrode but not necessarily a hydrogen electrode i.e., nobody is bubbling hydrogen gas over the counter electrode.
Now the reference electrode has no role in the cell. There is no redox reaction taking place at the reference electrode because the circuitry is such that no or very little current passes through the reference electrode. Electrochemists don't care what is happening at the counter electrode. The potential difference is measured between the reference electrode and the working electrode. This is the only purpose of the reference electrode.
When you are sweeping (=continuously changing) the potential of the working electrode with respect to the reference electrode, it can become a cathode or an anode during scanning of the potential. You can scan in any direction. You are right, isolated reduction or oxidation cannot occur. If the working electrode was a cathode at a particular moment, the counter electrode is the anode and vice versa.

You are correct that the potential of an isolated half-cell can't be measured. However the whole table of reduction potentials is measured relative to the standard hydrogen electrode which would be a third electrode in the system.


text:
How to separate a racemic mixture of 1,2-diols
[upjohn dihydroxylation]2I want to ask that i synthesise R-substituted-1,2-Diol with (ee=1:1) ratio of R-, S- with the help of Upjohn Dihydroxylation. i try to seprate it by different mobile phasese but failed to separate, then i try -OTBS at primary alcohol by wish that may it work and make two seprate points at TLC but i failed, i need both R- and S- enantiomers of Diol.
my question is that what possible method can i use to separate two diols completely or make enantio-selective diol by using different conditions like by Sharpless method.

Are the single isomer diols your final product, or are you reacting them further?

I can't open this paper but it looks as if it may be useful   https://www.researchgate.net/publication/244234526_Stereoisomer-differentiating_esterification_of_diols_with_methylboronic_acid_A_simple_method_for_the_separation_of_cis-_and_trans-12-diols

@Waylander: I looked at your link. It uses methyl boronic acid derivatives of 1,2-diols over 1,2-trans diols for cyclobutyl, cyclopentyl and cyclohexyl 1,2-diols. It does not look good for the OP's case.

You might consider using your racemic 1,2-diol to form the ketal of enantiopure camphor. These diastereomeric ketals may have different physical properties, e.g., chromatography BTW, the ee equals 0 not 1:1.

I need to react it further with fattyacids at both OH groups.

Actually when i react primary OH with OTBSCL then it give different polarity on TLC but that time structure was R-CH2-CHOH-CH2OH, but now I am using one carbon long chain. So that's why i tried it with wish it behaves same.

@user55119 can you explain ketal formation with enantiopure camphor? Any reference?

@Dr.usamafarooq typical conditions would be an organicsulfonoc acid and solvent that can separate from water.  So for example, toluene or MIBK with toluenesulfonic acid and Dean-Stark kit (or other water removal technique). Reflux to remove water as it forms. Or use a polymer bound sulfonic acid and you can filter that off afterwards.

The standard method for ketalization is illustrated in the reaction of cyclohexanone 1 with ethylene glycol to form the ketal 2. The reaction is reversible which requires irreversible removal of water shifting the reaction to the right. In this reaction the toluene/water azeotrope may be separated by a Dean-Stark trap. Because of an unfavorable equilibrium, hindered ketones, such as camphor 3, are more difficult to ketalize. Standard conditions may be employed in this instance using more efficient removal of water.1
Dauben, et al.,2 have demonstrated an alternative technique for ketalizing camphor and other hindered ketones by employing high pressure at ambient temperature as illustrated in 3 --> 4.3. To illustrate the resolution technique I have chosen (R)-camphor because it is hindered, inexpensive and recoverable. The full structure of your racemic 1,2-diol was not provided so I will employ racemic butan-1,2-diol to this end. There are four possible diastereomers that can be formed in the ketalization--namely-- 5, 6, 7 and 8. The (R)-diol can form ketals 5 and 6 while the (S)-diol can afford ketals 7 and 8. Because each pair is formed under equilibrating conditions, ideally one would hope that the less hindered of the 5/6 pair and 7/8 pair dominate such that a single ketal would arise for each of the enantiomers of the diol. The success of such a technique requires separation of diastereomers, presumably via chromatography, or, ideally, by selective crystallization. If the latter method were viable, an enantiopure diol can be liberated by acid hydrolysis. The enantiomeric diol would arise by utilizing (S)-camphor, which, unfortunately, is much more expensive than (R)-camphor. continued...

One need not be limited to camphor. Other hindered ketones may be considered. The methodology described here is not a guarantee of success. Conceptually, it is no different from Ladenburg's resolution of racemic coniine with tartaric acid in the 19th-century.
1) Estrone methyl ether was successfully ketalized by user55119 using a spinning band fractionating column to remove water. The Dean-Stark method had failed.
2) W. G. Dauben, J. M. Gerdes and G. C. Cook, J. Org. Chem., 1986, 51, 4964.
3) The role of ethyl orthoformate is to scavenge water thereby forming ethanol and ethyl formate. Ketalization was also accomplished at high pressure using TMSOCH2CH2OTMS/TMSOTf.


text:
Is kinetic energy proportional to temperature for non-gases?

If the molecules in a sample of ice at $\pu{-15 °C}$ and $\pu{1 bar}$ all have their kinetic energy doubled, then what is the final phase?

The answer is gas, but I am not sure how to get this answer.
I was wondering if the equation $\displaystyle k = \frac{3}{2}RT$ still holds in this case, even though the molecules are not a gas. It seems as though it does (it gives the right answer because $2 \times \pu{258.25 K} = \pu{516.3 K},$ which is a gas), but I am not too sure why.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Note that the mean kinetic energy is proportional to absolute temperature for monoatomic gases only. For other gases only in temperature region where vibration is frozen at the base level.


text:
How do you find the bond enthalpy from an equation?

Which equation represents the $\ce{N-H}$ bond enthalpy in $\ce{NH3}?$
$$
\begin{align}
&\textbf{A.} &\ce{NH3(g) &-> N(g) + 3 H(g)}\\
&\textbf{B.} &\ce{1/3 NH3(g) &-> 1/3 N2(g) + H(g)}\\
&\textbf{C.} &\ce{NH3(g) &-> 1/2 N2(g) + 3/2 H2(g)}\\
&\textbf{D.} &\ce{NH3(g) &-> .NH2(g) + .H(g)}
\end{align}
$$

From what I've learnt, I understand that the bond enthalpy is defined as the energy required to break one mole of a specific bond.
In the question above, I opted for answer C as it was the only one with the products in the form of $\ce{N2}$ and $\ce{H2}.$ But since there are three $\ce{N-H}$ bonds in $\ce{NH3},$ I am unsure about answer B.
Can you clarify this?

Option D is the only one corresponding to the definition "energy required to break one mole of a specific bond" and nothing more.

Do you mind explaining what the dots infront of NH2 and H represent?

The dots represent the fact that the species are free radicals (they have unpaired electrons). Strictly speaking, they could have put a dot in front of H in all 4 options and N in option 1.

If you want to have a "none of the above" answer, you could also consider
$$\ce{1/3 NH3(g) -> 1/3 N(g) + H(g)}$$
or written more conventionally, one third of the reaction enthalpy of
$$\ce{NH3(g) -> N(g) + 3 H(g)}$$
The atoms $\ce{N}$ and $\ce{H}$ are radicals but are shown without dots here.
This would be the average of the three N-H bond dissociation energies. The key is not to form new bonds (i.e. no $\ce{N#N}$ or $\ce{H-H}$ as products), and to cleave bonds in a homolytic manner (one electron of the bond remains on each atom).

