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Current Question (ID: 8062)

Question:

$\text{The correct order of increasing s-character (in percentage) in the hybrid orbitals of the following molecules/ions is:}$ $\text{(i) } \text{CO}_3^{2-}$ $\text{(ii) } \text{XeF}_4$ $\text{(iii) } \text{I}_3^{-}$ $\text{(iv) } \text{NCl}_3$ $\text{(v) } \text{BeCl}_2$

Options:

1. $\text{(ii)}<\text{(iii)}<\text{(iv)}<\text{(i)}<\text{(v)}$
2. $\text{(ii)}<\text{(iv)}<\text{(iii)}<\text{(v)}<\text{(i)}$
3. $\text{(iii)}<\text{(ii)}<\text{(i)}<\text{(v)}<\text{(iv)}$
4. $\text{(ii)}<\text{(iv)}<\text{(iii)}<\text{(i)}<\text{(v)}$ (Correct)

Solution:

$\text{HINT: Find out the hybridisation to calculate % s - character.}$ $\text{Explanation:}$ $\text{(i) } \text{CO}_3^{2-} = \text{sp}^2 \text{ (33.33 % character)}$ $\text{(ii) } \text{XeF}_4 = \text{sp}^3\text{d}^2 \text{ (16.66 % character)}$ $\text{(iii) } \text{I}_3^{-} = \text{sp}^3\text{d(20 % character)}$ $\text{(iv) } \text{NCl}_3 = \text{sp}^3 \text{ (25 % character)}$ $\text{(v) } \text{BeCl}_2 = \text{sp } \text{ (50 % character)}$ $\text{So, order of increasing s-character is :}$ $\text{ii } < \text{ iii } < \text{ iv } < \text{ i } < \text{ v}$

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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
}