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Predict the most preferable structure of $\ce{NCS-}.$

$$\Large\underset{(\textbf{1})}{\ce{(:\!\!\overset{\huge.\!\!.}{N}=C=\overset{\huge.\!\!.}{\underset{\huge.\!\!.}{S}})-}}\qquad\underset{(\textbf{2})}{\ce{(\overset{\huge.\!\!.}{\underset{\huge.\!\!.}{C}}=S=\overset{\huge.\!\!.}{\underset{\huge.\!\!.}{N}})-}}$$

My Attempt

A chart for the formal charges:

$$ \begin{array}{l|rr} \hline \text{Atom} & (\textbf{1}) & (\textbf{2})\\ \hline \ce{N} & -1 & -1 \\ \ce{C} & 0 & -2 \\ \ce{S} & 0 & +2 \\ \hline \end{array} $$

As a conclusion, in structure 2 the formal charges on $\ce{C}$ and $\ce{S}$ (i.g. bond becomes polar because of $+2$ and $-2$) make it weaker, whereas in structure 1 there are less charge separations. So, (2) must be more a accurate structure.

But according to my book, structure 2 is more stable. Why? Please explain. Is there any other approach to this question?


Edit

I got this statement in favor of structure #2 "is more stable because each atom has non-zero formal charge in the lowest energy state". But i can't understand this statement. Can anyone explain me?"

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    $\begingroup$ The name of your ion is the thiocyanate ion. Now,the word 'thio' basically refers to the replacement of a sulphur somewhere(usually an O) in a parent molecule to get the current one. The parent ion in this case will obviously be the cyanate ion with the molecular formula [OCN]- . Now,due to restrictions on the valency of O,you can't make it the central atom ,and therefore a structure like (2) will not be possible,and so a probable structure for [OCN]- would be something like [O=C=N]- . Now,simply replace O by S here to get the 'thio' derivative's structure,which will resemble (1) $\endgroup$ – Yusuf Hasan Sep 27 '19 at 21:44
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    $\begingroup$ I took a liberty to compact all lists of formal charges into one table for a better visual cue as well as corrected formatting. We prefer to use Markdown as a more lightweight markup alternative to plain HTML tags, and MathJax only if necessary. If you want to know more, pleasevisit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$ – andselisk Sep 27 '19 at 22:55
  • $\begingroup$ Did you book give any reasons? It is unlikely that statements are left unexplained in a chemistry textbook, although there may be a minority of such cases. $\endgroup$ – Tan Yong Boon Sep 28 '19 at 0:35
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    $\begingroup$ Did you mistype? Looks like you meant the book favors #1. They probably also said that structure has less formal charge separation, which you typically want with all-nonmetal molecules. $\endgroup$ – Oscar Lanzi Sep 28 '19 at 12:25
  • $\begingroup$ Just a note: it is an ill posed exercise. It should ask about stability not what is accurate. The fact that A is less stable than B it doesn't mske A inaccurate. The used terminology would be valid about mesomers, in which sometimes one is more closed to the the actual hybrid of resonance and as such can be called "a more accurate structure". Here we have different species so what? Perhaps is not the book fault but OP. $\endgroup$ – Alchimista Sep 28 '19 at 12:32
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This is a misprint. Here is a similar problem (OpenStax Chemistry, retrieved from https://opentextbc.ca/chemistry/chapter/7-4-formal-charges-and-resonance/) that makes sense and has a correct answer:

As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: CNS–, NCS–, or CSN–. The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:

enter image description here

Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).

The statement made rationalizing the incorrect structure is also flawed:

"is more stable because each atom has non-zero formal charge in the lowest energy state"

This is an ion. Some atom has to have a non-zero formal charge because the net charge should equal the sum of formal charges. Why having each atom have a non-zero formal charge should result in a more stable structure is not clear to me. Mentioning the lowest energy state is curious as well. Maybe they meant most relevant resonance structure.

One should also keep in mind that formal charges are mostly an accounting method and do not reflect actual charge distribution (see https://chemistry.stackexchange.com/a/119771).

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    $\begingroup$ Ah yes, the big issue between charges and formal charges. And the way we teach stupid, meaningless approximations to students. The reasoning, while fitting the interpretation, is incomplete at best. The the atom, linear molecules are a nightmare to describe it understand, if you're not considering resonance, you're having the wrong idea. The original question is bad, if it's not done at a proper background; and I guess this is high school, so it's even worse. $\endgroup$ – Martin - マーチン Sep 28 '19 at 22:34
  • $\begingroup$ @Martin-マーチン Oops, trigger warning - formal charges and stability. I think formal charges have their place (teaching acids and bases, maybe), but guessing the stability of fictitious species is probably not the best use of time when teaching an introductory course. $\endgroup$ – Karsten Theis Sep 29 '19 at 1:01

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