# Question 3d18e

Jun 21, 2016

$4.8 \cdot {10}^{24}$

#### Explanation:

The trick here is to realize that the nitride anion, ${\text{N}}^{3 -}$, is isoelectronic with neon, $\text{Ne}$, the noble gas that shares a period with nitrogen, $\text{N}$.

In other words, the nitride anion has the same number of electrons as neon.

Nitrogen is located in group 15 of the periodic table, so right from the start you know that it's $3$ electrons away from completing its octet.

Adding $3$ electrons to a neutral nitrogen atom will get you the nitride anion, ${\text{N}}^{3 -}$, which has a complete octet, i.e. it has $8$ electrons on its outermost shell.

As you know, one mole of any element is defined as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)color(white)(a)"atoms} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's number

In your case, one mole of nitride anions will contain $6.022 \cdot {10}^{23}$ nitride anions. If each nitride anion has $8$ valence electrons, i.e. electrons located on the outermost shell, it follows that one mole will contain

6.022 * 10^(23) color(red)(cancel(color(black)("N"^(3-)"ions"))) * "8 valence e"^(-)/(1color(red)(cancel(color(black)("N"^(3-)"ion")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(4.8 * 10^(24)color(white)(a)"valence e"^(-))color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs.

SIDE NOTE It's worth mentioning that this problem is aimed at testing your knowledge of valence electrons and of Avogadro's number; this problem is not very practical because the nitride anion cannot exist in solution.

That is the case because it acts as a base in aqueous solution, i.e. it gets protonated by water.