Calculate the oxidation number of

(i) `N` in `NO_(3)^(-)`,
(ii) `P` in `H_(3)P_(2)O_(7)^(-)`,
(iii) `C` in `CO_(3)^(2-)`,
(iv)`Cl` in `ClO_(4)^(-)`
(v)`Cr` in `Cr_(2)^(2-)`
(vi) `Mn` in `MnO_(4)^(-)`
(vii) `Fe` in `[Fe(CN)_(6)]^(4)`

Explanation

`N in NO_(3)^(-3)` Let the oxidaton number of N in `NO_(3)^(-)` be x writing the oxdation number of each atom above its symbol `overset(x)N O_(3)^(-3)`
`therefore` sum of the oxdatin number of all th eaoms in `NO_(3)^(-)=x+3(-2)=x-6`
But the sum of xodatin numbers ofall the atoms in `NO_(3)^(-)` ion is equal to the cahrge present on it i.e (Rule 8) `therefore x-6 =-1 or x=+5`
Thus the oxdation number of N in `NO_(3)^(-) is +5`
(ii) `P in H_(3)P+_(2)O_(7)^(-)` Let the oxdiationn number of P in `H_(3)P_(2)O_(7)^(-)` be x writing the oxidation number of each atom above its symbol
`overset(+1)H_(3) overset(x)P_(2) overset(-2)O_(7)`
Sum of hte oxdaion numbers of all the atoms in `H_(3)P_(2)O_(7)^(-)=3(+1)+2(x)+7(-2)` or 2x-11
But the sum of oxdiatin numbers of all the atoms in `H_(3)P_(2)O_(7)^(-)` is equal to the charge present on it i.e (Rule 8) `therefore` 2x-11=-1 or x=+5
Thus the oxdation number of P in `H_(3)P_(2)O_(7)^(-)` is +5
(iii) C in `CO_(3)^(2-)` Let the oxidatin number of C in `CO_(3)^(2-)` be x writing the oxidation number of each atom above its symbol `overset(x)C overset(-2)O_(3) therefore` sum of hte oxidation number ofll the aotms in `CO_(3)^(2-)` ion = x+3(-2)=x-6 But the sum of oxidation numbers of all the atoms in `CO_(3)^(2-)` ions is -2 (Rule 8) `therefore` x-6 =-2 or x = +4
thus the oxidaon state of cin `CO_(3)^(2-)` is +4
(iv) `CI` in `CIO_(4)^(-)` let the oxidation numbner of CI in `CI_(4)^(-)` ions is -2 (Rule 8) `therefore` x-6 =-2 or x=+4 above its symbol `overset(x)CI O_(4)^(-2) therefore` sum of oxidation numbers of al the atoms in `CIO_(4)^(-)` ion = x+4(-2)=x-87
But the sum of oxidaiotin numbers of all the atoms in `CIO_(4)^(-)` ion is equal to the charge present on it i.e -1 (Rule 8) `therefore x-8 =-1 or x -+7`
Thus the oxidation number of CI in `CIO_(4)^(-)` is +7
(v) Cr in `Cr_(2)O_(7)^(2-)` Let the oxidation number of Cr in `Cr_(2)O_(7)^(2-)` be x writing the oxidaitn number of each atom above its symbol `overset(x)Cr_(2) overset(-2)O_(7)`
`therefore` Sum of he oxidation numbers of all the atoms in `Cr_(2)O_(7)^(2-)` ions =2(x)+7 (-2)=2x-14
But the sum of oxidation numbers of all the atoms in `Cr_(21)O_(7)^(2-)` is equal to the charge on it i.e -2 (Rule 8)
`therefore` 2x-14=2 or x=6
Thus the oxidation number of Cr in `Cr_(2)O_(7)^(2-)` ion is +6
(vi) Mn in `MnO_(4)^(-)` Let the oxidation number of Mn in `MnO_(4)^(-)` be x writing oxidation number of each atom above its sysmbol we get
`therefore` sum of he oxidation numbers of all the aotms in `MnO_(4)^(-) =x(-2)=x-8`
But the sum of oxidation numbers of all the atoms in `MnO_(4)^(-)` is -1 (Rule 8) therefore x-8=-1 or x=+7 Thus the oxidation number of Mn in `MnO_(4)^(-)` is +7
(vii) Fe in `[Fe(CN)_(6)]^(4-)` Let the oxidation number of Fe in `[Fe(CN)_(6)]^(4-)` be x writing the oxidation number of each atom above its symbol and that of cyanide ion above its formula we get
`overset(x)Fe overset(-1)(CN)_(6)`
`therefore` Sum of oxidation numbers of all the atoms in `[Fe(CN)_(6)]^(4-)`=x+6(-1)=x-6
But the sum of oxidation numbers of all the atoms in `Fe(CN)_(6)]^(4-)` is equal to -4 (Rule 8)
`therefore` x-6 =-4 or x=+2
Thus the oxidation number of Fe in `[Fe(CN)_(6)]^(4-)` is +2