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Current Question (ID: 18772)

Question:
The hybridizations of $[\text{Ni(CO)}_4]$ and $[\text{Cr(H}_2\text{O)}_6]^{3+}$, respectively, are:
Options:
  • 1. $\text{sp}^3 \text{ and d}^3\text{sp}^2$
  • 2. $\text{dsp}^2 \text{ and d}^2\text{sp}^3$
  • 3. $\text{sp}^3 \text{ and d}^2\text{sp}^3$
  • 4. $\text{dsp}^2 \text{ and sp}^3\text{d}^2$
Solution:
$\text{STEP 1: In } [\text{Ni(CO)}_4], \text{ Ni} = 0 \text{ and configuration is Ni} = 4s^2 3d^8. \text{ CO is a strong field ligand it will pair up the electron.}$ $\text{The hybridization of } [\text{Ni(CO)}_4] \text{ is sp}^3 \text{ and geometry is tetrahedral.}$ $\text{It is a diamagnetic compound because it does not contain any unpaired electrons.}$ $\text{STEP 2: In } [\text{Cr(H}_2\text{O)}_6]^{3+}, \text{ Cr oxidation state is } +3. \text{ It is a } 3d^3 \text{ system. The hybridization is d}^2\text{sp}^3.$ $\text{It is paramagnetic because it contains three unpaired electrons and it is octahedral in nature.}$

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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
}