Nope, the answer is B from my understanding. This is because the definition of bond enthalpy is the energy required to break "one mole" of a bond. One mole of $\ce{NH3}$ has three N-H bonds, so for you to have one N-H bond you would need 1/3 moles of NH3

B is not correct because of the formation of new bonds in the products. For the correct (but not listed) alternative, see my answer.

oh yeah ur right the N2 forms a new triple bond. But this question was actually an IB Chemistry question and they marked B as the correct answer. So technically all of the given answers are wrong right?


text:
How does a halogen react in an electrophilic substitution reaction

From what I understand, electrophiles are electron deficient and therefore would be attracted to the electron-rich benzene ring. So would the correct answer be A?

The double bond in D is also electron rich

A is correct because the question specifically says an electrophilic substitution reaction (D will also react, but it's an electrophilic addition reaction). However, you also need a Lewis acid catalyst like FeCl$_3$. This is because, while the delocalised system of electrons in benzene is indeed electron rich, the fact that it's aromatic (there is a system of pi electrons delocalised throughout the ring) makes it very stable.
The mechanism of the bromination of benzene is shown below (image from Wikipedia as I don't have ChemDraw on this computer). The mechanism is the same for Cl$_2$.

The OP did not ask about electrophilic aromatic substitution.

No, but they asked which molecule reacts via an electrophilic substitution mechanism. In this case it happens to be an electrophilic aromatic substitution, and I thought providing the mecanism would help to explain how this happens.


text:
How do you determine if a compound can go through addition polymerisation

For a polymer to form from a monomer, the monomer requires open bonds to attach itself to another monomer. Logically for me, it requires a double bond so I would go for answer C. But can the benzene ring actually open up to undergo an addition polymerisation?

What do you think?

I think the question in the picture could use one additional letter.

Hint: Benzene has a delocalized electron cloud which is not exactly equivalent to a double bond.

hint: basically all polymerisation reactions are exergonic. Would that apply for C? (And would it for A, btw.?)


text:
Radius of hydride (H⁻) ion
What is the radius of the hydride $(\ce{H-})$ ion? There are several sources with conflicting reports on the same.
Inorganic Chemistry: Principles of structure and Reactivity cites Pauling’s 1960 study of the hydride ion, stating that

it has a radius of 208 pm compared to 216 pm for the iodide ion

This would imply that it is larger than the fluoride, chloride and even the bromide ion! This article in nature disproves the above and states:

the ionic radius of an $\ce{H-}$ ion is 134±2 pm, which is slightly larger than the radius of $\ce{F^-(II)}$ (128.5 pm) and almost the same as the radius of $\ce{O^2-(II)}$ (135 pm). This evaluation agrees with the generally accepted trend that the radius of an $\ce{H-}$ ion is nearly the same or slightly larger than that of an $\ce{F-}$ ion.

Which one of these is correct? Also, could additional citations be given to tell the exact size of the hydride ion?

Unfortunately, there is no caliper-like tool to measure ionic radii directly.
They are determined based on experimental data such as crystal structure or crystal lattice energy using empirical relationships.
The exact value would inevitably deviate from one method to another.
Pauling [1, pp. 150–152] used a hard-sphere model which doesn't account for overlap in crystal lattice and was predicting higher interatomic distances compared to the crystallographic data, hence an overestimated value of the anionic radii.
Several limitations of the hard-sphere model can be resolved by applying a soft-sphere model.
According to the recent work by Lang and Smith [2], who also compared applicability of both hard- and soft-sphere models, $r_\mathrm{i}(\ce{H-}) = \pu{139.9 pm}.$
CRC Handbook of Chemistry and Physics [3, p. 12-22] lists the value of thermochemical radius determined using Kapustinskii equation from a set of known lattice energies: $r_\mathrm{i}(\ce{H-}) = \pu{(148 ± 19) pm}.$
References

Pauling, L. The Chemical Bond; Cornell University: Ithaca NY, 1960.
Lang, P. F.; Smith, B. C. Ionic Radii for Group 1 and Group 2 Halide, Hydride, Fluoride, Oxide, Sulfide, Selenide and Telluride Crystals. Dalton Trans. 2010, 39 (33), 7786. DOI: 10.1039/c0dt00401d.
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.

Since the question is about the exact size — and additional citations are requested — it may be worth mentioning Randell Mills' model which calculates the hydride radius analytically as $1+\sqrt{s(s+1)}$ times the Bohr radius, where $s$ is the electron spin of $\frac{1}{2}$. This works out to 99pm and represents the radius of a free hydride ion.
This is different from the ionic radius in a crystal lattice. Depending on your needs, one value or the other may be useful. For example, Mills' radius allows him to calculate the ionization energy of $\ce{H-}$ to within experimental accuracy with a simple, analytical formula (see also here).
References:
Randell L. Mills, The Grand Unified Theory of Classical Physics; 2018 (ISBN: 978-0-9635171-5-9)

Well since the ion looks like the helium atom but without 1 proton and since the helium core  is approximately 4 times heavier than the hydrogen core and if we neglect the magnetic interactions between core and electrons i would expect it to be 4 times the radius of helium = roughly 125 pm

I wanted to say ion instead of atom in the first row.

I'm afraid that's not how it works.

You can edit your answer. You should certainly expand your answer and tell which formula and what theory you used.

@Andselisk how did he reach to that conclusion? I mean what (which theory or formula) did he use?

@Zenix How am I supposed to know? Maybe a linear proportion, from the look of it.


text:
Effect of impurities on Enthalpy of vapourization
How does the Enthalpy of vapourization change when impurities are added? I know that the boiling point of a substance increases on addition of impurities, but how does the Enthalpy of vapourization change? Is less heat needed to convert the liquid from a solution to a gas?
Related: This question deals with a similar problem, but for Enthalpy of fusion instead of Enthalpy of vapourization.

As long as we talk about colligative properties, this should be a simple piece of math.  And yes, delta H must be lower i guess.

Are you talking about non-volatile impurities?

Yes @ChetMiller

Adding impurities(which mostly have low heat capacity, like salt) decreases the enthalpy of vaporization. This is because the impurities lower the heat capacity of the solution as a whole, making the enthalpy of vaporization lesser than the original solvent.
This is why even though the boiling point of a solution increases when impurities are added it is heated up faster(boils faster) than the one without impurities. For example, saltwater (salt is a non-volatile solute) boils faster than pure water because adding salt lowered the heat capacity of water as a result of which the enthalpy of vaporization is lesser.
Sources:

https://www.studocu.com/en/document/western-governors-university/natural-science-lab/mandatory-assignments/task-1-boiling-salt-water/5667142/view
https://www.thoughtco.com/adding-salt-to-boiling-water-607427

The addition of impurities lowers the heat capacity since the solvent molecules bind to the solute particles in a specific manner and hence require less heat to vaporize than they'd require when they're randomly moving.
Source:

https://socratic.org/questions/how-does-salt-change-the-specific-heat-capacity-of-water#143950

I think this a bit of a circular reasoning. Why does adding impurities lower the heat capacity?

@Karl https://socratic.org/questions/how-does-salt-change-the-specific-heat-capacity-of-water#143950

Sorry, thats not the answer. What if the "impurity" is not so well soluble as table salt, say, ether? Also your answer should be understandable without linking to an external Q&A site.

