Oxidation number of $\mathrm{S}$ involved in coordination bond in $\mathrm{Na}_{2} \mathrm{~S}_{2}…

Oxidation number of $\mathrm{S}$ involved in coordination bond in $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ is
  1. $-2$
  2. $+2$
  3. $+5$
  4. $+6$

Solution

\(\mathrm{Na}^{+} \mathrm{O}^{-}-\underset{|| \atop \huge \stackrel{ }{\mathrm{O}}}{\stackrel{\huge \mathrm{S} \atop \uparrow}{\mathrm{S}}}-\mathrm{O}^{-} \mathrm{Na}^{+}\) There is a coordinate bond between two sulphur atoms. The oxidation number of acceptor \(\mathrm{S}\)-atom is -2. Let, the oxidation number of other \(\mathrm{S}\)-atom be \(\mathrm{x}\). \(\underset{\text { For } \mathrm{Na}}{2(+1)}+\underset{\text { For catoms }}{3 \times(-2)}+x+\underset{\text { For coordinate S-atoms }}{1(-2)}=0\) \(x=+6\) *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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