Chapter 8, Problem 8.41QE

(a)

Interpretation Introduction

Interpretation:

The neutral atom that is isoelectronic with ${\text{O}}^{2-}$ has to be determined.

Concept Introduction:

The group of atoms and ions that consists of the same number of electrons is known as isoelectronic series.

Answer to Problem 8.41QE

The neutral atom that is isoelectronic with ${\text{O}}^{2-}$ is $\text{Ne}$.

Explanation of Solution

The atomic number of $\text{O}$ is 8 and its electronic configuration is $1s^{2}2s^{2}2p^4$. ${\text{O}}^{2-}$ is formed by the gain of two electrons to the valence shell of oxygen atom so its configuration becomes $1s^{2}2s^{2}2p^6$. The number of electrons in ${\text{O}}^{2-}$ is 10.

The neutral atom that is isoelectronic with ${\text{O}}^{2-}$ contains the same number of electrons as in ${\text{O}}^{2-}$. The neutral atom with 10 electrons is neon. So $\text{Ne}$ is isoelectronic with ${\text{O}}^{2-}$.

(b)

Interpretation Introduction

Interpretation:

The neutral atom that is isoelectronic with ${\text{Fe}}^{2+}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 8.41QE

The neutral atom that is isoelectronic with ${\text{Fe}}^{2+}$ is $\text{Cr}$.

Explanation of Solution

The atomic number of $\text{Fe}$ is 26 and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^2$. ${\text{Fe}}^{2+}$ is formed by the loss of two electrons from the valence shell of the iron atom so its configuration becomes $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^6$. The number of electrons in ${\text{Fe}}^{2+}$ is 24.

The neutral atom that is isoelectronic with ${\text{Fe}}^{2+}$ contains the same number of electrons as in ${\text{Fe}}^{2+}$. The neutral atom with 24 electrons is chromium. So $\text{Cr}$ is isoelectronic with ${\text{Fe}}^{2+}$.

(c)

Interpretation Introduction

Interpretation:

The neutral atom that is isoelectronic with ${\text{Fe}}^{3+}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 8.41QE

The neutral atom that is isoelectronic with ${\text{Fe}}^{3+}$ is $\text{V}$.

Explanation of Solution

The atomic number of $\text{Fe}$ is 26 and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^2$. ${\text{Fe}}^{3+}$ is formed by the loss of three electrons from the valence shell of the iron atom so its configuration becomes $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$. The number of electrons in ${\text{Fe}}^{3+}$ is 23.

The neutral atom that is isoelectronic with ${\text{Fe}}^{3+}$ contains the same number of electrons as in ${\text{Fe}}^{3+}$. The neutral atom with 23 electrons is vanadium. So $\text{V}$ is isoelectronic with ${\text{Fe}}^{3+}$.

(d)

Interpretation Introduction

Interpretation:

The neutral atom that is isoelectronic with ${\text{In}}^{+}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 8.41QE

The neutral atom that is isoelectronic with ${\text{In}}^{+}$ is $\text{Cd}$.

Explanation of Solution

The atomic number of $\text{In}$ is 49 and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^1$. ${\text{In}}^{+}$ is formed by the loss of one electron to the valence shell of indium atom so its configuration becomes $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^2$. The number of electrons in ${\text{In}}^{+}$ is 48.

The neutral atom that is isoelectronic with ${\text{In}}^{+}$ contains the same number of electrons as in ${\text{In}}^{+}$. The neutral atom with 48 electrons is cadmium. So $\text{Cd}$ is isoelectronic with ${\text{In}}^{+}$.