Q. 6.31 - An Introduction to General, Organic, and Biological Chemistry

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

Step-by-Step Solution

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 1: 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 ${{\mathrm{HCO}}_3}^{-}$

Step 2: Explanation (Part b)

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

Step 3: 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 ${\mathrm{PO}}_3^{3-}$.

Step 4: Explanation (Part d)

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