Chapter 7, Problem 6PS

a)

Interpretation Introduction

Interpretation:

The electron configuration of  Strontium ($\text{Sr}$) has to be predicted using its orbital notation box and noble gas electron filling method.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons of a given molecule or respective atoms in atomic or molecular orbitals. The important there rules for electronic configuration given below:

Aufbau principle: This rule statues that ground state of an atom or ions electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. If consider the 1s shell is filled the 2s subshell is occupied.

Hund's Rule: The every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Pauli exclusion rule: an atomic orbital may describe at most two electrons, each with opposite spin direction.

Explanation of Solution

Let us consider the orbital filling method of Strontium (Sr)

1. 1. The spdf and orbital box notations, assign the 38 electrons for Sr to orbitals based on the order of filling.
2. 2. The noble gas notation: The first 36 electrons are identified by the symbol of the krypton (Kr) it is a one of noble gas. The remaining (2) electrons are assigning to the 5s, sub-shells. Thus, the Hund’s rule is followed as shown in the box notation.

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Strontium}\text{(Sr)}\text{= 38}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^{6}3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^{6}5{\text{s}}^{\text{2}}\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^{\text{3}}5{\text{s}}^{\text{2}}\\ spdf\text{with noble gas notation}=[\text{Kr}]5{\text{s}}^{\text{2}}\\ \text{Orbital}\text{box}\text{notation }\text{= [Kr]}\boxed{\uparrow \downarrow }\\ 5{\text{s}}^2\end{array}$

(b)

Interpretation Introduction

Interpretation:

Given the d-block element Zirconium ($\text{Zr}$) electronic configurations has to be derived using its orbital notation box and noble gas electron filling method.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons of a given molecule or respective atoms in atomic or molecular orbitals.

The important there rules for electronic configuration given below:

Aufbau principle: This rule statues that ground state of an atom or ions electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. If consider the 1s shell is filled the 2s subshell is occupied.

Hund's Rule: The every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Pauli exclusion rule: an atomic orbital may describe at most two electrons, each with opposite spin direction.

Explanation of Solution

Next, consider the electronic configuration of Zirconium ($\text{Zr}$) atom using (spdf) orbital notation method.

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Zirconium}\text{(Zr)}\text{= 40}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{(As)}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^{6}3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^{6}4{\text{d}}^{\text{2}}5{\text{s}}^{\text{2}}\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6\\ \boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{\uparrow \downarrow }\\ 4{\text{d}}^{2}5{\text{s}}^{\text{2}}\\ spdf\text{with noble gas notation}=[\text{Kr}]5{\text{s}}^{\text{2}}[\because \text{Atomic}\text{number}\text{of}\text{Kr}=36]\\ \text{Orbital}\text{box}\text{notation }\text{= [Kr]}\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{\uparrow \downarrow }\\ 4{\text{d}}^{2}5{\text{s}}^2\end{array}$

Sr is the alkaline earth metals, this element present in s-block element and 5^th period and group 2 in periodic table.

(c)

Interpretation Introduction

Interpretation:

The electronic configuration of Rhodium using its orbital notation box and noble gas electron filling method has to be predicted.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons of a given molecule or respective atoms in atomic or molecular orbitals.

The important there rules for electronic configuration given below:

Aufbau principle: This rule statues that ground state of an atom or ions electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. If consider the 1s shell is filled the 2s subshell is occupied.

Hund's Rule: The every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Pauli exclusion rule: an atomic orbital may describe at most two electrons, each with opposite spin direction.

Explanation of Solution

Let us consider the orbital filling method of Rhodium ($\text{Rh}$)

1. 1. The spdf and orbital box notations, assign the 45 electrons for $\text{Rh}$ to orbitals based on the order of filling.
2. 2. The noble gas notation: The first (36) electrons are identified by the symbol of the krypton (Kr). The remaining (2) electrons are assigning to the 5s sub-shell. Thus, the Hund’s rule is fallowed as shown in the box notation.

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Rhodium}\text{(Rh)}\text{=45}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{(Rh)}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^{6}3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^{6}4{\text{d}}^{8}5{\text{s}}^1\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\\ 4{\text{d}}^{8}5{\text{s}}^1\\ spdf\text{with noble gas notation}=[\text{Kr}]5{\text{s}}^{\text{2}}[\because \text{Atomic}\text{number}\text{of}\text{Kr}=36]\\ \text{Orbital}\text{box}\text{notation }\text{= [Kr]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\\ 4{\text{d}}^{8}5{\text{s}}^1\end{array}$

(d)

Interpretation Introduction

Interpretation:

The electronic configuration of Tin ($\text{Sn}$) atom has to be predicted using its orbital notation box and noble gas electron filling method.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons of a given molecule or respective atoms in atomic or molecular orbitals.

The important there rules for electronic configuration given below:

Aufbau principle: This rule statues that ground state of an atom or ions electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. If consider the 1s shell is filled the 2s subshell is occupied.

Hund's Rule: The every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Pauli exclusion rule: an atomic orbital may describe at most two electrons, each with opposite spin direction.

Explanation of Solution

Finally we consider the electronic configuration of Tin ($\text{Sn}$) atom.

1. 1. The (spdf) and orbital box notations, assign the (50) electrons for (Sn) to orbitals based on the order of filling.
2. 2. The noble gas notation: The first (36) electrons are identified by the symbol of the krypton (Kr) it is a one of noble gas. The remaining (14) electrons are assigned to the 4d, 5s, and 5p sub-shells. Hence the Hund’s rule is followed in the box notation.

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Tin}\text{(Sn)}\text{=50}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{(Sn)}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^{6}3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^{6}4{\text{d}}^{10}5{\text{s}}^{2}5{\text{p}}^{\text{2}}\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\\ 4{\text{d}}^{10}5{\text{s}}^{2}5{\text{p}}^{\text{2}}\\ spdf\text{with noble gas notation}=[\text{Kr}]4{\text{d}}^{10}5{\text{s}}^{\text{2}}5{\text{p}}^{\text{2}}[\because \text{Atomic}\text{number}\text{of}\text{Kr}=36]\\ \text{Orbital}\text{box}\text{notation }\text{= [Kr]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\\ 4{\text{d}}^{10}5{\text{s}}^{2}5{\text{p}}^{\text{2}}\end{array}$