Question

Calculate the O.N of Cr in K2Cr2O7 and P in H2P2O5​

Answer 1

Answer:

Oxidation number of Cr :

2(K) + 2(Cr) + 7 (O) = 0

2(K) + 2(Cr) + 7 (O) = 0K is a alkali matal therefore it's oxidation number is +1

2(K) + 2(Cr) + 7 (O) = 0K is a alkali matal therefore it's oxidation number is +1O is -2 in most of its compound

2(K) + 2(Cr) + 7 (O) = 0K is a alkali matal therefore it's oxidation number is +1O is -2 in most of its compound=> 2(1) + 2(Cr) + 7(-2) = 0

2(K) + 2(Cr) + 7 (O) = 0K is a alkali matal therefore it's oxidation number is +1O is -2 in most of its compound=> 2(1) + 2(Cr) + 7(-2) = 0=> 2 + 2Cr - 14 = 0

2(K) + 2(Cr) + 7 (O) = 0K is a alkali matal therefore it's oxidation number is +1O is -2 in most of its compound=> 2(1) + 2(Cr) + 7(-2) = 0=> 2 + 2Cr - 14 = 0=> 2Cr = 14 - 2

=> Cr = 12/2 = 6

P in H2P2O5:

2(H) + 2(P) + 5(O) = 0

Hydrogen has +1 in its nonmental compounds

Oxygen has -2

=>2(1) + 2(P) + 5(-2) = 0

=> 2 + 2P -10 =0

=>2P = 10-2

=> P = 8/2 = 4