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(D) The reaction involves the deprotonation of squaric acid $(A)$ by $2$ equivalents of $OH^-$ to form the squarate dianion $(B)$.$1$. Compound '$B$'

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$A, C$ and $D$ only

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(D) The reaction involves the deprotonation of squaric acid $(A)$ by $2$ equivalents of $OH^-$ to form the squarate dianion $(B)$.$1$. Compound '$B$' (squarate dianion) is aromatic because it is cyclic,planar,fully conjugated,and contains $6 \pi$ electrons ($4n+2$ rule,where $n=1$). Thus,statement $A$ is correct.$2$. The reaction is an acid-base reaction involving the removal of acidic protons from squaric acid. Acid-base reactions are typically very fast,not slow. Thus,statement $B$ is incorrect.$3$. Compound '$A$' (squaric acid) contains an enol group adjacent to a carbonyl group,which allows it to exhibit keto-enol tautomerism. Thus,statement $C$ is correct.$4$. Due to resonance in the squarate dianion $(B)$,the negative charge is delocalized over the entire ring,making all $C-C$ bond lengths equal. Thus,statement $D$ is correct.Therefore,statements $A, C,$ and $D$ are correct.

Correct statements for the given reaction are :
$A$. Compound '$B$' is aromatic
$B$. The completion of above reaction is very slow
$C$. '$A$' shows tautomerism
$D$. The bond lengths $C-C$ in compound $B$ are found to be same
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Correct statements for the given reaction are :$A$. Compound '$B$' is aromatic$B$. The completion of above reaction is very slow$C$. '$A$' shows tautomerism$D$. The bond lengths $C-C$ in compound $B$ are found to be sameChoose the correct answer from the options given below :