(I) Arrange NH3, H2O and HF in the order of increasing magnitude of hy

1.	(I) Arrange NH3, H2O and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement? (II) Can we use concentrated sulphuric acid and pure zinc in the preparation of dihydrogen?
Answer» 1. Increasing magnitude of hydrogen bonding among NH₃, H₂O and HF is HF > H₂O > NH₃ 2. The extent of hydrogen bonding depends upon electronegativity and the number of hydrogne atoms available for bonding. 3. Among N, F and O the increasing order of their electronegativities are N < O < F 4. Hence the expected order of the extent of hydrogen bonding is HF > H₂O > NH₃ (II) Conc. H₂SO₄ cannot be used because it acts as an oxidizing agent also and gets reduced to SO₂ . Zn + 2H₂SO₄ (Conc) → ZnSO₄ + 2H₂O + SO₂ Pure Zn is not used because it is non-porous and reaction will be slow. The impurities in Zn help in constitute of electrochemical couple and speed up reaction.

(I) Arrange NH3, H2O and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement? (II) Can we use concentrated sulphuric acid and pure zinc in the preparation of dihydrogen?

Answer»

1. Increasing magnitude of hydrogen bonding among NH₃, H₂O and HF is

HF > H₂O > NH₃

2. The extent of hydrogen bonding depends upon electronegativity and the number of hydrogne atoms available for bonding.

3. Among N, F and O the increasing order of their electronegativities are

N < O < F

4. Hence the expected order of the extent of hydrogen bonding is

HF > H₂O > NH₃

(II) Conc. H₂SO₄ cannot be used because it acts as an oxidizing agent also and gets reduced to SO₂ .

Zn + 2H₂SO₄ (Conc) → ZnSO₄ + 2H₂O + SO₂ Pure Zn is not used because it is non-porous and reaction will be slow. The impurities in Zn help in constitute of electrochemical couple and speed up reaction.

Discussion

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