Question

The total number of valence electrons in 4.2g of n3- ion is

Answer 1

the answer is that it contains 1.6NA no. of valence electr9ns

Answer 2

Answer: $14.45\times 10^{23}$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{4.2g}{14g/mol}=0.3moles$

Electronic configuration of nitrogen:

$[N]:7:1s^22s^22p^3$

Valence electrons of nitrogen = 5

$[N^{3-}]:10:1s^22s^22p^6$

Valence electrons of $N^{3-}$ = 8

1 mole of  $N^{3-}$ contains =$8\times 6.023\times 10^{23}=48.18\times 10^{23}$ electrons

0.3 moles of $N^{3-}$ contains =$\frac{48.18\times 10^{23}}{1}\times 0.3=14.45\times 10^{23}$ electrons

Thus total number of valence electrons in 4.2g of $N^{3-}$ ion is $14.45\times 10^{23}$ electrons.