/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 151 In \(\mathrm{O}_{2}\) molecule w... [FREE SOLUTION] | 91Ó°ÊÓ

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In \(\mathrm{O}_{2}\) molecule which of the following molecular orbitals are lower in energy than \(\sigma 2 \mathrm{p}_{\mathrm{z}}\) orbital (a) \(\sigma 2 \mathrm{~s}\) (b) \(\pi_{2 \mathrm{~m} \mathrm{x}}\) (c) \(\pi_{2 \mathrm{py}}\) (d) \(\sigma^{*} 2 \mathrm{~s}\)

Short Answer

Expert verified
(a) \(\sigma 2s\) and (d) \(\sigma^* 2s\) are lower in energy than \(\sigma 2p_z\).

Step by step solution

01

Understanding Molecular Orbital Theory

Molecular orbital (MO) theory explains the distribution of electrons in a molecule through molecular orbitals that result from the combination of atomic orbitals. These molecular orbitals can be bonding, non-bonding, or antibonding.
02

Identifying the Molecular Orbitals of Oxygen

The order of molecular orbitals for molecules like \(\mathrm{O}_2\) generally follows the pattern: \(\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z\). This order shows which orbitals are lower or higher in energy.
03

Comparing Energies of Orbitals with \(\sigma 2p_z\)

Given the general order, we see \(\sigma 2s\) and \(\sigma^* 2s\) are lower in energy than \(\sigma 2p_z\). In this order, \(\pi 2p_x\) and \(\pi 2p_y\) are normally higher in energy than \(\sigma 2p_z\).
04

Conclusion

Based on the order of the molecular orbitals, the options lower in energy than \(\sigma 2p_z\) are:\(\sigma 2s\) and \(\sigma^* 2s\). Thus, choices (a) \(\sigma 2s\) and (d) \(\sigma^* 2s\) are correct.

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Key Concepts

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

Oxygen molecule bonding
To understand the oxygen molecule, or \(\mathrm{O}_2\), let's explore how its atoms bond. In simple terms, bonding in an \(\mathrm{O}_2\) molecule involves the sharing of electrons between two oxygen atoms. Each of these atoms contributes electrons from their outermost energy level, which enables them to form a stronger and more stable unit.
In the context of Molecular Orbital (MO) Theory, the combination of atomic orbitals of these two oxygen atoms forms molecular orbitals. These are spread over both atoms rather than being localized like in valence bond theory. This means electrons are not just confined to one atom but are in a shared orbital space.
The \(\mathrm{O}_2\) molecule is a stable diatomic molecule where the oxygen atoms form a double bond. This bonding double bond comprises one sigma (\(\sigma\)) bond and one pi (\(\pi\)) bond, both derived from the overlap of atomic orbitals. Such a bond nature helps in decreasing the overall energy of the resulting molecule, contributing to its stability.
Molecular orbital energy order
The order of molecular orbital energy plays a crucial role in understanding the bonding structure in molecules like \(\mathrm{O}_2\). Molecular orbitals formed by the overlap of atomic orbitals are organized by energy levels.
In the case of \(\mathrm{O}_2\), the molecular orbitals are arranged in the following sequence of increasing energy levels:
  • \(\sigma 1s\)
  • \(\sigma^* 1s\)
  • \(\sigma 2s\)
  • \(\sigma^* 2s\)
  • \(\sigma 2p_z\)
  • \(\pi 2p_x = \pi 2p_y\)
  • \(\pi^* 2p_x = \pi^* 2p_y\)
  • \(\sigma^* 2p_z\)
These inclusions of bonding and antibonding orbitals determine the structure and characteristics of the oxygen molecule.
Noticeably, the bonding orbitals \(\sigma 2s\) and \(\sigma 2p_z\) are lower in energy than their respective antibonding orbitals, \(\sigma^* 2s\) and \(\sigma^* 2p_z\). Understanding this order helps predict which orbitals electrons prefer to occur naturally, which indeed informs chemical behavior of the molecule.
Bonding and antibonding orbitals
Bonding orbitals and antibonding orbitals are fundamental concepts in molecular orbital theory. They dictate how atoms interact within a molecule.
**Bonding Orbitals** - These orbitals form when atomic orbitals combine constructively, meaning they overlap in phase.
- Bonding orbitals have lower energy than the original atomic orbitals, which stabilizes the molecule.
- In \(\mathrm{O}_2\), bonding orbitals are like \(\sigma 2s\) and \(\sigma 2p_z\), contributing to the molecule’s stability.
**Antibonding Orbitals** - These appear when atomic orbitals combine destructively, meaning they overlap out of phase with a node between them.
- Antibonding orbitals have higher energy than the original atomic orbitals and are denoted with an asterisk like \(\sigma^*\) and \(\pi^*\).
- For \(\mathrm{O}_2\), \(\sigma^* 2p_z\) and \(\pi^* 2p_x = \pi^* 2p_y\) are the antibonding orbitals, making them less favorable for electrons compared to bonding orbitals in terms of molecule stability.
Understanding these concepts clarifies why certain orbitals are lower or higher in energy, influencing the chemical properties of \(\mathrm{O}_2\) and many other molecules.

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Most popular questions from this chapter

Consider the following type of energies and decide the correct order of magnitude. 1\. covalent single bond energy 2\. average translational kinetic energy of gases at room temperature 3\. rotational barrier energy in ethane between eclipsed and staggered forms 4\. ionization energy of hydrogen atom (a) \(4>1>3>2\) (b) \(4>1>2>3\) (c) \(2>1>4>3\) (d) \(4>3>2>1\)

The nodal plane in the \(\pi\) bond of ethene is located in (a) the molecular plane. (b) a plane parallel to the molecular plane. (c) a plane perpendicular to the molecular plane which bisects. The carbon- carbon \(\sigma\) bond at right angle. (d) a plane perpendicular to the molecular plane which contains the carbon- carbon bond.

In which of the following ionization processes, the bond order has increased and the magnetic behaviour has changed? (a) \(\mathrm{NO} \rightarrow \mathrm{NO}^{+}\) (b) \(\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}^{+}\) (c) \(\mathrm{N}_{2} \rightarrow \mathrm{N}_{2}^{+}\) (d) \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}^{+}\)

Match the following \begin{tabular}{ll} \hline Column-I & Column-II \\ \hline (a) \(\mathrm{KHF}_{2}\) & (p) Compound can exist \\ (b) \(\mathrm{KHCl}_{2}\) & (q) Compound is electrolyte in nature \\ (c) \(\mathrm{NF}_{3}\) & (r) Compound shows hydrogen \\ (d) \(\mathrm{PH}_{5}\) & (s) Compound is non electrolyte \\ & (t) Compound can not exist. \\ \hline \end{tabular}

Identify the correct sequence of increasing number of \(\pi\) bonds in the structures of the following molecules 1\. \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{6}\) 2\. \(\mathrm{H}_{2} \mathrm{SO}_{3}\) 3\. \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{5}\) (a) \(1,2,3\) (b) \(2,3,1\) (c) \(2,1,3\) (d) \(1,3,2\)

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