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Match the following: List I List II 1\. \(\mathrm{XeF}_{4}\) (i) distorted octahedral 2\. \(\mathrm{XeF}_{6}\) (ii) tetrahedral 3\. \(\mathrm{XeO}_{3}\) (iii) square planar 4\. \(\mathrm{XeO}_{4}\) (iv) pyramidal The correct matching is 1 2 3 4 (a) (iii) (i) (iv) (ii) (b) (ii) (i) (iii) (iv) (c) (i) (iii) (ii) (iv) (d) (iii) (iv) (i) (ii)

Short Answer

Expert verified
The correct matching is option (d): 1-(iii), 2-(i), 3-(iv), 4-(ii).

Step by step solution

01

Identify the structure of \( \mathrm{XeF}_{4} \)

\( \mathrm{XeF}_{4} \) has a steric number of 6 due to 4 bonded pairs of electrons and 2 lone pairs around the xenon atom. The lone pairs occupy positions to minimize repulsion, leading to a square planar geometry. Therefore, \( \mathrm{XeF}_{4} \) corresponds to (iii) square planar.
02

Identify the structure of \( \mathrm{XeF}_{6} \)

\( \mathrm{XeF}_{6} \) has 6 bonded pairs and 1 lone pair around the xenon atom, giving a steric number of 7. This results in a distorted octahedral geometry, which means \( \mathrm{XeF}_{6} \) matches with (i) distorted octahedral.
03

Identify the structure of \( \mathrm{XeO}_{3} \)

\( \mathrm{XeO}_{3} \) includes 3 bond pairs and 1 lone pair, giving it a steric number of 4. This leads to a pyramidal structure due to the presence of the lone pair. Therefore, \( \mathrm{XeO}_{3} \) matches with (iv) pyramidal.
04

Identify the structure of \( \mathrm{XeO}_{4} \)

\( \mathrm{XeO}_{4} \) has 4 bonded pairs of electrons and no lone pairs on the xenon. This results in a steric number of 4, giving a tetrahedral geometry. Hence, \( \mathrm{XeO}_{4} \) corresponds to (ii) tetrahedral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Geometry
Molecular geometry is a key aspect of understanding how atoms are arranged in a molecule. It describes the spatial arrangement of atoms bonded to a central atom. The shape of the molecule can greatly affect its properties and interactions with other molecules.
  • For example, the molecular geometry of a molecule like \( \mathrm{XeF}_{4} \) is "square planar," which means that the molecule has a flat geometric shape with four fluorine atoms bonded symmetrically around the xenon atom. This results from four bonded pairs and two lone pairs of electrons.

  • Another example is \( \mathrm{XeF}_{6} \), which forms a "distorted octahedral" shape. Here, the six fluorine atoms attempt to form an octahedral geometry, but a lone pair introduces some distortion, resulting in this specific shape.
Hence, understanding molecular geometry allows chemists to predict the behavior and reactivity of the compounds accurately.
VSEPR Theory
VSEPR theory, short for Valence Shell Electron Pair Repulsion theory, is a model used to predict the geometry of individual molecules based on the repulsion between electron pairs around a central atom. The idea is simple: electron pairs like to stay as far apart as possible in 3D space.
  • In the case of \( \mathrm{XeF}_{4} \), VSEPR theory helps explain its square planar structure. The lone pairs on the xenon are positioned to minimize repulsion, flattening the molecule.

  • For \( \mathrm{XeO}_{3} \), having three bonding pairs and one lone pair leads to a "pyramidal" shape as predicted by VSEPR. The lone pair occupies more space and pushes the bonded atoms closer together at the base, forming a pyramid-like shape.
By understanding how electron pairs repel each other to create distinct molecular shapes, VSEPR theory offers a powerful tool for visualizing and predicting molecular structures.
Xenon Compounds
Xenon is a noble gas that forms unusual compounds, often defying the expectations set by its inert nature due to a complete outer electron shell. It can engage in chemical reactions under specific conditions to form several stable compounds.
  • \( \mathrm{XeF}_{4} \) and \( \mathrm{XeF}_{6} \) are fluorides of xenon, demonstrating how xenon can bond with fluorine to create various structures like square planar and distorted octahedral forms, respectively.

  • \( \mathrm{XeO}_{3} \) and \( \mathrm{XeO}_{4} \) are xenon oxides. Here, xenon forms bonds with oxygen atoms leading to diverse molecular geometries such as pyramidal and tetrahedral.
These compounds of xenon illustrate the versatility and complexity of its chemical bonding, showing how it can deviate from being chemically inert to exhibit significant reactivity under certain conditions.

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Most popular questions from this chapter

The molecular shapes of \(\mathrm{SF}_{4}, \mathrm{CF}_{4}\), and \(\mathrm{XeF}_{4}\) are (a) different with 1,0 and 2 lone pairs of electrons on the central atoms respectively (b) different with 0,1 and 2 lone pairs of electrons on the central atoms respectively (c) the same with 1,1 and 1 lone pair of electrons on the central atoms respectively (d) the same with 2,0 and 1 lone pair of electrons on the central atoms respectively

For \(\mathrm{H}_{3} \mathrm{PO}_{3}\) and \(\mathrm{H}_{3} \mathrm{PO}_{4}\) the correct choice is (a) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is dibasic and reducing (b) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is dibasic and non-reducing (c) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is tribasic and reducing (d) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is tribasic and non-reducing

A compound A of S, Cl and O has a vapour density of \(67.5\). It reacts with water to form two acids and reacts with \(\mathrm{KOH}\) to form two salts \(\mathrm{B}\) and \(\mathrm{C}\). while B gives white precipitate with \(\mathrm{AgNO}_{3}\) solution, C gives white precipitate with \(\mathrm{BaCl}_{2}\) solution. Identify A (a) \(\mathrm{SOCl}_{2}\) (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{SO}_{2} \mathrm{Cl}\) (d) none of these

Which electrolyte is used in the electrolytic method of preparation of fluorine? (a) \(\mathrm{KHF}_{2}+\mathrm{LiF}\) (b) \(\mathrm{CaCl}_{2}+\mathrm{KF}\) (c) \(\mathrm{KF}+\mathrm{LiHF}_{2}\) (d) \(\mathrm{NaF}+\mathrm{CaCl}_{2}\)

Select the correct statements about diborane: (a) \(\mathrm{H}_{b} \ldots . \mathrm{B} \ldots \mathrm{H}_{b}\) bond angle is \(122^{\circ}\) (b) All hydrogens in \(\mathrm{B}_{2} \mathrm{H}_{6}\) lie in the same plane (c) \(\mathrm{B}_{2} \mathrm{H}_{6}\) has three centered bond (d) Each boron atom lies in sp \(^{3}\) hybrid state

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