If $N_A$ is Avogadro’s number, then the number of valence electrons in $4.2\text{g}$ of azide ions ($N_3^{-}$) are:

The correct option is C $1.6N_A$
$\text{Number of moles of azide ions}=\frac{\text{given mass}}{\text{molar mass}}=\frac{4.2}{42}=0.1\text{mole}$

Number of valence electrons in a nitrogen atom = 5
So, number of valence electrons in $N_3^{-}=3\left(5\right)+1=16$
Number of valence electrons in $0.1\text{mole}=16\times 0.1\times N_A=1.6N_A$