Problem 51
Question
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?
Step-by-Step Solution
Verified Answer
Paramagnetic: B2, O2. Bond order 1: Li2, B2, F2. Bond order 2: C2, O2. Highest bond order: N2.
1Step 1: 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.
2Step 2: 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 \( \mathrm{O}_2 \).
3Step 3: 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 = \( \frac{1}{2}(8-8) = 0 \).
4Step 4: 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 \(3\).
Key Concepts
ParamagnetismBond OrderDiatomic MoleculesSecond-Period Elements
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.
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.
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.
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.
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.
Other exercises in this chapter
Problem 49
The simple valence bond picture of \(\mathrm{O}_{2}\) does not agree with the molecular orbital view. Compare these two theories with regard to the peroxide ion
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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
View solution Problem 52
Which of the following molecules or ions are paramagnetic? What is the highest occupied molecular orbital (HOMO) in each one? Assume the molecular orbital diagr
View solution Problem 56
The elements of the second period from boron to oxygen form compounds of the type \(\mathrm{X}_{n} \mathrm{E}-\mathrm{EX}_{m}\), where X can be H or a halogen.
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