@Karl The OP mentions non-volatile impurities and ether is not one.
High volatility and less solubility (1.5g ether/100g water) would cause ether to vaporize much before water (93.95% of ether floats on water). Since we're talking about enthalpy of vaporization, much less heat would be required to vaporize all of the ether (because of the aforementioned reasons) than the same amount of water. Hence ether-water mix would have a lesser enthalpy of vaporization than water.

OK, replace "ether" by "propylene glycol".

@Karl Propylene glycol is soluble in water. The specific heat capacity of propylene glycol is 2.5kJ/kg K and that of water is 4.18kJ/kg K. Mixing 0.5 kg of both results in a mixture the heat capacity of which is 3.34kJ/kg K. Hence adding it lowers the heat capacity.

But you said it depends on the interaction, how the "the solvent molecules bind to the solute particles"!

@Karl NaCl in water does interact in that way. I can't confidently say that about other molecules since I'm not familiar with other examples but there has to be some interaction between the molecules for them to be soluble.


text:
Chemical Kinetics(Hypothetical elementary reaction)

For a hypothetical elementary reaction(image).
Initially, only 2 moles of A are present. The total number of moles of A, B, and C at the end of 75% of the reaction is:

$75$% of $2$ moles of A is $1.5$ mole. $1,5$ mol A produce twice as much of B or C. This makes $3$ mol of B and/or C. It remains $2$ - $1.5 = 0.5$ mol A. The total number of moles is $3 + 0.5 = 3.5$ mol. Where is the difficulty ?

Thank you and sorry if this question felt too easy.


text:
Does water in big cities get heated (boiled) for treatment?
I live in Bangkok and read how water comes to customers here:
I understand from the chart that water go through Thon Buri treatment plant (west Bangkok) and Mahasawat treatment plant (east Bangkok) before being pumped and transferred to the customer.
In the chart I clicked the links for these two treatment plants and read information in other parts of the site and understood water is treated at least either by chloride, fluoride and sulfate based chemicals but it wasn't clear to me if water is purified from bacteria by heating (boiling).
Does treated water in big cities get heated (boiled) to destroy bacteria as well as proteins, and if so, where does all the amino acid residue go to?

The most important parts in freshwater treatment is clarification (get the turbidity, caused by microscopic impurities, down, by sedimentation with the help of clarifiers). That already takes care of the greatest part of bacteria and other single cell organismns. Then comes filtration, and the tiny rest that still comes through there is killed via chlorine (or, more modern, ozone).

No. Heating usual water would be extraordinarily expensive. In all water treatment plant, all over the world, bacteria are destroyed by adding chlorine gaz Cl2 or similar substances delivering Cl2 when dissolved in water. The bacteria are killed by chlorine. So the customer drinks water containing dead bacteria. It does not hurt.

I don't want to be gross, but today I saw a man spitting on a paved road. I told myself, if someone spits or pisses in water reservations, shouldn't these be boiled at least from an aesthetic perspective? I guess I must go for "mineral water" (whatever that means); maybe these are indeed boiled...

@JohnDoea Isn't it against the law to spit on a paved road, aren't there heavy fines or stringent punishments for such violaters?

@Zenix I can't speak about the law of every state but I can tell you that I would prefer boiled plus well chemically treated water on not...

Chlorination is very oldschool (creates lots of toxic chlorinated compounds, ozone is used instead in modern installations), and before chlorination or other chemical treatment, you need to filter the water so good it would be drinkable already (the pipes would still foul over time with this water, hence the chlorination).

@johnDoea - LOL, you do realize where the waste from  fish, birds and turtles in a river goes, don't you?

@Karl In fact, ozonization is more aggressive than chlorination by creation a bunch of very suspicious compounds.

@Poutnik the problem is less acute toxicity but more environmental stability. have you got a link?

@Karl unfortunately not. Many of my knowledge comes from era the http protocols has not existed yet, or was not easily available. :-)

@JohnDoea Mineral water is definitely not boiled.

@Poutnik that is not less depressing for me --- in small countries like Cyprus, Jordan, Israel, Armenia, Tunisia so forth, I would assume residents are likely to put many body liquids into water reservations. Perhaps desalinated tap water are the best bet for me personally.

As mentioned above, boiling all the wastewater would be incredibly expensive. One technique not mentioned in the other answers is that the wastewater treatment center has portions of infrastructure where the water sits in open air pools for a certain amount of time. The UV rays from the sun kill some bacteria.

Joe, shouldn't such water be covered to prevent more bacteria to enter than to be destroyed as well as to prevent dirt - I mean, shouldn't there be some transparent (glass?) covering?

Any bacteria to enter from the air or particulate probably wouldn’t be pathogenic as would be some species in the wastewater. It probably comes down to cost though, the pools are rather large

They also bubble O2 in to kill anaerobic bacteria

UV irradiation for water sterilisation is done, but surely not in the open with sunlight. What a nonsense. You are aware what a perfect habitat for algae that would be?

Bubbling with oxygen (air, rather) is a common wastewater treatment method. For freshwater, you want 99.999% of the bacteria out, not just dead.

@Karl I am referring to wastewater treatment and if you look into it what I have said is true. The water is not stagnant but is rotated constantly

The original question is about freshwater. And for wastewater treatment, the irradiation by sunlight is completely irrelevant in terms of "killing bacteria". You want the algae to grow in wastewater treatment, because they eat a lot of unwanted compounds in the water and can easily be filtered out afterwards. Algae need light.


text:
NMR splitting of protons in a ring
I was presented with the following compound and asked to comment on the splitting of the proton $\ce{H_a}$ in $\ce{^1H}$ NMR.

The question presented today was:

What is the chemical shift (below) and splitting pattern for $\ce{H_a}$ up to and
  including $\ce{^4J}$?

I relabelled the molecule as shown:

but I was left stuck here. I know that $\ce{H_a}$ will be split by $\ce{H_d}$ and $\ce{H_e}$ to give a doublet of doublets (dd) but I was wondering, since the molecule is asymmetric. Now
are the protons $\ce{H_c}$ and $\ce{H_d}$ chemically equivalent but magnetically inequivalent
to yield the final splitting pattern as dddd?

And I find the wording of the question by your teacher a bit weird.The interesting point is wether a 4J coupling is large enough to become observable!

@Karl that was how the question was worded to me! I was just wondering if Hb and Hc are magnetically inequivalent, so they split the Ha signal with two different coupling constants?

@Karl done. I've shortened the question down to grasp the main point and clarified any grammatical errors.

Cool. ;) Ahem, a hint: you notice the methyl group at the bottom. I contains a lot of electrons. How do Hb and Hc feel about that?

@Karl Hb would feel a stronger magnetic field strength of the methyl then Hc as the methyl group and Hb are in the same plane so they would be at different chemical shifts as a result - That's what I would say is my answer to your question. Is that right?

I wouldn´t bet on which one is high/lowfield, but yes, they are certainly chemically inequivalent.

