Question

Give the formula of an ion or molecule in which an atom of (a) \(\mathrm{N}\) forms three bonds using \(\mathrm{sp}^{3}\) hybrid orbitals. (b) \(\mathrm{N}\) forms two pi bonds and one sigma bond. (c) O forms one sigma and one pi bond. (d) C forms four bonds in three of which it uses \(s p^{2}\) hybrid orbitals. (e) Xe forms two bonds using \(\mathrm{sp}^{3} \mathrm{~d}^{2}\) hybrid orbitals.

Short answer

Question: Identify the molecular formulas of the molecules based on their bonding and hybridization. a) N forms three bonds using sp³ hybrid orbitals. b) N forms two pi bonds and one sigma bond. c) O forms one sigma and one pi bond. d) C forms four bonds in three of which it uses sp² hybrid orbitals. e) Xe forms two bonds using sp³d² hybrid orbitals. Answer: a) NH₃ (ammonia) b) CH₂N₂ (diazomethane) c) CO (carbon monoxide) d) CH₂CHCl (vinyl chloride) e) XeF₆ (xenon hexafluoride)

Step-by-step solution

In this case, the nitrogen atom (\(\mathrm{N}\)) is forming three bonds using \(\mathrm{sp^3}\) hybrid orbitals. \(\mathrm{sp}^3\) hybridization occurs when one \(s\) and three \(p\) orbitals combine, forming a total of four \(\mathrm{sp}^3\) hybrid orbitals. #Step 2: Determine the bonding and molecular formula#

Since the nitrogen atom has 5 valence electrons, and it forms three bonds, it must share its three unpaired electrons with three other atoms. In this case, it must be forming three single bonds with hydrogen atoms (since hydrogen has one valence electron). So, the molecular formula is \(\mathrm{NH_3}\) (ammonia). #b)# N forms two pi bonds and one sigma bond. #Step 1: Identify the hybridization#

In this case, the nitrogen atom (\(\mathrm{N}\)) is forming two pi bonds and one sigma bond. This indicates \(\mathrm{sp}\) hybridization, where one \(s\) and one \(p\) orbital combine, forming two \(\mathrm{sp}\) hybrid orbitals. #Step 2: Determine the bonding and molecular formula#

One of the two pi bonds, along with the sigma bond, makes up a double bond with an adjacent carbon atom, and the remaining pi bond is shared with another carbon atom, creating a molecule called diazomethane. The molecular formula is \(\mathrm{CH_2N_2}\). #c)# O forms one sigma and one pi bond. #Step 1: Identify the hybridization#

In this case, the oxygen atom (\(\mathrm{O}\)) is forming one sigma bond and one pi bond. This indicates \(\mathrm{sp}\) hybridization, where one \(s\) and one \(p\) orbital combine, forming two \(\mathrm{sp}\) hybrid orbitals. #Step 2: Determine the bonding and molecular formula#

Since the oxygen atom forms a sigma bond and a pi bond, it must be forming a double bond with another atom. The most common occurrence of this is in a carbonyl group, where the oxygen atom forms a double bond with a carbon atom. The molecular formula, in its most simple form, is \(\mathrm{CO}\) (carbon monoxide). #d)# C forms four bonds in three of which it uses \(sp^2\) hybrid orbitals. #Step 1: Identify the hybridization#

In this case, the carbon atom (\(\mathrm{C}\)) forms four bonds, three of which involve \(\mathrm{sp^2}\) hybrid orbitals. \(\mathrm{sp}^2\) hybridization occurs when one \(s\) and two \(p\) orbitals combine, forming a total of three \(\mathrm{sp}^2\) hybrid orbitals. #Step 2: Determine the bonding and molecular formula#

Since three of the four carbon bonds use \(\mathrm{sp}^2\) orbitals, the carbon atom must form a single bond with another atom, like hydrogen, using its unhybridized p-orbital. In this case, we have a molecule called vinyl chloride, with a molecular formula of \(\mathrm{CH_2CHCl}\). #e)# Xe forms two bonds using \(\mathrm{sp}^3\mathrm{d}^2\) hybrid orbitals. #Step 1: Identify the hybridization#

In this case, the xenon atom (\(\mathrm{Xe}\)) forms two bonds using \(\mathrm{sp}^3\mathrm{d}^2\) hybrid orbitals. This hybridization involves one \(s\), three \(p\), and two \(d\) orbitals, forming a total of six \(\mathrm{sp}^3\mathrm{d}^2\) hybrid orbitals. #Step 2: Determine the bonding and molecular formula#

Since the xenon atom forms two bonds, it uses two of its six orbitals to create bonds, most commonly with fluorine atoms. In this situation, we have a molecule called xenon hexafluoride (\(\mathrm{XeF_6}\)).

Key concepts

Hybrid Orbitals

In chemistry, the concept of hybrid orbitals offers a descriptive explanation for how atoms combine their atomic orbitals to form bonds in molecules. This concept emerges from quantum mechanics and provides a basis for understanding the geometric arrangements seen in molecular structures.

When an atom forms covalent bonds, its atomic orbitals (such as the s and p orbitals) combine to create new orbitals that are identical to one another. These are known as hybrid orbitals, and they come in various forms—such as sp, sp², sp³, and sp³d²—based on which and how many atomic orbitals mix.

Significance of Hybridization

Hybridization helps explain molecular geometry. For example, the formation of four equivalent sp³ hybrid orbitals in a tetrahedral arrangement explains the structure of methane (CH₄). Likewise, sp² and sp hybridization explains the trigonal planar and linear structures, respectively.

Understanding the type of hybrid orbitals employed by an atom helps predict its bonding capacity and the molecule's overall spatial arrangement.

Bonding and Molecular Formula

The bonding and molecular formula of a substance indicate not only the types and numbers of atoms in a molecule but also provide clues to the actual shape and distribution of electrons within the molecule. For students to fully grasp the correlation between bonding and molecular formula, recognizing the valence electrons and types of bonds formed by an atom is essential.

Each element has a characteristic number of valence electrons that largely dictate its bonding behavior. Hydrogen, for example, has one valence electron and typically forms one bond. Carbon, with four valence electrons, often forms four bonds, as seen in organic compounds.

The molecular formula, such as NH₃ for ammonia, informs us that nitrogen uses its three sp³ hybrid orbitals to bond with three hydrogen atoms. Through exercises that require determining bonding types and corresponding molecular formulas, students can build a deeper understanding of molecular structures and chemical reactivity.

Sigma and Pi Bonds

Sigma (\f\(\f\)) and pi (\f\(\f\)) bonds are types of covalent bonds based on the overlap of atomic orbitals. Sigma bonds result from the end-to-end overlap of orbitals, such as one s orbital with another s or p orbital, or two p orbitals. They are characterized by their cylindrical symmetry around the bond axis and can rotate without breaking the bond, which allows for certain molecular flexibility.

Pi bonds, on the other hand, result from the side-to-side overlap of p orbitals and are found in double and triple bonds alongside sigma bonds. Pi bonds restrict rotational freedom, leading to fixed geometrical structures. In ethene (C2H4), for instance, the double bond between carbon atoms consists of one sigma and one pi bond, creating a planar structure.

Comprehending sigma and pi bonds is vital for students as it affects molecule geometry, stability, and reactivity. The creation of multiple bonds, as in the case of oxygen forming a double bond in CO (carbon monoxide), is critical in understanding complex bonding configurations and their implications in chemical reactions.