Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 9, Problem 114AP

Consider an N 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 N 2 + with N 2 , where the asterisk denotes the excited molecule. (c) Is N 2 + diamagnetic or paramagnetic? (d) When N 2 + loses its excess energy and converts to the ground state N 2 , it emits a photon of wavelength 470 nm. which makes up part of the auroras' lights. Calculate the energy difference between these levels.

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The molecular orbitals involved in the transition of an electron in a N2 molecule to its first excited state are to be identified. The bond order and bond length of N2 in ground and excited states are to be compared. The magnetic nature of the excited state of N2 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 σ (sigma) molecular orbital, and the orbitals parallel to each other combine to form π molecular orbitals.

The molecular orbital formed by the combination of 1s orbital forms the bonding molecular orbital designated as σ1s and the antibonding molecular orbital, σ1s. The 2s orbital forms the corresponding molecular orbitals.

Molecular orbital formed by the combination of 2px orbital forms a bonding molecular orbital designated as σ2px and an antibonding molecular orbital designated as σ2px.

Molecular orbitals formed by combining 2py and 2pz orbitals form bonding molecular orbitals designated as π2py and π2pz and antibonding molecular orbitals designated as π2py and π2pz.

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=hcλ

Here, h is Planck’s constant, c is the speed of light, and λ is the wavelength of the photon.

Answer to Problem 114AP

Solution:

a) One electron of N2 is promoted from σ2px to π2py or π2pz to attain the excited state.

Chemistry, Chapter 9, Problem 114AP , additional homework tip  1

b) Bond order of N2 is 3 and that of N2* is 2; bond length of N2* is longer than that of N2.

c) N2* is diamagnetic.

d) 4.23×1019J

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:

[ He ]2s22p3

The molecular orbital diagram for N2 is as follows:

Chemistry, Chapter 9, Problem 114AP , additional homework tip  2

The highest occupied molecular orbital in N2 is σ2px and the lowest unoccupied molecular orbital is π2py or π2pz

π2pz.

Thus, promote one electron of N2 from σ2px to π2py or π2pz, to attain the excited state as follows:

Chemistry, Chapter 9, Problem 114AP , additional homework tip  3

Explanation:

b) Comparison of bond order and bond length of N2* with N2.

In N2, there are 8 electrons in bonding molecular orbitals and 2 electrons in antibonding orbitals.

The bond order of N2 is calculated as follows:

Bond order =(822)=3

In N2*, one electron from bonding molecular orbital σ2px is moved to antibonding molecular orbital π2py or π2pz. Thus, there are 5 electrons in bonding molecular orbitals and 1 electron in antibonding orbitals.

The bond order of N2* is calculated as follows:

Bond Order=(512)=2

Hence bond order of N2 is 3 and that of N2* is 2. Higher the bond order, smaller is the bond length.

Thus, the bond length of N2* is longer than that of N2.

Explanation:

c) N2* is diamagnetic or paramagnetic

The excited state of N2 is formed by the transition of electron. During the first transition, the electrons do not change their spin. Thus, the electrons in σ2px and π2py or π2pz spin in opposite direction and the net spin is zero. So, the molecule N2* is diamagnetic.

Explanation:

d) The energy difference between two levels, when N2* loses energy and returns to the ground state; Wavelength of the photon emitted is 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=hcλ;

Here, h is Planck’s constant; c is the speed of light and λ is the wavelength of the photon.

Here, the wavelength (λ) is 470 nm;1nm=10-9m

Substitute 6.63×10-34Js for h; 3.00×108m/s for c and 470×109m for λ in the above expression and solve as

E=(6.63×1034Js)470×109m×(3.00×108)ms=4.23×1019J

Hence, the energy difference between the ground state and excited state is 4.23×1019 J.

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