The two protons Hb and Hc are also chemically inequivalent, because they have a different chemical environment with the methyl group, i.e. they don´t have the same chemical shift.
In a chiral molecule, the protons in a methylene group are rarely really chemically equivalent. There is usually a preferred orientation.


text:
Can 2‐chloropropane and acetyl chloride give the iodoform test?
Can $\ce{CH3-CHCl-CH3}$ and $\ce{CH3COCl}$ give the iodoform test?
I thought $\ce{CH3COCl}$ having the keto-methyl group would show the iodoform test, while $\ce{CH3-CHCl-CH3}$ will not since it's not a secondary alcohol but an alkyl halide.

acetyl chloride will become acetate under alkaline conditions.

@user55119 the keto methyl group would still be there(?)

Hydroxide adds to the carbonyl and eliminates chloride. The acetic acid formed is deprotonated by base to acetate. No iodoform reaction of acetate! All carbonyl groups are not created equal. Secondary methyl carbinols are oxidized by I2/NaOH to give the iodoform test via oxidation to acetone. To the extent that SN2 displacement of 2-chloropropane occurs, which is likely very little, 2-propanol is formed. Elimination (E2) is likely the major reaction.

@user55119 Okay, thanks. So, 2-chloropropane also doesn't give the iodoform test.

2-Chloropropane itself does not give a positive iodoform test. If you learn the mechanism, then the answer would be self-evident.

As mentioned by user55119, ethanoyl chloride will not give iodoform test, because acid derivatives will show nucleophilic substitution reactions. Resulting will be acetate ion, which is no more a good nucleophile. For a more detailed answer check out here.
2-chloropropane will show iodoform test, since through bimolecular substitution reaction $(\mathrm{S_N2}),$ chlorine will be replaced by $\ce{-OH}$ group. Propan-2-ol so formed will be oxidised to acetone, which will show iodoform test.
Remember that only $\ce{CH3COR}$, (where $\ce{R}$ can be alkyl or aryl part) or which can be converted to it, give haloform tests, until steric hindrance prevent even some of these compounds to react.

As also mentioned, isopropyl chloride generally won't turn into propanol.


text:
Oxidative phosphorylation analogue
Does the salt bridge in galvanic cells resemble the ATP pump in the inner mitochondrial membrane? And if yes than did someone use this pH difference in harvesting this energy?


text:
The number of optical isomers of 3,4-dichlorobutan-2-ol
The number of optical isomers possible for the given organic molecule is?
$\ce{CH_3-C^*H(OH)-CHCl-CH_2Cl}$ (the configuration is fixed at $\ce{C^*}$)
The given answer is 2.
The carbon marked with an asterisk is chiral, and the carbon to the right of it is also chiral. Doesn't that give 4 isomers? What does "configuration is fixed" mean?

The carbon marked with the asterisk cannot change its configuration, for it to be counted as a stereocentre, it must be able to. hence 2 isomers only

So we're considering the optical rotation of the carbon which lies next to the one marked with an asterisk?

Yes. The only chiral atom here is the third C, which is part of the CHCl group.

Yes, thats exactly what i meant


text:
Can I express the rate of a reaction as percent per second (%/s)?
Rate is calculated as change in concentration over change in time, so usually $\pu{M/s}$ $(\pu{mol L-1 s-1}).$
Say, I'm calculating it for $20\%$ solution of ammonium hydroxide, so my final rate is in $\pu{\%/s}.$ Is this a proper format, or should I calculate for $\pu{mol L-1}?$

According to IUPAC Recommendations for chemical kinetics [1], rates $\nu$ are defined for reactant $\ce{B}$ and product $\ce{Y}$ as a time $(t)$ derivative of the amount $n$ (in general) or per unit volume $V$ (for the closed system):
$$
\begin{array}{lll}
\hline
\text{System} & \text{Consumption} & \text{Formation} \\
\hline
\text{Open} & \displaystyle\nu(n_\ce{B}) = -\frac{\mathrm dn_\ce{B}}{\mathrm dt} & \displaystyle\nu(n_\ce{Y}) = \frac{\mathrm dn_\ce{Y}}{\mathrm dt} \\
\hline
\text{Closed} & \displaystyle\nu(c_\ce{B}) = -\frac{1}{V}\frac{\mathrm dn_\ce{B}}{\mathrm dt} & \displaystyle\nu(c_\ce{Y}) = \frac{1}{V}\frac{\mathrm dn_\ce{Y}}{\mathrm dt} \\
  & \displaystyle\nu(c_\ce{B})|_{V = \mathrm{const}} = -\frac{\mathrm d[\ce{B}]}{\mathrm dt} & \displaystyle\nu(c_\ce{Y})|_{V = \mathrm{const}} = \frac{\mathrm d[\ce{Y}]}{\mathrm dt} \\
\hline
\end{array}
$$
Rates of consumption and formation that are measured in $\pu{mol L-1 s-1}$ are defined for the closed system at constant volume, and molar concentration arises from its definition.
Whether you can or cannot substitute molar concentration $c_i$ with fractions can be answered by looking at the relationship between the two.
For example, molar concentration $c_i$ expressed in terms of mass fraction $\omega_i$ via the density of solution $\rho$ and the molar mass $M_i$ of the $i$th component:
$$c_i = \frac{\omega_i \rho}{M_i}\tag{1}$$
Similarly, molar concentration $c_i$ expressed in terms of mole fraction $x_i$ via the density of solution $\rho$ and the average molar mass $\bar{M}$ of the solution:
$$c_i = \frac{x_i\rho}{\bar{M}}\tag{2}$$
As you can see, there is a problem: as the reaction progresses, not only the amounts (or fractions) would change, but also the density and the molar mass of the solution may and will change in most cases, which leads to inapplicability of percent fractions as time derivatives.
All in all, in chemistry it's just easier to manipulate with the quantities based on amount of substance.
References

Laidler, K. J. A Glossary of Terms Used in Chemical Kinetics, Including Reaction Dynamics (IUPAC Recommendations 1996). Pure and Applied Chemistry 2009, 68 (1), 149–192. DOI: 10.1351/pac199668010149. (Free Access)

I don't fully understand. I get that % values change over the reaction (first order btw) , but if I have a rate that's in %/s, isn't that expressing the change in concentration in and of itself?

To summarize, and to answer the question, it is correct to say that the rate of a chemical reaction can be expressed in % per second if and only if the reaction is first order. This percentage is the first order rate constant $k$.

Can I get a source? That % is first order?

For a reaction like this: $\ce{aA +bB⟶cC +dD}$, the rate is given by
Rate=$−a\frac{d[A]}{dt}=−b\frac{d[B]}{dt}=c\frac{d[C]}{dt}=d\frac{d[D]}{dt}$ where the unit is usually $\ce{mol L^{-1}{s^{-1}}}$.
The rate of a reaction is different for different orders and has different units of the same. It's unit usually has the terms $\ce{mol, L, and s}$.

For a zero order reaction,

Rate= k, unit= $\ce{mol L^{-1}{s^{-1}}}$

For a first order reaction,

Rate= $\ce{k[A]}$, unit=$\ce{s^{-1}}$

For a second order reaction,

Rate=$\ce{k[A]^2}$, unit=$\ce{L mol^{-1}{s^{-1}}}$


text:
Molecular orbital diagram of hydroxide ion?
I would like to understand how to create a molecular orbital diagram for the hydroxide ion from scratch. This includes understanding the shape of the molecular orbital.
Here is an attempt I have come up with:

Where the top MO is sigma* and the bottom is simply sigma.
But this makes no sense. I haven't used up all the available electrons of the oxygen atom, and the diagram seems to suggest that there is a bond order of zero between the two atoms?
However, this goes against my understanding of MOs - surely only two molecular orbitals can be formed from a p orbital and an s orbital (sigma/sigma*) and none others? I couldn't form a pi molecular orbital, for instance.
I think I've made a mistake in that I need O$^{2-}$ instead of O, but even then I don't see how that helps.
Once I can understand that, hopefully I will be able to visualise the HOMO of the OH- ion.

