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P Block Elements question

2015 · Shift 2 · Q15
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P Block Elements question

2015 · Shift 2 · Q15

JEE AdvancedChemistryP Block ElementsMultiple correct+4 / −2
The correct statements regarding (i) HClOHClOHClO, (ii) HClO2HClO_2HClO2​, (iii) HClO3HClO_3HClO3​ and (iv) HClO4HClO_4HClO4​ is (are)
  1. A
    The number of Cl=O bonds (ii) and (iii) together is two.
  2. B
    The number of lone pairs of electrons on Cl in (ii) and (iii) together is three.
  3. C
    The hybridization of Cl in (iv) is sp3
  4. D
    Amongst (i) to (iv), the strongest acid is (i).
View written solutionFree

Correct answer: B, C

To determine the correct statements, we need to analyze the structure, bonding, and properties of each chlorine oxyacid.

Step 1: Determine the Lewis structures, number of Cl=O bonds, and lone pairs on the central Cl atom for each acid.

Chlorine (Cl) is the central atom and has 7 valence electrons.

  • (i) HClOHClOHClO (Hypochlorous acid):

    • Structure: H-O-Cl
    • The Cl atom forms one single bond with the O atom.
    • Electrons used by Cl in bonding = 1.
    • Lone pair electrons on Cl = 7 - 1 = 6.
    • Number of lone pairs on Cl = 6 / 2 = 3.
    • Number of Cl=O bonds = 0.

  • (ii) HClO2HClO_2HClO2​ (Chlorous acid):

    • Structure: H-O-Cl=O
    • The Cl atom forms a single bond with the -OH group and a double bond with the other O atom.
    • Electrons used by Cl in bonding = 1 (for Cl-O) + 2 (for Cl=O) = 3.
    • Lone pair electrons on Cl = 7 - 3 = 4.
    • Number of lone pairs on Cl = 4 / 2 = 2.
    • Number of Cl=O bonds = 1.

  • (iii) HClO3HClO_3HClO3​ (Chloric acid):

    • Structure: The central Cl atom is bonded to one -OH group and two O atoms via double bonds.
    • Electrons used by Cl in bonding = 1 (for Cl-O) + 2 * 2 (for two Cl=O) = 5.
    • Lone pair electrons on Cl = 7 - 5 = 2.
    • Number of lone pairs on Cl = 2 / 2 = 1.
    • Number of Cl=O bonds = 2.

  • (iv) HClO4HClO_4HClO4​ (Perchloric acid):

    • Structure: The central Cl atom is bonded to one -OH group and three O atoms via double bonds.
    • Electrons used by Cl in bonding = 1 (for Cl-O) + 3 * 2 (for three Cl=O) = 7.
    • Lone pair electrons on Cl = 7 - 7 = 0.
    • Number of lone pairs on Cl = 0.
    • Number of Cl=O bonds = 3.

Step 2: Evaluate each statement.

  • A: The number of Cl=O bonds in (ii) and (iii) together is two.

    • Number of Cl=O bonds in (ii) HClO2HClO_2HClO2​ is 1.
    • Number of Cl=O bonds in (iii) HClO3HClO_3HClO3​ is 2.
    • Total number of Cl=O bonds = 1 + 2 = 3.
    • Therefore, statement A is incorrect.

  • B: The number of lone pairs of electrons on Cl in (ii) and (iii) together is three.

    • Number of lone pairs on Cl in (ii) HClO2HClO_2HClO2​ is 2.
    • Number of lone pairs on Cl in (iii) HClO3HClO_3HClO3​ is 1.
    • Total number of lone pairs = 2 + 1 = 3.
    • Therefore, statement B is correct.

  • C: The hybridization of Cl in (iv) is sp3.

    • In (iv) HClO4HClO_4HClO4​, the central Cl atom is attached to 4 atoms (one O from -OH and three other O atoms).
    • The number of sigma (σ) bonds around Cl is 4.
    • The number of lone pairs on Cl is 0.
    • The steric number = (number of σ bonds) + (number of lone pairs) = 4 + 0 = 4.
    • A steric number of 4 corresponds to sp3sp^3sp3 hybridization.
    • Therefore, statement C is correct.

  • D: Amongst (i) to (iv), the strongest acid is (i).

    • The acidic strength of oxyacids of the same element increases with the increase in the oxidation state of the central atom.
    • Oxidation states of Cl:
      • HClOHClOHClO: +1
      • HClO2HClO_2HClO2​: +3
      • HClO3HClO_3HClO3​: +5
      • HClO4HClO_4HClO4​: +7
    • The order of acidic strength is HClO<HClO2<HClO3<HClO4HClO < HClO_2 < HClO_3 < HClO_4HClO<HClO2​<HClO3​<HClO4​.
    • This is because the stability of the conjugate base (ClOn−ClO_n^−ClOn−​) increases as the negative charge is delocalized over more oxygen atoms.
    • The strongest acid is HClO4HClO_4HClO4​ (iv), and the weakest is HClOHClOHClO (i).
    • Therefore, statement D is incorrect.

Conclusion: The correct statements are B and C.

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