Q: (Logics) If well-formed formula $q , \not\vDash q \text{, then } \vDash \lnot q$

Question

If well-formed formula without free variables $q , \not\vDash q \text{, then }\vDash \lnot q$. Is it true?

Answer 1

$\not\models q$ is equivalent to "$q$ isn't a tautology," or "In some interpretations, $q$ may not true".

$\models \neg q$ is equivalent to "$q$ is a contradiction," or "In all interpretations, $q$ can not true."

Answer 2

No. You have that $\vDash q$ if and only if $q$ is a tautology.

So, if you take $q$ to be a contingent statement (such as any atomic statement), then neither $q$ nor $\neg q$ are tautologies, and thus you have $\not \vDash q$ and $\not \vDash \neg q$. So, $\not \vDash q$ does not imply $\vDash \neg q$.