Chapter 12, Problem 89E

Interpretation Introduction

Interpretation: The number of electrons present in $n=5$ quantum number for element $115,\text{ Uup}$ needs to be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers. The four quantum numbers are:

• Principal quantum number, $\text{n}$ which indicates the size of the orbital.
• Azimuthal quantum number, $\text{l}$ which indicates the shape and angular momentum of the orbital.
• Magnetic quantum number, ${\text{m}}_{\text{l}}$ which indicates the orientation of orbitals.
• Spin quantum number, ${\text{m}}_{\text{s}}$ which indicates the angular momentum of the electron.

Answer to Problem 89E

The number of electrons present in $n=5$ quantum number of element $115,\text{ Uup}$ is $\text{32 electrons}$ .

Explanation of Solution

Electrons are distributed in the orbitals the of subshell. The specific region of space in which the movement of electrons are confined is said to be shells which are divided into subshells and are s-, p-, d-, and f-. Among these subshells, the electrons are grouped as orbitals.

The number of electrons that these subshells can hold are:

s-block - 2

p-block - 6

d-block - 10

f-block - 14

The increasing order of energy of shells, subshell is:

  

As the number of electrons for a neutral atom is equal to the atomic number of an atom. Thus,

Number of electrons in $\text{Uup}$ is $115$ . So, the electronic configuration of $\text{Uup}$ is:

  ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{5s}}^{\text{2}}{\text{4d}}^{\text{10}}{\text{5p}}^{\text{6}}{\text{6s}}^{\text{2}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6p}}^{\text{6}}{\text{7s}}^{\text{2}}{\text{5f}}^{\text{14}}{\text{6d}}^{\text{10}}{\text{7p}}^{\text{3}}$

The number of electrons present in $n=5$ quantum number are calculated as:

The electrons in $\text{5s, 5p, 5d and 5f}$ has $n=5$ quantum number so,

  $\text{number of electrons present in n = 5 = 2 + 6 + 10 + 14 = 32 electrons}$

Hence, the number of electrons present in $n=5$ quantum number of element $115,\text{ Uup}$ is $\text{32 electrons}$ .

Interpretation Introduction

Interpretation: The number of electrons in $l=3$ as one of their quantum number for element $115,\text{ Uup}$ needs to be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers. The four quantum numbers are:

• Principal quantum number, $\text{n}$ which indicates the size of the orbital.
• Azimuthal quantum number, $\text{l}$ which indicates the shape and angular momentum of the orbital.
• Magnetic quantum number, ${\text{m}}_{\text{l}}$ which indicates the orientation of orbitals.
• Spin quantum number, ${\text{m}}_{\text{s}}$ which indicates the angular momentum of the electron.

Answer to Problem 89E

The number of electrons in $l=3$ as one of their quantum number for element $115,\text{ Uup}$ is $\text{28 electrons}$ .

Explanation of Solution

The electronic configuration of $\text{Uup}$ is:

  ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{5s}}^{\text{2}}{\text{4d}}^{\text{10}}{\text{5p}}^{\text{6}}{\text{6s}}^{\text{2}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6p}}^{\text{6}}{\text{7s}}^{\text{2}}{\text{5f}}^{\text{14}}{\text{6d}}^{\text{10}}{\text{7p}}^{\text{3}}$

The number of electrons in $l=3$ as one of their quantum number are calculated as:

For $l=3$ , the orbital is $\text{f-orbital}$ . So, the electrons in $l=3$ are:

  $\text{number of electrons present in l = 3 = 14+ 14 = 28 electrons}$

Hence, the number of electrons in $l=3$ as one of their quantum number for element $115,\text{ Uup}$ is $\text{28 electrons}$ .

Interpretation Introduction

Interpretation: The number of electrons in $m_l=1$ as one of their quantum number for element $115,\text{ Uup}$ needs to be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers. The four quantum numbers are:

• Principal quantum number, $\text{n}$ which indicates the size of the orbital.
• Azimuthal quantum number, $\text{l}$ which indicates the shape and angular momentum of the orbital.
• Magnetic quantum number, ${\text{m}}_{\text{l}}$ which indicates the orientation of orbitals.
• Spin quantum number, ${\text{m}}_{\text{s}}$ which indicates the angular momentum of the electron.

Answer to Problem 89E

The number of electrons in $m_l=1$ as one of their quantum number for element $115,\text{ Uup}$ is $\text{23 electrons}$ .

