Question

The total number of valence electrons in 4.2 g of $N_3^{-}$ ion is [$N_A$ is the Avogadro's number)

• A. $2.1 N_A$
• B. $4.2 N_A$
• C. $1.6 N_A$
• D. $3.2 N_A$

Answer 1

Answer: (C)

$1.6 N_A$

Answer 2

Moles of $N_3^{-} \, ion \, = \frac {4.2}{42}= 0.1$ Each nitrogen atom has 5 valence electrons. Therefore, total number of electrons in $N_3^{-} \, ion \, = 16$ Total number of electrons in 0.1 mole or $4.2 \, g \, of \, N_3^{-} ion = 0.1 \times 16 \times N_A$ $= 1.6 N_A$