${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3}$ is reduced by ${{\rm{I}}_2}$ to

${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_4}{{\rm{O}}_6}$

${\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}$

NEET UG CHEMISTRY MCQ: S-BLOCK ELEMENTS

1 Medium

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Understanding +4 | -1

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}

is reduced by

{\mathrm{I}_2}

View answer and explanation

Correct option: D

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_4}{\mathrm{O}_6}

Option D is correct because sodium thiosulphate (

\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3

) reacts with iodine (

{\mathrm{I}_2}

) to form sodium tetrathionate (

\mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6

) and sodium iodide.