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Current Question (ID: 7462)

Question:

$\text{Which of the following has greater number of electrons than neutrons?}$

Options:

1. $\text{Al}^{3+}$
2. $\text{O}^{2-}$
3. $\text{F}^-$
4. $\text{C}$

Solution:

$\text{Hint: Count the number of neutron and electron in each given ions}$ $\text{Let us analyze each option:}$ $\text{1. } \text{Al}^{3+}\text{:}$ $\text{Aluminum has atomic number 13. Al}^{3+} \text{ has lost 3 electrons.}$ $\text{Number of electrons = 13 - 3 = 10}$ $\text{For most common isotope } {}^{27}\text{Al}\text{, number of neutrons = 27 - 13 = 14}$ $\text{10 electrons < 14 neutrons}$ $\text{2. } \text{O}^{2-}\text{:}$ $\text{Oxygen has atomic number 8. O}^{2-} \text{ has gained 2 electrons.}$ $\text{Number of electrons = 8 + 2 = 10}$ $\text{For most common isotope } {}^{16}\text{O}\text{, number of neutrons = 16 - 8 = 8}$ $\text{10 electrons > 8 neutrons } ✓$ $\text{3. } \text{F}^-\text{:}$ $\text{Fluorine has atomic number 9. F}^- \text{ has gained 1 electron.}$ $\text{Number of electrons = 9 + 1 = 10}$ $\text{For most common isotope } {}^{19}\text{F}\text{, number of neutrons = 19 - 9 = 10}$ $\text{10 electrons = 10 neutrons}$ $\text{4. } \text{C}\text{:}$ $\text{Carbon has atomic number 6.}$ $\text{Number of electrons = 6}$ $\text{For most common isotope } {}^{12}\text{C}\text{, number of neutrons = 12 - 6 = 6}$ $\text{6 electrons = 6 neutrons}$ $\text{Therefore, only } \text{O}^{2-} \text{ has more electrons than neutrons.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}