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Identify the hybrid or atomic orbitals that form the bonds and hold lone pairs in any specific molecule. Identify the hybrid orbitals on the central atom that form the bonds in the following species. (a) \(\mathrm{CF}_{4}\) (b) \(\mathrm{SbCl}_{6}^{-}\) (c) \(\mathrm{AsF}_{5}\) (d) \(\mathrm{SiH}_{4}\) (e) \(\mathrm{NH}_{4}^{+}\)

Short Answer

Expert verified
CF鈧, SiH鈧, NH鈧勨伜: sp鲁; SbCl鈧嗏伝: sp鲁d虏; AsF鈧: sp鲁d.

Step by step solution

01

Determine the Central Atom and Its Steric Number for CF鈧

In the molecule CF鈧, carbon (C) is the central atom. Carbon has four bonds and no lone pairs, giving it a steric number of 4 (4 bonds + 0 lone pairs).
02

Identify the Hybridization for CF鈧

A steric number of 4 corresponds to the hybridization of sp鲁. Thus, in CF鈧, carbon uses sp鲁 hybrid orbitals to form the four C-F bonds.
03

Determine the Central Atom and Its Steric Number for SbCl鈧嗏伝

In the molecule SbCl鈧嗏伝, antimony (Sb) is the central atom. It forms six bonds with chlorine atoms and has no lone pairs, yielding a steric number of 6 (6 bonds + 0 lone pairs).
04

Identify the Hybridization for SbCl鈧嗏伝

A steric number of 6 indicates sp鲁d虏 hybridization. Therefore, in SbCl鈧嗏伝, antimony uses sp鲁d虏 hybrid orbitals for the six Sb-Cl bonds.
05

Determine the Central Atom and Its Steric Number for AsF鈧

In AsF鈧, arsenic (As) is the central atom. Arsenic forms five bonds with fluorine atoms and has no lone pairs, giving a steric number of 5 (5 bonds + 0 lone pairs).
06

Identify the Hybridization for AsF鈧

A steric number of 5 corresponds to the hybridization of sp鲁d. Thus, in AsF鈧, arsenic uses sp鲁d hybrid orbitals to form the five As-F bonds.
07

Determine the Central Atom and Its Steric Number for SiH鈧

In SiH鈧, silicon (Si) serves as the central atom. Silicon has four bonds with hydrogen atoms and carries no lone pairs, giving it a steric number of 4 (4 bonds + 0 lone pairs).
08

Identify the Hybridization for SiH鈧

A steric number of 4 indicates sp鲁 hybridization. Therefore, in SiH鈧, silicon uses sp鲁 hybrid orbitals for the four Si-H bonds.
09

Determine the Central Atom and Its Steric Number for NH鈧勨伜

In NH鈧勨伜, nitrogen (N) is the central atom. Nitrogen forms four bonds with hydrogen atoms and has no lone pairs, resulting in a steric number of 4 (4 bonds + 0 lone pairs).
10

Identify the Hybridization for NH鈧勨伜

A steric number of 4 indicates sp鲁 hybridization. Thus, in NH鈧勨伜, nitrogen uses sp鲁 hybrid orbitals to form the four N-H bonds.

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Key Concepts

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

Steric Number
The steric number is a key concept in understanding molecular geometry and hybridization. It is calculated by adding the number of atoms bonded to the central atom and the number of lone pairs on the central atom. This number helps predict the hybridization of the central atom, which in turn affects the molecule's shape.

Here's a simple way to find the steric number:
  • Count the number of sigma bonds the central atom is forming.
  • Add any lone pairs of electrons present on the central atom.
For example, in \(\mathrm{CF}_{4},\) the carbon atom forms four sigma bonds with fluorine atoms and has no lone pairs. Thus, the steric number is 4. This number suggests that carbon is using sp鲁 hybridization to form these bonds. The steric number is a handy tool to predict and understand a molecule's electron arrangement.
Central Atom
The central atom in a molecule plays a pivotal role in determining the molecular structure as it is typically the atom to which other atoms are attached. Choosing the right central atom is important for analyzing a molecule's structure.

Here's how to identify the central atom:
  • It is usually the atom with the lowest subscript in the formula.
  • It is often the least electronegative atom, except for hydrogen, which is never a central atom.
For instance, in the molecule \(\mathrm{SbCl}_{6}^{-},\) antimony (Sb) is chosen as the central atom. Understanding the central atom allows us to calculate the steric number, essential for determining the molecule's shape and hybridization.
Bond Formation
Bonds are formed when atoms share or transfer electrons to achieve a full set of valence electrons, getting closer to the noble gas configuration. In terms of hybridization, bond formation involves the mixing of atomic orbitals to create new hybrid orbitals, which are used to form sigma bonds.

Consider the molecule \(\mathrm{AsF}_{5}.\) Arsenic forms five covalent bonds with fluorine atoms. For this, it uses sp鲁d hybrid orbitals. Hybridization helps explain the strength and directionality of these bonds, providing a deeper insight into molecular geometry. Understanding how hybrid orbitals contribute to bond formation is crucial for grasping molecular structure and behavior.
Molecular Structure
The molecular structure refers to the three-dimensional arrangement of atoms within a molecule. Determining this structure involves hybridization and the steric number, giving insight into how a molecule occupies space.

For instance, in the molecule \(\mathrm{SiH}_{4},\) with silicon as the central atom and sp鲁 hybridization, the molecular structure is tetrahedral. This means that the hydrogen atoms are as far apart as possible from each other in a symmetric shape, minimizing repulsion.
  • This dispersed arrangement facilitates stability.
  • Understanding the molecular structure allows for predicting physical and chemical properties.
Grasping molecular shape can aid in comprehending a molecule's interaction with its environment, predicting reactions, and understanding function in larger chemical systems.

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Most popular questions from this chapter

A 1.30-g sample of \(\mathrm{C}_{2} \mathrm{H}_{2}\) reacts with exactly \(1.22 \mathrm{~L} \mathrm{H}_{2}\) gas at \(27^{\circ} \mathrm{C}\) and 1.01 atm of pressure to yield a compound with the formula \(\mathrm{C}_{2} \mathrm{H}_{x}\). What is the value of \(x\), and what are the orbitals on the carbon atoms that form the \(\mathrm{C}-\mathrm{C}\) bond \((\mathrm{s}) ?\)

A Recently, the structure of an amine compound, \(\mathrm{NR}_{3}\) \((\mathrm{R}=\) large organic group), has been determined to have C-N-C bond angles of 119.2 degrees. It is believed that the bond angles of about 109 degrees expected from the VSEPR model are not observed because of the large substituents bonded to the nitrogen atom. Given this large bond angle, what type of orbital on the nitrogen atom makes the \(\mathrm{N}-\mathrm{C} \sigma\) bonds, and in what type of orbital is the lone pair located?

Identify the hybrid orbitals on the carbon atoms that form the \(\sigma\) bonds in the following species. (a) \(\mathrm{CO}_{3}^{2-}\) (b) \(\mathrm{CH}_{2} \mathrm{~F}_{2}\) (c) \(\mathrm{H}_{2} \mathrm{CO}\)

Give the bonded-atom lone-pair arrangement expected for a central atom that has (a) two bonded atoms and one lone pair. (b) three bonded atoms and two lone pairs. (c) four bonded atoms and two lone pairs. (d) five bonded atoms and one lone pair.

Following are drawings of two derivatives of acetylene. Indicate whether each is polar or nonpolar, and explain your answer. (a) \(\mathrm{F}-\mathrm{C} \equiv \mathrm{C}-\mathrm{F}\) (b) \(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{F}\)

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