Question
Explain the general structures of hydrides of group $16$ elements.
Give reason : Bond angle decreases from $H_2O$ to $H_2S.$

Answer

  • Group $16$ elements form hydrides of the type $H _2 M$ where $M = O, S, Se, Te$ and $Po.$
  • All these hydrides are formed by $sp ^3$ hybridisation of the central atom and have angular shape.
  • All hydrides have similar structure but differ in $H-M-H$ bond angle. This bond angle decreases from $H _2 O$ to $H _2 Te$.
  • All these hydrides have two $M-H$ bond pairs and two lone pairs of electrons.
  • Due to repulsion between lone pairs$-$lone pairs and lone pairs$-$bond pairs, and bond pairs$-$bond pairs the bond angle is reduced from regular $109^{\circ} 28^{\prime}$.
  • On moving down the group, atomic size increases, electronegativity decreases, electron density decreases.
  • Since among group $16$ elements, oxygen has the highest electronegativity, lowest atomic size and hence the highest electron density, the repulsion between electron pairs is maximum in $H _2 O$, hence $H-O-H$ bond angle is maximum.
  • The electron pair repulsion decreases down the group, hence bond angle also decreases down the group.
    Hydrides $H _2 O$ $H _2 S$ $H _2 Se$ $H _2 Te$
    $H-M-H$ angle $104^{\circ}$ $90^{\circ}$

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