Chapter 11, Problem 48A

(a)

Interpretation Introduction

Interpretation:

Short hand valence electron configuration of samarium with $Z=62$ should be determined.

Concept introduction:

Electronic configuration can be assigned to any elements in ground when they follow certain rules like Hund rule, Pauli Exclusion Principle and Aufbau rule. If atomic number of an element is $Z$ then $Z$ numbers of electrons are filled into the orbitals which are arranged in increasing order of energy.

No two electrons in an atom can have same group of four quantum numbers and this is Pauli Exclusion Principle.

While filling of orbital’s, electron first enters to each energy level with degenerate energy before paring of electron begins and this is Hund’s rules.

Answer to Problem 48A

Short hand valence electron configuration of samarium with $Z=62$ is as follows:

  $\left[\text{Xe}\right]4f^{6}6s^2$

Explanation of Solution

As per Aufbau rule electrons are filled in lower energy orbitals that are closer to the nucleus before they are filled in higher energy ones. The order of orbital arranged in their increasing energies is as follows:

  $\begin{array}{l}1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<\\ 6s<4f<5d<6p<7s<5f<6d<7p\end{array}$ .

So complete electronic configuration of samarium with $Z=62$ is as follows:

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

Hence the short hand valence electron configuration of samarium with $Z=62$ is

  $\left[\text{Xe}\right]4f^{6}6s^2$

(b)

Interpretation Introduction

Interpretation:

Short hand valence electron configuration of osmium with $Z=76$ should be determined.

Concept introduction:

Electronic configuration can be assigned to any elements in ground when they follow certain rules like Hund rule, Pauli Exclusion Principle and Aufbau rule. If atomic number of an element is $Z$ then $Z$ numbers of electrons are filled into the orbitals which are arranged in increasing order of energy.

No two electrons in an atom can have same group of four quantum numbers and this is Pauli Exclusion Principle.

While filling of orbital’s, electron first enters to each energy level with degenerate energy before paring of electron begins and this is Hund’s rules.

Answer to Problem 48A

Short hand valence electron configuration of osmium with $Z=76$ is as follows:

  $\left[\text{Xe}\right]4f^{14}5d^{6}6s^2$

Explanation of Solution

As per Aufbau rule electrons are filled in lower energy orbitals that are closer to the nucleus before they are filled in higher energy ones. The order of orbital arranged in their increasing energies is as follows:

  $\begin{array}{l}1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<\\ 6s<4f<5d<6p<7s<5f<6d<7p\end{array}$ .

So complete electronic configuration of osmium with $Z=76$ is as follows:

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

Hence the short hand valence electron configuration of osmium with $Z=76$ is

  $\left[\text{Xe}\right]4f^{14}5d^{6}6s^2$

(c)

Interpretation Introduction

Interpretation:

Short hand valence electron configuration of Iron with $Z=26$ should be determined.

Concept introduction:

Electronic configuration can be assigned to any elements in ground when they follow certain rules like Hund rule, Pauli Exclusion Principle and Aufbau rule. If atomic number of an element is $Z$ then $Z$ numbers of electrons are filled into the orbitals which are arranged in increasing order of energy.

No two electrons in an atom can have same group of four quantum numbers and this is Pauli Exclusion Principle.

While filling of orbital’s, electron first enters to each energy level with degenerate energy before paring of electron begins and this is Hund’s rules.

Answer to Problem 48A

Short hand valence electron configuration of Iron with $Z=26$ is as follows:

  $\left[\text{Ar}\right]3d^{6}4s^2$

Explanation of Solution

As per Aufbau rule electrons are filled in lower energy orbitals that are closer to the nucleus before they are filled in higher energy ones. The order of orbital arranged in their increasing energies is as follows:

  $\begin{array}{l}1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<\\ 6s<4f<5d<6p<7s<5f<6d<7p\end{array}$ .

So complete electronic configuration of Iron with $Z=26$ is as follows:

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

Hence the short hand valence electron configuration of Iron with $Z=26$ is

  $\left[\text{Ar}\right]3d^{6}4s^2$

(d)

Interpretation Introduction

Interpretation:

Short hand valence electron configuration of $\text{Pb}$ with $Z=82$ should be determined.

Concept introduction:

Electronic configuration can be assigned to any elements in ground when they follow certain rules like Hund rule, Pauli Exclusion Principle and Aufbau rule. If atomic number of an element is $Z$ then $Z$ numbers of electrons are filled into the orbitals which are arranged in increasing order of energy.

No two electrons in an atom can have same group of four quantum numbers and this is Pauli Exclusion Principle.

While filling of orbital’s, electron first enters to each energy level with degenerate energy before paring of electron begins and this is Hund’s rules.

Answer to Problem 48A

Short hand valence electron configuration of $\text{Pb}$ with $Z=82$ is as follows:

  $\left[\text{Xe}\right]4f^{14}5d^{10}6s^{2}6p^2$

Explanation of Solution

As per Aufbau rule electrons are filled in lower energy orbitals that are closer to the nucleus before they are filled in higher energy ones. The order of orbital arranged in their increasing energies is as follows:

  $\begin{array}{l}1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<\\ 6s<4f<5d<6p<7s<5f<6d<7p\end{array}$ .

So complete electronic configuration of $\text{Pb}$ with $Z=82$ is as follows:

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

Hence the short hand valence electron configuration of $\text{Pb}$ with $Z=82$ is

  $\left[\text{Xe}\right]4f^{14}5d^{10}6s^{2}6p^2$