If you have $x$ atomic orbitals to begin with, they form $x$ molecular orbitals. Two of your MOs are missing. My suggestion is to look up the MO diagram for HF in a textbook (or online). Hydroxide ion is the same as HF, except that you write O instead of F.

Your scheme has 9 electrons. Nine ! Why ?

@orthocresol unless we are excluding the oxygen 2s we need five total MOs not four.

@orthocresol HF's M.O. diagram shows 1s of H at a higher energy level than O's 2p orbitals. Why is that so? I would expect it to be of lower energy and more bound to the nucleus..

@napstablook HF, or O's 2p orbitals? Anyway, that might look surprising, but it shouldn't be. 1s is indeed lower than 2p, but only within the same atom. Oxygen's nucleus has 7 more protons than hydrogen. Fluorine has 8 more.

@orthocresol In both (hydroxide and fluorine), but I suppose you are right (the specifics are dependent on calculation for higher orbitals anyways). Thanks for the clarification!


text:
pi-type bonding interaction along the bond
I was given the diagram of $\ce{BF3}$'s LUMO and asked to assess its bonding character as shown below:

Below is the model answer for the annotated version.
I managed to correctly answer the HOMO MO with the same annotations but was confused here slightly.
The LUMO Interaction shows a $\ce{p}$ orbital coming out of the plane, where the B p orbital is out of phase with the $\ce{F3}$ fragmnet orbitals, but the model answer suggests there is a $\ce{\pi}$ type bonding interaction.
This does not make sense to me.
Unless mistaken, I thought the p-orbitals from the B and the F atoms (given they are out of phase) show along the bond $\ce{\pi}$ type antibonding properties, rather than $\ce{\pi}$-type bonding interaction.

Am I correct or not? How is there $\ce{pi}$-type bonding interaction possible along the bond here?


text:
Does calcium nitrate decahydrate dehydrate or decompose upon heating to 100 degrees Celsius?
Consider a sample of $\ce{Ca(NO3)2.10H2O}$. To remove water, I need to heat it to $\pu{100 ^\circ C},$ but will this high temperature decompose it or will I get the anhydrous salt?

This information is readily available on the internet, e.g. Wikipedia.

Are you sure it's a calcium nitrate decahydrate and not a tetrahydrate? Or are you asking about ammonium calcium nitrate decahydrate?

Calcium nitrate exists under three forms : 1) $\ce{Ca(NO3)}_2 $ which melts at 361°C; 2) $\ce{Ca(NO3)}_2$$·\ 3 $$\ce{H_2O}$, which is not commercial and melts at 51°C ; 3) $\ce{Ca(NO3)}_2$$·\ 4 $$\ce{H_2O}$ which is commercially available and  melts at 43°C. It looses its water at 132°C and is deliquescent. The decahydrate that you mention does not exist.

It would be even a better answer if it were backed up by a reputable source.


text:
What does it mean by " concentrations are (value) in relation to each other"?
This is not a home work help question or something like that, these days I self learn chemistry and when I was doing a model paper I came across this question. And the question itself is not difficult but I don't understand this term.
More accurately, I am not familiar with this term,

that part of the question says,

The cation Ag+(aq) and Hg2^2+ are in an aquesous solution.Their concentrations are 0.2 moldm-3 in relation to each other. Find which cation precipitates first as its iodide when a solution of NaI is added gradually to the above solution... and there are some other sub questions,

If you can explain to me if what it means by "Their concentrations are 0.2 moldm-3 in relation to each other", I think I can do the question easily, Thank you!

Post the full sentence and the context. As written, it does not mean $anything$.

I just did, thank you for pointing me out. Hope it makes sense now

Strictly one should refuse this exercise to be evaluated, in case of assignment. It is wrong.

@Alchimista this is a question from a model paper targetting Sri Lankan AL examination. ( And I attached an image of the given answer )

The cation Ag+(aq) and Hg2^2+ are in an aqueous solution. Their
  concentrations are 0.2 moldm^-3 in relation to each other. Find which
  cation precipitates first as its iodide when a solution of NaI is
  added gradually to the above solution

It is a poorly worded question. The author means that the ratio of Ag conc. to Hg conc. is 0.2. It is implied that the concentration of silver is lower. Which ion will ppt out first, if we add NaI?
The author should have written that the relative ratio of Ag:Hg concentration is 0.2.

Hm, If it is the relative ratio, why to mention the unit just next to the value? And you must mean "the concentration of silver is lower". I agree the question is worded very poorly.

I referred the marking scheme just now and it seems to be they have used that 0.2moldm-3 concentration in both Ksp equations, I just attached that image to question. Thank you!

Sujee0_0, Ignore the question it is meaningless written and solved. Tell your teacher that there are problems with it.

yes looks like it is my only option @M.Farooq Thank you for the support!


text:
Why do PBr3 and PCl3 have larger bond angles than PH3?
According to the Colby Chemistry Database, the bond angle of $\ce{PBr3}$ and $\ce{PCl3}$ are around $101$ degrees, but the bond angle of $\ce{PH3}$ is $92$ degrees. (Source: PCl3 PBr3 PH3)
There was a previous question on Stack Exchange about the bond angle difference in $\ce{PF3}$ and $\ce{PH3}$ here: PF3 and PH3, but the accepted answer mentions back bonding as a factor to increase the bond angle in $\ce{PF3}$. I was skeptical about this answer, because in $\ce{PCl3}$ and $\ce{PBr3}$, there should be less back bonding than in $\ce{PF3}$ because bromine and chlorine are both much larger than fluorine and there will be worse orbital overlap between the halogen. Less back bonding should make the bond angle of $\ce{PCl3}$ and $\ce{PBr3}$ smaller than that of $\ce{PF3}$, if back bonding were the major factor here? In reality, that is not the case. Why not?

The larger angles can simply be explained as a result of repulsion between the larger atoms of $\ce{Br}$ and $\ce{Cl}$. Hydrogen atoms in $\ce{PH3},$ as the are so small, experience less repulsion as compared to $\ce{Br}$ atoms in $\ce{PBr3}$ or $\ce{Cl}$ atoms in $\ce{PCl3},$ therefore the larger bond angles in $\ce{PBr3}$ and $\ce{PCl3}.$


text:
Does brewing tea with hard water reduce the amount of bioavailable fluorine?
I read that brewed tea contains a relatively large concentration of Fluorine.
Fluorine can have negative health effects such as Skeletal Fluorosis. EDIT: I understand some Fluorine is necessary and has benefits. This is just to inform why I have structured this hypothesis. I am interested in the chemistry not the health effects as that is a different topic.
Tap water and especially hard water contains lots of Calcium ions.
Would Calcium and Fluorine bond to form Calcium Fluoride? I'm obviously not a chemist, but I assume it would be a lower energy state than the soluble ion pair in the tea, correct? It forms Fluorapatite on tooth enamel.
This Calcium Fluoride SDS indicates that eating solid CaF2 straight won't immediately kill you, so I assume Calcium Fluoride does not decompose in HCl stomach acid, correct?
So with that logic would the following be true?
Mol bioavilable F-  =  Mol F- from brewed tea  -  2*(Mol of Ca2+ in hard water)
If that is the case, it brings up some other interesting questions, like if moderately hard municipal water actually has any Fluorine in solution or not...
EDIT: To select a best answer I need to know:

if Ca ions bond with F ions in solution to form CaF2
and to know the percent or time-rate solubility of CaF2 in stomach acid.

