Import Question JSON

Current Question (ID: 10623)

Question:

$\text{What is the correct order of oxyacids of phosphorous in terms of their acidic strength?}$

Options:

1. $\text{H}_3\text{PO}_2 > \text{H}_3\text{PO}_3 > \text{H}_3\text{PO}_4$ (Correct)
2. $\text{H}_3\text{PO}_4 > \text{H}_3\text{PO}_3 > \text{H}_3\text{PO}_2$
3. $\text{H}_3\text{PO}_3 > \text{H}_3\text{PO}_2 > \text{H}_3\text{PO}_4$
4. $\text{H}_3\text{PO}_2 > \text{H}_3\text{PO}_4 > \text{H}_3\text{PO}_3$

Solution:

$\text{Hint: As the phosphorus oxidation state increases the acidity of the compound decreases.}$ $\text{Acidity is directly proportional to free oxygen and inversely proportional to -OH group.}$ $\text{The double-bonded oxygen has a negative inductive effect (-I). And this electron-withdrawing effect is experienced by one H of the one -OH group in } \text{H}_3\text{PO}_2\text{. Hence it has greatest acidic strength.}$ $\text{Similarly, the -I effect of O is experienced by two H of the two -OH groups in } \text{H}_3\text{PO}_3\text{, hence the electron-withdrawing effect is reduced because it is getting distributed between two groups. Hence, it has lesser acidic strength. Similarly, in } \text{H}_3\text{PO}_4\text{, it is distributed between 3 groups and so the acidic strength is least.}$ $\text{p}K_a \text{ values are as follows:}$ $\text{H}_3\text{PO}_2: \text{ p}K_a = 1.2$ $\text{H}_3\text{PO}_4: \text{ p}K_a = 2.16$ $\text{H}_3\text{PO}_3: \text{ p}K_a = 1.3$

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
}