Why is NaBr only a reducing agent? Can you tell if something is a reducing/oxidizing/both agent?

I came across a question today that read "Which of the following can only act as a reducing agent?" The answer was $\ce{NaBr}$, and I knew this from a previous question I had seen before. Is there any way to determine these properties without rote memorization?


(If it helps to use as an example, the other answer choices were $\ce{I2,\;BrCl,\;HIO4}$)




Think about the state of the atoms in the compound and where they can go from there:



> 
> $\ce{I2}$
> 
> 
> 


In this molecule, the iodine atoms have $0$ oxidation state, and are easily reduced to the very stable $\ce{I-}$ ions, with $-1$ oxidation state. This means this compound is capable of reacting as an **oxidizing** agent.



> 
> $\ce{BrCl}$
> 
> 
> 


Likewise, in this molecule, the bromine and chlorine atoms have $0$ oxidation state, and are easily reduced to $\ce{Br-}$ and $\ce{Cl-}$ ions, which both have a $-1$ oxidation state. This means this compound is reacts as an **oxidizing** agent.



> 
> $\ce{HIO4}$
> 
> 
> 


The halogen oxyanions ($\ce{F^{(I)}O- ,\ Cl^{(III)}O- ,\ Br^{(V)}O3- ,\ I^{(VII)}O4-}$), and really any oxyanion for that matter, have the marked ability to act as either oxidizing or reducing agents. These anions can accept electrons on the central atom in exchange for bonds to oxygen, acting as an **oxidizing** agent. They can also also accept additional oxygen bonds to the central atom, acting as **reducing** agents. Of course there is a limit to how reduced and how oxidized an oxyanion can become, but what I am trying to get across is that as a whole these species can act either as an oxidizing or reducing agent.



> 
> $\ce{NaBr}$
> 
> 
> 


Here the ions exist as $\ce{Na+}$ and $\ce{Br-}$, which have $+1$ and $-1$ oxidation states respectively. Here, $\ce{Br-}$ is unable to accept any more electrons and can only donate them to form $\ce{Br2}$, making it an effective as a **reducing** agent. It can also be oxidized from $\ce{Br-}$ in the presence of $\ce{O3}$ to $\ce{Br^{(V)}O3-}$, and is still acting as a **reducing** agent.


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In the presence of something reasonably oxidizing, it could form $\ce{Br2}$ and reduce whatever oxidizes it. It could also oxidize to $\ce{NaBrO\_{x}}$. 


  

Those other things are not going to easily oxidize because that would put a formal positive charge on a highly electronegative atom. In the case of $\ce{HIO4}$, it would increase the charge on iodine from $+7$. This would be very hard.


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