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Concept explainers
(a)
Interpretation:
The proportion of time corresponding to the unpaired electron of NO2 which occupies 2s orbital has to be stated. The proportion of time corresponding to the unpaired electron of NO2 which occupies 2p orbital has to be stated. The total probability that the electron is found on N atoms has to be stated.
Concept introduction:
EPR stands for electronic paramagnetic resonance. It is used to study species that contain unpaired electrons. EPR can be used to study both solids and liquids. However, the study of the gas phase samples is difficult due to the free rotation of the molecules. The g− value represents the ease with which the applied field can generate currents through the molecule.
(a)
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Answer to Problem 12D.7P
The proportion of time corresponding to the unpaired electron of NO2 which occupies 2s orbital is 10.3 %_.
The proportion of time corresponding to the unpaired electron of NO2 which occupies 2p orbital is 38.2 %_.
The total probability that the electron is found on N atoms is 48.5 %_.
Explanation of Solution
The given hyperfine interaction of 2s orbital of N atom with nucleus is 55.2 mT.
The given isotropic hyperfine interaction of 2s orbital of N atom with nucleus is 5.7 mT.
So, the proportion of time corresponding to the unpaired electron of NO2 which occupies 2s orbital is calculated by the expression given below.
P(N2s)=Isotropic hyperfine interaction of 2s orbitalHyperfine interaction of 2s orbital
Substitute the values of hyperfine interaction of 2s orbital and isotropic hyperfine interaction of 2s orbital in the above expression.
P(N2s)=5.7 mT55.2 mT=0.103=0.103×100 %=10.3 %_
Therefore, the proportion of time corresponding to the unpaired electron of NO2 which occupies 2s orbital is 10.3 %_.
The given hyperfine interaction of 2p orbital of N atom with nucleus is 3.4 mT.
The given anisotropic hyperfine interaction of 2p orbital of N atom with nucleus is 1.3 mT.
So, the proportion of time corresponding to the unpaired electron of NO2 which occupies 2p orbital is calculated by the expression given below.
P(N2p)=Anisotropic hyperfine interaction of 2p orbitalHyperfine interaction of 2p orbital
Substitute the values of hyperfine interaction of 2p orbital and isotropic hyperfine interaction of 2p orbital in the above expression.
P(N2p)=1.3 mT3.4 mT=0.382=0.382×100 %=38.2 %_
Therefore, the proportion of time corresponding to the unpaired electron of NO2 which occupies 2p orbital is 38.2 %_.
The total probability that the electron is found on N atoms is calculated by adding P(N2p) and P(N2s) as follow.
P(N)=P(N2s)+P(N2p)
Substitute the values of P(N2p) and P(N2s) in the above expression
P(N)=0.103+0.382=0.485=0.485×100 %=48.5 %_
Therefore, the total probability that the electron is found on N atoms is 48.5 %_.
(b)
Interpretation:
The total probability that the electron is found on O atoms has to be stated. The hybridization ratio of N atom has to be stated. Whether the hybridization predicts that NO2 is angular or not has to be stated.
Concept introduction:
EPR stands for electronic paramagnetic resonance. It is used to study species that contain unpaired electrons. EPR can be used to study both solids and liquids. However, the study of the gas phase samples is difficult due to the free rotation of the molecules. The g− value represents the ease with which the applied field can generate currents through the molecule.
(b)
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Answer to Problem 12D.7P
The total probability that the electron is found on O atoms is 51.5 %_.
The hybridization ratio of N atom is 3.7_.
The hybridization predicts that NO2 is angular in shape.
Explanation of Solution
The total probability that the electron is found on N atoms is 0.485.
The total probability that the electron is found on O atoms is calculated by subtracting P(N) from 1 as follow.
P(O)=1−P(N)
Substitute the value of P(N) in the above expression
P(O)=1−0.485=0.515=0.515×100 %=51.5 %_
Therefore, the total probability that the electron is found on O atoms is 51.5 %_.
The expression to calculate the hybridization ratio of N atom is given below.
Hybridization ratio of N atom=P(N2p)P(N2s)
Substitute the values of P(N2p) and P(N2s) in the above expression.
Hybridization ratio of N atom=0.3820.103=3.7_
Therefore, the hybridization ratio of N atom is 3.7_.
As the shape of NO2 radical is non-linear, thus, NO2 is angular in shape.
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Chapter 12 Solutions
Atkins' Physical Chemistry
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