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?"