Chapter 4, Problem 44A

Interpretation Introduction

Interpretation:

The pairs in which name of binary compound is incorrect is to be chosen.

Concept introduction:

A binary compound is the compound having two components.

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.

To name binary compounds of various types, some rules are assigned:

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 first element.

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

Answer to Problem 44A

Options a, b, c and e are correct because the names of the binary compounds are incorrect.

Explanation of Solution

The name of $Si{I}_4$ should be silicon tetraiodide because Si is symbol for silicon not silver.

The name of $CoC{l}_2$ is cobalt (II) chloride because Co is symbol of cobalt not copper.

The name of $Ca{H}_2$ is calcium hydride and hydrogen present here is more electronegative than calcium. It can’t be released as proton.

The name of $P{H}_3$ is phosphorus tetrahydride because P is symbol of phosphorus.

Conclusion

Option D is an incorrect option because name of $Zn{\left(C_2{H}_3{O}_2\right)}_2$ is zinc acetate that is correct name for the given binary compound.