Thanks!

$\ce{CaF2}$ has a relatively low solubility in water, about 15 mg/L, with a $K_{sp}$ of about $4\times 10^{-11}$.  According to the paper you linked, the concentrations of fluoride in tea are around 5 mg/L, which corresponds to about $1.3\times 10^{-4}$ M. Thus, the concentration of $\ce{Ca^2+}$ required to precipitate the fluoride out is about 2.5 mM, which corresponds to 0.1 g/L. Hard water is defined as having a calcium concentration above about 0.6 mM, with very hard water being 1.8 mM calcium. Thus, it is unlikely that there would be high enough concentrations for $\ce{CaF2}$ to precipitate out of solution.
Because of the large solvation energy of fluoride ion, the molecules in solution are likely to be nearly completely ionized. That is, the fluoride anions and the calcium cation will each have their own solvation shell of water molecules. That means that they will not interfere with each other until the concentrations are high enough that solid $\ce{CaF2}$ begins to precipitate.
As long as the concentrations are below the solubility limit, the presence of calcium ions will therefore not affect the availability of the fluoride ions.
Furthermore, since HF is a weak acid, in the acidic environment of the stomach some of the fluoride ions will be protonated to HF, decreasing the concentration of free fluoride and increasing the apparent solubility of $\ce{CaF2}$.

Wish I could give an extra point for not just answering the question but ignoring all the irrelevant and misleading health woo triggered by any mention of fluoride.

Thanks! A much simpler answer than I expected. As a non-chemist I have the tendency to look at thing as soluble or non-soluble... but at low concentrations you definitely have to go by the numbers! Thanks again!

What Atharva and Ericmuch say is sometimes contrary to the truth. Calcium fluoride cannot be dangerous, because it is insoluble in water, and in diluted acidic solutions. It can produce negative effects such as skeletal fluorosis, but this happens only in cases where it is taken in huge amounts. Fluorides have no negative effects, if taken in small amounts.
It is even the contrary. Adding sodium fluoride in small amounts (90 - 250 mg/kg) to the usual salt NaCl has had a remarkable effect on the dental health of the general population. I can give you some values measured before and after the introduction of NaF in the salt (or in the tap water). This introduction has been introduced in USA during the war 1939 - 1945. It was adopted much later in Europe, where it was accepted with hesitation and reluctance.
I can give the results I know from my own country, Switzerland. In 1961, before the fluoration in NaCl, only 25.2% of the 7y children had no dental caries (tooth decay). And only 1.2% of the teenagers (15 y) had no dental caries. And an average teenager had 7.4 teeth with caries.
The fluoration of the salt was introduced in 1962. Fifteen years later, in 1977, 76.6% of the young children (7 y) had no dental caries at all. The tendency has been reversed. And 15,1% of the teenagers (15 y) had no dental caries, instead of 1.2% earlier. And the average teenager had 1.2 teeth with caries. It is a huge progress. It was not perfect, of course, but it shows that adding a small amount of fluorine improves the dental health.
The same measurements have been made in many if not all European and American countries. See the following references.
H. Dean, Fluorine in the Control of Dental Caries, Intern. Dent. Journal 4: 311-337,, 1954.
K. Binder, W.S. Driscoll, Caries-Preventive Fluoride Tablets Programs, Car. Res. 12 (Suppl. 1):22-30, 1978.
F.B. Glenn, The Rationale for the Administration of a NaF supplement, J. Dent. Child, 48:118-122, 1981.
I. Hellsröm, Y. Ericsson, Fluoride Reactions with Dental Enamel Following Different Forms of Fluoride Supply, Scand. J. Dent. Res. 84:225-267, 1976.
H.R. Mühlemann, K.G. König, T.M. Marthaler, Anleitung zur praktischen individuellen Kariesprophylaxie, Zahnärtl. Mitt, 16, 741 -743 1966. H.S. Horowitz et al., Perspectives on the use of Prenatal fluorides : A Symposium, J. Dent. Child 48, 102-134, 1981. T.M. Marthaler, Ph. de Crousaz, Médecine dentaire préventive, en particulier dans les écoles, Schweiz Mntschr. Zahnheilk., 89: 619-628, 1979. K. Toth, Ten Years of Domestic Salt Fluoridation in Hungary, Caries Res. 13:101, 1979.

Please see my EDITs above. I understand some Fluorine is necessary and has benefits. I noted negative health effects to inform why I have structured this hypothesis. I am interested in the chemistry not the health effects as that is a different topic. To select an answer I need to know if Ca ions bond with F ions in solution to form CaF2 and to know the percent or time-rate solubility of CaF2 in stomach acid. Thanks!

Even I am not a chemist, but according to my textbook there is already a counter ion present in hard water to the $\ce{Ca^2+}$ and $\ce{Mg^2+},$ which are mostly bicarbonates (temporary hardness) or fluorides, chlorides and sulfates (permanent hardness). These counter-ions would by themselves prevent any appreciable amount of fluoride forming a bond with the $\ce{Ca^2+}.$
Also, even though $\ce{CaF2}$ probably won't decompose fully in stomach acid, it would decompose enough to cause the negative effects of fluorides to come into effect, according to my professor. My professor also says that this the reason $\ce{EDTA^4-}$ is used to complex with $\ce{Ca^2+}$ and $\ce{Mg^2+}$ in hard water: since coordinate compounds don't dissociate in water, the $\ce{Ca^2+}$ and $\ce{Mg^2+}$ are not able to make the water hard. The anions are removed using organic ion exchangers.
Thus I believe making tea in hard water won't reduce the fluoride concentration very much, and in any case drinking hard water is generally a bad idea because it is not pure enough to be drunk.

Is this book ncert? @Atharva

Not sure how Mg influences, but Ca is higher on the activity series so I doubt it would be much. Though I can not do the calculation, the solubility of CaF2 in a certain concentration of HCl can be calculated. Maybe someone on here can do a more rigorous analysis of these for us.

Yes, it is the NCERT.

I would rather look if there is a known correlation between fluorine/fluorides health troubles and tea consumption. It should be evident comparing Russians or British people to Italians or Germans, or Indians to the rest of the world. As usual this kind of food related arguments can have too many facets. From the question we don't even know in which form is the F in tea. If would be Teflon if would likely go through. If it would Teflon nanoparticles it would likely be a disaster. These are hyperbola, of course.

Please see my EDITs above. I'm not interested in the health effects just the chemistry. To select a best answer I need to know if Ca ions bond with F ions in solution to form CaF2 and to know the percent or time-rate solubility of CaF2 in stomach acid. Thanks!

Your speculations don't help answer the question. "CaF2  probably won't decompose fully in stomach acid, it would decompose enough to cause the negative effects of fluorides to come into effect" is wrong despite what your prof says. And normal levels of fluoride won't have health effects, even the levels in tea. Also, there is nothing wrong with drinking hard water (and evidence exists for a small beneficial effect).

