The total number of valence electrons in 4.2 g of `N_(3)^(-) ` ions is (`N_A` is the Avogadro number):

Options

A. `1.6 N_A`
B. `3.2 N_A`
C. `2.1 N_A`
D. `4.2 N_A`

Answer

A

Explanation

Each N- atom has 5 valence electrons
` therefore ` Total number of electrons in `N_(3)^(-)` ion ` = 5 xx 3 + 1 = 16`
moles of `N_(3)^(-)` ions present ` = (4.2)/(42) = 0.1 ` (mol. Wt.of `N_(3)^(-)1 = 42 )`
` therefore ` No. of valence electrons in 4.2 g ` = 0.1 xx 16 xx N_A =1.6 N_A`