Question

Which of the following ions is in the lowest oxidation state?

A. P in H2PO−4
B. Cr in Cr2O2−7
C. Fe in Fe2O3
D. None of these

Answer 1

I believe it is a hope this helps

Answer 2

B. Cr in
`Cr_(2)O_(7)^(-2)`

Step-by-step explanation:

Let us first find the oxidation numbers in each of the following scenarios

A. The oxidation number of P in
`H_(2)PO_(4)^(-) \\` is +5

because 2H + 4O + P = -1

2(1) +4(-2) + P = -1

P = 5

B. The oxidation number of Cr in
`Cr_(2)O_(7) ]^(-2) \\` is

2Cr + 7O = -2

2Cr + 7(-2) = -2

Cr = 6

C. The oxidation number of Fe in
`Fe_(2)O_(3)` is +3

2Fe + 3O = 0

2Fe + 3(-2) = 0

Fe = 3

The final answer is Cr