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Which charge(s) for the \(\mathrm{N}_{2}\) molecule would give a bond order of 2.5\(?\)

Short Answer

Expert verified
None of the charge configurations for the N鈧 molecule (N鈧傗伜, N鈧偮测伜, N鈧傗伝, and N鈧偮测伝) give a bond order of 2.5. The calculated bond orders are 3.5, 2, 4.5, and 4, respectively.

Step by step solution

01

Determine nitrogen's atomic orbital configuration

Nitrogen has 7 electrons, so its atomic orbital configuration is \[1s^{2}2s^{2}2p^{3}\]. Each nitrogen atom contributes 7 electrons, and in the N鈧 molecule, there will be a total of 14 electrons. Step 2: Determine the molecular orbital diagram for nitrogen
02

Construct nitrogen's molecular orbital diagram

In molecular orbital theory, atomic orbitals of the same energy level combine to form molecular orbitals. For nitrogen, we have the following molecular orbitals: - One 蟽1s and one 蟽1s* from the combination of two 1s atomic orbitals - One 蟽2s and one 蟽2s* from the combination of two 2s atomic orbitals - One 蟽2p, two 蟺2p, one 蟽2p*, and two 蟺2p* from the combination of six 2p atomic orbitals Step 3: Determine N鈧 electronic configuration
03

Fill molecular orbitals for N鈧

Filling the molecular orbitals of N鈧 with 14 electrons, the electronic configuration for the neutral N鈧 molecule is: \[蟽_{1s}^{2}蟽_{1s*}^{2}蟽_{2s}^{2}蟽_{2s*}^{2}蟽_{2p}^{2}蟺_{2p}^{4}\] Step 4: Determine different charged N鈧 electronic configurations
04

List electronic configurations for different charged N鈧 molecules

- For N鈧傗伜 (13 electrons): \[蟽_{1s}^{2}蟽_{1s*}^{2}蟽_{2s}^{2}蟽_{2s*}^{2}蟽_{2p}^{2}蟺_{2p}^{3}\] - For N鈧偮测伜 (12 electrons): \[蟽_{1s}^{2}蟽_{1s*}^{2}蟽_{2s}^{2}蟽_{2s*}^{2}蟽_{2p}^{2}蟺_{2p}^{2}\] - For N鈧傗伝 (15 electrons): \[蟽_{1s}^{2}蟽_{1s*}^{2}蟽_{2s}^{2}蟽_{2s*}^{2}蟽_{2p}^{2}蟺_{2p}^{5}\] - For N鈧偮测伝 (16 electrons): \[蟽_{1s}^{2}蟽_{1s*}^{2}蟽_{2s}^{2}蟽_{2s*}^{2}蟽_{2p}^{2}蟺_{2p}^{6}\] Step 5: Calculate the bond order for each of these configurations
05

Calculate bond order with different charged N鈧 molecules

Remember that bond order is given by the following formula: Bond Order = \(\frac{(number\:of\:bonding\:electrons - number\:of\:antibonding\:electrons)}{2}\) Let's calculate the bond order for each of the charged N鈧 molecules: - For N鈧傗伜: \(\frac{(10-3)}{2}\) = 3.5 - For N鈧偮测伜: \(\frac{(8-4)}{2}\) = 2 - For N鈧傗伝: \(\frac{(12-3)}{2}\) = 4.5 - For N鈧偮测伝: \(\frac{(12-4)}{2}\) = 4 Step 6: Identify which charge(s) gives a bond order of 2.5
06

Find the correct charged N鈧 molecules

As we can see from our bond order calculations, none of the charged N鈧 molecules has a bond order of 2.5. Therefore, there is no charge configuration for the N鈧 molecule that would give a bond order of 2.5.

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Key Concepts

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

Bond Order
Bond order is an important concept when understanding molecular stability and bond strength. It provides insight into the number of chemical bonds between two atoms. The bond order is calculated using the formula: \(\frac{\text{number of bonding electrons} - \text{number of antibonding electrons}}{2}\). This indicates that the higher the bond order, the stronger and more stable the bond between atoms. For example, a bond order of 3 implies a triple bond, while a bond order of 2.5 suggests a bond strength between a double and a triple bond. In molecular orbital theory, bond orders are used to compare the stability of different molecular ion states by examining their electronic configurations.
Nitrogen Molecule
The nitrogen molecule, \( \text{N}_{2} \), is a stable diatomic molecule held together by a triple bond. It involves a strong interaction between the two nitrogen atoms, resulting in a bond order of 3 in its neutral form. Due to its high bond order, it is chemically inert under normal conditions. The triple bond in \( \text{N}_{2} \) accounts for its low reactivity and high dissociation energy, requiring significant energy to break the bonds. This property is critical in atmospheric nitrogen, making it essential for various industrial applications like fertilizers, and demonstrating why nitrogen serves as a critical component due to its stability.
Electronic Configuration
Electronic configuration refers to the distribution of electrons in an atom or molecule. For the neutral nitrogen molecule, \( \text{N}_{2} \), its electronic configuration is given by quadratic molecular orbitals: \( \sigma_{1s}^{2} \sigma_{1s*}^{2} \sigma_{2s}^{2} \sigma_{2s*}^{2} \sigma_{2p}^{2} \pi_{2p}^{4} \). The distribution of electrons among these orbitals determines the molecule's electronic properties and bond order. Changes in electron count, such as through ionization or reduction, alter the electronic configuration, affecting properties like bond length and stability. For charged states like \( \text{N}_{2}^{+} \) or \( \text{N}_{2}^{-} \), understanding the electronic configuration helps predict changes in molecular behavior and stability.
Molecular Orbital Diagram
A molecular orbital diagram is a visual representation of the molecular orbitals of a molecule from the combination of atomic orbitals. For nitrogen molecules, the diagram illustrates the bonding (lower energy) and antibonding (higher energy) orbitals formed by the overlapping atomic orbitals. Specifically, for \( \text{N}_{2} \), atomic orbitals such as \( 1s \), \( 2s \), and \( 2p \) from each nitrogen atom combine to form sigma and pi molecular orbitals. The molecular orbital diagram not only helps in predicting the bond order but also in determining the magnetic properties and overall electron configuration of the molecule. It's a key tool in molecular orbital theory that aids in understanding how molecules are bonded and their potential physical properties.

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Most popular questions from this chapter

Sodium can react with oxygen to form sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\) which is composed of \(\mathrm{Na}^{+}\) and \(\mathrm{O}_{2} 2-\) ions. Potassium can react with oxygen to form potassium superoxide \(\left(\mathrm{KO}_{2}\right)\) which is composed of \(\mathrm{K}^{+}\) and \(\mathrm{O}_{2}-\) ions. Does the peroxide ion or the superoxide ion have the shorter bond length? Explain.

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