Question

Assign oxidation number to the underlined elements in each of the following species:
a. NaH₂PO₄
b. NahSO₄
c. H₄P₂O₇
d. K₂MnO₄
e. CaO₂
f. NaBH₄
g. H₂S₂O₇
h. KAl(SO₄)₂.12H₂O

Answer 1

where is the under line mate

Answer 2

Answer:

Explanation:

(a) NaH2PO4 :- Oxidation number of P in NaH2PO4  1(+1)+2(+1)+1(x)+4(-2) =0

=  x = +5

(b) NaHSO4 :- Oxidation number of S in NaHSO4 is

= 1(+1)+1(+1)+x+4(-2)=0

= x = +6

(c) H4P2O7: - Oxidation number of P in H4P2O7 is

4(+1)+2(x)+7(-2)=0

= x = +5

(d) K2MnO4:- Oxidation number of Mn in K2MnO4 is

2(+1)+1(x)+4(-2) = 0

= x = +7

(e) CaO2 :- Oxidation number of O in CaO2 is

+2+2x = 0

= x = -1

(f) NaBH4:- Oxidation number of B in NaBH4 is

1(+1) + x + 4(-1) = 0

= x = +3

(g) H2S2O7 :- Oxidation number of S in H2S2O7 is

2(+1)+2(x)+7(-2)= 0

= x = +6

(h) KAl(SO4)2.12 H2O:- Oxidation number of S in KAl(SO4)2.12 H2O is   =1+3+2x-16=0

= x = +6