Chapter 8, Problem 50E

Write electron configurations for

a. the cations Sr²⁺, Cs⁺, In⁺, and Pb²⁺.

b. the anions P³⁻, S²⁻, and Br⁻.

Interpretation Introduction

Interpretation: The electronic configuration of given cation and anion is to be written.

Concept introduction: The electronic configuration is a method to explain the atomic orbitals in ground state. It is the representation of distribution of electrons of atoms and molecules. In ground state of an atom, the electrons fill in atomic orbitals according to the Aufbau principle.

Figure 1

To determine: The electronic configuration of cations ${\text{Sr}}^{2+},{\text{Cs}}^{+},{\text{In}}^{+}$ and ${\text{Pb}}^{2+}$.

Answer to Problem 50E

Answer

The electronic configuration of cations is as follows,

The electronic configuration of cation ${\text{Sr}}^{2+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}$.

The electronic configuration of ${\text{Cs}}^{+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}$. The electronic configuration of ${\text{In}}^{+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}$.

The electronic configuration of ${\text{Pb}}^{2+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}$

Explanation of Solution

• To determine: The electronic configuration of cation ${\text{Sr}}^{2+}$.

The atomic number of strontium $\left(\text{Sr}\right)$ is 38. The electronic configuration of strontium is,

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

It loses two electrons to get most stable inert gas configuration and form ${\text{Sr}}^{2+}$. The electronic configuration of ${\text{Sr}}^{2+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}$.

• To determine: The electronic configuration of cation ${\text{Cs}}^{+}$.

The atomic number of Cesium is 55. The electronic configuration of Cesium is,

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

It can lose an electron to get most stable inert gas configuration and form ${\text{Cs}}^{+}$. The electronic configuration of ${\text{Cs}}^{+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}$.

• To determine: The electronic configuration of cation ${\text{In}}^{+}$.

The atomic number of Indium is 49. The electronic configuration of Indium is,

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

It can lose an electron to get most stable state configuration and form ${\text{In}}^{+}$. The electronic configuration of ${\text{In}}^{+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}$.

• To determine: The electronic configuration of cation ${\text{Pb}}^{2+}$.

The atomic number of lead is 82. The electronic configuration of lead is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}{\text{6p}}^{\text{2}}$

It loses two electrons to get most stable state configuration and form ${\text{Pb}}^{2+}$. The electronic configuration of ${\text{Pb}}^{2+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}$.

Conclusion

In ionic compounds, cation loses electrons to get stable state by acquiring nearest noble gas electronic configuration.

The electronic configuration of cation ${\text{Sr}}^{2+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}$.

The electronic configuration of cation ${\text{Cs}}^{+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}$.

The electronic configuration of cation ${\text{In}}^{+}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}$.

The electronic configuration of cation ${\text{Pb}}^{2+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}$.

Interpretation Introduction

Interpretation: The electronic configuration of given cation and anion is to be written.

Concept introduction: The electronic configuration is a method to explain the atomic orbitals in ground state. It is the representation of distribution of electrons of atoms and molecules. In ground state of an atom, the electrons fill in atomic orbitals according to the Aufbau principle.

Figure 1

To determine: The electronic configuration of anions ${\text{P}}^{3-},{\text{S}}^{2-}$ and ${\text{Br}}^{-}$.

Answer to Problem 50E

Answer

The electronic configuration of anions is as follows,

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

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

            The electronic configuration of ${\text{Br}}^{-}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^6$.

Explanation of Solution

• To determine: The electronic configuration of anion ${\text{P}}^{3-}$.

The atomic number of Phosphorous is 15. The electronic configuration of Phosphorous is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{3}}$

It can gain three electrons to get most stable inert gas configuration and form ${\text{P}}^{3-}$. The electronic configuration of ${\text{P}}^{3-}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^6$.

• To determine: The electronic configuration of anion ${\text{S}}^{2-}$.

The atomic number of sulfur is 16. The electronic configuration of sulfur is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{4}}$

It can gain two electrons to get most stable inert gas configuration and form ${\text{S}}^{2-}$. The electronic configuration of ${\text{S}}^{2-}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^6$.

• To determine: The electronic configuration of anion ${\text{Br}}^{-}$.

The atomic number of bromine is 35. The electronic configuration of bromine is,

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

It can gain an electron to get most stable inert gas configuration and form ${\text{Br}}^{-}$. The electronic configuration of ${\text{Br}}^{-}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^6$.

Conclusion

In ionic compounds, anions gain electrons to get stable state by acquiring nearest noble gas electronic configuration.

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

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

The electronic configuration of anions ${\text{Br}}^{-}$ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^6$.