Chapter 2, Problem 2.57QP

Name these compounds: (a) Na₂CrO₄, (b) K₂HPO₄, (c) HBr (gas), (d) HBr (in water), (e) Li₂CO₃, (f) K₂Cr₂O₇, (g) NH₄NO₂, (h) PF₃, (i) PF₅, (j) P₄O₆, (k) CdI₂, (l) SrSO₄, (m) Al(OH)₃, (n) Na₂CO₃ · 10H₂O.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

The nomenclature of inorganic compounds involves following rules:

• The prefixes on each atom indicate the number of those 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 2.57QP

The given compound $N{a}_{2}Cr{O}_4$ is named as Sodium chromate.

Explanation of Solution

$N{a}_{2}Cr{O}_4$ has sodium Na, chromium Cr and oxygen O atoms.  According to the nomenclature rules the subscript 2 on Na indicates charge on $Cr{O}_4$  therefore anion of this compound is $Cr{O}_4^{2-}$ and cation is $N{a}^{+}.$  Hence, the name of $Cr{O}_4^{2-}$ is chromate and so name of the compound is sodium chromate.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $K_{2}HP{O}_4$ is named as Potassium hydrogen phosphate.

Explanation of Solution

$K_{2}HP{O}_4$ has potassium K, hydrogen H, phosphorus P and oxygen O atoms.  According to the nomenclature rules the subscript 2 on K indicates charge on $HP{O}_4$ therefore anion of this compound is $HP{O}_4^{2-}$ and cation is $N{a}^{+}.$  Hence, the name of $HP{O}_4^{2-}$ is Hydrogen phosphate and so name of the compound is Potassium hydrogen phosphate.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $HBr$ (gas) is named asHydrogen bromide.

Explanation of Solution

$HBr$ has hydrogen H, and Bromine Br atoms.  According to the nomenclature rules the subscript 1 on H indicates charge on $Br$ therefore anion of this compound is $B{r}^{-}$ and cation is $H^{+}.$  Hence, name of $B{r}^{-}$ is bromide ion and so name of the compound isHydrogen bromide.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $HBr$ (water) is named ashydrobromic acid.

Explanation of Solution

$HBr$ in water forms acid.  Binary acids are named based on the non-metal atom present in them.  Hence, name of the compound is given as hydrobromic acid.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $L{i}_{2}C{O}_3$ is named as Lithium carbonate.

Explanation of Solution

$L{i}_{2}C{O}_3$ has Lithium, carbon and oxygen.  The subscript 2 on Li indicates charge on $C{O}_3$ .  Thus the anion is $C{O}_3^{2-}$ and its name is carbonate.  Name of the compound is given as Lithium carbonate.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $K_{2}C{r}_2{O}_7$ is named as Potassium dichromate.

Explanation of Solution

$K_{2}C{r}_2{O}_7$ has Potassium, chromium and oxygen.  The subscript 2 on K indicates charge on $C{r}_2{O}_7$ .  Thus the anion is $C{r}_2{O}_7^{2-}$ and its name is dichromate with reference to table 2.3 in the text book.  Name of the compound is given as Potassium dichromate.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $N{H}_{4}N{O}_2$ is named as ammonium nitrite.

Explanation of Solution

$N{H}_{4}N{O}_2$ has Nitrogen, Hydrogen and oxygen.  The compound has cation $N{H}_4^{+}$ and anion $N{O}_2^{-}$ and their names are ammonium ion and nitrite ion respectively.  Hence, name of the compound is given asammonium nitrite.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $P{F}_3$ is named as phosphorus trifluoride.

Explanation of Solution

$P{F}_3$ has phosphorus and Fluorine.  It is not an ionic compound but covalent compound.  F has greater electronegativity than P and hence its name should end with ‘ide’.  Three F atoms are bonded to P and F gets the name trifluoride.  According to nomenclature rules the compound is named as phosphorus trifluoride.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $P{F}_5$ is named as phosphorus pentafluoride.

Explanation of Solution

$P{F}_5$ has phosphorus and Fluorine.  It is not an ionic compound but covalent compound.  F has greater electronegativity than P and hence its name should end with ‘ide’.  Five F atoms are bonded to P and F gets the name pentafluoride.  According to nomenclature rules the compound is named as phosphorus pentafluoride.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $P_4{O}_6$ is named as tetraphosphorus hexoxide.

Explanation of Solution

$P_4{O}_6$ has phosphorus and oxygen.  It is not an ionic compound but covalent compound.  O has greater electronegativity than P and hence its name should end with ‘ide’.  SixO atoms are bonded to P and O gets the name hexoxide.  4 P atoms are bonded to O and hence according to nomenclature rules the compound is named as tetraphosphorus hexoxide.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $Cd{I}_2$ is named as Cadmium iodide.

Explanation of Solution

$Cd{I}_2$ has cadmium and iodine.  The subscript 2 on I indicates charge on Cd thus $C{d}^{2+}$ is the cation.  The subscript 1 on Cd indicates charge on I thus $I^{-}$ is the cation and name of $I^{-}$ is iodide ion.  Hence name of the compound is cadmium iodide.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $SrS{O}_4$ is named as Strontium sulfate.

Explanation of Solution

$SrS{O}_4$ has Strontium, Sulfur and oxygen.  The compound has cation $S{r}^{2+}$ and anion $S{O}_4^{2-}$ and their names are strontium ion and sulfate ion respectively.  Hence name of the compound is strontium sulfate.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $Al{\left(OH\right)}_3$ is named as Aluminum hydroxide.

Explanation of Solution

The compound $Al{\left(OH\right)}_3$ has cation $A{l}^{3+}$ and anion $O{H}^{-}$ and their names are Aluminum ion and hydroxide ion are respectively.  Hence name of the compound isAluminum hydroxide.

Interpretation Introduction

Interpretation:

The given compound has to be named.

Concept Introduction:

Refer to part (a).

Answer to Problem 2.57QP

The given compound $N{a}_{2}C{O}_3.10H_{2}O$ is named as Sodium carbonate decahydrate.

Explanation of Solution

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

The compound $N{a}_{2}C{O}_3.10H_{2}O$ has cation $N{a}^{+}$ and anion $C{O}_3^2{}^{-}$ and the names of the ions are sodium ion and carbonate ion respectively.  10 water molecules are bonded to the molecule.  Hence name of the compound is sodium carbonate decahydrate.