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(A) The acidity of phenols is inversely proportional to their $pK_{a}$ values $(Acidity \propto \frac{1}{pK_{a}})$.Electron-withdrawing groups (like $

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$A > C > B > D$

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(A) The acidity of phenols is inversely proportional to their $pK_{a}$ values $(Acidity \propto \frac{1}{pK_{a}})$.Electron-withdrawing groups (like $-NO_{2}$) increase the acidity of phenol by stabilizing the phenoxide ion through $-I$ and $-M$ effects.$1$. Phenol $(A)$ has no electron-withdrawing group,so it is the least acidic (highest $pK_{a}$).$2$. meta-Nitrophenol $(C)$ exhibits only the $-I$ effect,making it more acidic than phenol but less acidic than ortho and para isomers.$3$. ortho-Nitrophenol $(B)$ and para-Nitrophenol $(D)$ exhibit both $-I$ and $-M$ effects. para-Nitrophenol $(D)$ is the most acidic due to the strong $-M$ effect of the $-NO_{2}$ group at the para position.ortho-Nitrophenol $(B)$ is less acidic than para-Nitrophenol $(D)$ due to intramolecular hydrogen bonding,which stabilizes the undissociated molecule.Thus,the order of acidity is: $D > B > C > A$.Since $pK_{a}$ is inversely proportional to acidity,the decreasing order of $pK_{a}$ is: $A > C > B > D$.

Arrange the following phenols in decreasing order of their $pK_{a}$ values:
$(A)$ Phenol
$(B)$ ortho-Nitrophenol
$(C)$ meta-Nitrophenol
$(D)$ para-Nitrophenol

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Arrange the following phenols in decreasing order of their $pK_{a}$ values:$(A)$ Phenol$(B)$ ortho-Nitrophenol$(C)$ meta-Nitrophenol$(D)$ para-Nitrophenol