# The oxidation number of N and Cl in NOClO4 ​

My doubt is not about the solution itself, actually I was able to find the solution myself.

We could split NOClO4 as (NO+) + (ClO4-) and then proceed, but here I "knew" this split. Is there another approach to this question? If I'm unaware of this split, Is it actually possible to solve this stuff?

• You are correct in the formulation Nov 21, 2021 at 14:46
• If I was unaware of this "split", is there a method to solve this? Nov 21, 2021 at 14:48
• No. The formula itself does not tell you whether it is (NO+)(ClO4-) or the other way around. You are supposed to deduce that using your knowledge of chemistry. Nov 21, 2021 at 15:02
• You have to remember things like ClO4- is a stable non-nucleophile anion that is often used as a counterion for reactive cations Nov 21, 2021 at 15:58

$$\ce{O=N\color{blue}{-}O-ClO3},$$
the electrons would be accorded to the oxygen end and thus the split $$\ce{NO^+, ClO4^-}$$ follows. You also split the bonds within each component "ion" with the same rule until you find that each oxgen atom ends with $$-2$$, the nitrogen with $$+3$$ and chlorine with $$+7$$ as you seem to already know. (The nitrogen is $$+3$$ instead of $$+5$$ because it retains a nonbonding pair.)
Nitrosyl bonded with perchlorate is easy. How the electrons are precisely arranged in a metal nitrosyl complex is nontrivial. For instance, my understanding is that the nitrosyl moiety used to be deemed $$\ce{NO^+}$$ in the iron-nitrosyl "brown ring" complex, but now it's rendered $$\ce{NO^-}$$, with iron in a correspondingly higher oxidation state, instead.