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Write the electron configuration for the first excited state for \(\mathrm{N}_{2}\) - that is, the state with the highest-energy electron moved to the next available energy level. (a) Is the nitrogen in its first excited state diamagnetic or paramagnetic? (b) Is the \(\mathrm{N}-\mathrm{N}\) bond strength in the first excited state stronger or weaker compared to that in the ground state? Explain.

Short Answer

Expert verified
The first excited state electron configuration for Nâ‚‚ is \(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^3 1\pi_g^1\), making it paramagnetic. The N-N bond strength in the first excited state is weaker compared to the ground state, as indicated by the bond orders (2.5 in the first excited state vs. 3 in the ground state).

Step by step solution

01

Write the ground state electron configuration of Nâ‚‚

The nitrogen atom has 7 electrons. Their electron configuration in the ground state can be expressed as follows: \(1s^2 2s^2 2p^3\). In a nitrogen molecule (Nâ‚‚), there are a total of 14 electrons. The electron configuration of Nâ‚‚ in the ground state can be written as: \(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^4\)
02

Write the electron configuration for the first excited state

To determine the first excited state of Nâ‚‚, we need to move the highest-energy electron to the next available energy level. The highest-energy electron resides in the 1\(\pi_{u}\) molecular orbital. We'll move one electron from the 1\(\pi_{u}\) MO to the next available energy level, which is the 1\(\pi_{g}\) MO. This gives us the following electron configuration: \(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^3 1\pi_g^1\)
03

Determine if the first excited state is diamagnetic or paramagnetic

A molecule is diamagnetic if all its electrons are paired, and it's paramagnetic if there's at least one unpaired electron. In the first excited state of Nâ‚‚, the electron configuration shows that there are unpaired electrons in the 1\(\pi_{u}\) and 1\(\pi_{g}\) MOs (\(1\pi_u^3 1\pi_g^1\)). Therefore, the first excited state of Nâ‚‚ is paramagnetic.
04

Compare the N-N bond strength in the first excited state to the ground state

To determine the bond strength, we look at the bond order. The bond order can be calculated using the formula: \(Bond\:Order=\frac{1}{2}(number\:of\:electrons\:in\:bonding\:orbitals - number\:of\:electrons\:in\:antibonding\:orbitals)\) For the ground state electron configuration (\(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^4\)): Bond Order = \(\frac{1}{2}(10 - 4) = 3\) For the first excited state electron configuration (\(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^3 1\pi_g^1\)): Bond Order = \(\frac{1}{2}(10 - 5) = 2.5\) Since the bond order for the ground state is higher than the first excited state (3 vs. 2.5), the bond strength in the ground state is stronger compared to that in the first excited state.

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Key Concepts

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

Molecular Orbitals
Molecular orbitals are the result of atomic orbitals overlapping when atoms bond together to form molecules. These orbitals can hold electrons that belong to the entire molecule rather than just one atom. When we talk about molecular orbitals, we classify them into bonding orbitals and antibonding orbitals.

  • Bonding orbitals are lower in energy and stabilize the molecule when they are filled with electrons. They are denoted with the sigma (\(\sigma\)) or pi (\(\pi\)) symbols followed by the subscripts \(g\) for "gerade" (even) and \(u\) for "ungerade" (odd), indicating symmetric or asymmetric properties.
  • Antibonding orbitals, on the other hand, are higher in energy and tend to destabilize the molecule. Antibonding orbitals are denoted with an asterisk, like \(\sigma^*\) or \(\pi^*\).
For the nitrogen molecule (\(\text{N}_2\)), we see the filling pattern as follows in the electronic configuration: bonding orbitals like \(1\sigma_g^2\) and antibonding sequences showing partial filling in its excited state electrons like \(1\pi_g^1\). This movement to higher energy levels creates what is referred to as the excited state, where electron excitations occur within these molecular orbitals.
Paramagnetism
Paramagnetism is a type of magnetism that occurs in materials due to the presence of unpaired electrons. When such a material is placed in a magnetic field, the unpaired electrons' magnetic moments align with the field, causing the material to be attracted to the magnet.

In the context of determining whether a molecule like \(\text{N}_2\) is paramagnetic, we analyze the electron configuration. For the ground state electron configuration, all electrons are paired, making the molecule diamagnetic. However, when the molecule is in its first excited state, as electrons are promoted to the next molecular orbital, you'll find unpaired electrons.

The excited electron configuration of \(\text{N}_2\) results in unpaired electrons in \(1\pi_u^3 1\pi_g^1\), indicating that the molecule will respond to a magnetic field and has paramagnetic properties because it has unpaired electrons. This change in the electron configuration from paired in the ground state to unpaired in the excited state is a classic example of how molecular orbital arrangements influence magnetic properties.
Bond Order
Bond order provides a numerical indication of the strength and stability of a bond between two atoms in a molecule. It is calculated as one-half the difference between the number of electrons in bonding orbitals and antibonding orbitals.

The formula for bond order is:\[ Bond\:Order = \frac{1}{2}(\text{number of electrons in bonding orbitals} - \text{number of electrons in antibonding orbitals}) \]In \(\text{N}_2\)'s ground state with an electronic configuration of \(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^4\), the bond order is 3. This indicates a robust triple bond between the nitrogen atoms. In its first excited state, however, with the configuration \(1\sigma_g^2 1\sigma_u^2 2\sigma_g^2 2\sigma_u^2 3\sigma_g^2 1\pi_u^3 1\pi_g^1\), the bond order drops to 2.5.

This decrease in bond order from 3 in the ground state to 2.5 in the excited state means that the bond is weaker when the molecule is excited. A lower bond order indicates that fewer electrons contribute to bonding, hence a diminished bond strength. Thus, bond order is a crucial concept in understanding molecular stability and electron configuration influence on chemical bonding.

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Most popular questions from this chapter

Propylene, \(\mathrm{C}_{3} \mathrm{H}_{6},\) is a gas that is used to form the important polymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylene molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the \(2 s\) orbital on each atom, (b) the \(2 p_{z}\) orbital on each atom (assume both atoms are on the \(z\) -axis), \((\mathrm{c})\) the \(2 s\) orbital on one atom and the \(2 p_{z}\) orbital on the other atom.

(a) If you combine two atomic orbitals on two different atoms to make a new orbital, is this a hybrid orbital or a molecular orbital? (b) If you combine two atomic orbitals on one atom to make a new orbital, is this a hybrid orbital or a molecular orbital? (c) Does the Pauli exclusion principle (Section 6.7\()\) apply to MOs? Explain.

(a) What is the probability of finding an electron on the internuclear axis if the electron occupies a \(\pi\) molecular orbital? (b) For a homonuclear diatomic molecule, what similarities and differences are there between the \(\pi_{2 p}\) MO made from the \(2 p_{x}\) atomic orbitals and the \(\pi_{2 p}\) MO made from the \(2 p_{y}\) atomic orbitals? (c) How do the \(\pi_{2 p}^{*}\) MOs formed from the \(2 p_{x}\) and \(2 p_{y}\) atomic orbitals differ from the \(\pi_{2 p}\) MOs in terms of energies and electron distributions?

How would you expect the extent of overlap of the bonding atomic orbitals to vary in the series IF, ICl, IBr, and \(I_{2}\) ? Explain your answer.

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