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Question

Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5,Cr2O72-andNO3-respectively.


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Solution

Oxidation state: The oxidation state, also known as the oxidation number in chemistry, is the hypothetical charge of an atom if all of its bonds to other atoms were entirely ionic.

Part 1: Calculating the oxidation state of Sulphur (S) in H2SO5

  • Let x be the oxidation number of Sulphur

2(+1)+x+5(2)=0x=+8

  • Sulphur can not have +8 oxidation state because S has only six valance electrons.
  • This fallacy is removed by calculating oxidation of S by chemical bonding method.
  • In H2SO5 two Oxygen (O) atoms are in -1 oxidation state.
  • Let us assume the oxidation number of Sulphur is x.

2(+1)+x+3(-2)+2(-1)=0+2+x-6-2=0x=+6

Part 2: Calculating the oxidation state of Chromium (Cr) in Cr2O72-

  • Let x be the oxidation number of Chromium in Dichromate ion
    2x+7(2)=2x=+6
  • Hence the oxidation number of Chromium in Dichromate ion is +6.

Part 3: Calculating the oxidation state of Nitrogen (N) in NO3-

  • Let x be the oxidation number of Nitrogen in the Nitrate ion (NO3-).
  • x+3(2)=1
  • From the structure:
  • x+1(1)+1(2)+1(2)=0x=+5
  • Thus, the oxidation number of Nitrogen in the Nitrate ion (NO3-) is +5

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