Chapter 2.8, Problem 2.9BFP

(a)

Interpretation Introduction

Interpretation:

The correct formula for potassium sulfide is to be determined.

Concept introduction:

The formula of an ionic compound represents the total number of ions of each element present in it.

The general rules for writing the systematic names of ionic compounds are as follows:

1) In binary ionic compounds, the name of the metal is written as the original name whereas the name of the non-metal has the root word with the suffix $\text{-}ide$.

2) In metals that form more than one ion, the Latin root of the metal is followed by a suffix $\text{-}ous$ for the ion with lower charge and suffix $\text{-}ic$ for the ion with a higher charge. In the systematic name, the name of the metal is followed by the charge on the metal within the parentheses in roman numerals.

3) In the family of two oxoanions, the ion with more oxygen atoms has the non-metal root and a suffix $\text{-}ate$ added to it. The ion with fewer oxygen atoms has the non-metal root and the suffix $\text{-}ite$ added to it.

4) For hydrated ionic compounds, the total number of water molecules are represented by the Greek numerical prefixes followed by the word hydrate.

Answer to Problem 2.9BFP

The correct formula for potassium sulfide is ${\text{K}}_2\text{S}$.

Explanation of Solution

The element potassium belongs to the $1\text{A}\left(1\right)$ group in the periodic table and therefore loses an electron to form potassium ion as follows:

$\text{K}\to {\text{K}}^{+}+{\text{e}}^{-}$

Sulfur is a non-metal from the $6\text{A}\left(16\right)$ group in the periodic table. The formation of sulfide ion occurs by the gain of two electrons by an atom of sulfur as follows:

$\text{S}+2{\text{e}}^{-}\to {\text{S}}^{2-}$

An ionic compound is neutral as a whole since the total charge on the positive ions is balanced by the total charge on the negative ions. In order to balance the total charge of $+2$ on the sulfide ion, two potassium ions must combine with it in a $2:1$ ratio as follows:

${\text{2K}}^{+}+{\text{S}}^{2-}\to {\text{K}}_2\text{S}$

Therefore the correct formula for potassium sulfide is ${\text{K}}_2\text{S}$.

Conclusion

The correct formula for potassium sulfide is ${\text{K}}_2\text{S}$.

(b)

Interpretation Introduction

Interpretation:

The correct formula for barium iodide is to be determined.

Concept introduction:

The formula of an ionic compound represents the total number of ions of each element present in it.

The general rules for writing the systematic names of ionic compounds are as follows:

1) In binary ionic compounds, the name of the metal is written as the original name whereas the name of the non-metal has the root word with the suffix $\text{-}ide$.

2) In metals that form more than one ion, the Latin root of the metal is followed by a suffix $\text{-}ous$ for the ion with lower charge and suffix $\text{-}ic$ for the ion with a higher charge. In the systematic name, the name of the metal is followed by the charge on the metal within the parentheses in roman numerals.

3) In the family of two oxoanions, the ion with more oxygen atoms has the non-metal root and a suffix $\text{-}ate$ added to it. The ion with fewer oxygen atoms has the non-metal root and the suffix $\text{-}ite$ added to it.

4) For hydrated ionic compounds, the total number of water molecules are represented by the Greek numerical prefixes followed by the word hydrate.

Answer to Problem 2.9BFP

The correct formula for barium iodide is ${\text{BaI}}_2$.

Explanation of Solution

The element barium is metal from the $2\text{A}\left(2\right)$ group in the periodic table and hence loses two electrons to form an ion as follows:

$\text{Ba}\to {\text{Ba}}^{2+}+2{\text{e}}^{-}$

Iodine is a non-metal from the $7\text{A}\left(17\right)$ group. Elements in this group gain an electron to form anions with $-1$ charge on them as follows;

$\text{I}+{\text{e}}^{-}\to {\text{I}}^{-}$

The charges on the barium and the iodide ions are $+2$ equal and $-1$ respectively. Therefore, barium and iodide ions combine in a $2:1$ ratio to form a molecule of barium iodide as follows:

${\text{Ba}}^{2+}+{\text{2I}}^{-}\to {\text{BaI}}_2$

Conclusion

The correct formula for barium iodide is ${\text{BaI}}_2$.

