Chapter 7, Problem 85A

(a)

Interpretation Introduction

Interpretation:

The correct formula for compound, AlCl needs to be determined.

Concept introduction:

The formula for ionic compound is written as based on the charge of cation and anion. The technique used mostly is crisscross method wherein the charges become subscript of the opposite ion. If the numbers are high, it is reduced to lowest ratio.

Answer to Problem 85A

The correct chemical formula for ionic compound AlCl is ${\text{AlCl}}_3$

Explanation of Solution

The ionic compound is AlCl.

Al is aluminum which has atomic number 13 with configuration as ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^1$. It has 3 valence electrons. Hence, it will lose three electrons to get octet and become ${\text{Al}}^{\text{3+}}$.

Cl is Chlorine having atomic number as 17 with configuration as ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^5$. It has 7 valence electrons. Hence, it will need one electron to get octet and becomes ${\text{Cl}}^{-}$

For the compound having ${\text{Al}}^{\text{3+}}$ and ${\text{Cl}}^{-}$ is

(b)

Interpretation Introduction

Interpretation:

The correct formula for compound, ${\text{Na}}_{\text{3}}{\text{SO}}_4$ needs to be determined.

Concept introduction:

The formula for ionic compound is written as based on the charge of cation and anion. The technique used mostly is crisscross method wherein the charge becomes subscript of the opposite ion. If the numbers are high, it is reduced to lowest ratio.

Answer to Problem 85A

The correct chemical formula for ionic compound ${\text{Na}}_{\text{3}}{\text{SO}}_4$ is ${\text{Na}}_{\text{2}}{\text{SO}}_4$

Explanation of Solution

The ionic compound is ${\text{Na}}_{\text{3}}{\text{SO}}_4$

Na is Sodium which has atomic number 11 with configuration as ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^1$. It has 1 valence electrons. Hence, it will lose one electron to get octet and become ${\text{Na}}^{+}$.

${\text{SO}}_4$ is sulphate, ${\text{SO}}_4{}^{2-}$, having charge of -2.

For the compound having ${\text{Na}}^{+}$ and ${\text{SO}}_4{}^{2-}$ is

(c)

Interpretation Introduction

Interpretation:

The correct formula for compound, ${\text{BaOH}}_2$ needs to be determined.

Concept introduction:

The formula for ionic compound is written as based on the charge of cation and anion. The technique used mostly is crisscross method wherein the charge becomes subscript of the opposite ion. If the numbers are high, it is reduced to lowest ratio.

Answer to Problem 85A

The correct chemical formula for ionic compound ${\text{BaOH}}_2$ is ${\text{Ba(OH)}}_2$

Explanation of Solution

The ionic compound is ${\text{BaOH}}_2$

Ba is Barium which has atomic number 56 with configuration as ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^1{\text{ 3p}}^6{\text{3d}}^{10}{\text{4s}}^2{\text{4p}}^6{\text{4d}}^{10}5{\text{s}}^{2}5{\text{p}}^{\text{6}}6{\text{s}}^2$. It has 2 valence electrons. Hence, it will lose two electrons to get octet and become ${\text{Ba}}^{\text{2+}}$.

OH is hydroxide ion and has charge of -1.

For the compound having ${\text{Ba}}^{\text{2+}}$ and ${\text{OH}}^{-}$ is

Interpretation Introduction

(d)

Interpretation:

The correct formula for compound, ${\text{Fe}}_2\text{O }$ needs to be determined.

Concept introduction:

The formula for ionic compound is written as based on the charge of cation and anion. The technique used mostly is crisscross method wherein the charges becomes subscript of the opposite ion. If the numbers are high, it is reduced to lowest ratio.

Answer to Problem 85A

The correct chemical formula for ionic compound ${\text{Fe}}_2\text{O }$ is ${\text{Fe}}_2{\text{O}}_3$ or $\text{FeO}$

Explanation of Solution

The ionic compound is ${\text{Fe}}_2\text{O }$.

Fe is Iron which has atomic number 26 with configuration as ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{ 3p}}^6{\text{4s}}^2{\text{3d}}^6$. It has 6 valence electrons. Hence, it will lose two electrons to get octet and become ${\text{Fe}}^{2+}$. It can also lose 3 electrons and become ${\text{Fe}}^{3+}$

O is Oxygen having atomic number as 8 with configuration as ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^4$. It has 6 valence electrons. Hence, it will need two electrons to get octet and becomes ${\text{O}}^{2-}$

For the compound having ${\text{Fe}}^{2+}$ and ${\text{O}}^{2-}$ is

For compound having ${\text{Fe}}^{3+}$ and ${\text{O}}^{2-}$ is

For compound having ${\text{Fe}}^{3+}$ and ${\text{O}}^{2-}$ is