Chapter 8, Problem 8.99QE

Arrange the elements lithium, carbon, and oxygen in order of

1. (a) increasing size.
2. (b) increasing first ionization energy.
3. (c) increasing second ionization energy.
4. (d) number of unpaired electrons.

Interpretation Introduction

Interpretation:

The elements lithium, carbon, and oxygen have to be arranged in the order of increasing size.

Concept Introduction:

Atomic size is half of the distance between the adjacent atoms in the molecule. Since the electron cloud has no fixed boundary so the size of the atom is measured with high difficulty.

Answer to Problem 8.99QE

The order of the increasing size is as follows:

  $\text{O}<\text{C}<\text{Li}$

Explanation of Solution

Lithium, carbon, and oxygen are present in the same period of the periodic table. $\text{Li}$ lies to the left side of both $\text{C}$ and $\text{O}$. The electrons are added in the same shell but the number of protons increases from left to right in a period. So the effective nuclear charge increases from left to right that results in the decrease in atomic size from lithium to carbon to oxygen.

The order of the increasing size is as follows:

  $\text{O}<\text{C}<\text{Li}$

Interpretation Introduction

Interpretation:

The elements lithium, carbon, and oxygen have to be arranged in the order of increasing first ionization energy.

Concept Introduction:

The energy that is needed to remove an electron from the isolated, neutral gaseous atom is known as ionization energy. It is represented by $I$. The energy required to remove the first electron from the neutral atom is called the first ionization energy, denoted by $I_1$. The energy required to remove the second electron from the monovalent anion is called the second ionization energy, denoted by $I_2$ and so on. Energy is always supplied to the atom so as to remove an electron from it so ionization energies have positive values.

Answer to Problem 8.99QE

The order of increasing ionization energy is as follows:

  $\text{Li}<\text{C}<\text{O}$

Explanation of Solution

Lithium, carbon, and oxygen are present in the same period of the periodic table. $\text{Li}$ lies to the left side of both $\text{C}$ and $\text{O}$. Atomic size decreases from left to right. $\text{Li}$ is the largest among the given elements, followed by $\text{C}$ and $\text{O}$ is the smallest. Ionization energy is inversely proportional to the size of the element. So $\text{Li}$ has the lowest ionization energy, followed by $\text{C}$ and $\text{O}$.

The order of increasing ionization energy is as follows:

  $\text{Li}<\text{C}<\text{O}$

Interpretation Introduction

Interpretation:

The elements lithium, carbon, and oxygen have to be arranged in the order of increasing second ionization energy.

Concept Introduction:

Refer to part (b).

Answer to Problem 8.99QE

The order of increasing second ionization energy is as follows:

  $\text{C}<\text{O}<\text{Li}$

Explanation of Solution

The atomic number of $\text{Li}$ is 3 and its electronic configuration is $1s^{2}2s^1$. For second ionization energy, two electrons are to be removed, one from $2s$ orbital and the other from $1s$ orbital. But after the removal of one electron, its configuration becomes $1s^2$ that is a stable noble gas configuration of neon. The removal of the second electron from this configuration is quite difficult and therefore $\text{Li}$ has the largest second ionization energy among the given elements.

The atomic number of $\text{C}$ is 6 and its electronic configuration is $1s^{2}2s^{2}2p^2$. After the removal of two electrons, its configuration becomes $1s^{2}2s^2$ that is also a completely filled stable configuration. So the removal of the second electron occurs with great ease and the second ionization energy of $\text{C}$ is the least among the given elements.

The atomic number of $\text{O}$ is 8 and its electronic configuration is $1s^{2}2s^{2}2p^4$. After the removal of two electrons, its configuration becomes $1s^{2}2s^{2}2p^2$. But the configuration of $\text{C}$ after the loss of two electrons is a stable configuration so the second ionization energy of $\text{O}$ is higher than that of $\text{C}$. But $\text{Li}$ achieves a stable noble gas configuration after the removal of one electron so its second ionization energy is even higher than that of $\text{O}$. Therefore the second ionization of $\text{O}$ lies in between $\text{Li}$ and $\text{C}$.

The order of increasing second ionization energy is as follows:

  $\text{C}<\text{O}<\text{Li}$

Interpretation Introduction

Interpretation:

The elements lithium, carbon, and oxygen have to be arranged in the order of number of unpaired electrons.

Concept Introduction:

The electronic configuration is defined as the distribution of electrons in various atomic orbitals of the atom. Electrons are filled in orbitals in accordance with three rules: Aufbau principle, Hund’s rule, and Pauli’s exclusion principle. Aufbau principle states that electrons are filled in the orbitals from lower to higher energy level as follows:

  $1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s$

Hund’s rule states that initially each orbital is singly occupied and then pairing occurs and Pauli’s exclusion principle states that the spin of two electrons in one orbital is always different.

Answer to Problem 8.99QE

The order of increasing number of unpaired electrons is as follows:

  $\text{Li}<\text{C}=\text{O}$

Explanation of Solution

The atomic number of $\text{Li}$ is 3 and its electronic configuration is $1s^{2}2s^1$. So it has only one unpaired electron in it.

The atomic number of $\text{C}$ is 6 and its electronic configuration is $1s^{2}2s^{2}2p^2$. So it has two unpaired electrons in it.

The atomic number of $\text{O}$ is 8 and its electronic configuration is $1s^{2}2s^{2}2p^4$. So it has two unpaired electrons in it.

The order of increasing number of unpaired electrons is as follows:

  $\text{Li}<\text{C}=\text{O}$