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P Block Elements question

2007 · Shift 1 · Q11
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P Block Elements question

2007 · Shift 1 · Q11

JEE AdvancedChemistryP Block ElementsMCQ+3 / −1
Statement 1 : In water, orthoboric acid behaves as a weak monobasic acid. Statement 2 : In water, orthoboric acid acts as a proton donor.
  1. A
    Statement 1 is True, Statement 2 is True, Statement 2 is a CORRECT explanation for Statement 1
  2. B
    Statement 1 is True, Statement 2 is True, Statement 2 is NOT a CORRECT explanation for Statement 1
  3. C
    Statement 1 is True, Statement 2 is False
  4. D
    Statement 1 is False, Statement 2 is True
View written solutionFree

Correct answer: C

Step-by-step Derivations:

Step 1: Analyze Statement 1

Statement 1 says: "In water, orthoboric acid behaves as a weak monobasic acid."

  • Orthoboric acid has the chemical formula H3BO3H_3BO_3H3​BO3​, which can also be written as B(OH)3B(OH)_3B(OH)3​.
  • When orthoboric acid is dissolved in water, it does not donate one of its own protons. Instead, it accepts a hydroxide ion (OH−OH^−OH−) from a water molecule. The reaction is as follows: B(OH)3+H2O⇌[B(OH)4]−+H+B(OH)_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+B(OH)3​+H2​O⇌[B(OH)4​]−+H+
  • This reaction releases one proton (H+H^+H+) into the solution for each molecule of orthoboric acid. An acid that furnishes one proton per molecule is called a monobasic acid.
  • The acid dissociation constant (KaK_aKa​) for this reaction is very small (Ka=5.8×10−10K_a = 5.8 \times 10^{-10}Ka​=5.8×10−10), which corresponds to a pKapK_apKa​ of about 9.25. A high pKapK_apKa​ value (or low KaK_aKa​ value) indicates that it is a weak acid.
  • Therefore, orthoboric acid is a weak monobasic acid in water. Statement 1 is True.

Step 2: Analyze Statement 2

Statement 2 says: "In water, orthoboric acid acts as a proton donor."

  • An acid that donates a proton (H+H^+H+) is called a Brønsted-Lowry acid.
  • Let's re-examine the reaction of orthoboric acid with water: B(OH)3+H2O⇌[B(OH)4]−+H+B(OH)_3 + H_2O \rightleftharpoons [B(OH)_4]^- + H^+B(OH)3​+H2​O⇌[B(OH)4​]−+H+
  • In this reaction, the H3BO3H_3BO_3H3​BO3​ molecule does not donate a proton. The proton that is released into the solution comes from the water molecule (H2OH_2OH2​O), which splits into H+H^+H+ and OH−OH^-OH−.
  • The central boron atom in B(OH)3B(OH)_3B(OH)3​ is electron-deficient (has an incomplete octet of electrons). It acts as a Lewis acid by accepting an electron pair from the hydroxide ion (OH−OH^−OH−).
  • Since orthoboric acid accepts an electron pair rather than donating a proton, it is a Lewis acid, not a Brønsted-Lowry acid (proton donor).
  • Therefore, the statement that orthoboric acid acts as a proton donor is incorrect. Statement 2 is False.

Step 3: Conclusion

  • Statement 1 is True.
  • Statement 2 is False.

Based on this analysis, we can evaluate the given options:

  • A: Statement 1 is True, Statement 2 is True... (Incorrect)
  • B: Statement 1 is True, Statement 2 is True... (Incorrect)
  • C: Statement 1 is True, Statement 2 is False. (Correct)
  • D: Statement 1 is False, Statement 2 is True. (Incorrect)

The correct option is C.

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