Import Question JSON

Current Question (ID: 8006)

Question:

The correct increasing order of the polarizing power of the cationic species $\text{K}^{+}$, $\text{Ca}^{2+}$, $\text{Mg}^{2+}$, $\text{Be}^{2+}$ is:

Options:

1. $\text{Be}^{2+} < \text{K}^{+} < \text{Ca}^{2+} < \text{Mg}^{2+}$
2. $\text{K}^{+} < \text{Ca}^{2+} < \text{Mg}^{2+} < \text{Be}^{2+}$ (Correct)
3. $\text{Ca}^{2+} < \text{Mg}^{2+} < \text{Be}^{2+} < \text{K}^{+}$
4. $\text{Mg}^{2+} < \text{Be}^{2+} < \text{K}^{+} < \text{Ca}^{2+}$

Solution:

$\text{HINT: Charge/size ratio is inversely proportional to the size of cation.} \text{Explanation:} \text{STEP 1:} \text{Polarizing power refers to an atom's ability to pull an electron toward it, polarizing the atom the electron comes from. Since cations are positive, they are able to attract electrons toward themselves.} \text{STEP 2:} \text{The increasing order of cation size is :} $\text{K}^{+} > \text{Ca}^{2+} > \text{Mg}^{2+} > \text{Be}^{2+} \text{So, its reverse order is applied for the charge size ratio. Higher is the charge/size ratio, more is the polarizing power i.e.,} $\text{K}^{+} < \text{Ca}^{2+} < \text{Mg}^{2+} < \text{Be}^{2+}$

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
}