Import Question JSON

Current Question (ID: 18854)

Question:

$\text{Match the properties given in Column I with the metals given in Column II:}$ $\begin{array}{|c|c|} \hline \text{Column I - (Property)} & \text{Column II - (Metal)} \\ \hline \text{A. Element with highest second ionisation enthalpy} & 1. \text{V} \\ \text{B. Element with highest third ionisation enthalpy} & 2. \text{Cr} \\ \text{C. M in M(CO)}_6 \text{ is} & 3. \text{Cu} \\ \text{D. Element with the highest heat of atomisation} & 4. \text{Zn} \\ \hline \end{array}$

Options:

1. $\begin{array}{cccc} 3 & 4 & 1 & 2 \end{array}$
2. $\begin{array}{cccc} 3 & 4 & 2 & 1 \end{array}$
3. $\begin{array}{cccc} 1 & 2 & 4 & 3 \end{array}$
4. $\begin{array}{cccc} 3 & 2 & 4 & 1 \end{array}$

Solution:

$\text{Hint: Half-filled and fully-filled configuration is stable, hence, high ionisation enthalpy is required.}$ $\text{Explanation:}$ $\text{A. Cu}^{+} = 3d^{10} \text{ which is a very stable configuration due to full-filled orbitals. Hence, the removal of the second electron requires very high energy.}$ $\text{B. Zn}^{2+} = 3d^{10} \text{ which is a very stable configuration. Hence, the removal of the third electron requires very high energy.}$ $\text{C. Metal carbonyl with formula M(CO)}_6 \text{ is Cr(CO)}_6. \text{ EAN = 24 + 2(6) = 36}$ $\text{Thus, Cr(CO)}_6 \text{ is highly stable. Compound that follows the EAN rule is highly stable}$ $\text{D. The enthalpy of atomization is the enthalpy change that accompanies the total separation of all atoms in a chemical substance.}$ $\text{Among the given options enthalpy of atomisation of V is highest.}$ $\text{Therefore, the correct option is (2).}$

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
}