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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 \pi bond is 4

In H2SO4 the number of \pibond is 2 and in H_{2}SO_{3} number of \pi bond is 1

Decreasing order of number of \pi bond is 

H_{2}S_{2}O_{7}> H_{2}SO_{4}> H_{2}SO_{3}

 


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 

Posted by

Vakul

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