Import Question JSON

Current Question (ID: 18828)

Question:

$\text{Which of the following species is colourless?}$

Options:

1. $\text{TiF}_6^{2-} \text{ and } \text{CoF}_6^{3-}$
2. $\text{Cu}_2\text{Cl}_2 \text{ and } \text{NiCl}_4^{2-}$
3. $\text{TiF}_6^{2-} \text{ and } \text{Cu}_2\text{Cl}_2$
4. $\text{CoF}_6^{3-} \text{ and } \text{NiCl}_4^{2-}$

Solution:

$\text{HINT: Species without unpaired electrons are colourless because transition of electron is not possible.}$ $\text{Explanation:}$ $\text{STEP 1: 1. In } \text{TiF}_6^{2-}, \text{ Ti is present as Ti}^{4+}$ $\text{Ti}^{4+} = [\text{Ar}]3d^04s^0$ $\text{Hence } \text{TiF}_6^{2-} \text{ is colourless.}$ $\text{Due to presence of unpaired electron, } \text{CoF}_6^{3-} \text{ is coloured.}$ $\text{3. In } \text{Cu}_2\text{Cl}_2, \text{ Cu is present as Cu}^{+}$ $\text{Cu}^{+} = [\text{Ar}] \begin{array}{ccccccccc} \uparrow & \uparrow & \uparrow & \uparrow & \uparrow & \uparrow & \uparrow & \uparrow & \uparrow \end{array}$ $\text{Due to absence of unpaired electron, } \text{Cu}_2\text{Cl}_2 \text{ is colourless}$ $\text{4. In } \text{NiCl}_4^{2-}, \text{ Ni is present as Ni}^{2+}$ $\text{Ni}^{2+} = [\text{Ar}] \begin{array}{ccccccccc} \uparrow & \uparrow & \uparrow & \uparrow & \uparrow & \uparrow & \uparrow & \uparrow & \uparrow \end{array}$ $\text{Since unpaired electron are present, } \text{NiCl}_4^{2-} \text{ is coloured.}$ $\text{STEP 2:}$ $\text{Hence } \text{TiF}_6^{2-} \text{ and } \text{Cu}_2\text{Cl}_2 \text{ are colourless species.}$

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
}