(c)

Interpretation Introduction

Interpretation:

The correct formula for cesium nitride is to be determined.

Concept introduction:

The formula of an ionic compound represents the total number of ions of each element present in it.

The general rules for writing the systematic names of ionic compounds are as follows:

1) In binary ionic compounds, the name of the metal is written as the original name whereas the name of the non-metal has the root word with the suffix $\text{-}ide$.

2) In metals that form more than one ion, the Latin root of the metal is followed by a suffix $\text{-}ous$ for the ion with lower charge and suffix $\text{-}ic$ for the ion with a higher charge. In the systematic name, the name of the metal is followed by the charge on the metal within the parentheses in roman numerals.

3) In the family of two oxoanions, the ion with more oxygen atoms has the non-metal root and a suffix $\text{-}ate$ added to it. The ion with fewer oxygen atoms has the non-metal root and the suffix $\text{-}ite$ added to it.

4) For hydrated ionic compounds, the total number of water molecules are represented by the Greek numerical prefixes followed by the word hydrate.

Answer to Problem 2.9BFP

The correct formula for cesium nitride is ${\text{Cs}}_3\text{N}$.

Explanation of Solution

Cesium is a metal from the $1\text{A}\left(1\right)$ group in the periodic table. Therefore an atom of cesium loses an electron to form the cesium ion as follows:

$\text{Cs}\to {\text{Cs}}^{+}+{\text{e}}^{-}$

The element nitrogen is a non-metal from the $5\text{A}\left(15\right)$ group in the periodic table. Therefore an atom of nitrogen gains three electrons to form the nitride ion as follows:

$\text{N}+3{\text{e}}^{-}\to {\text{N}}^{3-}$

The charges on the cesium and nitride ions are $+1$ and $-3$ respectively. Therefore to balance each other, the cesium and nitride ions combine in a $3:1$ ratio to form a molecule of cesium nitride as follows:

${\text{3Cs}}^{+}+{\text{N}}^{3-}\to {\text{Cs}}_3\text{N}$

Conclusion

The correct formula for cesium nitride is ${\text{Cs}}_3\text{N}$.

(d)

Interpretation Introduction

Interpretation:

The correct formula for sodium hydride is to be determined.

Concept introduction:

The formula of an ionic compound represents the total number of ions of each element present in it.

The general rules for writing the systematic names of ionic compounds are as follows:

1) In binary ionic compounds, the name of the metal is written as the original name whereas the name of the non-metal has the root word with the suffix $\text{-}ide$.

2) In metals that form more than one ion, the Latin root of the metal is followed by a suffix $\text{-}ous$ for the ion with lower charge and suffix $\text{-}ic$ for the ion with a higher charge. In the systematic name, the name of the metal is followed by the charge on the metal within the parentheses in roman numerals.

3) In the family of two oxoanions, the ion with more oxygen atoms has the non-metal root and a suffix $\text{-}ate$ added to it. The ion with fewer oxygen atoms has the non-metal root and the suffix $\text{-}ite$ added to it.

4) For hydrated ionic compounds, the total number of water molecules are represented by the Greek numerical prefixes followed by the word hydrate.

Answer to Problem 2.9BFP

The correct formula for sodium hydride is $\text{NaH}$.

Explanation of Solution

The element sodium is a metal from the $1\text{A}\left(1\right)$ group in the periodic table. Therefore an atom of sodium loses an electron to form the sodium ion as follows:

$\text{Na}\to {\text{Na}}^{+}+{\text{e}}^{-}$

The element hydrogen is a non-metal from the $1\text{A}\left(1\right)$ group in the periodic table. Therefore, an atom of hydrogen gains an electron to form the hydride ion as follows:

$\text{H}+{\text{e}}^{-}\to {\text{H}}^{-}$

In order to balance the charges on each other, aluminium and hydride ions combine in a $1:1$ ratio as follows:

${\text{Na}}^{+}+{\text{H}}^{-}\to \text{NaH}$

Conclusion

The correct formula for sodium hydride is $\text{NaH}$.