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Current Question (ID: 18727)

Question:

$\text{Which among the following complexes can exhibit isomerism?}$

Options:

1. $[\text{Ag(NH}_3)_2]^+$
2. $[\text{Co(NH}_3)_5\text{NO}_2]^{2+}$
3. $[\text{Pt(en)Cl}_2]$
4. $[\text{Co(NH}_3)_5\text{Cl}]^+$

Solution:

$\text{HINT: Square planar complexes can't show geometrical isomerism.}$ $\text{Explanation:}$ $[\text{Co(NH}_3)_5\text{NO}_2]^{2+}$ \text{ can show linkage isomerism because } \text{NO}_2 \text{ is an ambidentate ligand.}$ $[\text{Pt(en)Cl}_2] \text{ can't show optical isomerism due to plane of symmetry. The complex has square planar shape.}$ $\text{Due to bidentate ligand it can't show cis trans forms so, it will not show geometrical isomerism also.}$ $\text{In } [\text{Co(NH}_3)_5\text{Cl}]^{2+}, \text{ 5 ligands are same. Hence, it will not exhibit any kind of isomerism.}$

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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
}