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Current Question (ID: 21541)

Question:

$\text{Which of the following oxidizing agents is the strongest?}$

Options:

1. $\text{Ti}^{3+}$
2. $\text{Cr}^{3+}$
3. $\text{Fe}^{3+}$
4. $\text{Mn}^{3+}$

Solution:

$\text{Hint: If a less stable to the more stable configuration is obtained then the ion is strong oxidizing agent.}$ $\text{Step 1:}$ $\text{Oxidizing agent oxidized other and it self gets reduce. The oxidising power depends on the electronic configuration of the ions.}$ $\text{Ti}^{3+} \text{ is convert into Ti}^{2+} \text{ and } d^1 \text{ is converted into } d^2.$ $\text{Cr}^{3+} \text{ is converted into Cr}^{2+} \text{ and } d^3 \text{ is converted into } d^4.$ $\text{Fe}^{3+} \text{ is converted into Fe}^{2+} \text{ and } d^5 \text{ is converted into } d^6.$ $\text{Mn}^{3+} \text{ is converted into Mn}^{2+} \text{ and } d^4 \text{ is converted into } d^5. \text{ The } d^5 \text{ is a stable configuration hence, Mn}^{3+} \text{ acts as a strong oxidising agent.}$ $\text{Step 2:}$ $\text{The reduction potential value of ions as follows:}$ $E^\circ(\text{Ti}^{3+}/\text{Ti}^{2+}) = -0.37 \text{ V}$ $E^\circ(\text{Cr}^{3+}/\text{Cr}^{2+}) = -0.41 \text{ V}$ $E^\circ(\text{Mn}^{3+}/\text{Mn}^{2+}) = +1.57 \text{ V}$ $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77 \text{ V}$ $E^\circ(\text{Mn}^{3+}/\text{Mn}^{2+}) \text{ has the highest reduction potential value hence, it is a strongest oxidising agent}$ $\text{Hence, option 4th is the correct answer.}$

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
}