Match the List-I with List-II with respect to hybridization of Xe: List-I – List -II (i) $XeO_{3}$ – (a) $sp^{3}d^{2}$ hybridization (ii) $XeF_{6}$ – (b) $sp^{3}d$ hybridization (iii) $XeOF_{4}$ – (c) $sp^{3}d^{3}$ hybridization (iv) $XeF_{2 – (d) $sp^{3}$ hybridization

Q: Match the List-I with List-II with respect to hybridization of Xe: List-I – List -II (i) $XeO_{3}$ – (a) $sp^{3}d^{2}$ hybridization (ii) $XeF_{6}$ – (b) $sp^{3}d$ hybridization (iii) $XeOF_{4}$ – (c) $sp^{3}d^{3}$ hybridization (iv) $XeF_{2 – (d) $sp^{3}$ hybridization

Match the List-I with List-II with respect to hybridization of Xe: List-I – List -II (i) $XeO_{3}$ – (a) $sp^{3}d^{2}$ hybridization (ii) $XeF_{6}$ – (b) $sp^{3}d$ hybridization (iii) $XeOF_{4}$ – (c) $sp^{3}d^{3}$ hybridization (iv) $XeF_{2 – (d) $sp^{3}$ hybridization

Question

Accepted Answer

To justify why Option 3 is the correct answer, we need to analyze the hybridization of xenon in each compound listed in List-I and match it with the corresponding hybridization type in List-II.

1. **$XeO_{3}$**: 
   - **Hybridization**: In $XeO_{3}$, xenon is bonded to three oxygen atoms. Xenon has 8 valence electrons, and each oxygen forms a double bond with xenon, using 6 electrons. The remaining 2 electrons form a lone pair. This results in a total of 4 electron pairs (3 bonding pairs and 1 lone pair), which corresponds to $sp^{3}$ hybridization.
   - **Match**: (i)-(d)

2. **$XeF_{6}$**: 
   - **Hybridization**: In $XeF_{6}$, xenon is bonded to six fluorine atoms. Xenon has 8 valence electrons, and each fluorine forms a single bond with xenon, using 6 electrons. The remaining 2 electrons form a lone pair. This results in a total of 7 electron pairs (6 bonding pairs and 1 lone pair), which corresponds to $sp^{3}d^{3}$ hybridization.
   - **Match**: (ii)-(c)

3. **$XeOF_{4}$**: 
   - **Hybridization**: In $XeOF_{4}$, xenon is bonded to one oxygen and four fluorine atoms. Xenon has 8 valence electrons, and the oxygen forms a double bond while each fluorine forms a single bond, using 6 electrons. The remaining 2 electrons form a lone pair. This results in a total of 6 electron pairs (5 bonding pairs and 1 lone pair), which corresponds to $sp^{3}d^{2}$ hybridization.
   - **Match**: (iii)-(a)

4. **$XeF_{2}$**: 
   - **Hybridization**: In $XeF_{2}$, xenon is bonded to two fluorine atoms. Xenon has 8 valence electrons, and each fluorine forms a single bond with xenon, using 2 electrons. The remaining 6 electrons form 3 lone pairs. This results in a total of 5 electron pairs (2 bonding pairs and 3 lone pairs), which corresponds to $sp^{3}d$ hybridization.
   - **Match**: (iv)-(b)

Now, let's explain why the other options are incorrect:

- **Option 1**: This option incorrectly matches the hybridizations for all compounds. For example, it matches $XeO_{3}$ with $sp^{3}d^{2}$ hybridization, which is incorrect as explained above.

- **Option 2**: This option also incorrectly matches the hybridizations. For instance, it matches $XeF_{6}$ with $sp^{3}d$ hybridization, which is incorrect because $XeF_{6}$ has $sp^{3}d^{3}$ hybridization.

- **Option 4**: This option incorrectly matches the hybridizations as well. For example, it matches $XeF_{2}$ with $sp^{3}$ hybridization, which is incorrect since $XeF_{2}$ has $sp^{3}d$ hybridization.

Therefore, Option 3 is the only option that correctly matches each compound with its respective hybridization, making it the correct answer.