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

Answer:

I believe it is a hope this helps

Answer 2

Answer:

B. Cr in $Cr_{2}O_{7}^{-2}$

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