Find sum of oxidation state of central atom of anion in compound ( H ) and number of atoms in anion of compound (I) which are in zero oxidation state.

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TestHub · Accepted Answer

(A) is hypo; strongly heating anhydrous $Na_{2} S_{2} O_{3}$ gives $4 Na_{2} S_{2} O_{3} \to 3 Na_{2} SO_{4} + Na_{2} S_{5}$ . So (H) = $Na_{2} SO_{4}$ and (I) = $Na_{2} S_{5}$ (zigzag $S_{5}^{2 -}$ anion).

Anion of (H) = $SO_{4}^{2 -}$ : oxidation state of central S = +6.

In $S_{5}^{2 -}$ the two terminal S atoms carry -1 each, so the remaining 3 middle S atoms are in zero oxidation state.

Required sum $= 6 + 3 = 9$ .

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