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Current Question (ID: 20925)

Question:

$\text{The ion having the highest value of ionic radius among the following is:}$

Options:

1. $\text{Li}^+$
2. $\text{B}^{3+}$
3. $\text{O}^{2-}$
4. $\text{F}^-$

Solution:

$\text{Hint: For the same period element, anion size is always greater than cation size}$ $\text{Li, B, O, and F belong to the same period. For the same period elements, the cation size is smaller than the anion. As the negative charge on an anion increases the size of the anion also increases because of interelectronic repulsion.}$ $\text{As the positive charge increases, the size of the cation decreases because of the high effective nuclear charge.}$ $\text{Hence, oxygen has the highest negative charge hence, O}^{2-} \text{ has the highest value of ionic radius. and B has the highest positive charge hence, B}^{3+} \text{ has the lowest value of ionic radius.}$ $\text{Hence, option third is the correct answer.}$ $\text{The ionic radius value is as follows:}$ $1. \text{Li}^+ = 76 \text{ pm}$ $2. \text{O}^{2-} = 140 \text{ pm}$ $3. \text{F}^- = 133 \text{ pm}$

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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
}