Chapter 12, Problem 1E

Write the electronic configuration for the ions of the third period element that form monoatomic ions that are isoelectronic with a noble gas atom. Also, identify the noble gas atoms having those configurations.

Interpretation Introduction

Interpretation:

The electronic configuration for the ions of the third period element that form monoatomic ions that are iso-electronic with a noble gas atom is to be stated. The noble gas atoms having those configurations are to be identified.

Concept introduction:

The arrangement of electrons in the energy sublevels is explained by the electron configuration of an atom. The electron configuration of an atom can be written according to the arrangement of energy levels as shown below.

$1\text{s}<2\text{s}<2\text{p}<3\text{s}<3\text{p}<4\text{s}<3\text{d}<4\text{p}<5\text{s}<4\text{d}<5\text{p}<6\text{s}$

In the modern periodic table, the horizontal rows are known as periods, whereas vertical columns are known as groups. There are $7$ periods and $18$ groups in a modern periodic table.

Answer to Problem 1E

The electronic configuration of ${\text{Na}}^{+}$ is $1s^{2}2s^{2}2p^6$. It is isoelectronic with with neon, $\text{Ne}$. The electronic configuration of ${\text{Cl}}^{-}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$. It is isoelectronic with argon, $\text{Ar}$.

Explanation of Solution

An iso-electronic series is a group of ions that contain the same number of electrons.

The elements in the third period are $\text{Na,}\text{Mg,}\text{Al,}\text{Si,}\text{P,}\text{S,}\text{Cl,}\text{Ar}$. Among these elements, $\text{Na}$ and $\text{Cl}$ are monoatomic elements.

The atomic number of $\text{Na}$ is $11$.

The electronic configuration of $\text{Na}$ is $1s^{2}2s^{2}2p^{6}3s^1$. It can lose $1$ electron form the outermost shell. The resulting cation, ${\text{Na}}^{+}$ is isoelectronic with neon, $\text{Ne}$.

The electronic configuration of ${\text{Na}}^{+}$ is $1s^{2}2s^{2}2p^{6}3s^0$.

The electronic configuration of $\text{Ne}$ is $1s^{2}2s^{2}2p^{6}3s^0$.

The atomic number of $\text{Cl}$ is $17$.

The electronic configuration of $\text{Cl}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^5$. It can gain $1$ electron in the outermost shell. The resulting anion, ${\text{Cl}}^{-}$ is isoelectronic with argon, $\text{Ar}$.

The electronic configuration of ${\text{Cl}}^{-}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$.

The electronic configuration of $\text{Ar}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$.

Conclusion

The electronic configuration of ${\text{Na}}^{+}$ is $1s^{2}2s^{2}2p^{6}3s^0$. It is isoelectronic with with neon, $\text{Ne}$. The electronic configuration of ${\text{Cl}}^{-}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$. It is isoelectronic with argon, $\text{Ar}$.