Chapter 4, Problem 35A

Interpretation Introduction

Interpretation: The formulas of all possible compounds between given metal and nonmetals are to be written.

Concept introduction: A binary ionic compound is formed when a metal loses electrons and non metal gains the electrons..

Answer to Problem 35A

The possible formulas are $\text{MD,}{\text{M}}_{\text{2}}{\text{E,M}}_{\text{3}}{\text{F,MD}}_{\text{2}}\text{,}{\text{ME,M}}_{\text{3}}{\text{F}}_{\text{2}}{\text{,MD}}_{\text{3}}{\text{,M}}_{\text{2}}{\text{E}}_{\text{3}}\text{,MF}$

Explanation of Solution

Metallic	Cations	Non metals	Anions	Formula	Comments
M	+	D	1-	$\text{MD}$	${\text{M}}^{+}$ and ${\text{D}}^{\text{-}}$ are balanced.
		E	2-	${\text{M}}_{\text{2}}\text{E}$	2 ${\text{M}}^{+}$ to balance ${\text{E}}^{\text{2-}}$
		F	3-	${\text{M}}_{\text{3}}\text{F}$	3 ${\text{M}}^{+}$to balance ${\text{F}}^{\text{3-}}$
M	2+	D	1-	${\text{MD}}_{\text{2}}$	${\text{M}}^{2+}$and 2 ${\text{D}}^{\text{-}}$to balance.
		E	2-	$\text{ME}$	${\text{M}}^{2+}$ and ${\text{D}}^{\text{2-}}$ are balanced.
		F	3-	${\text{M}}_{\text{3}}{\text{F}}_{\text{2}}$	2 ${\text{M}}^{2+}$to balance 2 ${\text{F}}^{\text{3-}}$
M	3+	D	1-	${\text{MD}}_{\text{3}}$	${\text{M}}^{3+}$and 3 ${\text{D}}^{\text{-}}$to balance.
		E	2-	${\text{M}}_{\text{2}}{\text{E}}_{\text{3}}$	2 ${\text{M}}^{3+}$to balance 3 ${\text{F}}^{\text{2-}}$
		F	3-	$\text{MF}$	${\text{M}}^{3+}$ and ${\text{F}}^{\text{3-}}$ are balanced.

Conclusion

The use of a Roman numeral in a systematic name for a compound is required only when more than one ionic compound forms between a given pair of elements