Chapter 11, Problem 44A

(a)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of phosphorus with $Z=15$ 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 44A

Valence electronic configuration using previous noble gas of phosphorus with $Z=15$ is as follows:

  $\left[\text{Ne}\right]3s^{2}3p^3$

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 phosphorus with $Z=15$ is as follows:

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

Hence valence electronic configuration using previous noble gas of phosphorus with $Z=15$ is as follows:

  $\left[\text{Ne}\right]3s^{2}3p^3$

(b)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of chlorine with $Z=17$ 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 44A

Valence electronic configuration using previous noble gas of chlorine with $Z=17$ is as follows:

  $\left[\text{Ne}\right]3s^{2}3p^5$

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 chlorine with $Z=17$ is as follows:

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

Hence valence electronic configuration using previous noble gas of chlorine with $Z=17$ is as follows:

  $\left[\text{Ne}\right]3s^{2}3p^5$

(c)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of magnesium with $Z=12$ 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 44A

Valence electronic configuration using previous noble gas of magnesium with $Z=12$ is as follows:

  $\left[\text{Ne}\right]3s^2$

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 magnesium with $Z=12$ is as follows:

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

Hence valence electronic configuration using previous noble gas of magnesium with $Z=12$ is as follows:

  $\left[\text{Ne}\right]3s^2$

(d)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of zinc with $Z=30$ 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 44A

Valence electronic configuration using previous noble gas of zinc with $Z=30$ is as follows:

  $\left[\text{Ar}\right]3d^{10}4s^2$

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 zinc with $Z=30$ is as follows:

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

Hence valence electronic configuration using previous noble gas of zinc with $Z=30$ is as follows:

  $\left[\text{Ar}\right]3d^{10}4s^2$