What the formula of (lll)-meu-imido-meu- (lll) ion.

Can you please rephrase the question in more understandable terms? Let me know when you’ve done that and i’ll be glad to help!

You have to calculate the oxidation estates of the atoms in each compound.

I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.

In K2Cr2O7:

- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.

- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.

That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.

In Cr2O3:

- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6

- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.

So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.

Cr has a change in oxidation number of - 3.