ClO- + N2H4+ H2O2NH2OH + Cl- In the above reaction, the oxidation state of chlorine changes from blank to blank?

Oxidation state of Cl in ClO⁻ = (-1) - (-2) = +1
Oxidation state of Cl in Cl⁻ = -1

Ans: In the above reaction, the oxidation state of chlorine changes from __+1__ to __-1___.

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Consider the conversion of N₂H₄ to NH₂OH :
Oxidation state of N in N₂H₄ = (1/2) × {-[(+1)×4]} = -2
Oxidation state of N in NH₂OH = -[(+1)×3 + (-2)] = -1
The oxidation state of N changes from -2 to -1.
When 1 mole of N₂H₄ is converted to 2 moles of NH₂OH, no. of e⁻ transferred = 2 × [-1 - (-2)] = 2

Consider the conversion of ClO⁻ to Cl⁻ :
The oxidation state of Cl changes from +1 to -1.
When 1 mole of ClO⁻ is converted to 1 mole of Cl⁻, no. of e⁻ transferred = (+1) - (-1) = 2

Hence, no. of e⁻ transferred in the reaction = 2