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Current Question (ID: 18698)

Question:
$\text{A complex ion among the following can absorb visible light is:}$ $\text{(At. no. Zn = 30, Sc = 21, Ti = 22, Cr = 24)}$
Options:
  • 1. $[\text{Sc} (\text{H}_2\text{O})_3 (\text{NH}_3)_3]^{3+}$
  • 2. $[\text{Ti} (\text{en})_2 (\text{NH}_3)_2]^{4+}$
  • 3. $[\text{Cr} (\text{NH}_3)_6]^{3+}$
  • 4. $[\text{Zn} (\text{NH}_3)_6]^{2+}$
Solution:
$\text{HINT: } [\text{Cr} (\text{NH}_3)_6]^{3+} \text{ is a coloured complex.}$ $\text{Explanation:}$ $\text{By absorption of visible light, it simply means that the compound shows colour and is visible to us.}$ $\text{Also, we know that the compounds with unpaired electrons show d-d transition.}$ $\text{Hence, those metal complexes in which the d-subshell is incomplete are expected to absorb visible light.}$ $\text{(b) In } [\text{Ti} (\text{en})_2 (\text{NH}_3)_2]^{4+}, \text{ O.S. of Ti is +4.}$ $\text{Ti}^{4+} = [\text{Ar}] \ 3d^0 \ 4s^0$ $\text{It will also not absorb visible light as it has no electrons in d subshell.}$ $\text{(c) In } [\text{Cr} (\text{NH}_3)_6]^{3+}, \text{ O.S. of Cr is +3}$ $\text{Cr}^{3+} = [\text{Ar}] \ 3d^3 \ 4s^0$ $\text{Because of the presence of three unpaired electrons which will occupy lower } t_{2g} \text{ set of orbitals can easily undergo transition to higher } e_g \text{ set showing transition and hence colour i.e. it is expected to absorb visible light.}$ $\text{(d) In } [\text{Zn} (\text{NH}_3)_6]^{2+}, \text{ O.S. of Zn is +2.}$ $\text{Zn}^{2+} = [\text{Ar}] \ 3d^{10} \ 4s^0$ $[\text{Zn} (\text{NH}_3)_6]^{2+} = [\text{Ar}] \ 3d^{10}$ $\text{As the d subshell is completely filled, there is no room for transition to occur and no transition can occur resulting in no absorption of visible light.}$

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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
}