To find the formulas for these ionic compounds, we need to look at the charges of the ions each element forms. Ionic compounds are neutral, meaning the total positive charge must equal the total negative charge.
Here is the step-by-step breakdown for each pair:
a) Beryllium and Sulfur
*
Beryllium (Be) is in Group 2, so it loses 2 electrons to become
$Be^{2+}$.
*
Sulfur (S) is in Group 16, so it gains 2 electrons to become
$S^{2-}$.
* The charges (+2 and -2) balance perfectly with one of each atom.
*
Formula: BeS
b) Potassium and Fluorine
*
Potassium (K) is in Group 1, so it becomes
$K^+$.
*
Fluorine (F) is in Group 17, so it becomes
$F^-$.
* The charges (+1 and -1) balance with one of each atom.
*
Formula: KF
c) Magnesium and Oxygen
*
Magnesium (Mg) is in Group 2, so it becomes
$Mg^{2+}$.
*
Oxygen (O) is in Group 16, so it becomes
$O^{2-}$.
* The charges (+2 and -2) balance with one of each atom.
*
Formula: MgO
d) Hydrogen and Sulfur
*
Hydrogen (H) acts like a metal here and loses 1 electron to become
$H^+$.
*
Sulfur (S) is in Group 16 and becomes
$S^{2-}$.
* To balance the -2 charge from sulfur, you need two hydrogen ions (+1 + 1 = +2).
*
Formula: $H_2S$
e) Aluminum and Chlorine
*
Aluminum (Al) is in Group 13, so it becomes
$Al^{3+}$.
*
Chlorine (Cl) is in Group 17, so it becomes
$Cl^-$.
* To balance the +3 charge from aluminum, you need three chlorine ions (-1 -1 -1 = -3).
*
Formula: $AlCl_3$
f) Iodine and Sodium
*
Sodium (Na) is in Group 1, so it becomes
$Na^+$.
*
Iodine (I) is in Group 17, so it becomes
$I^-$.
* The charges (+1 and -1) balance with one of each atom.
*
Formula: NaI
g) Selenium and Strontium
*
Strontium (Sr) is in Group 2, so it becomes
$Sr^{2+}$.
*
Selenium (Se) is in Group 16 (same column as Oxygen and Sulfur), so it becomes
$Se^{2-}$.
* The charges (+2 and -2) balance with one of each atom.
*
Formula: SrSe
h) Gallium and Oxygen
*
Gallium (Ga) is in Group 13 (same column as Aluminum), so it becomes
$Ga^{3+}$.
*
Oxygen (O) is in Group 16, so it becomes
$O^{2-}$.
* These charges do not match directly. We need to find the least common multiple of 3 and 2, which is 6.
* We need two Gallium ions ($2 \times +3 = +6$).
* We need three Oxygen ions ($3 \times -2 = -6$).
*
Formula: $Ga_2O_3$
Final Answer:
a) BeS
b) KF
c) MgO
d) $H_2S$
e) $AlCl_3$
f) NaI
g) SrSe
h) $Ga_2O_3$
Parent Tip: Review the logic above to help your child master the concept of ionic bond formation worksheet.
