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Current Question (ID: 20970)

Question:

$\text{The ascending order of polarizing power of the following ions is:}$ $\text{K}^+, \text{Ca}^{2+}, \text{Mg}^{2+}, \text{Be}^{2+}$

Options:

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

Solution:

$\text{Polarizing power} \propto \frac{\text{Charge}}{\text{Size}}$ $\text{The ability of a cation to distort an anion is known as its polarization power and the tendency of the anion to become polarized by the cation is known as its polarising power.}$ $\text{As the charge density on a cation increases, its polarising power also increases. The charge/size ratio of a cation determines its polarising power.}$ $\text{So, the polarizing power of the cation increases as the charge of the cation increases or as the radius decreases.}$ $\text{The polarizing power of K}^+ \text{ is less than Ca}^{2+}, \text{Mg}^{2+}, \text{Be}^{2+} \text{ because of the less charge of potassium.}$ $\text{The order of size is as follows:}$ $\text{Ca}^{2+} > \text{Mg}^{2+} > \text{Be}^{2+}$ $\text{Hence, the correct order of polarizing power is: K}^+ < \text{Ca}^{2+} < \text{Mg}^{2+} < \text{Be}^{2+}$

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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
}