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?

  • $\begingroup$ You are correct in the formulation $\endgroup$
    – Waylander
    Nov 21, 2021 at 14:46
  • $\begingroup$ If I was unaware of this "split", is there a method to solve this? $\endgroup$
    – Fr0zen
    Nov 21, 2021 at 14:48
  • 8
    $\begingroup$ 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. $\endgroup$ Nov 21, 2021 at 15:02
  • $\begingroup$ You have to remember things like ClO4- is a stable non-nucleophile anion that is often used as a counterion for reactive cations $\endgroup$
    – Waylander
    Nov 21, 2021 at 15:58

2 Answers 2


The rule is simply to give the electrons to the negative end of any polar bond. In the case of the central nitrogen-oxygen bond in the formulation


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.


First, ClO4 takes minus 1. The rest of ion is NO. It is plus 1. You should equal ClO4 to -1. Cl is +7. Also, you should equal NO to +1. N is +3. Thus, you are right. There isn't any way that I know to solve this issue-Fr0zenF0x


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