Normal levels of fluorides? That is considered to be around 1ppm, if I remember correctly. While it may be true that CaF2 will not cause the negative effects of fluorides to come into effect, 4 to 5 ppm is more than enough to do so (of course, the form fluorine is in matters). Also, it depends on the place where you're drinking hard water from. In places in India, drinking hard water supplied to rural areas can easily cause brown mottling of teeth. So, the generalisation 'there is nothing wrong with drinking hard water' does not hold everywhere.


text:
Why does the electrical conductivity increase as the temperature decreases?
I am measuring the dependence of electrical conductivity of ( NaCl and $\ce{CH3COONa}$ solutions  with concentration 9g/L) with temperature using conductivity meter.
Theoretically the conductivity should decrease as the tempreture decrease, but unfortunately that does not seem to happen.
As you can see, as I start measuring conductivity at  $\pu{85 ^\circ C}$ it decreases with decrease in temperature till $\pu{65 ^\circ C}$ ( the first three data) then again start to increase.
I repeated the experiment again and again but am getting the same result .
The conductivity start to increase around $\pu{65 ^\circ C}$! Why is this happening?

I edited your question, just check that I made no error. Also what does this statement "....conductivity decreases with decrease in temperature the first three date" mean?

Oh sorry i mean data ( in the table)

I write the question at the end of the work so i can see know a lot of mistakes

The problem just that in my expermint the conductivity increase at the Temperature decrese after reaching 65 c

Did you verify/repeat the calibration of instruments used and double check the condition of any standard solutions used?

These are small changes... does it matter that density will decrease at higher temperatures? Also, is the electrode spacing constant as temperature changes?

I use KCl as standard solution i will try to calibrate the instrument one more time


text:
Which free radical is more stable, allyl free radical or benzyl free radical?
Can someone pls tell me which one is it? Benzyl free radical has 4 resonance structures compared to allyl free radical which has only 2. More the resonance structures should mean more stable the radical.

But in the book I am referring it is written that free radical stability depends on bond dissociation energy.

According to this Allyl free radical should be more stable but according to resonance structures Benzyl free radical should be stable. Which one is correct?

Yes, generally more resonance structures mean more stability but this case is an exception.
In three of the four resonance structures of the benzyl radical, the aromaticity is being broken which makes then less stable whereas in the allyl radical both the canonical forms are stable.
Therefore the allyl radical is more stable than the benzyl radical.

The reasoning provided here might not be entire accurate. See ron's answer here to know more


text:
How can an Organic chemist know the chemical formula of a natural product she isolated from an organism?
As a non scientist I several times wondered how can an Organic chemist successfully isolate a single molecule which a natural product of an organism;
Whether a plant-part/mushroom-part/algae-part or a coral-part or even from an animal or human organ in autopsy (neurotransmitter/hormone and so forth).
I also wondered how could she go further to know the natural product's (exact?) chemical formula and give such newly discovered molecule a name.
Given that molecules are such small entities, way smaller than the smallest uicellular organism and are hard to be photographed uniquely with current microscopal technologies (AFM/STM); it is unclear to me How could she know if a plant part contains a certain molecule (say caffeine or nicotine) so precisely that she could tell:

This is what I will now name Caffeine → not Theobromine
  or any other Methylxantine → just Caffeine, pure and exact and unequivocal.

Practical examples for my problem are:

How did organic chemists isolated THC from various Cannabis plant-parts and gave it a formula and a name, although today it is known that Cannabis plant parts has tens of cannabinoids?
How did organic chemists isolated Ganodermadiol in some Ganodrma mushroom part/s and gave it a chemical formula and name and claimed it is similar to steroidal hormones?
How did they know they weren't wrong?

My question
How can an Organic chemist know the chemical formula of a natural product she isolated from an organism?

Because they perform analysis such as LC-MS or NMR using accredited methods. For the high-precision structural models there is single-crystal XRD and neutron diffraction.

How do you know about AFM and STM, but not methods mentioned by Andselisk? Not to say such questions were already asked and also tended to be too broad https://chemistry.stackexchange.com/questions/32240/how-can-an-inexperienced-chemist-determine-the-chemical-structure-of-a-molecule

@Mithoron I saw about AFM/STM in Wikipedia article molecule; if the same question was asked then it should be marked as duplicate.

https://chemistry.stackexchange.com/questions/34048/can-molecular-structure-be-determined-without-modern-characterization-methods https://chemistry.stackexchange.com/questions/771/how-were-elements-and-molecular-formul%c3%a6-determined-in-the-old-days

Similar question https://chemistry.stackexchange.com/questions/86063/how-do-scientists-figure-out-the-chemical-structure-of-a-given-compound was closed as duplicate of https://chemistry.stackexchange.com/questions/57784/has-anyone-even-taken-a-picture-of-a-molecule-to-confirm-the-geometry-predicted still it's somewhat different...

This is so broad it should be trimmed to be good dupe target. https://chemistry.stackexchange.com/questions/7269/atomic-properties

How can an Organic chemist know the chemical formula of a natural product she isolated from an organism?

There are two levels of answers. One is historical and one is modern. Historically, determining the chemical formula for had been a trivial job for most small or medium sized molecules. You do a combustion analysis, determine how much carbon dioxide, and water were produced. By doing other classical element detection tests, you would determine if it had halogens, nitrogen and sulfur. These were the most common approaches. The determination of molecular weights was done by classical physical chemistry experiments.
Today you would still do combustion analysis and elemental analysis as a first step and determine the CHNS ratios. Then you run a mass spectrum of the compound, and assess the molecular weight of the compound.
Determining a structural formula may take a long time and others have already mentioned in the comments. Today advanced NMR techniques, high resolution mass spectrometry, X-ray of crystals (if they exist), all help in structure determination. These things have become routine now.
Nobody routinely determines the molecular structure by AFM or STM as far as I know.

Determining the structure of a medium sized molecule used to be a take years of determined work before xrd and nmr.

Yes, because only chemical methods existed. I can only praise those early German organic chemists of the 18th and 19th century who did extensive organic reactions to determine the structure of a molecule by bits and pieces.

Working out the structure of a compound (organic, organometallic or organomain group) is not a simple matter. It is something which can be very intellectually taxing. I have done it many times in my life and I can tell you that for some thing you can do it in a matter of minutes while for other things it can take weeks, months or even years)
The first problem is that you need to know that your substance is sufficiently pure that the peaks in spectra are due to your compound of interest rather than some impurity.
For an organic compound if it is a pure substance which you have plenty of, my advice is normally.

Get an infrared spectrum of the substance so that you can identify some of the functional groups. Look for multiple bonds, you can look for evidence of ketones, amides, esters, carboxylic acids, nitriles, metal carbonyls, benzene rings and alkenes. In my experience it is sometimes hard to see the C C stretch for acetylenes but you might get lucky. This will give you a list of mental jigsaw bits.
Get a proton NMR of the substance, this will either be a great help or close to useless. If you are lucky you will be able to identify parts of the structure. For example if you have a phenyl group, a methyl close to TMS and two multiplets which appear to be all coupled to each other. Then using the 3J couplings you could tell if you have a propyl group.

