Chapter 2, Problem 2.111QP

(a)

Interpretation Introduction

Interpretation:

The electronic configuration and shorthand electronic configuration of $I^{-}ion$ 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

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 $\left[Xe\right].$

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to Part (a)

Explanation of Solution

Barium is in Group $IIAor\text{ 2}$ and the atomic number is $56.$

$B{a}^{2+}ion$ Indicate that the ion has two lesser number electrons than a neutral atom of Barium.  The total number of electrons in $B{a}^{2+}ion$ can be calculated:

$\begin{array}{l}56e^{-}-2e^{-}=54e^{-}\\ electronsinBaelectronslosselectronsinB{a}^{2+}\end{array}$

The electron configuration is as follows:

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

Since $B{a}^{2+}ion$ is isoelectronic with $Xe$, Xenon, all $54$ electrons of $B{a}^{2+}ion$ are indicated by the symbol $Xe$ in square brackets.  The shorthand electronic configuration for $B{a}^{2+}ion$ is $\left[Xe\right].$

(c)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to Part (a)

Explanation of Solution

Selenium is in Group $VIAor16$ and the atomic number is $34.$

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

$\begin{array}{l}34e^{-}+2e^{-}=36e^{-}\\ electronsinSeelectronsgainedelectronsinS{e}^{2-}\end{array}$

The electron configuration is as follows:

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

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

(d)

Interpretation Introduction

Interpretation:

The electronic configuration and shorthand electronic configuration of $A{l}^{3+}ion$ has to be written.

Concept Introduction:

Refer to Part (a)

Explanation of Solution

Aluminum is in Group $IIIAor\text{ 3}$ and the atomic number is $13.$

$A{l}^{3+}ion$ Indicate that the ion has lesser three electrons than a neutral atom of Aluminum.  The total number of electrons in $A{l}^{3+}ion$ can be calculated:

$\begin{array}{l}13e^{-}-3e^{-}=10e^{-}\\ electronsinAlelectronslosselectronsinA{l}^{3+}\end{array}$

The electron configuration is as follows:

  $1s^{2}2s^{2}2p^6$

Since $A{l}^{3+}ion$ is isoelectronic with $Ne$, Neon, all $10$ electrons of $A{l}^{3+}ion$ are indicated by the symbol $Ne$ in square brackets.  The shorthand electronic configuration for $A{l}^{3+}ion$ is $\left[Ne\right].$