Question

6.31 Write the formula for each of the following ionic compounds:

a. cobalt(III) chloride

b. lead(IV) oxide

c. silver iodide

d. calcium nitride

e. copper(1) phosphise

f. chromium(II) chloride

Short answer

The formulas are

(part a) cobalt(III) chloride's formula is $c{l}_{3}co$

(part b) lead(IV) oxide's formula is $PbO2$

(part c) silver iodide's formula is $AgI$

(part d) calcium nitride's formula is $Ca3N2$

(part e) copper(1) phosphite's formula is $Cu3PO4$

(part f) chromium(II) chloride's formula is $CrCl2$

Step-by-step solution

Explanation (Part a)

One hydrogen atom added to the carbonate ion is the formula for hydrogen carbonate ion. As a result, the hydrogen carbonate ion's formula is ${{\mathrm{HCO}}_3}^{-}$

Explanation (Part b)

The ammonium ion's formula is ${{\mathrm{NH}}_4}^{+}$

Explanation (Part c)

Write the formula for the phosphite ion, which has one less oxygen atom than the phosphate ion. As a result, the phosphite ion formula is ${\mathrm{PO}}_3^{3-}$.

Explanation (Part d)

The chlorate ion formula is${\mathrm{ClO}}_3^{-}$.