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Current Question (ID: 7657)

Question:

$\text{The first ionization enthalpy values (in kJ mol}^{-1}\text{) of group 13 elements are:}$\n\n$\text{B} \quad \text{Al} \quad \text{Ga} \quad \text{In} \quad \text{Tl}$\n\n$801 \quad 577 \quad 579 \quad 558 \quad 589$\n\n$\text{The explanation for the deviation from the general trend can be -}$

Options:

1. $\text{Ga has lower ionization enthalpy than Al.}$
2. $\text{Ga has higher ionization enthalpy than Al.}$ (Correct)
3. $\text{Al has higher ionization enthalpy than Ga.}$
4. $\text{Ga has a lesser valence electron than Al.}$

Solution:

$\text{HINT: On moving down a group, ionization enthalpy generally decreases due to an increase in the atomic size and shielding.}$\n\n$\text{Explanation -}$\n\n$\text{Al follows immediately after s-block elements, whereas Ga follows after d-block elements.}$\n\n$\text{The shielding provided by d-electrons is not very effective.}$\n\n$\text{These electrons do not shield the valence electrons very effectively. As a result, the valence electrons of Ga experience a greater effective nuclear charge than those of Al.}$\n\n$\text{Gallium has a slightly higher ionization energy than aluminum due to poor shielding by its 3d electrons, which increases the effective nuclear charge. This makes it harder to remove an outer electron in Ga compared to Al.}$\n\n$\text{Further, moving from Ga to In, the ionization enthalpy decreases due to an increase in the atomic size and shielding. But, on moving from In to Tl, the ionization enthalpy again increases.}$\n\n$\text{In the periodic table, Tl follows after 4f and 5d electrons. The shielding provided by the electrons in both these orbitals is not very effective. Therefore, the valence electron is held quite strongly by the nucleus.}$\n\n$\text{Hence, the ionization energy of Tl is on the higher side.}$\n\n$\text{From the given data: Al = 577 kJ/mol, Ga = 579 kJ/mol. Therefore, Ga has higher ionization enthalpy than Al, which is option 2.}$

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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
}