Chapter 12, Problem 45CR

Interpretation Introduction

(a)

Interpretation:

Predict the simple binary compound that can be formed by the use of electron configuration of the simple ions which are formed by those elements.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. Depending on that the periodic table is created. Periodic table further defines the simple ion that can be formed by any element. And using those ions, simple binary compounds can be produced.

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

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

Simplest ion form by each group as follows:

If it’s a compound made by a combination of metal and nonmetal elements, called as ionic compounds.

Ionic chemical compound must have a zero charge. If a compound contains ions;

1. Both positive ions (cations) and negative ions (anions) must be there.

2. No: of cations and anions collectively should produce a zero-net charge.

Answer to Problem 45CR

$\text{A}\text{l}\text{C}{\text{l}}_3$.

Explanation of Solution

Al and Cl

$\text{A}\text{l},\text{Z}=13 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{1 }$ ( Group III A element ( Simple ion = $\text{A}{\text{l}}^{3+}$

$\text{C}\text{l},\text{Z}=17 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3\text{p}$ ( Group VII A element ( Simple ion = $\text{C}{\text{l}}^{-}$ $\begin{array}{l}\text{A}{\text{l}}^{3+ }+\text{C}{\text{l}}^{-} = \text{A}\text{l}\text{C}{\text{l}}_3 \\ +3+\left(-1\right)\times 3= 0 \end{array}$.

Interpretation Introduction

(b)

Interpretation:

Predict the simple binary compound that can be formed by the use of electron configuration of the simple ions which are formed by those elements.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. Depending on that the periodic table is created. Periodic table further defines the simple ion that can be formed by any element. And using those ions, simple binary compounds can be produced.

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

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

Simplest ion form by each group as follows:

If it’s a compound made by a combination of metal and nonmetal elements, called as ionic compounds.

Ionic chemical compound must have a zero charge. If a compound contains ions;

1. Both positive ions (cations) and negative ions (anions) must be there.

2. No: of cations and anions collectively should produce a zero-net charge.

Answer to Problem 45CR

${\text{Na}}_{\text{3}}\text{N}$.

Explanation of Solution

Na and N

$\text{N}\text{a},\text{Z}=11 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^1{}^{5 }$ ( Group I A element ( Simple ion = $\text{N}{\text{a}}^{+}$

$\text{N},\text{Z}=8 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^4$ ( Group V A element ( Simple ion = ${\text{N}}^{3-}$

$\begin{array}{l}\text{N}{\text{a}}^{+ }+{\text{N}}^{3-}= \text{N}{\text{a}}_3\text{N}\\ \left(+1\right)\times 3+-3= 0 \end{array}$.

Interpretation Introduction

(c)

Interpretation:

Predict the simple binary compound that can be formed by the use of electron configuration of the simple ions which are formed by those elements.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. Depending on that the periodic table is created. Periodic table further defines the simple ion that can be formed by any element. And using those ions, simple binary compounds can be produced.

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

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

Simplest ion form by each group as follows:

If it’s a compound made by a combination of metal and nonmetal elements, called as ionic compounds.

Ionic chemical compound must have a zero charge. If a compound contains ions;

1. Both positive ions (cations) and negative ions (anions) must be there.

2. No: of cations and anions collectively should produce a zero-net charge.

Answer to Problem 45CR

$\text{MgS}$.

Explanation of Solution

Mg and S

$\text{M}\text{g},\text{Z}=12 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^2$ ( Group III A element ( Simple ion = $\text{M}{\text{g}}^{2+}$

$\text{S},\text{Z}=16 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{4 }$ ( Group VI A element ( Simple ion = ${\text{S}}^{2-}$

$\begin{array}{l}\text{M}{\text{g}}^{2+ }+{\text{S}}^{2-} = \text{M}\text{g}\text{S}\\ +2+-2= 0 \end{array}$.

Interpretation Introduction

(d)

Interpretation:

Predict the simple binary compound that can be formed by the use of electron configuration of the simple ions which are formed by those elements.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. Depending on that the periodic table is created. Periodic table further defines the simple ion that can be formed by any element. And using those ions, simple binary compounds can be produced.

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

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

Simplest ion form by each group as follows:

If it’s a compound made by a combination of metal and nonmetal elements, called as ionic compounds.

Ionic chemical compound must have a zero charge. If a compound contains ions;

1. Both positive ions (cations) and negative ions (anions) must be there.

2. No: of cations and anions collectively should produce a zero-net charge.

Answer to Problem 45CR

CaO.

Explanation of Solution

Ca and O

$\text{C}\text{a},\text{Z}=20 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{2 }$ ( Group II A element ( Simple ion = $\text{C}{\text{a}}^{2+}$

$\text{O},\text{Z}=8 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^4$ ( Group VI A element ( Simple ion = ${\text{O}}^{2-}$ $\begin{array}{l}\text{C}{\text{a}}^{2+ }+{\text{O}}^{2-} = \text{C}\text{a}\text{O}\\ +2+-2= 0 \end{array}$.

Interpretation Introduction

(e)

Interpretation:

Predict the simple binary compound that can be formed by the use of electron configuration of the simple ions which are formed by those elements.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. Depending on that the periodic table is created. Periodic table further defines the simple ion that can be formed by any element. And using those ions, simple binary compounds can be produced.

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

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

Simplest ion form by each group as follows:

If it’s a compound made by a combination of metal and nonmetal elements, called as ionic compounds.

Ionic chemical compound must have a zero charge. If a compound contains ions;

1. Both positive ions (cations) and negative ions (anions) must be there.

2. No: of cations and anions collectively should produce a zero-net charge.

Answer to Problem 45CR

${\text{K}}_2\text{O}$.

Explanation of Solution

K and O

$\text{K},\text{Z}=19 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{1 }$ ( Group I A element ( Simple ion = ${\text{K}}^{+}$

$\text{O},\text{Z}=8 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^4$ ( Group VI A element ( Simple ion = ${\text{O}}^{2-}$

$\begin{array}{l}{\text{K}}^{+ }+{\text{O}}^{2-} = {\text{K}}_2\text{O}\\ +2\left(1\right)+\left(-2\right)= 0 \end{array}$.

Interpretation Introduction

(f)

Interpretation:

Predict the simple binary compound that can be formed by the use of electron configuration of the simple ions which are formed by those elements.

Concept Introduction:

Electron configuration describes the positions of electrons of an atom. Depending on that the periodic table is created. Periodic table further defines the simple ion that can be formed by any element. And using those ions, simple binary compounds can be produced.

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

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

Simplest ion form by each group as follows:

If it’s a compound made by a combination of metal and nonmetal elements, called as ionic compounds.

Ionic chemical compound must have a zero charge. If a compound contains ions;

1. Both positive ions (cations) and negative ions (anions) must be there.

2. No: of cations and anions collectively should produce a zero-net charge.

Answer to Problem 45CR

$\text{S}\text{r}\text{B}{\text{r}}_2$.

Explanation of Solution

Sr and Br

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

( Simple ion = $\text{S}{\text{r}}^{2+}$

$\text{B}\text{r},\text{Z}=35 1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^{2}3{\text{d}}^{10}4{\text{p}}^5$ ( Group VII A element ( Simple ion = $\text{B}{\text{r}}^{-}$

$\begin{array}{l}\text{S}{\text{r}}^{2+ }+\text{B}{\text{r}}^{-} = \text{S}\text{r}\text{B}{\text{r}}_2\\ +2+\left(-1\right)\times 2= 0 \end{array}$.