could any organic chemist check this suggested edit and see if it is correct? --^ i'll likely avoid approving edits that change the core of the answer in such radical manner
Huckels rule is derived from like how many electrons go in bonding molecular orbital and how many electrons go in antibonding. And on the basis of which one is greater, stability is known? So, why can't we apply the same for inorganic ones?
A triangle is inscribed in a circle. The vertices of triangle divide the circle into three arcs of length 3, 4 and 5 units, then find the area of triangle.
Editorial note: Voting will be open until the end of next weekend (29 April), at which point we will make a decision based on the usual number of upvotes criterion.
In light of the "unearthing" (or re-unearthing) of the Q&A e-books, I'd like us to consider as a community whether we are conte...
@DivyankaChaudhari Counting rules are an oversimplification. This one only works well for the cases where you actually have degenerate orbitals and orbital contributions are equal among the participating atoms (i.e. a reasonably high symmetry). If that does not apply any more, you'll have a special case and that means that these counting rules need to be modified.
You should rather focus on why these rules work for the cases for simple monocyclic hydrocarbons, and the underlying explanations, then you don't need to rely on an arbitrary number. And don't treat aromaticity as some property; it is only a word for an ideal bonding situation and a certain reactivity.
@Zhe But if $\alpha$ is the degree of dissociation , then wouldn't it be $a-a\alpha$ for PCl5 ... ? Your line will be correct if I say $\alpha mole is dissociated in equilibrium ...?
I recently came across a poorly written question and thought of suggesting some edits to correct some spellings, mathjax and format it for a better look.
But, after I was completed and clicked on suggest edits, this is what I came across:
Couldn't understand what and why this was? Any help w...
It is known that chromyl chloride exists (popularly known as the product of a test for presence of chloride ion) but chromyl bromide and chromyl iodide are unknown.
Why is it so? Why don't they exist? I'd guess it has something to do with the large size of bromine and iodine compared to chlorine...
Possibly the electrophile is molecular $\ce{HNO3}$. Nitric acid is strong but not super-strong ($\mathrm{p}K_\mathrm{a} = -1.3$), so solutions on the order of a mole per liter have a significant amount of molecular $\ce{HNO3}$. As $\ce{NO2(OH)}$ this can transfer its nitryl ($\ce{NO2+}$) ion moie...
This doesn't directly or indirectly answer his question. Why was my not an answer flag declined?
@GaurangTandon nice to see you generously open so many bounties these days.
@Abcd well most of my janitorial duties are well carried out by being above 3.1k-rep, and I really don't see a point in having excess rep beyond what I require, especially when there's such decent questions left unanswered. So, yeah, I like putting bounties :-)
@Abcd NAA flag is only for the extreme cases. This may be a vague attempt at answering the question. So, in such cases, it's better to downvote and leave a comment instead of flagging for straightway deletion.
Usually answering off-topic questions is frowned upon. Giving the answer without explaination to someone who seems to not understand is also frowned upon. Either give guidelines to finding the IUPAC name or give a partial answer and have Diwakar finish the job. — JavaScriptCoder34 mins ago
(1) We, meaning us moderators, do not mind complete answers to homework questions (2) Some people still discourage it, which is alright, I have nothing against that (3) However, please don't suggest that people write partial answers
@Abcd That depends on the context, so you should state a molecule. With transition metal-oxide complexes, for example, there is undoubtedly p-d backdonation (from oxide p orbitals to metal d orbitals). For main group stuff, I don't know what example you have in mind, but you generally want to replace the d orbital with some kind of σ* MO. Again, it depends on the exact case.
@JavaScriptCoder mainly because effort is not a useful criterion for judging whether a question is valuable or not to SE.
Anyway, coming back to answers to hw questions... if the answer is poor, then downvote it, discourage it. If the answer is good, then that's perfectly fine
@Abcd Exactly, so why does SF2 need to be hybridised?
@JavaScriptCoder sorry for the delay. Anyway, coming back to answers, if the answer is bad then downvote, discourage, sure. But if the answer is complete and useful to future visitors, then what's the issue?
@orthocresol In exam, we are just asked: "is there p pi-d pi back bonding in SO2?" answer has to be boolean value. So, from the discussion, it seems your answer is "no"?
Well, if we want to go down this route, then we need to note that orbitals are hybridised, not atoms.
People loosely say an atom is "sp3 hybridised" when what they mean is that it forms four bonds (or lone pairs) with four (roughly) sp3-hybridised orbitals.
But anyway, I don't see a reason why any logic for H2S cannot be applied to SF2. In both cases you have one lone pair on sulfur which has mainly s character, one lone pair on sulfur which has mainly p character, and two bonds to fluorine which have mainly p character. Ergo, bond angle close(r) to 90 degrees.
@orthocresol the way we were taught d pi-p pi bonding in SO3 was that S has accessible d subshell. Once its two p electrons are excited to 3d, those two electrons form two d pi p pi bonds with two oxygen atoms. Going by this logic, I'd say nitrate anion won't have d pi p pi bonding because N does not have acceesible d aubshell.
@orthocresol Unfortunately yes... Have a look at this YT video youtu.be/Y_U5zFsSPA8 at time 14:46. You won't understand what he says, but you can have a look at the blackboard to infer what's going on :-(
Just confirm that Blackboard image is wrong then I'll go to a sound sleep
Due to symmetry constraints ($D_\mathrm{3h}$) in $\ce{SO3}$ there 6 electrons in $\pi$ type orbitals. In a wider sense of the term this molecule is Y-aromatic, but the HOMO actually represents two electrons in in-plane lone pair orbitals of oxygen. The LUMO is an antibonding $\pi$ orbital with re...