Import Question JSON

Current Question (ID: 7848)

Question:

$\text{The following property/ies decrease from left to right across the periodic table and increase from top to bottom:}$ $\text{(i) Atomic radius (ii) Electronegativity (iii) Ionisation energy (iv) Metallic character}$

Options:

1. $\text{(i) only}$
2. $\text{(i), (ii), and (iii)}$
3. $\text{(i), (iii), and (iv)}$
4. $\text{(i), and (iv)}$ (Correct)

Solution:

$\text{HINT: Left to right effective nuclear charge increases and down the group it decreases.}$ $\text{Explanation:}$ $\text{Down a group, the number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital. This results in a larger atomic radius.}$ $\text{Atomic size gradually decreases from left to right across a period of elements. This is because, within a period or family of elements, all electrons are added to the same shell. The valence electrons are held closer towards the nucleus of the atom. As a result, the atomic radius decreases.}$ $\text{Metallic character decreases with increase in electronegativity.}$ $\text{Thus, atomic radius and metallic character decreases from left to right across the period and increases from top to bottom down the group.}$

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
}