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) Let the oxidation number of p be x writing the oxidation number of each atom above its symbol we
sum of oxidation numbr of various atoms in `NaH_(2)PO_(4)=1(+1)+2 (+1)+1(x)+4(-2)=x-5` but the sum of oxdation number of various atoms in `NaH_(2)PO_(4)` (neutral) is zero
Thus the oxidation nuber of p in `NaH_(2)PO_(4)=5`
(b) `overset(+1)na overset (x)H overset(x)S overset(-2) O_(4) therefore +1(+1)+x+4(+1)-2=0 or x=_6`
Thus the oxidation number of S in `NaHSO_(4)=+5`
(c ) `overset(+1)H_(4)overset(+1)P_(2)overset(x)S overset(-2)O_(4) therefore 4(+1)+2(x)+7(-2)=0 or x=+5`
(d) `K_(2)Mn O_(4)^(-2) therefore 2(+1)+1(x)+7(-2)=0 or x =+6`
thus the oxidation numbr of Mn in `K_(2)MnIO_(4)=+7`
(e ) let hte oxidation number of O be x since ca is an alkaline earth metal therefore its oxidation number is +2 thus
`CaO_(2) therefore +2+2(x)=0 or x=-1`
(f) In `NaBH_(4)` H is present as hydride ion therefore its oxidation number ois -1 thus
`Na BH_(4) therefore 2(+1)+x+(-1)=0 or x=+3`
thus the oxidation number of B in `NaBH_(4)=+3`
(g) `H_(2)S_(2)O_(7)^(2-) therefore 2(+1)+2(x)+7(-2)=0 or x_6`
thus the oxidation number of s in `H_(2)S_(2)O_(7)=+6`
(h) `K AI(SO_(4))12(H_(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 number of S
`therefore +1+3+2+x-16=0 or x=+6`
thus the oxidation number of S in KaI `(SO_(4))12H_(2)O=+6`