Question

The total no. of electrons in $4.2\mathrm{g}$ of ${\mathrm{N}}^{3-}$ ion is:

• A. $3{\mathrm{N}}_{\mathrm{A}}$
• B. $3.2{\mathrm{N}}_{\mathrm{A}}$
• C. $2.1{\mathrm{N}}_{\mathrm{A}}$
• D. $4.2{\mathrm{N}}_{\mathrm{A}}$

Answer 1

The correct option is A

$3{\mathrm{N}}_{\mathrm{A}}$

The explanation for the correct option

[a]$3N_A$

We can find$3N_A$ to be the number of electrons present in $4.2g$ of $N^{3-}$ion by the following steps

Step 1: Finding the number of moles.:

• $\mathrm{No}.\mathrm{of}\mathrm{moles}=\frac{\mathrm{Mass}\mathrm{given}}{\mathrm{Atomic}\mathrm{mass}}$
• $\mathrm{No}.\mathrm{of}\mathrm{moles}=\frac{4.2}{14}$
• [Atomic mass of Nitrogen atom$=$$14$]
• So, $\mathrm{No}.\mathrm{of}\mathrm{moles}=0.3$

Step 2: Finding the no. of electrons present in one $N^{3-}$ion.

• No. of electrons present $=$Atomic no. $+$ negative charge on the ion
• [Atomic no. of Nitrogen $=7$]
• [Negative charge present on ion $=3$]
• No. of electrons$=$$7+3$]
• No. of electrons$=10$

Step 3: Calculating the no. of electrons present in$0.3mol$ of$N^{3-}$ ion.

• Total no. of electrons$=$ in one $N^{3-}$ion $\times$ no .of electrons in $0.3mol$
• [No. of electrons in $0.3mol=0.3\times N_A$]
• [$N_{A=}$No. of electrons present in one mole or$\mathrm{Avagadro}\mathrm{number}$]
• Total no. of electro Total no. of electrons$=3N_A$

The explanation for the incorrect options

Since the correct number of electrons are $3N_A$. Therefore, options (b), (c), and (d) are incorrect.

Hence, option (A) is the correct answer.