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Current Question (ID: 18747)

Question:

$\text{An ion, among the following, that has a magnetic moment of 2.84 BM is:}$ $\text{(At. no. Ni = 28, Ti = 22, Cr = 24, Co = 27)}$

Options:

1. $\text{Ni}^{2+}$
2. $\text{Ti}^{3+}$
3. $\text{Cr}^{2+}$
4. $\text{Co}^{2+}$

Solution:

$\text{HINT: } \mu = \sqrt{n(n+2)} \text{ BM}$ $\text{STEP 1:}$ $\mu = \sqrt{n(n+2)} \text{ BM}$ $\mu = 2.84 \text{ BM}$ $\text{where, } \mu \text{ is Magnetic moment}$ $\text{STEP 2: On substituting the values and solving it,}$ $(2.84)^2 = n(n+2)$ $8 = n^2 + 2n$ $n^2 + 2n - 8 = 0$ $n^2 + 4n - 2n - 8 = 0$ $n(n+4) - 2(n+4) = 0$ $n = 2$ $\text{STEP 3: So, only } \text{Ni}^{2+} \text{ has 2 unpaired electrons as its atomic number is 28}$ $\text{and configuration is } 3d^8.$

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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
}