The total number of valence electrons in 4.2 g of N 3-ion is [NA. is the Avogadro's number]

Question

The total number of valence electrons in 4.2 g of N 3-ion is [NA. is the Avogadro's number] (A) 2.1 NA (B) 4.2 NA (C) 1.6 NA (D) 3.2 NA

Tardigrade Admin · Accepted Answer

Moles of N3-ion =(4.2/42)=0.1 Each nitrogen atom has 5 valence electrons. Therefore, total number of electrons in N3-ion =16 Total number of electrons in 0.1 mole or 4.2 g of N3- ion =0.1 × 16 × NA =1.6 NA

Q. The total number of valence electrons in $4.2 g$ of $N_{3}^{-}$ ion is $[N_{A}$ is the Avogadro's number]

$2.1 N_{A}$

$4.2 N_{A}$

$1.6 N_{A}$

$3.2 N_{A}$

Moles of $N_{3}^{-}$ ion $= \frac{4.2}{42} = 0.1$
Each nitrogen atom has 5 valence electrons.
Therefore, total number of electrons in $N_{3}^{-}$ ion $= 16$
Total number of electrons in $0.1$ mole or
$4.2 g$ of $N_{3}^{-}$ ion $= 0.1 \times 16 \times N_{A}$
$= 1.6 N_{A}$

Q. The total number of valence electrons in 4.2g of N3−​ion is [NA​ is the Avogadro's number]

2.1NA​

4.2NA​

1.6NA​

3.2NA​