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Current Question (ID: 10621)

Question:

$\text{In the nitrogen family, the H-M-H bond angle in the hybrides } \text{MH}_3 \text{ gradually becomes closer to } 90° \text{ if we move from N to Sb. This shows that gradually}$

Options:

1. $\text{The basic strength of hybrides increases.}$
2. $\text{Pure p-orbitals (almost) are used for M-H bonding.}$ (Correct)
3. $\text{The bond energies of the M-H bond increase.}$
4. $\text{The bond pairs of electrons become nearer to the central atom.}$

Solution:

$\text{Hint: In } \text{NH}_3\text{, hybridization occurs and after third-period element hybridization did not occur in hydrides}$ $\text{The hybrides have a pyramidal or tetrahedral shape with a lone pair of electrons in one of the orbitals. The H-M-H bond angle is less than the original } 109°28' \text{ tetrahedral bond angle (H-N-H in } \text{NH}_3 \text{ is } 106°45') \text{ because of greater repulsion between lone pair and a bond pair than between two bond pairs of electrons.}$ $\text{Because electronegativity of M decreases from N to Bi, the bond pair lie farther away from the central atom, and the lone pair causes greater distortion of bond angle.}$ $\text{Thus H-P-H bond in } \text{PH}_3 \text{ is } 94°\text{, while in } \text{AsH}_3 \text{ and } \text{SbH}_3 \text{ it is about } 91.8° \text{ and } 91.3° \text{ respectively (closer to } 90°\text{). This suggests that orbitals used for bonding are closer to pure p-orbitals.}$

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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
}