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Current Question (ID: 7473)

Question:

$\text{The unpaired electrons in Al and Si are present in the 3p orbital. The electrons that will experience a more effective nuclear charge from the nucleus are:}$

Options:

1. $\text{Electrons in the 3p orbital of silicon}$
2. $\text{Electrons in the 5d orbital of aluminium}$
3. $\text{Electrons in the 3p orbital of aluminium}$
4. $\text{Electrons in the 5p orbital of silicon}$

Solution:

$\text{Hint: Nuclear charge is defined as the net positive charge experienced by an electron in a multielectron atom.}$ $\text{Explanation:}$ $\text{The higher the atomic number, the higher is the nuclear charge.}$ $\text{Electronic configuration of Al} = [\text{Ne}]3\text{s}^2 3\text{p}^1; \text{Electronic configuration of Si} = [\text{Ne}] 3\text{s}^2 3\text{p}^2$ $\text{Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13). Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.}$

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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
}