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The maximum number of \(90^{\circ}\) angles between bond pair-bond pair of electrons is observed in (a) sp \(^{3} \mathrm{~d}^{2}\) hybridization (b) \(\mathrm{sp}^{3} \mathrm{~d}\) hybridization (c) \(\mathrm{dsp}^{3}\) hybridization (d) \(\mathrm{dsp}^{2}\) hybridization

Short Answer

Expert verified
(a) sp³d² hybridization exhibits the maximum number of 90° angles due to its octahedral geometry.

Step by step solution

01

Understanding Hybridization

Hybridization is a process of mixing atomic orbitals to form new hybrid orbitals. The type of hybridization determines the geometry of the molecule and the angles between bonds.
02

Analyzing sp³d² Hybridization

In sp³d² hybridization, six orbitals mix to form six equivalent sp³d² hybrid orbitals. These are arranged in an octahedral geometry, where each bond pair-bond pair angle is 90°. This structure allows six bonds to form around a central atom.
03

Analyzing sp³d Hybridization

In sp³d hybridization, five orbitals mix to form five equivalent sp³d hybrid orbitals. These are arranged in a trigonal bipyramidal geometry with two distinct bond angles: 120° in the equatorial plane and 90° between axial and equatorial bonds.
04

Analyzing dsp³ Hybridization

In dsp³ hybridization, five orbitals (one d, one s, and three p) combine to form five dsp³ hybrid orbitals, resulting in a trigonal bipyramidal shape, similar to sp³d hybridization. This geometry also has bond pair-bond pair angles of 90° and 120°.
05

Analyzing dsp² Hybridization

In dsp² hybridization, four orbitals (one d, one s, and two p) hybridize to form four dsp² hybrid orbitals, which arrange themselves in a square planar geometry. This results in bond pair-bond pair angles of 90° only.
06

Identifying the Maximum Number of 90° Angles

Upon reviewing all the hybridizations, sp³d² hybridization allows for all bond pair-bond pair angles to be 90° as it adopts an octahedral geometry. This results in six 90° angles between bond pairs, which is the maximum number possible compared to others.

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Key Concepts

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

sp3d2 hybridization
In chemistry, understanding how atomic orbitals mix together to form new hybrid orbitals is crucial to predicting the shape and properties of molecules. When we discuss the concept of sp³d² hybridization, we are referring to a specific configuration where one s orbital, three p orbitals, and two d orbitals from a central atom combine.
This hybridization results in six equivalent sp³d² hybrid orbitals.
These orbitals organize themselves in an octahedral geometry. This arrangement is quite symmetrical and allows for six bonding regions and can include lone pairs. When we look at examples featuring sp³d² hybridization, transition metals often display this kind of configuration.
The ability of a central atom to bond with six other atoms makes this a prevalent hybridization in coordination compounds.
molecular geometry
Molecular geometry is the three-dimensional arrangement of atoms within a molecule, ultimately determining its reactivity and properties. For molecules exhibiting sp³d² hybridization, the favored molecular geometry is octahedral. An octahedral geometry is characterized by its six-pointed shape, similar to an eight-sided die often used in board games. In an octahedral molecule, each point can be a bond to another atom, or, sometimes, a non-bonding pair of electrons. The central atom, usually a transition metal in these cases, sits at the center of the octahedron, forming ninety-degree angles between each adjacent bond.
The symmetrical arrangement reduces strain and creates fixed, uniform bond angles.
Knowing molecular geometry helps predict how a molecule interacts with other substances, influencing things like polarity and intermolecular forces.
bond angles
Bond angles play a vital role in shaping the physical and chemical properties of both simple and complex molecules. In the context of sp³d² hybridization, bond angles are particularly significant because they illustrate how the six hybrid orbitals spread out evenly to minimize repulsion between electrons. In an octahedral geometry, the bond angles are consistent at 90 degrees between any two adjacent bonds.
This angle is critical for reducing electron pair repulsions according to VSEPR (Valence Shell Electron Pair Repulsion) theory, which states that electron pairs orbiting a central atom tend to arrange themselves as far as possible to minimize repulsion. Understanding these bond angles is not only essential for predicting the behavior of individual molecules but also for determining how a molecule might interact with others in a larger system, playing a crucial role in fields ranging from material science to biochemistry.

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Most popular questions from this chapter

Hybridization of the underlined atom changes in which of the following transitions? (a) \(\mathrm{AIH}_{3}\) changes to \(\mathrm{AlH}_{4}^{-}\) (b) \(\mathrm{H}_{2} \underline{\mathrm{O}}\) changes to \(\mathrm{H}_{3} \mathrm{O}^{+}\) (c) \(\mathrm{NH}_{3}\) changes to \(\mathrm{NH}_{4}^{+}\) (d) in all cases

The correct order in which the O-O bond length increases in the following is (a) \(\mathrm{O}_{2}<\mathrm{H}_{2} \mathrm{O}_{2}<\mathrm{O}_{3}\) (b) \(\mathrm{O}_{3}<\mathrm{H}_{2} \mathrm{O}_{2}<\mathrm{O}_{2}\) (c) \(\mathrm{H}_{2} \mathrm{O}_{2}<\mathrm{O}_{2}<\mathrm{O}_{3}\) (d) \(\mathrm{O}_{2}<\mathrm{O}_{3}<\mathrm{H}_{2} \mathrm{O}_{2}\)

Species having the same bond order are (a) \(\mathrm{N}_{2}\) (b) \(\mathrm{N}_{2}^{+}\) (c) \(\mathrm{N}_{2}^{-}\) (d) \(\mathrm{N}_{2}^{2-}\)

Among the following species, identify the isostructural pairs \(\mathrm{NF}_{3}, \mathrm{NO}_{3}^{-}, \mathrm{BF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{HN}_{3}\) (a) \(\left[\mathrm{NF}_{3}, \mathrm{NO}_{3}^{-}\right]\)and \(\left[\mathrm{BF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\) (b) \(\left[\mathrm{NF}_{3}, \mathrm{HN}_{3}\right]\) and \(\left[\mathrm{NO}_{3}-\mathrm{BF}_{3}\right]\) (c) \(\left[\mathrm{NF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\)and \(\left[\mathrm{NO}_{3}-\mathrm{BF}_{3}\right]\) (d) \(\left[\mathrm{NF}_{3}, \mathrm{H}_{3} \mathrm{O}^{+}\right]\)and \(\left[\mathrm{HN}_{3}, \mathrm{BF}_{3}\right]\)

Select correct statement: (a) when a covalent bond is formed, transfer of electrons takes place (b) pure \(\mathrm{H}_{2} \mathrm{O}\) does not contain any ion (c) a bond is formed when attractive forces overcome repulsive forces (d) HF is less polar than \(\mathrm{HBr}\)

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