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Current Question (ID: 18684)

Question:
$\text{The correct IUPAC name for } \text{Mn}_3(\text{CO})_{12} \text{ is:}$
Options:
  • 1. $\text{Dodecacarbonylmanganate(0)}$
  • 2. $\text{Dodecacarbonylmanganate(II)}$
  • 3. $\text{Dodecacarbonyltrimanganese(0)}$
  • 4. $\text{Manganicdodecacarbonyl0}$
Solution:
$\text{STEP 1: The given complex is } \text{Mn}_3(\text{CO})_{12}$ $\text{Follow IUPAC rules:}$ $1.) \text{The central metal atom is Manganese (Mn) which is in zero oxidation state as CO is a neutral ligand. The oxidation state is written in parenthesis.}$ $2.) \text{Three Mn atoms are present. Hence, the prefix "tri" is used.}$ $3.) \text{For 12 carbonyl groups, the prefix "dodeca" is used.}$ $\text{STEP 2: Thus, the correct IUPAC name of } \text{Mn}_3(\text{CO})_{12} \text{ is dodecacarbonyltrimanganese(0).}$

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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
}