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Chapter 4: Problem 2

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

Short Answer

Expert verified
Hybridization changes in the transition (a).

Step by step solution

01

Understand Hybridization

Hybridization is the process of mixing atomic orbitals to form new hybrid orbitals. For example, in \(sp^3\) hybridization, one s and three p orbitals mix to form four equivalent \(sp^3\) hybrid orbitals.
02

Analyze Transition (a)

In the transition \(\mathrm{AlH}_3\) to \(\mathrm{AlH}_4^-\), aluminum (\underline{Al}) starts in an \(sp^2\) hybridization in \(\mathrm{AlH}_3\) because it forms three bonds. In \(\mathrm{AlH}_4^-\), after accepting an electron pair, it forms four bonds, changing its hybridization to \(sp^3\).
03

Analyze Transition (b)

In the transition \(\mathrm{H}_2\underline{\mathrm{O}}\) to \(\mathrm{H}_3\mathrm{O}^+\), oxygen starts in an \(sp^3\) hybridization in water because it uses one s and three p orbitals. However, adding a proton does not change hybridization because the electron arrangement allowing bonding stays \(sp^3\). This is due to the lone pairs also being in equivalent orbitals.
04

Analyze Transition (c)

In the transition \(\underline{\mathrm{N}}\mathrm{H}_3\) to \(\mathrm{NH}_4^+\), nitrogen is initially \(sp^3\) hybridized in ammonia, with one lone pair. When it forms \(\mathrm{NH}_4^+\), the lone pair is used for bonding, but the hybridization remains as \(sp^3\), since it still utilizes one s and three p orbitals.
05

Check for Consistency (d)

Review each case to check if the hybridization change occurs consistently. Transition (a) shows hybridization change, but transitions (b) and (c) do not, because hybrid orbitals remain the same despite positive ion formation in (b) and (c).

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Key Concepts

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

Atomic Orbitals
Atomic orbitals are regions around an atom's nucleus where the probability of finding an electron is the highest. These orbitals are defined by quantum mechanics and include s, p, d, and f orbitals, each with distinct shapes and energy levels. The simplest, the s orbital, is spherical. P orbitals, on the other hand, are dumbbell-shaped and oriented along the x, y, and z axes.

Each type of orbital can hold a certain number of electrons:
  • The s orbital holds up to 2 electrons.
  • P orbitals can collectively hold up to 6 electrons across their three orientations (p_x, p_y, p_z).
  • D orbitals can hold up to 10 electrons.
  • F orbitals, not usually involved in simpler bonding scenarios, can hold up to 14 electrons.
These orbitals determine how atoms bond and interact. By understanding the distribution of electrons in atomic orbitals, we grasp how atoms can combine to form different structures through chemical bonding.
sp3 Hybridization
The concept of hybridization, particularly sp3 hybridization, is central to understanding molecular shapes and bonding. Sp3 hybridization occurs when one s orbital and three p orbitals combine to create four new equivalent orbitals, termed sp3 hybrid orbitals. This mixing changes their orientation and energy to optimize bond formation.

Here's how sp3 hybridization works in simple steps:
  • The s orbital and three p orbitals around an atom merge.
  • This creates four sp3 hybrid orbitals.
  • These orbitals are spread out in a tetrahedral arrangement, maximizing distance and minimizing electron repulsion.
Practically, sp3 hybridization is observed in molecules like methane (CH4), where the central carbon forms four equivalent sigma bonds with hydrogen atoms. This results in a tetrahedral geometry, contributing to an optimal and stable molecular structure.
Bond Formation
Bond formation involves the interaction of atomic orbitals from two or more atoms to create a stable electron pair, effectively linking the atoms together. This interaction typically results in the formation of sigma (蟽) and pi (蟺) bonds.

Here's a quick overview:
  • Sigma bonds (蟽) are formed by the head-on overlap of atomic orbitals, often involving s orbitals or hybrid orbitals, offering strong and robust connectivity between atoms. They represent the "single bond" in a structure.
  • Pi bonds (蟺), on the other hand, form from the side-on overlap of p orbitals and usually accompany sigma bonds in double or triple bonds, bringing additional stability.
Understanding the mechanics of bond formation aids in grasping the geometry and reactivity of molecules. With hybridization, like sp3, four sigma bonds can connect to surround the central atom, as seen in methane, influencing every facet of molecular function and shape.
Hybrid Orbitals
Hybrid orbitals result from the mixing of different types of atomic orbitals within an atom, leading to new orbitals suited for bond formation. Unlike atomic orbitals that vary in shape and energy, hybrid orbitals are identical and directed towards bonding partners to maximize overlap and bond strength.

Here's a succinct breakdown of hybrid orbitals:
  • Formed via hybridization, such as sp, sp2, or sp3.
  • Provide the geometry and angle characteristics unique to molecule types.
  • Confer stability to molecules by maximizing electron cloud overlap in bonds.
For example, in the transition from AlH3 to AlH4-, the hybridization of aluminum changes from sp2 to sp3 as it forms an additional bond. This shift exemplifies how hybrid orbitals adapt to connect more effectively with other atoms, altering molecular shape and characteristics.

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Most popular questions from this chapter

The paramagnetic species among the following is: (a) \(\mathrm{KO}_{2}\) (b) \(\mathrm{SiO}_{2}\) (c) \(\mathrm{TiO}_{2}\) (d) \(\mathrm{BaO}_{2}\)

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

Both \(\mathrm{BF}_{3}\) and \(\mathrm{NF}_{3}\) are covalent but \(\mathrm{BF}_{3}\) molecule is non-polar while \(\mathrm{NF}_{3}\) is polar because: (a) Atomic size of boron is smaller than nitrogen (b) \(\mathrm{BF}_{3}\) is planar but \(\mathrm{NF}_{3}\) is pyramidal (c) Boron is a metal while nitrogen is gas (d) BF bond has no dipole moment while NF bond has dipole

The pair of species having identical shapes for molecules of both species is: (a) \(\mathrm{CF}_{4}, \mathrm{SF}_{4}\) (b) \(\mathrm{XeF}_{2}, \mathrm{CO}_{2}\) (c) \(\mathrm{BF}_{3}, \mathrm{PCl}_{3}\) (d) \(\mathrm{PF}_{5}, \mathrm{IF}_{5}\)

Compound \(\mathrm{X}\) is an anhydride of sulphuric acid. The number of sigma bonds and the number of pi bonds present in \(\mathrm{X}\) are respectively: (a) 3,3 (b) 4,2 (c) 2,4 (d) 4,3
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