Chapter 1, Problem 1D.24E

(a)

Interpretation Introduction

Interpretation:

Number of orbitals that can have the given quantum numbers has to be written.

    $n=3,l=1$

Concept Introduction:

There are four set of quantum numbers present for explaining the distribution of electron density in an atom.  They are principal quantum number, angular momentum quantum number, magnetic quantum number and spin quantum number.

    $\begin{array}{cccc}\text{Name}& \text{Symbol}& \text{Values}& \text{Specifies}\\ \text{Principal}& \text{n}& \text{1,2,}\mathrm{...}& \text{shell}\\ \text{angular momentum}& \text{l}& \text{0,1,}\mathrm{...}\text{n-1}& \begin{array}{l}\text{subshell:}\\ \text{l}\text{=}\text{0,1,2,3,4}\mathrm{...}\\ \text{s,p,d,f,g}\mathrm{...}\end{array}\\ \text{magnetic}& {\text{m}}_{\text{l}}& \text{l,l-1,}\mathrm{...}\text{,-1}& \text{orbitals}\text{of}\text{shell}\\ \text{spin}& {\text{m}}_{\text{s}}& \text{+}\frac{\text{1}}{\text{2}}\text{,-}\frac{\text{1}}{\text{2}}& \text{spin}\text{state}\end{array}$

For a given value of $\text{n,}$ there will be $\text{n}$ different values for $l.$  And in the case of magnetic quantum number, there will be $\left(2l+1\right)$ different values of $m_l$ present for each values of $l.$

Explanation of Solution

When the principal quantum number $\left(\text{n}\right)$ value is $3$ and the angular momentum quantum number $\left(l\right)$ value is $1,$ then the subshell notation is $3p.$

If the angular momentum quantum number $lis1,$ then there will be three different $m_l$ values are present, which means there will be three orbitals for the given subshell.

(b)

Interpretation Introduction

Interpretation:

Number of orbitals that can have the given quantum numbers has to be written.

    $n=5,l=3,m_l=-1$

Concept Introduction:

Refer to part (a).

Explanation of Solution

When the principal quantum number $\left(\text{n}\right)$ value is $5$ and the angular momentum quantum number $\left(l\right)$ value is $3,$ then the subshell notation is $5f.$

If the angular momentum quantum number $lis3,$ then there will be seven different values for  $m_l$ and they are $-3,-2,-1,0,+1,+2,+3.$  If  $m_l$ value is $-1,$ then there will be only one orbital.

(c)

Interpretation Introduction

Interpretation:

Number of orbitals that can have the given quantum numbers has to be written.

    $n=2,l=1,m_l=0$

Concept Introduction:

Refer to part (a).

Explanation of Solution

When the principal quantum number $\left(\text{n}\right)$ value is $2$ and the angular momentum quantum number $\left(l\right)$ value is $1,$ then the subshell notation is $2p.$

If the angular momentum quantum number $lis1,$ then there will be three different values for  $m_l$ and they are $-1,0,+1.$  If  $m_l$ value is $0,$ then there will be only one orbital.

(d)

Interpretation Introduction

Interpretation:

Number of orbitals that can have the given quantum numbers has to be written.

    $n=7$

Concept Introduction:

Refer to part (a).

Explanation of Solution

If the principal quantum number $\left(\text{n}\right)$ is $7$ then the possible angular momentum quantum number $\left(l\right)$ values are $0,1,2,3,4,5and6.$

If the angular momentum quantum number $lis0,$ then there will be only one $m_l$ value present, which means there will be only one orbital present.

If the angular momentum quantum number $lis1,$ then there will be three different $m_l$ values are present, which means there will be three orbitals present.

If the angular momentum quantum number $lis2,$ then there will be five different $m_l$ values are present, which means there will be five orbitals present.

If the angular momentum quantum number $lis3,$ then there will be seven different $m_l$ values are present, which means there will be seven orbitals present.

If the angular momentum quantum number $lis4,$ then there will be nine different $m_l$ values are present, which means there will be nine orbitals present.

If the angular momentum quantum number $lis5,$ then there will be eleven different $m_l$ values are present, which means there will be eleven orbitals present.

If the angular momentum quantum number $lis6,$ then there will be thirteen different $m_l$ values are present, which means there will be thirteen orbitals present.

So in total there will be forty nine orbitals.