Chapter 7, Problem 7.136AP

 (a)

Interpretation Introduction

Interpretation: The analysis of spectra of emission sources within the Milky Way by Far Ultraviolet Explorer satellite and the loss of five electrons by Oxygen atoms present in interplanetary clouds is given. Various questions based on this analysis have to be answered.

Concept introduction: Electronic configuration determines the electron distribution in an atom or a molecule. The entire periodic table is based upon the distribution of electrons in the atom.

To determine: The electronic configuration of highly ionized Oxygen atom.

Answer to Problem 7.136AP

Solution

The electronic configuration of highly ionized Oxygen atom is given as,

${\text{O}}^{+}:1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^3$

${\text{O}}^{2+}:1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^2$

` ${\text{O}}^{3+}:1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^1$

${\text{O}}^{4+}:1{\text{s}}^{2}2{\text{s}}^2$

${\text{O}}^{5+}:1{\text{s}}^{2}2{\text{s}}^{\text{1}}$

Explanation of Solution

Explanation

The atomic number of $\text{O}$ is $8$. It means there are eight electrons in the neutral Oxygen atom. The general electronic configuration of the atom is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^4$. Now the electronic configuration of various cationic species formed is given as,

${\text{O}}^{+}:1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^3$

${\text{O}}^{2+}:1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^2$

` ${\text{O}}^{3+}:1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^1$

${\text{O}}^{4+}:1{\text{s}}^{2}2{\text{s}}^2$

${\text{O}}^{5+}:1{\text{s}}^{2}2{\text{s}}^{\text{1}}$

With increase in positive charge the ionization energy keeps on increasing. Therefore, the sufficient amount of energy has to be supplied for the removal of electron.

(b)

Interpretation Introduction

Interpretation: The analysis of spectra of emission sources within the Milky Way by Far Ultraviolet Explorer satellite and the loss of five electrons by Oxygen atoms present in interplanetary clouds is given. Various questions based on this analysis have to be answered.

Concept introduction: Electronic configuration determines the electron distribution in an atom or a molecule. The entire periodic table is based upon the distribution of electrons in the atom.

To determine: The type of electrons that have been removed from the neutral atom.

Answer to Problem 7.136AP

Solution

The $2\text{p}$ electron has been removed from the neutral atom.

Explanation of Solution

Explanation

The general electronic configuration of the Oxygen atom is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^4$. It means the $2\text{p}$ electron has been removed from the neutral atom as the first electron that has to be removed for the first cationic species is present in $2\text{p}$ orbital.

(c)

Interpretation Introduction

Interpretation: The analysis of spectra of emission sources within the Milky Way by Far Ultraviolet Explorer satellite and the loss of five electrons by Oxygen atoms present in interplanetary clouds is given. Various questions based on this analysis have to be answered.

Concept introduction: Electronic configuration determines the electron distribution in an atom or a molecule. The entire periodic table is based upon the distribution of electrons in the atom.

To determine: The reason behind the requirement of high energy in the removal of additional electron from the previous electron.

Answer to Problem 7.136AP

Solution

Due to the increased influence of positive charge over the remaining electrons the removal of electron becomes difficult and hence the ionization energy also increases.

Explanation of Solution

Explanation

As the electron is removed from the neutral atom though the number of electron decreases but the number of protons remains the same. Now the hold of same positive charge over the less negative charge also increases and the outermost shell also gets closer to the nucleus. So, with removal of every single electron, the influence of positive charge over the remaining electrons increases and the removal of electrons also become difficult. This is the reason that with increase in positive charge over any species the ionization energy also increases.

(d)

Interpretation Introduction

Interpretation: The analysis of spectra of emission sources within the Milky Way by Far Ultraviolet Explorer satellite and the loss of five electrons by Oxygen atoms present in interplanetary clouds is given. Various questions based on this analysis have to be answered.

Concept introduction: Electronic configuration determines the electron distribution in an atom or a molecule. The entire periodic table is based upon the distribution of electrons in the atom.

To determine: The wavelength of the radiation that is used for the removal of fifth electron from an $\text{O}$ atom.

Answer to Problem 7.136AP

Solution

The wavelength of the radiation is $\underset{\_}{12.42\times 10^{-9}m}$

Explanation of Solution

Explanation

Given

The fifth ionization energy is $9391\text{kJ/mol}$.

The conversion of $\text{kJ}$ into $\text{J}$ is given as,

$1\text{kJ}=1000\text{J}$

Hence the conversion of $9391\text{kJ}$ into $\text{J}$ is given as,

$\begin{array}{c}9391\text{kJ}=9391\times 1000\text{J}\\ =\text{9391000}\text{J}\end{array}$

The energy of $6.022\times {10}^{23}$ atoms $=\text{9391000}\text{J}$

The energy of $1$ atom $\begin{array}{c}=\frac{\text{9391000}}{6.022\times {10}^{23}}\text{J}\\ \text{=1}\text{.6}\times {\text{10}}^{-17}\text{J}\end{array}$

The relation between energy, wavelength and speed of light is given as,

$\Delta E=h\frac{c}{\lambda }$

Where,

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{Js}\right)$.
• $\lambda$ is the wavelength of the light.
• $c$ is the speed of the light $\left(3\times {10}^8\text{m}/\text{s}\right)$.
• $\Delta E$ is the energy difference.

Substitute the value of $h$, $c$ and $\Delta E$ in the above equation as,

$\begin{array}{c}\Delta E=h\frac{c}{\lambda }\\ 1.6\times {10}^{-17}\text{J}=\text{6}\text{.626}\times {\text{10}}^{-34}\text{Js}\times \frac{3\times {10}^8\frac{\text{m}}{\text{s}}}{\lambda }\\ 1.6\times {10}^{-17}\text{J}=\frac{19.878\times {10}^{-26}\text{J}\text{m}}{\lambda }\\ \lambda =\underset{\_}{12.42\times 10^{-9}m}\end{array}$

The wavelength of the radiation is $\underset{\_}{12.42\times 10^{-9}m}$.