Problem 11
Question
Match each of the following species with one of these hybridization schemes: \(s p, s p^{2}, s p^{3}, s p^{3} d, s p^{3} d^{2} .\) (a) \(\mathrm{PF}_{6}^{-}\) (b) \(\operatorname{COS} ;\) (c) \(\operatorname{SiCl}_{4} ;\) (d) \(\mathrm{NO}_{3}^{-}\);(e) AsF \(_{5}\)
Step-by-Step Solution
Verified Answer
The hybridizations for the given species are - (a) \(PF_{6}^{-}\) is \(sp^{3}d^{2}\), (b) \(COS\) is \(sp\), (c) \(SiCl_{4}\) is \(sp^{3}\), (d) \(NO_{3}^{-}\) is \(sp^{2}\), (e) \(AsF_{5}\) is \(sp^{3}d\).
1Step 1: PF6-
Look at the central atom (P). The number of atoms bonded to Phosphorus is 6 and there are no lone pairs. This implies an octahedral shape, leading to hybridization of \(sp^{3}d^{2}\).
2Step 2: COS
In \(COS\), C is the central atom. It is bonded to two other atoms and there are no lone pairs. This indicates a straight-line structure with bond angles of 180°. This implies \(sp\) hybridization.
3Step 3: SiCl4
For \(SiCl_{4}\), the central atom is Si. The number of atoms Si is bonded to is 4 and there are no lone pairs. This corresponds to a Tetrahedral shape and hence the hybridisation is \(sp^{3}\).
4Step 4: NO3-
For \(NO_{3}^{-}\), the central atom is N. It is bonded to 3 atoms (oxygen atoms) and there is one resonance bond. This implies a Trigonal Planar shape and hence the hybridisation is \(sp^{2}\).
5Step 5: AsF5
Finally, in \(AsF_{5}\), As is the central atom. The number of atoms As is bonded to is 5 and there are no lone pairs. This implies a trigonal Bi-Pyramidal shape and hence the hybridisation is \(sp^{3}d\).
Key Concepts
Hybridization SchemesVSEPR TheoryMolecular GeometryChemical Bonding
Hybridization Schemes
Hybridization in chemistry is a concept explaining the structure and bonding of molecules. It is based on the combination of atomic orbitals to form new hybrid orbitals. These hybrid orbitals influence the shape and angles in a molecule. The most common types of hybridization include:
- **sp hybridization**: Seen in molecules with a linear structure, involving one s and one p orbital. An example is COS, where the carbon forms two bonds, creating a straight-line layout.
- **sp² hybridization**: Occurs in trigonal planar shapes, involving one s orbital and two p orbitals. For instance, NO₃⁻ shows this type with its three bonded oxygen atoms.
- **sp³ hybridization**: Found in tetrahedral molecules, using one s and three p orbitals. An example is SiCl₄ with four chlorine atoms symmetrically bonded.
- **sp³d hybridization**: Related to trigonal bipyramidal structures with one s, three p, and one d orbital, as seen in AsF₅ where five fluoro atoms bond to arsenic.
- **sp³d² hybridization**: Present in octahedral shapes, involving one s, three p, and two d orbitals. PF₆⁻ exemplifies this with its six fluorine atoms.
VSEPR Theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used to predict the geometry of individual molecules. This theory is centered on the idea that electron pairs in the valence shell of an atom will arrange themselves to minimize repulsion. This creates distinct molecular shapes. Some characteristics include:
- **Linear Geometry**: Related to sp hybridization, with bond angles of 180°, as seen in COS.
- **Trigonal Planar Geometry**: Correlates with sp² hybridization, featuring bond angles close to 120°, evident in NO₃⁻.
- **Tetrahedral Geometry**: Associated with sp³ hybridization, having bond angles of 109.5°, like SiCl₄.
- **Trigonal Bipyramidal Geometry**: Matches sp³d hybridization, with varied bond angles (90° and 120°), visible in AsF₅.
- **Octahedral Geometry**: Connects to sp³d² hybridization, marked by consistent 90° angles, as PF₆⁻ demonstrates.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It is closely linked to both hybridization and VSEPR theory. By understanding molecular geometry, we can infer many physical and chemical properties of the molecule, such as polarity and reactivity. Here are some common geometries with examples:
- **Linear Geometry**: In molecules like COS, atoms lie in a single straight line, suggesting non-polar characteristics when the surrounding atoms are identical.
- **Trigonal Planar Geometry**: In NO₃⁻, the atoms form a flat, triangle-like shape, allowing for resonance and stability.
- **Tetrahedral Geometry**: In SiCl₄, the molecules have a symmetrical arrangement, leading to a non-polar molecule due to equal distribution of charge.
- **Trigonal Bipyramidal Geometry**: Seen in AsF₅, this arrangement allows for flexibility in bonding angles and can create polar molecules depending on the surrounding atoms.
- **Octahedral Geometry**: As displayed in PF₆⁻, the atoms are symmetrically dispersed around the central atom, fostering stability and equal bond distribution.
Chemical Bonding
Chemical bonding is a fundamental concept in chemistry that describes the attractions between atoms that lead to the formation of molecules, and it comes in various types. Two of the most significant include:
- **Covalent Bonding**: Involves the sharing of electron pairs between atoms. For example, in the molecules discussed, many involve covalent bonds, where the electrons are shared to achieve stability.
- **Ionic Bonding**: Involves transfer of electrons from one atom to another, leading to attraction between oppositely charged ions. While our examples mainly focus on covalent structures, ionic bonds are prevalent in compounds like salts.
Other exercises in this chapter
Problem 5
In which of the following, \(\mathrm{CO}_{3}^{2-}, \mathrm{SO}_{2}, \mathrm{CCl}_{4}, \mathrm{CO}\) \(\mathrm{NO}_{2}^{-},\) would you expect to find \(s p^{2}\
View solution Problem 7
For each of the following species, identify the central atom(s) and propose a hybridization scheme for those \(\operatorname{atom}(\mathrm{s}):(\mathrm{a}) \mat
View solution Problem 12
Propose a hybridization scheme to account for bonds formed by the central carbon atom in each of the following molecules: (a) hydrogen cyanide, HCN; (b) methyl
View solution Problem 13
Indicate which of the following molecules and ions are linear, which are planar, and which are neither. Then propose hybridization schemes for the central atoms
View solution