Import Question JSON

Current Question (ID: 18833)

Question:

$\text{Which of the following statements is false?}$

Options:

1. $\text{V}^{2+} < \text{Cr}^{2+} < \text{Mn}^{2+} < \text{Fe}^{2+} : \text{Paramagnetic behaviour}$
2. $\text{Ni}^{2+} < \text{Co}^{2+} < \text{Fe}^{2+} < \text{Mn}^{2+} : \text{Ionic size}$
3. $\text{Co}^{3+} < \text{Fe}^{3+} < \text{Cr}^{3+} < \text{Sc}^{3+} : \text{Stability in aqueous solution}$
4. $\text{Sc} < \text{Ti} < \text{Cr} < \text{Mn} : \text{Number of oxidation states}$

Solution:

$\text{HINT: Paramagnetic character} \propto \text{Number of unpaired d electrons.}$ $\text{Explanation:}$ $\text{(1) } \text{V}^{2+} = 3 \text{ unpaired electrons}$ $\text{Cr}^{2+} = 4 \text{ unpaired electrons}$ $\text{Mn}^{2+} = 5 \text{ unpaired electrons}$ $\text{Fe}^{2+} = 4 \text{ unpaired electrons}$ $\text{Hence the order of paramagnetic behavior should be}$ $\text{V}^{2+} < \text{Cr}^{2+} = \text{Fe}^{2+} < \text{Mn}^{2+}$ $\text{(2) Ionic size decrease from left to right in same period}$ $\text{(3) As per data from NCERT.}$ $\frac{\text{Co}^{3+}}{\text{Co}^{2+}} = 1.97; \frac{\text{Fe}^{3+}}{\text{Fe}^{2+}} = 0.77; \frac{\text{Cr}^{3+}}{\text{Cr}^{2+}} = -0.41$ $\text{Sc}^{3+} \text{ is highly stable (It does not show } +2)$ $\text{(4) The oxidation states increase as we go from group 3 to group 7 in the same period.}$ $\text{Hence, option 1 is the correct answer.}$

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
}