What is the correct order of the oxidation st | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The oxidation state of $S$ is calculated as follows:For $SO_3^{2-}$: $x + 3(-2) = -2 \implies x = +4$For ${S_2}O_4^{2-}$: $2x + 4(-2) = -2 \implie

Vedclass · Accepted Answer

${S_2}O_4^{2-} < SO_3^{2-} < {S_2}O_6^{2-}$

Vedclass · Answer

(B) The oxidation state of $S$ is calculated as follows:For $SO_3^{2-}$: $x + 3(-2) = -2 \implies x = +4$For ${S_2}O_4^{2-}$: $2x + 4(-2) = -2 \implies 2x = +6 \implies x = +3$For ${S_2}O_6^{2-}$: $2x + 6(-2) = -2 \implies 2x = +10 \implies x = +5$Comparing the oxidation states: $+3 Thus,the order is ${S_2}O_4^{2-} < SO_3^{2-} < {S_2}O_6^{2-}$.

What is the correct order of the oxidation state of sulfur in the anions $SO_3^{2-}$,${S_2}O_4^{2-}$,and ${S_2}O_6^{2-}$?

Similar Questions

In the following reaction,assign the state for the underlined atom in the product of complete hydrolysis at $R.T.$:
$\underline{Cl_2}O + H_2O \longrightarrow 2HClO$

The oxidation number of $Mn$ in $KMnO_4$ is

The oxidation state of nitrogen in $N_3H$ is

The oxidation number of sulphur in $S_2O_3^{2-}$ is:

The oxidation state of sulphur in $H_2S_2O_7$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module