Chapter 9, Problem 114AP

Consider an ${\text{N}}_{\text{2}}$ molecule in its first excited electronic state, that is. when an electron in the highest occupied

molecular orbital is promoted to the lowest empty molecular orbital. (a) Identify the molecular orbitals involved, and sketch a diagram to show the transition. (b) Compare the bond order and bond length of ${\text{N}}_2^{+}$ with ${\text{N}}_{\text{2}}$, where the asterisk denotes the excited molecule. (c) Is ${\text{N}}_2^{+}$ diamagnetic or paramagnetic? (d) When ${\text{N}}_2^{+}$ loses its excess energy and converts to the ground state ${\text{N}}_{\text{2}}$, it emits a photon of wavelength 470 nm. which makes up part of the auroras' lights. Calculate the energy difference between these levels.

Interpretation Introduction

Interpretation:

The molecular orbitals involved in the transition of an electron in a ${\text{N}}_{\text{2}}$ molecule to its first excited state are to be identified. The bond order and bond length of ${\text{N}}_{\text{2}}$ in ground and excited states are to be compared. The magnetic nature of the excited state of ${\text{N}}_{\text{2}}$ is to be identified. The energy difference between the ground state and excited state is to be calculated.

Concept introduction:

Two atomic orbitals combine in order to form a bonding and an antibonding molecular orbital. The orbitals that lie on internuclear axis combine to form $\sigma$ (sigma) molecular orbital, and the orbitals parallel to each other combine to form $\pi$ molecular orbitals.

The molecular orbital formed by the combination of $\text{1s}$ orbital forms the bonding molecular orbital designated as ${\sigma }_{1s}$ and the antibonding molecular orbital, ${\sigma }^{\ast }{}_{1s}$. The $\text{2}s$ orbital forms the corresponding molecular orbitals.

Molecular orbital formed by the combination of $2p_x$ orbital forms a bonding molecular orbital designated as ${\sigma }_{2p_{\text{x}}}$ and an antibonding molecular orbital designated as ${\sigma }^{\ast }{}_{2p_{\text{x}}}$.

Molecular orbitals formed by combining $2p_{\text{y}}$ and $2p_{\text{z}}$ orbitals form bonding molecular orbitals designated as ${\text{π}}_{\text{2}{p}_{\text{y}}}$ and ${\text{π}}_{\text{2}{p}_{\text{z}}}$ and antibonding molecular orbitals designated as ${\text{π}}^{\ast }{}_{\text{2}{p}_{\text{y}}}$ and ${\text{π}}^{\ast }{}_{\text{2}{p}_{\text{z}}}$.

Electrons are filled in the molecular orbitals in increasing order of energy.

Bond order is determined by subtracting the number of electrons in antibonding orbitals from the number of electrons in bonding orbitals and dividing by two.

Higher the bond order, smaller is the bond length.

Paramagnetic substances possess net spin and a diamagnetic substance has zero spin.

The energy of the photon emitted is given by the expression as follows:

$E=\frac{\text{hc}}{\lambda }$

Here, $\text{h}$ is Planck’s constant, $\text{c}$ is the speed of light, and $\lambda$ is the wavelength of the photon.

Answer to Problem 114AP

Solution:

a) One electron of ${\text{N}}_{\text{2}}$ is promoted from ${\sigma }_{2{\text{p}}_{\text{x}}}$ to ${\text{π}}^{\ast }{}_{{\text{2p}}_{\text{y}}}$ or ${\text{π}}^{\ast }{}_{{\text{2p}}_{\text{z}}}$ to attain the excited state.

b) Bond order of ${\text{N}}_{\text{2}}$ is $3$ and that of ${\text{N}}_2^{*}$ is $2$; bond length of ${\text{N}}_2^{*}$ is longer than that of ${\text{N}}_{\text{2}}$.

c) ${\text{N}}_2^{*}$ is diamagnetic.

d) $4.23\times {10}^{-19}J$

Explanation of Solution

a) The molecular orbitals involved to be identified and a diagram to show the transition.

