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Chapter 9: Problem 51

Which of the homo-nuclear, diatomic molecules of the second-period elements (from \(\mathrm{Li}_{2}\) to \(\mathrm{Ne}_{2}\) ) are paramagnetic? Which have a bond order of \(1 ?\) Which have a bond order of \(2 ?\) Which diatomic molecule has the highest bond order?

Short Answer

Expert verified
Paramagnetic: B2, O2. Bond order 1: Li2, B2, F2. Bond order 2: C2, O2. Highest bond order: N2.

Step by step solution

01

Identify the diatomic molecules and electronic configuration

Diatomic molecules formed from second-period elements include \( \mathrm{Li}_2, \mathrm{Be}_2, \mathrm{B}_2, \mathrm{C}_2, \mathrm{N}_2, \mathrm{O}_2, \mathrm{F}_2, \mathrm{Ne}_2 \). To determine the bond order and magnetism, we need to consider the molecular orbitals formed by these molecules. Each of these molecules follows a similar pattern in molecular orbital filling according to their electronic configuration.
02

Determine paramagnetic molecules

Molecules are paramagnetic if they have unpaired electrons. Upon examining the molecular orbital configurations: - \( \mathrm{B}_2 \): Has two unpaired electrons making it paramagnetic. - \( \mathrm{O}_2 \): Has two unpaired electrons making it paramagnetic. Thus, the paramagnetic molecules are \( \mathrm{B}_2 \) and \( \mathrm{O}_2 \).
03

Calculate bond orders

Bond order is calculated using the formula: \[ \text{Bond Order} = \frac{1}{2}(n_b - n_a) \]Where \(n_b\) is the number of electrons in bonding orbitals, and \(n_a\) is the number of electrons in antibonding orbitals.- For \( \mathrm{Li}_2 \), \( n_b = 2, n_a = 0 \): Bond order = \( \frac{1}{2}(2-0) = 1 \).- For \( \mathrm{Be}_2 \), \( n_b = 2, n_a = 2 \): Bond order = \( \frac{1}{2}(2-2) = 0 \).- For \( \mathrm{B}_2 \), \( n_b = 4, n_a = 2 \): Bond order = \( \frac{1}{2}(4-2) = 1 \).- For \( \mathrm{C}_2 \), \( n_b = 6, n_a = 2 \): Bond order = \( \frac{1}{2}(6-2) = 2 \).- For \( \mathrm{N}_2 \), \( n_b = 8, n_a = 2 \): Bond order = \( \frac{1}{2}(8-2) = 3 \).- For \( \mathrm{O}_2 \), \( n_b = 8, n_a = 4 \): Bond order = \( \frac{1}{2}(8-4) = 2 \).- For \( \mathrm{F}_2 \), \( n_b = 8, n_a = 6 \): Bond order = \( \frac{1}{2}(8-6) = 1 \).- For \( \mathrm{Ne}_2 \), \( n_b = 8, n_a = 8 \): Bond order = \( \frac{1}{2}(8-8) = 0 \).
04

Identify bond order categories and highest bond order

From the bond order calculations, - Bond order of \(1\): \( \mathrm{Li}_2, \mathrm{B}_2, \mathrm{F}_2 \)- Bond order of \(2\): \( \mathrm{C}_2, \mathrm{O}_2 \)- Highest bond order: \( \mathrm{N}_2 \) with bond order of \(3\).

