Q. 6.115 - An Introduction to General, Organic, and Biological Chemistry

Question

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

a. tin(II) sulfide

b. lead(IV) oxide

c. silver chloride

d. calcium nitride

e. copper(I) phosphide

f. chromium(II) bromide

Step-by-Step Solution

Answer

(a) Tin(II) sulfide has the formula  $SnS$.

(b) $Pb{O}_2$ is the formula for lead(IV) oxide.

(c) Silver chloride has the formula AgCl.

(d) Calcium nitride formula is $C{a}_3{N}_2$.

(e) $C{u}_{3}P$ is the formula for copper(I) phosphide.

(f) Chromium(II) bromide has the formula $CrB{r}_2.$

Part(a) Step 1: Given information

We have been given,

Sn and S are the symbols used in this formula.

Part(a) Step 2: Explanation

Now, We must locate valency and exchange valency. 

So, The valency of Sn$=2$ and  S$=-2$.

Interchange valency in the subscript $S{n}_2$ and $S_2$

So, the formula is $SnS$.

Part(b) Step 1: Given information

We have been given,

Pb and O are the symbols used in this formula. 

Part(b) Step 2: Explanation

Now, We must locate valency and exchange valency.

So, The valency of $Pb=4$and $O=-2$,

Interchange valency in the subscript $P{b}_2$ and $O_2$,

So, the formula is $Pb{O}_2$

Part(c) Step 1: Given information

We have been given,

Ag and Cl are the symbols used in this formula. 

Part(c) Step 2: Explanation

Now, We must locate valency and exchange valency. 

So, the valency of Ag$=1$ and Cl$=-1$,

Interchange valency in the subscript $A{g}_1$and $C{l}_1$,

So, the formula is AgCl. 

Part(d) Step 1: Given information

We have been given,

Ca and N are the symbols used in this formula. 

Part(d) Step 2: Explanation

Now, We must locate valency and exchange valency.  

So, valency Ca$=2+$ and N$=-3$,

Interchange valency in the subscript $C{a}_3$ and $N_2$,

So, the formula is $C{a}_3{N}_2$.

Part(e) Step 1: Given information

We have been given,

Cu and P are the symbols used in this formula. 

Part(e) Step 2: Explanation

Now, We must locate valency and exchange valency.   

So, valency is Cu$=1$ and P$=-3$,

Interchange valency in the subscript $C{u}_3$ and $P_1$,

So, the formula $C{u}_{3}P$.

Part(f) Step 1: Given information

We have been given,

Cr and Br are the symbols used in this formula. 

Part(f) Step 2: Explanation

Now, We must locate valency and exchange valency.    

So, the valency Cr$=2$ and Br$=1$,

Interchange valency in the subscript $C{r}_1$ and $B{r}_2$,

So, final formula is $CrB{r}_2$.