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Using VB theory and the Lewis structure model, determine the bond order in (a) \(\mathrm{H}_{2},(\mathrm{b}) \mathrm{Na}_{2},(\mathrm{c}) \mathrm{S}_{2},(\mathrm{d}) \mathrm{N}_{2}\) and (e) \(\mathrm{Cl}_{2}\) Is there any ambiguity with finding the bond orders by this method?

Short Answer

Expert verified
Bond orders: H₂ = 1, Na₂ = 1, S₂ = 2, N₂ = 3, Cl₂ = 1. No major ambiguity for these molecules.

Step by step solution

01

Understanding Bond Order

Bond order refers to the number of chemical bonds between a pair of atoms. The bond order can be determined using the formula: \( \text{Bond Order} = \frac{1}{2}(\text{Number of bonding electrons} - \text{Number of antibonding electrons}) \). It indicates the stability of the bond; the higher the bond order, the more stable the molecule.
02

Analyzing H₂ Molecule

\( \mathrm{H}_{2} \) consists of two hydrogen atoms each sharing one electron, forming a sigma bond. According to the Lewis structure model, there are no antibonding electrons in \( \mathrm{H}_{2} \). Thus, the bond order is \( 1 \) (single bond).
03

Analyzing Na₂ Molecule

\( \mathrm{Na}_{2} \) involves two sodium atoms. Each sodium atom contributes one electron to form the bond. Similar to the hydrogen molecule, this results in a single sigma bond with no antibonding electrons, giving a bond order of \( 1 \).
04

Analyzing S₂ Molecule

\( \mathrm{S}_{2} \) has two sulfur atoms, each with six valence electrons. In \( \mathrm{S}_{2} \), two electrons are shared in a sigma bond and two in a pi bond, resulting in a double bond. Thus, the bond order is \( 2 \).
05

Analyzing N₂ Molecule

The nitrogen molecule \( \mathrm{N}_{2} \) involves a triple bond between the two nitrogen atoms. It involves one sigma bond and two pi bonds between the nitrogen atoms. Therefore, the bond order is \( 3 \).
06

Analyzing Cl₂ Molecule

In \( \mathrm{Cl}_{2} \), each chlorine atom shares one electron to form a sigma bond, with no antibonding electrons, creating a single bond. The bond order is \( 1 \).
07

Evaluating Ambiguity in Method

The bond order determination using VB theory and Lewis structures is straightforward for the given diatomic molecules because these molecules primarily involve simple single, double, or triple bonds, which align well within the model without ambiguity. However, more complex molecules or those with resonance structures could present challenges.

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Key Concepts

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

Lewis Structure
The Lewis structure model is a visual representation of the bonding between atoms within a molecule. It uses dots to symbolize valence electrons and lines to represent bonds between atoms. For example, in a diatomic molecule like \( \mathrm{Cl}_{2} \), each chlorine atom shares one electron to form a single covalent bond, depicted as a line between two chlorine symbols. The remaining valence electrons are non-bonding and are shown as dots around the atoms.

Lewis structures help in predicting
  • the number of bonds in a molecule,
  • the arrangement of electrons around atoms, and
  • resonance structures, if present.
A bond is represented by two dots or a line, giving a visual sense of molecular geometry and stability.
Valence Bond Theory
Valence Bond (VB) theory provides a deeper understanding of how atomic orbitals overlap to form bonds. It posits that bonds are formed when atomic orbitals of two atoms overlap and their electrons pair up to create a bond. For instance, in \( \mathrm{H}_{2} \), the 1s orbitals of two hydrogen atoms overlap to form a sigma bond.

This theory
  • explains bond strength and directionality,
  • emphasizes the role of hybridization in forming equivalent bonds, and
  • offers insight into the bond angles and molecular shapes.
VB theory is key in understanding the nature of the chemical bond and its impact on molecular energy levels and reactions.
Diatomic Molecules
Diatomic molecules consist of two atoms, which can be of the same element like \( \mathrm{O}_{2} \) or of different elements like \( \mathrm{CO} \). They are often straightforward in their bonding arrangements and typically exhibit simple single, double, or triple bonds, which are easily analyzed using Lewis structures and VB theory.

