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Current Question (ID: 8048)

Question:

A pair of species that have identical shapes is :

Options:

1. $\text{NO}_2^+$ and $\text{NO}_2^-$
2. $\text{PCl}_5$ and $\text{BrF}_5$
3. $\text{XeF}_4$ and $\text{ICl}_4^-$ (Correct)
4. $\text{TeCl}_4$ and $\text{XeO}_4$

Solution:

HINT: $\text{XeF}_4$ and $\text{ICl}_4^-$ have identical shapes. Explanation: According to VSEPR Theory, the shape of a molecule depends on the number of bonding pairs and lone pairs of electrons around the central atom. Let's analyze each pair: 1. $\text{NO}_2^+$: * Central atom: N * Valence electrons on N = 5 * Bonding atoms (O) = 2. Each O forms a double bond, contributing 2 electrons to the bond. So, 2 $\sigma$ bonds. * Charge: $+1$. So, 1 electron removed. * Total valence electrons involved in bonding = $5 - 1 = 4$. * Number of $\sigma$ bonds = 2. Number of lone pairs = $(4 - 2 \times 2)/2 = 0$. * Hybridization = $\text{sp}$ * Shape = Linear $\text{NO}_2^-$: * Central atom: N * Valence electrons on N = 5 * Bonding atoms (O) = 2. Each O forms a single bond, contributing 1 electron to the bond, with one O forming a double bond. Or more simply, count sigma bonds and lone pairs. $\text{NO}_2^-$ has 2 bonding pairs (2 $\sigma$ bonds) and 1 lone pair on the central Nitrogen atom. * Hybridization = $\text{sp}^2$ * Shape = Bent or V-shaped (due to lone pair repulsion) * Not identical shapes. 2. $\text{PCl}_5$: * Central atom: P * Valence electrons on P = 5 * Bonding atoms (Cl) = 5. All single bonds. So, 5 $\sigma$ bonds. * Lone pairs = 0. * Hybridization = $\text{sp}^3\text{d}$ * Shape = Trigonal bipyramidal $\text{BrF}_5$: * Central atom: Br * Valence electrons on Br = 7 * Bonding atoms (F) = 5. All single bonds. So, 5 $\sigma$ bonds. * Lone pairs = $(7 - 5)/2 = 1$. * Hybridization = $\text{sp}^3\text{d}^2$ * Shape = Square pyramidal * Not identical shapes. 3. $\text{XeF}_4$: * Central atom: Xe * Valence electrons on Xe = 8 * Bonding atoms (F) = 4. All single bonds. So, 4 $\sigma$ bonds. * Lone pairs = $(8 - 4)/2 = 2$. * Hybridization = $\text{sp}^3\text{d}^2$ * Shape = Square planar $\text{ICl}_4^-$: * Central atom: I * Valence electrons on I = 7 * Bonding atoms (Cl) = 4. All single bonds. So, 4 $\sigma$ bonds. * Charge: $-1$. So, 1 extra electron. * Total electrons for lone pairs = $(7 + 1 - 4)/2 = 2$. * Hybridization = $\text{sp}^3\text{d}^2$ * Shape = Square planar * Identical shapes. 4. $\text{TeCl}_4$: * Central atom: Te * Valence electrons on Te = 6 * Bonding atoms (Cl) = 4. All single bonds. So, 4 $\sigma$ bonds. * Lone pairs = $(6 - 4)/2 = 1$. * Hybridization = $\text{sp}^3\text{d}$ * Shape = See-saw $\text{XeO}_4$: * Central atom: Xe * Valence electrons on Xe = 8 * Bonding atoms (O) = 4. Each O forms a double bond, contributing 2 electrons to the bond. So, 4 $\sigma$ bonds. * Lone pairs = 0 (all valence electrons used in double bonds). * Hybridization = $\text{sp}^3$ * Shape = Tetrahedral * Not identical shapes. Thus, $\text{XeF}_4$ and $\text{ICl}_4^-$ have identical shapes (Square planar).

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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
}