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Current Question (ID: 18875)

Question:

$\text{The incorrect statement among the following is:}$

Options:

1. $\text{Zirconium and Hofnium have identical radii of 160 pm and 159 pm, respectively, as a consequence of lanthanoid contraction.}$
2. $\text{Lanthanoids reveal only a +3 oxidation state.}$
3. $\text{Lanthanoid ions, other than the } f^0 \text{ type and the } f^{14} \text{ type, are all paramagnetic.}$
4. $\text{The overall decrease in atomic and ionic radii from lanthanum to lutetium is called lanthanoid contraction.}$

Solution:

$\text{HINT: Lanthanides can also show +2 and +4 oxidation state.}$ $\text{Explanation:}$ $\text{STEP 1: The general outer electronic configuration of f block elements is } (n-2)f^{(0-14)}(n-1)d^{(0-1)}ns^2.$ $\text{Lanthanides show variable oxidation states. The most stable oxidation state of Lanthanides is +3. They also show +2 and +4 oxidation states due to the presence of either half-filled or completely filled or empty 4f sub shell.}$ $\text{STEP 2: Lanthanoid contraction, also called lanthanide contraction, in chemistry, the steady decrease in the size of the atoms and ions of the rare earth elements with increasing atomic number from lanthanum (atomic number 57) through lutetium (atomic number 71).}$ $\text{Because of this Zirconium and Hofnium have identical radii of 160 pm and 159 pm.}$ $\text{STEP 3: The lanthanoid ions other than the } f^0 \text{ types and the } f^{14} \text{ type are all paramagnetic because of presence of unpaired electrons.}$

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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
}