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Key Concepts

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

Paramagnetism
Paramagnetism is a phenomenon where certain materials are attracted by an external magnetic field. This occurs because the material contains unpaired electrons. Each electron has a magnetic moment, and if there is an imbalance in the number of paired versus unpaired electrons, the material becomes paramagnetic.
In the context of diatomic molecules, we can determine if a molecule is paramagnetic by looking at its molecular orbital (MO) configuration. If there are unpaired electrons in the molecular orbitals, the molecule is paramagnetic.
  • For example, in the case of \( \mathrm{B}_2 \) and \( \mathrm{O}_2 \), these molecules show paramagnetism because they each have two unpaired electrons in the \( \pi^* \) orbitals.
  • This means they react to magnetic fields, unlike diamagnetic molecules, which are slightly repelled by magnetic fields because all their electrons are paired.
Bond Order
Bond order is an indicator of bond strength in a molecule. It is calculated using the formula: \[\text{Bond Order} = \frac{1}{2}(n_b - n_a)\]where \(n_b\) is the number of electrons in bonding orbitals, and \(n_a\) is the number of electrons in antibonding orbitals.
Bond order tells us how strong and stable a bond is between two atoms.
A higher bond order implies a stronger, more stable bond.
  • For instance, \( \mathrm{N}_2 \) has a bond order of 3, which represents a triple bond. This is the highest bond order among the second-period diatomic molecules, making nitrogen gas extremely stable.
  • Conversely, \( \mathrm{Be}_2 \) has a bond order of 0, indicating no real bond forms between the two beryllium atoms under normal conditions.
  • Molecules like \( \mathrm{O}_2 \) with a bond order of 2 and \( \mathrm{F}_2 \) with a bond order of 1 offer other stability levels.
    Higher bond orders generally result in shorter bond lengths and increased bond energies.
Diatomic Molecules
Diatomic molecules are molecules composed of only two atoms, which may or may not be the same element. When speaking about homonuclear diatomic molecules, it refers to molecules consisting of two identical atoms.
In the second-period elements, from \( \mathrm{Li}_2 \) to \( \mathrm{Ne}_2 \), we see a range of diatomic molecules with different properties. These properties include bond order and magnetism which are influenced by their electron configuration and molecular orbitals.
  • Each molecule has unique characteristics. For example, \( \mathrm{Li}_2 \) features a simple sigma bond formed from the overlapping of \(2s\) orbitals.
  • As we move to \( \mathrm{N}_2 \) and \( \mathrm{O}_2 \), the type and number of bonds become more complex, involving \(\pi\) bonding and antibonding orbitals.
  • This diversity among diatomic molecules illustrates the variety of ways atoms bond and form stable or unstable molecular structures based on electron sharing.
Second-Period Elements
Second-period elements are the elements in the second row of the periodic table: lithium (\(\mathrm{Li}\)), beryllium (\(\mathrm{Be}\)), boron (\(\mathrm{B}\)), carbon (\(\mathrm{C}\)), nitrogen (\(\mathrm{N}\)), oxygen (\(\mathrm{O}\)), fluorine (\(\mathrm{F}\)), and neon (\(\mathrm{Ne}\)). These elements are known for their variety and are fundamental to constructing a wide range of compounds.
Among these, the homo-nuclear diatomic molecules from \( \mathrm{Li}_2 \) to \( \mathrm{Ne}_2 \) are significant because they serve as straightforward models for studying bonding theories such as Molecular Orbital Theory, which helps predict properties like bond length and magnetic behavior.
  • These elements aid in understanding how electron configuration influences the chemical and physical properties of molecules.
  • Molecular orbital diagrams for these diatomic molecules help visualize the bonding interactions and determine characteristics such as bond order and magnetic properties.
  • These insights are essential for chemists and material scientists when predicting the behavior of new materials or studying reaction mechanisms.

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Most popular questions from this chapter

What is the connection between bond order, bond length, and bond energy? Use ethane \(\left(C_{2} \mathrm{H}_{6}\right),\) ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right),\) and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) as examples.

Antimony pentafluoride reacts with HF according to the equation $$2\mathrm{HF}+\mathrm{SbF}_{3}\rightarrow\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}$$ (a) What is the hybridization of the Sb atom in the reactant and product? (b) Draw a Lewis structure for \(\mathrm{H}_{2} \mathrm{F}^{+}\). What is the geometry of \(\mathrm{H}_{2} \mathrm{F}^{+}\) ? What is the hybridization of \(\mathrm{F}\) in \(\mathrm{H}_{2} \mathrm{F}^{+} ?\)

Carbon dioxide \(\left(\mathrm{CO}_{2}\right),\) dinitrogen monoxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) the azide ion \(\left(\mathrm{N}_{3}^{-}\right),\) and the cyanate ion (OCN^-) have the same arrangement of atoms and the same number of valence shell electrons. However, there are significant differences in their electronic structures. (a) What hybridization is assigned to the central atom in each species? Which orbitals overlap to form the bonds between atoms in each structure. (b) Evaluate the resonance structures of these four species. Which most closely describe the bonding in these species? Comment on the differences in bond lengths and bond orders that you expect to see based on the resonance structures.

Nitrogen, \(\mathbf{N}_{2}\), can ionize to form \(\mathbf{N}_{2}^{+}\) or add an electron to give \(\mathrm{N}_{2}\). . Using molecular orbital theory, compare these species with regard to (a) their magnetic character, (b) net number of \(\pi\) bonds, (c) bond order, (d) bond length, and (e) bond strength.

Draw the Lewis structures of HSOs \(\mathrm{F}\) and \(\mathrm{SO}_{3} \mathrm{F}^{-}\). What is the molecular geometry and hybridization for the sulfur atom in each species? ( \(\mathrm{H}\) is bonded to an \(\mathrm{O}\) atom in the acid.)
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