Chapter 12, Problem 44CR

Interpretation Introduction

(a)

Interpretation:

Determine the electron configuration of atoms using the appropriate noble gas configuration of the core electrons.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. These positions were explained by several models developed by different scientists. Principal energy level, Type (shape) of orbital and No: of electrons in the orbital are respectively represented by an atom’s electron configuration.

Answer to Problem 44CR

$\text{S}\text{r}$, $\text{Z}=38$ ( $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6 }4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^2$

Using noble gas configuration( $\left[\text{K}\text{r}\right]$

$5{\text{s}}^2$.

Explanation of Solution

Order in which orbitals fill to produce the atoms in periodic table as follows:

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

There are only one $\text{s}$ orbital, three no: of $\text{p}$ orbitals, five no: of $\text{d}$ orbitals and seven no: of $\text{f}$ orbitals can be present and every one of above mentioned orbitals include, maximum two no: of electrons which are opposite in spin.

Noble gas configurations are as follows:

Period No

$\text{H}\text{e}$, $\text{Z}=2$ ; $1{\text{s}}^2$

$\text{N}\text{e}$, $\text{Z}=\text{ 10}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$

$\text{A}\text{r}$, $\text{Z}=\text{ 18}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^6$

$\text{K}\text{r}$, $\text{Z}=\text{ 36}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

$\text{X}\text{e}$, $\text{Z}=\text{ 54}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^6$

$\text{R}\text{n}$, $\text{Z}=86$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^{6}6{\text{s}}^{2}4{\text{f}}^{14}5{\text{d}}^{10}6{\text{p}}^6$

Thus, electronic configuration of Sr will be:

$\text{S}\text{r}$, $\text{Z}=38$ ( $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6 }4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^2$

Using noble gas configuration( $\left[\text{K}\text{r}\right]$

$5{\text{s}}^2$.

Interpretation Introduction

(b)

Interpretation:

Determine the electron configuration of atoms using the appropriate noble gas configuration of the core electrons.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. These positions were explained by several models developed by different scientists. Principal energy level, Type (shape) of orbital and No: of electrons in the orbital are respectively represented by an atom’s electron configuration.

Answer to Problem 44CR

$\text{A}\text{l}$, $\text{Z}=13$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^3$

Using noble gas configuration( $\left[\text{N}\text{e}\right]$

$3{\text{s}}^{2}3{\text{p}}^3$.

Explanation of Solution

Order in which orbitals fill to produce the atoms in periodic table as follows:

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

There are only one $\text{s}$ orbital, three no: of $\text{p}$ orbitals, five no: of $\text{d}$ orbitals and seven no: of $\text{f}$ orbitals can be present and every one of above mentioned orbitals include, maximum two no: of electrons which are opposite in spin.

Noble gas configurations are as follows:

Period No

$\text{H}\text{e}$, $\text{Z}=2$ ; $1{\text{s}}^2$

$\text{N}\text{e}$, $\text{Z}=\text{ 10}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$

$\text{A}\text{r}$, $\text{Z}=\text{ 18}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^6$

$\text{K}\text{r}$, $\text{Z}=\text{ 36}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

$\text{X}\text{e}$, $\text{Z}=\text{ 54}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^6$

$\text{R}\text{n}$, $\text{Z}=86$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^{6}6{\text{s}}^{2}4{\text{f}}^{14}5{\text{d}}^{10}6{\text{p}}^6$

Thus, the electronic configuration of Al will be:

$\text{A}\text{l}$, $\text{Z}=13$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^3$

Using noble gas configuration( $\left[\text{N}\text{e}\right]$

$3{\text{s}}^{2}3{\text{p}}^3$.

Interpretation Introduction

(c)

Interpretation:

Determine the electron configuration of atoms using the appropriate noble gas configuration of the core electrons.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. These positions were explained by several models developed by different scientists. Principal energy level, Type (shape) of orbital and No: of electrons in the orbital are respectively represented by an atom’s electron configuration.

Answer to Problem 44CR

$\text{C}\text{l},$

$\text{Z}=17$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^5$

Using noble gas configuration( $\left[\text{N}\text{e}\right]$

$3{\text{s}}^{2}3{\text{p}}^5$.

Explanation of Solution

Order in which orbitals fill to produce the atoms in periodic table as follows:

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

There are only one $\text{s}$ orbital, three no: of $\text{p}$ orbitals, five no: of $\text{d}$ orbitals and seven no: of $\text{f}$ orbitals can be present and every one of above mentioned orbitals include, maximum two no: of electrons which are opposite in spin.

Noble gas configurations are as follows:

Period No

$\text{H}\text{e}$, $\text{Z}=2$ ; $1{\text{s}}^2$

$\text{N}\text{e}$, $\text{Z}=\text{ 10}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$

$\text{A}\text{r}$, $\text{Z}=\text{ 18}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^6$

$\text{K}\text{r}$, $\text{Z}=\text{ 36}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

$\text{X}\text{e}$, $\text{Z}=\text{ 54}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^6$

$\text{R}\text{n}$, $\text{Z}=86$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^{6}6{\text{s}}^{2}4{\text{f}}^{14}5{\text{d}}^{10}6{\text{p}}^6$

Thus, the electronic configuration of Cl will be:

$\text{C}\text{l},$

$\text{Z}=17$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^5$

Using noble gas configuration( $\left[\text{N}\text{e}\right]$

$3{\text{s}}^{2}3{\text{p}}^5$.

