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Current Question (ID: 18763)

Question:
$\text{The expected number of unpaired electrons for the complex ion } [\text{Cr(NH}_3)_6]^{3+} \text{ are:}$
Options:
  • 1. $2$
  • 2. $3$
  • 3. $4$
  • 4. $5$
Solution:
$\text{HINT: Cr is in +3 oxidation state.}$ $\text{Explanation:}$ $\text{STEP 1: In } [\text{Cr(NH}_3)_6]^{3+}, \text{ the oxidation state of Cr in the given complex is } +3 \text{ as NH}_3 \text{ is a neutral ligand. Therefore, its electronic configuration will be } [\text{Ar}]3d^3 4s^0.$ $\text{Thus it shows the configuration } t_{2g}^3 e_g^0.$ $\text{STEP 2: } d^1, d^2, d^3, d^8, d^9, d^{10} \text{ configuration is the configuration that do not get affected whether the ligand is a strong field ligand or a weak field ligand.}$ $\text{Hence, the no. of unpaired electrons will be 3.}$

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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
}