n-factor for ammonium nitrite? The given reaction is $$\ce{NH4NO2 \rightarrow N2  +  2H2O}$$

The way I did it was
$$\ce{NH4+ \rightarrow \frac{1}{2} N2 + 4H+ + 3e-}$$
and
$$\ce{NO2- +3e- +4H+ \rightarrow \frac{1}{2} N2 + 2H2O}$$
Therefore n factor is 3

But the book I am referring to says 6. Can someone please help?
"n factor" of oxidizing or reducing agent = total number of electrons transferred per a mole of the reactant.

When more than one element in a substance is oxidized/reduced, the n-factor of the substance is taken as the sum of the n-factors of the individual elements of the substance.

This is the case here.You have  ammonium ion that undergoes oxidation reaction (n-factor= 3) and  nitrite ion that undergoes reduction reaction (n-factor= 3).
So, the n-factor of ammonium nitrite is 6.