N3−→(I)NO2→(II)NO3− .What will be the n-factor in step(I) and in step(II)?

N3−(−13)→(I)NO2(+4)→(II)NO3−(+5) Change in the oxidation state of nitrogen from azide to nitrogen dioxide is ( 4 + 1/3)×3 =13, hence n-factor = 13 Change in the oxidation state of nitrogen from nitrogen dioxide to Nitrate is +1, hence n- factor =1

N3−→(I)NO2→(II)NO3− .What will be the n-factor in step(I) and in step(II)?

N3−(−13)→(I)NO2(+4)→(II)NO3−(+5) Change in the oxidation state of nitrogen from azide to nitrogen dioxide is ( 4 + 1/3)×3 =13, hence n-factor = 13 Change in the oxidation state of nitrogen from nitrogen dioxide to Nitrate is +1, hence n- factor =1

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$N_{3}^{-} \overset{(I)}{\to} {NO}_{2} \overset{(II)}{\to} {NO}_{3}^{-}$ .What will be the n-factor in step(I) and in step(II)?

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13 , 1

13 , 2

11 , 5

13 , 3

answer is A.

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Detailed Solution

$N_{3}^{-} (- \frac{1}{3}) \overset{(I)}{\to} {NO}_{2} (+ 4) \overset{(II)}{\to} {NO}_{3}^{-} (+ 5) Change in the oxidation state of nitrogen from azide to nitrogen dioxide is (4 + 1 / 3) \times 3 = 13, hence n - factor = 13 Change in the oxidation state of nitrogen from nitrogen dioxide to Nitrate is + 1, hence n - factor = 1$