Chapter 8, Problem 8.76P

(a)

Interpretation Introduction

Interpretation:

The charge and full ground state electron configuration of the ion likely to be formed by $\text{Rb}$ is to be determined.

Concept introduction:

The electronic configuration of an atom describes how the electrons are ditributed in its atom among various shells and subshells. It is used to predict the chemical, physical, electrical and magnetic properties of the substance. The full electronic configuration of an atom tells about the distribution of electrons in the various atomic orbitals.

Answer to Problem 8.76P

The charge of monoatomic ion likely to be formed by $\text{Rb}$ atom is +1.

The full ground state electron configuration of the ion likely to be formed by $\text{Rb}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$.

Explanation of Solution

The atomic number of rubidium is 37 so its ground state electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^1$. Its configuration is just one electron more of the stable noble gas configuration of krypton so it loses one electron to become ${\text{Rb}}^{+}$ whose full ground state electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$.

The ion formation occurs as follows:

$\text{Rb}\left(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^1\right)-{\text{e}}^{-}\to {\text{Rb}}^{+}\left(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6\right)$

So the charge on the ion formed by $\text{Rb}$ is +1 and its full ground state electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$.

Conclusion

The charge of monoatomic ion likely to be formed by $\text{Rb}$ atom is +1 and the full ground state electron configuration of the ion likely to be formed by $\text{Rb}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$.

(b)

Interpretation Introduction

Interpretation:

The charge and full ground state electron configuration of the ion likely to be formed by $\text{N}$ is to be determined.

Concept introduction:

The electronic configuration of an atom describes how the electrons are ditributed in its atom among various shells and subshells. It is used to predict the chemical, physical, electrical and magnetic properties of the substance. The full electronic configuration of an atom tells about the distribution of electrons in the various atomic orbitals.

Answer to Problem 8.76P

The charge of monoatomic ion likely to be formed by $\text{N}$ atom is -3.

The full ground state electron configuration of the ion likely to be formed by $\text{N}$ is $1s^{2}2s^{2}2p^6$.

Explanation of Solution

The atomic number of nitrogen is 7 so its ground state electronic configuration is $1s^{2}2s^{2}2p^3$. Its configuration is just three electrons short of the stable noble gas configuration of neon so it gains three electrons to become ${\text{N}}^{3-}$ whose full ground state electron configuration is $1s^{2}2s^{2}2p^6$.

The ion formation occurs as follows:

$\text{N}\left(1s^{2}2s^{2}2p^3\right)+3{\text{e}}^{-}\to {\text{N}}^{3-}\left(1s^{2}2s^{2}2p^6\right)$

So the charge on the ion formed by ${\text{N}}^{}$ is -3 and its full ground state electron configuration is $1s^{2}2s^{2}2p^6$.

Conclusion

The charge of monoatomic ion likely to be formed by $\text{N}$ atom is -3 and the full ground state electron configuration of the ion likely to be formed by $\text{N}$ is $1s^{2}2s^{2}2p^6$.

(c)

Interpretation Introduction

Interpretation:

The charge and full ground state electron configuration of the ion likely to be formed by $\text{Br}$ is to be determined.

Concept introduction:

The electronic configuration of an atom describes how the electrons are ditributed in its atom among various shells and subshells. It is used to predict the chemical, physical, electrical and magnetic properties of the substance. The full electronic configuration of an atom tells about the distribution of electrons in the various atomic orbitals.

Answer to Problem 8.76P

The charge of monoatomic ion likely to be formed by $\text{Br}$ atom is -1.

The full ground state electron configuration of the ion likely to be formed by $\text{Br}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$.

Explanation of Solution

The atomic number of bromine is 35 so its ground state electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^5$ Its configuration is one electron short of the stable noble gas configuration of krypton so it gains one electron to become ${\text{Br}}^{-}$ whose full ground state electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$.

The ion formation occurs as follows:

$\text{Br}\left(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^5\right)+{\text{e}}^{-}\to {\text{Br}}^{-}\left(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6\right)$

So the charge on the ion formed by $\text{Br}$ is -1 and its full ground state electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$.

Conclusion

The charge of monoatomic ion likely to be formed by $\text{Br}$ atom is -1 and the full ground state electron configuration of the ion likely to be formed by $\text{Br}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$.