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Oxidation Numbers

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Hi,

 

what are the oxidation numbers of NaNi(IO6)?

 

Ni is supposed to get the oxidation number +IV

 

I hope someone can help me.

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Hi hypervalent_iodine,

 

I`m a teacher but not for chemistry, but it`s my hobby.

I`ve heard some of my colleagues talk about this compound, and I wanted to be able to talk with them about this, so I tried to manage it to get the right oxidation numbers. But it has been too long since I was at school by myself.

 

So please help me

Alright, well you can calculate it by looking at the net charge of the molecular and treating it algebraically and by knowing a few rules of thumb. The charge of the whole molecule is equal to the sum of the charge of its components and those charges correspond will to the oxidation numbers of each atom. The oxidation number for a sodium ion is +1 as it is a Group I element. Oxygen is always -2, except in peroxides (not what we're dealing with here), and you say you already know the charge of the Ni atom is +4. The net charge of the molecule is 0, so to calculate the charge of the iodide:

 

0 = 1 + 4 + x + (6 x -2)

 

0 = 5 + x - 12

 

x = 7

 

The oxidation number of the iodine is therefore +7.

 

If you didn't know the oxidation state of the Ni, which is often the case in these questions, you could have looked up the net charge of the periodate (which is -5) and used that value to work out the charge of the iodine, ignoring the Na Ni component, and then worked out the charge on the Ni.

To be fair, there is no way to know from just the formula whether that's Ni(IV) and I(VII) or Ni(V) and I(VI).

You also can't rule out a peroxy species unless you happen to know about orthoperiodates.

 

 

Assuming you knew that I(VI) probably doesn't exist, you could make the distinction fairly easily. However, that requires a degree of prerequisite chemical instinct I'm assuming the OP doesn't have, so you are correct in that sense.

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