Question

Give the number of each type of ion present on each of the ionic compounds formed from the following pairs of ions.

Answer 1

The following compounds and the number and type of ions are as follows:

Gold and chloride:

Gold (III) = Au 3+
Chloride = Cl 1-

Calcium and Sulfate:

Calcium Sulfate:

Calcium = Ca 2+
Sulfate = SO4 2-

Magnesium and Hydroxide

Magnesium Hydroxide: Mg(OH)2

Magnesium: Mg 2+
Hydroxide: (OH) 1-

Answer 2

Answer 1) : The number of ions present in $AuCl_{3}$ will have 3 Chloride ions and 1 Gold ion in the molecule.

Explanation : As the compound of $AuCl_{3}$ has only one Au atom which is in +3 oxidation state is bonded with 3 Cl atoms which are in (-1) oxidation state. Therefore, as a whole compound it is neutral in ionic charges. This is used as an antimicrobial.

Answer 2) Calcium Sulfate has one calcium ion and one sulfate ion with different ionic charge.

Explanation : As calcium sulfate has the molecular formula $CaSO_{4}$, where one atom of calcium has (+2) charge and one atom of sulfate has (-2) both combine and neutralize the molecule as a whole and form a compound which is used in plaster cast.

Answer 3) Magnesium hydroxide $Mg(OH)_{2}$ has one ion of magnesium with (+2) as the ionic charge and two ion of hydroxide which has (-1) charge.

Explanation : In milk of magnesia, the main compound which is magnesium hydroxide has 1:2 ratio of magnesium and hydroxide molecules attached to form a compound. Where magnesium atom is in (+2) state and each of hydroxide ion are in (-1) oxidation state. So,( -1 X 2) = -2  which neutralizes the ionic charge on magnesium ion and stabilizes the compound.