Chapter 12, Problem 23A

(a)

Interpretation Introduction

Interpretation:

The name of noble gas having same configuration as ions of given compound is to be written.

Concept introduction:

Ionic compound is formed when a metal loses its electrons and non-metal accepts that electrons. Metal loses and non-metal gains the electrons to attain nearest noble gas configuration.

Answer to Problem 23A

The electronic configuration of $B{a}^{2+}$ is same as $Xe$

The electronic configuration of $S^{2-}$ is same as $Ar$

Explanation of Solution

  $BaS$ is formed when barium loses two electrons and sulphur gains two electrons and both the ions attain noble gas configuration.

The electronic configuration of $Ba$ is:

  $\left[Xe\right]6s^2$

The electronic configuration of $B{a}^{2+}$ is:

  $\left[Xe\right]$

The electronic configuration of $S$ is:

  $\left[Ne\right]3s^{2}3p^4$

The electronic configuration of $S^{2-}$ is:

  $\left[Ar\right]$

(b)

Interpretation Introduction

Interpretation:

The name of noble gas having same configuration as ions of given compound is to be written.

Concept introduction:

Ionic compound is formed when a metal loses its electrons and non-metal accepts that electrons. Metal loses and non-metal gains the electrons to attain nearest noble gas configuration.

Answer to Problem 23A

The electronic configuration of $S{r}^{2+}$ is same as $Kr$

The electronic configuration of $F^{-}$ is same as $Ne$

Explanation of Solution

  $Sr{F}_2$ is formed when srontium loses two electrons and each fluorine gains one electron and both the ions attain noble gas configuration.

The electronic configuration of $Sr$ is:

  $\left[Kr\right]5s^2$

The electronic configuration of $S{r}^{2+}$ is:

  $\left[Kr\right]$

The electronic configuration of $F$ is:

  $1s^{2}2s^{2}2p^5$

The electronic configuration of $F^{-}$ is:

  $\left[Ne\right]$

(c)

Interpretation Introduction

Interpretation:

The name of noble gas having same configuration as ions of given compound is to be written.

Concept introduction:

Ionic compound is formed when a metal loses its electrons and non-metal accepts that electrons. Metal loses and non-metal gains the electrons to attain nearest noble gas configuration.

Answer to Problem 23A

The electronic configuration of $M{g}^{2+}$ is same as $Ne$

The electronic configuration of $O^{2-}$ is same as $Ne$

Explanation of Solution

  $MgO$ is formed when magnesium loses two electrons and oxygen gains two electrons and both the ions attain noble gas configuration.

The electronic configuration of $Mg$ is:

  $\left[Ne\right]3s^2$

The electronic configuration of $M{g}^{2+}$ is:

  $\left[Ne\right]$

The electronic configuration of $O$ is:

  $1s^{2}2s^{2}2p^4$

The electronic configuration of $O^{2-}$ is:

  $\left[Ne\right]$

(d)

Interpretation Introduction

Interpretation:

The name of noble gas having same configuration as ions of given compound is to be written.

Concept introduction:

Ionic compound is formed when a metal loses its electrons and non-metal accepts that electrons. Metal loses and non-metal gains the electrons to attain nearest noble gas configuration.

Answer to Problem 23A

The electronic configuration of $A{l}^{3+}$ is same as $Ne$

The electronic configuration of $S^{2-}$ is same as $Ar$

Explanation of Solution

  $A{l}_2{S}_3$ is formed when each aluminium loses three electrons and each sulphur gains two electrons and both the ions attain noble gas configuration.

The electronic configuration of $Al$ is:

  $\left[Ne\right]3s^{2}3p^1$

The electronic configuration of $A{l}^{3+}$ is:

  $\left[Ne\right]$

The electronic configuration of $S$ is:

  $\left[Ne\right]3s^{2}3p^4$

The electronic configuration of $S^{2-}$ is:

  $\left[Ar\right]$