Question

The total number of valence electrons in 4.2 grams of nitride ions is:
(A) 1.4${N_ \circ }$
(B) 2.4${N_ \circ }$
(C) 3.2${N_ \circ }$
(D) 4.2${N_ \circ }$

Answer 1

First find out the number of valence electrons present in a molecule of nitride ion and then moles associated with 4.2 grams of nitride ions. The total number of valence electrons will be the product of the number of electrons in 1 molecule and the number of moles.

Complete answer:
-First of all we will check what nitride ions are. It is a nitrogen atom carrying an oxidation state of (-3). It is represented as ${N^{ - 3}}$.
The question has given its weight to be 4.2g and we know that its molecular weight is 14. So, we will now find out the number of moles associated with it.
Moles = $\dfrac{W}{M}$
Where, W = given weight and M = molecular weight.
Moles of ${N^{ - 3}}$ = $\dfrac{{4.2}}{{14}}$
= 0.3 moles
-Now we will be calculating the number of valence electrons present in 0.3 moles of nitride ion (${N^{ - 3}}$).
For this we first need to find out the number of electrons present in 1 molecule of ${N^{ - 3}}$, they are = 5 (from outermost shell of N atom) + 3 (from the charge)
= 8
So, 1 mole of ${N^{ - 3}}$ ions will contain = 8 × ${N_ \circ }$ number of electrons; where ${N_ \circ }$is the Avogadro number = $6.023 \times {10^{ - 23}}$
And 0.3 moles of ${N^{ - 3}}$ ions will contain = (8 × ${N_ \circ }$) × 0.3
= 2.4${N_ \circ }$
So, the total number of valence electrons present in 4.2 grams of nitride ions will be 2.4${N_ \circ }$.

The correct option will be: (B) 2.4${N_ \circ }$

Note:
Do not get confused between nitride ion and nitrite ion. Nitride ion is ${N^{ - 3}}$ while nitrite ion is $NO_2^ -$. Also in the nitride compounds band gaps are usually large and hence they are usually insulators or the wide gap semiconductors, like the boron and silicon nitrides. The gallium nitride is used in LEDs to emit blue lights due to the large band gap.