Chapter 7, Problem 7.77QP

(a)

Interpretation Introduction

Interpretation: The numbers of orbitals in an atom for each of the given principal quantum number are to be stated.

Concept introduction: There are four quantum numbers that indicate 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 magnetic quantum number is an integer that takes the values from $\left(-l\text{ to }+l\right)$.

To determine: The number of orbitals in an atom for the given principal quantum number.

Answer to Problem 7.77QP

Solution

The number of orbital for $n=1$ is $\underset{\_}{1}$.

Explanation of Solution

Explanation

The given principal quantum number $\left(n\right)$ is $1$. For $n=1$, the value of $l$ is,

$\begin{array}{c}l=n-1\\ l=1-1\\ l=0\end{array}$

When $l=0$ then $m=0$ that is only one orbital is present. For $l=0$ indicates the presence of s orbital. Hence, there is only $1$ orbital in the first shell of an atom that is for $n=1$.

(b)

Interpretation Introduction

To determine: The number of orbitals in an atom for the given principal quantum number.

Answer to Problem 7.77QP

Solution

The number of orbital for $n=2$ is $\underset{\_}{4}$.

Explanation of Solution

Explanation

The given principal quantum number $\left(n\right)$ is $2$. For $n=2$, the value of $l$ is,

$\begin{array}{c}l=n-1\\ l=2-1\\ l=1\end{array}$

The allowed values of $l$ are $0\text{and}\text{1}$. For $n=2$, the value of $l$ represents $\text{2s}\text{and}\text{2p}$ sub shells. The possible values of $m_l$ take the values from $\left(-l\text{ to }+l\right)$ which are,

For $l=0$

$m_l=0$

The combination of $l$ and $m_l$ represents one $\text{2s}$ orbital.

For $l=1$

$m_l=-1,0,+1$

The combination of $l$ and $m_l$ represents three $\text{2p}$ orbitals.

Hence, the total number of orbitals for $n=2$ is $1+3=4$ orbitals.

(c)

Interpretation Introduction

To determine: The number of orbitals in an atom for the given principal quantum number.

Answer to Problem 7.77QP

Solution

The number of orbital for $n=3$ is $\underset{\_}{9}$.

Explanation of Solution

Explanation

The given principal quantum number $\left(n\right)$ is $3$. For $n=3$, the value of $l$ is,

$\begin{array}{c}l=n-1\\ l=3-1\\ l=2\end{array}$

The allowed values of $l$ are $0,1\text{and}2$. For $n=2$, the value of $l$ represents $\text{3s,3p}$ and $\text{3d}$ sub shells. The possible values of $m_l$ take the values from $\left(-l\text{ to }+l\right)$ which are,

For $l=0$

$m_l=0$

The combination of $l$ and $m_l$ represents one $\text{3s}$ orbital.

For $l=1$

$m_l=-1,0,+1$

The combination of $l$ and $m_l$ represents three $\text{3p}$ orbitals.

For $l=2$

$m_l=-2,-1,0,+1,+2$

The combination of $l$ and $m_l$ represents five $\text{3d}$ orbitals.

Hence, the total number of orbitals for $n=3$ is $1+3+5=9$ orbitals.

(d)

Interpretation Introduction

To determine: The number of orbitals in an atom for the given principal quantum number.

Answer to Problem 7.77QP

Solution

The number of orbital for $n=4$ is $\underset{\_}{16}$.

Explanation of Solution

Explanation

The given principal quantum number $\left(n\right)$ is $4$. For $n=4$, the value of $l$ is,

$\begin{array}{c}l=n-1\\ l=4-1\\ l=3\end{array}$

The allowed values of $l$ are $0,1,2\text{and}3$. For $n=4$, the value of $l$ represents $\text{4s,4p,4d}$ and $4\text{f}$ sub shells. The possible values of $m_l$ take the values from $\left(-l\text{ to }+l\right)$ which are,

For $l=0$

$m_l=0$

The combination of $l$ and $m_l$ represents one $\text{4s}$ orbital.

For $l=1$

$m_l=-1,0,+1$

The combination of $l$ and $m_l$ represents three $\text{4p}$ orbitals.

For $l=2$

$m_l=-2,-1,0,+1,+2$

The combination of $l$ and $m_l$ represents five $\text{4d}$ orbitals.

For $l=3$

$m_l=-3,-2,-1,0,+1,+2,+3$

The combination of $l$ and $m_l$ represents seven $4\text{f}$ orbitals

Hence, the total number of orbitals for $n=4$ is $1+3+5+7=16$ orbitals.

(e)

Interpretation Introduction

To determine: The number of orbitals in an atom for the given principal quantum number.

Answer to Problem 7.77QP

Solution

The number of orbital for $n=5$ is $\underset{\_}{25}$.

Explanation of Solution

Explanation

The given principal quantum number $\left(n\right)$ is $5$. For $n=5$, the value of $l$ is,

$\begin{array}{c}l=n-1\\ l=5-1\\ l=4\end{array}$

The allowed values of $l$ are $0,1,2,3\text{and}4$. For $n=5$, the value of $l$ represents $\text{5s,5p,5d,5f}$ and $5\text{g}$ sub shells. The possible values of $m_l$ take the values from $\left(-l\text{ to }+l\right)$ which are,

For $l=0$

$m_l=0$

The combination of $l$ and $m_l$ represents one $\text{5s}$ orbital.

For $l=1$

$m_l=-1,0,+1$

The combination of $l$ and $m_l$ represents three $\text{5p}$ orbitals.

For $l=2$

$m_l=-2,-1,0,+1,+2$

The combination of $l$ and $m_l$ represents five $\text{5d}$ orbitals.

For $l=3$

$m_l=-3,-2,-1,0,+1,+2,+3$

The combination of $l$ and $m_l$ represents seven $\text{5f}$ orbitals.

For $l=4$

$m_l=-4,-3,-2,-1,0,+1,+2,+3+4$

The combination of $l$ and $m_l$ represents nine $\text{5g}$ orbitals.

Hence, the total number of orbitals for $n=5$ is $1+3+5+7+9=25$ orbitals.

Conclusion

1. a. The number of orbital for $n=1$ is $\underset{\_}{1}$.
2. b. The number of orbital for $n=2$ is $\underset{\_}{4}$.
3. c. The number of orbital for $n=3$ is $\underset{\_}{9}$.
4. d. The number of orbital for $n=4$ is $\underset{\_}{16}$.
5. e. The number of orbital for $n=5$ is $\underset{\_}{25}$.