The average oxidation state of sulphur in `Na_(2)S_(4)O_(6)` is

Options

A. `+6`
B. `+(3)/(2)`
C. `+(5)/(2)`
D. `-2`

Answer

C

Explanation

Oxidation number of Na = +1
Oxidation number of O = -2
Let oxidation number of S= x
`therefore 2(On" of "Na) + 4(ON" of "S) +6(ON" of "O)=0`
`2(+1)+4x+6(-2)=0+2+4x-12=0`
`4x=+12-2`
`x=+(10)/(4)`
`rArr x=+(5)/(2)`