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- Need clarity, kindly explain! The correct sequence of decreasing number of π-bonds in the structures of H2SO3, H2SO4 and H2S2O7 is :
The correct sequence of decreasing number of π-bonds in the structures of H2SO3, H2SO4 and H2S2O7 is :
- Option 1)
H2SO3 > H2SO4 > H2S2O7
- Option 2)
H2SO4 > H2S2O7 > H2SO3
- Option 3)
H2S2O7 > H2SO4> H2SO3
- Option 4)
H2S2O7 > H2SO3 > H2SO4
Answers (1)
As we learnt in
Lewis structure -
1. pair of bonded electrons is by means of a dash (-) usually called a bond
2. Lone pairs or non-bonded electrons are represented by dots
3. Electron represent in the last shell of atoms are called valence electrons.
- wherein
In H2S2O7 the number of bond is 4
In H2SO4 the number of bond is 2 and in
number of
bond is 1
Decreasing order of number of bond is
Option 1)
H2SO3 > H2SO4 > H2S2O7
This option is incorrect
Option 2)
H2SO4 > H2S2O7 > H2SO3
This option is incorrect
Option 3)
H2S2O7 > H2SO4> H2SO3
This option is correct
Option 4)
H2S2O7 > H2SO3 > H2SO4
This option is incorrect
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