Chapter 2, Problem 96E

In the ground state of element 115, Uup,

a. how many electrons have n = 5 as one of their quantum numbers?

b. how many electrons have l = 3 as one of their quantum numbers?

c. how many electrons have m_l = 1 as one of their quantum numbers?

d. how many electrons have m_s = $-\frac{1}{2}$  as one of their quantum numbers?

Interpretation Introduction

Interpretation: The number of electrons in the ground state of element $115$ $\left(\text{Uup}\right)$ with their given quantum numbers.

Concept introduction: The electronic configuration of an atom is defined as the distribution of electrons in atomic orbitals. There are four quantum numbers that indicates the size, energy and shape of an atomic orbital. The quantum number $n$ is the principal quantum number that indicates the energy level and size of the orbital and $l$ is the azimuthal quantum number that determines the shape and angular momentum of an orbital. The azimuthal quantum number $\left(l\right)$ takes values $\left(l=n-1\right)$ and magnetic quantum number $\left(m_l\right)$ takes values from $\left(-l\text{ to }+l\right)$ and the spin quantum number $\left(m_s\right)$ indicates the spin of an electron $\left(m_s=+\frac{1}{2}\text{or}-\frac{1}{2}\right)$.

Answer to Problem 96E

The number of electrons with $n=5$ as one of their quantum numbers is $32$.

Explanation of Solution

To determine: The number of electrons with $n=5$ as one of their quantum numbers.

The number of electrons with $n=5$ as one of their quantum numbers is $32$.

The given element is $115$ $\left(\text{Uup}\right)$. The electronic configuration of $115$ $\left(\text{Uup}\right)$ is,

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

The given principle quantum number is $n=5$. The electrons in $5s,5p,5d\text{ and }5f$ has $n=5$ as one of their quantum numbers. Therefore,

$\text{Numberofelectrons}=2+6+10+14=3\text{2 electrons}$

Interpretation Introduction

Interpretation: The number of electrons in the ground state of element $115$ $\left(\text{Uup}\right)$ with their given quantum numbers.

Concept introduction: The electronic configuration of an atom is defined as the distribution of electrons in atomic orbitals. There are four quantum numbers that indicates the size, energy and shape of an atomic orbital. The quantum number $n$ is the principal quantum number that indicates the energy level and size of the orbital and $l$ is the azimuthal quantum number that determines the shape and angular momentum of an orbital. The azimuthal quantum number $\left(l\right)$ takes values $\left(l=n-1\right)$ and magnetic quantum number $\left(m_l\right)$ takes values from $\left(-l\text{ to }+l\right)$ and the spin quantum number $\left(m_s\right)$ indicates the spin of an electron $\left(m_s=+\frac{1}{2}\text{or}-\frac{1}{2}\right)$.

Answer to Problem 96E

The number of electrons with $l=3$ as one of their quantum numbers is $28$.

Explanation of Solution

Explanation

To determine: The number of electrons with $l=3$ as one of their quantum numbers.

The number of electrons with $l=3$ as one of their quantum numbers is $28$.

The orbital with $l=3$ is the $f$ orbital’s. So with the given element $115$ $\left(\text{Uup}\right)$ with electronic configuration

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

The number of electrons with $l=3$ is calculated by adding electrons present in $f$ orbital. Hence, the number of electrons with $l=3$ is $14+14=28$ electrons.

Interpretation Introduction

Interpretation: The number of electrons in the ground state of element $115$ $\left(\text{Uup}\right)$ with their given quantum numbers.

Concept introduction: The electronic configuration of an atom is defined as the distribution of electrons in atomic orbitals. There are four quantum numbers that indicates the size, energy and shape of an atomic orbital. The quantum number $n$ is the principal quantum number that indicates the energy level and size of the orbital and $l$ is the azimuthal quantum number that determines the shape and angular momentum of an orbital. The azimuthal quantum number $\left(l\right)$ takes values $\left(l=n-1\right)$ and magnetic quantum number $\left(m_l\right)$ takes values from $\left(-l\text{ to }+l\right)$ and the spin quantum number $\left(m_s\right)$ indicates the spin of an electron $\left(m_s=+\frac{1}{2}\text{or}-\frac{1}{2}\right)$.

Answer to Problem 96E

The number of electrons with $m_l=1$ as one of their quantum numbers is $23$.

Explanation of Solution

Explanation

To determine: The number of electrons with $m_l=1$ as one of their quantum numbers.

The number of electrons with $m_l=1$ as one of their quantum numbers is $23$.

The value of $m_l=1$ is possible for all values of $l$ equal to above $1$. So, all the $s,p,d\text{ and }f$ orbital’s carry electrons that has the value of $m_l=1$. Each orbital contain two electrons with $m_l=1$ except for the outermost orbital $7p$ which has only one electron with $m_l=1$. Hence, the number of electrons with $m_l=1$ is equal to the number of electrons present in $s,p,d\text{ and }f$ orbitals. There are five $p$, four $d$, two $f$ orbital’s containing two electrons each and one $p$ orbital with one electron present in element. Therefore, the number of electrons with $m_l=1$ as one of their quantum numbers is $10+8+4+1=23$ electrons.

Interpretation Introduction

Interpretation: The number of electrons in the ground state of element $115$ $\left(\text{Uup}\right)$ with their given quantum numbers.

Concept introduction: The electronic configuration of an atom is defined as the distribution of electrons in atomic orbitals. There are four quantum numbers that indicates the size, energy and shape of an atomic orbital. The quantum number $n$ is the principal quantum number that indicates the energy level and size of the orbital and $l$ is the azimuthal quantum number that determines the shape and angular momentum of an orbital. The azimuthal quantum number $\left(l\right)$ takes values $\left(l=n-1\right)$ and magnetic quantum number $\left(m_l\right)$ takes values from $\left(-l\text{ to }+l\right)$ and the spin quantum number $\left(m_s\right)$ indicates the spin of an electron $\left(m_s=+\frac{1}{2}\text{or}-\frac{1}{2}\right)$.

Answer to Problem 96E

The number of electrons with $m_s=-\frac{1}{2}$ as one of their quantum numbers is $59$.

Explanation of Solution

Explanation

To determine: The number of electrons with $m_s=-\frac{1}{2}$ as one of their quantum numbers.

The number of electrons with $m_s=-\frac{1}{2}$ as one of their quantum numbers is $59$.

In every completely filled orbital half of the total number of electrons has $m_s=-\frac{1}{2}$. The $7p$ orbital contains three unpaired electrons. They may have either $+\frac{1}{2}\text{or}-\frac{1}{2}$ spin. The total number of electrons present in the given element is $115$. On subtracting the three $7p$ electrons gives $112$ electrons. So the number of electrons with $m_s=-\frac{1}{2}$ becomes half of the $112$ electrons that are $56$ electrons. The three electrons in $7p$ orbital may or may not possess $m_s=-\frac{1}{2}$ spin. If they possess $-\frac{1}{2}$ spin then the number of electrons with $m_s=-\frac{1}{2}$ is $59$.