Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O7 (2-) and NO3 (-) .

  • H2SO4 by conventional method.

Let x be the oxidation number of

S 2(+1) + x + 5 (−2) = 0

x = +8

S can’t be 8 or more than 8 because S has only 6 valance electrons. This fallacy is removed by calculating oxidation of S by chemical bonding method.

in H2SO5 two oxygen atoms are in -1 oxidation state.

Let see the oxidation number of S is x.

2(+1) + x + 3(-2)+2(-1) =0

+2 + x -6 -2 = 0

x = +6

Hence, oxidation number of S is +6.

  • Cr2O72-

Let the oxidation number of Cr is x.

2x +7(-2) = -2

2x = 12

x = +6

Hence, the oxidation number of Cr in dichromate ion is +6.

  • Nitrate ion,

Let x be the oxidation number of N in nitrate ion.

x+3(−2)=−1

From the structure O−−N+(O)−O−

x+1(−1)+1(−2)+1(−2)=0

x=+5

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