Question

Write the formula for each of the following ionic compounds:

a. zinc bromide

b. iron(III) sulphide

c. manganese(IV) oxide

d. Chromium(III) iodide

e. lithium nitride

f. gold(I) oxide

Short answer

The formulas are

(part a)${\mathrm{NO}}_2^{-}$,

(part b)${\mathrm{SO}}_3^{2-}$,

(part c)${\mathrm{OH}}^{-}$,

(part d)${\mathrm{C}}_2{\mathrm{H}}_3{\mathrm{O}}_2^{-}$

Step-by-step solution

Formula (Part a)

Write the nitrite ion's formula as one oxygen atom fewer than the nitrate ion's formula, ${\mathrm{NO}}_3^{-}$. As a result, the nitrite ion formula is ${\mathrm{NO}}_2^{-}$.

Formula (Part b)

Sulphite ion has one less oxygen atom than sulfate ion in its formula, ${\mathrm{SO}}_4^{2-}$. As a result, the sulphite ion's formula is ${\mathrm{SO}}_3^{2-}$.

Formula (Part c)

The hydroxide ion's formula is ${\mathrm{OH}}^{-}$

Formula (Part d)

The formula of acetate ion is ${\mathrm{C}}_2{\mathrm{H}}_3{\mathrm{O}}_2^{-}$