Determine the oxidation state of Cr in Cr(SO4)2?

SO4 has a charge of -2, so the oxidation number is also -2. Since the overall oxidation number is 0. Then let Cr =x.

x + (-2)2 = 0

x -4 = 0

x = 4.

I took the overall charge of SO4, am I supposed to take the charge of each element individually, or was my right?

1 Answer

  • You were correct.

    Take whatever charge you can be most sure of. SO4 is always 2-. (On the other hand, S can have any number of different oxidation states, so you don't want to start there.)

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