Chapter 4, Problem 7A

Interpretation Introduction

Interpretation: The incorrect name of the compound has to be identified and then correct name of it must be provided.

Concept introduction: To name a compound, certain rules are followed. The given compounds are binary compounds.

Binary compounds are of three types:

• Type-I: Compound in which a metal forms ionic bond with a non-metal and the metal can form only one type of ions.
• Type-II: Compound in which a metal forms ionic bond with a non-metal and the metal can form more than one type of ions.
• Type-III: Compound in which a non-metal forms bond with a non-metal.

There is one another type of compound in which metal is bonded with a polyatomic ion

Rules for naming these compounds:

Type-I:

• Name of metal cation is written as name of element.
• Name of anion is written using root name and adding ‘ide’ to it.
• Writing these names in the same sequence gives the name of compound.

Type-II:

• Name of metal cation is written as name of element followed by charge carried by it in roman numerals in bracket.
• Name of anion is written using root name and adding ‘ide’ to it.
• Writing these names in the same sequence gives the name of compound.

Type-III:

• It is name by writing the name of first element as same as element.
• Second element in anionic form and prefixes are used to denote the number of atoms. Prefix mono is not used for one element.

When metal is bonded with polyatomic ion, rules are similar to ionic binary compounds.

Answer to Problem 7A

B is incorrect option and the correct name is lead (II) chloride.

Explanation of Solution

  ${\text{PbCl}}_{\text{2}}$ is dissociated as follows: ${\text{PbCl}}_{\text{2}}\to {\text{Pb}}^{2+}+2{\text{Cl}}^{\text{-}}$ .

So, charge on Pb is $+2$ not $+4$ .While naming in the bracket (II) must be put not (IV).

Conclusion

Hydrides are named similar to that of binary compound. So in case of ${\text{CaH}}_{\text{2}}$ , the name of the compound starts with metal calcium and ends with hydride.

In ${\text{CrI}}_{\text{3}}$ the metal Cr is in $+3$ oxidation state hence chromium (III) iodide is correct.

  ${\text{Na}}_{\text{2}}\text{S}$ can be named as either sodium sulfide or disodium sulfide both are correct.

When Cu is in $+1$ state it is called cuprous and when it is in $+2$ oxidation state it is called cupric.

In ${\text{CuBr}}_{\text{2}}$ the metal Cu is $+2$ oxidation state and hence it is named as cupric bromide.