Q. 6.116 - An Introduction to General, Organic, and Biological Chemistry

Question

Write the formula for each of the following ionic compounds: $(6.2,6.3)$

(a) nickel(III) oxide

(b) iron(III) sulfide

(c) lead(II) sulfate

(d) chromium(III) iodide

(e)  lithium nitride

(f) gold(I) oxide

Step-by-Step Solution

Answer

(a) nickel(III) oxide has the formula $N{i}_2{O}_3$.

(b) iron(III) sulfide has the formula $F{e}_2{S}_3$.

(c) lead(II) sulfate has the formula $PbS{O}_4$.

(d) chromium(III) iodide has the formula $Cr{I}_3$.

(e) lithium nitride has the formula $L{i}_{3}N$.

(f) gold(I) oxide has the formula $A{u}_{2}O$.

Part(a) Step 1: Given information

We have been given,

Ni and O are the symbols used in this formula. 

Part(a) Step 2: Explanation

Now, We must locate valency and exchange valency. 

So, the valency Ni$=3$ and O$=-2$,

Interchange valency in the subscript $N{i}_2$ and $O_3$,

Finally, the formula is $N{i}_2{O}_3$.

Part(b) Step 1: Given information

We have been given,

Fe and S are the symbol used in this formula.

Part(b) Step 2: Explanation

Now, We must locate valency and exchange valency. 

So, the valency Fe$=3$ and S$=2$,

Interchange valency in the subscript $F{e}_2$ and $S_3$,

Finally, formula is $F{e}_2{S}_3$.

Part(c) Step 1: Given information

We have been given,

Pb , O and S are the symbol used in this formula.

Part(c) Step 2: Explanation

Now, We must locate valency and exchange valency. 

So, the valency Pb$=2$ and $S{{O_2}^{-2}}_{}$,

Interchange valency in the subscript $P{b}_1$ and $S{O}_4$,

Finally, the formula is $PbS{O}_4$.

Part(d) Step 1: Given information

We have been given,

Cr and I are the symbol used in this formula.

Part(d) Step 2: Explanation

Now, We must locate valency and exchange valency. 

So, the valency Cr$=3$ and I$=-1$,

Interchange valency in the subscript is $C{r}_1$ and $I_3$,

Finally, the formula $Cr{I}_3$.

Part(e) Step 1: Given information

We have been given,

Li and N are the symbol used in this formula.

Part(e) Step 2: Explanation

We, We must locate valency and exchange valency. 

So,  valency is Li$=1$ and N $=-3$,

Interchange valency in the subscript $L{i}_3$ and $N_1$.

So final formula is $L{i}_{3}N$.

Part(f) Step 1: Given information

We have been given,

Au and O are the symbol used in this formula.

Part(f) Step 2: Explanation

Now, We must locate valency and exchange valency. 

So, valency Au$=1$ and O$=-2$,

Interchange valency in the subscript $A{u}_2$ and $O_1$,

So, Final formula is $A{u}_{2}O$.