Assign oxidation number to the underlined elements in each of the following species:
a.`NaH_(2)PO_(4)`
b. `NaHul(S)O_(4)`
c. `H_(4)ul(P_(2))O_(7)`
d. `K_(2)ul(Mn)O_(4)`
e. `ul(Ca)O_(2)`
f. `Naul(B)H_(4)`
g. `H_(2)ul(S_(2))O_(7)`
h. `KAl(ul(S)O_(4))_(2).12H_(2)O`

Explanation

(a) `NaH_(2)underlinePO_(4)` lt brgt Let the oxidation number of P be x. We know that, Oxidation number of Na = +1 Oxidation number of H = +1 Oxidation number of O = –2
`implies overset(+1)(Na) overset(+1)(H_(2)) overset(x)(P)overset(-2)(O_(4))`
Then , we have
`1(+1) + 2( +1) + 1 (x) + 4 (-2) = 0`
`implies 1 +2 + x - 8 = 0 `
`implies x = + 5`
Hence , the oxidation number of P is +5 .
(b) `NaHunderlineSO_(4)`
`overset(+1)(Na) overset(+1)(H) overset(x)(S) overset(-2)(O_(4))`
Then, we have
`1(+1) + 1(+1) + 1(x) + 4(-2) = 0`
`implies 1 + 1 + x - 8 = 0`
`implies x = +6`
Hence , the oxidation number of S is `+6` .
(c) `H_(4) underlineP_(2) O_(7)`
`overset(+1)(H_(4))overset(x)(P_(2))overset(-2)(O_(7))`
Then we have
`4 (+1) 2 (x) + 7 (-2) = 0 `
`implies 4 + 2x - 14 = 0 `
`implies 2x = +10`
`implies x = +6`
Hence , the oxidation number of Mn is + 6 .
(e) `CaunderlineO_(7)`
`overset(+2)(Ca)overset(x)(O_(2))`
Then , we have
`(+2) + 2(x) = 0`
`implies 2 + 2x = 0 `
`implies x = -1`
Hence , the oxidation number of O is `-1` .
(f) `NaunderlineB H_(4)`
Then , we have
`1 (+1) + 1(x) + 4(-1) = 0 `
`implies 1 + x - 4 = 0 `
`implies x = + 3 `
Hence , the oxidation number of B is +3 .
(g) `H_(2)underlineS _(2) O_(7)`
`overset(+1)(H_(2))overset(x)(S_(2))overset(-2)(O_(7))`
Then , we have
`2 (+1) + 2(x) + 7(-2) = 0`
`implies 2 + 2x - 14 = 0`
` implies 2x = 12`
`implies x = + 6`
Hence, the oxidation number of S is `+6` .
(h) `KAl(underlineSO_(4))_(2). 12 H_(2)O`
`overset(+1)(K) overset(3+)(Al) (overset(x)(S) overset(2-)(O_(4)))_(2) . 12 overset(+1)(H_(2))overset(-2)(O)`
Then , we have
`1 (+1) + 1(+3) + 2(x) + 8 (-2) + 24(+1)+ 12 (-2) = 0 `
`implies 1 + 3 + 2x - 16 + 24 - 24 = 0`
`implies 2x = 12`
`implies x = +6`
Or,
We can ignore the water molecule as it is a neutral molecule .Then , the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore , after ignoring the water molecule , we have
`1(+1) + 1(+3) + 2(x) + 8 (-2) = 0`
`implies 1 + 3 + 2x - 16 = 0`
`implies 2x = 12`
`implies = + 6 `
Hence , the oxidation number of S is `+6` .