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

Short Answer

Expert verified
Option (d) is the correct answer.

Step by step solution

01

Identifying Lone Pairs of Electrons

First, we must determine the number of lone pairs of electrons on the central atoms of each molecule. For this, we use the VSEPR theory. In \(\text{SF}_4\), sulfur has 6 valence electrons and forms 4 bonds with fluorine, leaving it with 1 lone pair. In \(\text{CF}_4\), carbon has 4 valence electrons, all involved in bonding with fluorine, resulting in 0 lone pairs. In \(\text{XeF}_4\), xenon has 8 valence electrons, forming 4 bonds with fluorine and 2 lone pairs.
02

Determining the Molecular Shapes

Analyze the molecular shapes based on the VSEPR theory using the lone pairs and bonds. The shape of \(\text{SF}_4\) is see-saw due to the presence of 1 lone pair. \(\text{CF}_4\) is tetrahedral as it has 0 lone pairs. \(\text{XeF}_4\) has a square planar shape due to the presence of 2 lone pairs reducing repulsion and shaping the molecule planar.
03

Matching with Given Options

Compare the identified number of lone pairs and molecular shapes with the options given. From Steps 1 and 2, \(\text{SF}_4, \text{CF}_4, \text{XeF}_4\) have 1, 0, and 2 lone pairs respectively and different shapes. Option (d) describes these conditions.

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Key Concepts

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

Molecular Shapes
Understanding molecular shapes is key to predicting a molecule's properties and behavior. The shape of a molecule is determined primarily by the arrangement of atoms around a central atom. This arrangement is influenced by the electron pair repulsion forces, a concept explained by VSEPR (Valence Shell Electron Pair Repulsion) theory.

VSEPR theory assumes that electron pairs, both bonding and lone pairs, repel each other. To minimize repulsion and achieve a more stable structure, these pairs arrange themselves as far apart as possible. The resulting spatial arrangement dictates the molecular shape, affecting things like molecular polarity and reactivity.

Different shapes arise from all bonding pairs, combinations of bonding with one or more lone pairs, or extensive lone pair presence, resulting in distinct geometrical configurations.
Lone Pairs of Electrons
Lone pairs of electrons, which are pairs of valence electrons not shared with another atom, play a significant role in shaping a molecule. These electrostatic lone pair-lone pair interactions strongly influence molecular geometry more than bonding pair-bonding pair interactions.

Lone pairs take up more space than bonded pairs because of their stronger repulsion, thus pushing bonded atoms closer together in the structure. This effect modifies molecular shapes, leading to geometries such as bent or trigonal pyramidal, depending on the number of lone pairs and bonded groups around the central atom.
  • The more lone pairs on the central atom, the more distorted the shape compared to those predicted by purely bonding interactions.
  • This distortion is key to predicting the actual three-dimensional arrangement of atoms in a molecule, which ultimately affects its physical properties and reactivity.
SF4 Molecular Structure
The molecule \( ext{SF}_4\) presents an interesting case with its see-saw shape, a direct outcome of VSEPR theory. Sulfur, the central atom, has a total of 6 valence electrons. Four of these electrons participate in bonding with the four fluorine atoms, while the remaining pair exists as a lone pair.

This lone pair occupies an equatorial position in a trigonal bipyramidal electron pair geometry, causing a geometrical push or distortion on the bonded fluorine atoms. This asymmetry results in the see-saw configuration.
  • See-saw shapes are less symmetric and therefore less stable than some other geometrical structures.
  • The presence of the lone pair is significant in influencing molecular behavior and properties, such as polarity and reactivity towards other molecules or ions.
CF4 Molecular Structure
For \( ext{CF}_4\), we find a classic example of a tetrahedral molecular shape, resulting from no lone pairs disrupting the geometric arrangement. Carbon, the central atom in \( ext{CF}_4\), has 4 valence electrons, all involved in forming single bonds with four fluorine atoms.

This arrangement places all electron pairs at equal distances in a symmetrical tetrahedral shape, which minimizes electron pair repulsion perfectly.
  • Tetrahedral geometry, due to its symmetry, often leads to non-polar molecules if all substituents are identical.
  • It provides a stable structure that has significant implications in understanding the molecule's physical properties such as low reactivity and higher stability compared to its isomers or counterparts.
XeF4 Molecular Structure
Unlike \( ext{SF}_4\) and \( ext{CF}_4\), the \( ext{XeF}_4\) molecule exhibits a square planar shape. Xenon at the core contributes 8 valence electrons, among which four electrons form bonds with the four fluorine atoms, leaving two pairs as lone pairs.

These lone pairs assume positions opposite each other, minimizing their repulsion while flattening the structure into a square planar form.
  • This geometry is typically symmetrical and can contribute to possible non-polar characteristics, despite potential dipoles in individual Xe-F bonds.
  • The presence of lone pairs and the subsequent planarity affects the behavior of \( ext{XeF}_4\) in different chemical environments, having significant roles in crystal formation and reactivity patterns.

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Most popular questions from this chapter

In which of the following molecules all the bonds are not equal? (a) \(\mathrm{AlF}_{3}\) (b) \(\mathrm{NF}_{3}\) (c) \(\mathrm{ClF}_{3}\) (d) \(\mathrm{BF}_{3}\)

The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, \(\mathrm{K}^{+}, \mathrm{Ca}^{2+}\), \(\mathrm{Mg}^{2+}, \mathrm{Be}^{2+} ?\) (a) \(\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2}+<\mathrm{Mg}^{2+}\) (b) \(\mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}\) (c) \(\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}\) (d) \(\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}\)

The acid having \(\mathrm{O}\) - \(\mathrm{O}\) bond is: (a) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{6}\) (c) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}\) (d) \(\mathrm{H}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\)

Iodine pentafluoride has which of the following hybridization? (a) \(\mathrm{d}^{2} \mathrm{sp}^{3}\) (b) \(\mathrm{dsp}^{3}\) (c) \(\mathrm{sp}^{3} \mathrm{~d}\) (d) \(\mathrm{sp}^{3} \mathrm{~d}^{2}\)

Stability of \(\mathrm{Ge}^{2+}, \mathrm{Sn}^{2+}\) and \(\mathrm{Pb}^{2+}\) is in order: (a) \(\mathrm{Ge}^{2+}>\mathrm{Sn}^{2+}>\mathrm{Pb}^{2+}\) (b) \(\mathrm{Sn}^{2+}>\mathrm{Ge}^{2+}>\mathrm{Pb}^{2+}\) (c) \(\mathrm{Pb}^{2+}>\mathrm{Sn}^{2+}>\mathrm{Ge}^{2+}\) (d) \(\mathrm{Sn}^{2+}>\mathrm{Pb}^{2+}>\mathrm{Ge}^{2+}\)

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