In diatomic molecules,
  • bond type determines bond strength and length,
  • bond order indicates stability, and
  • electronegativity differences can lead to polarity in heteronuclear diatomics.
Studying diatomic molecules helps in grasping basic molecular bonding, serving as a stepping stone to understanding more complex compounds.
Chemical Bond Stability
Chemical bond stability refers to the resilience of a molecular bond against breaking. Bond order, determined by the number of bonds between two atoms, is a good indicator of stability. In general, higher bond orders correlate with greater bond strength and stability.

Key factors impacting bond stability include
  • bond length (shorter bonds are typically stronger),
  • bond energy (the energy required to break a bond), and
  • resonance stabilization (delocalization of electrons increases stability).
Bond stability is crucial for predicting molecular behavior and reactivity. It helps chemists design molecules with desired properties and understand biological processes.

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Most popular questions from this chapter

One member of cach of the following sets of compounds is not isoelectronic with the others. Which one in each set is the odd one out? (a) \(\left[\mathrm{NO}_{2}\right]^{+}, \mathrm{CO}_{2},\left[\mathrm{NO}_{2}\right]^{-}\) and \(\left[\mathrm{N}_{3}\right]^{-}\) (b) \([\mathrm{CN}]^{-}, \mathrm{N}_{2}, \mathrm{CO},[\mathrm{NO}]^{+}\) and \(\left[\mathrm{O}_{2}\right]^{2-}\) (c) \(\left[\mathrm{SiF}_{6}\right]^{2-},\left[\mathrm{PF}_{6}\right]^{-},\left[\mathrm{AlF}_{6}\right]^{3-}\) and \(\left[\mathrm{BrF}_{6}\right]^{-}\)

Does VB theory indicate that the diatomic molecule He \(_{2}\) is a viable species? Rationalize your answer.

Draw Lewis structures to describe the bonding in the following molecules: (a) \(\mathrm{F}_{2} ;(\mathrm{b}) \mathrm{BF}_{3} ;(\mathrm{c}) \mathrm{NH}_{3} ;(\mathrm{d}) \mathrm{H}_{2} \mathrm{Se}\) (c) \(\mathrm{H}_{2} \mathrm{O}_{2} ;(\mathrm{f}) \mathrm{BeCl}_{2} ;(\mathrm{g}) \mathrm{SiH}_{4} ;(\mathrm{h}) \mathrm{PF}_{5}\)

(a) Draw possible stereoisomers for the trigonal bipyramidal \(\left[\mathrm{SiF}_{3} \mathrm{Me}_{2}\right]^{-}\) anion \(\left(\mathrm{Me}=\mathrm{CH}_{3}\right) .\) An X-ray diffraction study of a salt of \(\left[\mathrm{SiF}_{3} \mathrm{Me}_{2}\right]^{-}\) shows that two \(\mathrm{F}\) atoms occupy axial sites. Suggest why this stereoisomer is preferred over the other possible structures that you have drawn. (b) Account for the fact that members of the series of complexes \(\left[\mathrm{PtCl}_{4}\right]^{2-},\left[\mathrm{PtCl}_{3}\left(\mathrm{PMe}_{3}\right)\right]^{-},\left[\mathrm{PtCl}_{2}\left(\mathrm{PMe}_{3}\right)_{2}\right]\) and \(\left[\mathrm{PtCl}\left(\mathrm{PMe}_{3}\right)_{3}\right]^{+}\) do not possess the same number of stereoisomers.

Pick out pairs of isoelectronic species from the following list; not all species have a 'partner': HF; \(\mathrm{CO}_{2}\); \(\mathrm{SO}_{2} ; \mathrm{NH}_{3} ; \mathrm{PF}_{3} ; \mathrm{SF}_{4} ; \mathrm{SiF}_{4} ; \mathrm{SiCl}_{4} ;\left[\mathrm{H}_{3} \mathrm{O}\right]^{+} ;\left[\mathrm{NO}_{2}\right]^{+} ;[\mathrm{OH}]^{-}\) \(\left[\mathrm{AlCl}_{4}\right]^{-}\)

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