Question

Oxidation number of P in PO3−4, of S in SO2−4and that of Cr in Cr2O2−7 are respectively :

Answer 1

PO3-4 =》p + 3(-2)=-4

p=2

SO2-4 =》s+2(-2)=-4

s=0

Cr2O2-7 =》2Cr+2(-2)=-7

2Cr=-3

Cr=-3/2

this is your answer!!!

Answer 2

Answer:

+5, +6 and +6

Explanation:

Sum of oxidation states of all atoms = charge of ion

Oxidation number of oxygen (o2) = -2.

Thus,

Let the oxidation number of P in element PO43- be = x.

= x + 4(-2) = -3

= x = +5

Let the oxidation number of S in element SO42- be = y

= y + 4(-2) = -2

= y = +6

Let the oxidation number of Cr in element Cr2O72- be = z

= 2z + 7(-2) = -2

=  z = +6

Thus, the oxidation state of P, S and Cr are +5, +6 and +6.