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Current Question (ID: 21003)

Question:

$\text{Given below are two statements:}$ $\text{Statement I:}$ \text{Since electronegativity of F > H, so dipole moment of } \text{NF}_3 > \text{NH}_3.$ $\text{Statement II:}$ \text{The lone pair dipole in } \text{NH}_3 \text{ is not in the direction of the resultant bond dipole, while in the case of } \text{NF}_3, \text{ the lone pair dipole is in the direction of the resultant bond dipole.}$

Options:

1. $\text{Statement I is false but Statement II is true.}$
2. $\text{Both Statement I and Statement II are false.}$
3. $\text{Both Statement I and Statement II are true.}$
4. $\text{Statement I is true but Statement II is false.}$

Solution:

$\text{Both } \text{NH}_3 \text{ and } \text{NF}_3 \text{ have a trigonal pyramidal shape due to the lone pair on nitrogen.}$ $\text{However, the direction of individual bond dipoles differs:}$ $(\text{I}) \text{ In } \text{NH}_3, \text{ the N-H bond dipoles point toward nitrogen (more electronegative than hydrogen).}$ $\text{The lone pair's dipole moment adds to the resultant bond dipole, increasing the net dipole moment } (\mu = 1.47 \text{ D}).$ $(\text{II}) \text{ In } \text{NF}_3, \text{ the N-F bond dipoles point toward fluorine (more electronegative than nitrogen).}$ $\text{The lone pair's dipole moment opposes the resultant bond dipole, leading to significant cancellation and a smaller net dipole moment } (\mu = 0.24 \text{ D}).$ $\text{Thus, the dipole moment of } \text{NH}_3 > \text{NF}_3 \text{ because in the case of } \text{NF}_3 \text{ the lone pair dipole is in the direction of the resultant bond dipole.}$

Import JSON File

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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
}