Write the complete balanced equation for the reaction between barium metal (Ba) and nitrogen gas (N2). You do not need to make the subscripts smaller; just write them out as regular numbers.
Will someone help me with this? I think that its just BaN2 but some places are saying its not

Question

Write the complete balanced equation for the reaction between barium metal (Ba) and nitrogen gas (N2). You do not need to make the subscripts smaller; just write them out as regular numbers. 
Will someone help me with this? I think that its just BaN2 but some places are saying its not

anonymous · Answer

@chemENGINEER

aaronq · Answer

That's a strange reaction, but we can think about this logically, Barium metal (elemental) has 2 loosely held electrons (in the 6s orbital) easily it can lose (and in turn be oxidized). Nitrogen (being more electronegative) can obtain these electrons and be reduced.. question is how many?

Each N in $N_2$ has 5 electrons, to complete it's octet it will gain 3.

                $Ba\rightarrow Ba^{2+}+2e$

      $N_2+6e^-\rightarrow 2N^{3-} $
---------------------------------

        $3Ba\rightarrow 3Ba^{2+}+6e$

      $N_2+6e^-\rightarrow 2N^{3-} $
---------------------------------
       $3Ba+ 2N_2 +\cancel{6e^-}\rightarrow Ba_3N_2+\cancel{6e^-}$

the only way to get this formula to work (because of the electrons exchanged) is having the product be $Ba_3N_2$.

anonymous · Answer

Oh wow thank you so much!!!!

aaronq · Answer

no problem! next time just use the charges the elements obtain (when they gain or lose electrons to achieve a noble gas configuration (i.e. an octet)) to predict the compounds formed.