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Current Question (ID: 7674)

Question:

$\text{First three ionisation energies (in kJ/mol) of three representative elements are given below:}$ $\begin{array}{|c|c|c|c|} \hline \text{Element} & IE_1 & IE_2 & IE_3 \\ \hline \text{P} & 495.8 & 4562 & 6910 \\ \text{Q} & 737.7 & 1451 & 7733 \\ \text{R} & 577.5 & 1817 & 2745 \\ \hline \end{array}$ $\text{Then incorrect option is :}$

Options:

1. $\text{Q: Alkaline earth metal.}$
2. $\text{P: Alkali metal.}$
3. $\text{R: s-block element.}$ (Correct)
4. $\text{All three: P,Q & R belong to the same period.}$

Solution:

$\text{HINT: Alkali metals have less ionization energy than alkaline earth metal.}$ $\text{Explanation:}$ $\text{R is a p-block element because the difference between } IE_2 \text{ and } IE_3 \text{ is not very high as compared to between } IE_1 \text{ and } IE_2\text{; hence stable oxidation state of R will be higher than +2.}$ $\text{Analysis of ionization energies:}$ $\text{P: Large jump from } IE_1 \text{ to } IE_2 \text{ (495.8 to 4562) indicates P is an alkali metal (Group 1).}$ $\text{Q: Large jump from } IE_2 \text{ to } IE_3 \text{ (1451 to 7733) indicates Q is an alkaline earth metal (Group 2).}$ $\text{R: Gradual increase in ionization energies with no extremely large jump indicates R is a p-block element, not an s-block element.}$ $\text{Therefore, option 3 "R: s-block element" is incorrect.}$

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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
}