The electronic configuration of a nitrogen atom is as follows:

$\left[\text{He}\right]\text{2}{s}^{\text{2}}\text{2}{p}^{\text{3}}$

The molecular orbital diagram for ${\text{N}}_{\text{2}}$ is as follows:

The highest occupied molecular orbital in ${\text{N}}_{\text{2}}$ is ${\sigma }_{2p_{\text{x}}}$ and the lowest unoccupied molecular orbital is ${\text{π}}^{\ast }{}_{\text{2}{p}_{\text{y}}}$ or ${\text{π}}^{\ast }{}_{\text{2}{p}_{\text{z}}}$

${\text{π}}_{\text{2}{p}_{\text{z}}}$.

Thus, promote one electron of ${\text{N}}_{\text{2}}$ from ${\sigma }_{2p_{\text{x}}}$ to ${\text{π}}^{\ast }{}_{\text{2}{p}_{\text{y}}}$ or ${\text{π}}^{\ast }{}_{\text{2}{p}_{\text{z}}}$, to attain the excited state as follows:

Explanation:

b) Comparison of bond order and bond length of ${\text{N}}_2^{*}$ with ${\text{N}}_{\text{2}}$.

In ${\text{N}}_{\text{2}}$, there are $\text{8}$ electrons in bonding molecular orbitals and $\text{2}$ electrons in antibonding orbitals.

The bond order of ${\text{N}}_{\text{2}}$ is calculated as follows:

$\begin{array}{c}\text{Bond order }=\left(\frac{\text{8}-\text{2}}{\text{2}}\right)\\ =3\end{array}$

In ${\text{N}}_2^{*}$, one electron from bonding molecular orbital ${\sigma }_{2p_x}$ is moved to antibonding molecular orbital ${\text{π}}^{\ast }{}_{\text{2}{p}_{\text{y}}}$ or ${\text{π}}^{\ast }{}_{\text{2}{p}_{\text{z}}}$. Thus, there are $5$ electrons in bonding molecular orbitals and $1$ electron in antibonding orbitals.

The bond order of ${\text{N}}_2^{*}$ is calculated as follows:

$\begin{array}{c}\text{Bond Order}=\left(\frac{\text{5}-\text{1}}{\text{2}}\right)\\ =2\end{array}$

Hence bond order of ${\text{N}}_{\text{2}}$ is $3$ and that of ${\text{N}}_2^{*}$ is $2$. Higher the bond order, smaller is the bond length.

Thus, the bond length of ${\text{N}}_2^{*}$ is longer than that of ${\text{N}}_{\text{2}}$.

Explanation:

c) ${\text{N}}_2^{*}$ is diamagnetic or paramagnetic

The excited state of ${\text{N}}_{\text{2}}$ is formed by the transition of electron. During the first transition, the electrons do not change their spin. Thus, the electrons in ${\sigma }_{2p_x}$ and ${\text{π}}^{\ast }{}_{{\text{2p}}_{\text{y}}}$ or ${\text{π}}^{\ast }{}_{{\text{2p}}_{\text{z}}}$ spin in opposite direction and the net spin is zero. So, the molecule ${\text{N}}_2^{*}$ is diamagnetic.

Explanation:

d) The energy difference between two levels, when ${\text{N}}_2^{*}$ loses energy and returns to the ground state; Wavelength of the photon emitted is $\text{470 nm}$.

The energy difference between the excited state and the ground state is equal to the energy of the photon emitted.

The energy of the photon is given by the expression as follows:

$E\text{=}\frac{\text{hc}}{\lambda }$;

Here, $\text{h}$ is Planck’s constant; $\text{c}$ is the speed of light and $\lambda$ is the wavelength of the photon.

Here, the wavelength $\left(\text{λ}\right)$ is $\text{470 nm}$;${\text{1nm=10}}^{\text{-9}}\text{m}$

Substitute $\text{6}{\text{.63×10}}^{\text{-34}}\text{J}\cdot \text{s}$ for $\text{h}$; $3.00\times {10}^8\text{m/s}$ for $\text{c}$ and $470\times {10}^{-9}m$ for $\lambda$ in the above expression and solve as

$\begin{array}{c}E=\frac{(6.63\times {10}^{-34}J\cdot \cancel{s})}{470\times {10}^{-9}\cancel{m}}\times \frac{\left(3.00\times {10}^8\right)\cancel{m}}{\cancel{s}}\\ =4.23\times {10}^{-19}J\end{array}$

Hence, the energy difference between the ground state and excited state is $4.23\times {10}^{-19}J$.