Alternatively if you had a isobutyl group then the integration heights would be different and the coupling pattern different, for the isobutyl you would have doublets for the methyls instead of the triplets you see in the propyl.
You should also look for signs of symmetry, for example if you have benzopheone (diphenyl ketone) then you will have seen in the IR a ketone (which is slightly lower than 1700 as a result of conjugation) and only one carbon / hydrogen group. As a result of symmetry both phenyls are magnetically identical.
Equally if you have 1,3,5-trichlorobenzene you will have a single singlet in the aromatic part of the proton spectrum. Compare this with the multiplets you would have for 1,2,4-trichlorobenzene.
If you see a very large H H coupling then it is possible you have a ring or some other thing where the conformation is locked. This would be a substance with a 2J coupling. Note this it could be a great clue. For example in the structure of some penicillin derivatives you can see 2J couplings where two different hydrogen on the same carbon have different magnetic environments.
Look in the proton for anything which is not usual, for example cyclopropanes have a very distinctive shift which is close to zero ppm or even at - ppm.
If the proton was little use or you can not understand it then my advice is to try some more things.
If the proton NMR is very congested, then I would suggest running it again in a different solvent. Commonly people use CDCl3 for NMR. I always say that if the spectrum in chloroform is impossible to understand as a result of being congested then I run a sample dissolved in D6 benzene. The change of solvent will alter the solvation and the proton environments oftein change in their chemical shift, this can make some areas of the spectrum more easy to understand. In this way you may be able to get a clearer look at it and as a result you might be able to understand the structure better.
Another trick which makes proton NMR more simple is to use double irradation, this can allow you to make the spectrum more simple by removing one coupling effect at a time. This will make the spectrum more simple. If it solves you problem then be thankful otherwise note down any new insights you have into the substance and move on.
I would also say that getting a carbon-13 NMR can be a great help, this will help you understand the substance more. You will learn quickly how many unique carbon centres there are in the substance.
You should also consider the question of will a 2D NMR experiment help you, for example you can use a COSY or a CH correlation. For NOSEY you need to exclude all oxygen from the NMR sample. For this you will need to use deoxygenated chloroform and work on a vacuum line and seal up the tube with no air getting inside. This will take some skill and time to do.
If you are stuck you should consider getting a mass spectrum of the substance, this can give you an idea of how large the thing is. Be careful with mass spectra. It is sometimes possible to get a nice spectrum off a tiny impurity which is present in the sample while the main compound does not give a strong nice spectrum.
The mass spectrum will give you an idea of the formula, you can get the formula weight. One thing which is worth doing is to look at the isotope splitting pattern of the molecular ion. For example bromine is present in the form of two isotopes with about equal abundance so for a monobromo compound you should expect a doublet. If you can see such splitting in the molecular ion then it is a good sign you have a heavy atom in the substance. Keep in mind iodine only has a single stable isotope.
Next trick I have is to use a sodium fusion test, now this test has been banned in a lot of universities as undergrads has made loud bangs or started fires with it. But when used with care it is perfectly safe. I like the test as a method of detecting nitrogen, sulfur or halides in an organic substance. Heat a SMALL pellet of sodium until it melts. Then add some of your thing and carefully reheat. There is a great debate on how best to do this, and it takes some practise. Heat the mixture up until it glows and plunge it into cold water. It may go bang, pop, fizz etc when you do this. Then boil the water mixed with broken glass and filter it.
Test the filtrate for halide with silver nitrate, if it goes cloudy then add ammonia. If the solid redissolves then it is likely to be a chloride while if it stays cloudy then it is likely to be either a bromide or iodide.
Test the filtrate by adding ferrous sulfate, boil it and then add acid. If it goes blue on adding the acid then you have made Prussian blue and you had some nitrogen in the organic substance.
If you can detect either thiocyanate or sulfide in the water then you had sulfur in the organic compound.
Add these new facts to the list of facts you have and see if you can work out a structure which makes sense for the organic molecule. Sometimes you can work out several possible structures based on NMR and IR. For example toluene and methyl benzoate will look very similar in the proton NMR.
Next take the possible structures and see if you can distinguish between them by molecular weight and which heteroatoms you have found in them. For example diphenyl sulfide and 4,4'-dihydroxybiphenyl would have the same molecular weight. But one will have sulfur in while the other will not. Also the fine detail in the mass spectrum on the molecular ion will help sort these two out.
Now if you have stared at it for a while and you still can not solve it, then it might be time to get a UV / vis spectrum. This is good when you have a conjugated system present in the molecule. It can be used to give you an idea of the size of a pi system.
If you see any strange couplings in the proton NMR spectrum where you can see one thing which couples to something which can not be seen then you might have a phosphorus atom present. Some years ago a student showed me a strange NMR spectrum with a doublet they could not understand. I spotted that the sample contained some HMPA, the doublet was in the right place for HMPA.
If you have phosphorus or another heteroatoms which couples such as selenium or platinum then you can try getting more NMR. For example consider the phosphorus spectrum you would have if you had a Pt binding to a triphenyl phosphine and to a dppe (1,2-bis diphenylphosphino ethane) and a chloride. You would expect to have three different phosphorus environments. Also three different coupling constants.
If you have a Pt with two PPh3 ligands and a dppe, then you would have two P environments and a very complex splitting pattern. Keep in mind also that the Pt-195 will couple with the P nuclei giving sidebands on the P signals.
The key thing to do is to keep gathering more data and trying to look at the clues from the different bits of the data to see if it will fit together to give a structure. If you have multiple structures then you can attempt to use physical properties to distingish between them. If the melting points are very different then you can use a copy of the rubber book, a melting point rig and your sample to work out which it is.
In the old days the gold standard was to take an authentic sample of the thing and mix it with your suspected sample of the same thing. Then to go and do the melting point at the same time as pure samples of the authentic sample and your sample. This is the idea of a mixed melting point. If all three samples melt sharply at the same temperature then you have identified your thing.
Now if you do not have a winner, then there are several things you can do. You can try to form derivatives or degrade your compound in some predictable way. This might make new compounds which can be identified with greater ease. This will give you clues to what you have. For example treatment with ozone can cut up a molecule by cleaving alkenes, this will make it more easy to spot some fragments.
Now if you can grow a crystal of your thing, then my advice is to try for single crystal diffraction methods. Just be careful of the rouge crystal problem, even if you get a pretty picture of the atoms you will need to check that the spectroscopic properties of your bulk sample are right.
You can get some nasty shocks, I once handed in crystals of a organic phosphorous sulfur compound to discover to my horror that the crystals were of some sulfur (S8) which was present as an impurity. The S8 is impossible to see with NMR or IR, so I thought that I was handing over a pure sample to the crystallographer. That was embarrassing.
There are other tricks you can try but I think for most things that the rather long post I have given will sort most things out.

This question cannot be answered in such a Forum. Because there is no unique answer. The method for giving a formula is a huge process. It takes hours and days and weeks to the university students and professors to be able to answer this question. Sometimes a given process is specific for a given product, and not applicable to the next substance. It is a long and hard job. The chemist must perform a lot of tests and counter tests, each one giving only a part of the answer. Sometimes the substance may be irradiated, oxidized or hydrolyzed (treated by water), and by analyzing the irradiated, oxidized or hydrolyzed product, you can infer the structure of the original product. It is usually a long and hard process.
And of course, the chemical obtained after such a first treatment must often be then treated by an acid or chlorine or soda, then in series with acetone, acid and water, then with zinc powder, then with acid again, and so on. The number of successive operations may be higher than twenty.