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For H2O2, H2S, H2O and HF , the correct order of decreasing extent of hydrogen bonding is :
  • a)
    H2O > HF > H2O2 > H2S
  • b)
    H2O > HF > H2S > H2O2
  • c)
    HF > H2O > H2O2 > H2S
  • d)
    H2O2 > H2O > HF > H2S
Correct answer is option 'D'. Can you explain this answer?
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Explanation:

  • The extent of hydrogen bonding depends on the electronegativity and size of the atoms involved.

  • The correct order of decreasing extent of hydrogen bonding is H2O2 > H2O > HF > H2S.

  • H2O2 has the highest extent of hydrogen bonding because it has two oxygen atoms that are highly electronegative and small in size, making them capable of forming strong hydrogen bonds.

  • H2O has the second-highest extent of hydrogen bonding because it also has two highly electronegative and small oxygen atoms that can form strong hydrogen bonds.

  • HF has the third-highest extent of hydrogen bonding because it has a highly electronegative fluorine atom and a small hydrogen atom that can form strong hydrogen bonds.

  • H2S has the lowest extent of hydrogen bonding because sulfur is less electronegative than oxygen and has a larger size, making it less capable of forming strong hydrogen bonds.

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For H2O2, H2S, H2O and HF , the correct order of decreasing extent of hydrogen bonding is :a)H2O > HF > H2O2> H2Sb)H2O > HF > H2S > H2O2c)HF > H2O > H2O2> H2Sd)H2O2> H2O > HF > H2SCorrect answer is option 'D'. Can you explain this answer?
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