Question

The total number of valence electrons in 4.2g of \[{{N}_{3}}^{-}\]ions are:
A.2.2 \[{{N}_{A}}\]
B.4.2\[{{N}_{A}}\]
C.1.6\[{{N}_{A}}\]
D.3.2\[{{N}_{A}}\]

Answer 1

Hint: Total number of valence electrons in an ion is calculated by using the following formula
Number of valence electrons in an ion = Number of moles of the given ion
                    $\times$ Number of valence electrons in the given ion $\times$ Avogadro number

Complete step by step answer:
-Given mass of azide ion (\[{{N}_{3}}^{-}\]) is = 4.2g
-We have to calculate the molar mass of azide ion (\[{{N}_{3}}^{-}\]).
-Atomic weight of Nitrogen is 14.
-Number of nitrogen atoms in azide ion (\[{{N}_{3}}^{-}\]) is three.

-Step-1:
So, the molar mass will be\[3\times 14=42\].
Number of moles of azide ion (\[{{N}_{3}}^{-}\]) \[\begin{align}
  & =\dfrac{4.2}{42} \\
 & =0.1\text{ }moles \\
\end{align}\]

-Step-2:
Number of valence electrons in one nitrogen atom are five.
The number of valence electrons in azide ion (\[{{N}_{3}}^{-}\])
\[\begin{align}
  & =(3\times 5)+1 \\
 & =16 \\
\end{align}\]

-Substitute the number of moles and number of valence electrons in the following equation to get the number of valence electrons in 4.2g of \[{{N}_{3}}^{-}\]ions.

Number of valence electrons in an ion = Number of moles of the given ion
               × Number of valence electrons in the given ion ×Avogadro number
      \[\begin{align}
  & \text{ = 0}\text{.1 }\times \text{ 16 }\times \text{ }{{\text{N}}_{\text{A}}} \\
 & \text{ = 1}\text{.6 }{{\text{N}}_{\text{A}}}\text{ } \\
\end{align}\]
So, the correct option is C.

Note: Don’t be confused with the words valence electrons and number of moles.
Valence electron means the electron which is present in outer most orbit is called valence electron.
Number moles is the ratio of given weight of the atom or ion and atomic weight of the atom or ion.