The molecular mass of `Na_2S_2O_3` and `I_2` are the `M_1` and `M_2` respectively, then what will be the equivalent mass of `Na_2S_2O_3` and `I_2` in the following reactions ?
`2S_2O_3^(2-)+I_2 to S_4O_6^(2-)+2I^(-)`

Options

A. `M_1, M_2`
B. `M_2, M_2//2`
C. `2M_1, M_2`
D. `M_1, 2M_2`

Answer

B

Explanation

In NO, total number electrons = 7+8 =15
M.O. configuration of NO :
`therefore` Bond order
In total number electron =16+1=17
M.O. configuration of :
`therefore` Bond order
In `C_2^(2-)` total number of electron = 12+2=14
M.O. configuration of `C_2^(2-)` :
Bond order =`(8-2)/2=6/2=3`
In `He_2^(+)`, total number of electron =4-1 =3
M.O. configuration of `He_2^+`
Bond order =`(2-1)/2=0.5`
`therefore` Correct order of bond order is :