Import Question JSON

Current Question (ID: 21565)

Question:

$\text{Match the transition metal ion in List-I with its corresponding spin-only magnetic moment (in B.M.) in List-II:}$ $\begin{array}{|c|c|} \hline \text{List-I} & \text{List-II} \\ \text{(Transition metal ion)} & \text{(Spin only magnetic moment (B.M.))} \\ \hline \text{(A) } \text{Ti}^{3+} & \text{(I) } 3.87 \\ \text{(B) } \text{V}^{2+} & \text{(II) } 0.00 \\ \text{(C) } \text{Ni}^{2+} & \text{(III) } 1.73 \\ \text{(D) } \text{Sc}^{3+} & \text{(IV) } 2.84 \\ \hline \end{array}$ $\text{Choose the correct answer from the options given below:}$

Options:

1. $(\text{A})-(\text{III}), (\text{B})-(\text{I}), (\text{C})-(\text{II}), (\text{D})-(\text{IV})$
2. $(\text{A})-(\text{III}), (\text{B})-(\text{II}), (\text{C})-(\text{IV}), (\text{D})-(\text{II})$
3. $(\text{A})-(\text{IV}), (\text{B})-(\text{III}), (\text{C})-(\text{III}), (\text{D})-(\text{I})$
4. $(\text{A})-(\text{II}), (\text{B})-(\text{IV}), (\text{C})-(\text{I}), (\text{D})-(\text{III})$

Solution:

$\text{Hint: } \mu_{\text{spin-only}} = \sqrt{n(n+2)}$ $\text{Sc}^{3+} = 3d^0 \quad \therefore \mu_{\text{spin}} = 0$ $\text{V}^{2+} = 3d^3 \quad \therefore \mu_{\text{spin}} = 3.87 \text{ B.M.}$ $\text{Ni}^{2+} = 3d^8 \quad \therefore \mu_{\text{spin}} = 2.84 \text{ B.M.}$ $\text{Ti}^{3+} = 3d^1 \quad \therefore \mu_{\text{spin}} = 1.73 \text{ B.M.}$

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
}