Chapter 11, Problem 47A

(a)

Interpretation Introduction

Interpretation:

Set of orbital that is last being filled in plutonium with $Z=94$ should be indicated.

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 47A

Set of orbital that is last being filled in plutonium with $Z=94$ is $5f^6$ .

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 plutonium with $Z=94$ is as follows:

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

Hence set of orbital that is last being filled in plutonium with $Z=94$ is $5f^6$ and can be shown as follows:

  $\left[Rn\right]7s^{2}5f^6$

(b)

Interpretation Introduction

Interpretation:

Set of orbital that is last being filled in nobelium with $Z=102$ should be indicated.

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 47A

Set of orbital that is last being filled in plutonium with $Z=94$ is $5f^{14}$ .

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 nobelium with $Z=102$ is as follows:

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

Hence set of orbital that is last being filled in nobelium with $Z=102$ is $5f^{14}$ and can be shown as follows:

  $\left[\text{Rn}\right]7s^{2}5f^{14}$

(c)

Interpretation Introduction

Interpretation:

Set of orbital that is last being filled in praseodymium with $Z=59$ should be indicated.

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 47A

Set of orbital that is last being filled in praseodymium with $Z=59$ is $4f^3$ .

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 praseodymium with $Z=59$ is as follows:

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

Hence set of orbital that is last being filled in praseodymium with $Z=59$ is $4f^3$ and can be shown as follows:

  $\left[\text{Xe}\right]6s^{2}4f^3$

(d)

Interpretation Introduction

Interpretation:

Set of orbital that is last being filled in radon with $Z=86$ should be indicated.

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 47A

Set of orbital that is last being filled in radon with $Z=86$ is $5f^6$ .

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 radon with $Z=86$ is as follows:

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

Hence set of orbitals that is last being filled in with $Z=86$ is $6p^6$ and can be shown as follows:

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