Chapter 2, Problem 95E

In the ground state of mercury, Hg,

a. how many electrons occupy atomic orbitals with n = 3?

b. how many electrons occupy d atomic orbitals?

c. how many electrons occupy p_z, atomic orbitals?

d. how many electrons have spin "up" $\left(m_s=+\frac{1}{2}\right)?$

Interpretation Introduction

Interpretation:

The given questions with respect to the ground state of mercury $\left(Hg\right)$ are to be answered.

Concept Introduction:

According to the Aufbau principal, in the ground state of an atom or an ion, the atomic orbitals of the lowest available energy level are filled by the electrons first. Then the orbitals of a comparatively higher energy level are filled. Also, according to the Hund’s rule of maximum multiplicity, the orbitals having the same energy level are filled by one electron each before any one of them is filled with a second electron.

To determine: The number of electrons occupying atomic orbitals with $n=3$.

Answer to Problem 95E

Answer

The required number of electrons is $\underset{\_}{18}$.

Explanation of Solution

The atomic number of mercury is $80$. This is equal to the number of electrons present in the atom. Hence, the ground state electronic configuration of mercury is given as,

$\text{Hg}\left(80\right)={\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}\text{,}\boxed{{\text{3s}}^{\text{2}}}\text{,}\boxed{{\text{3p}}^{\text{6}}}{\text{,4s}}^2\text{,}\boxed{{\text{3d}}^{\text{10}}}{\text{,4p}}^{\text{6}}{\text{,5s}}^{\text{2}}{\text{,4d}}^{\text{10}}{\text{,5p}}^{\text{6}}{\text{,6s}}^{\text{2}}{\text{,4f}}^{\text{14}}{\text{,5d}}^{10}$.

In this case, three orbitals with $n=3$ are present.

The total number of electrons occupying the $n=3$ orbitals is,

$2+6+10=\underset{\_}{18}$

Conclusion

The number of electrons occupying the $n=3$ orbitals in mercury $\left(\text{Hg}\right)$ is $\underset{\_}{18}$.

Interpretation Introduction

Interpretation:

The given questions with respect to the ground state of mercury $\left(Hg\right)$ are to be answered.

Concept Introduction:

According to the Aufbau principal, in the ground state of an atom or an ion, the atomic orbitals of the lowest available energy level are filled by the electrons first. Then the orbitals of a comparatively higher energy level are filled. Also, according to the Hund’s rule of maximum multiplicity, the orbitals having the same energy level are filled by one electron each before any one of them is filled with a second electron.

To determine: The number of electrons occupying the $d$ orbital.

Answer to Problem 95E

Answer

The number of electrons that occupy the $d$ orbital is $\underset{\_}{30}$.

Explanation of Solution

The atomic number of mercury is $80$. This is equal to the number of electrons present in the atom. Hence, the ground state electronic configuration of mercury is given as,

$\text{Hg}\left(80\right)={\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,4s}}^2\text{,}\boxed{{\text{3d}}^{\text{10}}}{\text{,4p}}^{\text{6}}{\text{,5s}}^{\text{2}}\text{,}\boxed{{\text{4d}}^{\text{10}}}{\text{,5p}}^{\text{6}}{\text{,6s}}^{\text{2}}{\text{,4f}}^{\text{14}}\text{,}\boxed{{\text{5d}}^{10}}$.

In this case, three $d$ orbitals are present.

The number of electrons which occupy the $d$ orbital is,

$10+10+10=\underset{\_}{30}$

Conclusion

The number of electrons that occupy the $d$ orbital in mercury $\left(\text{Hg}\right)$ is $\underset{\_}{30}$.

Interpretation Introduction

Interpretation:

The given questions with respect to the ground state of mercury $\left(Hg\right)$ are to be answered.

Concept Introduction:

According to the Aufbau principal, in the ground state of an atom or an ion, the atomic orbitals of the lowest available energy level are filled by the electrons first. Then the orbitals of a comparatively higher energy level are filled. Also, according to the Hund’s rule of maximum multiplicity, the orbitals having the same energy level are filled by one electron each before any one of them is filled with a second electron.

To determine: The number of electrons occupying the $p_z$ orbital.

Answer to Problem 95E

Answer

The number of electrons that occupy the $p_z$ orbital is $30$.

Explanation of Solution

The atomic number of mercury is $80$. This is equal to the number of electrons present in the atom. Hence, the ground state electronic configuration of mercury is given as,

$\text{Hg}\left(80\right)={\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}\text{,}\boxed{{\text{2p}}^{\text{6}}}{\text{,3s}}^{\text{2}}\text{,}\boxed{{\text{3p}}^{\text{6}}}{\text{,4s}}^2{\text{,3d}}^{\text{10}}\text{,}\boxed{{\text{4p}}^{\text{6}}}{\text{,5s}}^{\text{2}}{\text{,4d}}^{\text{10}}\text{,}\boxed{{\text{5p}}^{\text{6}}}{\text{,6s}}^{\text{2}}{\text{,4f}}^{\text{14}}{\text{,5d}}^{\text{10}}$.

The number of $p$ orbitals is $4$.

In each $p$ orbital there is one $p_z$ orbital; and in each $p_z$ orbital, two electrons are present.

Therefore, number of electrons which occupy the $p_z$ orbital in mercury $\left(\text{Hg}\right)$ is,

$4\times 2=\underset{\_}{8}$.

Conclusion

The number of electrons that occupy the $p_z$ orbital in mercury $\left(\text{Hg}\right)$ is $\underset{\_}{8}$.

Interpretation Introduction

Interpretation:

The given questions with respect to the ground state of mercury $\left(Hg\right)$ are to be answered.

Concept Introduction:

According to the Aufbau principal, in the ground state of an atom or an ion, the atomic orbitals of the lowest available energy level are filled by the electrons first. Then the orbitals of a comparatively higher energy level are filled. Also, according to the Hund’s rule of maximum multiplicity, the orbitals having the same energy level are filled by one electron each before any one of them is filled with a second electron.

To determine: The number of electrons having spin “up” $\left(m_s=+\frac{1}{2}\right)$.

Answer to Problem 95E

Answer

The number of electrons having spin quantum number, $m_s=+\frac{1}{2}$ is $\underset{\_}{40}$.

Explanation of Solution

The atomic number of mercury is $80$. This is equal to the number of electrons present in the atom. Hence, the ground state electronic configuration of mercury is given as,

$\text{Hg}\left(80\right)={\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^{\text{6}}{\text{,3s}}^{\text{2}}{\text{,3p}}^{\text{6}}{\text{,4s}}^2{\text{,3d}}^{\text{10}}{\text{,4p}}^{\text{6}}{\text{,5s}}^{\text{2}}{\text{,4d}}^{\text{10}}{\text{,5p}}^{\text{6}}{\text{,6s}}^{\text{2}}{\text{,4f}}^{\text{14}}{\text{,5d}}^{\text{10}}$.

There are total $80$ electrons present; half of the electrons have positive spin, while the other half will have a negative spin.

Hence, the number of electrons having $m_s=+\frac{1}{2}$ is $\underset{\_}{40}$.

Conclusion

The number of electrons having positive spin in case of mercury $\left(\text{Hg}\right)$ is $\underset{\_}{40}$.