Chapter 11, Problem 41A

(a)

Interpretation Introduction

Interpretation:

Valence electrons of sodium with $Z=11$ 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 41A

Since $3s$ is the valence shell therefore valence electron of sodium with $Z=11$ is 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 sodium with $Z=11$ is as follows:

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

Since $3s$ is the valence shell therefore valence electron of sodium with $Z=11$ is 1.

(b)

Interpretation Introduction

Interpretation:

Valence electrons 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 41A

Since $4s^2$ is the valence shell therefore valence electron of calcium with $Z=20$ is 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:

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

Since $4s^2$ is the valence shell therefore valence electron of calcium with $Z=20$ is 2.

(c)

Interpretation Introduction

Interpretation:

Valence electrons of iodine with $Z=53$ 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 41A

Since $5s5p$ is the valence shell therefore valence electron of iodine with $Z=53$ is 7.

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 iodine with $Z=53$ is as follows:

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

Since $5s5p$ is the valence shell therefore valence electron of iodine with $Z=53$ is 7.

(d)

Interpretation Introduction

Interpretation:

Valence electrons of nitrogen with $Z=7$ 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 41A

Since $2s2p$ is the valence shell therefore valence electron of nitrogen with $Z=7$ is 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 nitrogen with $Z=7$ is as follows:

  $\text{1}{s}^2\text{2}{s}^{2}2p^3$

Since $2s2p$ is the valence shell therefore valence electron of nitrogen with $Z=7$ is 5.