Question

The total number of valence electrons in $4.2gm$ of $N_3^ - $ ion are:
(A) $2.2{N_A}$
(B) $4.2{N_A}$
(C) $1.6{N_A}$
(D) $3.2{N_A}$

Answer 1

Hint: A valence electron in an outer shell electron that is associated with an atom, and that can participate in the formation of a chemical bond of outer shell is not closed.

Complete step by step answer:
We know that,
Molecular weight of $N_3^ - $$ = 3 \times 14 = 42g$
One ion of $N_3^ - $ has number of valence electrons$ = 3 \times 5 + 1 = 16{e^ - }$
We can say that,
$42g$ of $N_3^ - $ contains ${N_A}$ ions of $N_3^ - $
${N_A} \to $ Avogadro number $ = 6.023 \times {10^{23}}$
Valence electrons in ${N_A}$ ions $ = 16{N_A}$
Also,
$42g$ of $N_3^ - $ will have valence electrons$ = 16{N_A}$
$1g$ of $N_3^ - $ will have valence electrons$ = \dfrac{{16{N_A}}}{{42}}$
We can write according to the question as:
Therefore, $4.2g$ of $N_3^ - $ will have valence electrons$ = \dfrac{{16{N_A}}}{{42}} \times 4.2 = 1.6{N_A}$
Hence, the answer is option (C).

Additional information:
In the Lewis structure for $N_3^ - $ you will need to place double bonds between the nitrogen atoms to achieve full outer shells on all atoms while only using the valence electrons available for the molecule.
$\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\cdot\cdot}$}}{\ddot N} ::N::\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\cdot\cdot}$}}{\ddot N} $
$\mathop N\limits^{ - 1} = \mathop N\limits^{ + 1} = \mathop N\limits^{ - 1} $
Formal charge$ = - 1$


Note:
The correct name of $N_3^ - $ is azide. Azide is a polyatomic anion with $ - 1$ charge and written like $N_3^ - $. It consists of $3$ nitrogen atoms. It is the conjugate base of hydrazoic acid.Sodium azide is best known as the chemical found in automobile airbags. An electrical charge triggered by automobile impact causes sodium azide to explode and convert to nitrogen gas inside the airbag. Sodium azide is used as a chemical preservative in hospitals and laboratories.