Assign oxidation number to the underlined elements in each of the following species:
(a)`NaH_(2)underlineP O_(4) " " (b) NaHunderlineSO_(4) " " (c) H_(4)underlineP_(2)O_(7) " " K_(2)underline(Mn) O_(4)`
`(e) CaunderlineO_(2)" " (f) NaunderlineBH_(4) " " (g)H_(2)underlineS_(2)O_(7) " " (h) KAl(underlineSO_(4))_(2). 12 H_(2)O`

Explanation

a. Let the oxidation number of `P` be `x`. Writing the oxidation number of each atom above its symbol, we have
`overset(+1)(Na)overset(+1)(H_(2))overset(x)(P)overset(-2)(O_(4))`
Sum of oxidation numbers of varios atoms in
`NaH_(2)PO_(4) =1(+1)+2(+1)+1(x)+4(-2)`
`=x-5`
But the sum of oxidation number of various atoms in `NaH_(2)PO_(4)` (netural) is zero
`:. x-5=0` or `x=+5`
Thus, the oxidation number of `P` in `NaHPO=+5`.
b. `overset(+1)(Na)overset(+1)(H)overset(x)(S)overset(-2)(O_(4))`
`:. 1(+1)+1(+1)+x+4(-2)=0`
`or x=+6`
Thus, the oxidation number of S in `NaHSO_(4)=+6`.
c. `overset(+1)(H_(4))overset(x)(P_(2))overset(-2)(O_(7))`
`:. 4(+1)+2(x)+7(-2)=0`
or `x=+5`
Thus, the oxidation number of `P` in `H_(4)P_(2)O_(7)=+5`.
d. `overset(+1)(K_(2))overset(x)(Mn)overset(-2)(O_(4))`
`:. 2(+1)+1(x)+4(-2)=0`
or `x=+7`
Thus, the oxidation number of `Mn` in `K_(2)MnO_(4)=+7`.
e. Let the oxidation number of `Ca` be `x`. Since `O` in peroxides has an oxidation of `-1`. Thus,
`overset(x)(Ca)overset(-1)(O_(2))`
`:. x+2(-1)=0`
or `x=+2`
Thus, the oxidation number of calcium in `CaO_(2)=+2`.
f. In `NaBH_(4), H` is present as hydride ion. Therefore, its oxidation number is `-1`. Thus,
`overset(+1)(Na)overset(x)(B)overset(-1)(H_(4))`
`:. 1(+1)+x+4(-1)=0`
or `x=+3`
Thus, the oxidation number of `B` in `NaBH_(4)=+3`.
g. `overset(+1)(Na_(2))overset(x)(S_(2))overset(-2)(O_(7))`
`:. 2(+1)+2(x)+7(-2)=0`
or `x=+6`
Thus, the oxidation number of `S` in `Na_(2)S_(2)O_(7)=+6`.
h. `overset(+1)(K)overset(+3)(Al)(overset(x)(S)overset(-2)(O_(4)))_(2).12(overset(+1)(H_(2))overset(-2)(O))`
or `+1+3+2x+8(-2)+12(2xx1-2)`
or `x=+6`
Alternatively, since `H_(2)O` is a neutral molecule, therefore, sum of oxidation numbers of all the atoms in `H_(2)O` may be taken as zero. As such water molecules may be ignored while computing the oxidation number of `S`.
`:. +1+3+2x-16=0`
or `x=+6`
Thus, the oxidation number of `S` in `KAl(SO_(4))2.12 H_(2)O=+6`.