Chapter 11, Problem 43A

(a)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of calcium with $Z=20$ 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 43A

Valence electronic configuration using previous noble gas of calcium with $Z=20$ is as follows:

  $\left[\text{Ar}\right]4s^2$

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 calcium with $Z=20$ is as follows:

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

Hence valence electronic configuration using previous noble gas of calcium with $Z=20$ is as follows:

  $\left[\text{Ar}\right]4s^2$

(b)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of francium with $Z=87$ 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 43A

Valence electronic configuration using previous noble gas of francium with $Z=87$ is as follows:

  $\left[\text{Rn}\right]7s^1$

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 francium with $Z=87$ is as follows:

  $\text{1}{s}^2\text{2}{s}^{2}2p^{6}3s^{2}3p^{5}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{6}7s^1$

Hence valence electronic configuration using previous noble gas of francium with $Z=87$ is as follows:

  $\left[\text{Rn}\right]7s^1$

(c)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of yttrium with $Z=39$ 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 43A

Valence electronic configuration using previous noble gas of yttrium with $Z=39$ is as follows:

  $\left[\text{Kr}\right]4d^{1}5s^2$

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 yttrium with $Z=39$ is as follows:

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

Hence valence electronic configuration using previous noble gas of yttrium with $Z=39$ is as follows:

  $\left[\text{Kr}\right]4d^{1}5s^2$

(d)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of cerium with $Z=58$ 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 43A

Valence electronic configuration using previous noble gas of cerium with $Z=58$ is as follows:

  $\left[\text{Xe}\right]4f^{1}5d^{1}6s^2$

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 zinc with cerium with $Z=58$ is as follows:

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

Hence valence electronic configuration using previous noble gas of cerium with $Z=58$ is as follows:

  $\left[\text{Xe}\right]4f^{1}5d^{1}6s^2$