Interpretation Introduction

(d)

Interpretation:

Determine the electron configuration of atoms using the appropriate noble gas configuration of the core electrons.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. These positions were explained by several models developed by different scientists. Principal energy level, Type (shape) of orbital and No: of electrons in the orbital are respectively represented by an atom’s electron configuration.

Answer to Problem 44CR

$\text{K},$

$\text{Z}=19$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^1$

Using noble gas configuration( $\left[\text{A}\text{r}\right]$ 4s¹

Explanation of Solution

Order in which orbitals fill to produce the atoms in periodic table as follows:

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

There are only one $\text{s}$ orbital, three no: of $\text{p}$ orbitals, five no: of $\text{d}$ orbitals and seven no: of $\text{f}$ orbitals can be present and every one of above mentioned orbitals include, maximum two no: of electrons which are opposite in spin.

Noble gas configurations are as follows:

Period No

$\text{H}\text{e}$, $\text{Z}=2$ ; $1{\text{s}}^2$

$\text{N}\text{e}$, $\text{Z}=\text{ 10}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$

$\text{A}\text{r}$, $\text{Z}=\text{ 18}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^6$

$\text{K}\text{r}$, $\text{Z}=\text{ 36}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

$\text{X}\text{e}$, $\text{Z}=\text{ 54}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^6$

$\text{R}\text{n}$, $\text{Z}=86$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^{6}6{\text{s}}^{2}4{\text{f}}^{14}5{\text{d}}^{10}6{\text{p}}^6$

Thus, the electronic configuration of K will be:

$\text{K},$

$\text{Z}=19$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^1$

Using noble gas configuration( $\left[\text{A}\text{r}\right]$ 4s¹

Interpretation Introduction

(e)

Interpretation:

Determine the electron configuration of atoms using the appropriate noble gas configuration of the core electrons.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. These positions were explained by several models developed by different scientists. Principal energy level, Type (shape) of orbital and No: of electrons in the orbital are respectively represented by an atom’s electron configuration.

Answer to Problem 44CR

$\text{S}$, $\text{Z}=16$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^2$

Using noble gas configuration( $\left[\text{N}\text{e}\right]$

$3{\text{s}}^2$.

Explanation of Solution

Order in which orbitals fill to produce the atoms in periodic table as follows:

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

There are only one $\text{s}$ orbital, three no: of $\text{p}$ orbitals, five no: of $\text{d}$ orbitals and seven no: of $\text{f}$ orbitals can be present and every one of above mentioned orbitals include, maximum two no: of electrons which are opposite in spin.

Noble gas configurations are as follows:

Period No

$\text{H}\text{e}$, $\text{Z}=2$ ; $1{\text{s}}^2$

$\text{N}\text{e}$, $\text{Z}=\text{ 10}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$

$\text{A}\text{r}$, $\text{Z}=\text{ 18}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^6$

$\text{K}\text{r}$, $\text{Z}=\text{ 36}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

$\text{X}\text{e}$, $\text{Z}=\text{ 54}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^6$

$\text{R}\text{n}$, $\text{Z}=86$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^{6}6{\text{s}}^{2}4{\text{f}}^{14}5{\text{d}}^{10}6{\text{p}}^6$

Thus, the electronic configuration of S will be:

$\text{S}$, $\text{Z}=16$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^2$

Using noble gas configuration( $\left[\text{N}\text{e}\right]$

$3{\text{s}}^2$.

Interpretation Introduction

(f)

Interpretation:

Determine the electron configuration of atoms using the appropriate noble gas configuration of the core electrons.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. These positions were explained by several models developed by different scientists. Principal energy level, Type (shape) of orbital and No: of electrons in the orbital are respectively represented by an atom’s electron configuration.

Answer to Problem 44CR

$\text{A}\text{s}$, $\text{Z}=33$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^3$

Using noble gas configuration( $\left[\text{A}\text{r}\right]$

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

Explanation of Solution

Order in which orbitals fill to produce the atoms in periodic table as follows:

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

There are only one $\text{s}$ orbital, three no: of $\text{p}$ orbitals, five no: of $\text{d}$ orbitals and seven no: of $\text{f}$ orbitals can be present and every one of above mentioned orbitals include, maximum two no: of electrons which are opposite in spin.

Noble gas configurations are as follows:

Period No

$\text{H}\text{e}$, $\text{Z}=2$ ; $1{\text{s}}^2$

$\text{N}\text{e}$, $\text{Z}=\text{ 10}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$

$\text{A}\text{r}$, $\text{Z}=\text{ 18}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^6$

$\text{K}\text{r}$, $\text{Z}=\text{ 36}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^6$

$\text{X}\text{e}$, $\text{Z}=\text{ 54}$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^6$

$\text{R}\text{n}$, $\text{Z}=86$ ; $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^{6}5{\text{s}}^{2}4{\text{d}}^{10}5{\text{p}}^{6}6{\text{s}}^{2}4{\text{f}}^{14}5{\text{d}}^{10}6{\text{p}}^6$

Thus, the electronic configuration of As will be:

$\text{A}\text{s}$, $\text{Z}=33$ → $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^3$

Using noble gas configuration( $\left[\text{A}\text{r}\right]$

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