Question

Which of the following contains maximum number of electrons?

• A. 3 g-atom of oxygen
• B. 1 g-ion of NO$_3^{\ominus}$
• C. 10 g SO$_4^{2-}$ ion
• D. 3 g molecule of CO

Answer 1

Answer: (D)

3 g molecule of CO

Answer 2

To find the option with the maximum number of electrons, we calculate the total number of electrons for each option, expressing it in terms of Avogadro's number ($N_A$).

1. 3 g-atom of oxygen:

   • 1 g-atom = 1 mole of atoms.
   • 3 g-atom of oxygen = 3 moles of O atoms.
   • Electrons per O atom = 8.
   • Total electrons = $3 \text{ mol} \times 8 \text{ electrons/atom} \times N_A = 24 N_A$ electrons.

2. 1 g-ion of NO$_3^{\ominus}$:

   • 1 g-ion = 1 mole of ions.
   • 1 g-ion of NO$_3^{\ominus}$ = 1 mole of NO$_3^{\ominus}$ ions.
   • Electrons in NO$_3^{\ominus}$: N (7 e$^-$) + 3 * O (8 e$^-$) + 1 (for -1 charge) = $7 + 24 + 1 = 32$ electrons.
   • Total electrons = $1 \text{ mol} \times 32 \text{ electrons/ion} \times N_A = 32 N_A$ electrons.

3. 10 g SO$_4^{2-}$ ion:

   • Molar mass of SO$_4^{2-}$ = $32$ (S) + $4 \times 16$ (O) = $96$ g/mol.
   • Moles of SO$_4^{2-}$ = $\frac{10 \text{ g}}{96 \text{ g/mol}} = \frac{10}{96}$ moles.
   • Electrons in SO$_4^{2-}$: S (16 e$^-$) + 4 * O (8 e$^-$) + 2 (for -2 charge) = $16 + 32 + 2 = 50$ electrons.
   • Total electrons = $\frac{10}{96} \text{ mol} \times 50 \text{ electrons/ion} \times N_A = \frac{500}{96} N_A = \frac{125}{24} N_A \approx 5.21 N_A$ electrons.

4. 3 g molecule of CO:

   • 1 g molecule = 1 mole of molecules.
   • 3 g molecule of CO = 3 moles of CO molecules.
   • Electrons in CO: C (6 e$^-$) + O (8 e$^-$) = $6 + 8 = 14$ electrons.
   • Total electrons = $3 \text{ mol} \times 14 \text{ electrons/molecule} \times N_A = 42 N_A$ electrons.

Comparing the number of electrons: $24 N_A$, $32 N_A$, $5.21 N_A$, and $42 N_A$. The maximum is $42 N_A$, corresponding to 3 g molecule of CO.