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Current Question (ID: 18771)

Question:

$\text{A paramagnetic ion among the following, forms an outer orbital complex is:}$ $1.\ [\text{Ni(NH}_3\text{)}_6]^{2+}$ $2.\ [\text{Zn(NH}_3\text{)}_6]^{2+}$ $3.\ [\text{Cr(NH}_3\text{)}_6]^{3+}$ $4.\ [\text{Co(NH}_3\text{)}_6]^{3+}$

Options:

1. $[\text{Ni(NH}_3\text{)}_6]^{2+}$
2. $[\text{Zn(NH}_3\text{)}_6]^{2+}$
3. $[\text{Cr(NH}_3\text{)}_6]^{3+}$
4. $[\text{Co(NH}_3\text{)}_6]^{3+}$

Solution:

$\text{HINT: In outer complex, 4d orbital participate in hybridization and in inner complex 3d participate in hybridization.}$ $\text{Explanation:}$ $\text{An orbital possessing unpaired d-orbital has the ability to exhibit paramagnetic behaviour. Since NH}_3\text{ is a weak field ligand, so it cannot force the electrons to pair up.}$ $(1)\ \text{In}\ [\text{Ni(NH}_3\text{)}_6]^{2+},\ \text{Ni is in +2 state, so its electronic configuration will be}$ $[\text{Ar}]3d^84s^0.$ $\text{Since there are two unpaired electrons in this complex and it has}$ $sp^3d^2\ \text{hybridisation.}$ $\text{So, this is an outer orbital complex which has the paramagnetic character.}$ $(2)\ \text{In}\ [\text{Zn(NH}_3\text{)}_6]^{2+},\ \text{Zn is in +2 state, so its electronic configuration will be}$ $[\text{Ar}]3d^{10}4s^0.$ $[\text{Zn(NH}_3\text{)}_6]^{2+}\text{has}\ sp^3d^2\ \text{hybridisation, it is an outer orbital complex.}$ $\text{However, due to the absence of unpaired d-orbitals, it is diamagnetic in nature.}$ $(3)\ \text{In}\ [\text{Cr(NH}_3\text{)}_6]^{3+},\ \text{Cr is in +2 state, so its electronic configuration will be}$ $[\text{Ar}]3d^34s^0$ $\text{In this complex, the (n-1)d, i.e., 3d orbital is involved in the hybridization.}$ $\text{Therefore, this complex is an inner orbital complex.}$ $\text{Due to the presence of three unpaired electrons, it is paramagnetic in nature.}$ $(4)\ \text{In}\ [\text{Co(NH}_3\text{)}_6]^{3+},\ \text{Co is in +3 state, so its electronic configuration will be}$ $[\text{Ar}]3d^64s^0.\ \text{In this complex, NH}_3\text{ is a strong field ligand and pairing of electron takes place in d orbital.}$ $[\text{Co(NH}_3\text{)}_6]^{3+}\text{ is an inner d-orbital orbital having }d^2sp^3\text{ hybridisation.}$ $\text{Due to the absence of unpaired d-orbitals, it is diamagnetic in nature.}$ $\text{Hence, option first is the correct answer.}$

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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
}