MCQ
The correct order of increasing bond angles is
- A$OF_2 < ClO_2 < H_2O < Cl_2O$
- ✓$OF_2 < H_2O < Cl_2O < ClO_2$
- C$OF_2 < H_2O < ClO_2 < Cl_2O$
- D$ClO_2 < OF_2 < H_2O < Cl_2O$
✓
Answer
Correct option: B.
$OF_2 < H_2O < Cl_2O < ClO_2$
b
Among $OF _2, H _2 O$ and $OCl _2$ bond angle will depend upon electronegativity of surrounding atom as central atom is same and their hybridisation is also same
Among $OF _2, H _2 O$ and $OCl _2$ bond angle will depend upon electronegativity of surrounding atom as central atom is same and their hybridisation is also same
. $OCl _2$ has unexpectedly greater angle due to bulky surrounding group $Cl$.
Angle of $OF _2$ is smaller than that of $H _2 O$ as bond angle is inversely to electronegativity.
All these are $sp ^3$ hybridized. And have $2$ lone pairs but $ClO _2$ has only $3$ free electrons. Therefore, has highest bond angle.
Therefore, the bond angle order is: $ClO _2\,>\, OCl _2\, >\, H _2 O \, >\, OF _2$.
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