Chapter 2.6, Problem 2.14Q

(a)

Interpretation Introduction

Interpretation:

The electronic configuration and shorthand electronic configuration of $R{b}^{+}ion$ has to be written.

Concept Introduction:

Electron configuration: The complete description of the atomic orbitals occupied by all the electrons in an atom or monoatomic ion is known as its electron configuration.

Explanation of Solution

Rubidium is in Group $IAor1$ and the atomic number is $37.$

$R{b}^{+}ion$ Indicate that the ion has one less electron than a neutral atom of Rubidium.  The total number of electrons in $R{b}^{+}ion$ can be calculated:

$\begin{array}{l}37e^{-}-1e^{-}=36e^{-}\\ electronsinRbelectronslosselectronsinR{b}^{+}\end{array}$

The electron configuration is as follows:

  $1s^{2}2s^{2}2p^{6}3s^{2}3p^6 4s^{2}3d^{10}4p^6$

Since $R{b}^{+}ion$ is isoelectronic with $Kr$, Krypton, all $36$ electrons of $R{b}^{+}ion$ are indicated by the symbol $Kr$ in square brackets.  The shorthand electronic configuration for $R{b}^{+}ion$ is $\left[Kr\right].$

(b)

Interpretation Introduction

Interpretation:

The electronic configuration and shorthand electronic configuration of $S{r}^{2+}ion$ has to be written.

Concept Introduction:

Refer to Part (a)

Explanation of Solution

Strontium is in Group $IIAor\text{ 2}$ and the atomic number is $38.$

$S{r}^{2+}ion$ Indicate that the ion has two less electrons than a neutral atom of Strontium.  The total number of electrons in $S{r}^{2+}ion$ can be calculated:

$\begin{array}{l}38e^{-}-2e^{-}=36e^{-}\\ electronsinSrelectronslosselectronsinS{r}^{2+}\end{array}$

The electron configuration is as follows:

  $1s^{2}2s^{2}2p^{6}3s^{2}3p^6 4s^{2}3d^{10}4p^6$

Since $S{r}^{2+}ion$ is isoelectronic with $Kr$, Krypton, all $36$ electrons of $S{r}^{2+}ion$ are indicated by the symbol $Kr$ in square brackets.  The shorthand electronic configuration for $S{r}^{2+}ion$ is $\left[Kr\right].$

(c)

Interpretation Introduction

Interpretation:

The electronic configuration and shorthand electronic configuration of $S^{2-}ion$ has to be written.

Concept Introduction:

Refer to Part (a)

Explanation of Solution

Sulfur is in Group $VIAor\text{ 16}$ and the atomic number is $16.$

$S^{2-}ion$ Indicate that the ion has gained two electrons than a neutral atom of Sulfur.  The total number of electrons in $S^{2-}ion$ can be calculated:

$\begin{array}{l}16e^{-}+2e^{-}=18e^{-}\\ electronsinSelectronsgainedelectronsin{S}^{2-}\end{array}$

The electron configuration is as follows:

  $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

Since $S^{2-}ion$ is isoelectronic with $Ar$, Argon, all eighteen electrons of $S^{2-}ion$ are indicated by the symbol $Ar$ in square brackets.  The shorthand electronic configuration for $S^{2-}ion$ is $\left[Ar\right].$

(d)

Interpretation Introduction

Interpretation:

The electronic configuration and shorthand electronic configuration of $I^{-}ion$ has to be written.

Concept Introduction:

Refer to Part (a)

Explanation of Solution

Iodine is in Group $VIIAor17$ and the atomic number is $53.$

$I^{-}ion$ Indicate that the ion has gained one electron than a neutral atom of Iodine.  The total number of electrons in $I^{-}ion$ can be calculated:

$\begin{array}{l}53e^{-}+1e^{-}=54e^{-}\\ electronsinIelectronsgainedelectronsin{I}^{-}\end{array}$

The electron configuration is as follows:

  $1s^{2}2s^{2}2p^{6}3s^{2}3p^6 4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^6$

Since $I^{-}ion$ is isoelectronic with $Xe$, Xenon, all $54$ electrons of $I^{-}ion$ are indicated by the symbol $Xe$ in square brackets.  The shorthand electronic configuration for $I^{-}ion$ is $\left[Xe\right].$