Chapter 2, Problem 2.61QP

(a)

Interpretation Introduction

Interpretation:

Formula for rubidium nitrate has to be written.

Concept Introduction:

The nomenclature of inorganic compounds involves following rules:

• The prefixes on each atom indicates the number of that atoms in the compound.
• The number written on the subscript of the anion is numerically equal to the charge on the cation and vice versa.
• Subscripts are discarded when the numerical charge on cation and anion are equal.
• When a metal has more than one positive oxidation state, its lowest oxidation number has name of the metal ion ending with ‘ous’ and highest oxidation number has name of the metal ion ending with ‘ic’.  This rule is applicable when oxidation number of metal is not more than three.
• If a metal ion has multiple number of oxidation states more than three, then Roman numeral has to be used within square brackets to indicate the oxidation number of metal ion.
• Nomenclature of binary acids and oxoacids are not similar.  Binary acids are named based on the non-metal atom present in them.  Oxoacids are named on the basis of polyatomic anion present in them.

Answer to Problem 2.61QP

Formula for rubidium nitrite is written as $RbN{O}_2.$

Explanation of Solution

Nitrite is $N{O}_2^{-}$ anion and Rubidium here is Rubidium cation $R{b}^{+}.$  Hence formula of the given compound is written as $RbN{O}_2$

(b)

Interpretation Introduction

Interpretation:

Formula for Potassium sulfide has to be written.

Concept Introduction:

Refer to part (a)

Answer to Problem 2.61QP

Formula for Potassium sulfide is written as $K_{2}S.$

Explanation of Solution

Sulfide is $S_{}^{2-}$ anion and Potassium here is Potassium cation $K^{+}.$  Charge on the anion has to be written on the subscript of the cation.  Hence formula of the given compound is written as $K_{2}S.$

(c)

Interpretation Introduction

Interpretation:

Formula for Sodium hydrogen sulfide has to be written.

Concept Introduction:

Refer to part (a)

Answer to Problem 2.61QP

Formula for Sodium hydrogen sulfide is written as $NaHS$.

Explanation of Solution

Hydrogen sulfide is $H{S}^{-}$ and it is anion.  The cation is sodium ion $N{a}^{+}$ and the formula is written as $NaHS$

(d)

Interpretation Introduction

Interpretation:

Formula for the given compound has to be written.

Concept Introduction:

Refer to part (a)

Answer to Problem 2.61QP

Formula of the compound Magnesium phosphate is written as $M{g}_3{\left(P{O}_4\right)}_2$

Explanation of Solution

The cation and anion in Magnesium phosphate are magnesium ion $M{g}^{2+}$ and phosphate ion $P{O}_4^{3-}$ respectively.  Charge on the anion is written on the subscript of the cation.  Charge on the cation is written on the subscript of the anion.

Thus the formula of the compound Magnesium phosphate is written as $M{g}_3{\left(P{O}_4\right)}_2$.

(e)

Interpretation Introduction

Interpretation:

Formula for the given compound has to be written.

Concept Introduction:

Refer to part (a)

Answer to Problem 2.61QP

Formula of the compound Calcium hydrogen phsophate is written as $CaH\left(P{O}_4\right)$.

Explanation of Solution

The cation and anion in Calcium hydrogen phosphate are calcium ion $C{a}^{2+}$ and hydrogen phosphate ion $HP{O}_4^{2-}$ respectively.  Charge on the anion is written on the subscript of the cation.  Charge on the cation is written on the subscript of the anion.

Hence the formula is $Ca{H}_2{\left(P{O}_4\right)}_2$

The formula can be simplified and written as the same number in subscript gets called and thus the formula of the compound Magnesium phosphate is written as $CaH\left(P{O}_4\right)$

(f)

Interpretation Introduction

Interpretation:

Formula for the given compound has to be written.

Concept Introduction:

Refer to part (a)

Answer to Problem 2.61QP

Formula of the compound Potassium dihydrogen phosphate is written as $K{H}_2\left(P{O}_4\right)$

Explanation of Solution

With reference to table 2.3 in the text book, anions and cations formed by various elements are named.

The cation and anion in Potassium dihydrogen phosphate are potassium ion $K^{+}$ and dihydrogen phosphate ion $H_{2}P{O}_4^{1-}$ respectively.  Charge on the anion is written on the subscript of the cation.  Charge on the cation is written on the subscript of the anion.

Hence the formula is $K_1{\left(H_{2}P{O}_4\right)}_1$

The formula can be simplified and written as the same number in subscript gets called and thus the formula of the compound Magnesium phosphate is written as $K{H}_2\left(P{O}_4\right)$

(g)

Interpretation Introduction

Interpretation:

Formula for the given compound has to be written.

Concept Introduction:

Refer to part (a)

Answer to Problem 2.61QP

Formula for the compound Iodine heptafluoride is written as $I{F}_7$.

Explanation of Solution

The prefix hepta means seven.  Thus seven F atoms are bonded to Iodine I.  Hence the formula is $I{F}_7$

(h)

Interpretation Introduction

Interpretation:

Formula for the given compound has to be written.

Concept Introduction:

Refer to part (a)

Answer to Problem 2.61QP

Formula for the compound Ammonium sulfate is written as ${\left(N{H}_4\right)}_{2}S{O}_4$.

Explanation of Solution

The cation and anion in Ammonium sulfate are ammonium ion $N{H}_4^{+}$ and sulfate ion $S{O}_4^{2-}$ respectively.  Charge on the anion is written on the subscript of the cation.  Charge on the cation is written on the subscript of the anion.

Hence the formula is ${\left(N{H}_4\right)}_{2}S{O}_4$

(i)

Interpretation Introduction

Interpretation:

Formula for the given compound has to be written.

Concept Introduction:

Refer to part (a)

Answer to Problem 2.61QP

Formula of the compound Silver perchlorate is written as $AgCl{O}_4$

Explanation of Solution

The cation and anion in Silver perchlorate are silver ion $A{g}^{+}$ and chlorate ion $Cl{O}_4^{-}$ respectively.  Charge on the anion is written on the subscript of the cation.  Charge on the cation is written on the subscript of the anion.

Hence the formula is $AgCl{O}_4$

(j)

Interpretation Introduction

Interpretation:

Formula for the given compound has to be written.

Concept Introduction:

Refer to part (a)

Answer to Problem 2.61QP

Formula of Boron trichloride is written as $BC{l}_3$

Explanation of Solution

Boron trichloride is three Cl atoms bonded with Boron.

Hence the formula is $BC{l}_3.$