Chapter 8, Problem 8.42QE

What neutral atoms are isoelectronic with the following ions?

1. (a) Pb⁴⁺
2. (b) Br^–
3. (c) S^2–
4. (d) Ni³⁺

Interpretation Introduction

Interpretation:

The neutral atom that is isoelectronic with ${\text{Pb}}^{4+}$ 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 8.42QE

The neutral atom that is isoelectronic with ${\text{Pb}}^{4+}$ is $\text{Pt}$.

Explanation of Solution

The atomic number of $\text{Pb}$ is 82 and its electronic configuration is $\left[\text{Xe}\right]4f^{14}5d^{10}6s^{2}6p^2$. ${\text{Pb}}^{4+}$ is formed by the loss of four electrons from the valence shell of the lead atom so its configuration becomes $\left[\text{Xe}\right]4f^{14}5d^{10}$. The number of electrons in ${\text{Pb}}^{4+}$ is 78.

The neutral atom that is isoelectronic with ${\text{Pb}}^{4+}$ contains the same number of electrons as in ${\text{Pb}}^{4+}$. The neutral atom with 78 electrons is platinum. So $\text{Pt}$ is isoelectronic with ${\text{Pb}}^{4+}$.

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 8.42QE

The neutral atom that is isoelectronic with ${\text{Br}}^{-}$ is $\text{Kr}$.

Explanation of Solution

The atomic number of $\text{Br}$ is 35 and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^5$. ${\text{Br}}^{-}$ is formed by the gain of one electron to the valence shell of the bromine atom so its configuration becomes $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^6$. The number of electrons in ${\text{Br}}^{-}$ is 36.

The neutral atom that is isoelectronic with ${\text{Br}}^{-}$ contains the same number of electrons as in ${\text{Br}}^{-}$. The neutral atom with 36 electrons is krypton. So $\text{Kr}$ is isoelectronic with ${\text{Br}}^{-}$.

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 8.42QE

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

Explanation of Solution

The atomic number of $\text{S}$ is 16 and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^4$. ${\text{S}}^{2-}$ is formed by the gain of two electrons to the valence shell of the sulfur atom so its configuration becomes $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$. The number of electrons in ${\text{S}}^{2-}$ is 18.

The neutral atom that is isoelectronic with ${\text{S}}^{2-}$ contains the same number of electrons as in ${\text{S}}^{2-}$. The neutral atom with 18 electrons is argon. So $\text{Ar}$ is isoelectronic with ${\text{S}}^{2-}$.

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 8.42QE

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

Explanation of Solution

The atomic number of $\text{Ni}$ is 28 and its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{8}4s^2$. ${\text{Ni}}^{3+}$ is formed by the loss of three electrons from the valence shell of nickel atom so its configuration becomes $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^7$. The number of electrons in ${\text{Ni}}^{3+}$ is 25.

The neutral atom that is isoelectronic with ${\text{Ni}}^{3+}$ contains the same number of electrons as in ${\text{Ni}}^{3+}$. The neutral atom with 25 electrons is manganese. So $\text{Mn}$ is isoelectronic with ${\text{Ni}}^{3+}$.