# O2N-N2O is homolysis or heterolysis? What is the product?

Oct 13, 2015

Do you mean homolysis of ${N}_{2} {O}_{4}$? This is not the formula that features above.

${N}_{2} {O}_{4} \left(g\right) \rightarrow 2 N {O}_{2} \left(g\right)$

#### Explanation:

$N {O}_{2}$ is a radical species; that is it features a single, unpaired electron conceived to be on the nitrogen centre. $N {O}_{2}$ necessarily has $17$ valence electrons to distribute around the 3 atoms $\left(2 \times 6 + 5\right)$. The nitrogen is formally cationic and bears a single electron, whereas one of the oxygen atoms bears a negative charge. Of course, we draw another resonance structure such that the negative charge is borne by the other oxygen:

$O = {\dot{N}}^{+} - {O}^{-}$. From left to right, each atom has 6, 4, and 7 valence electrons, and formally has a charge of $0$, ${1}^{+}$, and ${1}^{-}$ respectively. Two $\dot{N} {O}_{2}$ molecules can dimerize when the single electrons on the nitrogen centres couple to form a nitrogen-nitrogen bond, i.e. ${O}^{-} \left(O =\right) {N}^{+} - N {O}_{2}$. Of course, there is a still a formal positive charge on each nitrogen centre. Also, of course, the nitrogen-nitrogen bond of the dimer can homolyze to give 2 equiv $N {O}_{2}$, which is the answer to the question I presume you have asked.

Apologies if I am barking up the wrong tree. Woof!