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Current Question (ID: 18670)

Question:

$\text{A solution of } [\text{Ni(H}_2\text{O)}_6]^{2+} \text{ is green, but a solution of } [\text{Ni(CN)}_4]^{2-} \text{ is colorless because:}$

Options:

1. $\text{There are paired electrons in } [\text{Ni(H}_2\text{O)}_6]^{2+} \text{ while all electrons are unpaired in } [\text{Ni(CN)}_4]^{2-}$
2. $\text{There are unpaired electrons in } [\text{Ni(H}_2\text{O)}_6]^{2+} \text{ while all electrons are paired in } [\text{Ni(CN)}_4]^{2-}$
3. $\text{There are unpaired electrons in } [\text{Ni(H}_2\text{O)}_6]^{2+} \text{ and } [\text{Ni(CN)}_4]^{2-}$
4. $\text{None of the above.}$

Solution:

$\text{HINT: CN}^- \text{ is a strong field ligand and water is weak field ligand.}$ $\text{STEP 1: In } [\text{Ni(H}_2\text{O)}_6]^{2+}, \text{ H}_2\text{O is a weak field ligand. Therefore, there are unpaired electrons in Ni}^{2+}.$ $\text{In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d-d transition is present. Hence } [\text{Ni(H}_2\text{O)}_6]^{2+}, \text{ is colored.}$ $\text{STEP 2: In } [\text{Ni(CN)}_4]^{2-}, \text{ the electrons are all paired as CN}^- \text{ is a strong field ligand.}$ $[\text{Ni(CN)}_4]^{2-}$ $\text{The empty } 4d, 3s \text{ and two } 4p \text{ orbitals undergo } dsp^2 \text{ hybridization to make bonds with CN}^- \text{ ligands in square planar geometry. Thus } [\text{Ni(CN)}_4]^{2-} \text{ is diamagnetic.}$ $\text{Therefore, a d-d transition is not possible in } [\text{Ni(CN)}_4]^{2-}. \text{ Hence, it is colorless.}$

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
}