\(\text{Both (A) and (R) are correct and (R) is the correct explanation of (A)} \)
\(\text{Both (A) and (R) are correct But (R) is not the correct explanation of (A)} \)
\(\text{(A) is correct but (R) is incorrect} \)
\(\text{(A) is incorrect but (R) is correct} \)
Step 1: Understanding the Acidity of Carboxylic Acids vs. Phenols
- Carboxylic acids (\( R-COOH \)) are more acidic than phenols (\( C_6H_5OH \)) because they have a lower pKa.
- The carboxylate ion (\( R-COO^- \)) formed after deprotonation is highly stabilized by resonance.
Step 2: Resonance and Stability Comparison
- Carboxylate ion (\( COO^- \)):
- Resonance structures are equivalent.
- The negative charge is delocalized equally between two oxygen atoms, making the carboxylate ion highly stable. \[ \text{O=C-O}^- \leftrightarrow ^-O-C=O \] - Phenoxide ion (\( C_6H_5O^- \)):
- Resonance is not equivalent.
- The negative charge is not equally delocalized across the oxygen and the benzene ring.
- Oxygen retains a significant part of the negative charge, making it less stable than the carboxylate ion.
Step 3: Conclusion
- Since carboxylate ion is more stable than phenoxide ion, carboxylic acids are more acidic than phenols.
- The given reason correctly explains this fact.
Thus, the correct answer is: Option (1): Both (A) and (R) are correct and (R) is the correct explanation of (A)
Given below are two statements:
Statement (I): \(\textit{An element in the extreme left of the periodic table forms acidic oxides.}\)
Statement (II):\( \textit{Acid is formed during the reaction between water and oxide of a reactive element present in the extreme right of the periodic table.} In the light of the above statements, choose the correct answer from the options given below:\)
Consider the following oxides:
V_2O_5, Cr_2O_3, Mn_2O_7, V_2O_3, VO_2
A number of oxides which are acidic is \( x \).
Consider the following complex compound:
[Co(NH_2CH_2CH_2NH_2)_3](SO_4)_3
The primary valency of the complex is \( y \).
What is the value of \( x + y \) is?
If the roots of $\sqrt{\frac{1 - y}{y}} + \sqrt{\frac{y}{1 - y}} = \frac{5}{2}$ are $\alpha$ and $\beta$ ($\beta > \alpha$) and the equation $(\alpha + \beta)x^4 - 25\alpha \beta x^2 + (\gamma + \beta - \alpha) = 0$ has real roots, then a possible value of $y$ is: