Import Question JSON

Current Question (ID: 20923)

Question:

$\text{The increasing order of the first ionization enthalpies of the elements}$ $\text{B, P, S and F (lowest first) is:}$

Options:

1. $\text{F} < \text{S} < \text{P} < \text{B}$
2. $\text{P} < \text{S} < \text{B} < \text{F}$
3. $\text{B} < \text{P} < \text{S} < \text{F}$
4. $\text{B} < \text{S} < \text{P} < \text{F}$

Solution:

$\text{Hint: Left to right across the period ionization energy decreases}$ $\text{The ionization energy depends on the size of the element. As size}$ $\text{decreases, the ionization energy increases.}$ $\text{Left to right size decreases because effective nuclear charge}$ $\text{increases. If we compare B, P, S, and F. B is present at the extreme left}$ $\text{and F is present at the extreme right in the periodic table.}$ $\text{Hence, B has the least ionization enthalpy and fluorine has the}$ $\text{highest ionization enthalpy.}$ $\text{But P has high ionization enthalpy than S because it has a half-filled}$ $3p \text{subshell and the half-filled subshell is highly stable.}$ $\text{Due to size differences, P and S have less ionization energy than F.}$ $\text{Hence, the correct answer is option 4th.}$ $\text{B = 801 KJ}$ $\text{S = 999.6}$ $\text{P = 1011}$ $\text{F = 1681}$

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
}