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Current Question (ID: 7901)

Question:

$\text{Given below are four orders for the size of the species. Choose the correct ones:}$ $\text{(a) Al}^{3+} < \text{Mg}^{2+} < \text{Na}^+ < \text{F}^-$ $\text{(b) Al}^{3+} < \text{Mg}^{2+} < \text{Li}^+ < \text{K}^+$ $\text{(c) Fe}^{4+} < \text{Fe}^{3+} < \text{Fe}^{2+} < \text{Fe}$ $\text{(d) Mg} > \text{Al} > \text{Si} > \text{P}$

Options:

1. $\text{(a), (b) & (c)}$
2. $\text{(b), (c) & (d)}$
3. $\text{(a), (c)}$
4. $\text{(a), (b), (c) & (d)}$ (Correct)

Solution:

$\text{HINT: In isoelectronic species, the anion is larger than the cation.}$ $\text{Explanation:}$ $\text{(1) Mg}^{2+}, \text{Na}^+ \text{ and F}^- \text{ are isoelectronic and thus follow the order}$ $_{12}\text{Mg}^{2+} < _{11}\text{Na}^+ < _9\text{F}^-.$ $\text{In the case of isoelectronic species, the anion is always greater than cation and neutral species.}$ $\text{Na}^+ = 102 \text{ pm}; \text{Mg}^{2+} = 72 \text{ pm}; \text{Al} = 143 \text{ pm}, \text{F}^- = 133 \text{ pm}.$ $\text{(2) K}^+ \text{ has more number of shells than Mg}^{2+} \text{ and Al}^{3+} \text{ and Mg}^{2+} \text{ are isoelectronic but Al}^{3+} \text{ has higher nuclear charge so Al}^{3+} < \text{Mg}^{2+}. \text{Mg}^{2+} \text{ and Li}^+ \text{ has a diagonal relationship. But due to +2 charge in Mg}^{2+}, \text{ the Mg}^{2+} \text{ is smaller than Li}^+. \text{ Hence Al}^{3+} \text{ is the smallest one.}$ $\text{K}^+ = 1.38 \text{ Å}, \text{Li}^+ = 0.76 \text{ Å}, \text{Mg}^{2+} = 0.72 \text{ Å and Al}^{3+} = 0.535 \text{ Å}.$ $\text{(3) As the number of electrons is lost, the attraction between valence shell electrons and nucleus increases. As a consequence of this, the electrons are pulled closer to the nucleus leading to the contraction in the size of ions.}$ $\text{(4) Across the period the nuclear charge increases and thus the size of atoms decreases.}$ $\text{Mg} = 160 \text{ pm}; \text{Al} = 143 \text{ pm}; \text{Si} = 118; \text{P} = 110 \text{ pm}.$

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{
  "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
}