Question

The total number of valence electrons in $4.2g$ of $N_3^{-}$ ions are:

• A. $2.2N_A$
• B. $4.2N_A$
• C. $1.6N_A$
• D. $3.2N_A$

Answer 1

Answer: (C)

$1.6N_A$

Molecular wt of $N_3^{-}$=$3\times 14=42g$

One ion of $N_3^{-}$ has valence electrons = $3\times 5+1=16$

42 g $N_3^{-}$=$N_A$ $N_3^{-}$ ion

valence electrons in $N_A$ ions = $16N_A$ (where $N_A$ = Avogadro's number)

So, $1g$ of $N_3^{-}$ will have valence electrons = $16N_A/42$

therefore,4.2g of $N_3^{-}$ ion will have valence electrons = $16N_A\times 4.2/42=1.6N_A$.

Option C is correct.