Which of the following statement is true for H2S?

  1. A

    The sulphur atom is sp3 hybridized and bonding take place in excited state.

  2. B

    H2S and BF3 form adduct readily through coordinate bond between boron and sulphur

  3. C

    Its molecular geometry is non-linear and H-S-H bond angle is smaller than H-0-H

    bond angle in H2O

  4. D

    (b) and (c) both

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    Solution:

    The sulphur atom is sp3 hybridized and bonding takes place in ground state. Two filled orbitals 3s2 and 3px2 

    H2SBF3 has coordinated bond, S from H2S gives lone pair and B receives it.
    Bond angle difference leads to form hydrogen bonds in water.
    so that two lone pairs of oxygen in water engaged in hydrogen bond.
    Formation of hydrogen bonds reduce the lone pair - bond pair repulsion. That's why bond angle is greater in water when compare to hydrogen sulfide. In H2S  the sp3 hybrid, orbitals are bigger than H2O and so the bond pair-bond pair repulsion decreases. 

    Hence, the option D is correct.

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