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Chapter 4: Problem 19

The correct hybridisation state of sulphur atom in \(\mathrm{SF}_{2}, \mathrm{SF}_{4}\) and \(\mathrm{SF}_{6}\) molecules is respectively : (a) \(s p^{3} d, s p^{3}, s p^{3} d^{2}\) (b) \(s p^{3}, s p^{3} d, s p^{3} d^{2}\) (c) \(s p^{3} d^{2}, s p^{3}, s p^{3} d\) (d) \(s p^{3} d^{2}, s p^{3} d, s p^{3}\)

Short Answer

Expert verified
(b) \(s p^{3}, s p^{3} d, s p^{3} d^{2}\).

Step by step solution

01

Understanding Molecular Geometry

First, determine the electronic geometry around the sulphur atom in each compound by counting the regions of electron density around the sulphur.
02

Analyze SF2

In \(\text{SF}_2\), the sulphur atom is bonded to two fluorine atoms and has two lone pairs. Total regions of electron density (electron clouds) = 4. This corresponds to \(\text{sp}^3\) hybridization, which describes a tetrahedral electronic geometry.
03

Analyze SF4

In \(\text{SF}_4\), the sulphur atom is bonded to four fluorine atoms and has one lone pair. Total regions of electron density = 5. This corresponds to \(\text{sp}^3d\) hybridization, indicating a trigonal bipyramidal geometry.
04

Analyze SF6

In \(\text{SF}_6\), the sulphur atom is bonded to six fluorine atoms and has no lone pairs. Total regions of electron density = 6. This corresponds to \(\text{sp}^3d^2\) hybridization, indicating an octahedral geometry.
05

Choose Correct Answer

Based on the analysis, the hybridization states of sulphur in \(\text{SF}_2\), \(\text{SF}_4\), and \(\text{SF}_6\) are \(\text{sp}^3\), \(\text{sp}^3d\), \(\text{sp}^3d^2\) respectively. Compare these to the provided options to find the correct answer.

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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 the three-dimensional arrangement of atoms within a molecule. It's essential because it affects various properties of a compound, like polarity, reactivity, and color. Understanding molecular geometry involves recognizing how atoms are bonded and the overall shape they create. When considering sulfur compounds like SF鈧�, SF鈧�, and SF鈧�, each exhibits different molecular geometries due to their unique arrangements of electron pairs and bonding atoms. In SF鈧�, sulfur forms a bent shape due to its lone pairs, while SF鈧� adopts a seesaw shape, and SF鈧� forms a perfect octahedron.
Electron Density Regions
Regions of electron density around an atom include bonds and lone pairs of electrons. They are areas where electrons are likely to be found, and they play a crucial role in determining a molecule's shape. For sulfur compounds, we count these regions to predict molecular geometry. For example, in SF鈧�, the sulfur atom has 4 regions of electron density because it is bonded to two fluorine atoms and has two lone pairs. In contrast, SF鈧� has 5 regions due to one lone pair and four bonds. SF鈧�, having six bonds and no lone pairs, also has 6 regions.
Sulfur-Fluorine Bonding
Sulfur-fluorine (S-F) bonds are strong and polar due to the high electronegativity of fluorine. These bonds play a significant role in the stability and geometry of sulfur compounds. In SF鈧�, SF鈧�, and SF鈧�, the variations in bonding and electron pair arrangements significantly influence the geometry and hybridization of these molecules. The strength of these bonds also contributes to the compounds' distinct properties, such as their reactivity and interaction with other substances.
Octahedral Geometry
Octahedral geometry is a molecular shape that arises when a central atom is surrounded by six regions of electron density. In this arrangement, the regions strive to be as far apart as possible, leading to an equal 90-degree angle between any two bonds. SF鈧� is a classic example of octahedral geometry. The sulfur atom in SF鈧� forms six S-F bonds with no lone pairs, leading to a perfect octahedral shape. This geometry correlates with the \(\mathrm{sp}^3d^2\) hybridization of the sulfur atom, which combines s, p, and d orbitals to accommodate the bonding.
Trigonal Bipyramidal Geometry
Trigonal bipyramidal geometry is characterized by five regions of electron density around a central atom. In this shape, there are three equatorial and two axial positions, making it asymmetrical. SF鈧� is an example of this geometry; the sulfur atom forms four S-F bonds and retains one lone pair. The lone pair occupies one of the equatorial positions due to space considerations, resulting in a seesaw molecular shape. This geometry results from \(\mathrm{sp}^3d\) hybridization, where sulfur combines its s, p, and one d orbital to create five equivalent hybrid orbitals.

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