Explanation of Solution

The electronic configuration of $\text{Uup}$ is:

  ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{5s}}^{\text{2}}{\text{4d}}^{\text{10}}{\text{5p}}^{\text{6}}{\text{6s}}^{\text{2}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6p}}^{\text{6}}{\text{7s}}^{\text{2}}{\text{5f}}^{\text{14}}{\text{6d}}^{\text{10}}{\text{7p}}^{\text{3}}$

The number of electrons in $m_l=1$ as one of their quantum number are calculated as:

Since, the magnetic quantum number, ${\text{m}}_{\text{l}}$ have values from $\text{-l to +l}$ where $\text{l}$ is azimuthal quantum number. So, the possibility for $m_l=1$ is when the values of $\text{l}$ is equal to or greater than $\text{1}$ . For $l\ge 1$ , the orbitals are $\text{p, d, and f-orbitals}$ . The number of electrons present in each orbital is 2 having $m_l=1$ except for the $\text{7p}$ orbital which has 1 electron having $m_l=1$ .

So, the electrons in $m_l=1$ are:

  $\begin{array}{l}{\text{number of electrons present in m}}_{\text{l}}{\text{=1 = 2×5}}_{\text{p-orbitals}}{\text{+ 2×4}}_{\text{d-orbitals}}{\text{+ 2×2}}_{\text{f-orbitals}}+{\text{ 1×1}}_{\text{p-orbital}}\\ {\text{number of electrons present in m}}_{\text{l}}\text{=1 = 10 + 8 + 4 }+\text{1 = 23 electrons}\end{array}$

Hence, the number of electrons in $m_l=1$ as one of their quantum number for element $115,\text{ Uup}$ is $\text{23 electrons}$ .

Interpretation Introduction

Interpretation: The number of electrons in $m_s=-\frac{1}{2}$ as one of their quantum number for element $115,\text{ Uup}$ needs to be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers. The four quantum numbers are:

• Principal quantum number, $\text{n}$ which indicates the size of the orbital.
• Azimuthal quantum number, $\text{l}$ which indicates the shape and angular momentum of the orbital.
• Magnetic quantum number, ${\text{m}}_{\text{l}}$ which indicates the orientation of orbitals.
• Spin quantum number, ${\text{m}}_{\text{s}}$ which indicates the angular momentum of the electron.

Answer to Problem 89E

The number of electrons in $m_s=-\frac{1}{2}$ as one of their quantum number for element $115,\text{ Uup}$ is $\text{56 electrons}$ or $\text{59 electrons}$ .

Explanation of Solution

The electronic configuration of $\text{Uup}$ is:

  ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{5s}}^{\text{2}}{\text{4d}}^{\text{10}}{\text{5p}}^{\text{6}}{\text{6s}}^{\text{2}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6p}}^{\text{6}}{\text{7s}}^{\text{2}}{\text{5f}}^{\text{14}}{\text{6d}}^{\text{10}}{\text{7p}}^{\text{3}}$

The number of electrons in $m_s=-\frac{1}{2}$ as one of their quantum number are calculated as:

Since, the spin quantum number, ${\text{m}}_s$ have values $+\frac{1}{2}\text{ or }-\frac{1}{2}$ . For completely filled orbitals, half of the total number of electrons possess $m_s=+\frac{1}{2}$ and the other half possess $m_s=-\frac{1}{2}$ value. All the inner orbitals are completely filled for the given element except that the outermost orbital, $\text{7p}$ contains 3 unpaired electrons that can possess either $m_s=+\frac{1}{2}$ or $m_s=-\frac{1}{2}$ value.

• >For $m_s=+\frac{1}{2}$ values for $\text{7p}$ orbital: then the subtracting these three electrons from total number of electrons, 115 will be $\text{115 - 3 = 112}$ .

Thus, the number of electrons possessing $m_s=-\frac{1}{2}$ is:

  $\frac{\text{112}}{2}\text{ = 56 electrons}$

• For $m_s=-\frac{1}{2}$ values for $\text{7p}$ orbital then the number of electrons will be:

  $\text{56 + 3 = 59 electrons}$

Hence, the number of electrons in $m_s=-\frac{1}{2}$ as one of their quantum number for element $115,\text{ Uup}$ is $\text{56 electrons}$ or $\text{59 electrons}$ .