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Choose the most correct option · Q7

Q.vii. The reaction, 3 ClO−⟶ClO3−+2 Cl−\mathrm{3\ ClO^- \longrightarrow ClO_3^- + 2\ Cl^-} occurs in two steps,

(i) 2 ClO−⟶ClO2−\mathrm{2\ ClO^- \longrightarrow ClO_2^-}
(ii) ClO2−+ClO−⟶ClO3−+Cl−\mathrm{ClO_2^- + ClO^- \longrightarrow ClO_3^- + Cl^-} The reaction intermediate is a. Cl−^-
b. ClO2−_2^-
c. ClO3−_3^-
d. ClO−^-
[!NOTE]
Step
(i) is printed exactly as shown -- "2 ClO−^- ⟶\longrightarrow ClO2−_2^-" with no Cl−^- product -- so the two printed steps sum to only one Cl−^-, not the overall equation's two (the balanced first step is 2 ClO−^- ⟶\longrightarrow ClO2−_2^- + Cl−^-). The answer is unaffected: ClO2−_2^- is formed in step
(i) and consumed in step
(ii) either way.
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Step 1. Steps: (i) 2ClO⁻ -> ClO2⁻; (ii) ClO2⁻+ClO⁻ -> ClO3⁻+Cl⁻.

Step 2. Adding the steps: 2ClO⁻+ClO2⁻+ClO⁻ -> ClO2⁻+ClO3⁻+Cl⁻. Cancelling ClO2⁻ (present on both sides) gives 3ClO⁻ -> ClO3⁻+2Cl⁻, matching the stated overall reaction exactly. …

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